UC-NRLF B 3 112 Library FINANCIAL ENGINEERING FINANCIAL ENGINEERING A TEXT FOR CONSULTING, MANAGING AND DESIGNING ENGINEERS AND FOR STUDENTS BY O. B. GOLDMAN CONSULTING ENGINEER PROFESSOR OF HEAT ENGINEERING, THE OREGON STATE AGRICULTURAL COLLEGE, HONORARY MEMBER LOCAL 87, I. U. STEAM AND OPERATING ENGINEERS, MEMBER OREGON SOCIETY OF ENGINEERS, ETC. NEW YORK JOHN WILEY & SONS, INC. LONDON: CHAPMAN & HALL, LIMITED 1920 TAAT-3 Engineering Library COPYRIGHT, IQ20, BY O. B. GOLDMAN THE-PLIMPTON-PRESS-NOBWOOD'MASS-U'S-A PREFACE AN engineer must know the properties of all material with which he comes in contact and he must understand thoroughly the action and limitations of all machines and instruments which he is called upon to use or test, as well as the proper application of the same. More than this, he should know how to translate engineering factors into dollars and cents. In addition he must know how to install a service not necessarily at the highest mechanical or electrical efficiency, which may prove and often does prove far too expensive, but always with regard to the highest financial efficiency so the resulting service will be rendered with the least effort, in the preparation for the service and its actual rendition. This is Financial Engineering. Technical Engineering must be supplemented by Financial Engineering to make a complete and harmonious system. What is demanded of the Financial Engineer is a solution in terms of money, the standard measure of commerce. Every engineer in a responsible position has felt this and likewise the need of a definite, scientific method of determining the comparative value of all things which he must use and the value of systems and of investments in general. He has felt the need of a correct method of determining the financial efficiency of undertakings, not merely as a whole, but element by element, so that all losses might be discovered, and so that the size and design for best economy might be determined. It was because of this demand that the author devoted so much of his time, over a period of fourteen years, to the development of Financial Engineering. Financial Engineering does not invade the field of Economics. As a science it is founded on facts as all true science must be and each fact is thoroughly checked. Financial Engineering V ^47192 vi PREFACE is just as applicable to a farm as to a railroad, just as applicable to a store as to a power system. It extends engineering over business and administrative problems. This book is written primarily for the practicing engineer. All mathematical deductions are worked out in detail, leaving no gaps for the reader to bridge. Many examples are also fully worked out, to illustrate the practical applications of the technique. The author has found by experience, that with the aid of an instructor, students can master the subject well. Like the practicing engineer, they are greatly interested in it, and seek it with greater avidity than any other course within the author's experience. The work in its various stages of development has been repeatedly submitted to engineers, in articles, lectures and addresses, so that it has had the benefit of their criticisms and suggestions. The Author's obligation extends to so many that he is unable to do justice to all. Especial obligation is acknowledged to Messrs. Kremers, Johnson and Byrne, and to Professor Teeter; also to the General Electric Co., the Westing- house Co., and Gordon and Finkheimer, of Portland, Oregon, for authentic data on the performance and costs of engineering equipment beyond that which had been accumulated by the author. O. B. GOLDMAN. CONTENTS ARTICLE CHAPTER I. INTRODUCTION PAGE 1. Duties of the Financial Engineer i 2. Cost Segregation and Cost Analysis i 3-4. Profit 2 5. Value 3 6-9. Basis of Rates 3 10. Division of a Service 6 1 1 . Fixed Charges and Operating Costs 6 12-13. Individual and Integral Undertakings 7 14. Competition, Law of Supply and Demand. ... 10 15-18. Replacement Costs, Market Value of Outstand- ing Liabilities, Capitalization of Earnings . . n CHAPTER II. COST SEGREGATION 19-20. Interest and Rents 15 21. Depreciation and Appreciation 16 22. Obsolescence 17 23. Inadequacy 17 24. Uselessness 18 25-26. Natural and Operating Life 18 27. Materials Consumed 21 28. Attendance .- 22 29. Maintenance 22 30. Stock Account 23 31-32. Storage, Sale, and Distribution 23 CHAPTER III. FUNDAMENTAL FINANCIAL CALCULATIONS 33-34. Simple and Compound Interest 25 35. Principal 26 vii Vlll CONTENTS 36-39. Equity N 27 40-41. Term Factor 31 42-47. Depreciation Rate 36 48. Present Worth of a Depreciating Equipment. . 42 49. Vestance 44 50. Depreciation Vestance 45 5 1 . Operating Vestance 48 52. Total Vestance 49 53. Annual Operating Cost 52 54. Taxes and Insurance 53 CHAPTER IV. BASIC COSTS 55. Basic Costs 59 56-57. Records 60 Prices and Operating Costs of: 58-63. Steam Engines 62 64-65. Boilers 75 66-67. Buildings 78 68-7 1 . Centrifugal Pumps 80 72-77. Motors and Generators 92 78. Oil Engines 99 78. Gas Producer Engines 99 79. Diesel Engines 101 80-83. Illuminating Gas anpl Suction Gas Engines 103 84. Prices of Standard Wrought Iron Pipe, Casing, Wood Pipe, Riveted Steel Pipe. Pipe Fric- tion. Cost of Tunnels, Canals, Excavations, Hydro-electric Installations 107 85. Table of Prices of Various Equipments 116 CHAPTER V. VESTANCES 86-87. The Time Element in Vestance 126 88-89. Valuance 128 90-97. The Steam Engine 129 98-99. Change Points 140 loo-ioi . Vestances at Full and Fractional Loads of Steam Engines 141 CONTENTS 102. Oil Engines 103. Diesel Engines I 5 104-106. Induction Motors J 5 2 107. Generators I S^ 108. Comparison of Power Units i5 8 109. Vestances at Full and Fractional Loads 159 110-115. Centrifugal Pumps 162 116-117. Vestances of Pipe *74 118-121. Determination of Velocity in Pipe for Best Financial Efficiency i?7 122. Oregon Rates. Vestances of Induction Motors with the Oregon Rates 184 . CHAPTER VI. UNIT COST DETERMINATION 123. Unit Cost of Service 187 124. Time Element in Service Costs 187 125-127. Change Points. Unit Cost with Constant Load, and Nearly Constant Load 188 128-131. The Two-Part Load 200 132. The Three Part Load. The w-part Load 210 133-137. Service Modulus 213 CHAPTER VII. DETERMINATION OF SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY 138. Variable Operating Cost 222 139. Total Production Cost 224 139. Determination of Equation of Actual Load Curve 224 140. Analysis of Type of System for Heat Transmis- sion 232 140. Determination of Point of Best Financial Effi- ciency 232 CHAPTER VIII. DETERMINATION OF TYPE AND SIZE OF UNITS 141. Stand-by Units 242 142. Determination of Cost of Service of a Unit at Fractional Loads 243 x CONTENTS 143. Determination of Cost of Service of a Plant at Fractional Loads 244 144-145. Determination of the Number of Units in a Plant for Best Financial Efficiency 246 146-147. Design of Plants for Best Financial Efficiency. 255 INDEX 2 6 9 FINANCIAL ENGINEERING CHAPTER I INTRODUCTION 1. In designing a system for the generation and distribution of power, or in laying out a -factory for the manufacture of a certain article, or in the rendition of any other service, the duties of the engineer are twofold: In the first place, he must so design the system that it will operate with reasonable con- tinuity. In the second place, he must so design the system that it will operate, not only with reasonable economy, but with the best possible economy. In any system already built and in operation, the engineer should be able to determine the exact unit cost of production of the service as well as determine what parts of the system are operating with good, and which with low economy, so that all leaks may be closed. In the following pages we shall treat the definite and exact solution of such problems, both in theory and in practical application. 2. The entire problem' of financial engineering, naturally divides itself into two parts, namely: (a) Cost segregration and (b) Cost analysis. Cost segregation deals with the proper allocation of the various items of expense, and their division and arrangement as best suited for cost analysis. This is essentially the book- keeping phase of the problem, and will therefore be treated in the following pages only where absolutely necessary and no more. Cost analysis deals with the utilization of the data, obtained from cost segregation, together with all available engineering data, for the purpose of reaching conclusions as to the economy INTRODUCTION of a system a$ ^1dieie.ora|i part, whether the system is proposed or already exists. 3. In cost analysis, we start from the basis that cost 1 governs price; competition and utility commissions, only the profit therein contained. It is evident that if a servant corpo- ration sells for less than cost, it must eventually go into bank- ruptcy. If it sells just at cost, the service will be rendered without reward. The servant company must therefore sell at cost plus a certain margin called profit. We can therefore say that cost plus profit equals price or rate. So also, an equipment must earn its costs and profits, when in operation, for certainly it cannot do so when idle. 4. Profit. This word is used in two senses, namely that of gross profit and that of net profit. Thus if an article cost us $100 and we sell it for $130, then the gross profit is $30. If the cost of making the sale is $20, then the net profit is $10. It is evident that gross profit is not profit at all for it includes essen- tial cost. When we use the term profit in the following pages we shall mean only net or real profit. It should be borne in mind that a finished article almost invariably carries not merely just one profit but usually a whole series of profits. Thus, for example, a certain amount of iron is mined at a cost of $i to which is added, let us say, a profit of 10%, making the price $1.10. It is converted into steel at a cost of, say, $0.50, making the cost to the steel pro- ducer $1.60. To this a profit of 10% is added, making the price of the steel $1.76. The manufacturer converts this steel into a machine part at a cost of $1.24, making a cost to him of $3. To this he adds 10% profit, making the price $3.30. Finally the sales organization disposes of this part at a cost of $0.70, making their cost $4. To this they likewise add 10% profit, making the final selling price $4.40. Under such conditions, we would have a total cost of i.oo -f 0.50 + 1-24 -f 0.70 = $3.44 and total profits of o.io + 0.16 -f- 0.30 + 0.40 = $0.96. 1 Goldman, Trans. A. I. E. E. BASIS OF RATES 3 The over-all per cent of profit is then ,0.96 -r- 3.44 = 35.8%. 5. Value. While we can and shall determine comparative values, we have come to realize that " value " has no meaning at all in any exact sense, for the reason that we possess no absolute standard of measurement by which to measure it. Furthermore value is intensely variable with circumstances. What, for example, is the value of a meal to one that has just been fed as compared with the value of the same meal to a man who is starving? It was for some time attempted to use the value of service as the basis of rates, but it was found to be synonymous with " what the traffic will bear," a method of coining money from the wants or distress of others. 6. Basis of Rates. Each servant corporation must recover from its employers, i.e., customers, all its costs, to which is added a certain per cent of profit. According to present arrangements, it is not reasonable to expect a service as a whole, to be rendered without profit. And what applies to the service as a whole, applies equally well to each and every single item of the service. It is as unreasonable to expect any item of the service to be rendered without profit as to expect the service as a whole to be so performed. But while it is com- paratively easy to determine if the service as a whole is profit- able, by little more than consulting bank balances, it is entirely another matter to do this for any single item of service. While it is unreasonable to expect a servant company to render a service or any item thereof without profit, it is equally unreasonable to expect one customer or any one group of cus- tomers, to pay, besides the cost burden that they place on the undertaking, more than their pro-rata of the profit. Nor would it be equitable to make one group of customers pay, besides their cost burden and pro-rata of profit, the cost burden of any other group of customers. In other words, it is not just to take the cost or profit burden, or any part thereof, off of one group of customers and place it on another group. Although in this way, the profit of the undertaking as a whole could be maintained unaltered, it would, nevertheless, be nothing short 4 INTRODUCTION of discrimination in favor of one group as against another. For this reason, it has been held illegal according to decisions of the U. S. Supreme Court. It is not to be inferred from the above that each servant company should receive the same per cent of profit. For under such conditions, there would be no reward for efficiency, no incentive to engage the best engineering and administrative talent, and no punishment in the way of reduced incomes for poor economy. 7. Since cost governs price, the entire problem of rate or price determination is primarily that of engineering; that of design for maximum economy is entirely so. Knowing the exact unit cost, the price made therefrom is a matter of judg- ment, for it is founded on fact. But a price or rate made with- out knowledge of the unit cost, is an unadulterated matter of guesswork, if not worse. Very often one company will set its prices in accordance with those of its competitors. That is one company will copy the prices of its competitors and use these prices as its own. This looks right on the ground that " competition regulates the price/' But it is not right because competition does not regulate the price but only at best what we may demand in profit. If the competitor's prices are less than our costs, then we must obtain a higher price, turn our attention to some other line of endeavor or else go broke. 8. In the problem of rate determination for public utilities, altogether too much stress has been laid on the various deci- sions of our state courts, in spite of the fact that these deci- sions have been exceedingly contradictory. Nor could these decisions help being so, for the problems to be solved are those of engineering and not law. The latter is therefore helpless in reaching a solution until engineering shows the way. In other words engineering must lead in these matters and determine the correct solutions to which the law must conform. Courts base their decisions on the expert opinion of engineers in so far as these opinions are based on exact scientific knowl- edge. Beyond that the courts have as much right to guess as DIVISION OF SERVICE 5 anyone. If the opinion of the experts were crude and far from the truth, then the decisions based thereon are bound to be equally bad. It must be borne in mind that bad decisions of our courts, going as they do, counter current to the trend of natural development of human society, are a great hindrance to progress, while good ones are a great aid. While bad decisions cannot prevent, they most assuredly do delay advancement. After all it must be borne in mind that knowledge is all one and indi- visible. We have hypothetical divisions, not real ones. One so-called branch of knowledge can no more live alone and separate from all other knowledge than the hand or the heart can exist separate from the whole living being. A discovery or advancement in engineering or a good decision in law both aid one another in that both aid in the progress of human society 9. As we have pointed out above, there are primarily two divisions to our entire problem, namely that of cost segregation and cost analysis, as applied to (a) cost determination and (b) design for best economy. Each charge growing out of the rendition of a service must be recorded, charged against the proper account, while mixed charges, applying to more than one item, must be properly apportioned. The apportioning of mixed items of expense is a task requiring a broad knowledge of engineering and perfect common sense. Cost segregation must be made in accordance with the demands of cost analysis. For the cost segregation is of no value in and of itself, and is useful only as an absolutely necessary basis for cost analysis. Cost segregation alone, where no further use is made of it, as is so often done, represents just so much useless and undigested data. Drawing conclusions from data is the part of cost analysis. It is the business of determining exact unit cost for actual load conditions under which the 'system operates, as well as the aggregate effect of the load characteristics of each unit of the system. In the problem of design, the field is still broader. The system is not yet fixed, but is to be determined so that, 6 INTRODUCTION . under actual or anticipated load conditions, the greatest econ- omy will be attained. This part of the problem may be subdivided into: (a) The determination of the type and size of a system for best economy (b) The determination of the type and size of the units of a system for best economy (c) The determination of unit and total annual production cost. 10. The service that is rendered by every servant, whether of a private or public character, may be divided into four primary parts, namely: (a) Collection, (b) Production, (c) Transmission (Transportation) and (d) Distribution. A railway service, for example, consists in the collection of freight, its transportation and distribution, production being zero as nothing is produced as such. A power service consists in the collection (storage) of water for power, the production of power, its transmission and distribution. Again an engine factory service consists in the collection of materials of manu- facture, the production of the engines, their transportation and distribution. The actual rendition of one of these divisions of a complete service may be delegated to an agent. Thus, for example, in the case of an engine factory, the actual transportation of the engines is almost invariably left to a railway company, who, by specializing for this particular division of service, can render it at an incomparably lower cost. So also the actual distribu- tion of the engines may be left to a sales agency. Such is in fact very commonly done. 11. All cost pertaining to the rendition of a service may be divided into two parts, namely: (a) Fixed charges, and (b) Operating costs. FIXED AND OPERATING COSTS 7 Fixed charges are all those costs which continue when the operation of a system is discontinued. Thus, interest, taxes, and the like continue whether the plant is idle or in use, i.e., productive or nonproductive. Thus the total annual costs of reserve units, while standing idle, is merely the total annual fixed charges of these units, the operating costs being zero. Such stand-by costs must be considered an insurance against discontinuity of service and needs be as carefully determined for best economy as any other factor of the system. The operating cost consists of all costs over and above those of fixed charges. The costs of fuel, oil, water (materials con- sumed) , attendance, and the like are clearly all operating costs. But the salary of the engineer may be partly an operating and partly a capital cost. When the time and skill of the engineer is spent on the running of the system, or its maintenance, or repair, it is an operating cost. But when it is spent on the design of extensions or their construction, the salary must be charged to capital and bear fixed charges as any other item of investment. The same holds for supervision and the like of what may be called the overhead services. We may therefore segregate our costs as follows: (1) Fixed charges: (a) Interest and rents, (b) Depreciation, (c) Taxes, (d) Insurance. (2) Operating costs: (a) Materials consumed, power, etc., (b) Attendance, (c) Maintenance, (d) Repair. 12. Ordinarily, undertakings are divided into private and public utilities. But the production of steel, clothes, food or the like is as much a public necessity as transportation or power service. The above classification, though legal, is therefore 8 INTRODUCTION both arbitrary and temporary. Instead of the above, we shall classify all undertakings into individual and integral under- takings, as well as mixed. An integral system is one that is designed to serve one and only one one community or one given district and all of that community or district. Such a system is designed for the general needs of the community that it serves and not the particular needs of certain individuals of that district, except in so far as service connections are needed. These needs can be closely determined from previous experience with similar communities. An individual system is one that is designed to meet part or all of the needs of certain, definite individuals. The distinc- tion may be made clear by an illustration. Thus a telephone or power system is distinctly an integral system. You, indi- vidually, do not pay any charge unless you are actually receiv- ing service from the undertaking, although the necessary costs and profits for operating the undertaking are collected from the community, in particular from that part of the community which does receive service. The undertaking is 'ready' to serve you, but that costs you nothing. All public utilities and most private companies are integral undertakings. On the other hand if, let us say, six men join to put in a common pumping system, we have an individual system, for the size and type of the system are chosen to suit the needs of just these particular six men and no more or less. In such a system the readiness to serve must be paid for whether service is actually taken or not. Example i : Four men, each owning 40 acres of land, put in a common pumping system costing $4000. One man uses an average of 5 acre feet of water per acre per year, the second 4, the third 3, and the fourth no water at all. If the fixed charges on the system are 12%, and the operating costs are 50 cents per acre foot, determine the annual charges that must be assessed against each man. Solution: The fixed charges are 4000 X 12% = $480. INDIVIDUAL AND INTEGRAL SYSTEMS 9 Since the system is designed to serve each man equally and as each man owns the same amount of land, the fixed charges will be divided equally, each man paying one-fourth, or $120, per year. The total water pumped is 5 X 40 = 200 acre feet. 4 X 40 = 1 60 " " 3 X 40 = 120 " " o X 40 = OOP " " Total 480 of water, costing 480 X .50 = $240. This amount may be considered as divided into 5+4 + 3+0 = 12 parts. Of this the first man, using 5 acre feet, pays in operating costs T \ X $240 = $100; the second pays T 4 2 X $240 = $80; the third pays i 3 2- X $240 = $60; and the fourth, using no water, pays no operating cost. Evidently, then, the first man must pay a total in fixed and operating charges of 120 + 100 = $220; the second, 120 + 80 = $200; the third, 120 + 60 = $180; and the fourth 1 20.00 + oo.oo = $120. Although the fourth man receives no service, he must pay his pro-rata of fixed charges, his " readiness to serve " charge. For were he not in the system, the size and thus the total cost and fixed charges of the system could be proportionately re- duced. This, however, holds true only for individual systems. 13. The government has always exercised more or less con- trol over what are now classed as public utilities. But not until special commissions were formed did this control begin to be io INTRODUCTION effective. Even so, the commissions have been weak and unreasonable in many respects. But this must be expected to a certain extent in the incipient exercise of this authority. And it will continue until the commissions are composed of experts and experts only. It is now clearly established that actual, real costs must be used as the basis of rates in such undertakings, and not the hypothetical " value of service," or the piratical " what the service will bear " basis. "For," says Justice Hughes in the de- cision of the United States Supreme Court in the North Dakota coal case, " where it is established that a commodity . . . has a rate imposed, which would compel the carrier to transport it ... virtually at cost, and thus the carrier would be denied a reasonable reward for its (that particular) service ... it must be concluded that the state (commission) has exceeded its authority." This decision, at a single sweep, landed most of the work of our inexpert state commissions in the waste basket. 14. Competition is industrial disorganization. Under such conditions no service can be rendered economically, for it involves the wasteful expenditure of capital and labor. The formation of the so-called trusts is a natural attempt to evade such conditions. Under competitive conditions, the public gets poor and expensive service while the returns on the under- taking are often subnormal. It is true that in some cases better and cheaper service has been rendered under competitive con- ditions. But this was due either to lack of ability of those in control of the first undertaking, or abnormally large profits collected, due either to lack of authority, or ability, of the commission or both. Commissions should have the ability to know what is right, the courage to do what is right, and the authority to execute it. We emphasize this because it seems probable that the authority of the commissions will be extended eventually to all integral systems. Competition is always the short cut to higher costs. Economists have stated that there was such a thing as the law of demand and supply, which automatically regu- lates prices. Even the most elementary study shows that COMPARATIVE VALUES n there is no such law. In the first place the demand for any necessity is practically a constant but the amount used will depend, not so much upon the price, as upon the wealth (pur- chasing ability) of the consumer. In the second place, while it is true that the prices vary with the supply, this is due to the unregulated, unscientific methods of production, giving us alternate waves of oversupply and undersupply, alternate waves of waste and want. If there is such a law, it is the law of the wilds, where they stuff in summer and starve in winter. Certainly it is not the law of intelligent production. Cooperation and not competition, work and not war and waste is the keystone of our modern social structure. This accounts for the condensation of small industrial units into fewer large one, for the formation of the labor unions, fraterni- ties and the like. Much as capitalism has been condemned, one thing certainly may be said for it; that it compelled cooperation and coordination amongst workers long before there would have been voluntary cooperation and in so doing has served to advance civilization. But with the coming of voluntary association and cooperation, compulsory association and compulsory cooperation will be little tolerated. 15. The comparative value of an undertaking is in direct proportion to its size and economy of rendering the service. There are even yet, however, a number of different " methods " for determining the so-called " value" of an undertaking. These are based on (i) Original Cost, (2) Replacement Cost, (3) Market Value of Outstanding Liabilities, and finally that based on (4) the Capitalization of Earnings. The Original Cost includes not only the money spent on the undertaking at its inception, to bring it into being, but also all money spent for extensions and improvements, except in so far as such may be chargeable to depreciation. Original cost will however, not give the comparative value of an undertaking for, according to this basis, the more a given undertaking cost, the greater would be its worth. It is evident then that original cost will lead to the comparative value only if we take into consideration most fully the economy of the undertaking. In 12 INTRODUCTION other words we must determine exactly how well it serves the purpose for which it was constructed. The Replacement Cost is the amount of money that would have to be expended now, or at some other time subsequent to the original formation of the undertaking, to bring it into being. Replacement cost includes, besides the original cost, the depreciation of this sum of money in purchasing power during the interim and therefore tends to give the undertaking a speculative value. The " market value" of the outstanding liabilities of a com- pany should give some information as to the comparative value of an undertaking, but, with rare exception, they most cer- tainly do not. The real value is usually quite dilute. There are many reasons for this, but fundamentally, whenever a corporation has the power to create an artificial market, then these " market values" represent nothing but the ability of the companies along these lines of manipulation. Attempting to find the " value" of an undertaking from its earnings, so as to determine the earnings from the " value" so found, is like trying to find the end of a circle. A company might, for example, be guilty of exploiting the community that it serves. If the " value" of the undertaking under such conditions is determined from its earnings, it would give such a concern a "vested right" in such exploitation. It must be clearly kept in mind in discussing the above that the word " value " has no definite meaning. In view of this, the purposelessness of the above "methods" becomes more evident. In order to illustrate the enormous discrepancy in deter- mining so-called "value" by the various methods above, we give below some of these "values" for one certain company. (a) Market "Value" as per outstanding liabilities, $72,000,000 (b) Present "Value" as per utility commission. . . 36,000,000 (c) Present "Value" as per assessor 15,000,000 (d) "Value" as per capitalized earning 20,000,000 (e) Estimated real present value 16,000,000 The above speaks for itself. A guess is scientific in comparison. PROFITS 13 16. It took the great war to bring to full realization the fact that gold is not a necessity, that necessities are not measured by gold, but gold by necessities. Under the present era (1918) of "high prices," it takes no more sacks of potatoes or suits of clothes to build a house than under normal times although it takes twice as much gold. It is not the price of all commodi- ties that has increased, but the price of gold that has decreased. It is far more sensible to acknowledge this view than to take the opposite one that everything has changed but gold. When, for example, the price of wheat was set at $2 per bushel, it was in reality not the price of wheat that was set at all, but the price of gold that was set. This allowed a 50% depreciation in the price of gold and had this not been done, its depreciation would have been far greater, far, probably, below its cost of production. That is why the payment of wages in gold has proved unsatis- factory, because the " doubling" of wages under such conditions meant no raise at all. But were they based on necessary com- modities, wages would at once be stabilized. 17. Profits have been a fruitful source of political and economic discussion. The view generally taken is that profit is the reward that the servant gets fpr rendering a service, notwithstanding that the wages, that is the total costs, do pay for this service. The peculiar anomaly, therefore, exists that, if this servant is a human being, he gets only wages for rendering the service, but, if the servant is an undertaking, it gets not only full wages, but profit besides. Looking at it from the standpoint of the purchaser, i.e., the employer of the service, what does he, the purchaser (employer) , get for the profit that he may pay to the servant? It is evident that, if the costs (or wages) are the full reward for the service rendered, the employer (customer) obtains absolutely nothing for this profit that he pays above the cost (total wages) of a commodity. 18. It is evident from the above that all business organiza- tions, whether we choose to speak of them as companies, under- takings, utilities, corporations, or the like, have but one purpose and that is the rendition of a service for others. They are evidently just servant organizations and we shall in these pages 14 INTRODUCTION speak of them on occasion as servants. The relation of a customer, patron, or purchaser to an organization or individual from whom he obtains a service (or commodity) is that of employer to employee in that he employs the organization to render the service. The service of such an organization is rendered by the organized efforts of individual servants. The relation of the organization to the individual servants is that of employer to employee. But the relation of the public to a public utility is (or should be) that of Master to Servant. In the case of an individual servant, there is no question but that the price equals the total wages, or cost of the service. But in the case of an organization of Servants, as a corporation, the price is greater, often much greater, than the wages (costs) , and this excess or profit does not go to the individual servants of that organization but to others, who render no service whatever. It must be borne in mind clearly that all costs of a given service are exactly equal to all wages associated with that service. Usually such costs are spoken of as wages proper and cost of materials. But the cost of the materials themselves is only wages, so that, in fact, cost equals total wages. So-called overhead expenses may be divided into salaries paid for essen- tial service rendered and thus come under wages, rents which in the main come under profits and so forth. Wages as used herein is understood to mean all moneys paid for essential service rendered. Iron ore, for example, has no cost, being a natural resource. But a finished machine has for its cost as distinguished from its price all moneys paid for essential service of whatever kind, in the mining of the ore, its smelting, transportation, etc., and its manufacture finally into a finished machine. The cost of such a machine, or any other material we may buy, is the sum total of all wages associated with its production and manufacture. The best member of this, or any other country, is the one that renders the community the best and most service. It certainly is not the " successful man" who, instead of serving the community, has merely proven his ability to take and gather the most unto himself, instead of giving the most. CHAPTER II COST SEGREGATION Interest. Depreciation and Appreciation. Taxes. Insurance. Materials Consumed. Attendance. Maintenance Repair. Life of Structures and Equipment. Stock. Storage. Sale and Distribution. 