LIBRARY OF THE UNIVERSITY OF CALIFORNIA. Class BOILERS THE POWER HANDBOOKS The best library for the engineer and the man who hopes to be one. This book is one of them. They are all good and they cost $ 1.00 postpaid per volume. (English price 4/6 postpaid.) -SOLD SEPARATELY OR IN SETS BY PROF. AUGUSTUS H. GILL OF THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY ENGINE ROOM CHEMISTRY BY HUBERT E. COLLINS BOILERS KNOCKS AND KINKS SHAFT GOVERNORS PUMPS ERECTING WORK SHAFTING, PULLEYS AND PIPES AND PIPING BELTING BY F. E. MATTHEWS REFRIGERATION. (In Preparation.) HILL PUBLISHING COMPANY 505 PEARL STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E. C. THE POWER HANDBOOKS BOILERS COMPILED AND WRITTEN BY HUBERT E. COLLINS OF TM.; ' Of 1908 HILL PUBLISHING COMPANY 505 PEARL STREET, NEW YORK 6 BOUVERIE STREET, LONDON, E.G. American Machinist Power The Engineering and Mining Journal Copyright, 1908, BY THE HILL PUBLISHING COMPANY All rights reserved Hill Publishing Company, New York, U.S.A. INTRODUCTION THIS volume endeavors to furnish the reader with much new and valuable material on an old subject, together with much standard information which every engineer likes to have at his hand. A glance at the chapter headings will show the scope of the book. It will be seen that the subject is pretty fully covered from the working conditions inside of a boiler to simple talks on the various phases of boiler practice. It also covers the design of boiler furnaces for wood burn- ing, and much other useful material. One very important feature is the portion on the safety valve based on Mr. Fred R. Low's supplement to Power on that subject. The author is indebted to Mr. Low for permission to incorporate this material in the book, and to various other contributors, whose articles have been used as a whole or in part in the work. ^ ^ HUBERT E. COLLINS. NEW YORK, November ; 1908. 196487 CONTENTS CHAP. PAGE I WATCHING A BOILER AT WORK i II SIMPLE TALK ON EFFICIENCY OF RIVETED JOINTS n III SIMPLE TALK ON THE BURSTING STRENGTH OF BOILERS 17 IV SIMPLE TALK ON THE BURSTING STRENGTH OF BOILERS 24 V SIMPLE TALK ON THE BRACING OF HORIZONTAL RE- TURN TUBULAR BOILERS 30 VI CALCULATING THE STRENGTH OF RIVETED JOINTS . 40 VII To FIND THE AREA TO BE BRACED IN THE HEADS OF HORIZONTAL TUBULAR BOILERS 67 VIII GRAPHICAL DETERMINATION OF BOILER DIMENSIONS 70 IX THE SAFETY VALVE 75 X HORSE-POWER OF BOILERS 120 XI BOILER APPLIANCES AND THEIR INSTALLATION . 123 XII CARE OF THE HORIZONTAL TUBULAR BOILER . . 133 XIII CARE AND MANAGEMENT OF BOILERS .... 145 XIV SETTING RETURN TUBULAR BOILERS .... 150 XV RENEWING TUBES IN A TUBULAR BOILER ... . 156 XVI USE OF W'OOD AS FUEL FOR STEAM BOILERS . . . 161 XVII BOILER RULES 179 XVIII MECHANICAL TUBE CLEANERS 184 vii WATCHING A BOILER AT WORK 1 IF we take a test-tube filled with water nearly to the top and hold it over a Bunsen flame, the water boils violently and overflows the tube. This violent over-boiling is due to the conflicting action of the ascending and descending currents of steam and water in the tube. On the other hand, if we take a tube shaped like a U, the arms of which are connected together at the top, fill it with water^and place one leg of the U in the flame, a direct circulation soon com- mences. The water passes along in one direction and the steam is liberated at the surface. In this case there is very little violent ebullition, because there are no counter currents and the steam is discharged quietly over a liberal surface. Desiring to ascertain just how nearly a boiler could be designed to work upon the U-tube principle of circulation, after several trials the model boiler shown in Fig. i was produced. This model was built entirely of brass. It contained three drums four inches in diameter and 120 brass tubes one-quarter of an inch in diameter. The tubes were connected into headers and into the circum- 1 Contributed to Power by C. Hill Smith. I BOILERS ferences of the drums. The heads of all three drums were made of plate glass, for observation of the in- terior of the boiler when making steam. It will be noted that the design of the boiler closely resembles FIG. i. the U shape, only that one leg is considerably longer than the other, and there are two legs on the side of the U where the heat is applied. Each of the three drums serves a special function WATCHING A BOILER AT WORK 3 which will be noted from the description of the experi- ments. The two legs, instead of being connected together at the top, as was the case in the U-tube, are connected by two separate passages, one for the water to pass through and the other for the steam. In preparing for the tests the boiler was mounted on a stand, so that the tubes inclined from the hori- zontal 20 degrees, and the whole was enclosed on all sides by brass plates. Alcohol lamps were placed inside the casing at a point to correspond with the regular location of grates, or at about one-fourth of the distance between the front headers and the rear drum. The steam outlet was located on the rear drum, as was the safety valve, for experimental reasons, although in actual practice the safety valve would be located on the front drum. The feed-pipe was introduced in the rear drum, while the blow-off entered the lowest point of the lower drum, which we will call the mud-drum. The boiler was attached to an open condenser. The boiler being ready for test, it was filled with cold water until the upper drums were filled to one- half their volume. Candles were placed behind one head of each of the three drums for the purpose of lighting the inside. The alcohol lamps were then lighted and the boiler interior was ready to observe through the glass heads of the drums. The first action noted was in the front drum, which served as a discharge chamber for all the steaming tubes. The tubes of the lower bank discharged into it through headers, while those of the upper bank dis- 4 BOILERS charged into it independently. Many faint, oily-white streamers were seen to rise from the nipples connecting the headers to the front drum, passing upward to the surface of the drum. They resembled little streamers of white smoke. On reaching the surface of the water they passed into the horizontal circulating tubes which connected the two upper drums. These little streamers were heated water, which, being lighter than the water in the drum, rose to the surface. This same action soon appeared from the ends of the upper bank of tubes, the little streamers rising in a similar manner and passing into the horizontal tubes. By observing the rear drum, the little streamers could be seen coming into this drum from the front drum. Here they turned downward into the vertical tubes which connected the rear drum to the rear headers and the mud-drum. No action could be noted in the mud-drum, which fact seemed to indicate that these currents of water passed into the upper tubes and thence into the front drum again, as the action from these tubes appeared very much more decided than the action from the nipples, notwithstand- ing the fact that the lower tubes were nearer to the flame than the upper ones. This was undoubtedly due to the fact that the heated currents of water remained as near the surface as possible, while the colder water passed to the bottom of the boiler, hav- ing greater specific gravity. Particles of sediment could be seen coming down the vertical circulating tubes into the mud-drum, evidently precipitated from the water that was being WATCHING A BOILER AT WORK 5 heated. This sediment passed to the bottom of the drum, where it remained. A very gradual action was now noted in the mud-drum in the nature of similar currents of water coming down the vertical tubes. These currents acted strangely on entering the drum; they spread out on coming in contact with the colder and denser water lying at the bottom. By placing the finger on the upper portion of the glass head and then on the lower, quite a difference in temperature was noted. Little streamers of heated water soon commenced to pass into the lower bank of streaming tubes which were connected into this drum. They passed across the drum with a sort of shivering motion. A new and very interesting phenomenon was now noticed. Occasionally there would appear from the ends of the steaming tubes little rings of heated water, which shot across the drum with considerable velocity. The action in the front drum became very much more pronounced and air bubbles appeared from the nipples and tubes. The boiler was circulating water with great rapidity in the same direction and it was noticed, by placing the hand on different parts of the boiler, that all parts were of the same temperature. The air bubbles now discharged in great quantities from the tubes and nipples and rising to the surface disturbed the water level considerably. It was noted that they floated along under the surface of the water before they broke. Gradually these air bubbles ceased to appear and a new kind took their places. The latter were steam bubbles and they discharged into the drum with greater 6 BOILERS velocity than the former. On reaching the surface of the water they broke immediately, but they agi- tated the water level to a much greater extent. Foun- tains of water would shoot up into the drum for quite a distance and showed very vividly the conditions present in the shell type of boiler, where there are no defined paths for the water and steam to travel and nothing to prevent their conflict with each other. This also shows the cause for wet steam, and the great danger of entraining water with steam, as is the case where the steam is removed from the same place where violent ebullition is present. While the water level in the front drum was violently agitated, the water level in the rear drum remained perfectly calm. No steam was generated in this drum, as the horizontal tubes connecting it to the front drum only circulated water that had been freed of its steam. As the steam gage soon registered a pressure of 9 pounds, the main stop-valve was opened to allow the steam to flow to the condenser. The abrupt release of pressure caused the water to expand suddenly and the water level rose about one-quarter of an inch. This was evidently due to the sudden generation of steam caused by the drop in the pressure. This increased ebullition caused a very violent action in the front drum and the circulation of water through the boiler increased greatly in velocity. The nipples and tubes in the front drum discharged great quantities of bubbles. The water level in the rear drum during this increase in ebullition showed but a few ripples, WATCHING A BOILER AT WORK 7 which were evidently due to the vibration of the steam passing into the steam main, or the discharge into the water of the open condenser. The sudden generation of steam caused by the opening of the steam valve and subsequent reduction in pressure, it is believed, explains how the partial rupture of the shell of a return-tubular boiler is ad- vanced to a disastrous explosion by the unexpected increased generation of steam due to the lowered pressure. The steam main was now closed sufficiently to allow the boiler to operate on a constant pressure of about 6 pounds. It operated very smoothly under these con- ditions and made a very interesting sight with the steam generating in the front drum, where the nipples and tubes discharged great quantities of bubbles. The action in the mud-drum had in the meantime become well worth watching. In the other two drums the water showed clear in the candle light, but the color of the water in the mud-drum was very murky. Particles of sediment were noted settling to the bottom. The withdrawal of water from this drum by the steam- ing tubes did not appear to draw this sediment into the tubes, as the drum was of ample size so that the suction was not felt at the bottom where the sediment deposited. This emphasized clearly the advantage of a very large mud-drum to allow of the thorough settling of the sediment. The condition of the front drum was thought to be too violent for good practice, because the ebullition indicated restriction of circulation. The boiler was 8 BOILERS put out of operation for the purpose of making changes in this drum to prevent extreme ebullition. The glass heads were removed and other nipples inserted over the nipples that connected the headers into the drum, it being here that the most violent discharge of steam was discernible. These new nipples were cut long enough to reach to the water level, or just a trifle below it. The glass heads were replaced and the boiler put in operation again. The circulation was similar to that in the first test, and no real difference was noted until the boiler commenced to make steam. Then it was seen that the ebullition in this drum was considerably reduced, the agitation that remained being caused by the discharge from the tubes of the upper bank. This reduction was evidently due to having provided a channel through which the water and steam from the nipples might flow to the surface of the water and so prevent contact with the water in the drum. As the steam and water no longer had to force their way to the surface, the disturbance of the water level was naturally reduced entirely in this direction. The water rose from the nipples in little fountains, the steam disengaging from it in the upper part of the drum. The boiler was operated under very severe condi- tions to try the value of this addition of nipples. The main stop-valve was suddenly opened after a consider- able steam pressure was obtained. It had very little effect on the water level in this drum, only causing the nipples to discharge fountains of water quite a distance into the drum. No water was thrown into WATCHING A BOILER AT WORK 9 the superheating tubes, as the fountains of water dis- charged vertically and fell back immediately to the water level. The value of this attachment being proved, the boiler was blown down, and after the water was all withdrawn from the boiler considerable sediment was found in the bottom of the lower or mud-drum. No- where else was sediment found, as the drums offered no opportunities for the sediment to settle, being pierced at their lowest points by tubes and nipples. The tubes were inclined 20 degrees, which insured thorough draining of the boiler. From the foregoing experiments many points of great value for improvement in design of water-tube boilers can be derived. The violent ebullition in the front drum shows conclusively that steam should not be withdrawn from the boiler at a point where ebulli- tion is present, on account of the danger of getting water entrained with the steam. It also shows that any sudden reduction of the pressure causes violent ebullition and priming. The front-drum conditions show that this is a good place to locate the safety valve, as the sudden opening of it would cause no liability of priming if the steam is not withdrawn from this drum. The total lack of any ebullition in the rear drum shows that this is an ideal spot to remove the steam. It was noted that, owing to the large amount of sepa- rating surface provided, the opening of the steam valve caused no priming in this drum. Another fea- ture to be noted is the value of a large mud-drum to I0 BOILERS provide ample opportunity for the sediment to settle, and also to provide a large supply of water for the bottom tubes. It would be impossible to force the boiler hard enough to drain this drum of water, so the danger of burning out these tubes is eliminated. The provision of the long nipples in the front drum proved the advantage of providing separate passages to allow the steam and water to reach the surface of the water, thus obviating the necessity of their forcing their way to the surface through the large body of water in this drum and so cause violent ebullition. II SIMPLE TALK ON EFFICIENCY OF RIVETED JOINTS MATTER is conceived to be composed