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 4 5 6 
 

 ENGINEERS' 
 SUPPLIES 
 
 ot Every Description 
 
 fu°Sf".''h'?"J.IVI.T.«"-'"«'°^»' 
 
 Angle, Chc-kand Gate V/kLVES 
 
 The Standard for Valve Excellence 
 
 Iron Pipe and Fittings, Lubricators and Oiling Devices, 
 Steam Gauges, Steam Specialties and Appliances, 
 
 Inspirators, Tools, Packing, &c.; &c. 
 
 Canadian Headquarters for the Celebrated. 
 
 "HEiNTZ" Steam SAVER 
 
 The Most Valuable Device of 
 its kind in existence. 
 
 Saves your fuel and increases the 
 efficiency . 
 
 Write us for i'ooklet F — explains its 
 Simplicity anrl Effectiveness— 'Twill 
 interest yon . 
 
 The JIliS MOmilSON BIIII88 MFG. GO., Limited 
 
 89-97 Adelaide St. W., TORONTO 
 
wmsmmm 
 
 W.A. FLEmiNGcSiCO. 
 
 57 St. Francois-Xavier Street - MONTREAL 
 
CANADIAN 
 
 Hf\ND-BOOK 
 
 / 
 
 OF 
 
 STEAm 
 
 AND 
 
 ELECTRICITY 
 
 ADAPTED TO THE REQUIREMENTS OF PERSONS IN CHARGE OF 
 
 THE OPERATION OF STEAM AND EI.ECTRICAI. 
 
 APPLIANCES. 
 
 Bv Wm. Thompson. 
 
 Published hy 
 Ti.K C. H. MORTIMER PUBLISHING CO. of Toronto, Limited, 
 
 ToRo.N'To, Canada : 
 Canadian Electric: \i. Nkws and Engineering Journal Press. 
 
 1899. 
 
--J 
 
 Entered according to Act of the Parliament of Canada, in the 
 year iSgg, by Chas. H. Mortimer, at the 
 Department of Agriculture^ 
 
PREFACE 
 
 THK information contained in tlie following pages h.is been 
 specially prepared with a view to meeting- the requirements 
 of persons desirous of qualifying themselves to undertake the 
 successful management of electrical and steam appliances. The 
 preparatory chapters are devoted to a concise explanation of the 
 foundation principles of mathematics, a knowledge of which is 
 absolutely essential to the study of electricity and engineering. 
 In the succeeding chapters the student is lead by gradual stages to 
 a more complete acquaintance with these subjects, and is equipped 
 with knowledge sufficient to enable him to pursue his researches 
 to any further extent within the compass of his ability and oppor- 
 tunities. 
 
 Thk Author. 
 
* 
 
 INDEX TO ENGINEERS' MANUAL 
 
 Ad«liiion of Decimals 
 I' ractions 
 
 and 
 
 Area of a S.,uare, Rule to Find - ' ' ' 
 
 Cali"r.-.l!on „r Work Dons In I>u„,p, 
 
 »u^^P^J?:-ij^™ -«-. ■ . ■ , ■ 
 
 piameterof . ' ' • 
 
 Circular Measurements " " - - 
 
 Combustion Hoxes, Rue to F n.l Wo.^in"^^''"*^ ^'^'''''^ "^ « ^'ane 
 
 Plate of . '^'"" ^^Vo.Kmg Pressure on Supf^orting Holts a, 
 
 ^-ornrnon Fractions . ' ■ - 
 
 Common Denonuna'^;r'^'^'^='^''"«'« ' " " • * 
 
 Corrugated Stee, Fur^ces ^d Flues. Rule to ^.„d Wor.in. Pressure 
 
 S&?t"'^''^^'''^.^'^-°f Stays for . : ." 
 
 Cubical Measurements 
 
 Becimal Fractions . _ 
 
 Notation of . ' ' - 
 
 Principles of . ' " • 
 
 n • , Reduction of ' ' 
 
 Oecimais, Addition of 
 
 1^* Tjivision of . " - - 
 
 II ^Ii'Itiplication of . 
 
 Subtraction of . " ' - 
 
 Denominator, Common 
 
 n„. . . Least Common . " " 
 
 Determination of Mean Pressure 
 
 " Fractions . ' " 
 
 iSo?"'^ '° ' '"^ "-- ^°-' of " . ■ . ' 
 
 Mat Surfaces Pressure on Plates Forming 
 
 '• .. '^"'^^° '1?^ if >^«;'fin« Pressure on Plate Formin.. " 
 
 Required Fbickness of Plate F\>rming 
 
 It 
 
 14 
 
 II 
 
 »9 
 la 
 
 as 
 
 86-88 
 58 
 
 86- yl 
 
 io6 
 
 11 
 
 a? 
 
 69 
 69 
 
 8a 
 
 9 
 10 
 10 
 70 
 
 70 
 
 36-39 
 38 
 
 • *7 
 17 
 
 18 
 24 
 
 19 
 21 
 ao 
 19 
 10 
 II 
 94-103 
 68 
 21 
 14 
 
 103-105 
 33 
 
 82 
 82 
 83 
 
It 
 
 11 
 
 Fiartions Addition of • • _ 
 
 " Commo;i 
 
 " Principles Relating to • * ^ 
 
 " Decimal ■ ' ' 
 
 Notation of Decimal 
 
 Principles of Decimal 
 
 Reduction of Decimal • ^ * 
 
 Division of ■ * 
 
 MultiplicaiiMi t)f 
 •' Subtraction of • 
 
 Girders SupporlinR Tops of Combustion Hoxes, Rule to Find Wor 
 en RectanRular 
 
 Horse Power Calculations ,. " ,^- ' ♦ . 
 
 " " Calculating from Indicator Diagrams 
 
 " " What it is ' ... i " 
 
 • I <• of an ICni.f'np, Rule to P md 
 
 I mtierial Gallon, Contents of • ' 
 
 mficS'r l)iaRr;ms, Calculating He rse Power from 
 incuuu K^ Kule to Divide into Equal Parts 
 
 Least Common Denominator 
 
 Material, Tensile Strength of - " " 
 
 Mathematical Signs - ' ' ^ 
 
 Multiplication '^t Decimals 
 •< " Fractions 
 
 Notation of Decimal Fractions - ' 
 
 Planimeter, Uses of - ri, -i 
 
 Plate, Rule to Find Required Thickness "' V"''*^ >■,... 
 ,. .' " " Strain Per Sectional Inch on HoUei 
 
 Practical Measurements - • ' * : 
 
 Practical Rules Relating to Measurements - 
 
 Pressure of a Boiler, Rule to I-md Safe Working 
 Pressure Allowed on Plates I'ornung P lat Surfaces 
 Pumps, Calculation of Work Done 
 
 " Rule to Find Capacity of • , • 
 
 " " " Contents of • , 
 
 " " " Required Size of Boiler h eed 
 <( 11 n vVork Done 
 
 king Pressure 
 
 9 
 lO 
 
 «r 
 
 H 
 H 
 
 n 
 
 It 
 
 8i, 3j 
 
 (I 
 
 t« 
 
 • 103 
 103-105 
 
 29 
 
 106 
 107 
 
 11 
 
 20 
 M 
 
 17 
 
 106 
 68 
 68 
 25 
 
 32 
 
 67 
 82 
 86-91 
 86 
 - 86 
 86-88 
 89-90 
 
 Wivf'ts Rule to Determine Pitch of ,' », • r^ <• 1 
 
 Rule ;« De.enTdne Number of Equal Parts that Engine Cylmd 
 
 be Divided into ■ 
 
 Rule to Determine Pitch of Rivets ,^ ' , n . 
 
 Rule to Divide Cylinder of an Engine int.. Equal Paris 
 ic 1. •« Indicator Diagrams uito Equal 1 arts 
 
 Rule to Find Area of a Circle 
 
 <« " " " " Diagonal Stays - ; 
 
 M .. i< " for Square Steel, Diameter for Round Steel, 
 
 Outside Diameter of Spring Safety Valve 
 
 Rule to Find Area of a Square /or \t 1 u^u 
 
 KUie ^^ ^^^^^^ Effective Weight of Safety Valve Ball 
 
 " Atea of an Irregular Figure 
 
 " Capacity of a Square Tank 
 
 '« " Pumps 
 
 er Requires to 
 
 and Inside and 
 
 11 
 
 II 
 II 
 ii II 
 
 63 
 
 97 
 
 63 
 
 96 
 
 107 
 
 28 
 
 80, 81 
 
 5> 
 as 
 46 
 
 29 
 
 86 
 
II 
 II 
 
 <l 
 
 11 
 
 41 
 
 ^"i'* !l^ ''•.'J'^9'''^"'"'''^''*"^^ "f" Circle 
 
 Co-KfTitient of KflTiciency . . 
 
 Collapsing Pressure of Plane Circula. Furnace Tube ' . 
 L. impressKin Required on Safety Valve 
 Conlents of Pumps 
 
 r)iameter of a Cirf le . • . 
 
 ;; Safety V dive . ' . 
 
 ., |. " .'' " '' I" '5l'>w Offat a (;iv n Pressure * 
 
 ,. [?«Jwn ward Pressure on Spring Loaded Safety Valve 
 
 bxternal \Vork.nK Pressure on Circnlar Steel Furnaces and Flut-s 
 ^^ Kffective Monu-nt on Safety Valve I.ever 
 
 •' Crl? ^[""T"" ^V^''^'^ to Discharge Water from a (;iven Orifice 
 (»raduale Lever of Safety Valve . "v/iiutc 
 
 It 
 
 II 
 
 II 
 
 It 
 
 II 
 
 II 
 
 11 
 
 « I 
 
 II 
 
 II 
 
 II 
 
 II 
 
 II 
 
 II 
 II 
 II 
 II 
 II 
 
 i< 
 
 ii 
 II 
 11 
 •I 
 It 
 II 
 <i 
 
 Strok* 
 
 Horse Power of an Engine 
 Number of Coils for Safety Valv.; 
 
 »7 
 
 U 
 
 al 
 
 4* 
 
 47 
 
 50 
 
 69 
 
 88 
 45 
 
 103-105 
 54 
 
 Outside Diameter of Safety Valve Spiral Spring ' . '* 
 
 Fercenta«eorStrengthofPlateat Joint Compared with Solid Plate 50 I" 
 rres<iir#' at \V1.:.~K U,»;i^- ...:m ni /-mt 01 
 
 T'ressure at Whinh Boiler will Filow Off 
 
 "f c. " ^x^'T-^ Loa.jed S.-ifety Valve will Hbw Off' 
 °; '^'*^''»'" ^^'thui Lngine Cylinder at Any Point During 
 
 Rule ,0 Find Required Area u( Kscape for Steam r.„ a Spring Loaded Safety 
 
 Rule to Find Required Diameter of a IJoiler . " 
 
 "Stays 
 Length cf Lever of Safety Naive 
 Pilch of Stays 
 !! ^}ff "f Hoiler Fejd Pumps 
 
 ^' J'ickness of J^oiler Plato 
 
 ,.,.", '.' ]''«ite Forming Flat Surfaces 
 
 ^^ Weight of Safety Vabe Ba'l 
 
 c r t.' ,• ". "»". Lever of Safety Valve 
 vSt-ife V\ orkmg Pressure of Boilers 
 
 " "a Hoiler 
 
 Strenrr.h nrp!. , '. r "" ^''^'^ f^»''"""K ^'lat Surfaces 
 ^trength of Riveted Joint 
 
 Strain Per Sectional Inch on Boiler Plate 
 Size of Stays for Cottars 
 
 Total Weight Hohling Spring Loaded Safety Vahe in PI ,'ce 
 Stress on Stay . ^ m i i.u,e 
 
 Working Pressure on Corri.gated Steel Furnaces ...,d Tlues 
 II „ " '•'on " " ti 
 
 ofa Boiler with Stays of a Given Area 
 
 on Rectan^MiIar Girders Supporting Tops of 
 
 11 
 
 II 
 
 II 
 
 II 
 
 11 
 
 II 
 
 It 
 
 <l 
 
 II 
 
 It 
 
 It 
 
 if 
 
 1 1 
 
 II 
 
 tf 
 
 II 
 
 1 1 
 
 tt 
 
 II 
 
 It 
 
 f i 
 
 II 
 
 II 
 
 II 
 
 II 
 
 ■ 1 
 
 1 . 
 