19. As pointed out in the first chapter, all production costs may be divided into fixed charges and operating costs. The fixed charges are divided into the following divisions: (a) Interest and rents, (b) Depreciation, (c) Taxes, (d) Insurance, while the operating costs are divided into : (a) Materials consumed, power, etc., (b) Attendance, (c) Maintenance, (d) Repair. 20. Interest. The price of any service, or commodity, is made up of just two items, namely: the cost, or total wages and salaries, and the profit. To which of these does interest belong? It is a nonproductive charge, not being paid to the individual servants of the organization but to some one else who renders no service. It is evidently a profit charge and is so treated by both federal and state commissions. So-called public utilities are allowed all their profits in the form of interest on the " capital" invested, and no profit is allowed in any other form. This is from the actual standpoint of the customer (employer) of the service. An organization using 15 16 COST SEGREGATION" borrowea money sees this interest as a cost, and we so treat it in the following pages. Interest is charged on all money invested. This consists of the price paid for all machinery, materials, and wages, as well as unavoidable losses, such as loss of time during construction due to inclement weather, loss of tools, cost of temporary structures and the like. The rate of interest varies greatly, usually from two to eight per cent. Governments normally pay two to two and a half per cent, large industrial undertakings four to six per cent, while small firms and individuals of very moderate wealth pay seven to eight per cent. It is not clear what determines the rate of interest unless it is "all that the traffic will bear." It is often said that the "risk" determines the rate of interest, so that the greater the "risk" the greater the rate of interest. In other words, the smaller and weaker the firm (or individual) the greater the profit burden it must bear. But as # matter of fact, the greater the rate of interest, the greater the "risk" naturally becomes. So this is evidently not the basis of interest but rather the want or distress of the user. As a matter of fact, money cannot produce money any more than a reservoir can produce water. In general, what applies to interest, which is merely the .rent paid for capital, applies equally well to rent paid for the loan of any other commodity, in so far as the rent charged for any such commodity exceeds the actual costs associated with the making and collecting of the loan, the maintenance of the commodity in its original condition and the depreciation associated with this article. Net profit, net interest and net rent are all alike in that they are all net profits, i.e., charges for which no services are rendered. The fact that the profits are paid to others, besides the owners of the organization, does not reduce the burden that the customer must bear. 21. Depreciation and Appreciation. As things become older, their comparative value changes. If the comparative value decreases, we have depreciation, while if it increases, DEPRECIATION 17 we have appreciation, or negative depreciation. A given item depreciates due to its tendency to become (a) obsolete, (b) inadequate, (c) useless. What we speak of as land has really two meanings, namely, the land itself and location. For agricultural purposes we buy the land itself, and, unless we fertilize it, it depreciates rapidly. For marketing (selling) purposes we buy location and this invariably appreciates with increase in population. The price or rent of locations varies with the serviceability of the loca- tions to the public (the employer). By thus varying the price, the location is made to serve the owner (servant) instead of the public (the employer). Locations thus become the toll- gates of industry, though not the only tollgates. 22. Obsolescence. Progress in engineering results in im- provements that yield either increased efficiency or reduced first cost. For this reason, a unit that we have in service loses its comparative value or becomes obsolete. When the efficiency of new apparatus has been so far improved that it would pay to discard the unit in use entirely, then the unit in use has become completely obsolete and its comparative value zero or even much less, even though it is otherwise in perfect run- ning condition. Thus, for example, a uniflow engine will give as good efficiency as a triple expansion engine while costing only 40% as much. Had one just purchased a triple expansion engine, when the uniflow appeared, 60% of the purchase price of the former would have gone into depreciation at once, for the reason that the same service could be performed equally well with this reduction in capital outlay. 23. Inadequacy. It often happens in a new and fast growing business that certain items of the system may be replaced to advantage though not obsolete. Thus in a power plant, it often is an advantage to replace a number of smaller units, however modern, that have accumulated during growth, with one large unit. Again in the case of a railway system, i8 COST SEGREGATION the traffic may increase so that a single track bridge is incapable of handling the traffic, although it is still in good condition. In such cases the item affected has become inadequate. It may be argued that a larger unit should have been installed in the first place and in some cases that would have been right, but more often this would have resulted in reduced economy due to too early an outlay of capital. Only when the plan of installing smaller units first and replacing them with larger units later results is better economy than having installed the larger units at the start, have we a case of inadequacy of equipment. When conditions are reversed we have a case of inadequacy of the management. 24. Uselessness. A system or part thereof becomes use- less (without use) when the service for which it was intended is no longer demanded. Thus a temporary structure, used during construction, becomes useless and in fact often a nuisance, when the construction is finished. Again a logging railroad may be built into a certain district. When the logging is finished, the railroad becomes useless, though some parts of it may be salvaged. 25. Fundamentally, the object of cost calculation is to maintain the capital intact, neither permitting it to increase nor decrease. For this reason, it is necessary to set aside each year a certain sum of money, equal to the amount that the system has depreciated in that time. This money, so set aside, forms the Depreciation Reserve or Sinking Fund, the former term being preferable. This depreciation is an essential part of the cost of the production of the service being rendered. The amount of the depreciation reserve laid by should be such that, at any time that the equipment has depreciated to zero, it, together with the accumulated interest thereon, should amount to the original cost of the equipment, less whatever scrap value it may have. So also if the equipment appre- ciates in value, this amount must be deducted from the pro- duction cost for the very same reasons that depreciation must be added. The Depreciation Reserve is not a dona- tion to the Servant from the employer, but a trust fund that LIFE 19 must all be spent for the purposes for which it was allowed or returned. Primarily, the amount of the depreciation reserve that must be annually laid by depends upon the life of the structure or equipment. This life must be determined from experience and is now fairly well known. It is given as well as possible in the table below. TABLE i APPROXIMATE USEFUL LIFE OF STRUCTURES AND EQUIPMENTS DESCRIPTION LiFE,Yrs. DESCRIPTION LIFE,YRS. ASH CONVEYERS, steam jet. ... 50 ELEVATORS, bucket 10 BRIDGES, concrete Permanent FANS, centrifugal 25 " steel 40 GRATES 10 " wood 15 GENERATORS, A.C 30 BUILDING, concrete Permanent " D.C 20 brick. 50 HEATERS, feed water, closed. . . 20 wood or sheet-iron. . 15 " " " open 30 BOILERS, fire-tube 15 HOISTING MACHINERY 20 " water-tube 40 MOTORS same as generators BELTS, leather 8 PUMPS, plunger 15 BINS, steel 25 " centrifugal 25 " wood 10 PRODUCERS, gas 20 CHIMNEYS, concrete Permanent PLPING 35 brick 50 ROTARY CONVERTERS 30 steel, self-sustaining 35 STORAGE BATTERIES 4 sheet-iron 8 SWITCHBOARD 50 CONDENSERS, jet 20 TRANSFORMERS 50 surface 20 TRANSMISSION LINE, pole 12 CONVEYERS, bucket 20 " " steel 20 belt 7 TRAVELING CRANES 50 ENGINES, high-speed 15 TURBINES, steam 20 " low-speed 25 " water 25 ECONOMIZERS 15 WIRING, electric 30 The above list is necessarily very limited. Besides it must be borne in mind that the life given is necessarily rather approxi- mate. It depends greatly on the quality of the apparatus when purchased, together with a multitude of other conditions. The above table is for good substantial equipment, not the cheap stuff that is also found on the market. Other conditions that affect life are the surrounding ele- ments. Thus, the life of wood pipe is very great, if it is kept thoroughly impregnated with water, but deteriorates rapidly with disuse. So again soil conditions often are a deciding factor, iron pipe decomposing rapidly in alkaline soil or sea 20 COST SEGREGATION water. On the other hand the life of stacks is greatly influ- enced by the amount of sulphur in the fuel. This is due to the formation of sulphurous acid from the sulphur dioxide and its oxidation to sulphuric acid in the presence of water and free oxygen, when the temperature is not too high. This acid attacks the iron, causing very active corrosion. It seems to be most active during light or no load periods. So also, the life of boilers being fed with untreated water, depends more upon the chemical composition of the raw water than upon any one other item. The life of all apparatus depends upon the skill of the operat- ing engineers as well as the proper correlation of all parts of the system by the designing engineer. All such items must be taken into account in determining the probable life of a system. But the difference between normal and actual conditions should be charged to management. With unusually good operators a plant will tend to appreciate in value for quite a few years. This should be credited not only to the operators but to the management which had the courage and foresight to pay the price for first class men. On the other hand if the first cost of a system or the operating cost is abnormally high due to poor designing, the difference should be charged to bad management where it ultimately belongs. The reader is cautioned not to confuse depreciation with wear, as is often done in practice. Depreciation is a fixed charge, continuing unaltered whether the unit is in use or idle. Wear is an operating cost, the amount of wear being in direct proportion to the amount of use. When a machine is worn out, it has not depreciated to zero worth. It has merely worn out and we have merely a major repair item. Thus when a bearing is worn out, we replace the bearing and thus repair the engine. This is admittedly a repair item. So when an engine is worn out, we replace the engine and thus repair the power plant. The replacement of an entire worn- out engine is just as much a repair item as the replacement of any of its parts, the only distinction being the degree of the replacement. Whenever the cost of maintenance of an old MATERIALS CONSUMED 21 unit is greater than the total costs (fixed charges and mainte- nance) of a new unit, it is evident that it will pay us to discard the old unit. In such a case it is worn out. We may therefore distinguish two lives for all things, namely the natural life and the operating life. The natural life depends on the factors of obsolescence, inadequacy and uselessness. The operating life depends upon the amount of use that may be obtained from anything before it is worn out. The shorter of the two determines the true life of the unit. It would be senseless and very bad engineering to make the operating life of a unit longer than its natural life. Yet this very thing has been done in the past. The large, extremely low speed engines of forty and fifty years ago, some of which are still in use, refuse to wear out, though they are hopelessly obsolete. It is for this reason that practically none of this type of machinery is now built. 26. Taxes may be definitely determined in any community, usually running between one and two per cent of the assessed value. The assessed value is usually somewhat below the original cost of the undertaking and this must of course be taken into account. Insurance usually runs between 0.5 and 1.5% but this again depends upon the construction of the system. A concrete building that is absolutely fireproof need carry no insurance, thus tending to reduce the total cost of such a structure. If, for example, a building cost $100,000 and insurance is i%, then the annual insurance outlay is $1000 which, at 5%, interest, capitalizes at $20,000. A fireproof building would be worth this much more. But to all this we must add the cost of insurance on the equipment that is to be housed within the building, further increasing the value of the firepoof structure. 27. Materials Consumed. Any undertaking rendering a given service, whether a factory producing a given commodity or a power plant, uses up a certain amount of materials, which is either entirely consumed or else rendered into a more or less 22 COST SEGREGATION worthless condition. The amount of material so consumed is very closely in direct proportion to the amount of service produced. Thus a steam power .plant consumes fuel, lubricating oils and greases, waste, packing, paints, and the like. A hydro-electric power plant consumes all of these except fuel. A factory pro- ducing machinery consumes a great variety of supplies as well as iron, steel, brass, and the like in considerable quanti- ties, which are rendered into the more or less useless form of turnings and punchings. The cost of all such materials consumed should be charged under this head to operation, together with their cost of pur- chase and delivery, and the cost of waste removal, less the salvage if any. 28. Attendance. The salary or wages paid to all who are actually engaged in the production of a given service should be charged to operation under the head of attendance. In an iron works, for example, this includes the wages paid to machin- ists, molders, boiler makers, helpers, clerks, bookkeepers, col- lectors, and the like, together with such parts of the salary of the superintendent, engineer, and manager as are devoted to the actual production of the service. And so for other under- takings. 29. Maintenance. -- The cost of all materials and labor, engineering, and inspection service necessary to keep an under- taking in good running condition, so that its productive capacity does not decline, should be charged to operation under the head of maintenance. The largest item that comes under this head is repair and replacement of worn-out parts of equipment and the maintenance of grounds and buildings. But as pre- viously pointed out, the cost of equipment replaced because of obsolescence, inadequacy, or uselessness should be assessd to fixed charges under the head of depreciation. Attendance, maintenance, and materials consumed decrease with the increase in quality of the system, while its first cost increases therewith. Therefore as the fixed charges are allowed to increase, the operating costs decrease. This gives us one of STORAGE, SALE AND DISTRIBUTION 23 our most important problems, to find, under given load con- ditions, where the sum of the fixed charges and operating costs, i.e., the total costs of production of the service, are a minimum. 30. Stock Account. It is invariably necessary in any undertaking, whether large or small, to carry a special, tempo- rary account of materials on hand for use or sale. This account should be divided into two parts, namely raw .stock and fin- ished product. Thus in a steam-electric power plant, we have on hand usually considerable materials, such as fuel, supplies, and the like, which should be charged to capital until used and then transferred to operation. In this case we have no fin- ished product, for electricity is and must be used as produced, excepting what little is carried over in storage batteries. In the case of an engine factory, we have the raw materials account, covering materials consumed as well as those going into the finished product. This stands as a debit against the production of the engines, while the finished product stands as a credit account. 31. Storage, Sale and Distribution. As previously shown, the costs of rendering a given service may be divided into col- lection, production, transportation, and distribution, the latter including sales. The item of storage, sales, and distribution, while in some business the most important function, is no exception to the treatment given any of the other items in the complete rendition of a service. The more uniformly we operate throughout the year the more must be stored, i.e., we can decrease the production cost at the expense of the storage cost. To determine the conditions of best economy, in such a case, gives a typical problem in cost analysis. Selling is one of the most important services that is rendered, because it comprises not only the actual selling but the giving of advice and assistance to the purchaser in the successful use of the product being sold, as well as trouble shooting, that is the correction of troubles that appear in the use of the prod- uct whether due to more or less defect in the product or to difficulties that the purchaser gets himself into. 24 ^ COST SEGREGATION 32. In any actual plant all the above items of expense are found to be more or less mixed together since operation, main- tenance, replacement, and the like are taking place simul- taneously. They can, of course, be readily segregated by an engineer, not with absolute accuracy, but sufficiently so for the most exacting practical needs. CHAPTER III FUNDAMENTAL FINANCIAL CALCULATIONS Compound Interest. Principal. Equity. Term Factor. Operating Vestance. Depreciation Rate. Present Worth of a Depreciating Equipment. Depreciation Vestance. Total Vestance. Annual Operating Costs. 33. Interest. The loan of money is paid for by the annual payment of interest usually expressed as a certain per cent of the principal. The loan of money is always for a certain term of years only, at the end of which time the principal must be returned, the interest in the meantime being paid annually. This is so-called Simple Interest. On the other hand, instead of paying the interest when due, it may be added to the prin- cipal and become a part thereof, the sum of the two thereafter bearing interest. This is called Compound Interest. Thus we meet the problems of how much would a principal amount to in a certain number of years, bearing compound interest at a given rate, and conversely, knowing that a certain amount is due in a given number of years at a known rate, to find its present value. 34. Compound Interest Formula. Letting P = the principal, R = the rate of interest, N = the number of years, and A = the amount at compound interest, then A = P(i + /?) This can be easily shown as follows : Example i. To begin with we have AQ = P dollars. 25 26 FUNDAMENTAL FINANCIAL CALCULATIONS At the end of the first year, this has increased to A l = P(i + R), so at the end of the second year, it is A, =P(i +K) (i + R) =P(i+R)\ And at the end of the third year, we have A 3 = P(i + R)* (i + R) = P(i + tf)' 3 , and so on. At the end of (n) years, it is evidently A n = P(l +R)" (i) Example 2. How much will $100 amount to in 10 years at 8% compound interest? Solution: In this case A = ioo(i.o8) 10 or log A = log 100 -f 10 log 1.08 = 2 + 10 X 0.0334 = 2.334, so that A = $215.80. Example 3 . In how many years will a sum of money treble itself at 10% compound interest? Solution: In this case we have where p ~ 3 and R = o.io, so that 3 = (i.i) n , and n log i.i = log 3, Iog3 Q-477I or n = ; = , log i.i 0.0414 whence n = 11.52 years. 35. Principal. Knowing the sum of money that will be due in a certain number of years at a given rate of compound interest, we can find its present worth, by solving the above formula for (P), thus P = - -^- (2) ' EQUITY 27 Example 4. One thousand dollars is due in ten years, the compound interest rate being 7%. What is its present worth? Solution: In this case A = $1000, R = 7% and n = 10, 1000 so that P = -, TTT, or log P = log 1000 10 log 1.07 = 3 - 10 X 0.0294 = 2.706, whence P = $508.10. 36. Equity. If the operation of a certain equipment is guaranteed, and its actual performance falls below this, the cost of operation of this unit will be increased by a certain amount annually. In such a case a certain sum (the equity) must be deducted from the price of the unit to compensate the purchaser for this loss in guaranteed economy. On the other hand, if the guarantee is exceeded, a certain sum should be added to the price to compensate the manufacturer for the increased economy attained. In determining the equity, it is evident that if we capitalize this annual amount by which the guaranteed economy falls short (or is exceeded) at the usual rate of interest, and pay this sum to the purchaser, he will receive this annuity, as interest therefrom forever, whereas the equipment has a limited life only. We would thus be paying too much. Call- ing (a) the annuity, or the annual cost of operation in excess of the guarantee, and (R) the interest rate, then this capitalized amount (A) will equal merely Instead of deducting (A) dollars from the selling price (C) which, as pointed out above, would be too large a sum, we should loan this sum (A) to the purchaser during the life of the unit, without charge, the principal to be returned by him at the end of this time, the interest thereon, would compensate the 28 FUNDAMENTAL FINANCIAL CALCULATIONS purchaser for the reduced efficiency. This would be an equi- table adjustment. But this adjustment can be simplified. If instead of waiting for (n) years, the life of the equipment before the amount (^4) is returned, we deduct from this amount (A), due in (n) years, its present worth (P), then we can reach a complete adjustment at once. This difference (A - P) is the equity (), so that E = (A - P), but '-(Tng;, and A = ~ 37. Alternate Solution. We can obtain the above formula in a more direct but also more laborious way as follows: To begin with we lay aside E dollars. At the end of the first year, this has increased to E (i + R) dollars, from which we must pay the purchaser (a) dollars. So we have left [E(i + R)- a] dollars. This will increase to [(i + R) - a] (i + R) = E(i + R) 2 - a(i + R) dollars at the end of the second year, when we must pay out the second installment of the annuity. After this we have left E(i + R) 2 - a(i + R) - a which will increase to [E(i + #) 3 - a(i + R) 2 - a(i + P)] at end of the third year. So then at the end of the third year, we have left after paying the third installment of (a) dollars E(i + #) 3 - a(i + R) 2 - a(i + R) - a, EQUITY 29 and so at the end of (n) years we have left E(i + RY - a(i + RY' 1 - a(i + ^) n ~ 2 . . . a(i + R) 2 - a(i + R) - a. If (n) is the life of the equipment the term of the annuity - then the balance left on hand should be zero, so that - a(i + R) - a~] = o, or E(i + RY - fl[(i + #) n - ! +-(i + ^) n ~ 2 + (i + J?) + i]. We have next to find the sum of the series on the right hand. Calling this sum (S) , then (i + RY = aS y and S = (i + RY- 1 + (i + RY~ 2 ' ' ' + (i + R) + i. Multiplying through by (i + R), we get (i + R)S = (i + #)" + (i + tf)"- 1 . . + (i + ) 3 + (i + ) 2 +.(i + ^)- Now subtracting (5) from the left hand side of this equation, and its value, the foregoing series, from the right hand side, we get (i + K)S - S = (i + RY - i, or RS = (i + RY - i, (i + RY - i and o = - ^ But since E(i + i?) n = aS we can substitute for (S) its value just found and get or f(i + *> - il a f Ri (i+RY ] Rl (i as before. 38. Example 5. A 100 h.p. engine is guaranteed to consume not over 2O# of steam per h.p.h. A test shows its actual con- sumption to be 22#. If the engine is run at full load for 3000 30 FUNDAMENTAL FINANCIAL CALCULATIONS hours per year, and the steam costs 20 cents per iooo#, what will be the amount which should be deducted from the pur- chase price of the engine to compensate for this loss in econ- omy? The life of the engine is 20 years; interest rate 5 %. Solution: Under the above conditions, the engine evidently uses 2oo# of steam per hour in excess of the guarantee. In a year of 3000 hours it would use in excess 3000 X 200 = 6oo,ooo#, .. 600,000 ft costing - X 0.20 = $120, 1000 so we have that a = $120, R = 0.05, n = 20, whence ^ 1 20 (i.cx) 20 i E - - 24 x or E = $1239. 39. Example 6. A 300 h.p. motor is sold for $1800 with a guaranteed efficiency of 92 %. It is used at full load for 3000 hours per year. The life of the motor is 20 years and the power costs one cent per h.p.h. If the motor only develops 90% effi- ciency, how much should be deducted from the price to com- pensate the purchaser therefor, assuming an interest rate of Solution: The horse power that would be consumed by the motor under the guarantee is 300 -5- 0.92 = 326 h.p., while it actually consumes 300 -5- 0.90 = 333.3 h.p., a difference of 333-3 - 3 2 6 = 7-3 h.p., costing 7.3 cents per hour or 0.073 X 3000 = $219 per year. TERM FACTOR 31 So we have here that a = $219, R = 0.06, n = 20, so that 219 (i.o6) 20 -i " 0.06' (i.o6) 20 or E = $2450. The corrected price would then be 1800 2450 = $650. That is, assuming we can get a motor of 92 % efficiency from another firm, we could not afford to accept this motor as a present, unless we received therewith a cash bonus of $650. 40. Term Factor. The factor ^ + f^~ T we shall desig- (i + R) n nate for convenience the term factor (T), so that aT ( N ... = ' ' ' (4) In the table below, we give the values of this term factor (T) for various values of the interest (R) and the life (n). FUNDAMENTAL FINANCIAL CALCULATIONS II p ONOOM O\ O l- O ON to to O co to O\M M PO J>- O Tj- ON OO t^CN O M t^. (N ^ J>. rf O ON O -- ON O ,M M M OO O ON t^cs M t^ r>. co O co t^ OO O O * t^. fo ^* t^ vO HH ^ to vO 4(i + R) + A~] (i + R), to which we add another (A) dollars by deposit, making the reserve then F 3 = A[(i + R) + i] (i + R) + A or F, = A[(i + R)* + (i + R) + i], so at the end of the fourth year it is Ft = A[(i + tf) 3 + (i + R) 2 + (i+R) + i], and at the end of the (n) year, the depreciation reserve is F n = A[(i + R)"- 1 + (i + R)- 2 . . . H- (i + R) 2 + (i + R) + i] dollars. 38 FUNDAMENTAL FINANCIAL CALCULATIONS Again, calling S = (i + R)*- 1 + (i + R) n ~* . . . + (i + *) + i, then (i +*)-! R so that But at the end of (w) years, where (w) is the. life of the equip- ment, then the depreciation reserve (F n ) must equal (C), the first cost, whence c = |[d + *) - i], or A R C (i + R) - i But the depreciation rate (D) equals , by eq. (7), so that C_x D _ R . ,<* (8) Example n. The life of an engine is 10 years, interest rate 6%. What is the depreciation rate? Solution: Here n = 10 years, R = 0.06, so that D = - - = 7.5%, (i.o6) lu i instead of 10% usually assumed. Example 12. If in example n, the interest rate is 8% instead of 6%, what will be the depreciation rate? Solution: Then n = 10 years, R = 8%, 0.08 whence D = (i>o8)10 _ T = 6.9%. The larger the interest rate or the longer the life of the equipment, the smaller the depreciation rate, as illustrated below. APPRECIATION 39 Example 13. The life of a concrete building is 50 years, interest rate 6%. What is the depreciation rate? Solution: Here R = 6%, n = 5 years, 0.06 sothat ' D = 0.06) - That is, the depreciation rate is approximately one- third of one per cent instead of the 2% usually assumed, giving an error of some 600% as mentioned before. In order to facilitate calculations, we give below, in Table 3, the depreciation rate for various values of (n) and (R), accord- ing to equation (8). 45. We have spoken thus far only of depreciation. Machinery depreciates with age. But lands, live-stock, and the like almost invariably appreciate in price. Evidently appreciation is merely negative depreciation, so that if we charge depreciation as a cost, then we must charge appreciation as a profit. Lands invariably increase in value with the increase in popu- lation in their vicinity or in the development of their resources previously lying undeveloped and perhaps undiscovered. The increase in price of land with population is very pronounced. Live-stock appreciates by natural processes. But besides all this, all classes of things may apparently increase in price due in reality to a depreciation of the currency. Under such condi- tions, they really do not increase in worth at all. A pound of wheat has no more nutritive value now than a hundred years ago" though it now costs many times as much. Its nutritive value is a constant. Its increased cost is only apparent due to the decreased price of the money. So a piece of machinery may increase in price, but not worth, due likewise to the decreased price of gold. Where the price of the metals and labor has doubled, it is evident that the price of the machine must be doubled, which merely means, according to our present system of exchange, that it will take FUNDAMENTAL FINANCIAL CALCULATIONS \>~ ^ A i \ O^O co vo PO PO PO CM CM OO 10 PO O O O OO O O O w O M r^O w w rl- rt- oo r^^O t^Tft^M co o " M M Tj- CO \O Tf 'T \O r^. O f"O ^ *>. M \ c o oo M Hooooooo o o CS H M M 10 O HH Tj- Tj- O MOO cor^O\M O O O O^O UO ^O^OfMO CO . 10 W M SO CO . 10 CM o ts-Po^O OO CM^O M\O t^vO 10 10 Tf rj- co CO CM MHOOOOOO CO CS M H H M O v O<"O'-O cOOOO COM M CM POrfOvO t^OO OO H CM O >o O 10 O O Q CM CM PO CO Tf 10 VO APPRECIATION 41 twice as much gold to buy the same machine of the same worth as before. The increased production of gold results in a decrease in its price. The price is determined by the demand for the gold for such useless uses as ornamentation and exchange and for useful use in industry. The price that can be paid for gold for industrial use is strictly limited by its intrinsic worth. The price it may have for useless uses is a problem of mob psy- chology rather than one of science. As the production increases, the price falls, and with the fall in price, more gold is used in industry. When the price has fallen sufficiently, all that is pro- duced will be usefully used and the substance, gold or whatever it may be, will have fallen to its true comparative value, or industrial worth. It must be borne in mind that the prices of iron, copper and nickel, as well as silver and gold, went through the same stages, from high artificial price to real industrial worth. 46. In the case of public utilities, depreciation is not a sum of money paid outright to the companies, but is in reality a trust fund to insure maintaining the service at par. It is the duty of the commissions to see that it is all properly used. But in taking depreciation into account, we must also take into account appreciation. To charge the public for that which depreciates without allowing for that which appreciates in value is eminently unjust. In private companies the matter should be treated likewise if the true status of the business is desired. 47. While each item of a plant has usually a different depre- ciation rate, yet a mean rate for the entire plant may be ob- tained so that the plant may be treated as a whole. This is illustrated in the following illustrative problem. Example 14. A power plant of 1000 h.p. consists of the fol- lowing items, with costs and depreciation rates as set forth. Determine the mean depreciation rate. Solution: Summing up the costs gives us a total of $91,400 for the plant. So also summing up the depreciation amounts gives us $1682.10 $1280 = $402.10 net depreciation. The 42 FUNDAMENTAL FINANCIAL CALCULATIONS mean depreciation rate is evidently the net depreciation amount divided by the total cost, or 402.10 -T- $91,400 = 0.44% Ans. DEPRECIATION Item Cost Rate Amount Grounds. $16,000 oo 8 o% $1280 oo Buildings, concrete 9,000.00 O 3 27 OO Turbo-generators 1 8 ooo oo 2 7 486 oo Condensers 3 800 oo 40 1^2 OO Boilers 18,000 oo I O 180 oo Stokers 2 COO OO 6 o Stacks brick. . 3 OOO OO O T.T, 900 Coal bins, steel Coal conveyers 1,800.00 4,000 . oo 5-9 6.0 106. 20 240 oo Boiler feed and service pumps. . . . Feed water heaters, open Switchboard and wiring Exciters 800 . oo 1,600.00 2,400.00 1,700 oo 3-0 i .0 0-35 2 O 24.00 16.00 8.40 r T on Foundations (mach.). . 1,000.00 o OO OO Piping and conduits 5,600.00 3 .0 168 oo Crane . I 2OO OO O 3 3 00 48. Present Worth of a Depreciating Equipment. As an equipment depreciates, the sum of its present worth plus the depreciation fund together with accrued interest thereon, equals the first cost of the equipment. It is often desirable to know the worth of a depreciating unit at some given time. This we can determine as follows: As before let C = the first cost, D = the depreciation rate, R = the interest rate, n = the total life of the unit, and m = the actual age of the unit. At the end of each year we lay aside DC dollars. So at the end of the first year the depreciation fund (F) amounts to F l = DC. This increases to DC(i + R) at the end of the second year, to WORTH OF A DEPRECIATING EQUIPMENT 43 which we add another DC dollars. The depreciation fund therefore amounts to F 2 = DC(i + R) + DC = J9C[(i + R) + i]. So also at the end of the third year the depreciation fund is Fs = DC[_(i + R) 2 + (i + R) + i]. And at the end of (m) years, it is But (i + *)-i + (i + U)- 2 - . . + i = so that But since (i + R) n - i we get by substitution F m = The present worth (W) is the first cost, (C), less the total depre- ciation (F m ) , so that W = C - F m which, by substitution, becomes C[(l +*)"-!] (Q] Example 15. The original cost of a unit is $10,000 and its life is 30 years. The interest rate is 5%. What is the worth of the unit at the end of the 2oth year? Solution: In this case n = 3 years, m = 20 years, R = 5%, and C = $10,000, so that _ io,ooo[(i.Q5) - (1.05)*] (i.os) 30 - i . or W = $5023.90. 