of myriads of tiny molecules separated from each other by distances which are very considerable as compared with their diameters, and held in fixed relation to each other in solid bodies, by such an attraction as holds the earth to the sun or the moon to the earth. When we tear a piece of boiler sheet apart it is the attraction of these molecules which we are overcoming, and if the metal is uniform the force required to separate it will depend upon the surface which we expose. It will take twice as much force to pull the larger of the two bars in Fig. 2 apart as it will the smaller, because there is twice as much surface exposed at B as at A, and the attrac- tion of twice as many molecules to overcome. The force tending to pull a body apart in this way is called a "tensile" force, and the resistance to the force necessary to pull a piece apart is called its "ulti- mate tensile strength." This is usually given in pounds per square inch, and is for boiler iron around 45,000 and for boiler steel around 60,000 pounds. It should be found stamped on the sheets of which boilers are made. Suppose we have a single riveted joint like Fig. 3. We 12 BOILERS FIG. 2. can divide it into strips as by the dotted lines half- way between the rivets, and consider one of these strips, for since they are all alike, what is true of one will be true of all. The width of each strip will be the same as FIG. 3. EFFICIENCY OF RIVETED JOINTS 13 the distance from center to center of the rivets. This is called the "pitch." Let us suppose the pitch to be 2\ inches, the diameter of the rivet I inch, the thickness of the plate i inch, the tensile strength of the plate 60,000 and the shearing strength of the rivets 43,000 pounds. FIG. 4. There are two ways in which this joint can fail: by tearing the sheet apart where there is the least of it to break, as at a a a a, Fig. 3, or by shearing the rivet as in Fig. 4. If the strip were whole as at A in Fig. 5, it would have 2j X i = 1.125 square inches of section, and since it takes 60,000 pounds to pull one square inch apart it would take 1.125 X 60,000 = 67,500 pounds to separate it. I 4 BOILERS But i inch of the sheet has been cut out for the rivet, so that there are left only 2\ - i == ij inches of width to be separated, and il X i = 0.625 square inch of area. This would stand a pull of only 0.625 X 60,000 = 37,500 pounds. Whether the joint will part by tearing the sheet or shearing the rivet depends, of course, on which is the stronger. The rivet has i X i X 0.7854 = 0.7854 square inch of area, and it takes 49,000 pounds to shear each square inch, so that it would take a pull of 0.7854 X 49,000 = 38,484.6 pounds. to shear the rivet. It is evident that the rivets would go, then, long before the plate, and that the strength of. the joint would be o o /- /- 38,484.6 -f- 67,500 = 0.57 or 57 per cent, of the strength of the full plate. But we can add to the rivet strength without reducing the plate strength by putting in another row of rivets behind the first row. In Fig. 6 two rivets have to be sheared, doubling the rivet strength without reducing the plate strength, for the holes for these extra rivets do not reduce the plate section along any one line if there is space enough between the rows. In Fig. 7 the EFFICIENCY OF RIVETED JOINTS 15 sheet is no more apt to part along the line aaaa than it would be if the second row of rivets were not there, and no more likely to part on the line bbbb than on the other. Any strip of a width equal to the pitch will FIG. 6. contain two rivets, whether we take it through the rivet centers, as at A, Fig. 7, or at equal distance to either side of one rivet in each row,'as at B in the same figure. In the first case it includes one full rivet and two halves, and in the latter case two full rivets. FIG. 7. To find the efficiency of this joint, then, we calculate the efficiencies of the plate and use the lowest efficiency. To calculate the plate efficiency, divide the difference 1 6 BOILERS between the pitch and the diameter of the rivets by the pitch. This is simpler than the operation which we went through above, which was (pitch-diam.) X thickness X tensile strength pitch X thickness X tensile strength the numerator being the pull required to separate the sheet with the rivet holes cut out, and the denominator the pull required to separate the full sheet. As the thickness and tensile strength appear in both numerator and denominator, they cancel out. To find the rivet efficiency, multiply the diameter of the rivet by itself, by 0.7854, by the shearing strength per square inch and by the number of rows, and divide by the product of the pitch, thickness and tensile strength per square inch of section. These rules are applicable only to lap joints where the rivets are in single shear. Ill SIMPLE TALKS ON THE BURSTING STRENGTH OF BOILERS THERE are two ways that a shell, such as is shown in the sketches herewith, might break under internal pressure. The sheets might tear lengthwise, letting the shell separate, as in Fig. 8, or they might tear across, letting it separate endwise, as in Fig. 9. 8095 i't* 80956 1C V^ FIG. 8. FIG. 9. Which is it the more likely to do? To push it apart endwise, as in Fig. 9, we have the force acting on the heads. This force is the pressure per square inch multiplied by the number of square inches in the head. The area of a circle is the diam- eter multiplied by itself and by 3.1416 and divided by 4; or since 3.1416 divided by 4 is .7854, the area is the square of the diameter multiplied by .7854. 17 l8 BOILERS Suppose the internal diameter of the shell to be 48 inches, and the pressure 100 pounds per square inch, the pressure on each head would be 48 X 48 X .7854 X 100 = 180,956.16 pounds, or over 90 tons. This pressure would act on each head, and the effect would be the same as though two weights of 180,956.16 pounds each were pulling against each other through the boiler, as in Fig. 10. FIG. 10. If the shell were not heavy enough to stand the strain, it would tear apart along the line where the metal happened to be the weakest, as at A. At first sight it looks as though the metal had to sustain both these forces or weights, and that the stress upon the shell would be twice 180,956.16 pounds; but a little consideration will show that this is not so. One simply furnishes the equal and opposite action with which every force must bje resisted. A man pulling against a boy on a rope (Fig. 1 1) can pull no harder than the boy pulls against him. If he does he will pull the boy off his feet, and the strain on the rope will be only what one of them pulls, not the sum of both pulls. In order that the man may pull with a force of 50 BURSTING STRENGTH OF BOILERS 19 pounds, the boy must hold against him with a force of 50 pounds. Both are pulling with a force of 50 pounds, but the tension on the rope is 50 pounds, not 100. The boy might be replaced with a post (Fig. 12). Now, when the man pulls with a force of 50 pounds FIG. 12. against the post, you would not say that there was 100 pounds tension on the rope; yet the post is pulling or holding against him with a force of 50 pounds, 2O BOILERS just as the boy did. In Fig. 13 it is easily seen that the tension on the cord is 50 pounds. You would not say that it was 100, if the pull of the weight were resisted by another weight of 50 pounds, as in Fig. 14, instead of by the floor. v I 50 Ib 3 . I FIG. 13. FIG. 14. The shell is therefore in the case which we have imagined subjected to a force of 180,956.16 pounds, which tends to pull it apart endwise, as in Fig. 10. To resist this there are as many running inches of shell as there are inches in the circumference. The circumference is 3.1416 times the diameter, so that to pull the boiler in two 48 X 3.1416 = 1 50.7968 inches of sheet would have to be pulled apart. The force exerted upon each running inch of sheet would be the pressure acting endwise divided by the circumference, or 180,956.16 4- 150.7968 = i, 200 pounds. The area is diam. X diam. X 3.1416 4 BURSTING STRENGTH OF BOILERS 21 The circumference is Diam. X 3.1416. Dividing the area by the circumference we have diam. X diam. X 3.1416 _ diam. 4 X diam. X 3.1416 4 or the strain on each running inch of sheet per pound of pressure is one-fourth the diameter. -Diamoter- FIG. 15. Now let us see what it would be in the other direc- tion. If we consider the pressure acting in all directions as in the upper half of Fig. 15, we should, to get the total pressure on the area, have to multiply the pres- 22 BOILERS sure per square inch by the whole area, which would be the circumference for a. strip i inch wide; but if we are considering the effect of pressure in one direc- tion only, we must consider only the area in that direction. If we are studying the effect of the pres- sure in forcing the shell in the direction of the arrows in the lower half of Fig. 15, we must consider only the area which comes crosswise to that direction, the "projected area," as it is called; the area which the piece would present if we were to hold it up and look at it in the direction of the arrows or of the shadow which it would cast in rays of light running in the direction of the pressure. This, it will be easily recog- FIG. 1 6. nized, is the diameter of the boiler wide and i inch high, as shown in Fig. 15, so that the number of square inches upon which the pressure is effective in one direction is equal to the diameter for a strip i inch wide. There is therefore a force tending to pull each i -inch ring of the shell apart, as in Fig. 16, of 48 X 100 = 4800 pounds, and as this force is resisted by two running inches of metal, one at/4 and one at B (Fig. i 5), BURSTING STRENGTH OF BOILERS 23 the stress per inch will be 4800 ~ 2 = 2400 pounds. This is just twice what we found it to be in the other direction; and it is plain that this should be so, for the stress per pound of pressure tending to burst the boiler, as in Fig. 8, is, as we have just seen, diam. which is just twice the - - which we found it to be in the other direction. It is for this reason that boilers are double riveted along the side or longitudinal seams, while single riveting is good enough for girth seams. IV SIMPLE TALKS ON THE BURSTING STRENGTH OF BOILERS IN the preceding chapter we found that a cylinder equally strong all over, will split lengthwise with one- half the pressure which would be needed to tear it apart endwise. FIG. 17. Let us see how much pressure it would take to burst a shell of this kind. We will consider a strip i inch in width, as in Fig. 17, for the action upon all the similar strips into which the boiler can be imagined to be spaced off will be the same. We see that the pressure 24 BURSTING STRENGTH OF BOILERS 25 tending to pull the ring, i inch in width, apart is equal to the pressure per square inch multiplied by the diameter of the ring. The total pressure in all direc- tions, acts on the circumference as shown by the radial arrows at the upper portion of the cut, but when we come to consider the force acting in one direction we must take the projected area in that direction; the area of the shadow, as explained before, cast by rays of light flowing in that direction, and that area would be that of the strip as we see it at the top of Fig. 17, i inch wide and the diameter of the boiler long. It is sometimes hard for one to see why the diameter is used here, instead of the circumference, and a further illustration is here given. FIG. 18. Suppose you had a piston in an engine cylinder made in steps like Fig. 18. This would have a good deal more surface to rust or to condense steam than would a flat piston, but it would have no more effec- tive area for the production of power, would it? One hundred pounds behind it in the cylinder would push no harder on the crosshead with this than with a per- fectly flat piston; because the sidewise pressure against 26 BOILERS the steps is balanced by an equal pressure from the opposite side; only the pressure on the flat rings effec- tive to move the piston forward, and the area of all these rings added together, is just the same as that of a flat surface of the same external diameter, as seen by the projection at the right. FIG. 19. This would be just as true if the steps were a millionth or a hundred-millionth of an inch wide and high instead of an inch or more, so that it is just as true of a conical surface, like Fig. 19, as of Fig. 18, or of a concave surface, like Fig. 20, as of either; and it is evident that it is the flattened-out area which one sees in looking at the object in the line of the force considered, the projected area in that direction as it is called, and not the real superficial area which is effective. BURSTING STRENGTH OF BOILERS 27 We have then a force equal to the pressure per square inch multiplied by the internal diameter of the shell tending to pull each inch in length of it apart, and we have two sections, A and B, Fig. 15, where the sheet must part. The force tending to tear eacb of these is the pressure per square inch multiplied by the radius, or half the diameter of the shell. The resistance that the piece of shell will offer to being torn apart is the tensile strength per square inch multiplied by the number of square inches to be torn apart. FIG. 21. This area is one inch long and the thickness of the sheet in width. The area in square inches is therefore the same as the thickness in inches. If the plate were f of an inch thick, for example, its section per inch of length would be f of a square inch, as shown in Fig. 21. The two opposing forces, which must be equal, not only at the point of fracture, but at all times, are: Pressure per square inch X radius, and pull per square inch X thickness. The pull on the sheet is called the tensile force. If we want to find the tensile force on the sheet for any pressure per square inch, we multiply that pressure by 28 BOILERS the radius and divide by the thickness of the sheet in inches. If we want to find the pressure per square inch neces- sary to get up a given tensile force per square inch, we multiply the given pull per square inch by the thickness of the plate and divide by the radius in inches. We can find the pressure per square inch necessary to rupture the sheet by multiplying the ultimate tensile strength, that is, the tensile force required to pull a square inch of it apart by the thickness and dividing by the radius. Example. What pressure would be required to burst a tank 48 inches in diameter, made of steel \ of an inch in thickness, having a uniform tensile strength of 60,000 pounds per square inch? Tensile strength X thickness -^T. ' - = pressure, radius 60,000 X .25 , ., = 625 Ibs. per sq. in. But we cannot or do not in boiler practice get a shell of uniform strength. There have to be joints and these joints are not so strong as the plate itself. We will have a talk later about how to figure the strength of a riveted joint. Suppose the riveted joint was only 70 per cent, of the plate strength, then it would take only 70 per cent, of the force to pull it apart, and the result just found must be multiplied by .70 if the tank, instead of having a " uniform tensile strength of 60,000," has a sheet strength of 60,000 and a longitudinal seam of 70 per cent, efficiency. BURSTING STRENGTH OF BOILERS 29 The complete operation of finding the bursting strength of a boiler shell is Tensile strength X thickness X efficiency of joint radius RULE. Multiply the tensile strength of the weakest sheet in pounds per square inch by the least thickness in inches and by the efficiency of the longitudinal riveted joint, and divide by the inside radius of the shell in inches. The result is the pressure per square inch at which the shell should split longitudinally. The safe working pressure is found by dividing the above by the desired ''factor of safety/' usually from 3-5 to 5. This, it must be noted, is the pressure at which the shell should