 II 
 
 l> 
 
 It 
 
 11 
 
 II 
 
 It 
 
 II 
 
 tl 
 
 Combustion FSoxes 
 Rule to^f-i.^ Working Pressure on Supporting Bolts and Plate of Combustion 
 
 ^"}^ \? ^'"^ ^'^o'"'" Dene by a Pump 
 
 of a Pump'""^' '' "'''^' ''''''''' ^"' Travel through the Discharge Pipe 
 
 47 
 
 54 
 
 98, 99 
 
 II 
 
 75 
 
 46 
 75 
 86-88 
 68 
 83 
 47 
 
 ■ t, '♦^ 
 
 58, 59 
 
 si 
 
 61 
 68 
 83 
 54 
 74 
 70 
 70 
 75 
 
 Rules Relating to Measurements, Practical 
 bafety V\ilve Cal.ulations 
 
 "•'It 
 
 II 11 ''^^er lype 
 
 Rule lo Find Actual Effective Weight of Ball 
 
 8r, 82 
 
 82 
 89, <,n 
 
 88 
 32 
 
 58. 59 
 4» 
 
 • 42 
 4^ 
 
li 
 
 Safety Valve, Rule fo Find Area for Square Steel, Diameter for Round Steel, 
 
 and Inside and Outsic'e Diameter of Spring 
 Safely Valve, Ride to Find Diameter of - - - 
 
 " " " " " Valve to Blow rff at a Given Pressure 
 
 " " << " " Downward Pressure on Valve 
 
 " " " ' " Effective Moment of Lever 
 
 " " " Graduate Liver 
 
 ' " Find Number cf Coils 
 
 " " " Outside Diaiueter of a Spiral Spring 
 " ' '* " " Pressure at which Hoiler will blow off 
 
 t( II II II ii 11 t« li Valve " " " 
 
 " '■ " •• " Required Area of Escape for Steam 
 
 ' " Length of Lever 
 
 Weight of Ball 
 " Total VVei<;ht Holding Valve in Place 
 " " Spring Loaded .... 
 
 '* " " " Dimensions of Standard Spring - • 
 
 Signs, Math matical ..... 
 
 Signs that Represent Roots of Numbers .... 
 
 Spring Loaded Safety Valves - ■ - - '^ , 
 
 " *' •" " Dimensions of Standard Sprir.g • '• - 
 
 S(iuave Measure - . - - - - •« 
 
 Square Root ..... «• ' 
 
 Square Root oi a Number, Rule to Find - - - 
 
 rftandaid Boiler ..... * 
 
 Standard Spring, Dimensions of S.Tfely Valve Spring Loaded 
 Stays on Flat Siiifaces, Strength «,f 
 
 Stays, Rule to Find Total Stress .... " .. 
 
 Stays of a Given Area, Ru'e to Find Working Pressure of a Boiler with 
 Stays, Rule to Find Required Diatneter of - - 
 
 Pitch of 
 " " Area of Diagonal - - . 
 
 " " " Size of Stays for Cottars 
 
 Strength of Plaie at Joint Compared with Solid Plate, Ride to Find Percent- 
 age of - - - ... 
 Strength of Pavets, Rule to Find Percentage of " • , ■ 
 Strength c( Riveited Joints, Rule to Find - ? >• ' 
 !>trength of Stays on Flat Surfaces - , • • 
 Strength of Boilers - - - •'. ,^- ' ' ' .i 
 Subtraction of Decimals - ■ - * . , * 
 Subtraction of Fractions - - - . 
 Supporting Bolts and Plate ofCombusti:n Boxes, Rule to P'inil Working 
 "ressure on - - - - - - 
 
 Tank, Rule to Find Capacity of - - • ■ - - 
 
 Tensile Strength of Material - - ' > 
 
 Triangular Measurements ..... 
 
 U. S. '.iallon, Contents of ...... 
 
 Velocity at which Water will Travel Through the Discharge Pipe of a Pump, 
 Rule to Find ...... 
 
 Weight of a Cubic Foot of Wafer ... 
 
 5» 
 
 *'* 
 47 
 50 
 4^ 
 
 ■ 45 
 54 
 50 
 47 
 54 
 
 - 55 
 46 
 
 47 
 
 54 
 49-50 
 
 49 
 39.40 
 40, 41 
 
 49-55 
 49 
 25 
 33 
 
 3,'-3^ 
 58 
 49 
 
 74-78 
 74 
 75 
 75 
 75 
 8j. 81 
 
 83 
 
 59-60 
 
 61 
 
 - 62 
 
 74-78 
 
 53 
 
 • >9 
 12 
 
 29 
 58 
 25 
 
 29 
 
 88 
 29 
 
INDEX TO ELECTRICAL CALCULATIONS 
 
 Alternating Current Systems 
 
 Apphcation of Ohm's Law . . ' 
 
 Circuits, Portions of . ' " 
 
 Circuits Divided and Shunt 
 
 Circular Mil . * - . . 
 
 Conversion, Rule to Determine Ratio of * 
 Copper Wire, Resistance of . of . 
 
 ;; Rule to Determine Resistance of 
 
 Currei?Kr?^i:S^^^-^'-S>^""' '"'""'' ^"^ '^^ ^r" Multiple 
 lamp 
 
 D-vided and Shunt Circuits . . 
 
 F W.'"^?"^'"^y;. ^^""*^<imental P, inc iples of 
 i-lectro-Motive Force Rule to Find ' . 
 
 Fundamental Principles of Electric Energy 
 Lamp Current, Rule to tind 
 
 i-caas, Kule to I ind Percentage of Drcp on 
 
 Ohm's Law 
 
 Ohm's Law, Application of " 
 
 Portions of Circuits . ' ' ' 
 
 Potential, Drop in . * • - 
 
 Primary Mains. Rule to Determine Re.uired'si.e of ' 
 Resistance of Copper Wire 
 
 «< '< r ** „", f^"'e to Determine 
 
 .,■■ R&Ki?;''--"'^'"" Combined / . " . 
 
 Rule to Determine R„uo of Conversion ' . " 
 
 Rule ,»'^"°"^Thre'e^W^^|;^^r' ''""' "' Copper Wire for Multiple Con 
 Rule to De,e,;m,ne Required Sije „, Mains for Three-Wire Sys-e. " 
 " " '• .. „ ,. Primary Mair.s . 
 
 *' RpcU.oMr, f #- Secondary Mains 
 
 Kesistance of Copper .Vire 
 
 Page 
 
 lOI 
 
 - 63 
 
 65-66 
 
 71-74 
 
 92 
 
 102 
 
 92 
 
 92 
 
 lOI 
 
 64 
 
 79 
 71-74 
 
 47 
 
 65 
 
 • 47 
 
 79 
 - 80 
 
 80. 83, 84, 85, 86 
 
 Id 
 79 
 
 • 55-57 
 63 
 
 6s, 66 
 
 79 
 loa 
 
 92 
 
 - ' ro 
 
 loa 
 
 If 
 11 
 
 (I 
 
 lot 
 
 - I02 
 103, 103 
 92 
 
ass5s= 
 
 80 
 
 - 64 
 
 65 
 
 79 
 
 80, 83, 84. 85, 86 
 f'4 
 
 " Find Combined Resistance of Lamps 
 
 " " Cum nt . ■ 
 
 '■ " Electro- Motive torce 
 
 «< " Lamp Current " * 
 
 - " Percentage of Drop on Leads 
 
 .. " " Rtsistance 
 Seco„Ca,y Main,. Rule ,0 Be.er.m. R.o.»Uri S«. of ■ 
 
 Shunt and Divided Circuits , s, .,;„„„! \rea of Copper 
 
 Th, ee-Wire System, Rule to Determine Required SeUional Area 
 
 f^'-'/'^r'turDeSne Required Sue of Mains • ■»■ 
 
 Three- Wire System, Kuie lo x^ci _ ^p^ ^^ 
 
 Units of Measure 
 
 102, 103 
 71-74 
 
 '■"38 
 
ERRATA. 
 
 i'ajce 15 : Sixth line from top reads, - Divide 100-34^ = 400-4 = 
 •33M;- should read. " ,00-^=400-3= ,33./..- Eighth 
 line from bottom reads, " ^ - 3/= 7/ ^ 4 _7 X4 ,, ,. „ 
 should read, " H-^ U =1^ = M= i yi." 
 
 I'age 61 : Second line from bottom reads, 
 
 ^ ^x ,oo = lL^^54ili, ,oo-'393' 
 
 ,00-— =.3,44 X, 00 = 3, .44%; 
 
 ^•5>^-5 
 
 P T 
 should read, 
 
 ^-^ V 100 --^^'^ -7854x2 
 
 p ^ ^ I uo 7~r~. ~ X I 
 
 ^•25 
 
 ■393 
 
 2-5 X. 5 '"''°-7:25^'^'-*-^^'°°-^3'.44%. 
 
 '''^bJii; •^'""' '" '"" '"^^""' ^^^'^' " '^''^72 total area of 
 boiler -.30,753, number of stays required- ' .h^ 11 / 
 
 •• 16 >-- tr^t^i r. .. required, should read, 
 
 10,2^2, total area of bo er-io7- CI n..,«K^ «• 
 q„i^ ^- .. • 307-53. number of stays re- 
 
. .-^.iii:^iiii»>^ n-iyiJ»ii'|,iii <>i«ii)|Mti 
 
 mtm 
 
COMMON FRACTIONS. 
 