44 FUNDAMENTAL FINANCIAL CALCULATIONS Example 16. A building has a life of 50 years. Interest is 4%. What per cent of the first cost is its worth at an age of 40 years? Solution: In this case n = 5 years, m 40 years, R = 4%, C = 100%, whence 9. Vestance. The first cost of the same type of equip- ments varies greatly. The life of the various types and their cost of operation offer two more intensely variable conditions. Besides all these we must consider the different classes of equip- ment that are all designed to serve the same purpose. Thus, for example, we have to consider the various classes of pumps, such as reciprocating, centrifugal, and jet pumps. And, for any one class, i.e., reciprocating pumps, we must consider the various types, such as simplex, duplex, or triplex pumps, single or double acting, vertical or horizontal and so forth. With such a kaleidoscopic mass of variable factors, it would seem at first hardly possible to get a basis from which compara- tive values of such different classes and types may be deter- mined. Yet such a basis may be rather readily obtained in what we call vestance, the equivalent cost of a permanent service. We shall lead up to this with the following example. Example 17. If we make a certain crossing with a wooden bridge, it will cost $10,000 and its life will be ten years. If we make this crossing with a steel bridge, its cost will be $18,000 and have a life of thirty years. If the interest rate is 6 %, which of the two bridges will be the most economical investment? Solution: The first cost of the wooden bridge is $10,000. At the end of ten years we mus t again build the wooden bridge at the cost of another $10,000. Finally at the end of twenty VESTANCE 45 years, we must again rebuild the wooden bridge at the cost of still another $10,000, when, neglecting other considerations, we will have attained the life of the steel bridge, 30 years, and rendered with the wooden bridge, or rather with three wooden bridges, the same service as with the steel bridge. To begin with, then, we invest $10,000. At the end of ten years we again invest another $io,ooo,whose present worth (P) is 10.000 So at the end of twenty years we must invest another $10,000, whose present worth is n 10,000 The total cost of this service for thirty years, with the wooden bridges is 10,000 + 5585 +3119 = $18,704. The same service may be rendered with a steel bridge for $18,000, showing that the latter is the best by $704. Of course to these costs must be added that of maintenance, loss to traffic during reconstruction, and such other considerations as the physical nature of the specific problem may demand. Taking these into consideration, in the above case, would throw the decision very decidedly in favor of the steel bridge. 50. Depreciation Vestance. The depreciation vestance is the total of the present worths of the investment and reinvest- ments. We can determine this as follows: Let C = the first cost, n = the life in years, and R = the interest rate. To begin with, we invest (C) dollars. At the end of (n) years, we must invest another (C) dollars, whose present worth is p c " (i + R)" 46 FUNDAMENTAL FINANCIAL CALCULATIONS Again at the end of (2w) years, we must invest another (C) dollars, whose present worth is P 2n = _ c and so again at the end of (3^) years, we must invest another (C) dollars, whose present worth is p - c n ~~( I P\3n and so on. So in general at the end of (mn) years we must again, for the wth time, invest (C) dollars, with a present worth of p - c mn 7 T" pWn' Summing all these up gives , v r , C C C_ r Vd= t" ( T _i-^n + (.__\2n + f T _|_lttn ' ' ' + or Let (S) equal the sum of the series in the brackets. Then c _ T . I . I , i ** I / T-\ I / i -r>\ o_ I / T->\ .. Reducing to a common denominator, gives V- 1 */ i^V- 1 - i^- 1 *-/ I^V- 1 !^ **/ T^V- 1 I *-/ / v Now multiply through by (i + R) n and we get (i+R) n S= (i + I #)" m ~~' ^ And subtracting equation (a) from equation (b) gives (, D\nC O __ ( J + 1?) ( *+ 1 >- I and " (i + )*[( i + /?) - i]' DEPRECIATION VESTANCE 47 Now dividing through by (i + R) mn gives s = (i + ) n - i (i + #) w - i When (inn), the length of the service becomes infinite, then (i + R) mn also becomes infinite and -. becomes zero. So for the complete summation (m = oo) of the above series, we get (i + *) " (i + R) n - i' And since v d = cs we have But we had that -t ' ~ T is the term factor (T), so that F d = .......... (n) Example 18. A concrete building has a first cost of $100,000 and a life of 50 years. If the interest rate is 5%, what is the depreciation vestance? Solution: In this case C = $100,000, R = 5%, and n = 50 years, so that by Table 2 T = 0.91279, and the depreciation vestance is 100,000 fa V d = - - - = $109,550 Ans. 0.91279 If the interest rate were only 3 %, then the term factor would be, by Table 2, only 0.77189 and the depreciation vestance would be rr IOO.OOO ,, V d = - - = $129,540. 0.77189 48 FUNDAMENTAL FINANCIAL CALCULATIONS Example 19. A wooden building to take the place of the concrete building of problem (18), would cost $70,000 and have a life of 20 years. With interest at 5 %, which would be the best investment? Solution: Here C = $70,000, n = 20 years, R = 5%, so that T = 0.62312, and T7 70.000 a. V d = -t-z = $112,340. 0.62312 This would make a difference of 112,340 - 109,550 = $2790 in favor of the concrete building, even excluding the higher maintenance cost of the cheaper structure. If the interest rate were 3%, then the term factor (T) would be 0.44632 and the depreciation vestance would be Vd = . = $ I5 6,8 3 8. 0.44632 In this case the difference would be 156,838 - 129,540 = $27,298. Example 20. A permanent concrete viaduct costs $5000. What is its depreciation vestance? Solution: In this case n = oo so that the term factor is unity. The depreciation vestance is then = $5000, the same as the first cost. 61. Operating Vestance. With depreciation vestance, we get a method of comparing equipments having different first costs and different lives. We thus have a method of determin- ing comparative values as to fixed charges and those only. OPERATING VESTANCE 49 The operating costs are continuous, and are not limited to a term of years. By capitalizing the annual cost of operation, we get, therefore, the operating vestance. Letting a = operating cost per year, and R = interest rate, then V. = I ......... (12) This is on the assumption that the operating cost is constant. However, improvement in design is what renders equipments obsolete, and this refers particularly to reductions in the cost of operating. This more than compensates for obsolescence and both it and reduction in cost of operation must be taken into consideration. 52. Total Vestance. The total vestance (F) is evidently the sum of the depreciation and operating vestance, so we have V = V d + V a or or So that knowing the life, operating cost and vestance, we can determine what the cost should be. Also since (I3B) we can determine the life of the unit, knowing the first cost, vestance and operating cost. In a similar manner we can solve for any of the other quantities in equation (13). Example 21. A 1000 h.p. Diesel Engine plant costs $75,000. The life of the engine is 20 years. The interest rate is 5%. The cost of operation is 0.4 cent per horse power hour (h.p.h.). 50 FUNDAMENTAL FINANCIAL CALCULATIONS If the engine is operated at full load for 3000 hours per year, what will be the vestances? Solution: Here C = $75,000, R = 5%, n = 20 years, so that and V d = 0.62312 The total annual operating cost is 1000 X 0.004 X 3000 = $12,000, so that the operating vestance is T7 12,000 .., V a = - - -- = $240,000. 0.05 The total vestance is therefore 120,360 + 240.000 = $360,360. Example 22. If, in Example 20, we put in a steam plant, the first cost will be $50,000, and the life is likewise 20 years, while the cost of operation will be 0.7 cent per h.p.h. With the same interest rate, i.e., 5%, which will be the best investment? Solution: In this case Vd . -- = $80,240. The annual cost of operation is 1000 X .007 X 3000 = $21,000. The operating vestance therefore is ,, 21,000 4. V a = -- = $420,000, 0.05 giving a total vestance of 420,000 -f- 80,240 = $500,240. TOTAL VESTANCE 51 The difference between the total vestance of the steam and Diesel plant is 500,240 - 360,360 = $139,880 in favor of the Diesel plant. Example 23. If the power were purchased at one cent per h.p.h. and the motor cost $6000 and had 90% efficiency with a life of 30 years, how would the total vestance compare with that of the steam and Diesel plants? Solution: In this case n = 30 years, C = $6000, R = 5%, so that T = 0.76861, and V d = -- = $7810. 0.76861 The total annual cost of operation is loop X o.oi X 3000 _ * *^^s =^ ^Din ' . as nnec nera ed ors- ^ Er Si nple Hist Er -spet pcted "dEi gine 200 400 600 800 1000 1200 1400 1600 1800 20( I.H.P. FIG. i. Showing the variation in cost per i.h.p. of steam engines and turbines with size and type. 59. Operating Costs. The number of pounds of dry sat- urated steam used per indicated horse power hour (i.h.p. h.) for compound condensing engines with 26-inch vacuum and ioo# gage initial pressure is as given in table 8: OPERATING COSTS \ 65 TABLE 8 STEAM CONSUMPTION COMPOUND CONDENSING ENGINES I.h.p ............. 50 60 70 80 100 125 150 200 250 Lbs. per i.h.p ..... $20.2 $19. 6 $19.1 $18.7 $18.0 $17.5 $17.3 $16.7 $16.4 I.h.p ............. 300 400 500 600 700 800 900 1000 Lbs.peri.h.p ...... $16.1 $15.7 $15.4 $15.2 $15.1 $15.0 $14.9 $14.8 These values correspond to the equation 13-2, (24) where W = the pounds of steam per i.h.p.h., and M = the size in i.h.p. With fractional loads the steam consumption is as given in the following tables. In these tables, only the strokes of the engines are given. Since the allowable piston speed is well known, the stroke determines the r.p.m. The i.h.p. for any of these engines may therefore be readily determined, for any given bore, which may be anything from one-half to one times the stroke. A stroke equal to ij times the bore is most com- monly used, but good practice allows considerable deviation. Throughout, all data are given for well-designed and well- constructed machinery and not the cheap equipment that is made to sell, not to use. 66 BASIC COSTS TABLE Q STEAM CONSUMPTION SIMPLE HIGH SPEED ENGINES NONCONDENSING STROKE LOAD PRESSURE 8o-go# " IOO-IIO# I20-I3O# ^ 10" | 30.0 29.6 29-3 29.0 3 4 30.6 29.8 29.1 28.5 . I 34 - 32.2 30.5 29.1 i 40.0 38.4 37-0 35-6 l 30.8 30.5 30.2 30.0 12" 1 29.6 29.2 28.9 28.6 I 30.2 29.4 28.7 28.1 I 31-4 30.0 28.6 | 38.7 37-2 36.0 34-8 ll 30-3 30.0 29.7 29 5 14" 1 29.1 28.7 28.4 28.1 f- 29.7 28.9 28.2 27.6 ' i 32.0 30.6 29-3 28.1 J 37-5 36-1 35-o 34-o H 29.8 29-5 29.2 29.0 16" 1 28.5 28.1 27.8 27.5 i 29.2 28.4 27.7 27.1 I 31.0 29.8 28.6 27-5 4 36.2 35-o 34-o 33- if 29-3 29.0 28.7 28.5 18" 1 27.9 27-5 27.2 26.9 1 28.5 27.8 27.1 26.5 ' f 30.0 29.0 28.0 27.0 i 35.0 34-0 33.0 32-0 ** 28.8 28.5 28.2 28.0 STEAM CONSUMPTION 67 TABLE 10 STEAM CONSUMPTION HIGH SPEED TANDEM COMPOUND ENGINES STROKE LOAD PRESSURE ioo-no# I20-I3O# I40-iso# ioo-no# 120-130! I40-iso# Condensing 26" vacuum Noncondensing 10-12'' * 19-3 18-5 17.7 25-5 24.7 24.0 f 20.1 19-3 18.6 28.0 27.2 26. 4 1 22.0 21.0 20.4 32.6 31-6 30.6 i 26.5 26.O 25-5 44.0 43-0 42.0 i* 20.8 19.6 19.0 27.0 25-7 25-5 14" 1 18.8 18.0 17.2 24.8 24.0 23-3 f 19-5 18.7 18.0 27.0 26.2 25-5 I 21.2 20.4 19.7 32.0 31.0 30.0 1 26.0 25-5 25-0 43- 6 42.6 41.6 ii 19.9 18.8 18.2 26.0 25.0 24-5 16" 1 18.2 17-4 16.7 24.0 23.2 22.5 t 18.9 18.1 17-4 26.0 25-2 24-5 1 20.4 19.8 19.0 3i-5 30-5 29-5 4" 25-5 25.0 24-5 43-3 42-3 4i-3 ii I9.I 18.3 17-5 25.0 24.2 23-5 18' 1 I7-S 16.7 16.0 23.0 22.2 21-5 f 18.2 17.4 16.7 25.0 24.2 23-5 i 19-5 19.2 18.3 31.0 30.0 29.0 ^ 25.0 24-5 24.0 43-o 42.0 41.0 ij 18.3 17-5 16.8 24.0 23-4 22.5 68 BASIC COSTS TABLE ii STEAM CONSUMPTION SIMPLE NONCONDENSING FOUR VALVE ENGINES STROKE LOAD PRESSURE ioo-s# uo-S# I20-S# I30-S# I40-S# ISO-S# i6o-5# 18" | 26.0 25-7 25-4 25.2 24.9 24.7 24-5 1 27.0 26.3 25-6 25.0 24-3 23-8 23-3 I 29.7 28.5 27-5 26.5 25-5 24.7 23-9 1 4 35-5 34-5 33-6 32.7 31-7 31.0 30-3 ii 27.1 26.8 26.5 26.2 26.0 25-7 25-5 21" 1 25-8 25-5 25-2 25.0 24.7 24.5 24-3 i 4 26.8 26.1 25-4 2 4 .8 24.1 23-6 23.1 3 4 29-5 28.3 27-3 26.3 25-3 24-5 23-7 i 35-o 34-0 33-i 32.2 31.2 30.5 29.8 if 26.9 26.6 26.3 26.O 25-8 25-5 25-3 24" 1 25.6 25-3 25.0 2 4 .8 24-5 24-3 24.1 1 26.6 25-9 25.2 24.6 23-9 23.4 22.9 I 29-3 28.1 27.1 26.1 25-1 24-3 23-5 i 4 34-5 33-5 32 .6 31-7 30-7 30.0 29-3 ii 26.7 26.4 26.1 25-8 25-6 25-3 25.1 27" 1 25-4 25-1 24.8 24.6 24-3 24.1 23-9 i 4 26.4 25-7 25.0 24.4 23-7 23-2 22.7 1 29. i 27.9 26.9 25-9 24.9 24.1 23-3 i 34-0 33-o 32.1 31.2 30.2 29-5 28.8 I* 26.5 26.2 25-9 25-6 25-4 25.1 24.9 30" 1 25.2 24.9 24.6 24.4 24.1 23-9 23-7 f 26.2 25-5 24.8 24.2 23-5 23.0 22.5 1 28.9 27-7 26.7 25-7 24.7 23-9 23.1 i 33-5 32.5 31-6 30-7 29.7 29.0 28.3 ii 26.3 26.0 25-7 25-4 25.2 24-9 24.7 36" 1 25.0 24.7 24.4 24.2 23-9 23-7 23-5 f 26.0 25-3 24.6 24.0 23-3 22.8 22.3 I 28.7 27-5 26.5 25-5 24-5 23-7 22.9 i 33-0 32-0 3i-i 30.2 29.2 28.5 2 7 .8 if 26.1 25-8 25-5 25.2 25.0 24.7 24-5 STEAM CONSUMPTION 69 TABLE 12 STEAM CONSUMPTION CROSS AND TANDEM COMPOUND ENGINES PRESSURE STROKE LOAD IOO-IIO# I2O-I3O# 140-1 so# IOO-IIO# I20-I30# I40-iso# Condensing 26" vacuum Noncondensing 18" 1 15-6 15-15 14.8 24.0 23.0 22. 1 16.75 16.2 15-5 24-75 22.7 22.7 i 17-5 17-3 17.2 26.0 24.0 27.0 1 4 20.5 20.1 19.8 41-5 41 .0 40-5 Ij 16.2 15-8 15-5 25-0 24.0 22. 21" f 15-45 15.0 I4-65 23-5 22.5 21-5 4 16.5 16.0 15-4 24.2 22.6 21.6 1 17.5 17-4 17-3 26.4 26.8 27.1 1 22.0 21.0 20.3 41.2 41.0 40.7 I* 16.1 15-7 14.9 24-5 23-4 21.8 24" 1 15-35 14.9 14-55 23.0 22.0 21.0 1 16.2 15-7 15-2 23-9 22.6 21.4 t 17.6 17-5 17.4 26.8 27.0 27.2 i 24.0 22.5 20.8 40.9 40.9 40.9 ii 16.0 15-7 15-3 24.0 23.0 21.6 27" 4 4 15-1 14-75 14.4 22. 5 21-5 20.5 1 16.0 15-6 15-1 23-3 22.3 21.3 1 17.7 I 7 .6 17.6 27.2 27.3 27-3 1 4 26.0 24.0 21-5 40.6 40.9 4I.I J 15-8 15-5 15-1 23-3 22.3 21.4 30" 4 15.0 14.6 14-25 22. 21.0 20.0 a 4 15-8 15-4 14.9 22.6 21.8 21.1 1 17.7 17.7 17.6 2 7 .6 27.4 27.4 i 28.0 25-0 22. 40-3 40.8 41-3 ! 15-7 15-3 15-0 22.6 21.8 21.2 36" 1 14.6 14.4 14.0 21-5 20.0 I9.O f 15-6 15-2 I 4 .8 22. O 21-5 21.0 1 17.8 I 7 .8 I 7 .8 28.0 27-5 27-5 i 30-7 26.6 22.6 40.0 40.7 41-5 11 15-5 15-2 14.8 22. 21 .2 21. BASIC COSTS TABLE 13 STEAM CONSUMPTION HORIZONTAL SINGLE CYLINDER LENTZ ENGINES Referred to i.h.p. at normal load and dry saturated steam STROKE PRESSURE 85# IOO# 5# I2S# I40# ISO# i6o# Noncondensing 1 8" 24-6 24.1 23-8 23-4 23-0 22.6 22.2 23.6 23.1 22.8 22.4 22.0 21.6 21 . 2 22.7 22.2 21 .9 21-5 21. I 20.7 20.3 22 .1 21 .6 21.3 2O.9 20.5 2O. I 19.7 21-3 20.8 20.5 20. 1 19.7 19-3 18.9 20.8 20.3 20. o 19.6 19.2 18.8 18.4 20.3 19.8 I9-S I9.I I8. 7 I8. 3 17.9 14, 15, 16 and 17 X 21" 18, 19, 20 and 21 x 21" 24". . 27" '^*. *. . Ret =9MM urn ' ubul jrB lers ^_ ) 25 50 75 100 125 150 175 200 225 25C Bo. H.P. FIG. 4. Variation in prices of boiler with size and type. sion of costs in power plants in good practical form, some of which is given below, BUILDINGS 79 TABLE 21 COST OF BUILDINGS, 1906-10 Type Cost in dollars Cost in cents per sq. ft. per cu. ft. floor surface of contents Mill construction o. 80 to i . 10 6 . 5 to 8.5 Fireproof stores, factories and warehouses with brick, concrete, stone and steel construction. . . 2.00 to 3.00 14.0 to 25.0 Concrete and reiniorced concrete shops, factories and warehouses i . 25 to i . 75 8.0 to 16.0 Plain power houses, with concrete floors, brick and steel super-structure 2 . oo to 2 . 75 9 . o to 12 . o Power houses under city conditions with superior architecture 3.00 to 4. 50 15.01030.0 Exact data cannot possibly be given, because the cost of buildings varies enormously with local conditions. These must always be determined by the engineer before reaching any final decision. TABLE 22 COST OF STEAM POWER PLANT BUILDINGS 1906-10 PER ENGINE HORSE POWER Simple Noncondensing Engines H.p 10 12 15 20 30 40 50 75 Boiler house $37. 15 $33.00 $28.50 $24.50 $20.50 $18.00 $16.00 $13.00 Engine house 4.80 4.35 3.90 3.30 2.75 2.50 2.30 2.15 Coal bins 20.00 18.00 16.00 13.70 n.oo 9.80 8.30 6.00 Simple Condensing Engines H.p 10 12 15 20 30 40 50 75 100 Boiler house $33.70 $29.60 $26.20 $21.60 $18.20 $16.00 $14.80 $11.30 $9.70 Engine house. ... 14.40 12.60 10.90 8.60 7.75 6.40 5.35 4.00 4.30 Coal bins 19.00 17.90 15.80 13.60 11.00 8.70 8.50 6.30 5.70 Compounding Condensing Engines H.p 100 200 300 400 500 600 700 800 Boiler house 1 $ f 11.20 8.00 6.40 5.70 5.35 5.00 Engine house j \ii.2o 9.35 8.50 7.20 6.30 5.60 Coal bins 5.70 4.00 3.10 2.60 2.40 2.25 2.10 2.05 H.p 900 1000 1500 2000 Boiler house $4-7 $4-55 $4-i $3-95 Engine house 5.35 5.00 4.75 4.55 Coal bins 1.95 1.80 1.75 1.66 8o BASIC COSTS The total cost of engine and boiler house and coal bins for simple noncondensing engines is given by the equation Ci = 63.5 - J \/2ooM - M 2 - 1900. . . (44) For simple condensing, this is C 2 = 65.0 J V2ooM M 2 1900, . . (45) while for compound condensing engines the approximate equation is . C.- + 8.2.. . . . . . (46) 67. The cost of complete steam installations is given as follows : TABLE 23 H.p ................ 5000 10,000 20,000 30,000 40,000 50,000 Cost per h.p ........ $120.00 $100.00 $90.00 $80.00 $75.00 $70.00 These costs are found to be divided about as follows: TABLE 24 Percentage of total Buildings, real estate and excavations 14.6 Turbines and generators 23 . 5 Condensers 5.7 Boilers, superheaters, stokers and stacks 28. 7 Bunkers and conveyors 4.8 Boiler, feed and service pumps i . o Feed water heaters 1.6 Switchboard and wiring , 3.5 Exciters 2.1 Foundations machinery i . i Piping and conduit 7.1 Crane 1.5 Superintending and engineering 4.8 68. Centrifugal Pumps. The variation in price of one certain make of centrifugal pumps with size and type is given below. BELTED CENTRIFUGAL PUMPS 81 TABLE 25 Low HEAD HORIZONTAL BELTED CENTRIFUGAL PUMPS 1910-12 Size inches Normal capacity, g.p-m. Cost of pump Cost per 1000 g.p.m. Adjusted costs I 20 $27.00 $1350.00 if 50 33-oo 66o.OO 2 IOO 42.00 42O.OO 2\ 150 51.00 34O.OO m * 3 225 60.00 266.OO 3^ 300 69.00 230.00 4 400 75.00 187 .50 < 700 IOO.OO 142.86 O 6 900 I 20.00 133-33 7 I2OO 150.00 125.00 8 I600 180.00 112.50 $180.70 10 3000 240.00 So.OO 253-10 12 4500 300.00 66.67 330-75 14 6OOO 425.00 70.83 408 . 2O 15 7000 475.00 67.86 459-90 16 8000 525-00 65.62 511.60 18 IO,OOO 900.00 9O.OO 615.00 20 I4,OOO 1000.00 71.40 821.80 24 20,000 I2OO.OO 6o.OO 1132.00 30 30,000 1550.00 5I-67 1649.00 36 4O,OOO 2400 . oo 6o.OO 2l66.OO 40 50,000 2700.00 54-oo 2683.00 44 60,000 3200.00 53-33 32OO.OO The factory costs are somewhat erratic due to the lack of knowledge of exact costs on the part of the manufacturer. The prices on the larger sizes are no doubt based on " esti- mates." The equation based on the adjusted values, i.e., those that have been brought into consonance with themselves, is (47) where C cost in dollars per 1000 g.p.rn. of capacity, and M = the capacity in zooo's of g.p.m. BASIC COSTS 4 8 12 16 20 24 28 32 36 40 Size of Pump in Inches of Discharge Nozzle FIG. 5. Variation in price of centrifugal pumps, with size and type. TABLE 26 HIGH HEAD HORIZONTAL BELTED CENTRIFUGAL PUMPS 1910-12 Size inches Normal capacity, g.p.m. Cost, factory Cost per 1000 g.p.m. Adjusted costs II SO $67.20 $1344.00 $77-625 2 IOO 84.00 840.00 83-75 *\ 150 94.80 632.00 90-375 3 225 112. 2O 500.00 99.94 4 400 I34-40 338.00 122.25 5 700 150.00 214.30 160.50 6 900 198.00 220.00 186.00 7 1200 24O.OO 2OO.OO 224.25 8 l6oO 294.00 183.75 275.25 10 2500 390.00 156.00 390.00 12 4OOO 522.00 130.50 581.25 The equation based on the adjusted values is C= ^TF 5 + $ I2 7-5- (48) DIRECT-CONNECTED CENTRIFUGAL PUMPS 83 TABLE 27 DIRECT-CONNECTED HORIZONTAL CENTRIFUGAL PUMPS LESS MOTOR 1910-12 25 Ft. Head H.p. R.p.m. Size inches Capacity g.p.m. Cost factory Cost per looo g.p.m. i 1800 I IS $70.00 $4666.67 I 1800 3 70 IOO.OO 1428.00 2 1800 3* 130 125.00 960.00 3 1800 4 240 140.00 580.00 5 1800 5 420 170.00 405.00 7i 1200 5 610 230.00 376.00 7^ I2OO 6 650 310.00 476.00 10 I2OO 5 820 240.00 294.00 10 I2OO 8 IOOO 385.00 385-00 15 QOO 7 1300 310.00 237.00 20 I800 7 1350 330.00 245.00 20 720 10 2300 500.00 2l8.OO 25 720 10 2500 530.00 212. OO 30 1800 8 2000 385-00 I92.OO 30 900 10 3200 500.00 156.00 35 600 12 3500 630.00 iSo.OO 50 9OO 12 4800 675.00 144.00 50 600 12 4800 750.00 156.00 8 4 BASIC COSTS TABLE 28 DIRECT-CONNECTED HORIZONTAL CENTRIFUGAL PUMPS WITHOUT MOTOR 1910-12 50 Ft. Head H.p. R.p.m. Size inches G.p.m. Cost Cost per looo g.p.m. 2 1800 I 25 $80.00 $3200.00 2 1800 2 50 IOO.OO 2000.00 3 1800 H 100 115.00 1150.00 5 3600 3 200 135-00 675.00 5 1800 2^ 2OO 145.00 725.00 7i 1800 4 350 180.00 512.00 10 1800 4 500 185.00 370.00 i5 1800 4 700 210.00 300.00 20 I2OO 6 IOOO 27O.OO 270.00 20 I800 7 IIOO 330.00 300.00 25 QOO 7 1250 400.00 320.00 30 I800 8 1500 385-00 260.00 35 900 8 2OOO 420.00 210.00 40 900 8 22OO 420.00 191.00 40 900 12 2600 930.00 357-oo So 900 IO 25OO 530.00 212. OO 75 900 IO 3900 550.00 141 .OO 100 900 12 5500 650 . oo IlS.OO 125 600 15 7000 IOOO.OO 143-00 150 720 14 775 900 . oo n6.co DIRECT-CONNECTED CENTRIFUGAL PUMPS 85 TABLE 29 DIRECT-CONNECTED HORIZONTAL CENTRIFUGAL PUMPS WITHOUT MOTOR 1910-12 100 Ft. Head H.p. R.p.m. Size inches G.p.m. Cost Cost per 1000 g.p.m. 3 3600 I 43 $140.00 $3260.00 5 3600 3 135 135-00 IOOO.OO 7i I800 2 125 150.00 1 200 . 00 IS 3600 si 300 150.00 500.00 15 I800 2* 315 185.00 588.00 20 1800 4 470 210.00 446.00 25 1800 4 560 235.00 420.00 30 I800 5 700 26o.OO 37l-< 35 1800 4 700 24O.OO 342.00 40 I2OO 5 1040 325.00 313.00 So 1800 6 1275 27O.OO 212.00 75 I2OO 8 1800 42O.OO 233.00 75 1200 7 2OOO 530.00 265.00 IOO I2OO 8 2450 550.00 217.00 IOO QOO 8 2700 640.00 237.00 150 1800 12 3800 IOOO.OO 264.00 150 900 10 4OOO 680.00 170.00 2OO I2OO 12 5600 680.00 121. OO 250 900 14 7000 1100.00 157-00 $3500 r 3000 1 & 'o 9^nnl- ; | 2o00 r o 9000 1- \\ si _ l^nnL \\ v loooN \ s^ 5 \ 1 s X* ^^ cnn \ \, ^> ^ - ^s ~~Low Ef icien yj 1 r 25 50 75 100 125 150 J75 200 225 25 H.P. of Pump FIG. 6. Prices of direct-connected centrifugal pumps. 86 BASIC COSTS TABLE 30 DIRECT-CONNECTED HORIZONTAL CENTRIFUGAL PUMPS WITHOUT MOTOR 1910-12 150 Ft. Head H.p. R.p.m. Size, inches G.p.m. Cost Cost per 1000 g.p.m. cap. 3 1800 I 17 $230.00 $13,530.00 5 1800 I 30 230.00 7,670.00 7* 1800 ii 50 270.00 5,400.00 10 1800 2 90 270.00 3,000.00 IS 3600 3* 150 150.00 1,000.00 20 1800 2* 160 190.00 1,190.00 20 1800 4 300 430.00 1,430.00 25 1800 3 235 2OO.OO .850.00 35 1800 4 500 450.00 900.00 35 1800 5 550 575-oo 1,042.00 40 I2OO 4 43 415.00 965.00 50 I800 5 800 600.00 750.00 5 I800 6 850 725.00 850.00 50 I2OO 5 600 510.00 850.00 75 I2OO 6 1 200 500.00 416.00 75 1800 8 1400 900.00 643.00 IOO I2OO 8 1500 55o.oo 366.00 100 I800 8 1800 925-00 5i3-oo VERTICAL CENTRIFUGAL PUMPS OrtA / /^ I7f\f\ / / x Cost of Pump Complete s?^v // x v x x ~? x X x ' ^. ^> x / /^ , /^ ^ x^ ^ x / ? ^ x^ x x x ? / x x / x x 7 X ' X X X > X 2^ 456 Size of Pump in Inches of Discharge Nozzle.D 9 1( lameter FIG. 7. Prices of vertical centrifugal pumps, complete. TABLE 31 VERTICAL CENTRIFUGAL PUMPS 1910-12 Complete with frames, valves, shafting, boxes, pipe, collars, etc. SIZE, INCHES NORMAL CAPACITY PRICE With 25 ft. frame With 50 ft. frame With 75 ft. frame 2 100 g.p.m. $196.00 $265.00 $334-00 2^ 150 " 221 .OO 293.00 364.00 3 i 225 249.00 332.00 408.00 300 " 281.00 368.00 452.00 4 400 " 300.00 397.00 482.00 5 700 353-oo 462.00 561.00 6 900 423.00 550.00 658.00 7 1200 496.00 616.00 733-oo 8 1600 " 605.00 740.00 860.00 88 BASIC COSTS 69. Cost of Large Pumping Units. The following quotations were made on various types of complete pumping plant f .o.b. factory, between 1908 and 1910. 1. A 25,000 g.p.m. horizontal pump for a total head of 16 ft., direct- connected to a 150 h.p. 257 r.p.m. a.c. motor together with primer, etc., and 100 ft. of piping. Price quoted $6000. 2. Same as (i) above but for 20 ft. head. Price quoted $7500. 3. A 25,000 g.p.m. horizontal pump for a total head of 16 ft., direct connected to a 150 h.p. horizontal tandem compound condensing steam engine, complete with water tube boiler, jet condenser, and necessary piping. Price quoted $12,000. 4. Same as (3) but for 20 ft. head. Price quoted $15,000. 6. One 50,000 g.p.m. horizontal, double suction pump, direct-connected to one 300 h.p. 200 r.p.m. a.c. motor for 16 ft. total head, together with loo ft. of piping and electric primer. Price quoted $9000. 6. Vertical axial flow turbine with a capacity of 125 cu. sec. ft. against a head of 6 ft., with 69 in. runner, speed 65 r.p.m. direct geared to a 150 h.p. 485 r.p.m. vertical motor and having a guaranteed efficiency of 60%. Price quoted $6100. 7. Same as (6) but for 75 cu. sec. ft. of capacity and 57 in. runner and 100 h.p. motor. Price quoted $4900. 8. One 40 in. horizontal centrifugal pump having a capacity of 125 cu. sec. ft. against a total head of u ft., belted to a 250 h.p. motor. Pump efficiency guaranteed 50%. Price quoted $6100. 9. Same as (8) but for 6 ft. head, 50% efficiency, and belted to a 150 h.p. motor. Price quoted $3400. 70. Operating Costs of Centrifugal Pumps. The material consumed in the case of pumps is power together with a very little oil and waste. The amount of power used depends upon the pump efficiency and the cost of the power or its price. The efficiency of pumps varies both with the size of the pump and the head pumped against, larger pumps and greater heads giving higher efficiency, according to the following table. OPERATING COSTS OF CENTRIFUGAL PUMPS 89 TABLE 32 EFFICIENCIES CENTRIFUGAL PUMPS HEAD IN FEET Capacity, g.p.m. 10' 12' is' 20' 25' 30' 40' to 70' 50 22 23 24 25 29 30 33 75 25 26 27 29 33 35 36 IOO 28 30 32 34 36 38 40 150 32 34 35 37 40 42 45 200 36 37 39 42 45 48 50 300 40 4i 43 45 47 51 53 400 42 44 46 48 50 54 56 500 44 46 48 50 53 56 58 600 46 48 50 52 54 58 60 700 48 5 52 54 57 60 62 800 5 52 54 56 61 63 64 900 52 54 56 59 62 63 64 1,000 53 55 57 59 62 64 65 1,200 54 56 58 60 64 65 66 I,4OO 55 57 59 62 65 67 68 1, 60O 57 59 61 63 65 67 68 PUMP EFFICIENCY CONTINUED s' 10' 15' 25' 35' So' 6,OOO 50 6^ 65 70 70 70 7,OOO 50 61 65 70 70 70 8,000 50 62 66 70 70 70 10,000 50 63 67 70 73 15,000 50 63.5 68 70 74 20,000 50 64 69 70 75 30,000 to 60,000 50 65 70 70 BASIC COSTS 2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 G. P. M. FIG. 8. Efficiencies of centrifugal pumps. At 10 ft. head, the variation of pump efficiency with capacity y = 6 7 - -^=, (49) is where y = efficiency in per cent, and M = the pump capacity in thousands of gallons per minute capacity. With increased head the pump efficiency increases by the following per cents: Head ......................... 10 12 15 20 25 30 40 50 Per cent increase in efficiency. ... o 4 9 15 18.5 21 25 28 Corresponding to the equation 160 , x ...... (50) where y p = per cent increase in efficiency, and h = the head in feet. The lubricating oil used amounts to about o.ooi pint per water horse power hour. FRACTIONAL LOAD EFFICIENCY 91 71. It is often assumed that the cost of attendance of cen- trifugal pumps is zero. This is hardly correct, although this cost for centrifugal pumps is very small. It depends upon the general design of the installation normally being about <*' where A = cost of attendance in cents per h.p.h. as of 1910, and M = the h.p. required to drive the pump. It must be borne in mind that all costs are as of 1906-10. To get the attendance cost as of today divide by the relative value of gold now as compared with the date above, as indi- cated by the relative purchasing power. Thus if the purchasing power of gold now is only half of what it was then, all costs now would be double that given above. Based on the study of the efficiency of pumps at fractional loads covering the tests of several hundred pumps varying in size from i inch to 12 inches and consisting of four different makes and some twelve different types, we have obtained the following average values of the per cent of full load efficiency at fractional loads viz. : Per cent of full load ............ o 25 50 75 100 125 150 Per cent of full load efficiency. .. o 43.6 75 93.7 100 93.7 75 Corresponding to the equation y p = F(2 - o.oiF) ...... (52) where y p = per cent of full load efficiency, and F = per cent of full load efficiency, 9 2 BASIC COSTS 72. Motors and Generators. TABLE 33 INDUCTION MOTORS WITH STARTERS PRICES, 1912 2 and 3 phase 60 cycles 110-220 volts, squirrel cage type SIZE, R.P.M. H.P. 1800 I2OO 900 720 600 i $33-oo $40 . oo $47 oo i 40.00 50.00 63.00 $77-00 2 50.00 58.00 75-oo 91 .00 $110.00 3 57-oo 71 .00 94.00 115.00 140.00 5 70.00 90.00 117.00 143.00 170.00 1\ i35-oo 160.00 195.00 220.00 267.00 10 160.00 195.00 230.00 265 . oo 300.00 15 192.00 225.00 270.00 306 . oo 348.00 20 235-00 272.00 305.00 330.00 390.00 25 255-00 305.00 340.00 377-oo 430 . oo 30 305-00 340.00 375-oo 420 . oo 465 . oo 40 350.00 390.00 435-00 481 .00 535-oo 50 400 . oo 440.00 485.00 535-oo 595-oo 75 515-00 565.00 610.00 664 . oo 720.00 100 680 . oo 724.00 780 . oo 832.00 ISO 960.00 IOIO.OO 1070.00 $1200 1000 800 600 400 200 20 40 60 8< FIG. 9. Prices of squirrel cage motors. 100 120 140 160 180 200 H. P. 73. MOTORS TABLE 34 INDUCTION MOTORS WITH STARTERS PRICES, 1910-12 Slip-ring type 1 10-550 volts, 3 phase 60 cycle 93 SIZE, R.P.M. . H.P. 1800 I2OO goo 720 600 I $60 oo $77 .00 I 80.00 104.00 $113.00 2 IO4. OO 113 oo 136 oo 122 OO 138 oo 164 oo 5" I? 1 ? OO 175 OO 188.00 7! 210 do 224 oo 250.00 IO 22C OO 260 oo 280 oo $300 oo 15 20 275.00 300.00 34.0 oo 330.00 380 oo 360.00 410.00 $380.00 440.00 2C 380 oo 425.00 468.00 405 .OO 30 AQ 425.00 s;O5 OO 475-00 562 oo 515-00 608 oo 550.00 640 oo r. f '/ ' t >/x \ '* X ; I X ' |/ / .*_ = =E ^ X * ~~ ~f* t***^ 72 73 74 7.5 76 77 78 79 80 81 82 83 84 85 86 87 Per cent. Efficiency FIG. 13. Efficiencies of A-c. 60 cycle motors. This corresponds to the equation 1^.8 y = 92.2 89 90 91 (54) where y = the per cent efficiency, and M = the size in h.p. It should be borne in mind that such data as we present are based on the study of the results of tests of our very best and largest manufacturers. The data themselves, together with the curve of values adopted, is shown in Fig. 13. The discrepancies in the data are due to redesign of some units be- fore the like service could be given to others; and to some error in test, which, though small, is nevertheless quite signifi- cant and finally to commercial exigencies such as the use of a frame designed for one given size and speed, for a different size and speed. Such adaptations of one frame to several sizes generally give greater economy of production but at a re- duction in worth of the unit due to reduced efficiency and power factor. 9 8 BASIC COSTS The efficiency of generators at full load for 3 phase alter- nating current of 60 cycles at from 200 to 550 volts varies with size as follows: TABLE 39 EFFICIENCIES OF A-c. GENERATORS 60 cycles 220 to 550 volts Size, kv.a 50 75 100 125 150 200 300 500 Per cent efficiency 88.5 90.0 90.9 91.5 91.9 92.6 93.2 93.9 Size, kv.a 750 1000 1250 1500 2000 3000 4000 5000 Per cent efficiency 94.4 94.7 94.9 95.0 95.2 95.4 95.5 95.6 These values correspond to the equation y = 96.4 - 55 (55) 77. At fraction loads the per cent of full load efficiency obtained from motors and generators is as follows : 100 99 98 97 96 or 3 95 93 92 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 K. V. A. FIG. 14. Efficiencies of a-c. generators. INTERNAL COMBUSTION MOTORS 99 TABLE 40 PER CENT OF FULL LOAD EFFICIENCY OF INDUCTION MOTORS 3 phase 60 cycle 210-550 volts Per cent of full load .............. 125 100 75 50 25 Per cent of full load efficiency ...... 97 100 100 97 90 These values correspond to the equation I00 - . (56) 4OO where y p = per cent of full load efficiency, and F = per cent of full load efficiency. TABLE 41 PER CENT OF FULL LOAD EFFICIENCY OF A-c. GENERATORS 3 phase 60 cycle 220-550 volts Per cent of full load ....... 125 100 90 80 70 60 50 40 30 25 Per cent of full load efficiency 98 . 5 100 99.5 98.5 97.0 95.0 92.0 89.0 85.0 82.5 This corresponds to the equation (100 - FY , x y p = 100 - ^ - ' .... (57) 320 The attendance cost of motors and generators of 1910-12 is approximately A = > ........ (58) 78. Internal Combustion Motors. The prices of gas engines depends on whether they are two or four cycle, Otto cycle, Diesel, or semi-Diesel, as well as on the design and workmanship. Some engines are needlessly elaborate, others so unsubstantially built that they are made to sell rather than to use. The two stroke engine, except for the Diesel or semi- Diesel cycle, is not of practical value, and no data thereon is therefore given. The sizes and corresponding prices (1912) for one make of well-designed and well-built type of gas engines is given below: IOO BASIC COSTS TABLE 42 STATIONARY OIL ENGINES PRICES, 1910-12 Heavy Duty Otto cycle SIZE HORIZONTAL VERTICAL No. cy. Total Per h.p. No. cy. Total Per h.p. 2 $180.00 2 7O OO $9O.OO 90 oo I $l8o.OO $90 . oo 4 6 8 12 18 360.00 500.00 630 . oo 810.00 i 080 oo 90.00 83.33 78.75 67.50 60 oo I I I 270.00 360.00 72O.OO 67 . 50 6o.OO 60.00 25 OC 1350.00 54.00 I 2 1080.00 I3C.O.OO 43-25 ^4.00 22 i 1660 oo <2 .OO 36 /in 2 I 2160.00 1980 oo 6O.OO 4.O ^O 3 1980 . oo 55-oo So i or 2 2520.00 50.40 2 I 2160.00 2520.00 43-20 50.40 64 2 j 2AO OO CO CO 7e 3 3260.00 43.50 7t? I 3870 oo 51 .60 80 IOO 2 2 3870.00 5000 . oo 48.38 C.O.OO i or 2 5000 . oo 50.00 IOO 4 4320.00 43- 20 Price of Engines .^ ^ x^" s* ^^ ^^ x^ ** x^ S^~ ^ x^ Me E mH iffine rizo Cost ital- s X >^ >x x x^ ,X ^ <^ S^ x Mean Vertical ' Engine Costs -X"* ^ x* ^ j^r x^ ^ ^ ^ ^ ^ ^ X* x x ^ * ^ ^ ) 10 20 30 40 50 60 70 80 90 10 H.P. FIG. 15. Prices of oil engines. INTERNAL COMOTSTlON '< 101 Fernold and Orrok in their " Engineering of Power Plants " give the cost of producer gas engines as follows : TABLE 43 PRODUCER GAS ENGINES PRICES, 1908-10 PRICE F.O.B. FACTORY dL.., n.f. Total Per h.p. 20 $1,000.00 $50.00 60 2,800.00 46.67 75 3,6lO.OO 48 . 10 3,400.00 42.50 80 3,250.00 40.70 3,830.00 47.90 8s 4,150.00 48.90 o 3,500.00 41.80 IOO 4,925.00 49-25 (4,950.00 45-oo no { 4,960.00 45 -!0 1^ 4,200.00 37-5 130 5,250.00 40.40 135 6,600.00 48.80 160 / 5,500.00 \ 6,100.00 35-oo 38.10 250 6,650.00 26.60 400 (12,000.00 12,800.00 30.00 32.00 600 17,400.00 29.00 IOOO 33,750.00 33-75 2OOO 64,850 . oo 32-43 79. The prices of 4 cycle Diesel engines run as follows ; TABLE 44 DIESEL ENGINES, FOUR CYCLE 1910-12 Rated Price of engine Price per h.p. 60 $4,620.00 $77-oo 90 6,060 . oo 67.00 no 7,000.00 63 50 150 9,000.00 60.00 200 11,500.00 57-50 250 14,000.00 56.00 300 16,500.00 55-oo 400 21,200.00 52.80 500 25,800.00 51 .60 750 37,200.00 49.60 IOOO 48,300.00 48.30 102 SVr ! BASIC COSTS 9YU,UUW 60,0000 3 50,0000 24 O OO .lOOooO OO tOMOO 00 n ^ent, Efficiency 2 g Ml 1 . . - ., SS~* - * . " ^ r ^ ^ ,^-- 1 91 X / / 7 on " U 100 200 300 400 500 600 700 800 900 100 H.r. FIG. 1 8. Full load efficiency of gas and oil engines. The maintenance of gas and oil engines is usually expressed as between 2 % and 3 % of the first cost of the engine. This is not good practice because the first cost of a certain engine in use is a constant, while the cost of maintenance varies with the price of wages, or, if you please, with the value of the dollar. This method could be used provided that the first cost is first multiplied by the relative value of the dollar (indexed) before the 2 % or 3 % is taken. Otherwise, it would usually be too low, the error greatly increasing with time. 83. The variation of full load efficiency of Diesel engines with size is as follows: io6 BASIC COSTS TABLE 47 FULL LOAD EFFICIENCY OF DIESEL ENGINES H.p. per cylinder ....... 