fail in the manner described. The boiler may be weaker somewhere else, as upon some of the stayed surfaces, so that all these points should be con- sidered before the allowable pressure is fixed upon. SIMPLE TALKS ON THE BRACING OF HORIZONTAL RETURN TUBULAR BOILERS IN former chapters we have discussed the strength of a boiler so far as the parting of the shell is con- cerned, but even if the shell is heavy enough and the joint well proportioned the boiler may be weak in other respects. The head of a 6o-inch boiler has an area of 60 X 60 X 0.7854 = 2827 square inches. At 100 pounds per square inch there would be a pres- sure against the head of 2827 X 100 = 282700 pounds. or over 140 tons. Besides its tendency to pull the shell apart endwise, this pressure tends to bulge the heads, as shown in Fig. 22. In the case of a tank, or of the drums of water-tube boilers where there are no tubes in the heads, they can be made safe against change of shape under pressure by giving them in the first place the shape that the pressure tends to force them into; but the tube sheet of a horizontal tubular boiler, for in- 3 HORIZONTAL RETURN TUBULAR BOILERS 31 stance, must be flat to allow the tubes to enter square with its surface. The tubes themselves act as stays to the lower part, but the pressure on the part above the tubes tends to bulge the head and might cause the central tubes to pull out. FIG. 22. This is prevented by bracing the unsupported part of the head either by "through braces/' as in Fig. 23, or by "diagonal braces/' as in Fig. 24. In order to find how many braces are required, or if a given boiler is sufficiently braced, the area to be braced must be computed. This area may be taken as that included within lines drawn 2 inches above the top line of tubes and 2 inches inside of the shell, 3 2 BOILERS HORIZONTAL RETURN TUBULAR BOILERS 33 34 BOILERS as in Fig. 25, the area outside of these lines being con- sidered to be sufficiently braced by the shell and tubes. This figure is a "segment" of a circle and its area is found by dividing its height h by the diameter of the circle, of which it is a part, finding the quotient in the column of " versed sines" of the accompanying table, and multiplying the segmental area as given opposite that quotient in the next column by the square of the diameter. oooooooooo ! O O O O O O O O O O O'; FIG. 25. For example, suppose the hight b in Fig. 25 to be 1 8 inches and the diameter of the boiler 60 inches. The diameter of the circle of which the segment is a part is 56 inches, because we go 2 inches inside the shell on both ends of the diameter. Following the rule we divide the height by the diameter 18 ~ 56 = 0.3214. The values in the table are given to only three places of decimals, but the division should be carried out to HORIZONTAL RETURN TUBULAR BOILP:RS 35 four places. If the last figure is less than 5, drop it off. If it is 5 and the quotient comes out even, i.e., there is no remainder after the 5, drop it off, also. If the last figure is greater than 5, or in case it is 5, and the division did not come out square, drop it off, but raise the third figure one. In the case in hand the quotient, 0.3214, conies be- tween the 0.321 and 0.322 of the table, and is nearer the 0.321, being but 0.3214 0.321 = 0.0004 ff while it is 0.322 - 0.3214 = 0.0006 off from the higher value. If it were 0.3216, however, it would be nearer 0.322 than 0.321. The segmental area corresponding with 0.321 in the table is 0.2176. Multiplying this by the square of the diameter gives 56 X 56 X 0.2176 = 682.39 square inches as the area of the segment, and the force to be braced against is this number of square inches mul- tiplied by the pressure per square inch. A through-brace which pulls squarely on the plate has an effect in keeping it from bulging equal to the tensile strain in the brace itself i.e., if the brace were under a strain of 6000 pounds, it would tend to pull the head in and keep it from bulging with an equal force; but if a diagonal brace, as in Fig. 26, were under a strain of 6000 pounds, it would tend to pull the lug on the head in its own direction with that force, but would resist a force in a direction at right angles with the head of only .91 as much, or 5,460 pounds. This figure is found by dividing the length of the line b c by the length of the line a b. . In order to find if a boiler is sufficiently braced; 36 BOILERS Find the smallest cross-section of each brace in square inches. Multiply the cross-section of each diagonal brace by the quotient of the distance of its far end from the head in a line perpendicular to the head (b c, Fig. 26), divided by the length of the brace. Add all these results together. FIG. 26. For such braces as are all alike, as for through- braces of the same diameter of cross-section, you can of course compute one and multiply it by the number of similar ones. Divide the product of the area to be braced and the pressure per square inch by the sum of all these values, and you will have the strain on the braces per square inch of section. The rules of the United States Board of Supervising Inspectors allow a strain of 6000 pounds per square inch on the braces. If the computed stress does not exceed this amount, the boiler is sufficiently braced. HORIZONTAL RETURN TUBULAR BOILERS To determine what pressure a boiler will stand, so far as its bracing is concerned, multiply the minimum cross-section of each brace by the quotient of the dis- tance of its far end from the plate perpendicularly di- vided by the length of the brace. Add the results and multiply by 6000. Divide the produce by the number of square inches in the segment, and the quotient will be the pressure per square inch that the bracing is good for. AREAS OF SEGMENTS OF CIRCLES Versed Sine. Segmental Area. Versed Sine. Segmental Area. Versed Sine. Segmental Area. Versed Sine. Segmental Area. .1 .04087 .121 .05404 .142 .06892 .163 .08332 .101 .04148 .122 .05469 143 .06892 .164 .08406 .102 .04208 .123 05534 .144 .06962 .165 .0848 .103 .04269 .124 .056 145 07033 .166 08554 .104 .0431 125 .05666 .146 .07103 .167 .08629 .105 .04391 .126 05733 .147 .07174 .168 .08704 .106 .04452 .127 5799 .148 07245 .169 .08779 .107 .04514 .128 .05866 .149 .07316 .17 .08853 .108 4575 .129 05933 15 07387 .171 .08929 .109 .04638 13 .06 J 5i 07459 .172 .09004 .11 .047 131 .06067 .152 07531 173 .0908 .III .04763 .132 06135 153 .07603 .174 09155 .112 .04826 133 .06203 i54 .07675 175 .09231 113 .04889 134 .06271 -155 .07747 .176 .09307 .114 4953 J 35 .06339 156 .0782 .177 .09384 US .05016 .136 .06407 157 .07892 .178 .0946 .Il6 .0508 137 .06476 158 .07965 .179 09537 .117 .05145 .138 06545 159 .08038 .18 .09613 .118 .05209 i39 .06614 .16 .08111 .181 .0969 .119 .05274 .14 .06683 .161 .08185 .182 .09767 .12 .05338 .141 -06753 .162 .08258 .183 .09845 38 BOTLKRS AREAS OF SEGMENTS OF CIRCLES Continued Versed Sine. Segmental Area. Versed Sine. Segmental Area. Versed Sine. Segmental Area. Versed Sine. Segmental Area. .184 .09922 .216 .12481 .248 .15182 .28 .18002 185 .1 .217 .12563 .249 .15268 .281 .18092 .186 .10077 .218 .12646 25 .15355 .282 .18182 .187 IOI55 .219 .12728 .251 .15441 .283 .18272 .188 .10233 .22 .12811 252 .15528 .284 .18361 .189 .10312 .221 .12894 253 15615 .285 .18452 .19 .1039 .222 .12977 254 .15702 .286 .18542 .191 .10468 .223 .1306 255 15789 .287 18633 .192 .10547 .224 .i3 J 44 .256 .15876 .288 .18723 193 .10626 .225 .13227 257 .15964 .289 .18814 .194 .10705 .226 I 33n 258 .16051 .29 .18905 195 .10784 .227 J 3394 259 .16139 .291 .18995 .196 .10864 .228 .13478 .26 .16226 .292 .19086 .197 .10943 .229 .13562 .261 .16314 293 .19177 .198 .11023 23 .13646 .262 .16402 294 .19268 .199 .IIIO2 .231 I373I .263 .1649 .295 .1936 .2 .IIl82 .232 13815 .264 .16578 .296 I945r .201 .11262 2 33 .139 265 .16666 297 .19542 .202 II343 234 .13984 .266 .16755 .298 19634 .203 .11423 235 .14069 .267 .16844 299 19725 .204 .H503 .236 I4I54 .268 .16931 3 ' .19817 .205 .11584 .237 .14239 .269 .1702 .301 .19908 .206 .11665 .238 .14324 27 .17109 .302 .2 .207 .11746 239 .14409 .271 .17197 303 .20092 .208 .11827 .24 .14494 .272 .17287 304 .20184 ,2O9 .11908 .241 .1458 273 .17376 305 .20276 .21 .1199 .242 .14665 .274 .17465 .306 .20368 .211 .12071 243 i475i 275 17554 307 .2046 212 I2I53 244 14837 .276 .17643 308 20553 2I 3 12235 245 .14923 .277 17733 309 .20645 .214 .12317 .246 . 1 5009 .2 7 8 .17822 3i .20738 .215 .12399 .247 .i5 95 279 .17912 311 2083 HORIZONTAL RETURN TUBULAR BOILERS 39 AREAS OF SEGMENTS OF CIRCLES- Continued Versed Sine. Segmental Area. Versed Sine. Segmental Area. Versed Sine. Segmental Area. Versed Sine. Segmental Area. .312 .20923 343 23832 374 .26804 405 .29827 .313 .21015 344 .23927 375 .26901 .406 .29925 .314 .21108 345 .24022 376 .26998 .407 .30024 315 .21201 346 .24117 377 .27095 .408 .30122 .316 .21294 347 .24212 378 .27192 .409 .3022 3*7 .2I38 7 348 .24307 379 .27289 .41 30319 .318 .2148 349 .24403 38 .27386 .411 30417 3i9 21573 35 .24498 .381 27483 .412 30515 .32 .21667 351 24593 382 .27580 413 .30614 .321 .2176 352 .24689 383 .27677 .414 .30712 .322 21853 353 .24784 384 27775 .415 .30811 323 .21947 354 .2488 385 .27872 .416 30909 3 2 4 .22O4 355 .24976 386 .27969 .417 .31008 325 .22134 356 .25071 387 .28067 .418 .31107 .326 .22228 357 .25167 388 .28164 .419 31205 327 .22321 358 25263 389 .28262 .42 31304 .328 22415 359 25359 39 28359 .421 31403 329 .22509 36 25455 391 .28457 .422 31502 33 .22603 361 25551 392 28554 423 .316 331 .22697 .362 .25647 393 .28652 .424 .31699 332 .22791 363 25743 394 .2875 .425 31798 333 .22886 364 25839 395 .28848 .426 -3 l8 97 334 .2298 365 2593 6 .396 28945 .427 .31996 335 .23074 .366 .26032 397 .29043 .428 32095 336 .23169 367 .26128 .398 .29141 .429 .32194 337 .23263 .368 .26225 399 .29239 43 .32293 338 .23358 369 .26321 4 2 9337 431 32391 339 23453 37 .26418 .401 29435 432 3249 34 23547 371 26514 .402 29533 433 3259 341 .23642 372 .26611 .403 .29631 434 .32689 342 2 3737 373 .26708 .404 .29729 435 32788 VI CALCULATING THE STRENGTH OF RIVETED JOINTS 1 IN calculations relative to the strength of steam boilers and vessels of a similar character for withstand- ing high pressures, one of the most important points to be considered is the strength of the seams where the plates are joined. This is not only important to the designer of such vessels, but also to the operating engineer, who is often required to fix the limit of pres- sure which should be carried on the boilers under his charge, and frequently, owing to increased output without corresponding addition to the boiler capacity, it becomes necessary to carry the pressure as high as safety will permit, and in such cases it is important for the engineer to be able to fix this safe limit.' It is the purpose in this chapter to show how the strength of the various types of joint generally used in boiler construction may be calculated, and as only simple arithmetic is required for the calculations, any reader should find no difficulty in understanding how it is done, and applying the principles to calculate the strength of the particular joints which may be of interest to them. To avoid the use of formulas, which 1 Contributed to Power by S. F. Jeter. 40 STRENGTH OF RIVETED JOINTS 41 are confusing to many, numerical examples will be used to illustrate the methods of making the calcula- tions, and for the sake of uniformity the tensile strength of the sheets (which is the strength to resist being pulled apart) will be assumed as 55,000 pounds per square inch; the shearing strength of the rivets (which represents their resistance to being sheared through by the plates at right angles to their length) will be assumed as 42,000 pounds per square inch in single shear, as represented in Fig. 31, and 78,000 pounds per square inch in double shear, as represented in Fig. 34. The resistance of the rivets to crushing will be assumed at 95,000 pounds per square inch. For modern construction consisting of steel plates and steel rivets, the above values are average figures. It is customary to express the strength of a riveted joint as a percentage of the strength of the plates which are riveted together. Thus, if the joint illustrated in Fig. 35 has an efficiency of 62^ per cent., it would mean that any portion of its length that divides the rivet spaces symmetrically would be 0.625 times as strong as a section of the same length through the solid plate. POSSIBLE MODES OF FAILURE Before proceeding to calculate a practical boiler joint, the different ways in which two pieces of plate riveted together might fail should be noted. If a piece of boiler plate, f inch thick and 2| inches wide, is placed in the jaws of a testing machine, as illustrated in Fig. 27, and pulled apart, it would separate at some section as A A. If the tensile strength was 55,000 pounds per 42 BOILERS square inch, the force that would have to be applied to the jaws would be 55,000 times the area separated in square inches, which in this case is 2! X f = H = -9375 square inch, so that the pull would be pounds. 55-000X0.9375 = 5, ,562.5 If another piece of plate be taken, identical in every rf--j \ / - ; !l / SK V' 1 ( i B fl- B - r i S \ \ !i r -\ r \ hi: ( j i~J L J ~H u P FIG. 27. M FIG. 28. respect to the first, except that a hole i inch in diameter is drilled through it as illustrated in Fig. 28, and the plate be pulled apart in the testing machine as before, it is evident that it would fail along the line B B, as the area of the reduced section caused by drilling the hole would be only (2.5 i) X 0.375 = 0.5625 square inch, and the force necessary to pull it apart would be 55,000 X 0.5625 = 30,937.5 STRENGTH OF RIVETED JOINTS 43 pounds, the strength of the metal being the same in both instances. Now if the relation between the strength of the solid plate and the drilled plate be expressed by dividing the latter by the former, the result would be ^^ ^ 51,562.5 or, in other words, the drilled plate is capable of sus- taining 60 per cent, of the load that could be carried by the solid plate. If, instead of using a single piece of plate, two plates are drilled with i-inch holes in the ends and are joined together by a rivet, as shown in Fig. 29, and an attempt FIG. 30. should be made to pull them apart as before, there would be four probable ways in which failure might take place, all of which are considered in the calculation and design of riveted joints. First, the section of plate each side of the rivet hole might break, leaving the ends 44 BOILERS as shown in Fig. 30. Again, the plates might shear the rivets off, as illustrated in Fig. 31. Thirdly, it has been found by practical tests of joints that steel rivets can- not be subjected to a pressure much greater than 95,000 pounds per square inch of bearing surface without materially affecting their power to resist shearing, and therefore the joint might fail, as shown in Fig. 31, due to an excess crushing stress on the rivet. QD oh FIG. 31. FIG. 32. A fourth possible method of failure would be for the metal in the sheet in front of the rivet to split apart or pull out, as illustrated in Fig. 32. This latter mode of failure is erratic, and cannot be calculated, but it has been practically demonstrated in tests of joints, that if the distance from the edge of the plate to the center of the rivet hole is i \ times the diameter of the hole, this mode of failure is improbable, and in the fol- lowing calculations of joints it will be assumed that they are properly designed to render such failure im- possible. STRENGTH OF RIVETED JOINTS 45 To determine the actual strength or efficiency of such a joint as is illustrated in Fig. 29, the force re- quired to produce rupture must be calculated for each of the first three ways mentioned, and the weakest mode of failure taken as the maximum strength of the joint. To rupture the plate as illustrated in Fig. 30, the pull required would be the same as to rupture the drilled plate illustrated in Fig. 28, which