 A FRACTION is one or more of (he euuil m... • . 
 
 |,Noih.-rh„.s a iwo foot rulo is divided into ' . . , ,/ 
 
 and so on and consoquen.ly ll,i» slyle of n^nneraiion^s' ^^'.sA',,, 
 occurrituj^ m m,., •,-,.,:.,..! • . . ' '^ *-on.si.ini 
 
 Tl, 
 
 ng- 111 mechanical engineeiin^. ] 
 
 Jy 
 
 N 
 
 e terms of a fraction i 
 
 ue styled and dist 
 
 tl 
 
 umerator and the Denoniinatoi- 
 lie numerator of a fraction indicates tl 
 
 m 
 
 guished as the 
 
 o unit considered or faken. 
 The denominat 
 
 le number of parts of 
 
 unit is divided. 
 
 or indicates the number of part 
 
 ■^ into which the 
 
 T 
 
 o 
 
 express a fraction is to ind 
 
 parts in the nunierat 
 below a horizontal 1 
 
 icale 
 
 or and denominat 
 
 by fig- 
 
 ures the number of 
 
 me. 
 
 or respectively abov 
 
 c am 
 
 Tl 
 
 Common Fraction 
 
 uis 
 
 '• 'A, 
 
 plex. 
 
 s are known as Simple, Compound 
 
 A simple fraction is a fraction wl 
 
 nator consist oi' 
 
 simple number 
 
 pound and Com- 
 pose numerator and denonii- 
 
 Tl 
 
 lUS 
 
 A compound fract 
 
 5-f> JTiJ-j /S 
 
 iyC. I 
 
 f 72' 
 
 pressed, but must be .- 
 
 ion is a fraction whose val 
 
 irrived at by computat 
 
 ue 
 
 ion. 
 
 is not fully ex- 
 
 A complex f. action 
 
 Thus^of-ii^, y^ofi 
 
 54, etc, 
 
 numerator or denominat 
 by computation. 
 
 IS one which cont 
 
 or, or in both. Its val 
 
 ains a fraction 
 
 in th« 
 
 ue must be found 
 
 Th 
 
 us 
 
 ^f6 
 of I 
 
 of 
 
 I o 
 
 O 
 
 f, 
 
 Tl 
 
 of 6 
 
 le numerator or fignues above the 1 
 
 me correspond with ilu 
 
lo 
 
 dividend, and the dononiinalor or lijLfiires below the line willi the 
 divisor. 
 
 The chief prineiples relating' to fractions are : 
 
 1. Multiplying' the numerator by any number multiplies the 
 fraction. 
 
 2. Multiplying the denominator by any number divides the 
 fi-action. 
 
 3. Dividing the numerator by any luniiber divides the fraction. 
 
 4. Dividing the denominator by any number multiplies the 
 fraction. ■''''. '■■'w'- ''•'"■' v";.' -■ 
 
 5. Multiplying both numerator and denominator by the same 
 number does not affect the value of a fraction. 
 
 6. Dividing both numerator and denominator by the same 
 number does not affect the value of a fr.action. 
 
 Giving us tlie important general principle that any change in 
 the numerator produces a corresponding cliange in the value of 
 the fraction, and any change in the denominator produces an 
 opposite change in the value of the fraction. 
 
 COMMON i:)RNOMINATOR. 
 
 A common denominator is a denominator which is common to 
 two or more fractions. 
 
 Rule : To find a common denominator to two or more fractions, 
 multiply together all the denominator and the product will be the 
 common denominator. 
 
 Example: Find common denominator of j^, Y^, ^, ];5=:The 
 denominators multiplied together, as 2x3x4x16 = 320, which 
 becomes the denominator common to the whole of these frac- 
 tions. 
 
 This rule often entails a lot of work, consequently I prefer to 
 reduce denominators to their simplest form and multiply together 
 the final quotient and the divisors, which will give the common 
 denominator in lowest form. . 
 
1 1 
 
 ; in 
 3 of 
 
 i an 
 
 Kxaniplo : Find Uio common denominafoi of 
 
 •'> 7/ 
 
 » .12. 73. J 
 
 Reduce denominalors to lowest form, thus : 
 
 4 I I ♦ 4.16 .1.5 
 I.I. 4.1.5 
 TI.en by nu.Itiplyin^r fi„al quotient and divisors to^^ethor, thus, we 
 
 5x1x4x1x4x2x3 
 
 Since, however. sinKle units do not in anv way efTecl results we 
 wnti' it . J . 
 
 5x4x4x2x3 = 4800. D. 
 
 Rule: To reduce two or more fractions to equivalent fractions 
 
 ..ivms; a common denominator, divide the common denominator 
 
 by the denominator of the given fraction, multiply the quotient so 
 
 found by their numerators, and write the results over the common 
 
 clenommator. 
 
 Reduce to equivalent fractions having a common denominator. 
 
 ^' ^. Ih %y f 
 
 Since we have just found the common denon^inator of these 
 tractions to be 480 we place it thus 
 
 4^^£0_225^ 320_384 
 
 480^ ~" 
 
 and proceed as in rule to divide the common denominator by the 
 
 denonnnator of the nrst of our fractions, a, which equals 480 --6 = 
 
 |So. I roceeding again as by rule, we nu.ltiply this product by 
 
 ihe numerator, 5. thus 80x5 = 400, which result we place in first 
 
 posM.o,, ,1 , the common denominator. Proceeding similarly 
 
 v.th the other tractions in our example, we get a series of fractions 
 
 'avu,g a conunon denonnnator .tnd equivalent to the fractions 
 
 trom which we started conqnitation. 
 
12 
 
 !'■ 
 
 ADDITION OF FRACTIONS. 
 
 Addition is l!io process of fniding^ the sum of two or more 
 fractions. 
 
 Principle involved : Fi actions to be adiled must be re.'uced to 
 equivalent fractions havinj^ a common denominator. 
 
 Rule : To find the sum of two or more fractions havinj;^ different 
 denominators, reduce the g'iven fractions to equivalents with a 
 common denominator. Add the numerators so found, and if their 
 sum is greater than the common denominator divide the numerator 
 by the common denominator ami the result will be the sum of the 
 g'iven fractions. 
 
 Find the sum ot {\. 
 
 -I- U i 
 
 I'roceedinj^ as before we find the common denominator to he 
 
 i6, thus 
 
 4 I 16 . 8 . 4 . 2 
 
 2 I 4.2.1.2 
 
 2 . .1 . I . I = 16 
 
 Proceeding' exactly as before we find the equivalent fractions 
 having a common denominator to be expressed thus : 
 
 r\ 16 c. D. ~^* 
 
 Adding these numerators together, as in rule, we get the 
 fraction ■,';{. Now dividing the numerator 39 by the denominator 
 16 we get 39-=- i6 = 2,v,, then our question may be expressed thus : 
 
 5+ 14+ 12+8 
 
 j\+H + H+'A=' 
 
 r6 
 
 TC — 2fff 
 
 This principle underlies the whole subject of addition, and needs 
 no further demonstratior.. 
 
 SUBTRACTION OF FRACTIONS. 
 
 Subtraction of fractions is the process of finding the difference 
 between two fractions. 
 
 Principle involved : Fractions, to be subtracted, must be re- 
 duced to equivalents with a common tienominator. 
 
 Rule: Tt> find the difference between two simple fractions, 
 
>3 
 
 reduco the fractions to ocjuivaloiits li.iviiig^a (.•oninu>ii ikMiotniiiati>r 
 ami siihiiact the luiinerators, and ivtluce the lesiih to its simplest 
 form. 
 
 t n 
 
 I'Miul the diflforence between J4 'i"d 
 
 Proceedinj^ as already describeil in atltlilion of fractions, weg'et 
 
 ^ ■1 - 
 
 I (T 
 
 •4_ 
 
 i6 
 
 — 1 c 
 
 It will be observed thai the proiess is exactly similar to the 
 piocess of addition, and exceedinj^ly simple, requiring^ practically 
 no explanation w en we have mastered the principles. 
 
 MII.TFIM.ICATION OK KKACTIONS. 
 
 Multiplication is the process of fnulinj;- the product of two fac- 
 tors, one or both of which may be fractions. 
 
 Rule : To nniltiply a fraction by a fraction, multiply ihe 
 numerators together and also the denominators, and reduce to 
 simplest form. 
 
 Multiply together )4 x /4' Proceeding as per rules and multiply- 
 ing the numerators together and the denominators likewise, we get 
 
 I X I 
 
 2X2 
 
 :— v. 
 
 Following out the principle set forth in clause 6 of our introduc- 
 
 tion, multiply together 
 
 'S ^ ii ^ k ^ lH' 
 
 Following out this principle we proceed by a process of cancel- 
 lation to reduce to simplest form, thus 
 
 21 
 
 
 
 
 
 i6 64 
 
 By applying this system of cancellation, based upon the pri 
 
 ji- 
 
 ciple set forth in clause 6, you will observe we materially shorten 
 the process of calcula.ion without in any way affecting the result. 
 
'4 
 
 Rule: To nuiUiply a fraction by an iiUi'ger or an inlcy^er by a 
 fraction, 
 
 r. niviiU" the denoniinatoi" o( llie fraction by the integ'er n\n\ 
 place the resuh niuler the ninnerator, anvl rciUfc to simplest 
 form, or 
 
 2. Divide the intei^er by the denominator oi' the fraction, and 
 multiply the result by the ninnerator, or 
 
 3. Multiply the numerator of the fraction by the intej^er and 
 place the lesult over the denominator. (See clauses i and 4 of 
 principles). 
 
 Example (employing 1st method): 
 
 Multiply H^ x8 = 32-^8 = 4 7=6^4:. 
 
 ICxample (employi ij^ 2nd method) : 
 
 Multiply f,?i X 64 = 64-^32 = 2 X 27 = 54. 
 Example (employing- 3rd method) : 
 Multiply )^x8=V^ = 5. 
 It will be observed that the first of these methods can only be 
 employed when the integfer can be divided itito the denominator 
 an equal number of times, and the second when the denominator 
 can be divided into the integer similarly, and the third meUiod 
 can be used at any lime, but when either of the other methods 
 can be used, lengthens the process, as is evidenced by calculation 
 in this case. Employingf method i and enibracing' principle 6, 
 
 calculation would have been made a 
 
 s foil 
 
 ows 
 
 5 
 
 X!^ = 5. 
 
 DIVISION Ol' FRACTIONS. 
 
 nivisioii effractions is the process of finding the quotient when 
 either dividend or divisor is a fraction or mixed number, or when 
 both dividend and divisor are fractions or mixed numbers. 
 
 Rule : To divide a fraction by an integer, divide the numerator 
 or multiply the denominator of the fraction by the integ^er. The 
 result will be the quotient. (See principles, clauses 2 and 3). 
 
 Example : f, | -r 9 = rfV. 
 
»5 
 
 ator 
 
 Ltion 
 c6, 
 
 when 
 when 
 
 Rule : Ti) divide an into^or by a fraction, inuliiply ihe inloj^er 
 by the tlonoininalor of the fraction and divide the prtulucl by llie 
 nu!neratt>i, or diviili' (ho inlcjk'^er by the numerator and tnuhiply 
 'he quotient by the denominator. 
 
 Example (by ist nieth d) : 
 
 Divide I oo-r ^^ = 4004-4= I33|j. 
 Example (by 2nd method) : 
 
 3o-f;j4: = Y= '0x4 = 40. 
 