10 15 20 25 30 35 40 50 60 70 Full load efficiency, per cent ................ 24.0 26.3 27.7 28.5 29.1 29.5 29.8 30.2 30.5 30.7 H.p. per cylinder ....... 80 100 125 150 200 Full load efficiency, per cent ................ 30.9 31.1 31.2 31.3 31.4 These correspond to the equation i , = 5-tan-M ). ...... (62) 20 40 60 80 100 120 140 160 180 200 H.P. FIG. 19. Full load efficiency of Diesel engines. At fraction loads, the per cent of full load efficiency obtained is: TABLE 49 PER CENT OF FULL LOAD EFFICIENCY OF DIESEL ENGINES AT FRACTIONAL LOADS Per cent of full load . .100 90 80 70 60 50 40 30 20 10 Per cent of full load 100 99 97 93 88 82 74 65 54 42 The equation for these values is y = 100 72(100 F) 2 (63) All the above data is based on the guarantees of manufac- turers together with a careful study of the most reliable tests reported. The accuracy should be better than can be obtained from any single test. STANDARD PIPE 107 84. We give below prices of various sized pipe as of 1912, inasmuch as we shall have use of such data hereafter. TABLE 50 STANDARD PIPE PRICE (1912) PER 100 FEET, RANDOM LENGTHS S^ZE, INCHES STANDARD EXTRA STRONG DOUBLE EXTRA STRONG FLANGE Black Galv. P.E. C.&T. Plain ends Threads 1 $5 25 $c, en i T oC $2 7C, 5 2 5 5 5 2 7C c . co i 2 7 C. 7 85 c. 25 5.50 $18.00 * 1 3-20 4-45 5-50 5-8o 2O.OO I 4-55 6.36 8.10 8.50 25.00 $0.06 jl 6 2C, 8.70 ii .00 ii .60 35-oo if 7-5 10.40 13.20 13.90 45-oo 0.08 2 9-55 I3-50 17-75 18.65 60.00 O.IO $0.90 2? 15-75 22. 2O 29.15 30.60 90.00 0.15 1.20 3 20.70 29.10 37-75 39-65 125.00 O.2O 1.35 3^ 28.15 38.75 51.40 54-oo 165.00 0.25 1.50 4 32.00 44-05 58.00 60.90 190.00 o-35 I. 80 4l 40.00 54-50 99.40 0-55 5 44.60 60.80 IIO.OO 260.00 0-55 2. 2O 6 57-8o 78.80 148.00 360.00 0.70 2.40 7 80.00 153-00 240.00 0.85 3-45 8 85 25 163 .OO 271. oo I OO 300 116.00 221 .OO 4QCJ 10 i 20.00 227.OO 394 . oo I . C.O 5C. C. ii 292 .OO 12 i53-oo 292.OO 466.00 2.50 7.50 io8 BASIC COSTS TABLE 51 STANDARD CASING PRICE (1912) PER 100 FEET, RANDOM LENGTHS O.D. I.D. Wt. per ft. Casing C-it Thread Flange 2|* ti- 2.82 $12.90 O.IO 0-15 1-35 3 i ll 3-45 15-57 O.IO O.2O 1.42 3* 4-45 18.72 0.15 0.25 1. 80 4 3l 5-56 22.68 O.2O o-35 2.16 4! 4* 6.36 25-25 0.25 o-55 2.30 5 4f 7.80 31-59 0.30 0.60 2.50 6 Si 10.46 37-00 0-35 0.70 3-io 7 6| 12.34 47-54 0-45 0.85 4.60 8 7f I5-4I 58.25 0.50 i .00 5-25 10 9l 21 .90 84.75 0.75 i-5o 6.90 12 n| 30-35 131.00 1-25 2.50 9.00 $200 & LJ 3 160 >!-St Pipe 120 1 Utd Cas i. Wood 6 8 10 12 U 16 18 20 Dia. of Pipe, Inches FIG. 20. Prices of pipe. WOOD PIPE log TABLE 52 WOOD PIPE (1912) PI UCE PER 100 FEET FOR HEA DS OF SIZE So' 100' ISO' 200' 2 S o' 300' 350' 4 oo' 2" $9-75 $10.00 $10.28 $11.11 $11-95 $12.75 $14-45 $16.40 3 ii-QS 12.50 12-75 13-35 13.90 14-75 17.22 18.90 4 14-45 15.00 15.28 18.60 J 9-45 21. II 23-33 25-55 6 i8.33 21 .40 23-33 25.28 27.50 30.00 32-33 33-33 8 22.50 27.78 3I-40 35-55 38.60 41.40 44-45 51-67 10 29.72 35-83 40.83 46.95 50.85 55-83 59-17 68.33 12 33 -61 43.06 50-55 56.11 63.06 7O.OO 81.40 90.00 14 44.00 6O.OO T 6 56.00 67.00 18 72.OO 20 87.00 24 83.00 IO2 .OO no BASIC COSTS TABLE 53 W. I. PIPE FRICTION FEET PER 100 FEET OF PIPE G.P.M. SIZE OF PIPE IN INCHES OF DIAMETER i U i 2* 3 4 5 6 8 IO 12 5 10 15 20 25 30 40 SO 70 100 120 ISO 175 200 250 300 350 400 500 750 1000 1250 1500 2OOO 2500 3000 3500 4000 4500 5000 2.32 8.40 18.9 30.1 45-5 64.0 109.0 0.28 1.02 2.25 3-70 5-60 7 .8 13-3 20.2 37-6 73-0 0.08 0.36 0.81 1.29 i .96 2-73 4.68 7.10 13.2 25-6 36.0 54-0 0.06 0.12 0.25 0-43 0.66 0.92 1-57 2.38 4.42 8.6 I2.O 18.7 23-7 30-9 O.II 0.18 0.27 0.38 0.65 0.98 1.83 3-52 4-97 7.72 9-75 12.8 19.7 27.1 0.16 0.24 o-45 0.88 I . 22 1.82 2.4O 3.12 4.80 6.70 8.80 U-3 17.2 0.15 0.29 0.41 0.63 0.84 1. 06 I. 60 2-25 2.99 3-8l 5-8 12.3 ::::: 0.18 0.23 0-34 0.44 0.66 0.92 I. 21 1.58 2-33 4.87 10.3 13.0 o. 16 o. 26 0.29 0.40 0.58 I . 21 2-51 3-18 4.48 7.65 1 0.19 o-39 0.83 1.03 1.49 2.50 3-8i 5-3 7.20 0.17 o-34 0-43 0.61 1.02 I.S6 2.42 2.80 3-80 4.82 5.82 Harding & Willard: * When slightly rough add 15%. When very rough add 30%. , * John Wiley & Sons, Inc. WOOD AND RIVETED STEEL PIPE in For large sized pipe the following data are given at Chicago, including the cost of laying but less haulage. TABLE 54 WOOD STAVE PIPE COST (1912) PER FOOT HEAD IN FEET SIZE, INCHES, DIAM. 25 SO TOO 200 12 $0.42 $0.49 $0.63 $0.85 18 0.69 0.8o 1.02 I .46 24 0.79 0.91 I.I4 1.61 30 0.96 I. 12 1.44 2.06 36 1.19 1.40 1.82 2.65 42 1.40 1.68 2.23 3-33 48 i-S5 1.85 2.46 3.67 54 2.23 2.62 3-43 5-02 60 2.85 3-35 4-37 6.40 66 3-21 3-8i 5-oo 7.38 72 3-65 4-38 5-83 8-73 Wood pipe has made good in practice and has come into very general use. The friction of unplaned wood pipe is about equal to riveted steel pipe, but that of planed wood pipe is given at only 60% to 80% of that of riveted steel pipe. TABLE 55 RIVETED STEEL PIPE COST (1912) PER FOOT Size, GAGE THICKNESS inches, diam. 14 12 IO 8 6 i A I 2 $0 ^2 $0 18 $0.44 18 24 0-57 0.65 o.St; $0.78 1 .04 $0.98 1.28 $1-55 $i .99 ?o I 27 I $Q I Q3 2 46 $? 04 l6 I . CC I 03 2 . ^O 2 O2 3 58 4.2 I.6l 2.18 2.66 2 . 27 4.12 48 2.48 3.03 3.83 4.66 C4 2.80 3.41 4- 20 5-21 60 3 70 4 7^ c . 74 66 4- 3^ < . 21 6. 29 72 4- ^2 s.66 6.83 112 BASIC COSTS TABLE 56 TABLE OF PIPE FRICTION FOR LARGE RIVETED STEEL PIPE IN FEET PER TOO FEET OF PIPE Cu SEC FT SIZE c >F PIPE IN INC HES OF DlAME PER 12 18 24 30 36 42 48 54 60 66 72 er T 6 o ? IO 1 8 i .0 o 3 O. I 1C T 8 O. i? O. 2 0.06 2O "? 2 o 8 o. 3 O. 2 o.o<; 30 CO 1.9 o-S I A 0-3 O 7 0.17 O 4 O.I o 16 o 08 IOO 200 I .O 0.56 0.30 I .OO O.2 0.66 O. I 0.47 0.07 O 20 2QO o 84 O ^O The same authorities give the approximate section of dams not used as weirs, and their approximate cost as follows, where (h) is the height of the dam in yards. TABLE 57 APPROXIMATE COST OF DAMS 1910-12 Common Common Type of dam up stream batter, down stream batter, Approx. sec. in sq. yds. Approx. cost per cu. yd. Approx. cost per lineal yd. hor. to vert. hor. to vert. Earth .- 2 to I 3 to I 2- 5 A 2 $0.50 $I.2 S # Crib if to if \\ to i i . 5/? 2 i -5 2.25/J 2 Rock fill 2 tO I 2 tO I 2.0/f 2 2.00 4 . ooA 2 Masonry, straight or o to i vertical 0.8 to i o.4/f 2 12 .OO 4 Soh z Masonry, arched. 0.15 to i 0.3 to i 15.00 3- So/* 2 Tunnels. On the average, the cost of tunnels including timbering and lining is $15 per cubic yard. The costs and capacities of tunnels is given below. CANALS TABLE 58 COST OF TUNNELS 1910-12 Size in ft. Sectional area, sq. ft. Allowable velocity, ft. per sec. Capacity, sec. ft. Approx. slope per 1000 ft. Approx. cost per lih. ft. Cost per lin. ft. per 100 sec. ft. 4X7 28 3-6 IOO 0.46 $16.00 $16.00 7X7-25 5 10 - 500 2.O 28.00 5.6o 10 X 10 IOO 10 1,000 1-5 56.00 5.6o 12 X 12.5 150 IO 1,500 I.I 85.00 S-6 7 14 X 14-25 2OO IO 2,OOO 0.9 115.00 5-75 20 X 25 500 IO 5,000 0.6 280.00 5-6o 30 X 33 1000 10 10,000 o-3 500.00 5.00 Canals. In ordinary earth a velocity of 2 feet per second is commonly used. On this basis the following data is given. TABLE 59 COST OF CANALS IN ORDINARY EARTH 1910-12 CAPACITY, SEC. FT. AREA WET, SEC. SQ. FT. DEPTH OF WATER SLOPE IN FT. PER MILE APPROX. COST PER RUNNING FT Low High 50 25 2-5 4 $0.375 $0.75 IOO 50 3-5 2 0-75 1-50 200 300 400 IOO 150 200 5 6 7 1-5 1 .0 0-75 1-50 2.25 3-00 3.00 4-50 6.00 500 250 7 0-75 3-75 7-50 1000 500 IO 0.50 7-50 15.00 1500 75 12 0.50 11.25 22.50 2OOO IOOO 12 o-33 15.00 30.00 3000 1500 15 0.25 22.50 45 oo In rock the allowable velocity is 8 feet per second if the canal is lined. Under such conditions the following data is given. The costs may vary greatly with local conditions. H4 BASIC COSTS TABLE 60 COST OF CANALS IN ROCK 1910-12 CAPACITY, SEC. FT. AREA WET, SEC. SQ. FT. WATER, DEPTH SLOPE IN FT. PER MILE COST PER RUNNING FT. Low High 50 6.25 2-5 40 $0.32 $1.28 100 12.5 3-5 25 0.63 2.50 200 25.0 S-o 16 1.25 5.00 300 37-5 6.0 12 1.87 7-50 400 50.0 7 10 2.50 IO.OO 500 62.5 7 9 3-25 13.00 IOOO 125.0 10 6 6.00 24.00 1500 175.00 12 4-5 8.75 35-oo 2OOO 250.0 2 3-5 12.50 50.00 3000 375-o IS 3-o i8.75 75-oo The Reclamation Service gives the following costs of excavation. TABLE 61 COST or EXCAVATIONS 1910-12 Low High Average Plowable with 4 horses $o 098 $1 OO $0 18 Plowable with 6 horses Indurated material Loose rock o. 1225 o. 29 O 3? 2.OO 2.OO 3OO 0.30 O.6o O 7"s Solid rock o 60 r OO 2 OO Excavation below plane of saturation . o 20 1 OO I 80 Solid rock below water A CO COST PER Cu. YD. The same authorities, the General Electric Company, give the following valuable data on hydro-electric installations. HYDRO-ELECTRIC INSTALLATIONS . I w ** I-H PM cq o <* ^ ^ O Q o O I ON tO ON t^ rf | | M M 04 t> -<*-OO . tf> C4O IOI>. ^- O 1 vo 10. O 10 o o 10 T HMMCS -CSt>. O O I i O\MD 10 68 88 VO O M 'o O MM O O t>. ON O O O 1 1 o o o O O l ;S S " ..T -Hu < . O O NO cl Q Q O O O O t^ CO ' ' oo 00 V> H t-t **> " O O O O oo oo I : i U H H CJ UU n6 BASIC COSTS In the above Low head, 50 to 200 ft. Medium head, 200 to 600 ft. High head, 600 ft. and above. Small capacity, 200 to 1000 kw. Medium capacity, 1000 to 5000 kw. Large capacity. 5000 kw. and over. The per cent of total costs of various items of construction in hydro-electric plants is given as follows: SMALL Low HEAD PLANT Hydraulic work not including power house. 55 % Power house building 6 % 1 Water wheels '. . . . 9 % \ Power house fully equipped 45 % Electric equipment .3% J 100% MEDIUM Low HEAD PLANT Dam 43%) Low pressure pipe 20 % > Hydraulic work less power High pressure pipe 3 % j house 66% Power house and machinery 34 % SMALL MEDIUM HEAD PLANT Hydraulic work not including power house building 76 % Power house building 8 % j Turbines 3 5 % f Power house fully equipped 24% Electric equipment 12.5 J LARGE MEDIUM HEAD PLANT Hydraulic work 38 % Power house building 10 % ] Hydraulic machinery 31 % \ Power house fully equipped 62 % Electric equipment 21 % j LARGE HIGH HEAD PLANT Dam 22 % 1 Hydraulic work not including Low pressure pipe 23% , r house fi High pressure pipe 16 % J Power house fully equipped 39 % 85. More complete data on the cost of buildings, excava- tions, fills, foundations, bridges, tunnels and the like are given in Gillette's "Handbook of Cost Data" (Myron C. Clark REFERENCES 117 Publishing Co.). For .similar data on power plants, the reader is referred to Fernold and Orrok's "Engineering of Power Plants" (McGraw Hill Book Co.), Harding and Wil- lard's " Power Plants and Refrigeration" (John Wiley & Sons, Inc.), and Guldner's "Internal Combustion Motors" (D. Van Nostrand Co.). The equations of cost as given by Harding and Willard in their "Power Plants and Refrigeration" (John Wiley & Sons, Inc., 1917) are as follows: 118 BASIC COSTS ill e 6 . J S "^ *O .**!* .s *j "^ ** 3 d 5" c L, In O .43 X x 5 .3 d -5 x-d v -c 3 vx u U v A v u X v to * ' .. X x ' ^ ? & Q 1 . ^oc oc ^1" M " "5 if) sO ^ to ^ ' CO ^h P^ co\O c CS ^ -T 3 <*M 3 + ;? + M 43 CO to vc 00 ^ +0 S - + X + + to + * ^ ^ X + + -f T M O* - ^ O ~l xo VO O O vO M M o \O CO M HI e^ O\ ^O !> (J \O U ^ O 1* M to CO co cs rt *- -* M o ^o to N H C N H H M H : J^' j j j j a ^ i' ; Scj ^ S S : : D 1 f\ CL CXi OH ^4 i d to tn | "o "o 1 : 10 ds2 u ^33333^ d j * S C-. ^ Cu ^ -d g g o o gc^oo^ggQ 43 ^ ^ S "S g J^ -, > > ^OOioco^ 01 ^^ O u o -) M o . o s H d M o o . . OOOOOofe n C o^o c 3 o ?- >S ^4) -4-J -*-J -4-J - ' 4-J ,^ O 4- *M ^ ! P {* 8^^ w N M | ' I ' o ' o .i S3 S3 : : :.S :.S Ij o ; : : S* : g* *3 tn | 1 1 S : : S : g I : ; ? ; ^ | : 1 1 g 1 : 8 : 8 yitiiiii llllPlI ;i a 2 tf J 5 -tf*! -5 K P O K o O > 5 fe Horizontal, fire-tube cylindri tubular, 100 Ib. per sq. in. Portable locomotive Vertical, water-tube, press 12? Ib. ner so. in. . Horizontal, water- tube, pres 125 Ib. per sq. in Barometric (28 in. vacuum) . Jet condensers m j "8 1 1 e 1 8 i 1 M *3 1 COST DATA 119 g o i 8 o | g x J ^ ^ 10 \O * ft cx 8 % x; | o ^ do. w cx^ ^ ^ H v;c 'x ' to to ) a S o ~> ON 1 + X-N O M 0^ + o X + + X + + +V? +4 \o oo to s -|- - 10 ro co vo O H OO H Tj" ^ CS M ON O t^ NO T MCO^OcOO CO OO OO NC ^ COMCOVO^ VO MOO co H CO T T _O M ' -i ^ S oooooo o oo o c & _ U CX OH CX CX OH OH OH OH OH C -4- t> P U WPPt>P t> PP P t CO I i >H M i.i . u . i ' ' ' " : c ' 3 ^ j g : : ^ -5 : ^ : j- 4n i 8 ^ W) W) ! fn 'rt r2 '. i --3 | 1 | * : s :> ^y* ^ flJ * C3'di o~ H O o 11 |4 ^ ^ : 1 II & *J CX *H 4> 0) ^ 1 i C^ ^ 14:? 1 1 !! ill s c 1 a 1 1 M 'g -y 5 w U c* c< ^^ ^ ~*^ T'c^o^fll ^ ^^n'-^'~ i o 5 ||||>ll^l3|jsl4rl-a S^^ 1 | C' 3 *7SC ) o3Oj-n:>'~-'^'"CJ K Oo53't3CXHu>-iQ^ | ?CX x*^ C- ^ c rt ^ 8 6c_,'-^ N Di3 S'^ w I i 1 8 *0fl *00 120 BASIC COSTS j ^ j, .a . t d * JH ^ 8 ^ 4< d d 3* & d c - -^ Jts > < ^ . "o a x x .d ^^ y * ^ .S v y x * A v Tf S x -s < ? t^^ IO CJ ^ s.-^ Q< i Xd i ON *fr 00 00 ON <* * f 1 d : ^43 0, a 4 a 6 - U d d d a ex a a, a ex c u d ex j ip ^ -d -3 43 J^ ,d 43 ^3 rd X 1 O O O O O C O rO O O O O T o o h O O i 0^8 vo ^ CN H O ^ O O C O < D - ^a a g ^a, a a ^a, ^a a ^c ^ a a t "& P P t i J^ D I | " H t S'g 8 ^ 'C ^ ? :3 : 1 -8 fO jTv C^ S^ v i' : 1 Flywheel governor, Corliss n leasing valve, horizontal Corliss governor and valves, zontal f III 1 f 1 i 8-ii 1 1 ^ 6 151 I i 1 _ i ~^ zJ iz3 _ i i c/i M o||p& 5 9&Bi ^ ^pqpqtiH co ^^j h Flywheel governor, Corliss, n leasing valves, horizontal Flywheel governor, multiple valves Sizes 70 to 140 hi r\^ n . "g 1 5 3 n rt E j o a 1 o - S ^ j a3 T C3 8 % u ^ Z M t/3 * 1 W P ^ COST DATA 121 1 o ation of cost in c JS i'/ j i 4 *4* * XI > > 44 X 43 X 43 X X \/ O^ + M r- *.* ^O i/> '^fe X X ^ ^ ^^.t^io < 7 2d, c? ^ S CN 2^0 d d 43 43* X X IO 04 i + 5 +x S ^ + + + +0 + + M ^}- M , H^-O-j- crjiOCO "I MW 1 CN M ?O ^ OO C<) CN M IO M CN VO ^O M s-^ o s~^ O O O O O '""' ^"^ 6 i : a : < . a : s> : 8> : I g g : 1 1 d H :d^ M : G : w H dd: M M S* Sla * * B' a jjj || j^lj I l| | 1 3 8 01 8^ o ^' islt ^ =i :^ ^ go ^ M ^ : 1 |^ : ^^ J : S : d d : &| ^ M g j ' J4 j * -43 43 : ^^ . d d : 43 | S ^ g o _ 8 _ ^ ^_ ^_^| 1 !'! J? o : ^^??Q S?Q?0?Q?3?S h S S H B^dS-^^d^a^d^a aoa^S^ o ^ 8> ^ , ^ a ^ ^ ^ ^ |{a j j P M p coi > *> M O 1 ' HH & ?0? a^ a 10 ; ; ; '. S '^ : 3 'o g g> ^ S o *4J -M ^ (N S k 1 "8 CN O c P ? i 11 1 !i 1 1 ! ! 1 ternating cu ingle phase ' 1 1 S ^ S 5 > ^^ M i 1 i "o a ^ * | | 122 BASIC COSTS vo 00 M O VO ^" lO V O + + 4- (N O O X X XX XX O O . a a a ex a, a cxaRpHcxao, 111 O 0) ^ CA! Pn cr P *~ "? ^ ^ ^ J^ fe ^cScS X X X^cS X^ X^ ^ X X ^ ^ ^^^"^ XXtoiooo Xt->. ioXtooo\ ^o 'rt **5 OOHOt^ l OVOOO'^-' -OOCOt^O CO M M r< CQ ^ M Oi > O0 1 O v O| w_i_ ,*^,^<)H + ^++++^^;+++^++ +g tit X+ O O fO n t^* O c^j O co cs o 10 w OO O co t > * t^ O O O w MOO'4"COCOHlO'd''4-lO VESTANCES Total Vestance. The Time Element. Change Point. Vestances of Steam Engines of Various Types at Full and Fractional Loads. Vestances at Full and Fractional Loads of Oil Engines. Diesel Engines. Induction Motors and Generators. Comparison of Power Units of Various Size and Types at Full and Fractional Loads. Ves- tances at Full and Fractional Loads of Centrifugal Pumps and of Standard and Wood Pipe at Varying Capacities. 86. As previously shown, the total vestance gives us a basis of comparing two or more units, or two or more systems that are designed to render similar service but each differing from the other in one or more of their cost elements, such as first cost, life, efficiency, attendance and the like. The question that both the manufacturer (producer) and user must face con- tinually is one of economy. The user desires to purchase that unit which will give him the desired service at the least ulti- mate cost. The manufacturer's problem leads to the same end, namely to produce a unit which will represent the best possible investment for the purchaser. But the purchaser's needs are variable. What may be best for one, may be a very poor investment for another purchaser, whose conditions may be entirely different. It is therefore necessary to produce various classes and types of apparatus, to meet these variable conditions so that a best investment for each condition may be offered. That is the problem in general. Specifically the producer may face such a problem as this: a more efficient engine can be built, but at a greater first cost. If produced, will this engine represent a better investment than the unit now being produced and will it sell? Whether the x new engine will represent a better investment for any given service, can be readily determined by comparing the total Vestances. If that of the new engine is the smaller, greater 126 THE TIME ELEMENT 127 financial efficiency has been attained. Whether the engine will sell, depends on whether it is really a better investment or not, and by how much, and second whether we can prove to the user that it is a better investment. Of course if we can prove it for ourselves, we can prove it to the user, but when it comes to a matter of guess, one man's guess is quite as good as another's. At all times we must bear in mind that we do not purchase the machine, but the service that it will render us. Manufacturers therefore produce service in just the same sense that the so-called public utilities do. 87. The Time Element in Vestance. The total vestance depends upon the amount of use that is made of the apparatus. This may be shown as follows: The total vestance is neglecting insurance and taxes, where C = the first cost, T =* the term factor, R = the interest rate, and A = the annual cost of operation. The annual cost of operation (^4) varies with the number of hours of use per year and the cost of operation per hour. If a = the mean operating cost per hour of the entire unit or system and N = the number of hours of use per year, then A = aN and the total vestance becomes ' . ' J V - + !. ....... (64) Dividing equation (16) through by (N) gives the vestance per hour Vn of service or By equation (65) it is evident that as (N), the number of hours of use increases, the vestance per hour decreases. The 128 VESTANCES maximum value that (N) can attain is 8760, the total num- ber of hours in a year. For this value of N, the vestance per hour of service is a minimum. When, on the other hand, ( N) decreases, the vestance per hour increases, becoming infinite when N = o. Evidently then as ( N) becomes smaller, and the term ( J becomes larger, the constant term ( ) becomes of less and less consequence. We can interpret this for practical use by saying that where the unit is but little used per year, the first cost is of primary consideration, and the operating cost per hour (efficiency and energy cost) of secondary importance only, while for many hours of operation per year conditions are reversed. That is, when the apparatus is much used, we can afford to buy expensive equipment in order to get low cost of operation, but where little used, cheap apparatus of 'comparatively low efficiency is the best. 88. Valuance. If we take the reciprocal of the vestance per hour (Vh), we get by equation (65) TT X "=" TRN h CR+TaN We shall call (U) the valuance. It is a measure of the worth of the unit. When (N), the number of hours of use, is zero, the valuance is zero, i.e., the unit is worthless (useless). The valuance increases as (N) increases, becoming a maxi- mum when N = 8760, i.e., the apparatus is worth most (most useful) when it is used continually. 89. We shall now determine the vestances of various types of equipment for a normal year of 3000 hours, assuming a constant load during the entire period. In this way we can get a first approximation of the comparative values of such equipment. Further on, we shall consider more variable and exact conditions. 90. The Steam Engine. From the previous chapter, we find that the first cost of a 100 h.p. low speed compound steam THE STEAM ENGINE 129 engine is $3300. The efficiency with dry saturated steam at ioo# initial steam pressure is 12 %. The cost of oil, attendance, etc., is about $0.05 per hour for this size of engine, and the maintenance about $0.02 per hour. The life of low speed compound engines is 25 years. If the interest rate is 5%, then by Table 2, the term factor is 0.70469, so that the depreciation vestance is Vd = 33 = $4682.90. 0.70469 If the coal used for fuel contains 13,743 B.t.u. per pound, its energy content in h.p.h. is 2545 If this coal costs $2 per ton, or $0.001 per pound, then its cost per h.p.h. is o.ooi -s- 5.4 = $0.000185. If the combined efficiency of the boiler, furnace, and grate is 70%, then the cost per h.p.h. of heat energy delivered to the engine is 0.000185 -r- 0.7 = $0.000264. To this we must add the other costs of producing steam, amounting to, let us say, $0.000236, giving a total cost of the heat delivered to the engine of 0.000264 + 0.000236 = $0.005 per h.p.h. Since the efficiency of the engine is 12%, the cost of the heat per i.h.p. is 0.0005 -4- o.i 2 = $0.00417, and for 100 h.p. this amounts to 0.00417 X 100 = $0.417. To this must be added the other operating costs of $0.07, giving a total of 0.417 + 0.07 = $0.487 per hr. For a year of 3000 hours, this amounts to 0.487 X 3000 = $1461, giving an operating vestance at 5 % interest, of 1461 -T- 0.05 = $29,220. 130 VESTANCES Since the depreciation vestance is $4682, the total vestance is V t = 4682 + 29,220 = $33,902, or $339.02 per i.h.p. Note that the operating vestance is over six times as great as the depreciation vestance. In other words the first cost is of relatively small importance as compared with the costs of operation, a matter not always fully appreciated. 91. As compared with the above, let us determine the total vestance of the above engine when the initial pressure of the steam is 2oo#, other things remaining the same. Since for compound engines, the efficiency is y = 0.0675 ( ~ i45)i and the steam temperature () at 2oo# is 388, we find that y = 0.0675 (388 - 145) = 16.4% instead of 12% previously taken. The heat will cost therefore 0.0005 -f- 0.164 = $0.00305 per h.p.h., or 0.00305 X 100 = $0.305 per hr., giving a total cost per hour of 0.305 + 0.07 = $0.375, or O-375 X 3000 = $1125 per normal year. The operating vestance is then 1125 -T- 0.05 = $22,500, giving a total vestance of 22,500 + 4682 = $27,182. This shows a reduction of 33,902 - 27,182 = $6720, or $67.20 per h.p. If we add to the 2oo# initial gage pressure 200 of super- heat, then the efficiency will be increased by 20%, amounting to 1.20 X 16.4 = 19. The cost of heat per hour will then be THE STEAM ENGINE 131 and the total cost per hour will be 0.254 + 0.07 = $0.324, or 0.324 X 3000 = $972, giving an operating vestance of 972 -f- 0.05 = $19,440, and a total vestance of 4682 + 19,440 = $24,122. This represents a saving of 27,182 24,122 = $3060, as compared with the same unit, not using superheat but trie same pressure, and 33,902 - 24,122 = $9780, as compared with the engine using dry saturated steam at ioo# initial pressure. The per cent saving by use of 2oo# pressure as compared with ioo# initial gage pressure is 6720 ^ 33,902 = 20%. The saving in total vestance by the use of 2oo# pressure together with 200 superheat as compared with the use of ioo# dry steam pressure is 97 8 + 33.902 = 28.8%. Note the enormous saving especially with the use of higher pressures. Modern practice tends towards continually higher pressures. That this is good practice is too clearly shown above to need further emphasis. 92. Let us compare with the above the total vestance of a 1000 h.p. compound engine, the life, rate of interest and cost of coal being assumed the same as that for the 100 h.p. engine. For this engine the first cost is $18.60 per h.p. and the de- preciation vestance is then 18.60 -f- 0.70469 = $26.394 per i.h.p. At ioo# initial gauge pressure, and dry saturated steam, the efficiency of the engine is 14.7%. The heat cost per h.p.h. will then be 0.0005 * - I 47 = $0.0034. I 3 2 VESTANCES Allowing $0.0006 as the cost of attendance, maintenance, etc., per h.p.h., the total operating cost per h.p.h. will be 0.0034 + 0.0006 = $0.004, or 0.004 X 3000 = $12 per h.p. year. The operating vestance is then 12-f- 0.05 = $240 per h.p., and the total vestance per h.p. year is 240 + 26.394 = $266.394, as compared with a total vestance of 339.02 per h.p.h. for the 100 h.p. engine. This gives a difference of 339.02 - 266.394 = $72.626. The per cent saving as compared with the 100 h.p. engine of 72.626 -T- 339.02 = 21.4%. This is of course not a clear saving due to the greater dis- tribution area, and corresponding greater distribution cost for the larger plant. In fact the vestance of this increased cost of distribution may, if the system is too large, more than off- set the decreased vestance of the engine itself. 93. We shall now determine the total vestance of a 100 h.p. low speed simple noncondensing engine operating on dry saturated steam at ioo# initial gage pressure, the life, cost of coal, and interest rate remaining the same as above. The first cost of such an engine is $21.50 per h.p. The depreciation vestance is therefore 21.50 -r- 0.70469 = $30.51 per h.p. This engine will use 42% more steam than a compound condensing engine. The efficiency of the simple noncon- densing engine of 100 i.h.p. will therefore be 12 -T- 1.42 = 8.45%. The cost of the heat energy per h.p.h. will therefore be 0.0005 -7- 0.0845 = $0.0059. Allowing as before $0.0007 for the other operating costs, we get a total of 0.0059 + 0.0007 = $0.0066 per h.p.h., or 0.0066 X 3000 = $19.80 per h.p. year. THE STEAM ENGINE 133 The operating vestance is then 19.80 -f- 0.05 = $396, and the total vestance is 396 + 30.51 = $426.51, as compared with 339.02 for the 100 h.p. condensing engine. The difference is 426.51 - 339.02 = $87.49, or 87.49 * 339.02 = 26.8%. 94. For a 100 i.h.p. simple noncondensing high speed en- gine operating on dry saturated steam at ioo# initial gage pressure, the first cost is $12.75 per h.p. but the life now is only 15 years instead of 25 years for the low speed engines. The depreciation vestance is then 12.75 * 0.51897 = $24.58. Since the simple high speed noncondensing engine will use 64% more steam than a compound condensing engine and the efficiency of a 100 i.h.p. engine of the latter type is 12%, the efficiency of the high speed engine will be 12 -T- 1.64 = 7.3%. . The cost of the heat energy per h.p:h. is then 0.0005 -T- 0.073 = $0.00685. Adding thereto the other operating cost per h.p.h. we get for the total operating cost per h.p.h. 0.00685 + 0.0007 = $O- OO 755, or 0.00755 X 3000 = $22.65 P er h-P- year. The operating vestance is therefore 22.65 + -5 = $453, and the total vestance is 453.00 + 24.58 = $477-58, as compared again with 339.02 for the compound engine. This gives a difference of 477-58 - 339-0 2 = $138.56, or 138.56 -5- 339-02 = 41%. 95. In order to bring out the differences more clearly, we tabu- late the vestances of the three classes of 100 i.h.p. engines below. 134 VESTANCES TABLE 64 VESTANCES 100 I.H.P. ENGINES WITH DRY SATURATED STEAM AT IOO# G.P. Engine Vd Va F Percent Vt. Va Vd A 46.82 292. 2O 339.02 I .OOO 6.23 B 30-51 396 . oo 426.51 1.268 13.00 C 24.58 453-oo 477-58 1 .410 18.40 Here engine A = low speed compound condensing; B = low speed simple noncondensing; and C = high speed simple non- condensing. Note now that between engines A and B we save $16.31 in depreciation vestance at a cost of $103.80 in operating ves- tances, and between A and C we save $22.24 in deprecia- tion vestance. Neither of these represent very good bargains. Yet how often are such bargains entered into! 96. We tabulate below the vestances of steam engines, based on 5% interest, $2.50 cost per ton of coal of fuel value as above, i.e., 5.4 h.p.h. per pound and 25 years' life for low speed engines, and 15 years for high speed engines, all as taken above. The effect of increased cost of operation is to increase the operating vestance by a similar amount. Thus if coal were $5 per ton and other operating costs increased in like pro- portion, the operating vestance would be doubled, amounting to some $400 increase. If on the other hand the first cost is doubled, the depreciation vestance will be doubled, but this will amount only to some $30. Evidently then the effect of increased prices is to make efficiency relatively more im- portant and first cost, though itself increased, of relatively less importance. The effect of increased interest rate is to decrease the vestances. THE STEAM ENGINE 135 TABLE 65 VESTANCES PER H.P. or COMPOUND CONDENSING Low SPEED STEAM ENGINES AT FULL LOAD VESTANCES RATIO SIZE, I.H.P. Depreciation Vd Operating Va Total V Va Vd IOO $46.829 $292.00 $338-83 7-2 * . 20O 35.476 271 .00 306.48 8.7 3 00 31.688 262.00 293.69 9-3 4OO 29 . 800 256.00 285.80 9.6 500 28.736 251 . 2O 279.94 9-7 600 27.913 247 40 275-5I 9.8 700 27.572 244. 10 271.67 99 800 26.962 242.00 269 . oo 10. QOO 26.600 241 .10 267 . 70 IO.O IOOO 26.394 240 . oo 266 . 39 IO.I 1500 25.642 236.00 261 . 64 IO.2 2OOO 2S-259 234 . oo 259.26 10.2 ^ cv300 >100 ting V _Uei jrec ation Ves 200 400 600 800 1000 1200 1400 1600 1800 2000 H.P. FIG. 21. Vestances per h.p. of compound low speed condensing steam engines. 136 VESTANCES TABLE 66 VESTANCES PER H.p. OF SIMPIE Low SPEED NON CONDENSING STEAM ENGINES VESTANCE AT FULL LOAD RATIO SIZE, I.H.R Depreciation Vd Operating Va Total V Va v d IOO $30-SIO $398 . oo $428.51 J 3-i 125 28.381 386 . oo 414.38 13-7 ISO 26.962 382.00 408 . 96 14.2 175 25.898 374-oo 399.90 14.4 200 25.188 368 . oo 393-19 14.6 250 24.124 362 . oo 386.12 15.0 300 23-4I5 358.00 381-41 J5-3 350 22.918 354-oo 376.92 J5-4 400 22.521 350.00 372.52 15-5 450 22.237 346 oo 368.24 iS-6 500 21.996 342.00 364.00 15-5 TABLE 67 VESTANCES PER H.P. OF SIMPLE HIGH-SPEED NONCONDENSING STEAM ENGINES VESTANCE AT FULL LOAD RATIO SIZE, I.H.P. Depreciation Vd Operating Va Total V Va Vd 20 $49-3l3 $874 . oo $923-3i 17.7 30 36.328 618.00 654-33 17.0 40 29 . 800 550.00 579.80 18.5 50 25-898 503 oo 528.90 19.4 75 20.718 474.00 494.72 22.9 IOO 18.093 454-oo 472.09 25-0 150 15.462 434-00 449.46 28.5 200 14.191 418.00 432.19 29.6 250 13-339 41 2 . OO 425-34 31-4 THE STEAM ENGINE 137 , I l\ \ ^ -- = * - ~ - - ^ mplt =^ - Hig 2om] ^9^1M d- sp< ounc | on-co jle-L -spee ndea 3W--SI ^.HM sing eed._ idem Ingi Sons ing] _^__ ies onde . nsint ~ ^_ JEng ^__ ines. S^ 50 100 150 200 250 300 350 400 450 50( H. P. $1000 800 400 200 FIG. 22. Total vestances per kp. of steam engines. A comparison of these tables shows that the total vestance decreases with increased size of unit. The decrease is rapid for units less than 100 h.p., rather small between 100 and 500 h.p., and almost insignificant beyond that. Again the approximate mean vestance ratio is 10 for compound con- densing engines, 15 for simple low speed noncondensing en- gines and about 25 for simple high speed noncondensing engines. That is as the first cost is reduced, the relative amount of money that must be spent for operation is enor- mously increased. In as much as the vestance is proportional to the total amount of labor that it takes to produce a given service, including not only the labor of operation but of the construction of the units and the system, it appears that in cheap layouts, too little effort has been spent in the design and construction of the units and system and too much must therefore be spent in operation. Less total labor would be required if it were used in the right place, in properly prepar- ing for the rendition of the service first, before rendering it. On the other hand, too much may be spent on preparation. 97. The above tables were figured on the basis of a normal year of 3000 hours of use. If the number of hours of use are changed, then the cost of operation is changed, since the latter 138 VESTANCES is in direct proportion to the hours of use. But the deprecia- tion vestance remains constant. We have then, that F= Fz> + V a 'N, (67) where Va = the operating vestance per hour, and N = the hours of use, VD = the depreciation vestance, and V = the total vestance. From the above, it is evident that the vestance per hour is V k =f+V a > (68) Example 27. Determine the total vestance per hour of a 500 h.p. compound condensing steam engine for a variable number of hours of use per year. Solution: By Table 65, we have ' V = $28.736, and V a = $251.20 for N = 3000 hrs. Then F ' = *** = $0.08373. 3000 Therefore the total vestance per hour is Vj = ^ + 0.08373. From this, we tabulate as follows: N V h 5000 hrs. $0.08948 1000 hrs. 0.11247 100 hrs. 0.3711 i hr. 28.82 ohr. oo The total vestance of this engine is F = 28.736 + 0.08373 N. Example 28. Determine likewise the total vestance per h.p. for (a) a 100 h.p. compound condensing engine, (b) for a 100 h.p. simple low speed noncondensing engine and (c) for a 100 h.p. simple high speed noncondensing engine. THE STEAM ENGINE Solution: (a) In this case, we have by table 64 _ 46.829 292.00 Vh ~~ '' so that V h " = + 0.09733, V" = 46.829 + 0.09733 In this case we have by Table 65 V h '" = ^ -22*12 + 0.13*67, i\ 3000 ^OXIO . or so that V" = 30.510 + 0.13267 N (b) (c) In this case, we have by Table 66 T/ 18.093 , I/ '"' S*-' I f*. -T T <} 1 ** -ft - I 5 I 33> or V"" = 18.093 + 0.15133 N (c) Plotting these equations (a), (6), and (c) gives straight lines as shown in Fig. 23. It may be that for a certain number of hours of use per $280 & 200 & w / \T& 1 160 I 3 120 I 40 200 400 600 800 1000 1200 1400 1600 1800 2000 Hours of Use per Year FIG. 23. Variation of total vestance with hours of use for 100 h.p. steam engines. 140 VESTANCES year, that the total vestance for the compound and simple low speed engines are equal. In such a case then V" = V", or 46.829 + 0.09733 #1 = 3-5 10 + 0.13267 Ni, whence Ni = 461.1 hrs. So again the total vestance per year (V SL } for the simple slow speed engine is VSL = 3-5 10 + 0.13267^, and for the simple high speed engine it is VSH = 18.093 + 0.15133^. These two become equal when 18.093 + 0.15133 N 2 = 3-5 10 + 0.13267^2, that is when N z = 661.5 hrs. So again the vestance of the compound engines equals that of the simple high speed engine when 18.093 + 0.15133^3 = 46-829 + 0.09733^8, or Ns = 532.1 hrs. 98. Change Points. The meaning of these three points is clearly shown in Fig. 23 where the total vestance per horse power year has been plotted against the hours of use. The equations for vestance in terms of hours of use give straight lines as shown. From this it is evident that when the hours of use are less than 461.1, then the simple low speed engine, in spite of its lower efficiency, is more economical in use, i.e., has better financial efficiency than the compound engine. When used for more than 461.1 hours, however, the compound engine is best. So also when the hours of use are less than 532.1, the simple high speed engine has better financial efficiency than the com- pound engine. But this number of hours is greater than that of the simple low speed engine. So then for more than 532.1 hours per year use the compound engine; for less than this number of hours use the simple high speed engine and do not use the simple low speed engine at all. This is for the reason that for no hours of use during the year is it best, being always exceeded, either by the compound or simple high speed engine. CHANGE POINTS 141 We shall call the point of equality as NI, Nt, and Na above the Change Points. 