was found to be 30,937.5 pounds. To shear the rivet off, as in Fig. 31, would require a force equal to the area to be sheared in square inches, times the shearing strength per square inch; or since the area of a i-inch rivet is 0.7854 square inch, the force required would be 0.7854 X 42,000 = 32,987 pounds. The pressure required to cause failure by crushing was stated to be 95,000 pounds per square inch, and in calculating the area exposed to pressure for pins and rivets, it is figured as equal to the diameter of the pin or rivet, times the thickness of the plate; therefore, we have i X 0.375 = o-375 square inch of area to withstand crushing, or 0.375 X 95,000 - 35,625 pounds would be required to produce rupture of the joint in this manner. From these figures it is evident that the method of failure first considered is the weakest of the three, BOILERS and, therefore, determines the efficiency of the joint, which would be 60 per cent, as found for the drilled plate. If one plate is riveted between two other plates, as illustrated in Fig. 33, the several methods of failure are calculated in the same way, except for the shearing of the rivet, which would occur as shown in Fig. 34, and 61261 Lbs. FIG. 33. FIG. 34. is described as double shearing. While the metal sheared in this case would be just twice as much as in single shear, it has been found by test that the force required is not exactly twice as much, but 1.85 to 1.90 times the amount in single shear; so as stated at the beginning, 78,000 pounds per square inch is .assumed for rivets in double shear and 42,000 pounds per square inch when in single shear. Calculating the strength of a joint with the dimen- sions as illustrated in Fig. 34, the strength of the solid plate would be 3 X 0.75 X 55,000 = 123,750 pounds, The strength of the center plate through the STRENGTH OF RIVETED JOINTS 47 rivet hole, the failure being assumed similar to that illustrated in Fig. 30, would be (3 - i) X 0.75 X 55,000 = 82,500 pounds. The crushing strength of the rivet would be i X 0.75 X 95,000 - 71,250 pounds. The shearing strength of the rivet, or failure assumed as in Fig. 34, would be 0.78154 X 78,000 = 61,261 pounds. From these figures it is evident that failure would most likely occur as shown in Fig. 34, and the relative strength of the joint as compared with the solid plate is 61,261 - = 0.495, 123,750 or 49 J per cent. As will be shown later, the foregoing simple calculations are all that are required to estimate the strength of the most complicated joints. THE UNIT SECTION In calculating the strength of a practical boiler joint, the strength for the entire length of a sheet could be estimated, but this would be laborious owing to the number of figures involved in the calculations, and the same result can be obtained by considering any length that divides the rivets symmetrically. For convenience, the shortest length that thus divides the rivets is the one used in such calculations, and this length is called a unit section of the joint. When the lines dividing the 48 BOILERS joint into unit sections pass through a rivet, only one- half of the rivet is considered in the calculation, and when rivets thus divided are lettered for reference, the two halves on opposite sides will be lettered the same, so that referring to the letter will indicate a whole rivet. Thus, if the rivet A, in Fig. 43, is spoken of, it would mean the combined halves of the two rivets on the outer row. In measuring joints already constructed to obtain the length of a unit section, or the pitch, it should be remembered that rivet heads do not always drive fairly over the center of the rivet holes, and the rivet holes themselves are sometimes irregular distances apart; so it is more accurate to measure a number of pitches and divide the distance by the number measured to obtain the average pitch. It will be found most con- venient, where space permits, to measure ten pitches, and then placing the decimal point one figure to the left will give the average unit length. SINGLE-RIVETED LAP-JOINT First to be considered is the single-riveted lap-joint illustrated in Fig. 35. In a unit section of 2 inches one rivet is in single shear and J inch has been cut out of the plate by the rivet hole. The calculation for strength is the same as has been made for Fig. 38, and the three methods of failure to be considered are: (1) Breaking of the section of plate between the rivet holes, which is called the net section. (2) Shearing of a f-inch rivet in single shear. (3) Resistance of one rivet to crushing. STRENGTH OF RIVETED JOINTS 49 Using the numerical values given in Fig. 35 the following results are obtained: (1) (2 - 0.75) X 0.25 X 55,000 = 17,187.5 pounds. (2) 0.4418 X 42,000 = 18,556 pounds. (3) 0.75 X 0.25 X 95,000 - 17,812.5 pounds. - rfDU-HoU^ |^l 6< ^CJ)@(pi c L 4x FIG. 35. Of the three methods of failure, the first is seen to be the most probable, and since a unit section length of the solid plate would have a strength of 2 X 0.25 X 55,000 = 27,500 pounds, the efficiency of the joint would be 17,187.5 = 62.5 27,500 per cent. DOUBLE-RIVETED LAP-JOINT Next in line is the double-riveted lap-joint illustrated in Fig. 36. There is one feature connected with this joint which should be considered before proceeding with the calculation of its strength. It would evi- dently be possible to have the two rows of rivets form- ing this joint so close together that the combined net 50 BOILERS sections between rivets A B and B C would be less than between rivets AC. It has been found in prac- tical tests of joints that it is necessary to have the com- bined area of these two sections 30 to 35 per cent, in excess of that between rivets A and C in order to be sure that the joint will fail along line A C. This would FIG. 36. correspond to a diagonal pitch of two-thirds of the pitch from A to C plus one-third of the diameter of the rivet hole, or 1.9 inches in the joint shown in Fig. 36. Ordinarily, if the rows are much closer than this, the joint has an abnormal appearance which would be noted at once. In further calculations it will be assumed that the joints are proportioned so that this method of failure will not be possible. Proceeding with the calculation of the strength of the joint illustrated in Fig. 36, the methods of probable failure to be calculated are the same as for the single- riveted joint: (1) Failure of net section between rivet holes. (2) Shearing of two rivets in single shear. (3) Crushing strain on two rivets. STRENGTH OF RIVETED JOINTS 51 Using the values given in Fig. 36, we have for the above : (1) (2.5 - 0.75) X 0.3125 X 55,000 = 30,078 pounds. (2) 2 X 0.4418 X 42,000 = 37,112 pounds. (3) 2 X 0.75 X 3.5120 X 95,000 = 44>53 I pounds. It is evident that the first method of failure is the most probable, and since the strength of the solid P at< 2.5 X 0.3125 X 55,000 = 42,969 pounds, the efficiency of the joint will be 30,078 7-= 70 42,969 per cent. TRIPLE-RIVETED LAP-JOINT In Fig. 37 is illustrated a triple-riveted lap-joint. Here the length of unit section is 3 inches, and the different probable modes of failure are identical with those of the single- and double-riveted lap-joints except in rivet strength. It will be noted that in this case there are three rivets contained in each unit section, which are subjected to shear and crushing. The several methods of probable failure to be inves- tigated are as follows: (1) Failure of net section between the rivet holes of outer rows. (2) Shearing of three rivets in single shear. (3) Crushing strain on three rivets. 52 BOILERS Using the numerical values specified in Fig. 37, we Would have: (1) (3 0.75) X 0.375 X 55,000 = 46,406 pounds. (2) 3 X 0.4418 X 42,000 = 55,667 pounds. (3) 3 X 0.75 X 0.375 X 95,000 = 80,156 pounds. JiDia. Hole FIG. 37. The first method of failure assumed is the most likely, and as the strength of the solid plate for a unit section of length is 3 X 0.375 X 55,000 = 61,875 pounds, the efficiency of joint is 46406 _ 6i,8 75 ~ 75 per cent. The triple-riveted joint represents about the maximum strength that can be obtained in prac- tice from simple lap-riveted joints, as in this form the maximum pitch distance that permits proper calking of the edge of the plates is reached, and still leaving STRENGTH OF RIVETED JOINTS 53 the net section of metal between the rivet holes the weakest portion of the joint, so that further addition of rivets would not add to its strength. CHAIN RIVETING Joints illustrated in Figs. 36 and 37 have the rivets arranged so that the rivets in one row come opposite the FTG. 39. spaces in the adjacent rows, and this arrangement is termed staggered riveting. The same forms of joint are sometimes made with the rivets placed in straight rows across the joint, is illustrated in Figs. 38 and 39, which is known as chain riveting. The calculations for joint 54 BOILERS efficiency in chain-riveted joints are identical in every respect to those for staggered riveting, and with equal diameters and spacing of rivets and equal thicknesses of plate, the efficiencies are the same for either type. LAP-RIVETED JOINT WITH INSIDE STRAP While the lap-riveted joint with inside strap is not extensively used in the manufacture of new boilers, it affords a ready means of strengthening simple lap FIG. 41. seams on boilers already constructed, and it is quite extensively used for this purpose. This joint is illustrated in Figs. 40 and 41, and it will be seen that it is equally applicable to single- or double-riveted STRENGTH OF RIVETED JOINTS 55 lap-joints; it could also be applied to triple-riveted joints. The joint illustrated in Fig. 40 is identical in every respect with the one shown in Fig. 36, excepting the addition of the fVmch cover strip and the outer rows of rivets, these dimensions being selected to facili- tate comparison between the strengths of the two joints. A unit section of this joint is 5 inches long, and five methods of failure present themselves for consideration in determining the strength of the joint: (1) Breaking of the plate along the section between the rivet holes A A. (2) The separation of the plate along the section on line of rivets CD and shearing the rivet A. (3) Separation of the plate along the section on line of rivets C D and the crushing of rivet A. (4) Crushing of rivets A B C D E by the shell. (5) Shearing of rivets B CD E F in single shear. The pulling out of the upper plate, which would shear rivet A single, and E CD E double, need not be considered, since it would evidently be stronger than the first method considered above. Calculating the value of the possible methods of failure by using the dimensions given in Fig. 40, we have: (1) (5 - 0.75) X 0.3125 X 55,000 = 73,040 pounds. (2) [5 - (2 X 0.75)] X 0.3125 X 55,000 + 0.4418 X 42,000 = 78,726 pounds. (3) [5 " (2 X 0.75)] X 0.3125 X 55,000 + 0.3125 X 0.75 X 95,000 = 82,436 pounds. (4) 0.3125 X 0.75 X 95,000 X 5 = ii 1,330 pounds. (5) 0.4418 X 42,000 X 5 = 92,780 pounds. 56 BOILERS Evidently the next section between the rivet holes A A is the weakest portion of the joint, and since a section of the solid plate 5 inches long has a strength of 5 X 0.3 125 X 55,000 = 85,937 pounds, the efficiency of the joint is S^o _ 8 percent. 8 5'937 Calculation of the joint illustrated in Fig. 41 is pro- ceeded with in the same manner as for Fig. 40. It will be noted that to aid comparison the dimensions have been assumed the same as in Fig. 35 with the strap added. The methods of possible failure to be compared are: (1) Separation of the plate along net section G G. (2) Separation of plate along section H I and shear- ing of rivet G in single shear. (3) Separation of plate along section H I and crush- ing of rivet G. (4) Crushing of rivets G H I. (5) Shearing of rivets H I J in single shear. According to the dimensions given in Fig. 41 the numerical values would give the following results. (1) (4 - 0.75) X 0.25 X 55,000 = 44,687 pounds. (2) [4 - (2 X 0.75)] X 0.25 X 55,000 + 0.4418 X 42,000 = 52,931 pounds. (3) [4 - (2 X 0.75 )]X 0.25 X 55,000 + 0.75 X 0.25 X 95,000 = 51,187 pounds. (4) 0.75 X 0.25 X 95,000 X 3 == 53,436 pounds. (5) 0.4418 X 42,000 X 3 = 55,668 pounds. STRENGTH OF RIVETED JOINTS Since the strength of the solid plate is 4 X 0.25 X 55,000 = 55,000 pounds, the efficiency would be = 81.25 57 5 5 ,OOO per cent. It is thus apparent that by adding a strap to the joint illustrated in Fig. 35 and making it like Fig. 41, the efficiency has been increased from 62.5 per cent, to 81.25 P er cent., which would permit an increase in steam pressure of 30 per cent, on the boiler after such change. SINGLE-RIVETED DOUBLE-STRAPPED BUTT-JOINT In describing all forms of butt-joints it is customary to refer to the rivets on one side of the butt only; thus, in Fig. 42 there are actually two rows of rivets, but FIG. 42. the joint is only single-riveted, for the strength of the joint along either row is in no wise dependent on the other row. If the two rows should not be riveted alike, it would be necessary to consider each side as a separate joint to find which was the weaker, in order to deter- 58 BOILERS mine the strength of the combination. This, how- ever, is not necessary in practical boiler joints, since they are constructed alike on each side of the butt. In the joint illustrated in Fig. 42 it will be noted that all of the rivets are in double shear, and only three methods of possible failure are presented for calcula- tion: (1) Breaking the net section. (2) Shearing of one rivet in double shear. (3) Crushing of a rivet by the shell. With the dimensions given in the figure we have: (1) (2.25 -- 0.75) X 0.3125 X 55,000 == 25,781 pounds. (2) 0.4418 X 78,000 = 34,460 pounds. (3) 0.75 X 0.3125 X 95,000 = 22,230 pounds. The strength of the solid plate is 2.25 X 0.3125 X 55,000 = 38,672 pounds, and since the weakest portion of the joint is the resistance to crushing of the rivets, the efficiency is 22,230 _ 38,672 ~ 57 ' 5 per cent. DOUBLE-RIVETED DOUBLE-STRAPPED BUTT-JOINT Double-riveted butt-joints can be made in two forms, the one generally used being illustrated in Fig. 43. The calculations for the efficiency of this joint are the same as for the single-riveted joint, except that there STRENGTH OF RIVETED JOINTS 59 are two rivets to be considered in each unit section of the joint instead of one. The three methods of pos- sible failure are : (1) Pulling apart of the sheet along net section A A. (2) Shearing of rivets, A B, in double shear. (3) Crushing of rivets A B. r . i %$ , B , \ w---7s-r A A FIG. 43. Substituting the values given in Fig. 17, we have: (1) (2.5 - 0.75) X 0.3125 X 55,000 = 29,085 pounds. (2) 0.4418 X 78,000 X 2 = 68,920 pounds. (3) 0.75 X 0.3125 X 95,000 X 2 = 44,532 pounds. The strength of the solid plate is 2.5 X 0.3125 X 55,000 = 42,969 pounds, and the weakest portion of the joint is the net section between rivets A A. Therefore, the efficiency is per cent. 42,969 = 6 7 .6 6o BOILERS In Fig. 44 is illustrated the second type of double- riveted butt-joint. This form of joint, if proportioned properly, can be made considerably stronger than the one illustrated in Fig. 43. There are six methods of possible failure to be considered: FIG. 44. (1) Pulling apart of the sheet along net section A A. (2) Pulling apart of the sheet along section B C and shearing rivet A. (3) Pulling apart of sheet along section- B C and crushing of rivet A. (Note that in calculating the crushing of rivet A the thickness of the strap is to be used instead of the plate, owing to the strap being thinner than the plate.) (4) Shearing of rivet ^single and B, C double shear. (5) Crushing of rivets B C in the plate and A in the strap. (6) Crushing of rivets B C in the plate and shear- ing of rivet A. STRENGTH OF RIVETED JOINTS 61 Substituting the numerical values from Fig. 44, we have: (1) (4 - 0.75) X 0.3125 X 55,000 = 55,859 pounds. (2) [(4 - 1.5) X 0.3125 X 55,000] + (42,000 X 0.4418) - 61,525 pounds. (3) [(4 - 1.5) X 0.3125 X 55,000] + (0.75 X 0.25 X 95,000) == 60,781 pounds. (4) (42,000 X 0.4418) -f (78,000 X 0.4418 X 2) = 87,476 pounds. (5) (o./5 X 0.3125 X 95,000 X 2) + (0.75 X 0.25 X 95,000, = 62,272 pounds. (6) (0.75 X 0.3125 X 95,000 X 2) + (42,000 X 0.4418) == 63,087.25 pounds. From these figures it will be seen that the net sec- tion between the rivet holes A A is the one most likely to fail, and since the strength of a unit section of the solid plate is 4 X 0.3125 X 55.000 = 68,750 ponuds, the efficiency of the joint is - 81.25 per cent. 68 >75O TRIPLE-RIVETED DOUBLE-STRAPPED BUTT-JOINT The joint illustrated in Fig. 45 is known as the triple-riveted butt-joint. The methods of failure to be investigated are the same as those in Fig. 44, and are as follows : (i) Pulling apart of sheet at net section A A. 62 BOILERS STRENGTH OF RIVETED JOINTS 63 (2) Pulling apart of sheet along section C E and shearing rivet A. (3) Pulling apart of sheet along section C E and crushing rivet A. (4) Shearing rivet A single and B C D E double. (5) Crushing of rivets B C D E in the plate and A in the strap. (6) Crushing of rivets B C D E in the plate and shearing of rivet A. Substituting the values given in Fig. 45 : (1) (7.5 -- i) X 0.5 X 55,000 == 178,750 pounds. (2) [(7.5 - 2) X 0.5 X 55,000] + (42,000 X 0.7854) = 184,237 pounds. (3) [(7-5 -- 2) X 0.5 X 55>o] + X 0.5 X 95,000) = 198,250 pounds. (4) (0.7854 X 42,000) + (0.7854 X 78,000 X 4) = 278,027 pounds. (5) (i X 0.5 X 95,000 X 4) + X 0.375 X 95,000) = 225,625 pounds. (6) (i X 0.5 X 95,000 X 4) + (0.7854 X 42,000) = 222,987 pounds. For a unit length the strength of the solid plate is 7.5 X 0.5 X 55,000 = 206,250 pounds. The net section between rivets A, A is the weakest portion of the joint, so that the efficiency is per cent. _ 86 . 