 Rule : To ilivide a fraction by a fraction, nuiltiply the numera- 
 tor of the diviilend by the denominator of the divist>r and set 
 down the product as a new numerator, then multiply the denomi- 
 nator of the dividend by the numerator of the divisor and set 
 down the protluct as the new ileiiominator reduce new fraition 
 to simplest form, or 
 
 Invert the terms of the divisor and proceed as in multiplication 
 of fractions. 
 
 Example: Divide % ^ ?4 (following ist rule). Since numerator 
 of dividend is 7 and denominator of divisor 4, we ^et 7x4=28, 
 which becomes new numerator or dividend. 
 
 Since denominator of dividend is 8 and numer: tor of divisor 3, 
 we get 8x3 = 24, which liecomes new denominator or divisor and 
 givinj^ 11=1^*^=: 1 1^, that is, ^4 is contained in % i|6 times. 
 
 Employing 2nd method : 
 
 t- 
 
 b X3 
 
 Again employing 2nd method and applying clause 6 of prin- 
 ciples : 
 
 'A-^'X = 
 
 7X 4 
 
 = !>€. 
 
 It occasionally occurs in computation of formula that a frac- 
 tional part of a fraction requires to be divided by a fraction or a 
 fractional part of a fraction. 
 
 Rule : To divide a compound fraclion by a compound fraction. 
 
u li 
 
 l¥ 
 
 If I 
 
 i6 
 
 fiisl rt'cliui' \hc cotnpDiind fraclion lo a simple frartioii, and then 
 follow i"nlt' la'ul ilovvn for tli\iiliug' a fiarlion by a frariion. 
 
 Kxaiiipio : Diviilo )4 of ys by )^ of j ;;. 
 
 To reduce compound fraclions % oi J/s x ^4^ of [ f; lo simple 
 frai lions, proceed as in multiplication of fraclions. 
 
 Tben Yz oi }i^ >^x% /,,, and U of Vh -H<M 
 
 4.. . 
 4 > 
 
 I lien 
 
 c|iiosiion becomes a simple matter, since we proceed exactly as 
 in division i>f fraclions from this point. 
 
 Since dividend % of J^ - ,V„ 
 
 Since divisor % oi [% -Yl^ 
 
 - " -' ' 4 
 
 , 7 ^ 45 _ 7 ^ 0^ _ 28 
 we get ^ .-^-^^x--_ 
 
 f6 
 
DECIMAL FRACTIONS. 
 
 A DFCIMAL fraction is one whose iVaciional units are tenths, 
 hu"dredliis, thonsatuhhs, etc. 
 
 [NoTK. — Tlie clonDininator of a decimal fraction is net expressed 
 as in conunon fractions ; instead of expressing- the fractions ,',,, 
 T?.r,» I,-,',!,-., they would be decinially expiessctl as .i, .01, .001. | 
 
 The decimal point is the character (.)• It sijjnifies the decima; 
 when placed on the left of the fractional units expressed, and 
 separates the integer from tin' decimal, as it is on the right of the 
 former and on the left of the latter. 
 
 Decimal places consist of the number of figures on the right of 
 (he decimal piiint. 
 
 The value of each place is shown in table of decimal notation 
 
 DECIMAL NOTATION. 
 
 
 
 
 
 
 
 1/! 
 
 
 
 
 
 
 
 .a 
 
 
 
 
 
 
 
 *i4 
 
 
 s 
 
 "2 
 
 
 
 
 ■A 5 
 
 
 
 
 h 
 
 
 
 
 j= 5 
 
 
 F5- 
 
 
 
 
 c 
 
 lA 
 
 Ten thousandt 
 Hundred thou 
 Millionths 
 
 
 indred W 
 
 n 
 
 llions 
 
 indred ih 
 
 n 
 
 ousands 
 
 Hundreds 
 
 Tens 
 
 Units 
 
 B 
 
 ''■J 
 
 nths 
 
 ndredths 
 ousandth 
 
 
 khS 
 
 X^^ 
 
 
 ^X^ 
 
 
 401 
 
 235 
 
 409 
 
 • 
 
 478 
 
 342 
 
 ""'''■'■'^'' 
 
 ^r^ 
 
 ^.^^ 
 
 w.— / 
 
 
 ^--,— ' 
 
 ^^.^ 
 
 
 jzjsj: 
 
 j-_-^ 
 
 -z-o ^ 
 
 
 (A C u 
 
 X — s:. 
 
 
 ^-1 4-t 4^ 
 
 
 
 
 
 .-■ ■' ■-.- 
 
 tyoa c^ 
 
 vo 1/1 ■>«• 
 
 ro N "-I 
 
 
 w ,■. ro 
 
 ■* ^'O 
 
 ><_ 
 
 Integers 
 
 Decimals 
 
 [Note. — It will be observed that the integers are numerated 
 from rigiit to left, and the decimals from left to right. Thus the 
 figures on the left of the decimal point express the ntmiber of 
 
ii 
 
 units or integers, and those on tlie riglU of tlie decimal point the 
 niimberof tenths or decimal parts of a unit.] 
 
 A pure decimal is one in which no integer or common fraction 
 
 is expressed. 
 
 Thus .5, .i25, .0625. 
 
 A mixed decimal is one which contains an integer and a de>^i- 
 
 mal. 
 
 Thus 4.5, 3.125, 84.0625. 
 
 The chief principles of decimal notation are : 
 
 1st. Annexing ciphers to a decimal does not effect its value. 
 Thus, if a cipher be annexed to the decimal .1, it would then be 
 .10, and changed from tenths to hundredths, but the fractional 
 imits are increased tenfold; hence no change in value takes place. 
 
 2r.d. Prefixing a cipher to a decimal fraction and moving deci- 
 mal point to the left decreases the value of the decimal tenfold, or 
 is equivalent to dividing by 10. 
 
 3rd. Moving the decimal point to the right one point increases 
 the viilue of the decimal tenfold, moving two points one hundred- 
 fi/ld. etc. 
 
 4th. Moving the decimal point to the left one point decreases 
 the value of the decimal tenfold, moving two points hundred-fuid, 
 and so on. 
 
 5th. The numerator of any decimal is the number of fractional 
 units it contains, and its denominator is i followed by as many 
 ciphers as there are places after the decimal point. 
 
 [Note. — Thus, in the decimal .i?5 the numerator is 125, and 
 since there are three places after the decimal point, the denomi- 
 nator is I followed by three cyphers, and fraction is read iWff«] 
 
 It is important to the reader that the principle of decimal nota- 
 
 ion be thoroughly understood, since our scientists almost without 
 
 exception adopt the metric system and express their formula and 
 
 observations by decimal notation, and since micrometer gauges, 
 
 calipers and rules are now in universal use, it is just as important 
 
 I 
 
«9 
 
 the 
 
 Lion 
 
 le^i- 
 
 •alue. 
 Ill be 
 lional 
 place, 
 deci- 
 old, or 
 
 reases 
 ndred- 
 
 rreases 
 td-foid, 
 
 tciional 
 many 
 
 Icnomi- 
 
 1 2 ^ 1 
 
 ll nota- 
 kvUhout 
 Ilia and 
 
 jauge*^* 
 kportant 
 
 that 1 lie principles of conversion of common fractions to decimal 
 fractions and vice versa be eqiu'illy well understood. 
 
 ADDITION OF DECIMALS. 
 
 Addition of decimals is the process of finding- the sum of two or 
 more decimals, and becomes an easy process once principles of 
 decimal fractions are well undv?rstood. 
 
 Rule : To find the stun of two or more decimals, so place the 
 decimals that the figures oC same order shall fall in same column. 
 Then add as in simple additioti, placing- the decimal point in the 
 result between t'.ie integer and decimal. 
 
 Find the sum of 1.0625, .45, 87.5, .0007. 
 
 Proceed thus, 1.0625 
 
 ■ ' -45 
 
 "... 87-5 .:.:;,-: -:^' 
 .0007 
 
 8g.o> 32 = Total. 
 
 SLBTRACTIOxN OF DKCIMALS. 
 
 Subtraction of decimals is the process of finding the difference 
 'uetween two decimals. 
 
 Rule : To find the difference between two decimals, place the 
 less decimal or subtrahend under the greater decimal or minuend, 
 so that the figures of any order or denomination in the subtrahend 
 shall fall under those of the same order or denomination in the 
 minuend. Subtract as in simple numbers, placing the decimal 
 point in the result between the integer and decimal. 
 
 Example : Find the difference between .6504 and 75.9735. 
 
 Process as by rule : 
 
 75-9735 (minuend) 
 
 .6^04 (subtrahend) .-^ ^■' ■■■■^--■^-■'■---''•"'^■ 
 
 7S';\2^l (difference) 
 Example : Find the difference between .007S5 and .625. 
 
 Process 
 
 .625 
 00785 
 
 .6i7t5 
 
20 
 
 I 
 
 MULTIPLICATION OF DECIMALS. 
 
 Multiplication of cU'cIni.ils is (he process of finding" the product 
 when either the multiplier ov multiplicand, or both, are decimals. 
 
 Rule : To multiply an integer by a decimal, or vice versa. 
 
 Set the factor containing" the least niunber of figures as a multi- 
 plier and other factor as the multiplicand, proceed as in simple 
 numbers. Point off as many decimal places in the product as are 
 contained in BOTH FACTORS. If the product does not contain so 
 many places, prffix ciphers to supply the deficiency. 
 
 Example : Multiply 147. by. 75, proceed as per rule. 
 
 '47- 
 
 ^•.v;... :■.... i.\': :..;:, ,,^ 7 35 
 
 ;:;. 102 9 
 
 I 10. 2: 
 
 Since both factors contain but two places to the right of the 
 decimal point, we require to point off this number in product. 
 Example (2) : Multiply .75 by .625. 
 
 ■;' -■•>^- ..-■..:.. .... . '75 
 
 3125 
 
 .4«^75 
 Since both factors contain five decimal places we point off this 
 
 number in pi'oduci. 
 
 Example (3) : Multiply .0625 by .075. 
 
 .0625 
 •075 
 
 3125 
 4375 
 
 .0040875 
 
 Since both factors contain seven decimal placers and product but 
 five, we require to prefix two ciphers, and set decimal point to the 
 left. 
 
DIVISION OF DKCIMALS. 
 
 Division of decimals is the process of fiiuling- the o,uotient when 
 cither the dividend or divisor or both are decimals. 
 
 Rule : To find the okcimal (jronrcNT when the divisor and 
 dividend are both whole numbe-s, and (he divisor is g-reater than 
 the dividend, or when the divisor is not contained in the dividend 
 an exact number of limes. 
 
 Add as many ciphers to the dividend as there are decimal 
 places required in the quotient, divide as in simple numbers, and 
 point off from the RICH r of the quotient as many decimal places 
 as there have been ciphers added to the dividend, and LSEI>, pre- 
 fixing' ciphers to the qutUient if necessary. 
 
 Example : Divide 75 by i^oO to 41I1 decimal place. 
 