99. For the 200 i.h.p. engines, we have a total vestance for the compound engine of V* = 35476 + 0.09033^, (a) for the simple low speed engine of VSL = 25.188+0.12267^, ...... (b) and for the simple high speed engine of VSH = 14-iQ 1 + 0-13733^ (c) Equating equations (i) and (2) gives 25.188 + 0.12267 #1 = 3S-47 6 + 0.09033^1, so that Ni = 318 hrs., and so also equating equations (i) and (3) gives 14.191 + 0.13733 # 2 = 35-47 6 + 0.09033^2 so that N 2 = 452 hrs. Again the value of the change point is greater for the high speed than for the low speed engine with the compound engine. So again we find that the simple low speed engine possesses no period of superiority and should therefore under the condi- tions assumed not be used for this size. For this size and these conditions, the compound engine should be used, if the period of use is over 452 hours, and when the period of use is less than this number of hours, then the simple high speed engine gives the highest financial efficiency. 100. So far we have discussed only the condition of an engine operating at full load. If instead the engine is used at a constant fractional load, the efficiency will be decreased and thereby the cost of heat will be increased. The other operating costs such as attendance and the like will be changed to correspond to the actual load carried. The depre- ciation vestance will change in inverse proportion to the per cent of full load carried. Thus at half load it will be doubled; at one and one-quarter load it will be four-fifths as great, etc. This is evident since, for example, at half load we have pur- chased two horse power to get one. We give below the vestances for each of the three types of engines at 1.25, 0.75, 0.50 and 0.25 of full load, the per cents 142 VESTANCES of full load efficiency at fractional loads being assumed the same for simple low speed engines for want of authentic data on the former. The vestances are per actual horse power developed and not per rated horse power. A normal year of 3000 hours is again assumed. TABLE 68 VESTANCES PER H.P OF Low SPEED COMPOUND CONDENSING STEAM ENGINES AT FRACTIONAL LOADS DRY SATURATED STEAM AT IOO# G.P. RATED H.P. TOTAL VESTANCE AT LOADS OF 125% 100% 75% 5% 25% IOO $345 5 $338.8 $361.4 $41 i . 6 $544-3 200 313-4 306.5 321.3 361.0 469.9 300 3oi-4 293-7 308.3 345-4 444-8 400 291.8 285.8 298.3 334-2 424.0 500 284.9 280.0 291.8 326.4 4I5-6 600 280.3 275-5 286.6 321.8 407.6 700 278.2 271.7 283.0 318.5 403-7 800 276.6 269.0 281.1 3I5-4 400.0 900 275.0 267.7 279.0 3i3-i 397-6 IOOO 274.0 266.6 277.6 3*1-4 395-6 1500 270.5 261.6 273:2 306.0 387.2 2000 267.5 259-3 270.5 302.2 383-8 1800 200 FIG. 24. Vestances per h.p. of low speed compound condensing steam engines. CHANGE POINTS 143 TABLE 69 VESTANCES PER H.P. OF SIMPLE Low SPEED NONCONDENSING STEAM ENGINES AT FRACTIONAL LOADS DRY SATURATED STEAM AT IOO# RATED H.P. TOTAL VESTANCES AT LOADS OF 125% 100% 75% 50% 25% IOO $442.2 $429.5 $442.6 $496-4 $609.0 125 433-1 416.2 431-9 478.2 590.6 150 424.4 406.4 422.O 466.4 574-o 175 417.1 398.9 412.9 456.8 562.6 2OO 4IO. 2 393-6 405.9 448.4 554-8 250 401.9 386.1 395-8 440.0 543-0 300 396.3 379-4 390.1 434-2 531-6 35 390-7 374-7 384.2 427.8 524.6 400 384.8 369-9 379-6 421 .6 5i7.o 45 380.6 364.6 375-7 416.4 510.4 500 375-6 360.6 370.9 412.0 55-4 TABLE 70 VESTANCES PER H.P. OF SIMPLE HIGH SPEED NONCONDENSING STEAM ENGINES AT FRACTIONAL LOADS DRY SATURATED STEAM AT IOO G.P. RATED VESTANCES AT LOADS OF H.P. 125% 100% 75% 50% 25% 20 $742.3 $701.3 $725.6 $806.6 $1005.2 30 647.0 618.3 638-5 706.6 867.2 40 587.8 561.8 579-8 639-6 780.2 50 555-7 529-9 544-5 602.8 731.6 75 518.5 494-7 506.8 558.0 673.6 IOO 499.1 474-1 485.2 532.4 637-4 150 474-8 449.8 460.2 504.0 602.4 200 457-4 432.8 442.3 487.4 574.8 250 448.4 425-9 434-1 474-4 560.2 144 VESTANCES $1000 FIG. 25. Vestances per h.p. of simple high speed noncondensing engines at fractional loads. 101. It will be noticed that especially in the simple low speed noncondensing engines the total vestance per h.p. at three-quarters load is but little greater than at full load. The total vestance at half and one and one-quarter loads is not very great as compared with that at full load, while that at one-fourth load is comparatively very great. The latter is evidently then a very uneconomical load at which to operate. Since the above tables are based on 3000 hours of opera- tion, we can again get the variation of total vestance with the number of hours of operation per year, and thus determine the change point between two different types of engines at frac- tional loads. Example 29. Determine the change point between a 100 h.p. compound engine running at full load and a 200 h.p. compound engine running at half load. Solution: The depreciation vestance of a 100 h.p. engine at full load is $46.80, and the operating vestance is $292 per year or 292 -s- 3000 = $0.09733 per hr. The total vestance ( V) is then V = 46.80 + 0.09733 N. CHANGE POINTS 145 For the 200 h.p. compound engine operating at half load, the depreciation vestance is 35-5 X 2 = $71.00 per h.p., and the operating vestance is the difference between the total and depreciation vestances or 361 - 71 = $290 per yr., or 290 -T- 3000 = $0.09667 per hr. The total vestance is then F 20 o = 71 + 0.09667 N. The change point is obtained by equating Fioo and F 20 o or 46.80 + 0.09733 A 7 "]. = 71 + 0.09667^1, or Ni = 36,120 hrs. Since there are only 8760 hours in a year, the change point does not come within the year. At all times then the 100 h.p. engine is the most economical to use at 100% load, the decreased operating vestance of the 200 h.p. engine being insufficient to offset the increased depreciation vestance over that of the smaller engine. Example 30. Determine the change point of a 250 h.p. simple high speed engine running at full load and a 200 h.p. engine of the same type running at 25% overload'. Solution: In this case we have for the 250 h.p. engine TV = $13.30, and Va = 412.60 per yr., or 412.6 -r- 3000 = $0.13753 P er hr -> so that VI = 13.30 + 0.13753 N. So also for the 200 h.p. engine at 125% load VD" = 14.2 -f- 1.25 = $11.40, and V a " = 4574 - n-4 = $446 per yr., or 446 -r- 3000 = $0.14867 per hr., so that V" = 11.4 + 0.14867 N. I 4 6 VESTANCES Equating V I and V " gives 11.4 + 0.14867^1 = 13.3 + 0.13753 Ni so that Ni = 170 hrs. This shows that if an overload of 25% is to be carried less than 170 hours per year, it can be done more economically with the use of the 200 h.p. engine at 25% overload than by using a 250 h.p. engine at full load. 102. Oil Engines. We give in Table 71 the vestances per h.p. of oil engines per normal year of 3000 hours, at 5% interest rate. These are also based on a fuel cost of 0.511 cent per pound of oil having a heat value of 18,600 B.t.u. per pound. This is equivalent to a heat value of 7.3 h.p.h. per pound and a cost of heat energy of 0.07 cent per h.p.h. The life of the engines is taken as 15 years, although this is usually exceeded. OIL ENGINES TABLE 71 VESTANCES PER H.p. OF OIL ENGINES C T7 c. LOADS H.P. TYPE 100% 75% 50% 25% 2 i cy horiz VD $174.00 $231 . so $348 . oo $696 oo V a 44O.OC 485.00 607.00 948.OO V t $614 oo $7l6 'CO $0 "? ^ OO $164.4. oo 2 i cy. vert. VD $174.00 V a 44O.OO $231.50 485 . oo $348 . oo 607.00 $696.00 948.00 Vt $614 oo $716 50 $o ^ < . oo $1644. oo 3 i cy. horiz. VD $174.00 V a 394-50 $231.50 437-00 $248.00 540.00 $698.00 850 . oo Vt $568 50 $668 50 $788 oo $1546 oo i cy horiz VD ... $174 oo $231 $84.00 F a 223.50 $361 .00 $112. OO 242.OO $483 . oo $168.00 301.00 $867.00 $336.00 499.00 V t $307 50 $3=14 OO $469 . oo $83< 00 80 2 cv. horiz. VD $93-5 F a 217.00 $134.00 23I-50 $186.50 292.50 $373-00 482.00 80 2 cy. horiz. Ft $310.50 V D $96-50 F a 211 .OO $365-50 $128.50 224.OO $479-oo $193-00 283.50 $855-00 $386.00 470.00 V t . $307 jgf\ VD .. . . . $96 oo $128.00 $192.00 $384 . oo 75 V a 105.00 III. 00 130.00 179.00 v t $2OI OO $239.00 $322.00 $563.00 VD ... . $03.OO $124.00 $186.00 $372.00 V a 103.00 108.00 127.00 175.00 Vt $196.00 $232.00 $313.00 $547.00 152 $900 VESTANCES 80( .700 C-, EC 1 600 500 100 200 300 400 700 800 900 1000 500 600 H. P. FIG. 27. Total vestances per h.p. of Diesel engines at fractional loads. 104. Induction Motors. Since the efficiency of induction motors is independent of the speed, the only difference in the vestance of induction motors of the same size and type will be in the depreciation vestance. Since the depreciation vest- ance is but a small part of the total vestance, there will be but little difference in the cost of service even though the first cost of low speed motors may appear to be very large in comparison with the first cost of high speed motors. We illustrate this below. Example 31. Compare the total vestance of a 10 h.p. 1800 r.p.m. 3 phase 60 cycle 220 volt induction motor with another motor of the same size and type but of only 600 r.p.m. speed. Assume the normal year at 3000 hours, interest 5%, power cost one cent per h.p.h. and neglect other operating costs besides those of power. Solution: The first cost of the 1800 r.p.m. motor is $160 or $16 per h.p. With 25 years of life and interest at 5%, the term factor is 0.70469, so that the depreciation vestance per h.p. is 16 -5- 0.70469 = $22.75. INDUCTION MOTORS 153 The efficiency of this sized motor is 87.9% so that the power used per h.p. year is 3000 -T- 0.879 = 3420 h.p.h., costing 3420 X o.oi = $34.20. The operating vestance is then 34.20 -r- 0.05 = $684, and the total vestance is 22.75 + 684 = $706.75. For the 600 r.p.m. motor the first cost is $300 or $30 per h.p. so that the depreciation vestance per h.p. is 30 -T- .70469 = $42.50, while the operating vestance is the same as for the 1800 r.p.m. motor or $684. The total vestance is then 42.50 + 684 = $726.50. The difference is 42.50 - 22.75 = $19-75 out of a total of $706.65 or 19.75 -* 7 6 -75 = 2.66%. The only object in using low speed motors is for the pur- pose of combining them with other low speed apparatus and where this cost of combination may be reduced by an amount equal to or greater than the additional cost of the lower speed motor. We have chosen above an extreme difference in speed. Where the difference in speed is not so great, the difference in total vestance will be even less than this small amount of 2.66%. 105. We give below the vestances of three phase, 60 cycle, 1800 r.p.m. induction motors based on a normal year of 3000 hours, power cost one cent per h.p.h. and interest at 5 %. VESTANCES TABLE 73 VESTANCES PER H.p. OF INDUCTION MOTORS 3 PHASE, 60 CYCLES, l8oO R.P.M. AT FULL LOAD SIZE H.P. VESTANCES Ratio Va VD Depreciation Operating Total I $56.80 $767.00 $823 . 80 13-5 2 35-55 728.00 763.55 20.5 3 26.95 713.00 739-95 26.5 5 19.80 700.00 719.80 35-4 7-5 25.60 689.00 714.60 26.9 10 22.75 684.00 706.75 30.0 15 18.20 677.00 695 . 20 37-2 20 16. 70 674.00 690 . 70 40-3 25 14.50 671 .OO 685 . 50 46.3 30 14-45 669.00 683.45 46-3 40 12.43 667.00 67Q-43 53-6 SO n-37 665 . oo 676-37 58-5 75 9.76 663 . oo 672.76 68.0 Note the very great ratio of operating to depreciation vestance, and the very great total vestance. This we shall discuss further. Two phase motors usually run one to two per cent lower in efficiency, increasing the total vestance between $7 and $14 per h.p., a very appreciable amount. The operating vestance is very large though the power rate assumed is far lower than that which can ordinarily be obtained. The ten- horse power rate is usually 2 cents per h.p.h. On this basis the operating vestance would be doubled, increasing the ratio to 60. 106. At fractional loads, the splendid full load efficiency of induction motors is well maintained, so that there is no great variation of vestances at fractional loads. This is shown in the table below. INDUCTION MOTORS 155 TABLE 74 VESTANCES PER H.P. OF INDUCTION MOTORS 3 PHASE. 60 CYCLES, l8oO R.P.M. SUE, LOAD I H.P. 125% 100% 75% 50% 25% V D ' . . . $45 .40 $56.80 $75.75 $113.60 $227 20 V a ...- 7QT-00 767.00 767.00 791-00 852.50 V t $836.40 $823 . 80 $842 . 75 $904 . 60 $IO7Q 7O 2 V D $28.40 $35-55 $47 30 $71 .IO $142. 2O V a 750-00 728.00 728.00 750.00 809.00 V t $778.40 $763.55 $775-3 $821.10 $051 2O VD - $25 50 $6 /yu. /^ HUc 18 V a 685.00 665 . oo 665 . oo 685 . oo NP45 4 739-oo V t $694 10 $676 37 $680 ic $7O7 7A $78,, A Q 75 VD $7-80 V a 684 OO $9.76 66^ OO $13-00 66^ oo $I9-S2 684 oo sP/04-4 $39-04 /v5/ o u V t ..- $691.80 $672.76 $676.00 $703.52 $776.54 80 FIG. 28. Total vestances per h.p. of induction motors at fractional loads. There is comparatively little change of total vestance at fractional loads because the efficiency is so well maintained. This is a very valuable feature for maintaining good economy under actual load conditions, such as are invariably met in practice. 107. Generators. The vestances for alternating current generators of the direct connected type are given below based GENERATORS 157 on a normal year of 3000 hours, cost of energy cent per kv.a., interest 5% and a life of 25 years. TABLE 75 VESTANCES PER KV.A. OF D-c. GENERATORS 3 PHASE, 60 CYCLES, 220-550 VOLTS SIZE, LOADS KV.A. 125% 100% 75% 50% 25% IO VD $28.40 V a 208.10 $35-50 205.00 $47-30 209 . 80 $71.00 222.80 $142.00 257.00 V t . . $236.50 $240. so $257. 10 $2(H 80 $300 OO 2< V D $22.70 $28.40 $37 - 80 $56 . 80 $113.60 V a 182.00 178.00 182.50 193.00 217.00 Vt $204 70 $206 40 $220 30 $249 80 $230 60 V D $18.80 $23 . i V t $171 17 $171 3S Sl7O S S $198 70 $24. e, I 1 ? 158 VESTANCES $1200 1100 1000 900 700 600 700 800 900 1000 FIG. 29. Total vestances per kv.a. of electric generators at fractional loads. 100 200 300 400 500 600 "K. V. A. Like motors, there is comparatively little variation in total vestance with the size of the unit or the load. This is one of the immensely valuable characteristics of electrical machinery that it is so well adapted to meet the load variations of prac- tice with remarkably good economy. 108. Comparison of Power Units. By power units we mean, primarily, such units as steam engines, turbines, inter- nal combustion motors, and electric motors. In the fore- going, we have determined the vestances of these units based on a normal year of 3000 hours, constant load, and 5 % interest. The fuel costs assumed were coal at $2 per ton with a fuel value of 13,743 B.t.u. or 5.4 h.p.h. per Ib. Allowing for boiler service, we assumed a cost of $0.005 P er h.p.h. of energy de- livered to the engine. In comparison with this we assumed for oil engines, oil at 0.511 cent per pound, and allowing 320 Ibs. per barrel, this would amount to $1.635 P er barrel, or 0.0^7 cent per h.p.h. For Diesel engines we assumed the cost of the heat energy at 0.05 cent per h.p.h. corresponding to a cost of $1.168 per barrel. And finally for motors we assumed a cost of one cent per h.p.h. At 15 cents per gallon COMPARISON OF POWER UNITS 159 for gasoline, energy would cost 0.2675 cent per h.p.h. For clearness we tabulate these as follows: TABLE 76 COST OF ENERGY PER H.P.H. ASSUMED COST OF ENERGY PER UNIT H.P.H. IN CENTS Induction motor i . oooo Gasoline engine o . 2675 Oil engine 0.0700 Diesel engine 0.05 Steam engine 0.05 With these assumed costs clearly in view, for they vary greatly with different localities, we found comparisons as follows: At full load, the Diesel engine has the best economy, with the oil engine second, the compound condensing steam engine third, the simple low speed noncondensing steam engine fourth, the simple high speed noncondensing steam engine fifth, the electric motor sixth and the gasolene engine seventh. Below 30 h.p. the electric motor is far superior in economy to the simple high speed steam engine, and no doubt the other types of steam engines as well, but it is itself far inferior to the oil engine, somewhat inferior to the distillate engine at 7^ cents per gallon for the distillate, and far superior to the gaso- line engine. The sharp bending of the curve (Figure 30) of the Diesel to the horizontal for small sizes would lead one to expect the oil engine to be superior to the Diesel below 40 h.p. 109. We give below in tabular form the total vestances of these units for both full and fractional loads, together with the relative valuance based on unity for the machine which has the best economy, so as to show clearly their compara- tive values. Where the values are not given in the tables they have been interpolated from the curves. In the table, V t is the total vestance and U, the comparative value. 160 VESTANCES TABLE 77 TOTAL VESTANCES AND COMPARATIVE VALUE OF POWER UNITS SIZE, H.P. UNIT LOADS PEK CENT OF FULL LOADS 125 IOO 75 50 25 Vt U Vt V Vt U V U 0.98 I .00 0.80 0-75 0.99 0.90 i .00 0.83 0.77 0.98 I .00 0.82 0.70 1 .00 0.42 0.66 o. 70 Vt U 0.7 I .OC 0.8. 0.8 o.7< o.6( I .OC 0.9 0.8 o.8c o.8 v I.OC 0-9, 0.91 0.3^ o.8( I .OC 250 IOO 50 25 Diesel e Comp. < S. L. S. S. H.S. Diesel e Oil engi Comp. s S. L. S. S.H. S. Diesel t Oil engi S. H. S. Electric Oil engi Gasoline S. H. S. Electric ngii ;tea stes stej ngii ne. ie 223 300 386 426 254 300 339 429 474 300 321 530 676 33 2 864 660 686 i .00 0.74 0.58 0.52 I.OO 0.85 o-75 o-59 0-54 i .00 0.94 0-5.7 0.44 I.OO 0.38 0.50 0.48 267 315 396 434 306 345 < 361 < 443 < 485 < 365 373 < 545 < 680 c 383 960 c 682 c 690 c i .00 3.85 3.67 D.62 [ .00 3.89 3.85 3.69 3.63 [.00 3.98 1.67 3-54 c .00 5.40 3.56 5.55 36c 44c 474 4M 41: 49^ 532 497 603 503 1212 757 721 631 457 543 560 728 839 544 609 637 910 885 732 785 883 2089 936 804 m engine. . . . im engine . . . im engine. . . ie. . . 307 4.02 448 I.OO 0.76 0.69 teai StCc ste -ngi ne. "n engine. . . . im engine. . . im engine. . . ne 345 14 2 199 I.OO D. 7 8 3.69 ste mo ie. am engine . . . tor 556 594 [.OO 3.80 ! enj ste mo T ine un engine. . tor >95 ro4 [ .00 3.99 G - solin - . (Em fiflO 700 V W X _\ Klec trie Motoq a 600U \ I S \. 1 J\ ^^ V 400h- \ 1 - . ~ i - . -a^*i i ^^ S 1 r* MS. - 5.L.S. i ^on t - ion < mmmtfst mde nde jiin _sinj ' __ 300H ^. Ov Oil F npin - = C )iese trip _ET 5 nd ] S. S 1 iam r--i ^ *^ 1 - -t- ii *0 25 50 75 100 125 JL50 1.75 200 225 25C H. P. FIG. 30. Total vestances of power units at full load. COMPARISON OF POWER UNITS 161 It is evident from the above table that at full load the Diesel engine leads in economy by a wide margin, with the oil engine second and the compound steam engine third. For sizes over 50 h.p., the oil engine is worth only 85% as much as a Diesel and the compound steam engine about 75%. In other words, taking the cost of service of the Diesel as unity, then that of the oil engine is 1.15, i.e., 15%, greater and the cost of service with compound condensing engines is 1.33 or 33% greater. At 75% load the Diesel still leads but at 50 h.p. the oil engine has practically become equal to it. Diesel engines are seldom built below this size, though a few have been made as small as 8 h.p. It is certainly remarkable how well the designers of engines have chosen the inferior limit of the sizes of their various types. At 50% load, the Diesel and compound engines have reached equality in sizes of 100 h.p. and over, with the oil engine third. But below 50 h.p. the oil engine is best. At 25% load, the compound steam engine leads by a wide margin above 100 h.p. At 50 h.p. the simple high speed steam engine is best, with the electric motor second, while below 25 h.p. the electric motor leads with the oil engine second, based, however, on costs of energy as given. In all this the steam engines have the peculiar advantage that they can carry with comparative good economy 25% overload. The range of loads of these engines is therefore 25% greater than internal combustion motors. The advan- tages of this greater range of load will be shown later. The gasoline engine is inferior to the electric motor at all loads on the basis chosen. But if, for example, 2 cents per h.p. is paid for electric power as compared to 15 cents per gallon for gasoline, then conditions would be reversed. In the face of the great superiority of the Diesel engines, it seems remarkable that they are not more generally used. There are two reasons for this. First the general resistance of engineers to anything new to them, and second the fact that there are many cases where the undertakings are under- l62 VESTANCES financed, so that they are unable to put in the most economical installations. Above all this looms the element of uncer- tainty, so that there is a lack of incentive to build for permanency. 110. Centrifugal Pumps. In the case of centrifugal pumps, the costs of attendance, oil, and the like, are very small, as compared with the cost of power. In the following tables of vestances of centrifugal pumps, we have, therefore, left these items out of consideration. The vestances are given not only at full and fractional loads, but at variable heads as well. We have assumed a cost of power at one cent delivered at the pump shaft, interest rate 5% as before, and a normal year of operation at 1500 hours, with 25 years as the life of the pumps. TABLE 78 VESTANCES PER 1000 G.P.M. OF CAPACITY OF Low HEAD BELTED CENTRIFUGAL PUMPS RATED DEPRECIA- TION VES- OPERATING VE STANCES AT HE, \DS OF CAPACITY , G.P.M. TANCE PER IOOO G.P.M. 10' 20' 30' 50' 50 $936.60 $3400 . oo $6000 . oo $7500.00 $11,360.00 100 596.10 268O.OO 4410.00 592O.OO 9,360.00 150 482.50 2350.00 4050 . oo 5370.00 8,320.00 225 377-50 2O25 .OO 3625.00 4670.00 7,430.00 300 326.40 1875.00 3330.00 4420 . oo 7,070.00 400 265.40 i 790 . oo 3120.00 4170.00 6,690.00 700 2O2 . 80 1565.00 2770.00 3750.00 6,040.00 900 189. 2O 1445 . oo 2540.00 3570.00 5,850.00 I,2OO 177.20 1390.00 2500.00 3460 . oo 5,680.00 1, 600 160.00 1320.00 2380 . oo 3360 . oo 5,500.00 3,000 119.50 1295.00 2320.00 3310.00 5,460.00 4,5o 106.40 1270.00 2270.00 3260.00 5,400.00 6,000 96.70 1250.00 2230.00 3220.00 5,340.00 7,000 93-40 1230.00 2220.00 3170.00 5,310.00 8,000 91 . 2O I2IO.OO 22OO.OO 3120.00 5,280.00 10,000 87.30 i i 90 . oo 2l8o.OO 3080 . oo 14 ooo 8? AQ i 185 . oo 2 170 . OO 304.0 oo 20,000 O * T" So.OO i i 70 . oo 2160.00 O ^ ' 3000 . oo 7Q OOO 78.00 I j 5 tr OO 2 I 40 . OO O'-'j 1 -"-"-' 40,000 76.80 I l6o.OO 2135.00 50 ooo 76 . 2O 1 1 ee OO 2130 . oo 60 ooo 7^ 7O * L o o ^^ i 150 . oo 2125 .00 / o p* CENTRIFUGAL PUMPS 163 Total Vestance ^ 11 \ "v __ I \ s^ 9,0-ft -Tfri d V 10-ft H d 2.000 400.0 6000 8000 10000 12000 14000 16000 1SOOO 20000 Capacity in G.P..M. FIG. 31. Total vestances of low head belted horizontal centrifugal pumps at full load. $16000 200 400 600 800 1000 1200 1400 1600 1800 Capacity in G.P.M. FIG. 32. Total vestances of vertical centrifugal pumps. i6 4 VESTANCES TABLE 79 VESTANCES PER 1000 G.P.M. OF CAPACITY OF VERTICAL CENTRIFUGAL PUMPS CAPACITY, G.P.M. VESTANCES AT HEADS OF ,5' 50' 75' IOO ISO 225 300 400 700 900 1 200 1600 Va $5210 oo $9,370.00 3,760.00 $13,700.00 4,740.00 VD 2780.00 Vt $7QOO OO $13,130.00 $8,330 . oo 2,770.00 $18,440.00 $12,300.00 3,440.00 Va $4690 oo VD v t 2090 . oo $6780.00 $11,100.00 $7,500.00 2,100.00 $15,740.00 $II,O5O.OO 2,670.00 Va $4150.00 1570.00 VD v t $s72O OO $9,600.00 $7,070.00 1,740.00 $13,720.00 $IO,42O.OO 2,I4O.OO V a $3980 . oo VD v t 1330.00 $5310 oo $8,810.00 $6,690.00 i ,405 . oo $I2,56o.OO $9,890.00 I,7IO.OO Va $3750.00 1063 oo VD Vt $4813 oo $8,095 . oo $6,040.00 938 . oo $II,60O.OO $8,930.00 I,I4O.OO V a $3290 oo VD 717.00 v t $4007 oo $6,978.00 $5,860.00 866.00 $IO,O7O.OO $8,650.00 i ,040 . oo V a $3020.00 670 oo VD v t $3690 oo $6,726.00 $5,680.00 727.00 $9,690.00 $8,410.00 866.00 V a .. . . $202^ OO VD S87-00 V t $3512 oo $6,407.00 $5,510.00 657.00 $9,276.00 $8,160.00 762.00 $8,922.00 V a $2880 oo VD . . 378 oo Vt $ 3 2 tj 8 oo $6,167.00 CENTRIFUGAL PUMPS 165 1 8OO OOO 000 OOO OOO 00 OOO OOO OOO OOO OOO OOO OOO OOO OOO OOO O ^o to O ^O ^O O *o l o O CN (N O c^ & : ; Ufc^tT ^t-u t^^tT ^^tT fc. u- fcT ^ ^ fck i t^- O O O to I t CO to ON M CO to H|N O to ! , 888 888 888 888 888 OOO OOO O ^^ l/ ^ Q O O O "^t" ^" O co co Oc^ <& J5 eg . * > : : : : : : : : : : : : : ^^i^r ^^u u^^r t^^^r u^^r Capacity h.p. g.p.m. to to o to ^O Ococo Oto^o OOOOO O^OO toO^"^ toQto O 00 00 ' . ' . t?^u ^^ ;^^ ^sf tT t^u ^T fc^fcS fcT Capacity h.p. g.p.m. to O M to M * CO 10 10 O w 8. 8 8 O^ CO * 10 o C^\ co O4 t^* O O 10 00 t/1 JJ O M CS 0 i 888 888 888 888 84^ !> O CO OO O *O OO O *^ Tf" O\ O^> *o ex *>* O oo ^O *o ^" o\ rO co c^T co *O *O HH **" 888 88 odd do 10 O l o O co r^ co O 10 co 8 00 i| m & & & ^ . ^ ^ : ^ ^ & ^ 1 > : Q : Q : ' ^ : : Q : ; t^^^ ^^^ t.^^ ^^^" ^. U U ^ fc U Capacity h.p. g.p.m. o O O O Tt- NO O O O IO t^ f*> H 10 O 10 O CS CO CO "^ 10 o t^- o N 00 O 10 IO t^* T) i 888 888 888 888 O r^ O t > * i> O i>* *> 888 8 8 d oc 06 d - O t"* -rj- CO J^ IO CS 8 ft 33 3 S S 1 :* * : 4 i ; i ; ^ ^ ^T ^ ^ ^ Capacity h.p. g.p.m. o o CO 1 r r H* O JJ o o 10 o co O M 0 r I"- O O M M 1 8888888888888888888 o v o vo o OG oo o 'O ^o o ^* ^ o *-* ^ o ^ ^ c to r^* 01 to O to O co co to J^* t^^ C s. M OO t* \ 8 8 i U-) to ) M CN . CM Qv ) O To %Z 2 8 '; : . ^ t^ 'u d O O Q O O J O 10 O O O O C 5 ^- t- oo o o M OIO IOIO 1O1- 10 10 10 c^ '. ' '. ' ' . . . ^^^^^tT^^^^^^^^H^^^^t Capacity h.p. g.p.m. o o o o o o c o o o o o o c O N IO vO O\ !O C t> > t^. i 1 10 H M 888888888888888888 1 O<^O^O i68 VESTANCES 25000 30000 15000 0200001 10000 5000 1 \ l\ f L A \ \ \ \ 150 f Che .Res P Pu d raps indl add isign 11 V \ V \ X ft.H dde ad sign \ \ 100- trH ad K Dhea Pui tips s . 25 ft. Head 800 ;L0Q0 240.0 3200 40.00 4800 5600 6400 7200 8000 Capacity in G.P.M. FIG. 33. Total vestances of horizontal direct connected centrifugal pumps. 111. Nothing could show better how vestance indicates di- rectly the relative values of various equipment Than table 80. Where there are so many different sizes, speeds, capacities, etc., each with a different efficiency and price, it would seem at first that there are so many variables that a basis of compari- son could not be found. When, however, all these variables are reduced to the basis of a permanent rendition of the service, the problem simplifies at once and we have the total vestance. That unit which, under given conditions, has the lowest total vestance is the most economical, i.e., has the highest financial efficiency. Thus at 25 ft. head, the 20 h.p. 2300 g.p.m. pump has the lowest total vestance of $2919 per 1000 g.p.m. of capacity, the next lowest being the 30 h.p. 3200 g.p.m. pump, having a total vestance of $3042. Each of these pumps has a first cost of $500, apparently, a rather high price. Yet the first cost represents in this case only about one-sixth of the total cost of rendering the service. In the case of the 20 h.p. 2300 g.p.m. pump, the first cost per 1000 g.p.m. of capacity is $218 and the total depreciation CENTRIFUGAL PUMPS 169 vestance is $309. The difference of $91 is due to limited life. If then this pump could be so constructed that its life would be permanent, our total vestance would be reduced only by this amount of $91, not a very significant per cent of the total vestance. It is evident then that increasing the life of pumps is not nearly as important as increasing the pump efficiency. On the other hand, since the operating vestance is $2610, a saving of 10 per cent of the power would reduce our total vestance $261, a very significant amount. The question naturally arises as to how much the first cost could be in- creased under the above conditions for a saving of ten per cent of the power used. This is obtained by multiplying the saving in total vestance by the term factor which is 0.70469 in this case. This gives 261 X 0.70469 = $183.92. If, then, a pump of 2300 g.p.m. requiring 20 h.p. is worth $500, then a pump of the same capacity requiring 10 per cent less power or 18.0 h.p. will be worth 500 -f 183.92 = $683.92. If, for example, the cost of the latter were $600, it would still save the customer $83.92 and this is for each 10 per cent of the power saved. Such improvement by refinement in design and workmanship and by the use of better materials is attainable to the extent of some 5 % on the average pump built. It should be noted that according to the above list at 25 ft. head, it would be more economical to use five 20 h.p. 2300 g.p.m. pumps than two 50 h.p. 4800 g.p.m. pumps on a 100 h.p. installation. In the former case the pump cost will amount to 5 X 500 = $2500, and in the latter case to 2 X 675 = $1350. The saving in use of the 5-20 h.p. pumps in total vestance over that of two 50 h.p. pumps as above, will be $3324 - 2919 = $405 170 / VESTANCES for each 1000 g.p.m. of capacity or a total of $4657.50, cer- tainly a significant sum. At 50 ft. head the 40 h.p. 2600 g.p.m. pump is best, the next best being the 35 h.p. 200 g.p.m. pump, the difference in total vestance being $5558 - 5i35 = $423- At 100 ft. head the 250 h.p. 7000 g.p.m. pump is best with the 200 h.p. 5600 g.p.m. pump a close second. Finally at 150 ft. head, the 75 h.p. 1400 g.p.m. pump is best with the 100 h.p. 1800 g.p.m. pump as a close second. 112. It will be noted that the largest sized pump does not always represent the best investment considering the way pumps are now being manufactured. However, if it is possible at 50 ft. head to make a 40 h.p. pump with a total vestance of only $5135 it certainly is possible for the manufacturer to make a 150 h.p. pump with an even lower total vestance instead of $830 higher per 1000 g.p.m. of capacity. It is apparent that the design of the latter needs revision. In the future when a new design has been made and tested, it will have to be valuated, the results of which will determine whether it will be adopted or junked. In the list of pumps at 25 foot head are two 10 h.p. pumps, one of which costs $240 and delivers 820 g.p.m., while the other costs $385 and delivers 1000 g.p.m. The difference in costs is $145 and the difference in capacity 180 g.p.m. Usually the latter pump is considered to be so expensive that its use is prohibitive except in " fancy" installations. Yet the differ- ence in vestance of $520 or nearly 15% shows that in reality the use of the cheap pump is prohibitive. At high heads, good efficiency is still more important than at low heads. At 150 ft. head there are listed two 100 h.p. pumps, one having a capacity of 1500 g.p.m. and the other a capacity of 1800 g.p.m. The latter has a first cost of $925, and the former a cost of $550, a difference of $375. That is the latter costs 68% more than the former. It is often said that such nne pumps as the latter are installed merely for CENTRIFUGAL PUMPS 171 show. Yet the difference in total vestance in favor of the latter is 20,520 - 17,380 = $3140 or 3140 + 20,520 = 15.3%. This means that in the latter case the service can be rendered 15-3% cheaper. It should be borne in mind that the cost of power assumed is very low, much lower than can usually be obtained. The effect of increased power cost is to make an increase of effi- ciency more valuable, and to increase the difference in vestances of units whose efficiencies differ. 113. It is ordinarily assumed that pumps are either used at full capacity, or not at all. This is true only to the extent that pumps are not throttled. But almost in- variably the total head pumped against varies with the season, i.e., with the change in level between the supply water and the point of discharge. The point of discharge is usually constant, but the level of the supply water almost invariably changes. In the case where the water is pumped from a river, the water level may change as much as 15 or 20 ft. The effect of this is that if the head is decreased below normal, the capacity of the pump is increased above normal, i.e., the pump is over- loaded, while if the head is increased above normal, the pump will be underloaded. The amount of load variation is very small in high head pumping plants and usually large in very low head plants. It is evident, then, that the effect of fractional loading can- not be neglected since the efficiency of pumps falls off quite rapidly with either increase or decrease of capacity, especially the former. Unless we take into consideration the amount of water to be pumped at each stage of water level, the cost of service will be inexplicably increased. And it must be borne in mind that a very few feet variation in head will change the pump capacity very greatly, this being especially true of increased heads above normal. An idea of the effect of all this may be obtained from the following example. 172 VESTANCES Example 32. A centrifugal pump delivers 1000 g.p.m. against 66 ft. head at 60% efficiency, this being its point of best efficiency. Assuming a normal irrigation year of 1500 hours, cost of power at i cent per h.p.h. and interest at 5%, and neglecting other operating costs, determine the operating vestance at full, three-quarters, half, and one-fourth loads. Solution: The power used at full load is 66 X 1000 _ , 4000 X 0.60 ~~ costing 27^ cents per hour or 0.275 X 1500 = $412.5 per year. The operating vestance is then 412.5 -r- 0.05 = $8250. At three-fourths, or one and one-fourth load, this would be 8250 -T- 0.937 = $8804.70, since at that load the pump only gives 93.7% of full load efficiency. So at half or one and one-half loads, the operating vestance is 8250 -h 0.75 = $11,000, and at one-fourth load it is 8250 -T- 0.436 = $18,920. 114. It is evident that at less than three-fourths or over one and one-fourth load, the operating vestance becomes prohibi- tive, unless the pump is used but a few hours per year. The effect of the number of hours of use per year and the location of the change point is similar to that of steam engines, except that in this case we have assumed the normal year at 1 500, instead of 3000, hours. Thus if Fo is the operating vestance of pumps per normal year as listed in the preceding tables, then the operating vest- ance per hour is and the total vestance per year of ( N) 1500 hours is V=V D+ . 1500 CENTRIFUGAL PUMPS 173 Equating the total vestances of any two pumps will deter- mine the change point. Example 33. At 150 ft. head, two 75 h.p. direct connected pumps are listed, one having a capacity of 1200 g.p.m. and costing $500 and the other having a capacity of 1400 g.p.m. and costing $900. Determine the change point on the basis previously assumed. Solution: The operating vestance of the 1200 g.p.m. pump per hour is 18,750 -;- 1500 = $12.50. The total veatance per year of (N) hours is Fi = 592 + 12.5 Af. For the 1400 g.p.m. pump, the operating vestance per hour is 16,100 -4- 1500 = $10.733, so the total vestance per year of (N) hours is F 2 = 915 + 10.733 N. The change point is determined by equating the total vest- ances, i.e. 592 + 12.5 #1 = 915 + 10.733 #1, or Ni = 182.8 hrs. If the pumps are to be used less than 182.8 hours per year the cheaper pump is best; for more than this number of hours of use the more efficient pump is best. The difference in total vestances of two units operating at different efficiencies increases with the number of hours of use. Thus when N = 5000 hours, Vi = 592 + 12.5 X 5000 = $63,092, and F 2 = 915 + 10.733 X 5000 = $54,580, giving a difference of 63,092 - 54,580 = $8512. The difference at 1500 hours was only 19,342 - 17,015 = $2327. It is evident then that for very few hours of use per year a cheap pump should be used. But as the number of hours of use per year increases not only should we use the more 174 VESTANCES expensive pump, but the comparative value of the more effi- cient unit continually increases. What we pay for first cost is usually only a very insignificant part of the cost of service. The user, under normal conditions, can afford to pay much more in first cost to the manufacturer than is now commonly done for more efficient units. 