7 206,250 64 BOILERS QUADRUPLE-RIVETED DOUBLE-STRAPPED BUTT-JOINT The last type of joint to be considered is the quad- ruple-riveted butt-joint illustrated in Fig. 46. This joint is now used on nearly all high-grade boilers of the horizontal return-tubular type, and it marks about the practical limit of efficiency for riveted joints connecting plates of uniform thickness together. The methods of failure to be considered are practically the same as in the two preceding joints, except that there are more rivets concerned in the calculations: (1) Pulling apart of the sheets along net section A A. (2) Pulling apart of the sheet along section D E F G and shearing rivets ABC. (3) Pulling apart of sheet along section D E F G and crushing of rivets A B C in the strap. (4) Shearing rivets A B C in single shear and D E F G H I J K in double shear. (5) Crushing of rivets D E F G H I J K in plate and A B C in the strap. (6) Crushing of rivets D E F G H I J K in 'the plate and shearing rivets ABC. Using the numerical values of Fig. 46, we have: (1) (15.5- i) X 0.5625 X 5 5, ooo = 448, 580 pounds. (2) [(15.5 - 4) X 0.5625 X 55,000] + (3 X 42,000 X 0.7854) = 454,739 pounds. (3) [(15.5 - 4) X 0.5625 X 55>] + (3 X 0.4375 X i X 95,000) = 480,465 pounds. F X RIVETED JOINTS OF THS UNIVERS 66 BOILERS (4) (3 X 42,000 X 0.7854 ) + (8 X 78,000 X 0.7854) = 589,050 pounds. (5) (8 X 0.5625 X i X 95,000) + (3 X 0.4375 X I X 95,000) == 552,187 pounds. (6) (8 X 0.5625 X i X 95,000) + (3 X 42,000 X 0.7854 = 526,461 pounds. The strength of the solid plate is 15.5 X 0.5625 X 55,000 = 479,528 pounds, and the failure of the sheet by pulling apart along the net section A A is the one that determines the efficiency of the joint, which is 448.580^ per cent. 479.528 From the foregoing calculations it may be observed that estimating the efficiency of riveted joints, while very simple, is a rather tedious process, particularly if many joints are to be calculated VII TO FIND THE AREA TO BE BRACED IN THE HEADS OF HORIZONTAL TUBULAR BOILERS FOR the purpose of determining the number of braces to be used, it is not necessary to figure the area of a boiler head to a fraction of a square inch, and a simple rule, the reason for which is so plain that it can never be forgotten, will be helpful to the candidate before the examiner, or when a table of circular segments is not to be had. The diameter of the boiler and the hight above the top row of tubes are the only measurements which are ordinarily given. The flange is considered good for three inches around the outside, and the tubes for two inches above their top edges, so that the area to be braced is a part of a circle having a diameter six inches less than the given diameter of the boiler and a hight 5 inches less than that of the undiminished segment, which area is represented by the shaded area in Fig. 47. The area of a circle is the diameter multiplied by itself and by 0.7854. It is easy, then, to find the area of the circle of which the shaded area is a part. Sup- pose We are dealing with a 72-inch boiler. Allowing for 3 inches on each end of the diameter, the diameter 67 68 BOILERS of the circle of which the segment to be braced is a part would be rr . , 72 6 = 66 inches, and its area would be 66 X 66 X 0.7854 = 3421 square inches; and the area of the half circle abode would be one- half of this, or 1710 square inches. FIG. 47. Now, if from this area the area a b d e is subtracted, the remainder will be the required area of the (shaded) portion to be braced. The hight / g is the radius, or one-half the given diameter less the given hight plus 2, and it will be near enough if we consider its length equal to the diameter, as the length of the chord b d is not usually given. Suppose the hight h i to be 26 inches, then the hight / g of the portion to be subtracted would be - 26 + 2 = inches, HORIZONTAL TUBLAR BOILERS 69 and if its length be taken at 66 inches its area will be 12 X 66 = 792 square inches. This is too great by the area of the two little dotted triangles at a b and d e, but this is so small a proportion of the total area that it may be neglected, especially if it is borne in mind when deciding upon the number of braces that the area as determined is a little small. Subtracting this area from that of one-half the 66- irich circle, as above found, we have 1710 792 =918 square inches as the area to be braced. If the pressure is 100 pounds per square inch, the force to be braced against is 918 X 100 = 91,800 pounds, and if the braces used are good for 8000 pounds apiece, it will take 91,800 ~ 8000 =11.5 braces. We should have to use 12 braces, anyway, and these would be good for 12 X 8000 , . , - = 960 inches, 100 while the actual area is 936, instead of 918, as the above approximate method made it. Unless the number of braces comes out very nearly square in the calcula- tion, there will be enough leeway in using a whole brace for the fraction to make up for the shortness of the area. When this fraction exceeds, say, 0.9, safety would be insured by putting in an extra brace. VIII GRAPHICAL DETERMINATION OF BOILER DIMENSIONS * THE variables entering into the design of a steam boiler shell are the working pressure, the diameter of the shell, the thickness and tensile strength of the plate, the diameter, spacing and shearing value of the rivets, the efficiency of the joints and the factor of safety. The usual working pressures are 80, 100, 125, and 150 pounds per square inch. The standard diameters of shell are 44, 48, 54, 60, 66, 72, 78, 84, 90 and 96 inches. The tensile strength of the plate is 52,000 to 62,000 pounds per square inch for flange steel and 55,000 to 65,000 pounds per square inch for fire-box steel. The average assumed for calculations is 60,000 pounds per square inch. The shearing value of steel rivets is 38,000 to 42,000 pounds per square inch. Until recently 38,000 pounds per square inch was used for all calculations, but this value has been gradually increasing with the improved quality of steel rivets, until 42,000 pounds per square inch is now the more generally accepted value. 1 Contributed to Power by N. A. Carle. 70 BOILER DIMENSIONS 71 This has resulted in an increased spacing of rivets, together with an increase in the efficiency of the joints, and a consequent reduction in the thickness of plate. Rivet holes are usually punched T V-inch larger than the rivets and calculated as J-inch larger than the rivets. In marine practice, holes are specified as drilled or punched T V~inch small, the shell assembled and the holes then reamed to full size. The shearing value of the rivet is calculated for the stock size before driving. The crushing value of steel rivets has been practi- cally eliminated from the problem, because in practice the sizes selected give values in excess of the shearing value. No consideration is given to the friction of the joint, it being assumed that this is all destroyed before rup- ture, so that it is not a factor of the ultimate strength. The kind of joints and size and spacing of the rivets are governed by accident insurance companies' re- quirements and shop practice. The size of rivets and spacing used necessary to insure good calking usually make the horizontal joint the weakest point in the boiler and therefore the governing factor. It is desirable to get a high efficiency of the joint for high pressures and thick plates. Different types of joints are designated as single lap-riveted, double lap-riveted, triple lap-riveted, double butt-strap- riveted, triple butt-strap-riveted and quadruple butt- strap-riveted. 72 BOILERS The single lap-riveted joint is used on girth seams generally, as the stress is only one-half that on the horizontal joint, and on the horizontal seams only for very small diameters and pressures. The quadruple butt-strap-riveted joint is used only on very heavy plate, large diameters and high pres- sures. The efficiencies depend upon the rivet spacing, diameter of rivets and the allowances and assumptions made. Design conditions reduce the problem to the effi- ciency of the joint based on tearing between the outer row of rivets. The usual efficiencies used in calculations in the shell formula are double lap, 70 per cent.; triple lap, 75 per cent.; double butt-strap, 80 per cent., and triple butt-strap, 86 per cent. The factors of safety ordinarily used are 4, 4! and 5, with 6 sometimes specified in marine practice. The shell formula is D. X W. P. X F. S. = 2 X 5. X E. X *. D. = Diameter of shell in inches. W . P. == Working pressure in pounds per square inch. F. S. = Factor of safety. E. = Efficiency of horizontal joint. /. = Thickness of plate in inches. These are shown graphically in the calculating diagram (Fig. 48). The use of this diagram is probably best illustrated by an example: Given. The boiler shell 66 inches in diameter BOILER DIMENSIONS 73 for a working pressure of 125 pounds with a factor of safety of 5. What thickness of plate is required for the shell? Assume that a double butt-strap joint will be used with an efficiency of 80 per cent. Starting with 66 inches "diameter of shell," read across to 125 pounds 74 BOILERS "working pressure/' then up to a "factor of safety" of 5, and then across to its intersection with a vertical line through 80 per cent, "efficiency of joint." This gives a value slightly less than T 7 ^ inch for "thickness of plate." Hence use T 7 e inch and by reading back it will be found that this gives about 5.1 as factor of safety. Usually the designer has shop practice to follow, so that instead of using approximate values for the efficiency, the usual spacing and diameter of rivets can be selected and the actual efficiency obtained. As an example, assume that for a double butt-strap-riveted joint the shop spacing was 4J inches and 2\ inches, using ^-inch rivets. Read across from 4^ inches "spacing of rivets" to ^-inch rivets and then up to 82 per- cent, "efficiency of horizontal joint." The boiler heads are made T V inch thicker than the shell, as the metal is decreased about this amount in dishing and flanging the head. The spacing of the girth- seam rivets is according to shop practice and does not require a high efficiency, as the stress is only one- half that of the shell. It is, therefore, a dependent factor in the design. IX THE SAFETY VALVE THE study of the safety valve has been the first step of many a man in scientific engineering. Induced to its study by the necessity of solving its problems be- fore the examiner, his consideration of this simple device has led him into the computation of areas, into a study of the principle of the lever, of moments of forces, of the velocity of flow of steam and other fundamental principles of mechanics. This applies to those who have studied the subject intelligently, not to those who have attempted to get over it by learning a rule by rote, simply to be confounded when con- fronted by another rule, or a case to which their rule would not apply. The whole subject is so simple that an hour's study will put a man in possession of the fundamentals so that he can make his own rules or solve any problem without a rule, from a sheer under- standing of the principles involved, PRESSURE PER SQUARE INCH A cubic foot of water weighs, in round numbers, 62 pounds. If you can imagine ten cubic feet packed one above the other, as in Fig. 49, they would make a column weighing some 206 pounds, supported on a 75 7 6 BOILERS \ FIG. 40. THE SAFETY VALVE 77 base one foot square, so that the pressure would be 620 pounds per square foot. The water in a tank or pond may be conceived to be divided into columns of this kind, and it will be seen that there will be a pres- sure on the bottom of 62 pounds per square foot for every foot of depth. But, in the square foot support- ing this weight there are 144 square inches; and as the pressure is evenly distributed, each square inch carries: , 62 -T- 144 = 0.43 of a pound. for each foot in depth, and the pressure in the case of the column 10 feet in hight would be 620 pounds per square foot, or 4.3 pounds per square inch. Just as the tank or pond could be conceived to be divided into columns of one square foot section, each square foot can be conceived to be divided into 144 columns of one square inch section, as shown in Fig. 49, and each foot in hight of such column, like the piece marked A, would weigh T | of the whole weight of the cubic foot of which it is the T | part, and press upon its square inch of base with a pressure of: 62 -f- 144 = 0.43 of a pound. As this pressure in a liquid or gas is exerted in all directions, it is evident that the pressure on the hori- zontal piston in Fig. 50 would be 4.3 pounds per square inch, and if it has an area of 30 square inches there would be a force of: 4.3 X 30 = 129 pounds, forcing the piston to the right; and since there is at a BOILERS 116.1 Lbs. a j b I Depth in Pressure Ibs. Feet persq.in. r-0 0.43 0.86 1.29 5 10 1.72 2.15 2.58 3.01 3.44 3.87 4.30 FIG. 50. THE SAFETY VALVE 79 depth of 9 feet a pressure of 3.87 pounds per square inch the valve at the left would have 3.87 pounds pushing upward on each square inch of its exposed area, i.e., the area corresponding with the diameter a b, and if that area were 30 square inches it would t"i ICP * 30 X 3.87 = 116.1 pounds to hold the valve closed against that pressure. The steam gage shows the pressure per square inch. If the gage points to the 100 mark it indicates that if the pressure existing in the boiler Were exerted upon one square inch of area, Fig. 51, it would push with a force of one hundred pounds. If exerted upon an l.Inch Unch- Unch J FIG. 51. FIG. 52. FIG. 53. area of one-half a square inch, Fig. 52, it would push with a force of 50 pounds; upon an area J inch square, or 1 of a square inch, Fig. 53, 25 pounds; upon an area of one square foot, or 144 square inches, 14,400 pounds, etc. The force exerted by the steam to lift a safety valve depends then upon the area of the valve as well as upon the intensity of the pressure. 