 Process as per rule : i5o())75oooo(.0498 
 
 6024 
 
 1 4760 
 
 12060 
 1 2048 
 
 12 
 
 Since we wish to extend to fourth decimal place only, we annex 
 to dividend four ciphers and proceed as in division of simple num- 
 bers, and get 49;^. Since, however, we annexed four cyphers to 
 the dividend, and tlierefore require four decimal places in the 
 quotient, we must prefix one cypher, and c|uotienl will then read 
 .049S, proving that the factor 1506 is conlainetl in the factor 75 
 .0498 timer, or that 75 is 0498% of 1506. 
 
 Example (2) : Divide 743 by 125. 
 
 ' 25- )743- 000(5. 944 
 
 n8o 
 
 1 1 
 
 ^ ■> 
 
 550 
 500 
 500 
 qoo 
 
!7^ 
 
 22 
 
 Here we have an example wlieie two wliole numbers are to be 
 divided an exact number of times. Since we find the divisor is 
 contained in the dividend 5 and a fractional times, we add to the 
 dividend a cypher and bring' this down and annex to t!ie right of 
 the remainder, as in division of simple numbers, repeating this 
 process until there is no remainder. We then proceed to ascer- 
 tain the number of ciphers annexed to the dividend, and point off 
 a corresponding number of places in the quotient, counting from 
 the right. In example the 125 is contained in 743 exactly 5-944 
 times. 
 
 Example (3) : Divide .9735 by 50. 
 
 5o)-9735o(.oi947 ^^^^^^^ v ^^.^^^ ^ 
 
 50 '■■- --''■\-'-' '■■■'■y':-:-'::^^--::--^'': 
 
 473 
 
 450_ 
 
 235 
 200 
 
 350 
 
 J.-> 
 
 o 
 
 Here we divide a decimal by a whole number, proceeding ex- 
 actly as before, and since the decimal contains fou** places to the 
 right of the decimal point, and we require to add a cipher so as to 
 have no remainder, we necessarily require quotient to contain five 
 decimal places, and have to prefix a cipher to the quotient to bring' 
 about this result. 
 
 To divide when both divisor and dividend are decimals, and the 
 divisor contains more decimal places than the dividend — 
 
 Rule : Add ciphers to the dividend until it shall have as many 
 places as the divisor ; then proceed as in simple numbers. 
 
23 
 
 Example : Divide .125 by .0515. 
 
 .0515). 12500000(2.4270 
 
 '030 
 2200 
 2060 
 
 ,;;.;'; ■, - ■ 1400 -, :-■■■■'■ - 
 
 '■'--:■'" .;-■.. 3700 ■ ■ : ' ■ 
 
 '.;-:;■■,;..,-;:;;■ .,';;..■.■, \. ..'./ ^ 3605 :'_:::■■ :, .':r.^.'V 
 
 950 ,. , ■....•:-/:■; 
 
 This is an example of the application of principle i. While we 
 
 * annex to the dividend a cipher we in no way change its value, and 
 
 since we find .0515 to be contained in .125 two and a fractional 
 
 times, we pr*. i by process already illustrated to extend to the 
 
 fourth decimal place. 
 
 To divide when both divisor and dividend are decimals, and the 
 dividend contains more decimal places than the divisor — 
 
 Rule : Divide as in simple numbers, and point off from the right 
 of the quotient as many decimal places as the decimal places in 
 the dividend are greater in number than those in the divisor. 
 
 Example: Divide .5000 by . 125 
 
 .125). 5000(4.0 
 
 OCJ ^ ■;;/■■. \- 
 
 The application of both these rules is demonstrated in previous 
 example. Since, however, we want to make the principle clear, 
 we repeated in another form. The writer wishes especially io im- 
 press on the student the great importance of thoroughly master- 
 ing the principles of calculation contained in these exercises in 
 common and decimal fractions. A thorough knowledge of the 
 principles involved is of immense benefit in computing much of the 
 formula we shall at a later date meet with ; in fact, unless our 
 knowledge of these principles is thorough we cannot hope to 
 
24 
 
 il 
 
 properly appreciate all the good things that the scientist has priv 
 pared for us. 
 
 There remains for illustration yet what concerns many of us, 
 i.e., the principles involved during the conversion of common to 
 decimal fractions and vice versa, which I will briefly illustrate be- 
 fore closing this number. 
 
 Since the denominator of a simple fraction is always larger than 
 the numerator, to reduce or convert a common fraction to its 
 equivalent decimal fraction we have to follow one of the rules al- 
 ready laid down for the division of decimals, dividing the numer- 
 ator by the denominator and adding ciphers as required. 
 
 Example : Reduce [ i; to a decimal fraction. 
 
 i6)i5oooo(.9375 " ' 
 
 ^r, ■ :'"■■■■'•,:'" i44_ . .-:.-^V/-.;,;'.'^ v; 
 
 .;■■-■.."■ ;. ■ .. ■ ■ . 60 y ■■/--;:: ':■■■■:■- ^^-^ 
 
 120 
 
 I 12 
 
 So 
 So 
 
 Sii.ce we had to add to dividend four ciphers, we pL'ice the 
 decimal four places to the left, counting from the right, hence 
 
 ! ;"^ reduced to decimals = .9375. 
 
 Example : Reduce .9375 10 common fractions. 
 
 Since the numerator is 9375 and the denominator i, with as many 
 ciphers added as there is decimal places in the numerator, then 
 fraction will be set thus i'^tt/^o* which we reduce to simplest form : 
 
 lii$$ _ ht$ _ h^ 
 
 h _ 15 
 
 40()()() m)0 400 80 16 
 
PRACTICAL MEASUREMENTS. 
 
 -cs. of surface, of vo„„„. ..aZ:;:;^:;;"'""^"'''"^'' ■•■■^ ■■"■•■'- 
 
 MKAStRES OP srRP.UK OR SO, .v„„: ,„,.s, „„■ 
 
 o.>-.c.::^r:;;t:;-,--/--e,o,.,,,,,:^^ 
 
 Square 
 Rectang-le. 
 
 Oblong- Koctang-le. 
 
 ar..a. ^"'" '<'"Slh. and llic. producl will be ll, 
 
 Exa,„p,e. Kndthelengthofasu 
 and breadth 5 in. 
 
 rface whose area is o- «., • 
 "•v,rt IS 9^ sq. in. 
 
 95 -f 5-/9 in. = length required. 
 
26 
 
 IRREGl'LAR OR TRIANGULAR MEASIREMENTS. 
 
 A trianjjle is a surface having three sides and three angles. 
 An Isoceles tria ,le is a triangle having two equal sides. 
 
 IsocELEs Triangle. 
 An Equilateral triangle is a triangle having three equal sides. 
 
 EgiiLATERAL Triangle. 
 
 A Scalene triangle is a triangle which has no two sides equal to 
 each other. 
 
 Scalene Triangle. 
 
 The base of a triangle is the line which intersects the other two 
 lines at the point greatest distant from the vertex. 
 
27 
 
 The lines representing the base and altitude are perpendicular 
 to each other, and if a triangle be placed vertically, the base 
 will be the side on which it appears to stand. 
 
 4J 
 
 9 
 
 ■':^C": Base. 
 
 Rule : To find the area i>f a triangle, multiply the base by one- 
 half the altitude, and the prod.^ct will be the area. 
 
 Rule to find either the altitude or base of a triangle, atea, and 
 one side being known : Divide the area by one-half the given di- 
 mension, and quotient will be required dimension of other side. 
 
 Example: Find the required length I'f the base of a Scalene 
 triangle whose altitude is 6 inches and area equals 36 inches. 
 
 36^ by one-half altitude = 36^ 3= 12", required length of base. 
 
 CIRCULAR MKASUREMENT. 
 
 A circle, as geometrically defined, is a plain figure bounded by 
 a. curved line, all parts of which are equally distant from the 
 centre. 
 
 The circumference of a circle is the curve which bounds it. 
 
 The diameter of a circle is a straight line that passes through 
 its centre. 
 
 The radius of a circle is a straight line that joins the centre to a 
 point in the circumference. 
 
 [Note : Referring to this, it follows that the radius is one-half 
 of the diameter.] 
 
 Rule : To find the circumference ofa circle, the diameter being 
 
 
; t 
 
 28 
 
 j;iven, multiply the given diameter by 3. 1416, and the product 
 will be llie ciieumferenee. 
 
 Rule : To find the diatiieter of a circle, the circumference being- 
 given, multi[>lv the given circinnference by the decimal .31S3, and 
 the product will be the diameter. 
 
 Rules : To find the area of a circle, the circumference, radius 
 or diameter being known. 
 
 Multiply the circinnference by one-fovuth the diameter, and the 
 proiluct will be the .area ; 
 
 Or, multiply the square of the radius by 3.1416, and the product 
 will be the area ; 
 
 Or, multiply the square of the diameter by .7854, and the 
 pioduct will be the area. 
 
 Example : Find the area of a circular bolt 3.1416 inches in cir- 
 cumference. 
 _ Circumference x .3183 equals dia. 
 
 3.1416 X .31^3 — '"diameter, 
 
 1-^4 = .25 = one-fouith dia. 
 
 3. 1416 X .25 = '7854 sq. in. area. 
 
 Find the area of a circle whose radius is one-half 
 
 Exatnple 
 
 inch. 
 
 Radius squared x bv3.i4i6= ' 
 
 . r^ X .5 = .25 X 3. 1416 = . 7854 sq. in. area. 
 
 Example : Find the area of a circle whose diameter is i inch. 
 
 Diameter squared x .7854 = 
 
 I X I = 1 X .7854 = . 7854 sq. in. area. 
 
 A review of these examples will prove clearly to the student 
 why constant .7854 is used in formula when it is necessary to find 
 area of any circle, and needs no further comment. 
 
 MEASURES OF VOLIME OF SOLID BODIES OR CUBICAL MEASUREMENTS. 
 
 Cubical measurement includes length, breadth and thickness 
 being taken into consideration. 
 
 The process of finding the cubical contents of a solid body is 
 
g 
 
 id 
 
 us 
 I Vic 
 met 
 
 \ clr- 
 
 ,e-\uvlf 
 
 lnc\i« 
 
 [sludent 
 
 :MENTS. 
 
 uckness 
 
 body 
 
 IS 
 
 29 
 
 similar to JiiuHiij^ the area ; in fact, area multiptieil by tliickness 
 will ^\\c cubical contents. 
 
 MEASURES OF CAPACITY AND CIBICAL CONTEXTS OF' Si^JlARK 
 AND CYl.iNMJRKAL HOOIES. 
 
 Measures of capacity embrace meai-uremenis of barrels, tanks, 
 bins, boxes, cylinders, etc., etc. 
 
 The liquid gallon of the United States contains 231 cubic inches. 
 
 The Imperial ov Hritish jjallon contains 277.274 cubic inches. 
 
 A cubic foot of water contains approxi.nately 6.25 Imperial 
 gallons and weijfhs 62.5 pounds avoiriiupois at 62 F. under at- 
 mospherical pi'essure. 
 