115. A considerable decrease in the life of units does not greatly increase the cost of service. Consequently increased efficiency should be obtained even if it is necessary to greatly decrease the life of the unit. Yet too often the reputation of a certain make of apparatus is based on the length of service, rather than on the quality and cost. On the other hand, there is a tendency to continue to use units when they have become inefficient due to wear or other causes, so that the service is no longer rendered economically. Under such conditions, the pumps should either be brought back to full efficiency, or else discarded for new apparatus. Because under such conditions, the excess operating vestance has become so great that the continued use of the worn equipment would be far more expensive than the cost of replacement. 116. The vestances of such items of a system as belts, pipe, and the like may easily be determined, as illustrated below. Example 34. A belt for a 10 h.p. unit costs $20 or $2 per h.p. The belt efficiency is 85%. Assuming power at one cent per h.p.h., interest at 5%, and the life of the belt at 5 years, and neglecting other operating costs, determine the vestance per h.p. of the belt per normal year of 3000 hours. How does it compare with the vestance of a flexible, steel link coupling costing $45? Solution: With interest at 5% and the life at 5 years, the term factor is 0.21649. The depreciation vestance per h.p. is then 2 -T- 0.21649 = $9.25. For each h.p.h. transmitted 0.15 h.p.h. is lost. The loss per h.p. year is then PIPE 175 3000 X 0.15 = 450 h.p.h., costing 450 X o.oi = $4.50. The operating vestance is then 4.50 -T- 0.05 = $90, and the total vestance is 9.25 + 90 = $99.25. If a 10 h.p. flexible coupling costs $45, then the cost per h.p. is $4.5- For all practical purposes, the life of the coupling is per- manent. The depreciation vestance per h.p. is therefore 4.50 -r- i = $4.50. The efficiency is 100%. The operating vestance, neglecting attendance and repair as we did for the belt, is then zero and the total vestance is 4.50 + o = $4.50 as compared with $99.25 for the belt. On this basis, it ap- pears therefore that we could spend as much as $94.75 per h.p. for direct connection and still render the service as cheap as the belt-driven unit in cost. If the power costs 2 cents per h.p.h., the operating vestance of the belt would be doubled ($180) and its total vestance per h.p. would increase to $189.25. We could then afford to spend this greater amount per h.p. for direct connection. As a matter of fact the difference between the belt and coupling is much greater than is indicated above. For, while the cost of attendance and repair, that we neglected to take into consideration above, is very small for the coupling, it is very large for the belt. Furthermore the belt places heavy lateral strains on the bearings of both the driver and driven unit. The loss of power and increased repair on the units that this occasions must be charged against the belt. In a well-made coupling, these losses are eliminated, the power being transmitted without such strains. 117. Pipe. Neglecting attendance and repair which is inappreciable in a well-designed installation, the cost of opera- i 7 6 VESTANCES tion of pipe is the cost of the power lost in transmitting the fluid, usually water, in friction. This loss varies with the velocity at which the fluid is transmitted, increasing rapidly with increased velocity. The life of pipe has been the subject of much argument. If improperly used, the life may be very short, due to corro- sion. If, however, the pipe is covered with some noncon- ductor, such as asphaltum, or if it is taped and asphalted, this corrosion is eliminated. On the whole, it seems safe to assume a life of 50 years. We give below the vestances of pipe based on 5% interest, cost of power one cent per h.p.h. and also based on a normal year of 3000 hours. It must be borne in mind, however, that where the pipe is used for irrigation, the normal year should TABLE 81 VESTANCES OF STANDARD IRON PIPE PER 100 FT. SIZE, INCHES DIA. VELOCITY, FT. PER SECOND 0-5 I.O 2.0 5.0 10 20. o l 4 1 1 I I I* VD . . . $2 14 0-54 $2.14 ii .46 $2.14 93.60 $2.14 660 . oo $662.14 $2.14 810.00 V a Vt VD $2.68 $2.14 I. 21 $13.60 $2.14 15.40 $95 74 $2. I^ 106.00 $108.14 $3-01 144-00 $147-01 $3-51 202.50 V a v t 13-35 $3-oi i-5i $17-54 $3-oi 19.30 $812.14 $3-01 965.00 V D V a $3-oi 0.19 V t $3-20 $3-51 0.26 $4-52 $3-51 1.87 $22.31 $3-51 27.30 $968.01 $3-5i 1,340.00 VD . $3 Si V a 0.04 V t . $1 w $3-77 $4.99 0-39 $5.38 $4.99 2.30 $30.81 $4.99 31.00 $2O6.OI $4.99 232.54 $i,343-5i $4-99 i,707.75 VD $4. oo V a 0.05 V t $<; 04 $5-38 $8.23 0.40 $7-29 $8.23 3.00 $35-99 $8.23 44.60 $237-53 $8.23 351-00 $1,712.74 $8.23 2,300.40 VD . . . . $8 23 V a 0.06 Vt . $8 29 $8.63 $11.23 $52-83 $359-23 $2,308.63 PIPE TABLE 81 Continued 177 SIZE, INCHES DIA. VELOCITY, FT. PER SECOND o.S I.O 2.O 5-0 IO.O 20. o 2 2j 3 4 5 6 8 10 12 V D $10 45 $10.45 0-51 $10.45 3.83 $10.45 56.60 $10.45 404.00 $10.45 2,767.50 V a 0.07 V t $10.52 V D $17 25 $10.96 $l7-25 0.64 $14.28 $17.25 4-73 $67.05 $l7-25 65.00 $414.45 $17.25 496.12 $513-37 $22.70 573-75 $2,777-95 $17-25 3,213.67 Va 0.0 9 Vt $17-34 VD $22.70 V a O.II $17-89 $22.70 o-73 $21.98 $22.70 5-40 $82.25 $22.70 77.20 $3,230.92 $22.70 3,712.50 Vt $22 8l $23-43 $35-15 0.92 $28.10 $35-15 6.71 $99 . 90 $35-15 92.30 $596.45 $35-15 714-15 $3,735-20 $35-iS 4,606 . 87 VD . $35 15 V a O.I 3 V t . . . $-?<; 28 $36.07 $48.95 1.09 $41.86 $48.95 7.80 $127-45 $48.95 i i i . 03 $749-30 $48.95 810.00 $4,642.02 $48.95 5,675-o6 VD $48 95 V a 0.16 Vt" . $4-0 II $50.04 $63-25 1.22 $56.75 $63 25 8.91 $159.98 $63.25 121 .50 $858.95 $63.25 918.00 $5,724-01 $63-25 6,345-QQ $6,408.25 $93-70 7,830.00 Vn $6* 2$ V a 0.18 V t . . . $63.43 $64.47 $93 70 1.56 $72.16 $93 70 ii .50 $184.75 $93 - 70 153-56 $981 . 25 $93 70 1181.25 Yn . &CH 7o V a O. 2O Vt $93-90 '/ ) $132 OO $95 26 $132.00 1.86 I5l05 . 20 |l32.00 13-97 $247.26 $132.00 i 89 . oo $1274.95 $132.00 1398.60 $7,923.70 $132.00 10,266.75 V a 0.24 ft $1^2 24 $133-86 168.00 2.17 $145-97 $168.00 I4.70 $321.00 $168.00 202.00 $1530.60 $168.00 1620.00 $10,398.75 $168.00 10,800.00 7 D $168.00 7 'a 0.30 7 t $168.30 $170.17 $182.70 $370.00 $1788.00 $10,968.00 be assumed at only half of this amount or 1500 hours. Cost of delivery, laying, etc., is not included. 118. Galvanized pipe, running much higher in first cost than the black pipe, can be justified only by a proportionate increase in the life of the pipe, unless other advantages are forthcoming. This can hardly be done. Conducting cover- 178 VESTANCES $1400 1200 1000 Z 800 600 400 200 6 8 10 Diameter in Inches FIG. 34. Showing the variation of total vestance with the velocity, and size of pipe. ings have generally been displaced with nonconducting coverings. The above table shows that at very low velocity the operat- ing vestance is practically negligible in comparison with the depreciation vestance. For a 12 in. pipe, at 0.5 ft. per second of velocity, for example, the depreciation vestance is $168 while the operating vestance is only $0.30, or about 0.18% of the total vestance. But at high velocities condi- tions are reversed. For example, at 20 ft. per second ve- locity, in the case of the 12 in. pipe, the operating vestance is $10,800 while the depreciation vestance is still $168. The latter, under these conditions, is only 1.53% of the total vestance and is thus almost negligible. 119. Somewhere between extremely IOW T and extremely high velocity is the point of best economy. The above table does not show this. But this can be shown^if the table is reduced to vestances per cubic feet per second (cu. sec. ft.) of capacity. This we have done in the table below. PIPE 179 TABLE 82 VESTANCES PER 100 FT. OF STANDARD PIPE PER Cu. SEC. FT. OF CAPACITY SIZE, INCHES DIA. VELOCITY, FEET PER SECON 0.5 1.0 2.O 5-o 10. 20. o i I i f i i| 2 *i 3 4 5 V D 53150.00 795 . oo [!i 260.00 6750.00 $630 . oo 27,550.00 $315-00 97,000.00 Vn v t $3945.00 $1430.00 717.00 $8010.00 $570.00 4110.00 $28,180.00 $285.00 14,140.00 $97,315.00 $143.00 54,000.00 VD V a v t $2147.00 1132.00 566.00 $4680.00 $453-00 2890.00 $14,425.00 $227.00 IO,8OO.OO $54,143.00 $113.00 36,200.00 VD V a $2265.00 143.00 V t $2408.00 $II7O.OO 87.00 $1698.. oo $585.00 311.00 $3343.00 $234.00 1820.00 $11,027.00 $117.00 7,500.00 $36,313.00 $58.00 22,350.00 VD . . .$2340.00 V a 28.00 V t $2368.00 VD . . .$1870.00 V a IQ.OO $1257.00 $935 .00 73.00 $896.00 $468.00 159.00 $2054.00 $187.00 i 160.00 $7,617.00 $94 oo 4,360.00 $22,408.00 $47-00 16,000.00 V t $1889.00 VD . . .$1320.00 V a 10.00 $1008.00 $660.00 32.00 $627.00 $330 . oo I 20.00 $1347-00 $132.00 714.00 $4,454.00 $66.00 2,810.00 $16,047.00 $33-00 9,200.00 Vt $1330.00 VD . . $1045 -o V a 7-00 $692.00 $522.00 26.00 $450.00 $261 .00 96.00 $846.00 $104.00 566.00 $2,876.00 $52.00 2,020.00 $9,233.00 $26.00 6,919.00 Vt . $105^ oo $548 . oo $518.00 19.00 $357-00 $259.00 71.00 $670.00 $103.00 391.00 $2,072.00 $51.00 1,490.00 $6,945.00 $25.00 4,670.00 VD . . .$1035.00 V a 5-00 V t $1040.00 VD . . . $908.00 V a 4-00 $537-oo $454.00 16.00 $330.00 $227.00 54.00 $494 . oo $91.00 309.00 $1,541.00 $45-oo 1,150.00 $4,695.00 $23.00 3,712.00 V t $912.00 V D . . . $803.00 V a 3-00 $470 . oo $402.00 15.00 $281.00 $201 .00 38.00 $400.00 $80.00 211.00 $1,195-00 $40.00 816.00 $3,735-00 $20.00 2,635.00 Vt $806.00 VD . . . $734-oo Va 2.00 $417-00 $367-00 8.00 $239.00 $184.00 29.00 $291 .OO $73-00 167.00 $856.00 $37-oo 607.00 $2,655.00 $18.00 2,120.00 V t $736.00 $375-oo $213.00 $240.00 $644.00 $2,138.00 i8o VESTANCES TABLE 82 Continued SIZE, INCHES DIA. VELOCITY, FEET PER SECOND 0.5 I.O 2.0 S-o 10. 20. O 6 8 10 12 VD . . . $632.00 V a 2.00 $318.00 6.00 $159.00 22.OO $63.00 122. OO $32.00 459.00 $l6.00 I,586.OO $1,602.00 $I4.OO 1,175.00 Vt $634.00 VD . . . $562.00 V a 1. 00 $324.00 $281 .00 5.00 $181.00 $140.00 17.00 $185.00 $56.00 92.OO $491 .00 $28.00 355-oo V t $563.00 VD . . . $494.00 V a I. 00 $286.00 $247.00 4.00 $157-00 $148.00 13.00 $148.00 $49 . oo 71.00 $120.00 $45-oo 54.00 $383.00 $25.00 262.00 $1,189.00 $I2.OO 963 . oo Vt $495 -oo VD . . . $448.00 V a 1. 00 $251.00 $224.00 3.00 $161 .00 $112.00 10.00 $287.00 $22.00 216.00 $975.00 $11 .00 721 .00 V t $449.00 $227.00 $122.00 $99 . oo $238.00 $732.00 $2800 2400 7 2 Pipe 2000 I u &1600 d I I 800 g 400 Z 1 Ppe Pipe 12 Pipe 8 10 12 Velocity in f.p.s, 16 18 20 FIG. 35. Total vestances of standard iron pipe per 100 ft. of length and per cu. sec. ft. of capacity. We thus see at a glance that starting with low velocity the total vestance decreases as the velocity increases until it reaches PIPE 181 a minimum, and thereafter increases. This minimum point is the point of best economy under the conditions assumed. If the cost of power is in excess of one cent per h.p.h., lower veloci- ties must be used, otherwise higher velocities for best economy. There are instances in the piping of water where some power is available but not sufficient for economical development. Under such conditions small pipe may be used, wasting this power in friction but reducing thereby the cost of the pipe. But there are other instances where the power is wasted merely because of a lack of knowledge of its worth. We have in mind one such instance where the power wasted was afterward put to use and found sufficient for a town of 3000 people. 120. To give a better idea of the variation of total vest- ance per cu. sec. ft., with velocity, we give below more de- tailed data for the 6 inch pipe by way of illustration. TABLE 83 VELOCITY FT. PER SECOND 0.5 0.6 0.8 i.o i. "5 2.0 3.0 4.0 5.0 6.0 8.0 10 15 20 6" VD- $632 $527 $395 $316 $211 $159 $105 $79 $63 $53 $39 $32 $21 $16 V a . 2 2 3 6 14 22 48 81 122 180 298 459 1000 1586 Vt. .$634 $529 $398 $322 $225 $181 $153 $160 $185 $233 $337 $491 $1021 $1602 This shows that with the. conditions assumed a velocity of three feet per second is the point of best economy while a velocity of four feet per second is nearly as good. At a velo- city of 20 ft. per second, the total vestance for the transmis- sion of water is over ten hundred per cent greater. The friction of riveted pipe is considerably greater than smooth pipe, such as standard wrought iron or wood pipe. According to Mark's Mechanical Engineer's Handbook, that of 6 inch riveted pipe is 1.8 ft. per 100 ft. of pipe at 3 ft. per second velocity, while that of smooth pipe of the same size and at the same velocity is only 0.7 ft. under the same condi- tions; the operating vestance of the riveted pipe will be $122 as compared with only $47 for the smooth pipe. Since the total vestance of the latter is only $152, it would be neces- sary, for equal economy, to have only 152 - 122 = $30 182 VESTANCES TABLE 84 VESTANCES OF WOOD PIPE PER 100 FEET PER CUBIC SECOND FEET AT THREE FEET PER SECOND VELOCITY SIZE, IN. DIA. HEAD IN FEET 50 IOO ISO 200 250 300 350 400 2 3 4 6 8 10 12 16 20 24 ? D . $178.00 V a 203.00 182.00 203 . oo 5i88.oo 203 . oo 203.00 203 . oo r2i8.oo 203 . oo 1232.00 203 . oo 5263.00 203 . oo 299.00 203 . oo P,.... $381.00 VD . . $87.00 K a .... 115.00 $385.00 $91.00 115.00 $391.00 $93-oo 115.00 >4o6 . oo $97-00 115.00 421 .00 101 .00 115.00 $435-oo 5io8.oo 115.00 5466 . oo 126.00 115.00 502 . oo $138.00 115.00 V t. . . .$202.00 7 D . . $61.00 F a .... Sl.OO ^206 . oo $63.00 81.00 $208.00 $64 . oo 81.00 212.00 $78.00 81.00 216.00 $82.00 81.00 223.00 $89 . oo 81.00 241.00 $98.00 81.00 $253.00 108.00 81.00 V t $142.00 VD . . $33 - 00 Va.... 48.00 144.00 $39-oo 48.00 5145.00 $42.00 4 8_.oo $90.00 $31.00 34-oo 159.00 $46 . oo 48.00 163.00 $50.00 48.00 170.00 $55-oo 48.00 $179.00 $59-00 48.00 $189.00 $61.00 48.00 V t ... . $81.00 V D . . $22.OO F a .... 34-oo $87.00 $28.00 34-oo $94-00 $35-00 34-oo $98 . oo $38.00 . 34-oo 103.00 $41.00 34-00 $107.00 $44 oo 34-00 109.00 $51.00 34-oo Ft.... $56.00 VD . . $20.00 F a .... 26.00 $62.00 $26 . oo 26.00 $65-00 $28.00 26.00 $69 . oo $32-00 26.00 $72.00 $35-00 26.00 $75-oo $38.00 26.00 $78.00 $41.00 26.00 $85.00 $47-oo 26.00 V t .... $46.00 VD . . $15-00 F a .... 2O . OO $52.00 $20.00 20.00 $54 oo $23.00 20.00 $58.00 $25.00 20.00 $61.00 $29.00 20.00 $64 . oo $32.00 20.00 $67.00 $3.7-00 20 oo $73-00 $41.00 20.00 Vt.... $35-00 Fz> . . $14.00 F a . . . . 14.00 $40.00 $17.00 14.00 $43 oo $45-oo $49 oc $52.00 $57-oo $61.00 V t .... $28.00 VD $31.00 $15.00 11.00 $26.00 $11.00 9.00 V n V t V D .. $9-00 F a .... 9-00 Vt.... $18.00 $20.00 PIPE 183 $600 500 SI 400 5 300 * a 1C 200 100 2 4 6 8 10 12 14 16 18 20 Inches Diameter of Pipe FIG. 36. Total vestances of wood pipe at 50 and 200 ft. of pressure. of depreciation vestance. Assuming then that the riveted pipe has as long a life as the wrought iron pipe, i.e., 50 years, so that its term factor will be 0.91279, it would be necessary to purchase the riveted pipe at 30 X 0.91279 = $27.38370 per 100 ft. as compared with $57.80 for the wrought iron pipe, i.e., the riveted pipe would be worth only 47.4% as much as the latter. The use of extra strong, or double extra strong, pipe is war- ranted where, and only where, the pressure to be borne is too great for standard pipe. For still lower pressures, casing may be used to considerable advantage due to its lower first cost. 121. We give below the vestances of wood pipe on the same basis as the standard pipe above. This pipe has come into very considerable use. It is particularly well adapted for low heads. It cannot, however, be allowed to stand empty without ruining it. The friction is assumed equal to that of standard pipe. Since the operating vestance of wood and standard pipe has been assumed equal, while the first cost, and therefore the depreciation vestance, of the former is much lower, the total 184 VESTANCES vestance of wood pipe is far the lower. However, it must be remembered that there are some duties for which wood pipe is not adapted. But in such service as the distribution of water for irrigation, wood pipe has proved an inestimable aid, allowing much development that would have been pro- hibitive with more expensive pipe. It would be a simple matter in any concrete case to com- pare the total vestances of pipe with that of open ditches or flumes. However, both the first cost and the cost of main- tenance of such conveyances is highly variable, varying greatly with the character of the ground, the conformation of the soil, and local weather conditions. In any concrete case, where these items are definitely known, exact comparisons can be readily made, so that we can determine which is the best. 122. It must be borne in mind that, primarily the prices of all commodities are based upon the cost of production of these commodities. If a system is so installed that it does not render the service at best economy, then labor is being wasted. If the installation is too cheap in design and construction, then labor is wasted in operation. If, on the other hand, the installation is too expensive, then labor has been wasted in producing a system more refined and efficient than the amount and nature of the service warrants. Waste of brains, skill or labor is a dead loss; a waste that no com- munity, whether state or nation, can afford. In the face of international competition, such a waste leads with certainty to national impoverishment. It is a mark of incompetence. In the calculations of this chapter on motors, we have as- sumed a cost of i cent per h.p.h. This is, as we have pointed out, far lower than can be obtained in practice. In order to indicate what may actually be expected we publish the actual rates in Oregon for 1919, as announced by the public service commission. These are as follows: PIPE 185 TABLE 85 PRIMARY RATE First 500 kw.h. per mo. at 5 cents per kw.h. Next 500 " " " " 4 " " " 4,000 " " " " 3 " " " 10,000 " " " " 2 " " " AU excess " " " " ij " " " SECONDARY RATE First 4,000 kw.h. per mo. at i| cents per kw.h. Next 100,000 " " " " i.o" " " Excess kw.h. " " "0.8 " " Assuming a load factor of 40%, i.e., assuming that the motors run at full load 40% of the time on the basis of a 24- hour day and are idle 60% of the time, and also allowing for the unavoidable motor losses, these rates are, for various sized motors, as follows: TABLE 86 Corresponding mean rate Size of motors at 40% load factor Up to if h.p 5 cents per h.p. of motor if to 3 h.p 4* " " " " 3 to 17. 5 h.p 3.3 " " " " " 17. 5 to 52. 5 h.p 2.433 " " " 52. 5 to 105 h.p 2.004 " " " " To determine the total vestance of a motor under such a rate, multiply the operating vestance by the rate for the sized motor under consideration and add thereto the depreciation vestance. For example in Table 74 we gave for a one h.p. motor at full load: Depreciation vestance $56 . 80 Operating vestance 767 . oo Total vestance $823 . 80 The operating vestance was based on i cent per h.p.h. Ac- 1 86 VESTANCES cording to the above rate this should be 5 cents. For this rate, the operating vestance would be 767 X 5 = $3835> giving a total vestance of 3835 + 56.80 = $3891.80 instead of only $823.80. This would not compare well with engines of any type. We give in the table below the vestances of induction motors at full load with the rates as per the table above, other things as in the previous table on induction motors. TABLE 87 VESTANCES PER H.P. OF INDUCTION MOTORS AT FULL LOAD AND 40% LOAD FACTOR AS PER THE 1919 .OREGON RATE SIZE OF MOTOP, H.P. VESTANCES Depreciation Operating Total I $56.80 $3835-00 $3891.80 2 35-55 3276.00 33II-55 3 26.95 2353-00 2379-95 5 19.80 2310.00 2329.80 7l 25.60 2273.70 2298.30 10 22.75 2257.20 2279.95 15 18.20 2234.10 2252.30 20 16.70 1640. 10 1656.80 25 14.50 1632.80 1647.30 30 14-45 1627.90 1642.35 4 12.43 1623.00 I635-43 SO n-37 1618. 20 1629.57 75 9.76 1326.00 . I335-76 CHAPTER VI UNIT COST DETERMINATION Unit Cost. Time Element. Change Points. Unit Cost at Constant Load. Nearly Constant Load. Two Part Load. Three Part Load. N-part Load. Instantaneous Unit Cost of Actual Load Curve. Service Modulus and Replot. 123. We have seen in the foregoing chapters that cost governs price, so that our real problem is primarily that of cost determination. But in what way shall we apportion the costs, when all elements that enter into the problem are vari- able? The answer to this is: That an equipment must earn its costs, and profit, while in operation for certainly it cannot do so while idle. This is the fundamental basis of cost analysis. 124. Let us then, to begin with, take the simplest condi- tions of operation possible, that in which the undertaking runs at full load for a variable number of hours per year. Let C = the cost per h.p. of the system, p = the per cent of fixed charges, a = the operating cost per h.p.h., L = the load carried and N = the number of hours of operation then, evidently pC = the fixed charges per h.p. year and LpC = the total fixed charges per year. The greatest value that (N) can have is 8760, the number of hours in a year. Again since the operating cost is assumed to be constant at (a) dollars per h.p.h. (or other unit of service), then La = the total cost of operation per hour, and LaN = the total cost of operation per year. 187 i88 UNIT COST DETERMINATION Evidently then the total production cost (Ki) per year is the sum of the total annual cost of operation and the fixed charges or Ki = LpC + LaN = L(pC + aN) . . . (69) If now we divide this total annual cost by the load (L) and the hours of operation (N), we will get the cost per h.p.h. (# 2 )or Ki L(pC + aN) KZ ~TN = ~~LN~ S~1 or K 2 = ~ + a ............ (70) As (N) becomes less, the unit cost (K 2 ) becomes greater, until when N = o, the unit cost is infinite, the operating cost per h.p.h. being assumed constant. On the other hand as ( N) increases, the unit cost becomes less, tending to decrease to the value of the operating cost, when (N) = . But the largest value that ( N) can have is 8760, the number of hours in the year, and therefore the minimum value that the unit cost can reach is a 8760 when the plant is operated at full load throughout the entire year. 125. If we have two plants, one of which has a first cost of Ci dollars, a depreciation of pi per cent, and an operating cost of #i dollars, while the second has corresponding values, C 2 , pz, and #2 and each carry the same load L, then the total annual cost of the first plant is and of the second is If we plot the total annual production cost vertically as in Fig. 37 and the hours horizontally, then these equations will give straight lines AB and A'B'. It may be and often is true that two such lines as AB and A'B' intersect between N = o THE CHANGE POINT Hours of Operation per Year 8760 FIG. 37. Variation of production cost with use and the change point. and N = 8760, as at (C), Fig. 37. At this change point Ki and K" are equal and therefore or piCi + 01 N = p 2 C 2 + a 2 N, and solving for N gives the change point (71) For values of (N) less than N , Ki is less than Ki" For values of (N) greater than N , Ki" is less than Ki. Evi- dently then when the hours of operation are less than A^ , the first plant is the most economical, for a greater number of hours of operation, the second plant is best. The application of this will be illustrated in the problems below. Example 35. A pumping plant, having a capacity of ten thousand gallons per minute costs $8000 and has 14% fixed charges. The cost of operation of the plant is 20 cents per hour. What is the cost per hour per 1000 gallons of capacity? Solution: Under the above conditions the fixed charges are 8000 X 14% = $1120, UNIT COST DETERMINATION or $ii2 per 1000 gallons of capacity. The total cost of opera- tion is 20 cents per hour or 2 cents per 1000 gallons of capacity. For (AO hours this cost would be o.o2A r dollars. The total production cost per thousand gallons of capacity per year is then KI = 112 + O.O2N (a) The cost per hour is obtained by dividing (K\) by the num- ber of hours (N), getting K = 2 = N 0.02 We can get values for K\ and K^ from equations (a) and (b) for various values of (N) as follows; TABLE N Kt K , o $112.00 infinite IOO 114.00 $1.14 1000 132.00 0.132 3000 172.00 -573 5000 2 I 2 . OO 0.0424 7000 252.OO o . 0360 8760 287. 20 0.03278 Plotting the values of KI vertically against the time ( A^) horizontally as in Fig. 38 gives a hyperbola. Note the enor- mous variation of the unit cost (Kz) with time. The number of hours of operation per year is the most important factor in the cost of a service. Example 36. A power plant costs $100 per h.p. and is of 2000 h.p. capacity, running at full load. If the fixed charges are 15% and the cost of operation 0.5 cent per h.p.h. (a) what is the total annual cost; and (b) what is the cost per h.p.h.? Solution: The total cost of the plant is 2000 X ioo = $200,000, and the total fixed charges are 200,000 X .15 = $30,000. CONSTANT LOAD 191 The operating cost per year is 2000 X 0.005 X N = 10 N . . . The total annual production cost per h.p. is then $1.40 - = r + 0.005 2000 N N (fr) 1000 2000 3000 4000 5000 6000 7000 8000 8760 Hours per Year FIG. 38. Variation of total and unit production cost with use, according to example 35. The values of KI and K% for various values of (N) are given below, in order again to illustrate the enormous varia- tion in cost with the time of operation. TABLE 89 .V Ki Kt o $30,000 . oo infinite IOO 31,000.00 o-i55 IOOO 3000 5000 40,000.00 60,000 . oo 80,000 . oo O.O2 O.OI 0.008 7000 8760 100,000.00 117,600.00 0.007143 O.OO67I2 192 UNIT COST DETERMINATION Note that for 1000 hours of operation, the cost per h.p.h. is 30 times as great as for continuous ( N = 8760 hours) opera- tion during the year. Example 37. A pumping plant of 1000 g.p.m. capacity costs $2400. The fixed charges are 12 % and the cost of opera- tion is 0.3 cent per thousand gallons delivered. Each acre irrigated requires 43,200 gallons of water per month, so that with 12 hours of operation per day, we can irrigate 50 acres. We wish, however, to irrigate 100 acres, and in as much as night irrigation is not feasible, the question is, shall we put in another pumping unit, or shall we build a reservoir, pumping into it during the night? If we put in the reservoir, the pumping plant will operate 24 hours per day and therefore the power company will reduce our rates so that the cost of the water will be reduced to 0.2 cent per 1000 gallons of capacity. The reservoir costs 25 cents per yard to excavate and the land costs $200 per acre. The reservoir permits of only 10 ft. depth so as to keep the bottom above the land level. The fixed charges on the reservoir are 8% and the length of the irrigation season is 5 months. Shall we install another pumping plant, or put in the reservoir? Solution: The question can only be answered by determin- ing the costs in each case and comparing them, the one ren- dering the service cheapest being the one we shall decide upon. Where we are operating 12 hours per day our costs are as follows : Fixed charges per year are 2400 X 0.12 = $288, and the operating costs are $0.003 X 60 = $0.1 8 per hour, and since 5 months = 5X30X12 = 1800 hours the annual cost of operation is .18 X 1800 = $324. The total annual cost is then K\ = 288 + 324 = $612 per year. CONSTANT LOAD 193 If we install two units, this cost will be doubled, or Ki = 2 X 612 = $1224. To hold the water pumped during 12 hours, the capacity of the reservoir will have to be 1000 X 60 X 12 = 720,000 gal. or 720,000 -r- 7.5 = 96,000 cu. ft., or 96,000 -T- 27 = 3556 yards, costing 3556 X 0.25 = $889. If the reservoir is 10 ft. deep, it will cover 96,000 -j- 10 = 9600 sq. ft., which, allowing for some wastage of ground will use up about ^ acre, costing 200 X i = $100, so that the total reservoir costs are 889 + 100 = $989, on which the fixed charges are 989 X 0.08 = $79.12 per year. With one pumping unit in combination with the reservoir, the total fixed charges per year are 288 + 79.12 = $367.12 and the operating costs will be, at the reduced power rates, 0.002 X 60 = $0.12 per hour, or 0.12 X 3600 = $432, five months of operation amounting to 3600 hours now, based on 24 hours per day of operation. The total annual costs are then 367.12 + 432 = $799.12. To summarize then, (a) With two pumping units running 12 hours per day for 5 months, each of 1000 g.p.m. capacity, the total annual cost is $1224. (6) With one pumping unit of 1000 g.p.m. capacity running 24 hours per day for 5 months, and thus pumping an equal i 9 4 UNIT COST DETERMINATION amount as the two units under (a) above, in combination with the reservoir, the total annual costs are only $799.12, showing a saving of 1224 799.12 = $424.88 per year or 424.88 -i- 799.12 = 53%. 126. Example 38. A centrifugal pumping unit of 1000 g.p.m. capacity costs $1100 and is guaranteed to operate at 60% efficiency. A triplex plunger pumping unit of the same size costs $2400 and is guaranteed at 75% efficiency. If in either case the fixed charges are 15% and the power charges are i cent per h.p.h., other operating costs being i cent per hour, which equipment should we use, the head in either case being 80 ft.? Solution: (a) In the case of the centrifugal pump: The fixed charges are noo X 0.15 = $165 per year, . , . 1000 X 80 , the power consumed is 4ooo x p 6p = 33-3 h-P, The cost per hour for power is then 33.3 X o.oi = $0.333, and the total operating cost per hour is -333+ o-oi = $0.343, or -343 N P er year of N hours. The total annual cost is therefore K! = 165 + 0.343^. (a) (b) In the case of the triplex pump: the fixed charges are 2400 X .15 = $360, , . 1000 X 80 * the power consumed is = 26.7 h.p., 4000 X 0.75 so that the cost per hour for power is 26.7 X o.oi = $0.267, or 0.277 N per year of N hours. The total annual cost is therefore KS = 360 + 0.277^ :, (b) . CONSTANT LOAD 195 For N = 1000 hours per year, Ki = 165 + 343 = $508 for the centrifugal pump, and KI = 360 + 277 = $637 for the triplex pump, showing the centrifugal pump to advantage. For N = 5000 hours per year, then Ki = 165 + 1715 = $1880 for the centrifugal pump, and Ki = 360 + 1385 = $1745 for the triplex pump, show- ing the advantage to rest with this pump. We can determine for what number of hours per year the two costs will be equal, by equating (a) and (6), thus 165 -f 0.343^1 = 360 + 0.277^1 so that 0.066 Ni = 195 or Ni = 2955 hrs. per year. That is, if we operate less than 2955 hours per year, then under the conditions assumed above, the centrifugal pump is best; if we operate more than 2955 hours per year, then the triplex pump is best. Example 39. We wish to generate power under full load conditions and find that a hydro-electric plant will cost $100 per h.p. and the cost of operation will be o.i cent per h.p.h. A Diesel engine plant will cost $60 per h.p. and cost 0.2 cent per h.p.h. for operation, while a steam plant will cost $45 per h.p. and 0.3 cent per h.p.h. to operate. If the fixed charges in all cases above are 14% for what hours of operation would each of the above plants be best? Solution: The total annual fixed charges per h.p. for the hydro-electric plant are 100 X 0.14 = $14. For the Diesel plant, they are 60 X 0.14 = $8.40, and for the steam plant, they are 45 X 0.14 = $6.30. The annual cost of operation for the hydro-electric plant per h.p. is $0.001 N. For the Diesel plant, it is $0.002 N, and for 196 UNIT COST DETERMINATION the steam plant, it is $0.003 N, where in each case N is the number of hours of operation per year. The total annual cost per h.p. for the hydro-electric plant is then KI = 14.00 + o.ooiN. For the Diesel plant, it is KI = 8.40 + o.oo2N, and for the steam plant, it is KI" = 6.30 + 0.003^. Equating KI and KI', we get 6.30 + 0.003^1 = 8.40 + 0.002^1, so that o.oo i Ni = 2.10, and N! = 2100 hours per year. So also equating KI and KI, we get 8.40 + 0.002^2 = 14.00 + o.ooi7V 2 so that o.ooiA^ = 5.60 or N% = 5600 hours per year. The conclusions then are: (a) If the plant is to operate less than 2100 hours per year, use the steam plant. (b) If the plant is to operate more than 2100 hours per year, but less than 5600 hours per year, use the Diesel plant. (c) If the plant is to operate more than 5600 hours per year, use the hydro-electric plant. 127. If the plant does not operate at full load but at a con- stant fractional load, then calling M the capacity of the plant, and L the load, as before, we get that the annual fixed charges are MCp, the annual operating cost is LaN, the total annual cost is KI = MCp + LaN, and the cost per h.p.h. is KI MCp , f , -. -'- *-Zff--ffL +fl (72) and the application is as above. PROBLEMS 197 Example 40. A steam generating plant running at 80 % capacity costs $50 per h.p. If the fixed charges are 12% and the operating costs are 0.3 cent per h.p.h., what will be the annual cost per h.p. and the cost per h.p.h.? Solution: Here M = i, L = 0.8, p = 0.12, a = $0.003 an< 3 C = $5, whence KI = i X 50 X 0.12 + 0.8 X 0.003 A" or KI = 6 + 0.0024N (a) 11 TT i X 50 X 12 , and also K z = ^ h 0.003 or K z = ^ + 0.003 (b) While in the above we have shown the application of the cost analysis, under special conditions, only to pumping and power generation, it is equally applicable to railroad or wagon road analysis as to dry goods or grocery business. PROBLEMS 1. A pumping plant of 2000 g.p.m. capacity costs $2100. The fixed charges are 13%, and the cost of operation is 35 cents per hour. The plant operates for 2000 hours per year at full load. (a) What is the total annual cost of operation per year? (b) What is the total cost of operation per hour per thousand g.p.m. of capacity? 2. If in problem (i), the plant operates for (N) hours per year: (a) What is the total cost of operation per year? (b) What is the total cost of operation per hour per thousand g.p.m. of capacity? (c) Plot the curves. 3. An electric pumping unit of 10 h.p. size costs $240, while a gas engine driven unit costs $600 complete. The fixed charges in either case are 12%. The cost of operation of the electric pumping unit is 2 cents per h.p.h., while that of the gas engine driven unit is 1.2 cents per h.p.h. Determine for what number of hours of operation at full load each is best. Plot the curves. 