8o BOILERS To FIND THE AREA OF A CIRCLE The area of a i-inch circle is 0.7854 of a square inch, the difference, 0.2146, between this and the full square of the diameter being taken up by the corners, Fig. 54. If the side of the square is doubled the area V llnch - -2 Inches- -- FIG. 54. FIG. 55. of the square will be multiplied by four, as is plainly shown by Fig. 55, which obviously contains four squares of the area of that shown in Fig. 54, although its side is but twice as long; and it is equally evident that the inclosed circle bears the same proportion to the total area in both cases and that the shaded area of the circle in Fig. 55 is four times that in Fig. 54, although its diameter is but twice that of the smaller circle. If we treble the length of the sides the area of the square will be multiplied by nine, always the square of the side, i.e., the side multiplied by itself. THE SAFETY VALVE 81 The area of any circle may be found by multiplying the area of a i-inch circle (0.7854) square inch by the square of the given diameter. In Fig. 55 the diameter is 2 inches and the area is: 2 X 2 X 0.7854 = 3.1416 square inches. The area of a 4-inch circle would be: 4 X 4 X 0.7854 = 12.5664 square inches. It may aid in remembering the factor 0.7854 to know that it is one-fourth of 3.1416, the number by which the diameter is multiplied to get the circumference. The area of a triangle is obviously one-half the product of its base and night. In Fig. 530 the product FIG. 530. FIG. 536. FIG. 5 3 c. of the base A B and the hight A C would be the area of the rectangle A B C D and the shaded area of the triangle is obviously one-half of this, for the two unshaded portions put together would make a similar triangle. This is just as true if the base is an arc of a circle as in Fig. 53^, and just as true if the base incloses the apex as in Fig. 53^. The circle is therefore a triangle with a circular base 3.1416 times the diameter 82 BOILERS or 2 X 3.1416 times the radius, and with a hight equal to the radius, and its area (one-half the product of hight and base) is: A radius X 2 X 3.1416 X radius Area = - -| = 3.1416 radius 2 , so that the area equals 3.1416 times the square of the radius, and since the radius is one-half the diameter, the square of the radius is the square of the diameter divided by four: Area = 3.1416 r 2 = 3.1416 = D 2 X ^^ 4 4 = 0.7854 D 2 . EFFECT OF PRESSURE IN LIFTING A VALVE Suppose the 3-inch valve in Fig. 56 to be loaded with six weights of 100 pounds each and that the valve and steam weighed 30 pounds, what would the pressure per square inch have to be to lift it? The total weight to be lifted is 630 pounds. The total upward pressure must equal this, and if 630 pounds is exerted on 7.0686 square inches (the area of a 3-inch valve, see table) the pressure on each square inch will be: ^ 630 -T- 7.0686 = 89. i pounds. How much load would have to be put upon the same valve to allow it to blow off at 75 pounds per square inch? If the pressure exerts 75 pounds on one square inch, it would exert on the 7.0686 square inches of the valve which is exposed to it: THE SAFETY VALVE 83 75 X 7.0686 = 530 pounds, which must be the combined weight of the valve and the weights with which it is loaded. Fig. 56 does not show a practicable valve, but is sufficient to illustrate the point that the force tending FIG. 56. to lift the valve must equal that holding it to its seat (in this case the dead weight of the valve itself and the weights with which it is loaded), and that this upwardly acting force is the area of the valve in square inches, multiplied by the pressure per square inch. Such a dead-weight valve is ponderous and impracticable 8 4 BOILERS and the usual practice is to use a lighter weight, increas- ing its effect by leverage, or to hold the valve to its seat with a spring. THE PRINCIPLE OF THE LEVER Suppose a strip of board balanced over a sharp edge as in Fig. 57. If equal weights be placed upon it at h ' A FIG. 57. equal distances from the center it will still be in bal- ance. If one of the weights be moved in half of the distance to the point at which they are balanced, as in Fig. 58, the other weight will have to be halved to U 12 >!*- 6 *\ FIG. 58. preserve the equilibrium. If one of the weights be moved to one-third of its distance from the balancing point, as in Fig. 59, the other weight will have to be THE SAFETY VALVE 85 reduced to one-third of its original magnitude to pre- serve the balance at the original distance. A FIG. 59. Notice that in each case the product of a weight by its distance from the point over which they balance is the same as the product of the weight which balances it and its distance from the same point. Suppose the weights in Fig. 57 to be each 20 pounds, each at 12 inches from the center. Here obviously the weights and distances being the same their products are equal: 20 X 12 = 240 and 20 X 12 = 240. When the right-hand weight is moved in to 6 inches from the center the other had to be reduced to 10 pounds: , 10 X 12 = 1 20 and 20 X 6 = 120. When the left-hand weight was moved in to 4 inches from the center the other had to be reduced to 6: 6 X 12 = 80 and 20 X 4 = 80. The same principle applies in Fig. 60, where the force exerted by the man, multiplied by the distance A B, must, if he lifts the machine, equal the pressure with which the load bears on the bar at the point C, 86 BOILERS multiplied by the distance B C of that point from the point B around which the lever turns. In mechanics, this point, the B of Fig. 60, is called the "fulcrum" and the product of the load, weight or force by its distance from the fulcrum is called its "moment." In the case described by Fig. 57 the moment of each FIG. 60. weight is 240; in that of Fig. 58, 120; in that of Fig. 59, 80; in that shown in Fig. 60 the moment of the load is the weight or force with which the load bears on the point C, multiplied by its distance from the fulcrum B, and the moment of the force is the force which the man exerts upon the bar at A, multiplied by the dis- tance of that point from the fulcrum. Notice that in Fig. 61 the fulcrum is at one end of THE SAFETY VALVE 87 the lever instead of between the load and force as in the other examples. The principle is the same. The fulcrum is the stationary point about which the load and the force move. In Figs. 60 and 61 it is evident that the shorter the distance between the load and the fulcrum the less the man will have to exert himself. \ FIG. 61. The point to grasp and remember is that the mo- ments must be equal in order for the force to balance or lift the load. Equal Moments Produce Equilibrium There are four important things about a lever: 88 BOILERS L -the load. F = the force applied to balance or overcome the load. D t = distance of the load from the fulcrum. D f = distance of the force from the fulcrum. If any three of these are known the third can be easily determined, for, as has been just explained, Force X distance of force = load X distance of load. P X D = L X DI Moment of force = moment of load. To find the force required to lift a given load: FORMULA: T vx ^ RULE. Multiply tie load by its distance from tie fulcrum, and divide by tie distance at which tie force is applied from tie fulcrum. To find the distance at which a given force must be applied from the fulcrum to balance a given load: FORMULA: T n Df= ^- RULE. Multiply tie load by its distance from the fulcrum and divide by tie given force. To find the load which may be lifted with a given force : FORMULA: r n T r \ Uf THE SAFETY VALVE 89 RULE. Multiply the given force by the distance of its point of application from the fulcrum and divide by the distance of the load from the fulcrum. To find the distance at which a given weight or load must be placed from the fulcrum to balance a given force : FORMULA: D = F X D/ Li RULE. Multiply the given force by the distance of its point of application from the fulcrum and divide by the load. THE LEVER SAFETY VALVE Effect of the Leverage of the Ball Suppose the weight instead of setting directly upon 1 i FIG. 62. the valve, as in Fig. 56, is applied through a lever, as in Fig. 62. From what has preceded it will easily go BOILERS be seen that the weight multiplied by its distance from the fulcrum will equal the force which it will exert upon the valve stem multiplied by the distance of its point of application from the fulcrum. Weight f Distance of ball Pressure of ball i 1 Distance of stem of ball X from = on from L/dll fulcrum . stem I fulcrum . Let the weight equal 75 pounds, distance of weight from fulcrum 32 inches, distance of stem from fulcrum 2f inches, what will be the force exerted by the ball to hold the valve to its seat? Weight of ball X Distance of ball from fulcrum _ Distance of stem from fulcrum Pressure of ball on stem. Then the moment of the ball is: 75 X 32 = 2400 inch-pounds, and: 2400 -f- 2.75 = 872.727 pounds will be the pressure on the valve stem due to the ball and the moments will be equal: 75 X 32 = 2400 and 872.727 X 2.75 = 2400. Suppose this to be a 4-inch valve, the area of which is 12.5666 square inches. The pressure per square inch upon the under side of the valve necessary to balance the effect of the ball would be: THE SAFETY VALVE 91 872.727 ~ 12.5666 = 69.4 pounds. This is the pressure at which the valve would blow off if nothing but the ball were holding it to its seat. It takes a little additional pressure to lift the valve and to overcome the weight of the lever, as will be explained later, but this is a comparatively small affair and in usual approximate calculations is not taken into account. Neglecting these we can make the follow- ing simple Rules for lever safety valve, neglecting weight of valve, stem and lever: Let W = weight of the ball, D = distance of ball from fulcrum, A = area of valve in square inches, d = distance of stem from fulcrum, P = pressure per square inch on valve which will balance ball. To determine the pressure on a valve of given diameter required to balance a ball of given weight at a given distance from the fulcrum. F RMULA: W XD P = Axd RULE. Multiply the weight of the ball by its dis- tance from the fulcrum. Multiply the area of the valve in square inches by the distance of its stem from the fulcrum. Divide the first product by the second and the quotient will be the pressure per square inch required to overcome the weight of the ball. 92 BOILERS EXAMPLE. The stem of a 4-inch safety valve is 2 J inches from the fulcrum. Supposing the valve will blow when the gage shows 7 pounds without any weight upon the lever (i.e., that it takes 7 pounds per square inch on the area of the valve to overcome its own weight, that of the stem and the bearing effect of the empty lever), at what pressure would it blow with a weight of 75 pounds (Fig. 62) 32 inches from the fulcrum? BY THE FORMULA: P W A^ D i = 7 77^ = 69-4 + 7 = 76.4 pounds. A X d 12.5666 X 2.75 BY THE RULE: Area of valve 12.5666 75 Weight of ball Distance of stem 2.75 32 Distance of ball 628330 1 50 879662 225 251332 34.558^)2400.00)69.4 pounds. This is the pressure required to lift the ball.' Adding the 7 pounds required to blow the valve without the ball, the' answer would be 76.4 pounds. Scratching out the last three figures of the first product saves hand- ling large numbers and does not materially affect the result. If we called this 34.6 (nearer right than 34.5 because the 58 rejected is over one-half) the quotient would still be 69.36. To find the weight required to hold a given pressure on a given valve: THE SAFETY VALVE 93 FORMULA: D RULE. Multiply the area by the pressure and by the distance of the stem from the fulcrum and divide by the distance of the ball from the fulcrum. The quotient will be the weight of ball required to balance the steam pressure on the valve. EXAMPLE. What weight of ball would be required to allow the valve in the above example to blow off at 80 pounds? The ball must provide for 73 pounds per square inch, the lever valve and stem taking care of the other seven, so that P = 73 pounds. BY THE FORMULA: . y8 . 8 pounds . BY THE. RULE: 73 Pressure Distance of stem 376998 879662 917-3618 ?75 45868090 64215326 18347236 Distance of ball, 32)2522.744950(78.8 pounds. To find the position of the weight in order that it may exert a given pressure on the stem : 94 BOILERS FORMULA: _ AxPxd W RULE. Multiply the area by ihe pressure and by the distance of the stem from the fulcrum and divide by the weight of the ball. The quotient will be the distance at which the ball must be from the fulcrum in order to produce a given pressure on the stem. EXAMPLE. If the original 75-pound weight had been used, at what distance from the fulcrum would it have had to have been placed to have allowed the valve to blow off at 80 pounds? BY THE FORMULA: D . ^y = ".566x^73x3.75 = 3? 6 jnches . BY THE RULE. The product of the factor in the numerator is 2522.74495 as before, and dividing this by 75, the weight of the ball: 75)2522.74495 (33.6 inches. These simple rules will serve all practical purposes, especially if it is borne in mind that P represents the pressure with which the ball only bears upon the stem, not including the weight of the valve, lever, etc., and an allowance be made for these other effects as has been done in the examples. A general idea of what the pres- sure per square inch required to lift the valve, stem and lever may be is given in column 8 of the table on page 119. It is well, however, to know how to make these allowances accurately, and they will now be considered. THE SAFETY VALVE 95 EFFECT OF THE WEIGHT OF THE VALVE AND STEM The pressure acts directly upon the valve and stem without leverage, and must exert a force to balance their weight equal simply to that weight, just as was the case in Fig. 56. Suppose the valve and stem of a 3-inch valve to weigh 1.5 pounds, how much pressure per square inch would be required to lift the valve from its seat? Comparing Figs. 56 and 63, it will be seen that this case is the same as the first example given in describing the earlier cut. The total pressure on the valve must be 1.5 pounds, and if 1.5 pounds is to be exerted on 7.0686 square inches, the pressure per square inch will be: 1.5 -f- 7.0686 = 0.212 pound. Column 3 of the table on page 119 gives roughly the weights of valve and stem used on valves of the standard diameters of three makers, and in connection with column 4, which gives the pressure per square inch required to lift the valves of the given weights, serves to indicate the relative importance of this factor of the problem. THE EFFECT OF THE LEVER The weight of the lever tends to hold the valve upon its seat. It is evident that it would take a considerable pull to lift the lever of a large safety valve with a cord attached at the point at which the pin bears, as in Fig. 64, and this pull as measured upon a scale would be 9 6 BOILERS the force which the valve would have to exert to push the lever up. Every successive particle in the length of the lever is acting with a different leverage, so that it FIG. 63. would at first appear a complicated process to calculate this force; but a body acts in this respect just as though its whole mass were concentrated at its center of gravity and this makes the problem very simple. THE SAFETY VALVE 97 If the lever be taken off and balanced over an edge, as in Fig. 65, the center of gravity will be at the point FIG. 64. above the knife edge when the lever is balanced, and the effect of the lever would be the same as if all the mass were concentrated at that point. FIG. 65. Now find the distance of the center of gravity from the fulcrum, from the point around which the lever turns. This will be from the center of the hole when it turns upon a pin, as in Fig. 66, or from the point where it bears if a knife edge is used, as in Fig. 67; the dis- tance a c in each case. 9 8 BOILERS In measuring for moments the distances must be taken on a line passing through the fulcrum and at o- FIG. 66. right angles to the direction of the force. In the case of the lever safety valve the