 Rule: To find capacity of a square tank, dimensions of wliich 
 are known, multiply lenjj^th by breadih and product by depth, and 
 result will be cid^ical contents. 
 
 Example : Find the number of Imperial gallons of water a 
 rect.anyfular tank will contain, inside measurement of lank being : 
 length 2"' feet, width 15 feet, depth 3 feet 6 inches. 
 
 F'ormuh 
 
 Length x width x depth x 1728 
 
 277.27 ' 
 
 20' = length. 
 15' = width. 
 
 100 
 20 
 
 I 
 
 m 
 
 perial gallons 
 
 300 
 
 3.5 = depth. 
 
 1 500 
 900 
 
 1050.0 = cubical contents in cu. feet. 
 1728 = cubic inches in one cu. foot. 
 
 8400 
 2100 
 
 7350 
 1050 
 
 i.Si4,4oo = cubic inches = 
 1.814.400-^277.27 = 6543.8 Imp. gallons, 
 
 i: 
 
 i 
 
 iiJ-it 
 
 ft- 
 

 30 
 
 or, when absolute eoireetness is not required, short method for- 
 mula as follows may be adopted. 
 
 Leng^lh in feet x width in feet x depth in feet x O.25 Imp. 
 
 gallons. 
 
 20' X 15' X 3' 6"- 1050 X 6.25 6562.5 Imp. gals. , 
 
 Example (2) : How many jjjallons (U. S.) of oil will cylindrical 
 oil tank hold whose internal dimensions are : Depth 4 feet 6 Miches, 
 diameter 3 feet. 
 
 Formula : 
 Diameter squared x .7854 x depth x 172S 
 
 3'x3' 
 
 9'x 
 
 231 
 
 U. S. gallons. 
 
 7S54 7.0686 — area in sq. feet — 7.0686 
 
 4.5 depth. 
 
 31.80 
 1728 
 
 cap. in cu. It, 
 
 54950.4 cap. m. cu. m. 
 54950-4-^ 231 -237.9 gallons oil. 
 
 Example (3) : It is required to fit a tatik into a recess 6 ft. wide 
 and 14 ft. 6 in. long. Tank to be made of 3 in. plank and contain 
 5000 Imperial gallons of water. What height should tank be so as 
 to allow sides to project > in. above the water line? 
 
 First find exact internal measurement of tank, then exact quan- 
 tity of water contained in each inch of depth; divide required 
 quantity of water by this product, and add 6. Result will be re- 
 quired height. 
 
 Dimensions of recess are given as 6' x 14' 6". Since, however, 
 tank is to b.^ constructed of 3 in. plank, twice this thickness must 
 be vI_uuc*od to get internal measurement of tank, 
 
 .-. 6'-6"=5' 6" internal width of tank. 
 i4'6"-6"=i4' internal length of tank. 
 14' X 5' 6"= 1 1.08.S sq. inches. 
 1 1.088 -f 277.27 -39.99 Imp. gals, per inch. 
 5000-^39.99= 125 inches to water line. 
 ( 1 25 + 6) -i- 12=^ 10' 11" height of tank. 
 
3» 
 
 Example (4) : A tank i ft. y in. deep, 2 ft. long^ and 4 ft. \vide» 
 is found to contain 70 Imperial gallons of oil. How far from the 
 top is the oil, and what is capacity of tank when full ? 
 
 Since one g^allon of oil is equal to .16 cu. ft., we can here ma- 
 terially shorten our calculation by workint^ decimally. 
 
 First find how many cubic feet 70 gallons are equal to by mul- 
 tiplying by . 16. 
 
 70 ■ . 16 I i.j cu. feet, 
 
 and as the tank is 2 feet by 4 feet, its base will equal 8 sq. ft. 
 
 Then divide 1 1.2 cu. feet by S sq. feet to get iieight iM c',\. 
 ii.2-r8 1.4 feet height of oil. 
 The tank is given as being 1.75 feet high, 
 ^ and the oil is 1.40 " " 
 
 then oil is 0.35 feet from the top. 
 0.35 feet X !2 4.2 inches distance from top of tank to oil. 
 
 To find capacity of tank when full, we first find cubical con- 
 tents of tank in feet and then divide by .16. 
 
 1.75 X 2 X 4^ 14 ft., cubical contents of tank. 
 14 feet -r .16 87.5 gallons of oil in tank when full. 
 
 Example (5) : If one cubic foot of coal weighs 84 pounds, what 
 weight of coal will a car contain 30 feet long, 6 feet wide and 5 
 feet deep? 
 
 First find cubical contents of car in feet, then multiply by weight 
 per cu. ft. Result will be total weight of coal. 
 
 .*. 30 ft. X 6 ft. X 5 ft. ^=900 cu. feet. 
 900 X 84 = 75600 pounds of coal. 
 
 Example (6) : How many cubic feet of steam will a cylinder 
 contain, 18 in. x 36 in., with 5 per cent, clearance, and how many 
 cubic feet will be used per hour if engine runs 75 revolutions per 
 minute. 
 
 First find cubical contents of cylinder, including clearance, then 
 multiply by strokes per minute, and this product by 60. Result 
 will be number of cubic feet of steam consumed per hour. 
 
( — — -" --^ J^i-us 
 
 I 
 
 1 1 
 
 , '. 
 
 4 
 
 
 32 
 
 18 in. X 18 in. x .7854 = 254.47 sq. in., area of cylinder. 
 
 254.47x36111. = 9160.92 r^ cnbical contents of cylinder nithout 
 clearance. 
 
 9160.92 -r95x 100=9653.6 cu. in.. = contents of cylinder, clearance 
 included. 
 
 9653.6 -r 1728 = 5.58 cu. ieet. 
 
 5.58 X 75 X 2 = 837 ^^^' ^t. steam per minute. 
 
 837x60 = 50.220 cu. ft. of steam used per hour. 
 
 PRACTICAL RLLHS WORIFI REMEMBKRING. 
 
 The diameter of a circle x by 3. 1416 = the circumference. 
 
 The circumference of a circle -^ by 3.1416 = diameter. 
 
 The radius af a circle x by 6.283185 = circumference. 
 
 The circumference of a circle -f 6.283185 = radius. 
 
 The square of the radius of a circle X3. i4i6 = the area. 
 
 The square of the diameter of a circle x 0.7854 = the area. 
 
 The square of the circumference of a circle x 0.07958-- the area. 
 
 The circumference of a circle x o.i59i55 = the radius. 
 
 The square root of the area of a circle X 1.12838 = the diameter. 
 
 A semi-circle is ^ of a circle, or 180 deg^rees. 
 
 A quadrant is ^ of a circle, or 90 degrees. 
 
 A sextant is y^ of a circle, or 60 degrees. 
 
without 
 jarance;, 
 
 ce. 
 
 a. 
 
 the area. 
 
 iiameter. 
 
 EVOLUTION. 
 
 Evon TiON is llu" process of P-uliiig the root of any minibcr, 
 and is of frequent occurrence in eng^ineers" calcuhitions. 
 
 The most important and only cases wliich I sliall refer to is the 
 process of findiii^ the square or cube root of any number. 
 
 Example ( i ) : Find the square root of 1521. In formula this is 
 written \ 1521. 
 
 69. 
 
 621 
 621 
 
 Beg"inninj>- at the right hand figure i, count two Hgures to the 
 left and mark the second as shown in the example, lake the 
 figures to the left of this mark 15 and find what luuuber multiplied 
 by itself will give fifteen ; there is no number that will do this, 
 since 3 x 3 = 9 is too small and 4 x 4= 16 is too large ; we therefore 
 take the one that is too small, viz., 3, and place it in the quotient, 
 placing its square 9 under the 15 marked oft" in the number and 
 subtract, and bring down the next two figures 21, making new 
 dividend 62t. To get the new divisor multiply the quotient 3 by 
 2 = 6, and place as a trial divisor at ihe left of the dividend 621; 
 find how many times this is contained in the dividend, discarding 
 the last figure on the right ; 6 is then contained in 62 nine times. 
 Since we cannot pass this point, we |ilace 9 in the quotient and 
 also in the divisor ; then we multipl) the whole divisor 69 by this 
 number (9) and place the product under the dividend and subtract. 
 Having no remainder, root is now complete and found to be 39. 
 
 ■;*SM 
 
1 1< 
 
 mmm 
 
 mmm 
 
 r 
 
 '^ "■" 34 
 
 If we require proof of this we simply square the number, thus 
 
 39x39=1521 
 then 39- = 1521 : 
 
 and v^F52i=39 
 
 Example (2) : Find the square root of 366. 
 
 3.66{ 19. 131 126 
 1 
 
 ^9 
 
 381 
 
 266 
 261 
 
 3823 
 
 S.oo 
 
 I I goo 
 I 1469 
 
 38261 43foo 
 
 382621 { 483900 
 I 382621 
 
 3826222 
 
 1 01 27900 
 765 -'444 
 
 38262246 247445600 
 
 _^29573476 
 
 17872124 
 
 In this example we proceed to mark off as before and get as 
 our quotient the wliole number 19, but since the e is still a re- 
 mainder our roo. cannot be complete; we proceed as before, but 
 since there remains no more figures in the dividend we annex to 
 new dividend two ciphers and place a decimal point in the quotient 
 lo the right of the nineteen. Then proceeding as before we get 
 our trial divisor by doubling present quotient 19 x 2 = 38 and ' oC-2 
 lo the left of the dividend 500. 38 is contained in 50 once 
 therefore place a i in the quotient to the right of the deciuii 
 point and annex to the new divisor, multiplying as before and sub- 
 tracting, we proceed in an exactly similar mannet until we have 
 no remainder or until decimal begins to repeat itself or is ex- 
 tended sufficiently to answer our purpose. 
 
 m 
 
35 
 Ex,-uii[)Ie (3) : I-"iiui tho square root of 15227.56. 
 . 1.52.27.56(1^3.4 
 
 •52 
 44 
 
 243 
 
 827 
 729 
 
 2464 I 9S56 
 
 get as 
 ill a re- 
 re, but 
 Innex to 
 |uolient 
 we get 
 Id T' cvce 
 Ice -> •" 
 leciu)' I 
 md sub- 
 Ive have 
 )r is ex- 
 
 In a decimal quantity like the foiej>oiiig, the marking off differs 
 from the previous examples. Instead of counting twos from right 
 to left begin at the decimal point and count twos towards the left 
 and towards the right. Note : when the first two figures to the 
 right of the decimal is brought down we must place a decimal 
 poijil in our quotient before extending it. 
 
 Briefly, the process of finding the square root of a whole num- 
 ber may for the guidance of the student be described as lollows : 
 
 P'irst mark off in twos commencing from the right. Then find 
 the number whose square is next less than the figure or figures on 
 the left of the number as the case may be, place this figure in the 
 quotient and its square under liie figure or figures already marked 
 off on the left of the number, subtract and annex to the right the 
 next two figures of the number ; this then becomes a new divi- 
 dend. To secure a new divisor, douhle the quotient by multiply- 
 ing by 2 and place on the left of the dividend as a trial divisor ; 
 fiiul how manv times this is contained in the dividend, discarding 
 the right hand figure, place this result in the quotient and also in 
 the new divisor and multiply by same number, placing product 
 under dividend .'ind subtract as before ; continue the operation for 
 a new divisor. Bear in mind, however, you must always piace 
 two figures to the right of the new dividend. 
 