4. A 10 h.p. gas engine electric lighting unit costs $90 per h.p. and runs at full load for 3000 hours per year. The load increases during the period of operation to 15 h.p. We can take care of this additional load, either by installing another 5 h.p. unit as above, or by putting in a storage battery 198 UNIT COST DETERMINATION and running the 10 h.p. unit 4500 hours per year. The battery costs $75 per kw. of capacity and has an efficiency of 65%. The fixed charges on the above units are 15%. The operating cost of the engine unit is 1.3 cents per h.p.h. and of the battery 0.2 cent per kw.h. Should we install the extra engine unit or the storage battery for best economy? 5. In problem (4), if the load increased to 20 h.p., which should we install? 6. In problem (4), if the load increased only to 12 h.p., which should we install for best economy? 7. Given the following data: A horizontal 4-valve engine generating unit costs $22,700. The engine runs with steam at 150$ pressure, 100 superheat and with a 26 in. vacuum, and uses 16.5$ steam per kw.h. A steam turbine generating" unit of the same size costs $12,250 and operates under the same conditions as the steam engine unit except with a 28 in. vacuum. It consumes 17.7$ steam per kw.h. The plants are run at full load. If the fixed charges on either plant are 14% and the steam costs 18 cents per looojf, and other operating costs are 0.06 cent per kw.h., for what length of service per year is each best adapted? 8. A hydro-electric power plant costs $120 per kw.h. and operates at 60% of full load, for (N) hours per year. Determine: (a) The total annual cost per kw. of the service; (b) The cost per kw.h. of the service; (c) Plot the curves for (a) and (b). 9. We find that: a hydro-electric plant will cost $100 per kw. to install and 0.2 cent per kw.h. to operate; a Diesel plant will cost $70 per h.p. to install and 0.3 cent to operate; while a steam plant will cost $50 per h.p. to install and 0.4 cent to operate. The fixed charges on the hydro- electric plant are 14%, on the Diesel plant 15% and 16% on the steam plant. The plants are to operate at full load. For what number of hours of operation is each of the above plants best? 10. A pumping plant of 1000 g.p.m. capacity delivers water against 120 ft. head with a guaranteed efficiency of 70% at full load and 40% effi- ciency at half load. The plant costs $850 to install. The power costs one cent per h.p.h., while fixed charges are 12%. The plant operates at full load for 2000 hours per year and at half load for 1000 hours per year. Deter- mine the total annual cost of operation. 11. If in problem (10), the plant operated for (Ni) hours per year at full load and for (7V 2 ) hours per year at half load, what would be the total annual cost of operation? 12. In problem (n), plot the curve for the variation of the total annual cost of production with (Ni) if Ni + N z = 3000 hours. 13. We are irrigating 100 acres of land for 5 months in the year with a 25 h.p. direct connected electric centrifugal pumping unit costing $2200. The pump efficiency is 60%, motor efficiency 85%, while the head pumped PROBLEMS 199 against is 40 ft. total. The electric power costs 1.5 cents per kw.h. The plant runs at full load for 8 hours each day. We wish to irrigate 300 acres of land and can do so with the unit already installed by running 24 hours per day, in which case the power will cost us only one cent per kw.h. But in that case, we must install a reservoir to avoid night irrigation, costing 35 cents per cubic yard to excavate and having a maximum depth of 12 ft. The land costs $100 per acre. Fixed charges on the pumping unit are 12%, and other operating costs are 3 cents per hour. The fixed charges on the reservoir are 10%, with an annual cost of $78 for maintenance and repair. With the reservoir in use the head on the pump will be increased by 4 ft., with a corresponding increase in power consumed. Instead of using the above layout, we can install two more 25 h.p. pump- ing units, the three units operating for 8 hours per day, or we can install one more such 25 h.p. pumping unit together with sufficient reservoir to avoid more than 8 hours of operation. In the case of the three 25 h.p. units, our power cost will be 1.3 cents per kw.h. and 1.2 cents for the two 25 h.p. units. In the latter case the head will be increased by 2 ft., with a corresponding increase in power consumed. Determine which of the three possible types of installation are best, and by what per cent. 14. A 10 h.p. direct connected centrifugal pump operates at 60% effi- ciency at full load, for 2000 hours per year. We wish to operate for another 1000 hours per year with a load of 5 h.p. If we use the above pump, its efficiency will drop to 45%, while if we install a 5 h.p. unit it will give 52% efficiency. The power costs 2 cents per h.p.h. while fixed charges are 12%. Should we install the additional unit or use the 10 h.p. unit at half load? By what per cent would the one exceed the other? The current is a.c. 60 cy. 3 phase 220 volts. 15. We wish to install a 10 h.p. electrically driven centrifugal pumping plant. Which would be best, (a) a belt driven unit or (b) a direct connected unit? Take into consideration cost of units, building and foundations. How much would we lose by installing the poorer one? Assume a power cost of 2 cents per kw.h., fixed charges 10% and 1500 hours per year of operation at full load pumping against a 50 ft. head. 16. Determine the same as for problem (15) but for a 100 h.p. plant with a power cost of i cent per kw.h., other things being equal. 17. Determine the same as for problem (15) but for a one h.p. plant with a power cost of 3 cents per kw.h., other things being equal. 18. Determine the same as for problem (15) but for a 50 h.p. plant with a power cost of 1.5 cents per kw.h., other things being equal. 19. A 250 h.p. pumping plant is to operate against a head of 100 ft. Into how many units should we divide this plant for best economy, taking into account cost of building and foundations? Power cost one cent per kw.h., hours of operation 2000 per year and fixed charges 10%. 200 UNIT COST DETERMINATION 20. Determine the same as in problem (19) for a 150 h.p. plant operating against 50 ft. head, other things being the same. 21. Determine the same as in problem (20) but for a 30 h.p. plant operat- ing against 25 ft. head. 128. Thus far we have considered an equipment operating at one uniform load during the entire year. If the load varies with the time, as shown by the curve, OO, Fig. 39, but not sufficiently so to necessitate the discontinuance of opera- tion of part of the plant, then for any given period we determine the mean load (L m ), and base our cost analysis thereon, instead of ^ on the actual, variable load (M). In an electric gener- ating plant, the kw.h. produced, divided by the time in hours, gives the mean load. In general, the total production divided by the time in hours gives the mean load or average production per hour, i.e., the mean rate of production. But now, suppose we have a twin load as illus- trated in Fig. 40, i.e., a load of (Lz + Li) for (A 2 ) -Nr hours per year and follow- o ing that, a load of (Li) for (Ni Nz) hours per te- N t - year. It is Our problem FIG. 40. Diagram for a two-part load. then to determine the cost of production for a given plant, already in operation, or else the more difficult and important problem to so design the plant that the production will be at a minimum cost. In this case the TWO-PART LOAD 201 difference between L% and LI is assumed to be so great that it would be uneconomical to operate the entire plant for the lesser load, the entire plant operating for the load (L\ + Lz) and ( 7 ^j- ) per cent operating for the load LI. \LI -f- LZ/ The difficulty in the solution of this problem comes in the equitable distribution of fixed charges, primarily. But this difficulty is removed if you bear in mind that an equipment or unit thereof must earn its costs and profits while in opera- tion, for certainly it cannot do so when idle. Suppose then that this plant consists of two units, one of (Li) size and the other of (L 2 ) size. Evidently then the unit (Li) will operate for (Ni) hours per year, while the unit (Lz) operates for only (Nz) hours per year. During the balance of the year, above (Ni) hours, i.e., for A^ hours (where Ni + N = 8760 hours) the load is assumed to be zero. For the unit (Li) we have LiCp = total fixed charges per year and LidNi = total operating costs per year, where LI = the load, C = cost of plant per unit of load, p = per cent of fixed charges, a = operating cost per hour per unit of load and Ni = number of hours of operation. Whence the total annual cost ( Kn) is Kn = LiCp 4- L&Ni and the cost per unit of load (h.p.h., kw.h. etc.) per hour (^21) is K n = Q + a. ' So, also, if only the unit (Lz) were in operation for the hours, then the total annual cost KU being the cost per unit of load per hour is 202 UNIT COST DETERMINATION The total cost for operating the unit (L 2 ) for (N 2 ) hours is K i2 = L^Cp + I^aN 2 and for operating the unit (Li) for N 2 hours the total cost is obtained by multiplying the cost per hour (Eq. 68) by the load (Li) and the hours N 2) i.e., by LiA^, getting , L,N 2 Cp A i2v = -- -f- Lial\ 2 . The total cost of both units operating for A" 2 hours is then the sum of Kn and K i2 ff or Kn = LtCp + - ^ i.e., and the cost per unit of load per hour is obtained by dividing by the total load (Li + L 2 ) and the hours A T 2 or /L 2 N 1 + L 1 N 2 \ Cp ~ V N,N 2 , . a ' ' (73) for the period (N 2 ), while for the period (N\ N 2 ) it was, by equation (68) Cp . ^ = ^ +a and nothing for the period (N ) , since it is useless for this period according to the assumptions made in the problem. 129. Example 40. A steam power plant costs $50 per h.p. and is of 2000 h.p. size. It carries a load of 2000 h.p. for 3000 hours of the year and 500 h.p. for an additional 2000 hours. If the cost of operation is 0.3 cent per h.p.h. and fixed charges are 12%, determine the cost per h.p.h. for each period, the load being considered integral. Solution: We can determine the costs by substitution in the equations, or work them out in detail as follows: TWO-PART LOAD 203 The 500 h.p. unit will operate for 3000 H- 2000 = 5000 hours. The annual operating cost of this unit is 500 h.p. X $0.003 X 5000 hrs. = $7500, and the fixed charges are 500 h.p. X $50 X 0.12 = $3000, so that the total annual cost is KI = 3000 H- 7500 = $10,500. The cost per h.p. year is 10,500 ^ 500 = $21, and the cost per h.p.h. is $21 -f- 5OOO = $O.OO42. During the first 3000 hours we have an additional load of 2000 500 = 1500 h.p. whose operating cost per year is 1500 h.p. X 3000 hrs. X $0.003 = $13^500, and the fixed charges are 1500 h.p. X $50 X 0.12 = $9000. The total cost of this part of the unit is KI" = 9000 + 13,500 = $22,500. This is for the 1500 h.p. unit during the first 3000 hour period. But during this period we have in use the 500 h.p. unit as well, costing 500 X 0.0042 X 3000 = 6300, so that the total cost during this 3000 hour period is 22,500 + 6300 = $28,800, and since the load is 2000 h.p., the cost per h.p. is 28,800 -f- 2000 = $14.40, and since this period is 3000 hrs., the cost per hour during this period is 14.40 -r- 3000 = $0.0048 per h.p.h. 204 UNIT COST DETERMINATION Summary. Cost per h.p.h. during ist period $0.0048. " " " " 2nd " 0.0042 By substitution in the equations, we get Kz = h 0.003 = 0.0012 + 0.003 = $0.0042. 3000 X 2000 for the second period, and _ 20OO X 3000 + 2OOO X 20OO 500 X 2OOO 2000 X 3000 (3000 + 2000) X 0.12 X 50 + 0.003, or KZ 0.0018 + 0.003 = $0.0048 per h.p.h. for the second period, as before. Example 41. A factory for the manufacture of a gas en- gine costs $4,000,000 on which the fixed charges are 15%. The capacity of this plant is 600 engines per month. It sup- plies this number of engines per month during the first five months, and then 300 engines per month for the balance of the year, due to reduced demand. Exclusive of fixed charges, the cost of manufacture of each engine is $150. Determine the total cost per engine (a) during the first period, and (b) during the second period. Solution. If we had a factory capacity of only 300 engines per month, its cost would be $2,000,000 on which the fixed charges would be 2,000,000 X 0.15 = $300,000 per year, or 300,000 -T- 12 = $25,000 per month; and since 300 engines are turned out per month, this gives 2500 -f- 300 = $83.33 P er engine, wherefore, for the slack period, the total engine cost is 150 4- 83.33 = $233.33. The second $2,000,000 is required for the extra demand of 300 engines per month during the first 5 months. The fixed charges thereon are 2,000,000 X 0.15 = $300,000, TWO-PART LOAD 205 which must be paid during this 5-month period, together with -f% of the other $300,000 of fixed charges or T 6 2 X 300,000 = $125,000. Total fixed charges assignable to this first period are 300,000 + 125,000 = $425,000 or 425,000 -T- 5 = $85,000 per month. During this period, 600 engines per month are manufac- tured, wherefore the fixed charges per engine during this rush period are 85,000 -t- 600 = $183.33, and the total cost per engine during the maximum demand period is 183-33 + 150 = $333-33, or $100 per engine more than during the normal demand period. If these engines are sold for $400 per engine, the net profit per month during the rush period is (400 - 333.33) X 600 = $40,000, and during the normal period it is (400 - 233.33) X 300 = $50,000. It is no doubt surprising that the net profits should be less during the rush period than during the normal demand period, but this is due to the large investment to accommodate this peak period. In this case it would be far better to reduce the size of the plant so that it could meet the total annual de- mand by producing engines at a uniform rate throughout the year. This would necessitate the tying up of capital on prod- uct finished ahead of the demand. The interest thereon must be charged against these off-season engines plus the cost of storage, but this would hardly amount to more than $5 extra per engine as compared with $100 per engine extra on the first 3000 engines per year. 130. The application to railway transportation problems is exactly similar as illustrated in the following problem: Example 42. A branch railway 60 miles long costs $1,800,000 for ways and structures. Freight cars cost $3500 2 o6 UNIT COST DETERMINATION each and locomotives, one for each 20 cars, cost $30,000 each. Freight cars empty weigh 5 tons each and carry 30 tons net weight. Locomotives weigh 30 tons. Cost per ton-mile for freight haul is 0.4 cent. The freight load amounts to 120 cars per day during July, August, and September, and to 20 cars per day during the balance of the year. Besides this, passenger service is rendered. Passenger cars weigh 7 tons, costing $5000 each, and carry, on the average, 40 passengers. Cost per ton-mile of passenger service is 0.6 cent. The pas- senger load amounts to 4 cars per day during June, July, and August, and 2 cars per day during the balance of the year. Determine the total cost per ton-mile during rush and normal demand periods for passenger service, also the cost per pas- senger-mile if the fixed charges on ways and structures are 10%, cars 14%, and locomotives 14%. Solution: The first problem is one of cost segregation as between passenger and freight service. The proper basis is evidently the ratio of the maximum demand. Thus the maxi- mum demand of passenger service consists of 4 passenger cars @ 7 tons each ...... 28 tons 4 X 40 X i5o# = 24,000 (wt. of passengers) 12 " and i locomotive @ 30 tons ......... ' 30 " Totalweight ............ 70 " for 2 X 60 = 1 20 miles per day or 70 X 1 20 = 8400 ton-miles per day. For the freight service, the maximum demand consists of: 1 20 cars per day @ 35 tons (one way) ..... 4200 tons 60 empties returned for loads assumed @ 5 tons (one way) ................ 300 " 6 locomotives 30 tons (two ways) ...... 180 " so that (4200 + 300) X 60 = 270,000, 180 X 1 20 = 21,600 or 291,600 ton-miles per day. Of the $1,800,000 for ways and structures 4o 291,600 $ I? 8oo,ooo = $52,000 approximately TWO-PART LOAD 207 is to be charged to passenger service, and the balance, or 1,800,000 52,000 = $1,748,000, to freight service. Of structures it should be understood that certain items must be charged to the particular service at once instead of being apportioned. Freight sheds and yards should be charged to freight service and passenger depots to passenger service directly, and so forth. The segregation of investments is then as follows: TABLE 90 PASSENGER SERVICE Cost F. C. rate F. C. total Ways and structures $52,000 10% $5,200 4 passenger cars 20,000 14% 2,800 1 locomotive. 30,000 14% 4,200 Total $12,200 FREIGHT SERVICE Cost F. C. rate F. C. total Ways and structures. . $1,748,000 10% $174,800 180 cars 630,000 13 % 81,900 5 locomotives 150,000 14% 21,000 Total $277,700 SEGREGATION OF PASSENGER FIXED CHARGE COSTS NORMAL PERIOD Ways and structures $2600 2 passenger cars 1400 i locomotive 4200 Total , $8200 $I2,2OO 8200 = $4000. The normal load continues for 12 months, so that the fixed charges per month are 8200 -4- 12 = $683.33. The excess load continues for 3 months during which the excess fixed charges are 4000 -r- 3 = $1333.33 P er month. Total fixed charges per month during rush period are J 333-33 + 68 3-33 = $2016.66 or 2016.66 -r- 30 = $67.22 per day. 208 UNIT COST DETERMINATION During normal period it was found to be only $683.33 P er month, or 68 3-33 *- 3 = $22.77 P^ day. During the normal period, the train weight is Locomotive ................. 30 tons 2 cars ...................... 10 " 80 passengers @ i5o# ........ _6 " Total ................... 46 " or 46 X 120 = 5520 ten-miles per day. Whence the fixed charge per ton-mile is for the normal period 22 -77 * 55 20 = $0.0041. The total cost per ton-mile is 0.0041 + 0.006 = $0.0101 during normal period. (Ans.) (a) Passenger miles per day during normal period are 2 X 40 X 1 20 = 9600. Ton-miles per passenger-mile are 5520 ~ 9600 = 0.575, whence the total cost per passenger-mile is o.oioi X 0.575 = $0.00581. (Ans.) (b) . During the rush period, the train weight is Locomotive ................. 30 tons 4 cars ...................... 20 " 160 passengers @ 150$ ....... i " Total ................... 62 " or 62 X 120 = 7440 ton-miles per day. Whence the fixed charge per ton-mile during the rush period is 67.22 -f- 7440 = $0.00903, and the total cost per Ion-mile is '0.00903 + 0.006 = $0.01503. (Ans.) (c) Passenger miles per day during the normal period are 4 X 40 X 120 = 19,200, VARIABLE LOAD TWO PART 209 so that the ton-miles per passenger-mile are 7440 -* 19,200 = 0.3875, and the total cost per passenger-mile during the rush period is 0.01503 X 0.3875 = $0.005824. (Ans.) (d) In a similar way the freight costs per ton-mile during each period may be determined. Certain simplifications, hardly warranted in practice, have been made in the above problem in order to illustrate the application of the principles without too much detail. 131. In former equations, it was assumed that the size of the unit and the load were equal, i.e., the unit operated at full load. This is hardly ever the case. Suppose then (Li + Lz) = size of load during the rush period, (Mi + Mz) = " " units for the rush period, Nz = time in hours (days or months) in the rush period, LI = size of load during the normal period, MI = " " unit for the normal period, (Ni Nz) = time in hours (days or months) in the normal period, C = cost per unit of load, p = fixed charges in per cent, and a = operating cost per unit of load per hour. Then evidently the total annual cost for the M i unit is Kn = MiCp + LidNi .......... (74) and the cost per unit of load is The total cost of the excess load during the rush period is Kzi' = MzCp + L 2 aN 2 . During this same period, we also operate the (Mi) unit whose cost for Nz hours is _ iz A 12 = TV 210 UNIT COST DETERMINATION whence the total cost for the rush period is /[ ,Nz\ c 7 N #1 / P * a 2 + iaN *' and the cost per unit of load per hour (h.p.h., kw.h. or the like) is \sfr t f v : r + <* (77) during the rush period as compared with KZI = h & Ni Li for the normal period. 132. Limitations on the value of (Ni) must of course be ob- served. If the time is in hours, then (Ni) cannot be greater than 8760; if in days, then it cannot be greater than 365; and if in months, then (A'l) cannot be greater than i 2 for evident reasons. In a great many cases it is hardly sufficient to divide the load into two parts. Electric service naturally divides itself into three distinct periods, the day, peak, and night periods. So the load on a great many undertakings naturally divides itself into three periods, the rush, normal, and slack periods. Under these conditions, we qan determine the costs during each period as follows: Li = the mean load during the slack period, Mi the capacity of unit during the slack period, NI Nz = duration of the slack period, (Li + LZ) = mean load during the normal period, (Mi + Mz) = capacity of units during the normal period, Nz A^s = duration of the normal period, (Li + LZ -f 3) = mean load during the rush period, (Mi + Mz + M^) = capacity of units during the rush period, (Nz) = duration of the rush period, C = unit cost of the undertaking, p = per cent of fixed charges, a operating cost per unit of load per hour, THREE-PART LOAD then for the unit Mi, the total annual cost is K n = Mp + L&Ni . . . and the cost per unit of load per hour is 211 (78) (79) This is for the slack period. L -NT N FIG. 41. Diagram for a three- part load. The excess load during the normal over the slack period is (L 2 ) with a corresponding difference in size of unit of (M 2 ), the total annual cost of which is Ku' = M 2 Cp + and the cost per hour is = During this period it operates for (A^ Na) hours, for which time the cost is v" = Cp(N, - N,) to which must be added the cost of operating the (Mi) unit for the same period, or 212 UNIT COST DETERMINATION giving a total for this normal period of - N 9 ) (So) Dividing through by the load (Li -f L 2 ) and the hours ^2 Na) gives the cost per unit of load per hour, or ^T- + (81) which is identical with equation (77). For the rush period the total cost of operating the excess unit (Ms) is Kb' = M*Cp + L 3 aN z . The cost of operating the units (Af 2 + MI) for N s hours is obtained by multiplying equation (77) by the load ( LI + L 2 ) and the hours N 3 , or /Mi Jl -tvis .= I ~^r + - The total cost of operation during the rush period is the sum of KM and K^", or v -1 (82) so that the cost per hour, ..obtained by dividing equation (82) ~by (Li + L 2 + Lz)N z is /Mx M 2 M 3 \ Cp ( , } A 23 = I r + -T7- + -rr- 17-= ; ; r-r- + a . (d?,) If we had (If) divisions in our load instead of only three as above, then the costs during the nth period can be written at once by symmetry, getting (84) N-PART LOAD 213 total cost during the period, and therefore _ 1 ,M 1 M.\ Cp -~ + a . . .............. (85) 133. We wish to call particular attention to the term /Mi 4- M* \Ni + N, 4 i- "* N which we shall call the service modulus, and denote by (5), so that c M l Sl '-" If? . 5. = + + ... +;... (86) We shall also denote by L n ' the total load during any period, so that L n ' = L, + L 2 + - ' - + L nj whence we may rewrite equations (79) to (85) above as follows: K zl = S^+ a, ........ (87) L\ K a =StQ + a, ........ (88) Li KK = S, Q + a , ......... (89) and in general K, n = S n + a (90) 214 UNIT COST DETERMINATION 134. If we plot the time (AO vertically and (5"), the service modulus, horizontally as in Fig. 42, then the area is the size of the unit (M) required for the period. Thus ~ M, Sl -~- T; so that SiNi = Mi, etc., the size unit required. In determining the costs of an undertaking, the work is greatly simplified by first determining the service modulus when the costs for each N period may be written down at once. Example 43. A cer- tain plant carries a load as follows: first period, 4000 hours per year, mean load 400 kw., maximum 500 kw.; second period, 2000 hours, mean load 800 kw., maximum 1000 kw.; third period, maximum 2000 kw., mean load 2000 kw., time 2000 hrs. First cost per kw. $70, fixed charges 10%, and operating cost 35 cents per kw.h. Determine the cost per kw.h. during each period. Solution: In this case the service modulus (S 3 ) for the third period, where Mi = 500, Ni = 8000, MZ = 1000 500, NZ = 4000, Ms = 2OOO IOOO, NS = 2OOO, Z 3 = 2000 kw.h., is equal to c 5OO IOOO 5OO 2OOO IOOO ^3 o I I J 8OOO 4OOO 2OOO or S 3 = 0.0625 + 0-125 + -5 = 0.6875. Whence the cost per kw.h. during this period is FIG. 42. #23 #23 0.6875 X 70 X o.io 2OOO 0.0035, 0.0024 H~ -35 = $0.0059 per kw.h. (Ans.) SERVICE MODULUS 215 Similarly the costs during the first and second periods may be determined. 135. In general, we had for the service modulus, _ Mi .Mt , , M n -i . M_ ~ Ni + N 2 " ^ N n -i "*" N n for (n) periods. For the (n + i)st period, it is _ Mi Ml M n -l . Mn , M n +i kj*+i TT h "77 r * ' * ~r , 7 ~r AT ~r Ar > ^i N 2 N n -i N n N n +i the difference between the two being o, _ c ._ M n +i ^ n+l ^n "77 ' N n+1 When the number of periods is indefinitely increased, then this difference becomes indefinitely small and we may write S n +l S n = dS. At the same time M n +i becomes very small, so that we may write M n+1 = dM, dM whence M = ........ (91) and therefore S = + B . . . . (92) when B = a constant of integration. The determination of the constant of integration may offer some difficulty. Since the limits of integration are from 8760 hours to o hours, we must use one of these limits in determin- ing this constant. This must be the former. For we do not know the value of (S) when N = o, but when N = 8760, then S = Si = l = -, where (Mi) is the load when N = 8760. Ni 8760 So then B = ^-, and our formula becomes 8760 CdM M\ / x 5 = j -T+^o (93) 2l6 UNIT COST DETERMINATION The integration is really carried out from right to left instead of the usual order. At the same time the difference between the maximum de- mand (M) during a given period and the mean load vanishes, so that L = M and the cost per unit of load per hour becomes a . (94) 136. If we plot the hours of operation horizontally as in Fig. 44A and the load (M) vertically, then we get a load-hour curve as OO'. On the~other hand, if we plot (N) vertically N FIG. 43- as before and the service modulus (S) horizontally as in Fig. 43, then the area under the curve thus obtained equals (M). This we can prove as follows. The area (^4) under the curve is dA = NdS. But or so that and dS- dM ' W dM = NdS, dA = dM, A = M, as stated. REPLOT 217 The N-S diagram is of special interest because it shows at once the load, hours of service, and service modulus. Ordinarily the load curve is given as in Fig. 44A. This is the load curve as it is obtained from a graphical recording watt- meter, the horizontal distance being the calendar time. But 730 1460 2190 2920 3650 4380 5110 58.40 6570 7300 8030 8760 Hours FIG. 44. Replot of a typical power plant load. to obtain even a roughly approximate equation between the load (M) and the time (N) would be impracticable for such a curve even if it were not impossible. For use in cost analysis, a replot is necessary in order to simplify the curve. The replot is made in descending values of (M). In order to do this, first plot on the ordinate (OM), Fig. 446, the point (Mo) being the maximum instantaneous value of the load (M) for the year. Then draw a hori- zontal line as AB, in Fig. 44 A, at a height (Mi). Then the number of hours that the system operates at a load MI or over is given by the sum of the intercepts as shown, having a total value of (Ni). With coordinates (Mi) and (N^) we can now plot the point (P) in Fig. 446. By shifting the horizontal line AB, Fig. 44A, we can thus get as many points 2i8 UNIT COST DETERMINATION as we desire on the replot of Fig. 446 and thus determine this curve. The equation for this is comparatively very simple, a common algebraic equation of the third degree usually giving a sufficiently close approximation. In Fig. 45, 10 20 30 40 50 60 70 80 90 100 110 120 FIG. 45. Load curve and replot of the Portland Central Heating Co. * we show the load curve and replot of the Portland Central Heating Co. from August to December 1914. 137. Example 44. For a certain power plant, the load hour equation is found to be M = 12,000 o.oooi5./V 2 . The cost per kw. of the installation is $70 and the fixed charges 10%. If the operating costs are 0.2 cent per kw.h. deter- mine the total cost per kw.h. Solution: Here dM = and so that When = 0.0003 S = - o.ooo^N + B. . N = 8760 hrs., fNdN J IT' (a) EXACT UNIT COSTS 219 then M = 12,000 0.00015 (8760) 2 = 490 kw., for which S = Substituting these values in equation (a) above, we get 0.056 = 0.0003 X 8760 + B B = 0.056 + 2.628 = 2.684. So that equation (a) becomes S = 2.684 ~~ 0.0003 A 7 ", and the cost per kw.h. is v _ 2.684 o.ooo$N or ^ K 2 = M 18.788 - 0.002IN 70 X o.io.-f- 0.002. + O.OO2. 12,000 0.00015^2 We may tabulate values of the cost per kw.h. for various values of A^, as follows: TABLE 91 N Ki N Kt 0.0035 5000 0.00300 1000 o . 0034 6000 o . 00294 200O 0.0033 7000 0.00288 3000 0.0032 8000 0.00283 4000 0.0031 8760 0.00280 You will note that the cost of production of power is least during the periods of least load and greatest during the periods of greatest loads. If now the price or rate is made throughout proportionate to the cost, as it should and must be, the rate will be lowest during the periods of lightest load, thus encourag- ing the filling of the valleys in the power plant's load curve, reducing throughout thereby the cost and rate of the power service. 220 UNIT COST DETERMINATION PROBLEMS 1. A 10 h.p. pumping plant runs at full load for a variable number of hours per year. It costs $500 with 10% of fixed charges. The cost of opera- tion is 20 cents per hour. Determine the total annual and unit cost of production of the service. Plot the curves. 2. A 25 kw. electric lighting plant carries a load of 25 kw. for 1000 hours per year and a load of 15 kw. for an additional 3000 hours per year. Deter- mine the total annual and unit cost of production of the power, if the plant costs $80 per kw. of capacity, with 12% fixed charges and operating costs of 1.2 cents per kw.h. delivered. 3. A power plant cost $60 per h.p. installed, with fixed charges of 12%. The operating costs are 0.3 cent per h.p.h. The plant carries a load of 1800 h.p. for 400 hours per year. During this 400 hour period the maximum demand is 2200 h.p. During the balance of the year, the plant carries a mean load of 600 h.p. with a maximum demand of 700 h.p. Determine the total annual, and the unit cost of production of the service during each period. 4. A loo mile branch railroad is built having a capacity of 200 cars of freight per day, and costing $2,000,0x30. The fixed charges are 13%. The cost of operation is 15 cents per car-mile loaded and 10 cents per car-mile empty. During two months in the year, the system carries an average of 1 80 loaded cars out, with a maximum of 200, and an average return of 50 loaded cars with a maximum of 60. During the balance of the year, an average of 80 cars per day is carried out to the main line loaded, with a maximum of 100 and there is an average return of 20 cars with a maxi- mum of 30. Cars cost $3000, locomotives $20,000, with a capacity of 20 cars. The weight of each locomotive is the same as a loaded freight car. Determine the total annual and unit cost of service per car-mile during each period. 6. A steam engine generating plant costs $75 per kw. of capacity, with 10% fixed charges. It carries a mean load of 5000 kw. for 60 hours per year, with a maximum of 8000 kw., a mean load of 3000 kw. for 1500 hours per year with a maximum of 4000 kw. and a mean load of 1000 kw. for the balance of the year, with a maximum of 1500 kw. The cost of operation is o.i cent per kw.h. with a mean load of 5000 kw., 0.12 cent per kw.h. when the mean load is 3000 kw.h. and 0.14 cent when the mean load is 1000 kw. Determine the total annual, and unit cost of production of the service during each period. 6. What could we afford to pay per kw. of capacity for a turbo-generating plant to take the place of the steam engine generating plant of problem (5), if its costs of operation are, for the mean load of 5000 kw., 0.12 cent, for the mean load of 3000 kw., 0.14 cent and for the mean load of 1000 kw., 0.17 cent, other things being equal? PROBLEMS 221 7. A power plant has an annual load-hour curve as given by the equation M = 5000 - 2oN 4- 0.0024 N* and the cost of operation per h.p.h. is 0.2 cent. The first cost of the plant is $50 per h.p. of capacity installed, with 12.5% fixed charges. Determine the total annual, average unit, and true unit cost of production. Plot the curve of the variation of true unit cost with the variation of the load. 8. A power plant carries a load as given by the equation M = 50,000 5./V. The first cost of the plant is $40 per h.p. of capacity, installed, with 10% fixed charges. The cost of operation per h.p.h. varies according to the equation A = 0.3 0.00004 M. Determine (a) The total annual cost of production per h.p. (b) The average cost of production per h.p.h. (c) The true cost of production per h.p.h. (d) Plot the curve of the variation of the true cost of production of the service per h.p.h. with the variation of the load. CHAPTER VII DETERMINATION OF SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY Variable Operating Cost. Total Production Cost. Determination of Equation of Actual Load Curve. Analysis of Heat Transmission. Point of Best Financial Efficiency. Determination of Size of System for Best Financial Efficiency. 138. In the past we have assumed that the operating cost per h.p.h. was a constant. This is far from true, especially in smaller sized plants and this consideration must modify our conclusions considerably where best economy is aimed at. So also within any large plant, the unit operating cost at full load will be considerably less than at fractional loads. If then we call ai, #2, #3, etc., the operating costs, during given periods and corresponding loads, we can determine our costs as before, but modified by this variation in operating cost. For the first period, then, the total annual cost K n is K u = Micp + and the cost per h.p.h. is i i During the second period, the total cost for the excess load is where 0% is the excess operating cost per h.p.h. required over that of the first period. The cost of operating the unit LI for the N z hours is = cp N 2 + 222 VARIABLE OPERATING COST 223 so that the total cost of the second period is the sum of K i2 ' and Kn ", or cp + (L,a, + Z, 2 a 2 )tf 2 . (95) If we now call a the mean operating cost for the load (Li + Lz) then (Lidi + L 2 a 2 ) = (Li -f 1,2)0, and #12 = N* + M 2 # + (Li + La) The cost per h.p.h. during this period is then So in general -^2n = -5"n * -J-f- H~ ^nj -^ n where L n ' = LI + L^ + + L n , and the service modulus Ml , ^2 , , Mn ^ w ~^ 1 " i "]V 2 H h ^V n ' And so, also, when the length of the period becomes in finitesmal, then where 5=1 as before, and (a) is variable. Knowing then M = f(N) , and a = F(Af ) either by exact or approximate equation, a complete solution of the costs can be made, either of an existing system, or in advance of its construction. 