holding-down force is FIG. 67. gravity, which acts vertically. A line at right angles to the vertical is horizontal, so that distances should A B FIG. 68. be measured in a horizontal direction as through ABC, Fig. 68, and not on the lines x x or y y. THE SAFETY VALVE 99 In determining the distance a c, Figs. 66 and 67, do not get bothered about the piece of lever which extends back of the fulcrum. The more metal there is back of this point the nearer the center of gravity is to the fulcrum. If there were as much weight to the left of the pin in Fig. 65 as to the right, the center of gravity would be at the pin; the lever would balance over the pin as it did over the knife edge and not bear on the stem at all. To apply this, suppose that the lever of a 3-inch valve v/eighed six pounds, that the distance a c, Fig. 66, be- tween the fulcrum and the center of gravity was found to be 15 inches, and the distance a b from the fulcrum to the point at which the pin bears 2\ inches. The moment of the lever must be : 6 X 1 5 = 90 inch-pounds. The moment of the lifting force must equal this, and that moment is i\ times the force. Then the force must be: 90 -f- 2\ = 40 pounds, 2-J- X 40 = 90 and 15X6 = 90. Since a force of 40 pounds is to be exerted upon 7.0686 square inches, the force per square inch would 40 -v- 7.068 = 5.66 pounds. The combined effect of the valve and stem and of the lever of the 3-inch valve in question would be: 0.212 + 5.66 = 5.87 pounds. Columns 5 and 6 of the table already referred to give zoo BOILERS the weights of levers and the distances of their centers of gravity from the fulcrum as ordinarily found, and column 7 gives the pressure per square inch on the valve necessary to lift such levers. Column 8 gives the sum of the respective values in columns 4 and 7, i.e., the pressure per square inch required to lift the valve and stem and the lever. It will be seen that the values run fairly even for all sizes of valves, and that by using seven or eight pounds as an allowance as in the above examples, results can be attained with the simple rules which will be within a pound or two of right. SPRING-LOADED OR POP SAFETY VALVES A rule for calculating the pressure at which a spring- loaded valve will blow off is sometimes asked for. There are none reliable that do not involve the deter- mining by experiment of the force required to com- press the spring, and if you are going to do this you may as well determine by experiment at what pressure the valve will, blow off. In practice nobody thinks of com- puting the spring-loaded valve. If they want it to blow off at 1 20 pounds they procure a suitable spring from the makers and turn down upon the binding nut until the valve will blow experimentally at the desired pres- sure. The pressure at which a spring will yield depends not only upon the shape and size of the material of which it is made, the diameter, number, and pitch of the coils, all of which are measurable and determinable, but upon the nature and condition of the material itself. You can readily appreciate that a spring of brass would compress with less pressure than one of steel, similar in THE SAFETY VALVE 101 every other respect, and that there is such a wide differ- ence in steels that there will be a great deal of difference in the action of steel springs according to the kind of metal, degree of temper, etc. The best rule known is the following: To find at what pressure a valve will lift with a spring of given dimensions and compression: Multiply the compression in inches by the fourth power of the thickness of the steel in sixteenths of an inch, and by 22 for round or 30 for square steel. Product I. FIG. 69. Multiply the cube of the diameter of the spring, measured from center to center of the coil (as on the line d, in Fig. 69) in inches, by the number of free coils in the spring, and by the area of the valve in square inches. Product II. Divide Product I by Product II and the quotient will 102 BOILERS be the pressure per square inch at which the valve will blow off. The weight of the valve and of the spring should in strictness be added to Product I, when the construction is such that the valve supports the spring; but inasmuch as the values 22 and 30 are guessed at it will not pay to go into refinements in other directions. The result of this rule has never been compared with an actual valve. It is based on a formula adopted by a com- mittee of Scotch engineers and shipbuilders. Corre- spondence with the manufacturers of pop safety valves as to the accuracy of the formula brings out the fact that they proportion and calibrate their springs only by experience and experiment. However, this rule is given for what it is worth. If you have a spring- loaded valve calculate it by this rule and see how nearly it comes to the point at which the valve will blow off. With a dead weight or a lever-loaded valve the force required to lift it remains the same, no matter how high the valve lifts. The weights weigh no more if they are raised an inch or two, and the leverage does not change, but with the spring-loaded valve the more the valve lifts, the more the spring is compressed, and the more force is required to compress or hold it. It follows then that if an ordinary valve were loaded with a spring it would simply crack open and commence to sizzle when the pressure equaled the force at which the spring was set, and that if this were not enough to relieve the boiler the pressure would have to increase, opening the valve more and more until the steam blew of? as fast as it was made. THE SAFETY VALVE 103 COMPLETE SAFTEY-VALVE RULES It is evident that any complete rule for the safety valve must include the separate treatment of the valve and stem, the lever and the ball as factors in holding the valve to its seat. moment of ball Moment of the lifting force moment of lever moment of valve and stem. The lifting force consists of the pressure per square inch into the area of the valve, and its moment is the product of the force by its distance from the fulcrum. Expanded, then, the above becomes: Weight of ball X distance of its center of gravity from the fulcrum Pressure X Area X Distance of stem from fulcrum weight of lever X distance of its center of gravity from fulcrum weight of valve and stem X distance of their center of gravity from the fulcrum. In order to find one of these qualities we must know all the rest, and consequently since the missing quantity can be but on one side of the equal mark we can figure the combined value of the quantities on one side of the 104 BOILERS equation (that is, in one set of brackets). Then we can work out the operation indicated on the other side as far as we can go. If the missing quantity is on the left-hand side of the equation it can be found by divi- ding the value of the other side of the equation by the product of the two known factors on the left-hand side. To find the pressure at which a certain valve will blow off: Multiply the weight of the ball, of the valve and stem and of the lever, each by the distance of its center of gravity from the fulcrum and add the products. Multiply the area of the valve by the distance of its center from the fulcrum and divide the sum above found by the product. The quotient will be the pressure required. Or more briefly: Divide the sum of the moments of the valve, lever and ball by the product of the area of the valve and distance from the fulcrum. EXAMPLE. At what pressure will a 3-inch valve blow off with stem 2\ inches from the fulcrum, valve and stem weighing i J pounds, lever weighing 6 pounds, having its center of gravity 1 5 inches from the fulcrum and weighted with a 48-pound ball 24 inches from the fulcrum? Pressure X Area = 7.0686 X Distance 2j Product = 15.90435 j 48 X'24 = 1 1 52 6x15= 90 i.5X2i= 3.375 Sum of moments = 1245.375 1 2 45-375 - i5-9 435 = 7 8 -3 pounds. THE SAFETY VALVE 105 This is all that we shall be likely to wish to find on this side of the equation, for the distance of stem is fixed and the area determined by other considera- tions. The other two things that interest us are the weight of the ball and its distance from the fulcrum. To find weight of ball or its distance from fulcrum : Multiply the pressure by the area and by the distance of the stem from the fulcrum. The product is the moment of the force. Multiply the weight of the valve by the distance of the stem from the fulcrum; multiply the weight of the lever by the distance of its center of gravity from the fulcrum, and add the products. Subtract 'the sum of the products just found from the moment of the force, and the difference is the moment of the ball. Divide the moment of the ball by the weight of the ball and the quotient is its distance from the fulcrum. Divide the moment of the ball by the distance from the fulcrum and the quotient is the weight of ball required. EXAMPLE What weight of ball at the same dis- tance would be required to allow the valve given in the previous example to blow at 75 pounds, and at what distance would the 48-pound ball there given have to be placed from the fulcrum to produce the same result? io6 BOILERS Moment 75 X 7.0686 X 21 of force 1 192.826 93-375 1099.451 mo- rn e n t of ball Weight X distance of ball + 6X15 - 90.000 + 1.5 X 2.25 = 3.375 93.375 sum of moments, valve and lever. = inches, distance of ball. 4b 1099.451 . . - = pounds, weight of ball. 24 But the ideal valve should stay on its seat until the pressure reaches the desired limit, then open wide and discharge the excess. This result is accomplished by the construction shown in Fig. 70. With the first opening of the valve the steam passes into the little "huddling chamber" made by the cavity near the overhanging edge of the valve and a similar cavity surrounding the seat. The pressure which accumulates here, acting on the additional area of the valve, raises it sharply with the "pop" which gives the valve its name, and it is sustained by the impact and reaction of the issuing steam until the pressure has subsided sufficiently to allow the spring to overcome these actions. The outside edge of the lower trough in the valve shown is composed of an adjustable ring which may be THE SAFETY VALVE 107 FIG. 70. io8 BOILERS screwed up or down so as to diminish or increase the dis- tance between the overhanging lip of the valve and its own inner edge, controlling the outlet from the chamber; and diminution of pressure or the "blow back" required to allow the valve to seat so that the valve opens wide at a given pressure and seats promptly without sizzling or chattering when the pressure has been reduced a cer- tain amount depending upon the adjustment of the ring. The various makers have adopted different devices for adjusting the ring or other device for con- trolling the outflow from the huddling chamber. THE CAPACITY OF SAFETY VALVES Let us next consider the capacity of valves; how large a valve is required for a given boiler. Most of the rules deal with grate surface and the area of the valve; the rule adopted by the U. S. Board of Supervising In- spectors being one square inch of valve area for each two feet of grate area. That the valve should be pro- portioned to the grate surface seems proper because it is the grate surface, and not the heating surface, which determines and limits the capacity of a boiler. To a given grate surface, however, we should apportion a sufficient amount of area of opening, and this area of opening is not proportional to the area of the valve but to the diameter and lift. A valve i inch in diameter has an area of 0.7854 of a square inch, but that does not mean that there will be an opening of 0.7854 of a square inch for the steam to escape. If the valve is flat, as in Fig. 71, the area opened for the discharge of steam THE SAFETY VALVE 109 will be the circumference of the valve multiplied by the lift. The circumference is Diameter X 3.1416 (i) and the area of the complete circle is ^. ,. Diameter Diameter X 3.1416 X - (2) 4 and the area for the escape of steam is Diameter X 3.1416 X Lift. (3) When the lift is one-quarter the diameter, or Diameter the area for the escape of steam is the same as the area of the circle; formula 3 is the same as formula 2. FIG. 71. When a flat valve has lifted a quarter of its diameter it has reached the limit of its capacity to discharge steam. It doesn't do any good to lift higher, for the area around the edge of the valve is already as large as the area of the valve itself and the capacity of the valve is proportional to the area or the square of the diameter. no BOILERS In practice, however, the lift of valves is much less than one-quarter of their diameter, and for a given lift the area for the escape of steam is proportional to the cir- cumference or the diameter rather than to the area. Most of the rules, however, as above stated, allow a given amount of valve area to a square foot of grate surface, and make the allowance liberal enough to include all conditions. For instance, the rule of the U. S. Board of Supervising Inspectors calls for one-half a square inch of valve area for each square foot of grate surface. A 4-inch valve has about 12 square inches of area and would thus take care of 24 square feet of grate. It would not be possible to burn over 25 pounds of coal per square foot of grate per hour with natural draft, nor to evaporate over 12 pounds of water with a pound of coal, so that the boiler could not possibly make more than 25 X 12 X 24 = 7200 pounds of steam per hour, or 7200 -T- (60 X 60) = 2 pounds of steam per second. Now the weight of the steam which will escape through a given aperture per second is given by the following f rmUla: wt P^ssure X Area 70 that is, the weight in pounds which will escape in a second is equal to the absolute pressure in pounds per square inch multiplied by the area in square inches and divided by 70. THE SAFETY VALVE in On the other hand, the area required to discharge a given weight is Wd h Area = - Pressure that is, the weight in pounds to be discharged per second multiplied by 70, and divided by the absolute pressure equals the required area. Now we have found that with a rate of combustion practically impossible, with natural draft, and a practically unattainable evaporation per pound of coal, the most steam that the boiler with 24 square feet of grate suface could furnish is 2 pounds per second. The area required to discharge this at 70 pounds pressure, absolute, is 2 X 7O = 2 square inches. 70 The 4-inch valve which this boiler would require would have a circumference of practically 12 inches, and would need to lift only one-sixth of an inch to furnish the two square inches of opening necessary to discharge the steam, for 12 X J = 2. One-sixth of an inch is only one twenty-fourth of the diameter of the valve. You see that this simple rule gives an ample margin, requiring but a small lift to discharge more steam than the boiler can possibly make. It is altogether useless and nonsensical to figure the areas of opening to four places of decimals involving with beveled seats complicated operations with sines and cosines, in a calculation which involves no accuracy 112 BOILERS but which requires simply a result which shall be amply large to cover any emergency likely to be encountered in practice. It is like trying to measure the distance to the next town in feet and inches, in order to answer a man who would be abundantly satisfied to know that it was about three-quarters of a mile. You may be sure that a valve which has a square inch of area for each two square feet of grate surface will liberate all the steam that can be made by the coal that you can burn on that grate surface, so long as the valve is free and in good condition. It is quite probable that a smaller valve would do, but in a matter of this kind we want to provide not the smallest that will possibly do but enough capacity to be absolutely safe. For all purposes of ordinary practice, therefore, divide the grate surface by 2, which will give you the valve area required and you can find the corresponding diameter by multiply- ing the square root of the area by 1.128. Don't carry your decimals out too far because you will have to take the nearest commercial size after all. Here is a rule which will give you the diameter of the valve in inches at once: Multiply the square root of the grate surface by 0.8. This would be particularly handy when the grate is square, or nearly so, for then the length would be the square root of the area. You can see how the rule is made, or rather, makes itself. By the supervising inspector's rule the valve area required equals the grate surface divided by 2. THE SAFETY VALVE 113 A grate surface Area - . The diameter is the square root of the quotient of the area divided by 0.7854. TV , / area Diameter = 0.7854* And since in this case the area equals one-half the grate surface the diameter will be the square root of one-half the grate surface divided by 0.7854. . /grate surface. Diameter = \/ - 5 V 2 x 0.7854 Diameter = ./grate surface V 1.5708 We can get rid of the square root in the denominator by finding it once for all. It is 1.25 very nearly. So our formula becomes TV Vgrate surface Diameter = - 1.25 Dividing by 1.25 is just the same as multiplying by T.