 It will occasionally occur as in our first two exartjples that the 
 
36 
 
 product of llie trial divisor multiplied by the new divisor will 
 exceed the dividend. In this case take the next lowest term anil 
 proceed as described. 
 
 TO FIND I UK CI i;k root. ' 
 
 Example (i): Find the cube root of 1331. 
 
 3' 
 
 300 
 
 •3' 
 
 331 
 
 331 
 331 
 
 First mark off three figures from the rig-Jit towards the left of 
 the nun.ber, and we have 1 left. Now, 1 cubed equals one. 
 Place the cube of i under the one of the number and subtract. 
 There is no remainder. Bring- down the next three figures as a 
 new dividend. Next nniltiply tlie one placed in the cjuotient by 3, 
 and place the product well to the left, as in the example. Next 
 multiply this 3 by the quotient figure 1, and place the result, 3, to 
 the right of where tiie 3 was placed and to the left of the dividend, 
 and add two ciphers to it, as shown in the example. 
 
 This 300 is called the trial divisor. Now see how often it will 
 go into the dividend 331, which we find to be once. Put 1 in the 
 quotient to the right of the one already there, and place 1 also to 
 the right of the nuinbe.- 3 at the extreme left of the example. 
 Now, nuilti[>ly this tnunber 31 by the figure last placed in the 
 quotient, and place the result under the trial divisor 300 and add. 
 This now gives us the correct divisor, which we multiply by the 
 last figure placed in the quotient and place result vmder and sub- 
 tract from the dividend. There being no remaintler, our root is 
 
 now com 
 
 pleti 
 
 and we 
 
 find 
 
 \ 
 
 «i.^ 
 
 1 — I 
 
37 
 
 visor will 
 tcfin and 
 
 the left of 
 equals one. 
 d subtract, 
 ^ures as a 
 lotient by 3, 
 pie. Next 
 result, 3, to 
 je dividend, 
 
 ,t'len it will 
 Put I in the 
 ce I also to 
 he example, 
 placed in the 
 ;oo and add. 
 liply by the 
 ler and sub- 
 , our «oot is 
 
 Kxarnple {2) : I'iiid the cube root of S067756S161. 
 Process : 
 
 123 
 
 
 80,677, :;68, 161(4321 Ans. 
 
 4S00 
 
 369 
 
 16677 
 
 5 '69 
 
 •5507 
 
 554700 
 •^5«4 
 
 1 17056S 
 
 5.S7-'«4 
 
 \ 1114568 
 
 559S7200 
 1 296 1 
 
 
 5fcoooi6i 
 
 s6ooo 1 6 1 
 
 
 56000161 
 
 1292 
 
 I 296 1 
 
 The process to the second fig^ure of the answer is a reproduction 
 of last example. 
 
 TO IIN'I) TIIK TIIIKH FKU'RK OF THE ANSWKR. 
 
 Multiply the _I3 in the quotient by 3, equals 43x3-129; put 
 this weU ou. to the left as before. In the middle coluinu of figures 
 you will see the figures 369 antl 5169 ; add these together, and to 
 their sum aild the square of the last figiue in the quotient— that is 
 in the present ease 3. Then 
 
 5547 
 
 This nuuiber, with two I'iphers added, is our new trial divisor 
 554700. It goes into the dividend 1 170568 twice. Place 2 .... the 
 third figure of the quotient, and place 2 to the right of the .lumber 
 129 on the left. Multiply 1292 by this number 2 and place the re- 
 sult, 2584, under the trial divisor and add. Result, 557284 is now 
 the correct divisor. Multiply this number by 2 and place uPider- 
 nealh dividend and subtract. Bring ilown the next three figures. 
 We now have 56000161 as a new ilividend. 
 
 a 
 

 38 
 
 TO FIND Till-: FOl KTII FICHRE OF THK ANSWER. 
 
 Multiply llie quotient 432 by 3— 432 x 3— 1296, and place this 
 number to the left. Again in the middle column we have the 
 figures 25<S4 and 5572S4 ; add together and add the square of the 
 'ast figure in the quotient. 
 
 2584 --"::■' ":■'/■"-.,-,,.. 
 ■ ': 557284 ... 
 
 2 + 2= 4_ 
 
 559872 
 This number, with (wo ciphers annexed, 559S7200, is oin- new 
 trial divisor, and is contained in the dividend once. Add one to 
 the quotient, also to the number on the left. Multiply and subtract 
 as before. And since we have no remainder our root is now com- 
 plete, and ^80677^,68161 =4321. 
 
 There is .mother method of extracting the cube root which to 
 the student may be simpler and process clearer, since it has the 
 advantage over method just described that when the student has 
 mastered the process of working out the first figure of the quotient 
 he has the rest at his command, as the ptocess is but a repetition, 
 and is the same for four figures as for two. - ,. ., 
 
 Find the cube of 1728. 
 
 1.728(12 
 
 3 X I o- = 300 
 3x10 X 2 = 60 
 
 ^' , = 4 
 364 
 
 728 
 
 728 
 
 I'irst we mark ofl' the number into quantities of three figures as 
 in previous examples, and selectitig the next lowest cube for the 
 first figure of our quotient in the case 1, which is put down in the 
 quotient and as a divisor. The cube oi i is placed under the 
 dividend and subtracted and next penod 728 broiight down as a 
 new dividend. Now take the number in the quotient and add a 
 
 u — 
 
39 
 
 cipher to it, making" it lo, square this and multiply by constant 3 ; 
 we then g^et 3 x ( io-') = 30o which is our trial divisor ; for the next 
 lin*' multiply 10 by the constant 3 and multiply the product 30 by 
 the figfure placed in the quotient from the trial divisor 30 x 2 = 60; 
 place this 60 below the 300 aheady obtained and atld, to the sum 
 of these numbers add the square of the last fig"ure of the quotient, 
 thus, 300 + 60 4 2- =364. This is now our correct divisor. Mul- 
 tiply by the last fig"ure of the quotient, place underneath the divi- 
 dend and subtract ; beingf no remainder our root is complete. 
 
 APPLKATIl>N OF FORMKR STL DIKS TO Mi:CMANKAI, AND 
 KLKCTI^ICAI. KNC'.INKKRINc;. 
 
 Since both numerical and alg'abralcal formula will now be con- 
 stantly used, it is absolutely r.eeessary that students should en- 
 dejivor to obtain facility in reading and solving; formula g"ener- 
 ally. 
 
 The following signs must therefore receive particular atttMition, 
 and be committed to memory. 
 
 + is read plus, and means that the number following it is to be 
 added to the number before it ; thus, 4 + 3 is 7. 
 
 - is read minus, and means that the number after it is to be 
 subtracted from the number before it ; thus, 5 - 2 is 3. 
 
 X is read multiplied by, find means that the number before it is 
 be multiplied by the number following' it ; thus, 3x3 are 9. 
 
 ~ is read divided by, and means that the number before it is to 
 
 be divided by the number following it ; thus, 6^3 — 2. 
 =: is read equal to, and means that the quantity after it is of 
 
 same value as quantity before it ; thus, 7x3 = 21. 
 9- is read as 9 squared, and means that the number is to be 
 
 multiplied by itself; thus, 9- =9x9 = 81. 
 9"' means same number cubed ; thus, 9 x 9 x 9 = 729. 
 ^>-' is read the difference between any two numbers, and means 
 
 the less munber is to be subtracted from the gfreater. 
 
 
■ IHMV 
 
 I ,!i ! 
 
 40 
 
 ( ) are called brackets, and iiieati tliat all llie c|iiantitles within 
 lliem are to be put toj^fcMier first ; thus, 5 (4 - 2 f 3 x 9) >ieans 
 that 2 must be subtracted fVoni 4 = 2, and 3x9 = 27+2 = 29, 
 and tluMi this 2«) is to lie multiplied by 5 = 29X 5= 145. 
 
 NoTK — When 1.0 sij^n is placed between a cjuantity and a brack- 
 el or a letter, it means that the quantity within the bracket is to 
 be multiplied by the quantity outside. Thus in the tbres^'oinj^ the 
 quantitx- within (he bracket — 29 is to be multiplied by 5, tlu* quan- 
 tity outside. 
 
 SIGNS THAT RT'.PItESENT ROOTS OF NIMHKRS. 
 
 The sig'n known as the radical sis<n is common to all numbers, 
 and is expressed thus, \'or \/ . When it is required to ex- 
 
 press the square root of a number we simply put this sij^n before 
 it, as \ 25, but if the nvunber for which we desire the square root 
 of is made up of two or more terms then we express the square 
 root by the same sig-n in front, but with a line as far as the square 
 
 root extends, as V 1 5 + 10 = 5, or \ 5(2 + 3) = 5. 
 
 The cube root is expressed in a similar matiner, but with, a 
 
 n 
 
 small 3 in the elbow of the sig-n, as \' ; all other roots in exactly 
 
 5 (S 
 
 same maimer ; as \/ \/ and so on. 
 
 In mecha ical text books the power and root are often com- 
 
 bined and expressed thus 4-'. This is read as the square root of 
 4 cubed. Therefore, the numerator represents the power and 
 the denominator the root. In this case the square root of 4 = 2, 
 and 2 cubed equals 8. 
 
 The commonest form in which this is met with in en^ineerinj^ 
 
 calculations is when the cube root of a number is squared, as 27^ 
 which is read cube root oi' 27 squared. The cube root of 27 is 3 
 
 and 3 squared equals 9, which is the value of 27 •^. 
 
 One of the most frequently occurring- errors in algabraic calcu- 
 lations is the misunderstandingof the proper use of the multiplica- 
 
4» 
 
 tioti sii;"!! in ali^-abraio quantities and fonmila, and requires speeial 
 mention liere. I'oi* instance, we have the quantity 25 10x2. 
 It is a eoininon error ti> say 23 10 -15; this inullipHed by 2 
 equals 30. Tliis is entire!}' wrv^nj^', as a im>inent's refleetion will 
 l^rove to the student, as it clearly reads 25 minus 10 x 2, and the 
 first step should be to multiply 10x2 = 20; then, 25- 20 — 5, which 
 is correct, and not 30, as at first appears. 
 
 The student recjuires io remember the followinj^" rule, which 
 applies in .'dl cases : * 
 
 Multiplication and division sig-ns CONNKCT the numbers tog-ether 
 between which they occur; plus and minus signs SKl'ARATK them. 
 
 For simplicity a!id,accuracv, it is .v good rule when woi'king- 
 with complex formula, to first g;et rid of bracket signs, ami then 
 multiplication signs; the opi-ration of ci>mputation cif fc>rmula then 
 becomes an eas\- task. 
 
J*^ 
 
 
 SAFEl'Y VALVE CALCULATIONS. 
 