139. One of the advantages in using instantaneous costs per h.p.h. as given by the equation 224 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY is that the conditions for minimum costs can be determined by the usual method of equating the first differential to zero, and the system designed to meet these conditions. If in the cost determination, the entire system is treated as a unit, so that the costs include the transmission and distribu- tion as well as plant costs, then the above determination for minimum costs will determine both the size of plants and the corresponding areas of distribution for any given set of con- ditions, such that maximum financial efficiency will be attained, i.e., so that the service will be rendered at least cost. In determining the best size of a system for minimum cost, we must consider the unit first cost (c) of the system as vari- able, since this varies with its size. Once this size is deter- mined, (c) becomes fixed and has a definite constant value. Since the instantaneous production cost per unit of size (h.p.) per hour is the cost per hour for the size (M) will be K 2 ' = Scp + Ma, and the total production cost per year will be /87<5o /"876o ScpdN + I MadN . . . (96) The condition for minimum total production cost is then dK _ dM = The operating cost varies not only with the size of the units but also with the per cent- of full load that they are carrying, because their efficiencies vary with fractional loads often very greatly. The handling of the entire subject is quite complex but this should give peculiar satisfaction to engineers for it insures that in future the matter must be taken out of the hands of amateurs. Example 45. A power plant carries a load as shown in Fig. 46. If the cost of the plant per h.p. is $50, fixed charges MINIMUM PRODUCTION COST 225 are 12%, and the operating cost per h.p.h. is 0.8 cent, determine . (a) The cost per h.p.h. (ft) The total annual cost. Solution: The first step necessary is to determine the equa- tion of the curve shown in Fig. i. In order to do this, we as- sume the equation to be M = M + AN + BN 2 + CN* + DN* . . . . (i) The longer the equation, i.e., the more constants, the more nearly we can approximate to the true curve. For each 6000 5000 4000 3000 2000 1000 1000 2000 3000 4000 5000 6000 7000 8000 8760 Hours FIG. 46. Load curve for example 45. constant in the equation we can make the approximate curve run through one point on the true curve. Thus in the above equation a there are five constants, and we can therefore make the approximate curve run through five points on the true curve given. If we had another term (EN 5 ) then we could make it run through a sixth point and therefore approximate to the true curve just so much more closely. If we express (M) in thousands of kilowatts and (N) in thou- sands of hours, then we can choose the following five points 226 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY of the true curve through which to make the approximate curve as given by equation i run. Thus, when N = o, M = 5, # = 2, M = 4, N = 4, M = 2, N = 6, M = i, AT = 8, M = 0.9. Substituting in equation i, the value N = o and If = 5, we get M 5, so the equation becomes M = 5 + AN + BN* + CA T3 + IW 4 . . . (2) Our task is now to determine the values oi A, B, C, and D. We shall follow this through in considerable detail. Substitute in equation 2 the value N = 2 and M = 4, and we get 4 = 5 + 2^+ 4^+ 8C + i6D, or o = i + 2 A + 4B + 8C + i6Z) .... (3) Then substitute in equation 2 the values A r = 4 and M = 2, and we get 2 =*= 5 + 4,4 + i65 + 6 4 C + 256^ or o = 3 + 4A + i6B + 6^C + 256!) ... (4) The substitution in equation 2 of the values N = 6, M = i gives o = 4 + 6^4 + 36^ + 2i6C + 1296!), or o = 2 +3^ +i85 + io8C + 648!). . . (5) And finally the substitution of the values N = 8, M = 0.9 gives o = 4.1 + 8^4 + 64$ + 5i2C + 4096,0 . (6) We have thus the four following equations: o = i + 2A + 4B +8C + i6D (3) o = 3 + ^A + i6B + 6 4 C + 2$6D . . . (4) o = 4 + 6A + 36,8 + 2i6C + i2 9 6D . . (5) and o = 4.1 + 84 + 64.8 -f- 5i2C + 4096^) . (6) We can now eliminate (A) between equations 3 and 4 by mul- tiplying equation 3 by 2 and subtracting from equation 4, thus = 2+4^+ SB + i6C + 32Z> o = 3 + 4A + i6B + 64C + 256!) ... (4) subtracting, o = i + SB -f 486' + 224!) ... (7) LOAD-HOUR CURVE 227 We can also eliminate (A) between equations 3 and 5 by multiplying the former by 3 and subtracting from the latter, thus o = 3 + 64 + i2B + 24C + 4 8> o = 4 + 6A + 36^ + 2i6C + 1296!) subtract o = i + 24$ + iQ2C + 12487) . . (8) And finally, we can get a third equation with (A) eliminated, by multiplying equation 4 by 2 and subtracting from equation 6, thus = 6 + SA + 32^ + I28C + 512!) o = 4.1 + 8.4 + 64^ + 5i2C + 4096!) subtract o = -- 1.9 + 32$ + 384C + 3584!) (9) This gives us the three following equations: o = i + SB + 48C + 2240 ...... (7) = 1 + 24^ + iQ2C + 1248!) ..... (8) o = -- 1.9 + 3 2# + 3 8 4 C + 3584,0. . . (9) We now proceed to eliminate (B) from these equations. Multiply equation 7 by 3 and subtract from equation 8. This gives = 3 + 24$ + i44C + 672!) 0=1 + 24# + IQ2C + I24&D subtract o = -- 2 + 48C + 576^ ..... (10) Also multiply equation 7 by 4 and subtract from equation 9. This gives o = 4 + 32$ + I92C + Sq6D o = -- 1.9 + 32^ + 3 84C + 3584^ subtract o = - 5.9 + 1926' + 2688> . . (n) We have now only the two equations following, with only two unknowns, C and D. o = - 2 + 4 8C + 576!) ..... (10) o = - 5.9 + i92C + 2688Z). Multiply equation 10 by 4 and subtract from equation n. This gives o = 8 + I92C + 2304!) o = -- 5.9 -f ! 92 C + 2688D subtract o = -- 2.1 + 384!) -r So that D = -- = 0.00547 228 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY By substituting this value of D in equation 10, we get o = - 2 + 4 8C - X 576, 34 or o = - 2 + 48C - 3.15. So that C = -S^S = 0.1073. 48 We have now the values of both C and D. Substituting these in equation 7 gives 48 384 or o = i + SB + 5.15 1.225. So that o = SB + 4.925, and B = - =^p = - 0.6156. 8 Dividing equation 3 by 2 gives o = 0.5 + A + 2B + 4 C + 8J9, and substituting the values found for B, C, and D gives o = 0.5 + -4 1-23125 + .429167 - .04375, or o = 0.929 1.275^4, so that A = 0.346. Substituting these values in equation 2, we get M = 5 + 0.346A 7 " 0.6156^ + 0.1073^ 0.00547^ (12) as the approximate equation of the curve. The crosses (x) in Fig. 46 indicate points calculated from this equation. The conformation is very close. We can now determine the service modulus (5), since by equation 12 dM = (0.346 - 1.2312^ + 0.3219^2 - o.o2iSSN 3 )dN, . '^dM O I ~rr~j N ' 76 (0.346 - 1.2312^ + 0.3219^ - o.Q2i88# 3 ) dN, N LOAD-HOUR CURVE 229 or 5 = 0.346 log e Ar-i.23i2A r -f-o.i6ioA r2 -o.oo729/V' 3 -f-C. 1 (13) Where C' is a constant of integration. By equation 12, when N = 8.760, M = 0.71, so that when N = 8.760 S = jl = 0.08105. 8.760 Substituting this value of (S) and of (N) in equation 13, gives 0.08105 = 0.346 X 2.1702 1.2312 X 8.76 + 0.161 X (8.76) 2 - 0.00729 X (8.76) 3 + C 1 , so that C' = 2.6612 and 5 = o.3461og e ^V i. 2$i2N+o.i6iN 2 0.00729^ + 2.6612 (14) The cost per kw.h., is * = ir + M S S or K = 50 X o.i2 - + 0.008= 6 + 0.008. M M Substituting the values of (S) and (M) found, we get _ 6(0.346 logeN 1.2312^ + o. i6iN 2 0.00729^ + 2.6612) 1000(5 + 0.346^ 0.6156^ + o.iojsN 3 o. 00547^ + 0.008 ................ (15) as the cost per kw.h. in terms of N. (Ans. a.) The total cost per hour for the load (M) is K l = 6ooo5 -f o.ooSM X iooo 2 = 6oooS + SoooM, where (M) is expressed in thousands of kilowatts and N in thousands of hours. The total production cost per year is /8.7<5o /*8.76o MK'dN = 6000 / SdN Jo /. 7 6o MdN . . (16) 230 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY The integral f*-ff%-f%x* =/'".(, = 5000, dM since dS = -7- N The total fixed charges per year are 6 / S dN = 6000 X 5000 = $30,000. The number of millions of kilowatt-hours (thousands of kilo- watts multiplied by thousands of hours) produced per year is given by = r + 0.346^ - 0.6156^2 + 0.1073^ _ o. = 5 X 8.76 0.173 X 76.7376 0.2052 X 672.2214 . . '+ 0.02682 X 5888.6595 0.001094 X 51584.66 = 43.80 - 13,276 - 137.94 + 157-934 ~ 56.434 = 20.636 millions of kilowatt-hours, costing = 20.636 X 8000 = $165,088 whence the total annual cost is 30,000 + 165,088 = $195,088. (Ans. ft.) Example 46. If in the above example the operating cost per thousand kw.h. is given by the equation a = (10 - M) between 500 and 5000 kw., where M is again expressed in thousands of kw., (a) Determine the cost per kw.h. and (b) Determine the total annual cost. Solution: In the previous problem, the operating cost (A) was assumed constant. In this problem, it is assumed variable TOTAL PRODUCTION 231 in accordance with the above equation. It is evident that the value of (S) , the service modulus, will not be affected by varia- tions in the operating cost. Equation 15 above will not be changed except in the last term, which must be changed to the new value. We therefore get, for the cost per kw.h. ( K) K _ 6(0.346 \ogeN 1.2312^ + o.i6i7V 2 0.00729^ -j- 2.6612) 1000(5 + 0.346^- o.6i56N 2 + 0.1073^ ~ 0.00547^) (10 M) /A v + i - '- (Ans. a), 1000 where as before M = 5 + 0.346^ 0.6156^ -f 0.1073^* o.ooStfN*. The total cost per year is then /.76o _ /*8.7 SdN + iooo 2 I aMdN. /o /8.76o S dN = $30,000 as before /8.y6o r 8.760 aMdN = iooo / (10 - M)M dN Jo /8.?6o (loM - M 2 )dN. Substituting for M its value in terms of (N) as given above, we get 1000^(25 - o.i2A T2 + 0.426N 3 - o.4527A r4 + 0.1359^ o. 01825 N 6 + 0.001174^ o.oooo3A rs )^A 7 ' = iooo(25A r 0.04^ + o.io65A r4 o. i = ioooA r (25 O.O4A 7 " 2 + 0.1065^ o.o9O54A^ 4 + O.O3365M 0.00261^ + 0.000147 N 7 0.0000033 A' 8 ),, 8 - 76 = 8760(25 - 3.07 + 71.6 - 533.0 + 1166.1 - 1177-1 + 580.7 - 115.4) = 8760 X 14.2 = $124,392 operating costs, so that the total annual production costs are Ki = 30,000 + 124,392 = $154,392. (Ans. 6.) 2 3 2 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY While, as may be seen from the above, the work of solving such cost problems requires skill and considerable painstaking effort, it certainly permits of real and conclusive deductions. 140. In the following problem of determining the size of a heating system for best financial efficiency, we have assumed the heat to be transmitted in water instead of steam. It may be of interest to note the comparative cost of transmitting heat energy by steam and by water. We can allow approximately a velocity of 100 ft. per second for steam and 5 feet per second for water, the velocity of steam being twenty times as great as that of water. The heat capacity of water is one B.t.u. per pound, or 62.5 B.t.u. per cu. ft. The heat capacity of steam varies with the pressure (density) in accordance with the following table. TABLE 92 HEAT CAPACITY FOR STEAM ABOVE 150 F. Pressure, Temp., Heat Heat per cu. ft. Heat per cu. ft. Ibs. abs. dry sat. per cu. ft. per degree F. per degree X 20 10 193 26.5 0.62 B.t.u. 12.4 15 213 39-o 0.62 12.4 2O 228 5i-5 0.66 13.2 SO 281 124 0-95 19.0 IOO 328 240 i. 35 27.0 150 358 357 1.71 34-2 250 401 583 2.32 46.4 350 432 814 2.90 58.0 The third column is obtained by dividing the total heat per pound of the steam above 150 F. by the number of cubic feet per pound of steam, thus getting the B.t.u. per cu. ft. The fourth column is then obtained by dividing the heat per cu. ft. by the temperature of the steam above 150. But inas- much as steam may be allowed to travel 20 times as fast as water, we have multiplied this column by 20, giving the last column. This shows 12.4 to 58 B.t.u. for steam as compared with water at 62.5 B.t.u. per cu. ft. per degree. A number of important factors must be borne in mind. If the pressure TYPE OF HEATING SYSTEM 233 is over ioo#, then extra strong pipe and fittings must be used. Since these are far more expensive than standard, the cost of transmission is greatly increased, which does not permit of a fair comparison with that of hot water at nominal pressures. Therefore the real capacity of steam is really only 12.4 to 27 B.t.u. as compared with water at 62.5. On the other hand, while water transmission requires as large a return pipe as outgoing pipe, that for steam may be very small, and if the condensate is wasted, the return pipe may be done away with altogether. But in the latter case, the cost of purchasing or pumping the water thus wasted must be added to the total service cost. But even without a return pipe for the steam, and allowing nothing for the cost of water thus necessarily wasted, we have 12.4 to 27 B.t.u. for steam as compared with half of 62.5, or 31.25 B.t.u., for water transmission. So that even under these conditions the cost of steam transmission is at least 16% greater than for water. The cost of water wasted usually runs from one to ten cents per 1000 Ibs., a fairly significant item itself. The conclusion cannot be avoided, then, that in spite of the fact that most central heating stations are designed for steam transmission, heat transmission by water is far more economical. If, in the hot water system, instead of heating water .in the boilers, we generate steam, using the steam for the generation of power, and using the exhaust steam (atmospheric pressure, 212) for the heating of the water, we have an incomparably more economical system. Example 47. Assume that the cost per bo.h.p. of a central heating plant varies as curve A, Fig. 47, and that the cost of transmission pipe per foot complete in place varies as curve A, Fig. 48, and that the operating cost is one cent per bo.h.p. hour, neglecting radiation loss in transmission, deter- mine the area of transmission for maximum economy; in blocks of 200 ft., if the maximum demand is 268,000 B.t.u. per block per hour and the mean annual load 67,000 B.t.u. per block per hour, fixed charges 10%. 234 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY $120 100 80 ffi I 60 40 20 GOO 700 800 900 1000 FIG. 47. First cost per horse power of heating plant assumed in example 47. .100 200 300 400 500 Bo. H.P. $2.40 2.00 1 |l.60 M d |l.20 ! 0.80 - 0.40 k s> J,s ^ ^^ .>* ^ ^ ^ s* ^ s* ^ ^ ^ ^ ^ ^ ^ ^ ^ Cross-sectional Area of Pipe U FIG. 48. Variation of cost of pipe with size assumed in example 47. SIZE OF HEATING SYSTEM 235 Solution: The total production cost per year is Ki = Mcp H- LaN, where p = o.io, a = o.oi, N = 8760. Assuming the shape of the area served to be square, the number of blocks on each side being (B) blocks from the plant, then the total number of blocks served will be 2B(B + i), in which B is always odd. Since each block requires 268,000 B.t.u. maximum, or 8 bo.h.p., the total maximum load will be M = 2B (B + i) X 8 = i6B (B + i) bo.h.p., and the mean annual load (L) will be L = J X i6B (B + i) = 4B (B + i) bo.h.p. whence the annual operating cost will be LaN = 4B (B + i) X o.oi X 8760 = 350.4^ (B + i) dollars. We must now determine (C), the cost of the system per bo.h.p. Curve A, Fig. 47, gives (Ci) the cost per bo.h.p. of the heating plant. So we must only determine (Cz)j the cost per bo.h.p. of the transmission according to the cost data given in Fig. 48. Allowing 200 F. as the maximum temperature of the out- going water and 150 as the temperature of the return water we can carry 50 B.t.u. per Ib or 50 X 8j = 417 B.t.u. per gallon. Since each block requires 268,000 B.t.u. maximum per hour, or 4470 B.t.u. per minute, there will be required 4470 -4-417 = ii gallons per minute per block. This will require, for a friction head of about one foot per hundred feet of pipe, a ij" pipe. The length and size of distribution pipe may be computed as follows: (1) For one square block transmission area* Pipe required 4 X i (/) of n g.p.m. capacity, where I = 200 ft. (2) For a square of 3 blocks on a side Pipe required is 4 (/) (i + 3) of 11 g.p.m. capacity. 4 (/) X 2 of 22 g.p.m. capacity. 236 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY (3) For a square of 5 blocks on a side Pipe required is 4 (/) (i + 3 + 5) of n g.p.m. cap. 4 (/) X 4 of 22 g.p.m. cap., 4 (0 X 2 of 44 g.p.m. cap. (4) For a square of 7 blocks on a side Pipe required is 4 (/) (i + 3 + 5 + 7) of n g.p.m. cap. 4 (/) X 6 of 22 g.p.m. cap, 4 (I) X 4 of 44 g.p.m. cap, 4 (/) X 2 of 66 g.p.m. cap. (5) In general for a square B blocks on a side Pipe required is 4 (/) (i + 3 + 5 + 7 + 9 + + B) ft. of ii g.p.m. cap, 4 (/) X (B - i) ft. of 22 g.p.m. cap, 4 (/) X (B - 3) ft. of 44 g.p.m. cap, 4 (/) X (B 5) ft. of 66 g.p.m. cap, 4 (/) X (B - 7) ft. of 88 g.p.m. cap, etc. The cost of the i|" pipe according to Fig. 48 is 34 cents per foot, and since the cost in place is assumed in direct propor- tion to the area of the pipe, the cost per foot of the 22 g.p.m. pipe is 34 X 2 = 68 cents per ft.; of the 44 g.p.m. pipe is 34 X 4 = $1-36 per ft, etc. The total length of i|" pipe required is 4 X 200 (i + 3 + 5 + 7 + + B) ft, 800 X ( B + ] = 200 (B + i) 2 ft, \ 2 / and its cost is 200 (B + i) 2 X 0.34 = 68 (B + i) 2 dollars. The cost of the 22 g.p.m. pipe is 4 X 200 (B - i) X 0.68 = 544 (B - i) dollars; of the 44 g.p.m. pipe, it is 4 X 200 (B - 3) 2 X 0.68 = 544 X 2 (B - 3). SIZE OF HEATING SYSTEM 237 Of the 66 g.p.m. pipe, it is 4 X 200 (B - 5) 3 X .68 = 544 X 3 (B - 5), etc., up to 4 X 200 X 2 X $7000 11 Gpm. X - 68 = 544 X 2, FIG. 49. Heat distribution layout for example 47. whence the total transmission cost is C 2 = 68 (B + i) 2 + 544 (B -i) + 2 (B -3) + 3 (B - 5) + 4 (B - 7) + 5 (B - 9) + Letting X 2 we get C 2 = 68 (B + i) 2 + 5447. 238 SIZE OF SYSTEM BEST FOR FINANCIAL EFFICIENCY Dividing through by M = i6B(B + i) gives the cost of transmission per bo.h.p., or _ , _ (B + 1) 34 J B B(B + i) The variation of (J) and C 2 ' with (B) is as follows: B i 3 5 7 9 ii 13 *5 i7 19 21 / o 2 8 20 40 70 112 168 240 330 440 Cz' 8.5 11.3 1.42 17.0 IQ.8 22.6 25.6 28.3 31.2 34.0 36.8 Assuming now that we can represent the cost of the heating plant (Ci) per bo.h.p. by the following equation between 600 and 1 200 bo.h.p., Ci = 61.66 O.O35-M" + O.OOOOH33M 2 . We get, since M = i6B(B + i), that Ci = 61.66 - 0.035 X i6B(B + i) + 0.00001133 X 16 X i65 2 (5 + i) 2 or Ci = 61.66 - 0.56^(5 + i) + o.oo2 9 5 2 (5 + i) 2 . Where the total first cost (C) of the system per bo.h.p. is /~* S~** I /""* / or C = 61.66 - 0.565(5 + i) + o.oo2 9 5 2 (5 + i) 2 B the annual fixed charges per bo.h.p. are Cp = 6.166 - o.o$6B(B + i) + o.ooo29^ 2 (5 -f i) 5 5(5 + i) Since the total operating costs per year are LaN = 350.45(5 + i), the cost per installed bo.h.p. LaN = 350.45(5 4- i) M i65(5 + i) = 2I ' 9 ' SIZE OF HEATING SYSTEM the total production cost per bo.h.p. per year is therefor . K = 21.9 + 6.166 - o.o$6B(B + i) 239 0.425 or K = 28.066 - o.o$6B(B + i) + o.ooo29# 2 ( + i) 2 Differentiating (K) with respect to (B) and equating to zero, will give very complex results. We can, however, make a solution by plotting. Thus B ............. i 3 5 7 ii 13 15 17 19 K 28.7 29. i 28.1 27.5 27.4 28 30 32.8 41.2 52 67.8 The maximum financial efficiency evidently takes place at B = 9, an area of service of 180 blocks. Under this condition M = i6B(B + i) = 16 X 9 X 10, M = 1440 bo.h.p. The curve for the variation of the production of service cost per bo.h.p. per year with the length of area served is given in 5 21 19 17 15 9 5 3 1 ' P - -^z* ^ ~ =T *^ ^. ,* s^> t *^- ^ ^ * / /' / \ \. \ . ^ ^* $30 $40 $50 $60 Cost per Bo.H.P. Year FIG. 50. Variation of the unit cost of service with the size of the system according to example 47. 2 4 o SIZE OF SYSTEM BEST FOR FINANCIAL EFFICIENCY Fig. 50. Note that after the point of best economy is passed the increase in cost is very rapid. The cost of service of a heating system that was considerably too large would be far in excess of the individual plant costs. However, the results obtained above must not be generalized, as they apply only under the conditions assumed in the problem. But under these conditions, a steam heating system is still more expen- sive and capable of serving only a smaller area economically. Another factor that must be borne in mind is load density. We have in the above problem assumed uniform load density. The load density may be very much greater or less than that assumed, while on the other hand the load density, instead of being uniform, may vary over the area in a great many essen- tially different ways. The effect of these variations on the system's size and economy afford study of the most valuable kind in practice. PROBLEMS 1. Assume the annual load curve to be as in Fig. 51, the first cost of the power plant to be $60 per h.p. installed., fixed charges 12%, and operating costs to be 0.6 cent per h.p.h. 5000 4000 ^3000 2000 1000 \ 1DOO 2000 3000 4000 5000 6000 7000 8000 876.0 Hours Fig. 51. Load curve for problem i. PROBLEMS 241 (a) Determine the total annual cost of production, (b) Determine the cost of production per h.p.h. and (c) Plot the curve of variation of cost per h.p.h. with the load. 2. If in problem (i), the operating cost in dollars per h.p.h. is given by the equation a = 0.01125 0.00000125 M and all other things are equal, determine (a) the total annual production cost, (b) the number of h.p.h. produced per year, (c) the cost of service per h.p.h. and (d) plot the curve of variation of cost of service per h.p.h. with the size of the load. 3. In a power plant the load varies with time according to the equation M = 5OOO + 2N O.OO02JV 2 , and the cost of operation per kw.h. varies according to the equation a = 0.015 o.oooooiAf. The first cost of the plant per kw.h. is $75, fixed charges 10%. Determine: (a) the total annual production cost, (b) ' kw.h. produced per year, (c) " average cost per kw.h., (d) " true cost per kw.h., (e) " maximum cost per kw.h. and (f) "minimum " " kw.h. CHAPTER VIII DETERMINATION OF TYPE AND SIZE OF UNITS Stand-by Units. Number of Units in a Plant for Given Load for Best Financial Efficiency. Economy of Units at Fractional Loads. 141. In considering the matter of the proper selection of the size of units in any given plant, too much emphasis has been placed on the necessity of a reserve unit. Like most things, a reserve unit is desirable in inverse proportion to its cost. When this cost rises above the probable gain, it is no longer good judgment to provide it. The first question is, how long will a given piece of ma- chinery run, without a shutdown, under conditions of aver- age care? While this is a very difficult matter to decide on, yet as a rule the number of hours of run without shutdown will be in direct proportion to the life of the equipment. But with this, we must consider the nature of the equipment as well. For example", the life of a motor is twice that of an engine, therefore we may expect twice as long a run. But as a matter of fact, we will get more than that, because the shock of reciprocating parts is absent in the motor. So as a whole, the period of continuous run is far greater for all rota- tive classes of machinery, such as generators, turbines, centrif- ugal pumps and the like, than for the reciprocating classes, as engines, compressors and such. < That shutdown will occur is certain, for no machine is built to run continuously during its entire life, without certain parts being either adjusted or repaired, or entirely replaced. Among such parts are bearings, crosshead guides, stuffing boxes and the like. However, the period of necessary shutdown may be anticipated, and for that matter usually must be an- ticipated to avoid severe damage to the machine, due to break- .242 STAND-BY UNITS 243 down. And its repair may therefore be accommodated to the load being carried. But this is not the primary consideration in the matter of dividing our load among a number .of units. The menace of shutdown is not nearly so great as one would be lead to believe by the emphasis placed on it by so many authors. First-class machinery, run with average care, under normal loads, may be depended on to run at least a certain period without any danger of shutdown. They will then require certain repair, after which they may again be relied upon with reasonable certainty. 142. The reason for the design and installation of any system is the rendering of a given service. The primary object for the subdivision of the plant into several units must be to render this service at as low a cost as possible. We must therefore take into consideration a number of primary factors that con- trol this. These are as follows: 1. For larger units: (a) Decreased unit first cost of the equipment with in- creased 1 size. (b) Increased efficiency under similar conditions for the large units. 2. Against larger units: (a) Decrease in efficiency at fractional loads. The larger the unit, the smaller per cent of full load the unit will run at, under average conditions. For example, if the maximum load on a certain power plant is 10,000 kw. and the mean load is 3000 kw., then if we have only one unit, it will run on the average at only 30% load, with corresponding low efficiency. If on the other hand we put in two or more units, we can shut down one unit after an- other as the load decreases, maintaining much nearer full load conditions on those units that are kept in operation, and get- ting correspondingly much higher efficiency under actual running conditions. 244 TYPE AND SIZE OF UNITS Calling (^4) the attendance cost per h.p.h., (F) the fuel cost per h.p.h. at full load and (P) the per cent of full load efficiency at (X) per cent of full load then at full load, the operating cost (a) is a = A + F per h.p.h ...... (98) and at fractional loads, the cost of operation is -p (99) the load being under these conditions ( XM) . If we are actually carrying some load as may be given by some equation as N =f(M), then the operating cost per hour is A + 4 and the total operating cost per year is 8 7 6o / and thus the production cost per h.p.h. is and the total production cost per year is /8 7 6o /*8 7 6o / p\ ScdN + / \A +j 5 )MdN. (101) 143. The above are the conditions for a single unit or where all the units are operated, each carrying a uniform proportion of the entire load. But it certainly is not good practice, where there are a number of units, to operate them all at small loads, when it would be possible to shut one or more of them down and operate the balance at more nearly full load. Assuming then that we have a number of units, MI, Mz, M S) ... in our plant all of the same size, and that we operate them so that all which are in operation will be at full load SERVICE COST 245 except one, then for any load (M) we will have, let us say, ( 7) units of (Mi) size operating at full load and one at fractional load. Under these conditions the total attendance cost per hour is (AM) and the total fuel per hour is where UMi + XMi = Mi(U + X) = M, . . (102) and (X) is the per cent of full load on the unit that is run- ning light and (P) is a function of (X), which is obtained as an approximate equation from the actual test (or factory guarantee) curves. Whence the total operating cost per hour for the load (M) is AM + UMiF + M 1 = AM + 17 But since M 7 + X and P=f(X), the total operating cost is = M A + (U+ X) (104) and the operating cost per h.p.h. (a) is u \ a = A +F- .-..". (105) (U+ X) So under these conditions, the unit production cost is 1 .Y _ Scp w K,=^ + A FlU + f(X) (U + and the total annual production cost is F . .(106) /876o /*87 ScdN+ I M A+- (U + -L-} \ f(x)) 246 TYPE AND SIZE OF UNITS If the size is already determined, then (c) is a constant and the first part of (K^) above is simply pc I S dN. But fsdN=fNdS since both give the area under a curve the equation of which is o i So also Whence sdN= N dS = tf = *M = M. fsdN= f* Bearing this in mind will always simplify this part of the calculation. 144. In the above, the fixed charges must be based on the total installation, while the operating cost depends on the units in operation. In practice we should have to determine the total annual operating costs (a) by a series of integration instead of only one. If, for example, we have ( U + i) units in the plant, we would integrate from ( U + i) to (/)> then from U to ( U i), and so forth, i.e., always from X = 100% to X = o, or if the units will carry say 25% overload safely, and run with better economy than below 25% load, then our integration steps would have to be between X = 125% and X = 25%. The reason for this is that at the point of shut- down of a unit a point of discontinuity is introduced in our curve. It is evident then that units like generators, steam engines, and the like, that are technically capable of carrying overloads, should be built mechanically well to carry this overload for a reasonable per cent of their running time without giving trouble, in order to allow the operator to get the lowest pro- duction costs. Invariably better economy is obtained at reasonable per cent overload than near zero load. For ex- ample, a steam engine will use about ij% more steam be- tween full load and 25% overload, at J load it will use 60% DETERMINATION OF NUMBER OF UNITS 247 more steam, and below that still more. A steam turbine will use i% more steam at 25% overload than at full load, while at J load it will use 15 to 40% more steam, and so on. In the determination of the number of units (of equal size) required in a plant for best economy, we assume that there are (n) units and then find what the total cost of production is in terms of (n). By plotting this equation, or by differen- tiating it and equating it to zero, we can determine the value of (n), the number of units required for minimum production cost, and thus the size of each unit. This is comparatively simple. But we can obtain better efficiency usually by using dif- ferent sizes of units rather than a number of units all of the same size. In such cases, fractional determination will give the desired results. That is, when the load curve consists of two or more distinct parts, one, let us say, of short duration but a heavy load, and the other of long duration and com- paratively light load, we can divide the curve into two parts, considering each separately. After we have determined thus the units best for each part of the load, we can then make such adjustments as will bring the two or more parts into harmony. 145. Example 48. A producer gas power plant carries a load in kw. as given by the equation M = 1000 o.i AT". The cost per kw. of the units is given by the equation C = 50 0.005 Uj where U is the size of the unit. The fuel cost per kw.h. at full load varies with the size of the unit thus: F = 0.007 0.0000006 U. At fractional loads, the per cent of full load efficiency ob- tained is P = 0.5(1 + X), where (X) is the per cent of full load. The attendance cost is constant at $0.003 P er kw.h. 248 TYPE AND SIZE OF UNITS Assuming fixed charges at 10%, how many units of the same size will be required to give minimum production cost? Solution: (i) Total fixed charges per year are M cp = 1000 X o.io X (50 0.005 U) = 5000 - 0.5 U, where M U = -, (n) being the number of units employed in the plant, n , TT 1000 whence U = , n , ,, . 1000 ^oo and M cp = 5000 0.5 H = 5000 - n n (2) The total attendance cost per year is A = 0.003 I M dN where I M dN gives the total number of kw.h. produced per year /8?6o (1000 o.iN)dN = 0.003 (loooN - o.o5N 2 )o 8760 or A = 0.003 X 4,923,120 = $14,769.36. (3) Fuel Cost: If the unit runs at full load, the fuel cost is (F) per kw.h. given above. If it runs at x% of full load, the /F\ fuel cost is | - l = f. (a) Case of one unit: T . . v M M In this case X = = M 1000 since U = M = 1000 kw., so that P = 0.5(1 + o.ooiM) and F = 0.007 ~~ 0.0000006 X 1000 = 0.007 ~~ 0.0006, or F = 0.0064, and the actual fuel cost is , _ F _ 0.0064 0.1028 , , ~ P 0.5(1 + o.ooiM) ~ (i + o.ooi'AO PC DETERMINATION OF NUMBER OF UNITS For the load (M) the fuel cost per hour is TUft _ O.OI28 M (i 4- o.ooiM) and the total annual fuel cost is >.oi28 MdN 249 But since and whence so that + o.ooi M) M = 1000 o.iNj 1000 M , r N = O.I dN = lodM, ( , r v = 10(1000 M) /MfdN = -0.128 r J 1000 (.1 MdM + o.ooi M) o.ooiM) - (O.OOIj = $38,400. Summary of Case (a), one unit. The fixed charges are ^ 5000 - - = $4500. The attendance costs are $14,769.36, and the fuel cost is ,400, whence the total annual production cost is 4500 + 14,769.36 + 38,400 = $57,669.36. (b) Case of two units: In this case we have two units each of 500 kw. The first unit runs at fractional load from C to B, then at full load from B to E. The second unit cuts in at B, carrying the load area BDE. Or the units may carry the load area ABDO, both being equally loaded. Since M = 1000 o.i^V, when M = 500, N = 5000 hours. The fuel cost for the load BC is 760 /8 7 6o p MfdN = M-dN. o / 5000 * 250 TYPE AND SIZE OF UNITS But when U = 500, F = 0.007 0.0000006 X 500.0 = 0.0067 and X = = 0.002 M , whence BC 500 13760 M X 0.0067 5000 0.5(l + 0.002 M) MdM - 0.134 -L r Jsoo I 0.002 M or BC = - 0.134 (0.002) 2 $939340. So also BD = - o. FIG. 52. Case of two units. MdM + 0.002 M whence and finally - log e (l + 0.002M) Jsoo BD = $10,281.15 /Sooo 500 X 0.0067 1000 = $16,750. soo UNITS IN A PLANT FOR GIVEN LOAD Summary of Case (b) : The total fuel cost is 16,750 + 10,281.15 + 9393-40 = $36.424.55, and the total fixed charges are 251 * 5000 - ^ = $475- The total annual production cost is therefore 36,424.55 + 4750 + 14,769-36 = $55,943-91. (c) Case of Three Units: In the case of three units, each is of 333.33 kw. size. The first unit runs at fractional load until it reaches full load, after M FIG. 53. Case of three units. which it continues to run at full load, while the excess is taken care of by cutting in a second unit. When the load exceeds 666.67, the third unit is cut in. In this case the fixed charges are 5000 ^ = $4833.33. And as before, the attendance charges are $14,769.36. With units of 333.33 kw. size, the full load fuel cost per kw.h. is F = 0.007 ~ 0.0000006 X 333-33 = 0.0068 and P = 0.5(1 + X), M where X = 333-33 = 0.003^. 252 TYPE AND SIZE OF UNITS So that P = 0.5(1 + o.oo3M), and / = - = - - 68 = - 136 P 0.5(1 + o.oo3M) (i + 0.003 For the load area EDHJ, the fuel cost is ED = flffdN = f, M *- 013 * X dN, J J (i + 0.003^) ED ' f MdM - o.i 3 6/ since M = 1000 o.iN. We obtain the limits of the above integration for the value of Mj when N = 8760, or M = 124, whence zrn * f 124 MdM -136 f/ ED = - 0.136 / - = - ^ (i-f .oojJ) ^333.33(1 + -oo 3 M) (.oo 3 ) 2 L r4 = $3795-91, 3-33 when M = 333-33, N = 6666.67 hours and M = 666.67, N = 3333-33 hour s- The fuel cost during the load area 707 and GFIK is therefore respectively El = 333-33 X 6666.67 X 0.0068 = $15,111.11 and FG = 333.33 X 3333-33 X 0.0068 = $7555-55- The fuel cost during the load area FEK is r Mdu FE = - 0.136 / - -, ./333. 33 ( J + 0.003