^y, and as T is = 0.8, the multiplication is easier, so we have / Diameter = V grate surface X 0.8. The grate surface will never be so large that the square root cannot be easily determined with sufficient accuracy mentally. If it is between 25 and 36 the root is between 5 and 6. The square of 7 is 49, of 8, 64, etc., so that by trial the root can be determined approxi- H4 BOILERS mately. Here is an easy trick to get the square of a number with two figures ending in 5 : Multiply i plus the left-hand figure by the left-hand figure, and annex 25 to the product. What is the square of 35? The left-hand figure is 3. Three plus i is 4, and 4 X 3 = 12. Annex 25 and get 1225. This rule works just the same when the 5 is a decimal, only in that case the annexed 25 is a decimal too, and will enable you to determine instantly by inspection the nearest number advancing by halves to the square root. As the sizes of safety valves advance by half inches, the nearest root determined in this way will be sufficiently accurate, as we have to take the nearest commercial size anyhow. What is the square of 6.5? Six plus i = 7; 7 X 6 = 42; add 25, which in this case will be a decimal fraction, there being two places to point off, and get 42.25. In this way you can square 1.5, 2.5, 3.5, etc., and this is as near as it is ever necessary to get a root in the above formula. Suppose, for instance, you had 58 square feet of grate surface. What is the square root? Seven times 7 = 49, and 8 X 8 = 64. It must be between 7 and 8; 7.5 X 7.5 = 56.25. That is near enough to 58. The square root of 58 is really 7.61 5. Multiplying this by 0.8 we get 7.61 5 X 0.8 = 6.092, which is practically a 6-inch valve. We should have got at the same result if we had taken the square root as 7.5, for 7.5 X 0.8 = 6. When the grate surface is over 30 or 40 feet it is THE SAFETY VALVE 115 better to get the required capacity by putting on two valves than by using one large one. In fact it is a pretty good plan to have two safety valves anyway. There is a great deal of responsibility on that little appliance, and many of the most destructive of boiler explosions would have been avoided by an operative safety valve of sufficient capacity. So many little things can occur to make it hold against a destructive pressure, even when the attendant follows the usual directions to raise it from its seat daily, that prudence dictates the use of an auxiliary valve. It would be a remarkable coincidence if both stuck at the same time without criminal negligence. The amount of opening of an ordinary lever safety valve is determined by the amount of surplus steam to be delivered. If the boiler is making more steam than is to be taken out of it the pressure will increase, and when it reaches an amount sufficient to overcome the weight of the ball, etc., the valve will be raised a little from its seat and the steam will escape. If the opening thus afforded is sufficient with the other drafts on the boiler (such as the supply to the engine, etc.) to allow all the steam the boiler is making to escape, the valve will not open any wider, but if not the pressure will continue to increase and force the valve open until the steam can escape as fast as it is made. As the surplus production of steam decreases, as by closing the dampers or a greater demand by the engine, the valve gradually settles down to its seat again. On account of its greater lift and effective discharging area the pop valve is allowed by the Board of Super- Ii6 BOILERS vising Inspectors three square feet of grate surface per inch of area instead of two, as with the ordinary lever valve. We have seen that the escape of steam through an opening of given size is proportional to the absolute pressure. Twice as much steam will go out of an inch hole in a minute with 190 pounds behind it as with 95 pounds. It is presumed, for this reason, that the inspectors only require a square inch of valve area for every 6 feet of grate surface on boilers carrying a steam pressure exceeding 175 pounds gage. It has been said that although the area effective for the escape of steam is not proportional to the area due to the diameter of the valve, and although the latter area is that used in the formula for capacity, the allow- ance is so liberal that it is practically useless to figure the former. It may be interesting, however, to know how to figure it, and a treatise on the safety valve would hardly be complete without directions for so doing. With a flat valve we have already seen that the area for the escape of steam is the lift of the valve. multiplied by its circumference. With a bevel-seated valve in which the valve does not lift out of the seat the area A A, Fig. 72, is that of a frustum of a cone, Fig. 73. Now to find this area the rule is to add the circumfer- ence of the greater circle to the circumference of the lesser CD; divide by 2, and multiply by the slant hight C A. In other words, to multiply the average length of the strip which would be made by flattening this surface out by the width of that strip. To work this THE SAFETY VALVE 117 rule out would take us too far into trigonometry, but the rule follows: (i) Multiply the diameter of the valve by the lift, by the stine of the angle of inclination and by 3.1416. FIG. 72. (2) Multiply the square of the lift by the square of the sine of the angle of inclination, by the cosine of this angle and by 3.1416. (3) Add these two products. The U. S. rules require a bevel of 45 degrees, and most valves are made with seats of that degree of in- clination. For such a valve the rule becomes: (i) Multiply the diameter of the valve by the lift and by 2.22. n8 BOILERS (2) Multiply the square oj ihe lift by i . 1 1 . (3) Add these two products. When a valve with a beveled seat lifts clear of the seat as a valve with a slight bevel may, the area of the opening is computed by the above rule for a lift which would raise it to the upper level of the seat, and to this is added the circumference of the valve multi- plied by the lift above the seat level. THE SAFETY VALVE 119 2 3 4 1,1 3 > Area of Valve Weight of Valve and Stem Pressure Required to Valve and Stem Lift In. Sq. In. Pounds Pounds per Sq. In I o.i 104 0.125 0.131 I 0.1963 0.156 O. < 4 o-7947 0.713 a 0.441 8 0.187 0. 23 0.423 c 521 I 0.7854 0.187 o. 34 0.238 0.432 i\ 1.2272 0.312 0.60 0.254 0.488 I 2 1.76" i 0-437 0. 75 0.247 C 424 2 3-M 6 0-542 i.. 6a 5 0. )7 0.172 40 C .308 2* 4.9087 0.8395 2.75 1.69 0.171 0.560 0-344 3 7.o6 16 1-339 3-. o 2. 53 0.189 40 c 329 3-V 9.62 i 1.8 4-' r,5 2. io 0.187 o 40 I c .270 4 12.5666 2-371 5-' .s 4- t2 0.189 0.458 0.327 4* 15.904 I 3-0 6.' s 5. [8 0.189 o .42, I c .326 5 19-635 o 4.125 9.' s 6. 4S 0.210 o 49 c 324 6 28.2744 5.87 11-875 8.62 0.208 o 420 0.305 5 6 7 8 Distance of Center of Weight of Lever Gravity from Fulcrum Pressure Required to Raise Lever Pressure Required to Raise Valve, Stem and Lever Pounds Inches Pounds per Sq. In. Pounds per Sq. In. 0.125 3-25 5-89 6.003 0.140 0.20 3.0 6 25 2 Ss 8.49 3-6^ 47 9-203 0-343 0.38 4.812 9.0 4.98 10.32 5-403 10.841 l.O 0.48 7-75 9 o 8 32 5-50 8.5 ; 5-932 1.125 0.65 7.312 8 Si 2 5 5 OS 3-73 5-9<: >4 4.218 0-875 0.87 6.875 IO 125 2.87 3-63 3-II7 4-054 2-5 2.O 1. 12 I3-3I2 14.56 i i so 7 37 7-42 2-43 7-5' .2 7-92 2-738 3-5 3-562 4-0 15-25 16.37 tS O 5 70 5-59 7.24 s.o< )I 6.15 7-584 4-75 5-8i2 6.0 18.625 17-37 10 25 6 07 6.35 7.07 6.8 SO 6.84 7-399 5-75 8.25 10.50 21.75 19.0 10 25 5-78 6,52 9.08 5-967 7.01 9-350 6.0 12.875 13.0 23.0 22. 23 4 7,S 8.20 8-75 4-9< )0 8.66 9.077 7-o 13-125 i8.o 22.125 23.0 26.75 3-89 6,33 11.09 4.079 6.75 11.416 10.75 18.25 20.0 25.25 25-5 31 25 5 S i 7.22 10.62 5-7' I 7.72 10.944 15.0 21.25 32.0 27.5 29.62 37-25 5-56 6.36 12.05 5.768 6.78 12.355 9 10 Length of Lever Distance of Stem from Fulcrum Weight of Ball Ordinarily Furnished Inches Inches Pounds 6.31 0.62 1-56 5-62 12.5 0-75 o-75 3-2 2.0 9-50 18.0 0-75 c 75 5-5 2.6 14-87 18.0 1.19 1.0 8.12 4-8 14.4 17-625 1.19 ] 25 9.62 I l.O 13-4 20.25 1.19 ] 37 15.37 14.0 26.1 29.50 23-0 1.44 I 87 1.687 19.0 30 24.0 30.0 33.0 30.0 1.87 2 i l ] .687 29.0 45 34-5 37-12 35.0 38-5 1.87 2-25 2.312 38.0 63 50-5 43-i 38-3 75 38-5 2.25 s 3 2 .312 48.5 ss 67.0 45-5 44-5 46-5 2-37 2-75 2-75 70.0 110 86.5 43.75 46.1 25 53-5 2-5 3 o 2 75 83.0 I 40 86.5 49.87 51.5 62.5 2-5 2 "> " 4 .0 98.0 I 68 103.0 54-5 59-50 74-5 2.625 3-50 3-5 139.0 220 139.0 X HORSE-POWER OF BOILERS 1 IN a recent catalog of a well-known maker of engi- neering specialties the following approximate rules for calculating the horse-power of various kinds of boilers were noticed and copied. The rules are in- tended for use in determining the proper sizes of injectors and other apparatus when the exact dimen- sions or heating surface of the boiler is unknown or hard to obtain: KIND H. P. Horizontal Tubular = Dia. 2 X Length -r- 5 Vertical Tubular . . == Dia. 2 X Hight -T- 4 Flue Boilers == Dia. X Length H- 3 Locomotive Type = Dia. of Waist 2 X Length over all -r- 6. All dimensions to be in feet. In the first and third cases the length is the length of the tubes or that of a "flush-head" boiler and does not include the extended smoke-box. In the second case, the hight is that of a plain vertical boiler in which the upper part of the tubes is above the water line; it is not the hight of a boiler with submerged tubes. 1 Contributed to Power by C. G. Robbins. 120 HORSE-POWER OF BOILERS 121 The extreme simplicity of the rules aroused curiosity as to their accuracy, and comparisons were made between manufacturers' ratings and ratings calculated by the formulas above. The results are given in the accompanying table. They agree very closely, except in a few of the larger sizes of tubular boilers, where the calculated rating falls below that of the manufacturer. And in these sizes it will be noticed that the heating surface per horse-power is less than in the smaller sizes where the two ratings practically agree. It is quite possible that the ratings of other manu- facturers would show a better or worse agreement. In any event, the rules prove to be valuable for just what is intended and will save considerable trouble in measuring up and calculating the power of existing boilers when ordering injectors, feed pumps, and the like. 122 BOILERS M *t O O X ^ ^ f^ CO Qs * 10 *I5 to ^ fO O o vS^ ||8& o 00 * vO 00 00 Tft xjvo R ^CC g^> X 0*0 o X ^^ ^ 00 10 00 2-00 00 PI ? 00^^^ ^P) ^ X i-&4 *Pn? S|RR il"l o ^ * *.. V 0*00 {^ XoO S " HN *s- ^* H Ov N S^IO* $" 8 * 00 "> "5 ^ 2 n t^. ^^s ?>oo-> ro M \o x 2> * -0 " x S ^^ 't X^ I x^2o ^2 5 W ^ 5- x^-^ 00 "GO ^ Q oo r*> w ^ ^ ^ O> O on f*3 TJ- Hn K P 5"S 2Ts (4 5 Xoo 10 10 CO P4 S O W PH > xiq hJ o ^ I 2 P t> H O w PI PI \o N H M |g8l H I H I.B S < o H p hJ fn i o oo <3 O QJ M N * M ^ s 9 internal 17 on boilers, maximum 179 valve to lift ball 91 per square inch 75 safe working 29 to burst a shell 24, 28 lift valve and stem 119 raise lever 119 raise valve, stem and lever 119 Priming 9 Projected area 22, 25, 26 Quadruple butt-strap-riveted joint 72 -riveted double-strapped butt-joint 64 Rear drum, action in 4 Reduction of pressure, effect 9 Refuse for fuel 169 Return tubular boilers, setting 150 Rivet efficiency 16 holes 71 resistance to crushing 41, 44 Riveted joints, strength n, 28, 40, 41 plate, possible modes of failure 41, 43 Robbins, C. G 1 20 Rupturing plate 42, 43, 45 Safety valve 75, 1 23 194 INDEX PAGE Safety valve, capacity 108 care I4 8 force necessary to lift 79 position 3, 9 rules 103 Scale, avoiding 146 Sediment . 4, 7, 9 Segmental area, finding 34 Segments of circles, areas 37 Separators 140 Setting return tubular boilers 150 Sexton, J. E 156 Shearing of rivets 13, 45, 48, 50, 51, 55, 56, 58, 59, 60, 61, 64 strength of rivets 13, 41, 44, 46, 47, 70, 71, 180 Sheet strength 28 Shell formula 72 Single lap-riveted joints 72 -riveted double-strapped butt-joint 57 lap-joint 48 shearing 46 Sizes and weights of columns and I-beams 154 Sleeve for blow-off pipe 141 Smith, C. Hill i Solid plate, strength '. 43 Spacing of rivets 70, 71 Spring-loaded safety valves t 100 Square of a number, finding 114 Stack area, wood-burning furnace 175 Staggered riveting 53 Steam, amount escaping through opening 116 bubbles 5 gage 79, 127 wet, cause 6, 9 withdrawing 9 Steel frames and nozzles 130 lugs supporting boiler 131 INDEX 195 PAGE Stem, distance from fulcrum ............................. 119 Stop valves in column connection ......................... 126 Strength, ultimate tensile ................................. 1 1 Tensile force ....................................... 1 1, 27, 28 strength ............................................ 41 of cast iron and forgings ............................ 129 of plate ........................................... 70 ultimate ................ . .......................... 1 1 Test, boiler ............................................ 3 Testing boiler plate ..................................... 41 machine ............................................. 41 Thickness of plate ...................................... 70 Through braces ...................................... 31, 35 Top feed ......................................... . ..... 142 Triangle, area .......................................... 81 Triple butt-strap, efficiency ............................... 72 lap, efficiency ........................................ 72 -riveted double-strapped butt-joint ...................... 61 lap-joint .......................................... 51 Try-cocks ............................................. 127 Tube cleaners, mechanical ............................... 184 Tubes, cleaning ........................................ 137 renewing in tubular boiler .............................. 156 U-tube, circulation ...................................... i Ultimate tensile strength ................................. 1 1 Unit section of joint .................................... 47 United States Board of Supervising Inspectors, rules. . . .36, 108, no, Valve area ................................. 108, no, 112, 119 auxiliary ............................................. 115 diameter ............................................. 119 lifting ............................................... 82 weight with stem ...................................... 119 196 INDEX PACE Waste pipe for safety valve 1 24 Water column '. 125 -tube boilers 9 Weight of valve and stem of safety valve, effect 95 to hold pressure on valve 9 2 Wet steam 6 > 9 Wood as fuel for steam boijers 161 Working pressure 7 pressure, safe - 2 9 UNIVERSITY OF CALIFORNIA LIBRARY This is the date on which this book was charged out. OCT. 21911 [30m-6,'ll] YB 10714 196487