 All boilers should be fitted with two safety valves, one of which 
 should be a lock-up valve, and set by the Boiler Inspector under 
 whose miniediate control it is. 
 
 The Canada Steamboat Act provides that every safety valve 
 must have a lift equal to at least one-fourth of its diameter ; the 
 openings for the passage of steam to and from the valve must 
 each have an area not less than the area of the valve, as must also 
 all waste steam pipes, etc., and the area of a safety valve must 
 equal one-half inch for each square foot of grate surface in or 
 under the boiler. 
 
 LEVER TYPE. ' 
 
 Find the diameter of a safely valve required for a boiler whose 
 grate bars are 5 feet long and furnace 3 feet wide. 
 
 We first find number of square feet of grate surface; then divide 
 by 2, which gives area of valve in inches ; square root of area 
 divided b}- .7854 equals diameter, 
 
 LxW , 
 
 -,...;, ,,.;,,;;.•;,/-':;: then -=A 
 
 a nd \/-\ "^" • 7854 = I^' , , ' :■ ■ 
 
 Where L equals length of grate bars, 
 
 W equals width of furnace, 
 
 A equals area of valve in inches, 
 
 D equals required diameter of valve. 
 
 S ^ "? . . 
 
 " — ^^ = 7'5 ^q- inches. 
 
 7-5 ^-7^54 = 9- 549- 
 Square root of 9.549 equals 3.09 inches ; then required diameter 
 of valve is 3.09 inches, say 3 's inches. 
 
43 
 
 Kxainple (2) : What woi^lit i^ re(|iiired to bo placed 2 inches 
 from the end of a safety valve lever to eipial a boiler piessure of 
 .50 pouiuls to the sijiiare iMch, the diameter of valve being' 3 's 
 inches, the distance from fulcrum to valve 6 inches, and total 
 Ieng"th of lever from fulcrum 16 inches? The wei^-ht of valve .md 
 stem is 15 pounds, and elTective moment of 'ever So inch poimds. 
 
 
 y\ 
 
 
 
 n 
 
 Since the steam pressure within the boiler and consequently 
 pressing- ag^ainst the lovve** tVce of the valve and trying- to unseat 
 it, equals so many pounds per square inch, we must first find the 
 area of the valve in square inches, to ascertain the whole force 
 tending to raise the valve, but since the weight of the valve, 
 spindle and lever acts downwards and against the upward pressure 
 of the steam, we must make allowance for this, and the remaining 
 force is that which we have to counteract by weight. 
 
 In diagram No. i I'^ is the fulcrum, V is the point where pressure 
 
44 
 
 is I'xerleti, \\" is llu' \v»'i>fhl, l"\' is 6 iiuMios, \'\\' is lo iiu'lu's, and 
 F\V is 1 6 inches. 
 
 The priiuiple of (he lever is : The wei^^ht or force multiplied hy 
 its distance from tlie fulcrum is etpial to the \vei>;"ht or pressure on 
 the valve multiplied by its distance from the fulcrum, 
 
 or, \Vx K\V = VxFV. 
 
 Hut the steam pressure has, in addition to the actual weig'ht of 
 W, also to overi'ome the moment of the lever, which is found by 
 wi'i^-hing' the lever and (indinj^ how far its bal incing" point is from 
 I'" ; tlien this distance multiplied by its weight is the effective 
 monjenl of the lever, which we will call A. 
 
 Then VVxIWf A^VxFV ; 
 
 that is, the total force at work keeping- the valve down is equal to 
 the force or pressure endeavoring- to lift it off its scat. 
 
 In accordance, then, with these principles, we get the following 
 rules : 
 
 (i) Find the area of the valve and multiply it by the pressure 
 per square inch. 
 
 (2) P'rom the product take th(^ weight of the live and stem ; 
 the remainder is the \' of the formula. 
 
 (3) Multiply "the rcm.under by the distance from the fulcrum to 
 the valve," FV, then subtract the moment of the lever, and divide 
 by " the distance from the fulcrum to the weight," FW, found by 
 adding "the distance from the fulcrum to the valve " to that 
 " from the valve to the weiglil." 
 
 T,. I 25 X 3. 1 25 X . 7854 = 7.67, area oi' valve. 
 
 7.67 X 1:50 = 383.50, pressure against valve. 
 
 383.5- 15 = 368.5, effective upward weight. 
 
 368.5x6 FV = 2211.0, effective moment lifting lever. 
 
 221 1 - 80 (efHeclive moment of lever acting downwards) = 221 1 - 80 
 
 =^2131. 
 
 213T -^ 16= 133.-?, required weight of \V. 
 
 Note : For extreme accuracy it is necessary to take note of the 
 
:'f 
 
 Nveij^-ht of the v.'ilvo rind its parts, and also o( the monuMit of the 
 lever , hut in a ^reat many eases tliis is entirely omitted. Then 
 a question of this kind becomes simplifu'tl thus : 
 
 W 
 
 area x 
 
 pre 
 
 ssiu'e 
 
 rv 
 
 F\V 
 
 ro liKAin AIK A SAFKTY VAL\ K I K\ T.U. 
 
 We have a safety valve 4 inches in diameter, anil spimlle presses 
 ajjainst a point in the lever that is 4 inches from the fukrutn ; how 
 
 fa 
 
 ar must a weiglit ot 120 pounds be placeil from the fulcrum to 
 equal a boiler pressure of 60 pounds lo the square inch, when the 
 valve weighs 8 pounds and effective moment of the lever is 50 
 inch pounds. Also g'ive the gfraduation marks on the lever for 40 
 and 50 pounds pressure with the same weight. 
 
 Formula V'-= area X pressure 8. 
 
 V X FV - 50 ■ 
 
 W 
 
 FW 
 
 In this question we require to find distance, l'*\V. We first luul 
 area of va' e ; mullipl\- this by pressure per square inch, and then 
 subtract \^L'ig-ht of valve and parts b«Mrinji^ downward to ^cl total 
 effective iipward pressure ; this equals V of formula. Xow mul- 
 tiply total upward pressure by distance fulcrum to valve and sub- 
 tract effective moment of lever ; result, divided by weig"ht, equals 
 distance fulcrum to weig^ht. • 
 ; 4 X 4= 16 X .7854=: 12.5664, area valve. 
 
 1 2.5664 X 60 = 753.98, total pressure. 
 * 753.98 -8:= 745.98, \', total effective pressure. 
 745.98 X 4=^2983.92, lot.al moment of valve. 
 2983.92-50^:2933.92, total effective moment of valve. 
 2933.92^ 120=24.45 inches distance FW. 
 Since the distance FW = 24.45 inches for a pressure of 60 pounds 
 to the square inch, we can find w'hat distance repr€\sents a pres- 
 sure oi' 10 pounds by dividing this distance by 6, the number ot 
 times 10 is contiiined in 60. 
 
 -'4-45-^6 = 4-07. 
 
 . J 
 
 r 
 
 :| 
 
 w 
 

 wr 
 
 n 
 
 ^K^BHBBlfv 
 
 i 
 
 Then for each additional 10 pounds pressure we require to move 
 the weiifht a distance of 4.07 inches further from the fulcrum, and 
 to find the distance from the fulcrum equal to a pressure of 40 
 pounds, this distance multiplied by 4^FW ; therefore, 
 
 FW at 50 pounds pressure = 4.07 x 5 = 20.35 Ins. 
 FW at 40 pounds pressure = 4.0} x 4= 16.28 ins. 
 
 Required to find distance from fulcrum to valve, boiler pressure 
 
 being 50 pounds to square inch ; diameter of valve, 4 inches ; 
 
 weii^ht of valve and spindle, 10 pounds ; distance valve to weight, 
 
 12 inches, and weight 150 pounds; moment of valve lever, 40 
 inch pounds. 
 
 Formula : 
 
 VW X W f moment 
 
 FV 
 
 V 
 
 4- X .7854= 12.5664 square inches. 
 12.5664x50 = 628.32, total pressure. 
 
 628.32- 10 = 618.32, total effective upward press ire, or V. 
 1 2 x i 50 = 1 800 pounds. 
 1800 + 40= 1840, total weight acting downwards at V. 
 1840 + 618.32 = 2.975 inches distance FV ; or distance FV is 
 nearly 3 inches. 
 
 SUMMARY. 
 
 To FINI1 effective moment of lever, multiply weight of lever by 
 distance from lis balanciig point or centre of gravity to fulcrum, 
 and divide by distance fr.^r! centre of valve stem to fulcrum. Re- 
 sult will be effective moment of lever in inch pounds, or the weight 
 required to raise valve off its seat with nothing but lever holding 
 against steam. 
 
 To FIND actual effective weight cf ball, divide weight of ball by 
 dlstcance from fulcrum to valve stem, and multiply quotient by dis- 
 tance from ball to fulcrum. 
 
 To FIND length of lever, add together effective moment of lever 
 and weight of valve and stem, and subtract from total pressure 
 
47 
 
 Ir 
 le 
 
 acting' upwards against valve. Divide rem.'iinder by weight of 
 ball, and multiply quotient by distance from stem to fulcrum. > 
 To FIND vveig^ht of ball, add together the efTective moment of 
 iever and weight of valve and stem, and subtract this sum from 
 total pressure agfainst valve at blowing-olV point. Multiply re- 
 mainder by distance from fulcrum to stem, and divide quotient by 
 length of lever from fulcrum to weight. 
 
 To FIND diameter of valve to blow off at g'iven pressure, add to- 
 g-ether effective mument o( lever, weig^ht of valve and stem, and 
 effective weight of ball. Divide this sum by gauge pressure, and 
 result will be required c»rea of valve. 
 
 Square root of area divided by .7854 equals diameter. 
 
 To FIND pressure at which a boiler will blow off, add together 
 effective moment of lever, weigiit of valve and sten>, and effective 
 weight of ball. Divide this sum by area of valve in square inches. 
 Result will be gauge piessure at which safety valve will act. 
 
 FUNDAMENTAL PRINCIPLES OF ELECIRIC ENERGY. 
 
 COMMON I NITS OF MEASIRF. 
 
 The units commonly met with by eng;ineers in (he study of the 
 principles underlying the generation, transmission and uue of 
 electricity are the volt, ampere, ohm and watt. 
 
 In diagram No. i A is a dynamo or electric battery, and (he 
 source of electric energy, and the purpose of which is to produce 
 a difference in potential between the terminals B and B,. This 
 difference in potential is measured in volts, and we say that be- 
 tween the terminals B and B, there exists an electric potential or 
 pressure of so many volts, written symbolically (E.M.F.) 
 
 Let us now suppose that a difference in potential exists between 
 B and B,, and that B is the point of higher pressure ; if we con- 
 nect B and Bj together by a substance capable of conducting- 
 electricity, there will be a flow from B to B,. This flow of elec- 
 tricity is known as a current and measured as amperes. 
 
.^.-. ...J 
 
 The rale at which cunetit will flow iVom B lo H, when joined 
 together by a conductor as at C depends upon following- condi- 
 tions : I St, upon the difference in pressure or electric potential 
 between B and B, ; 2nd, upon cross section or area of conductor 
 
 F/q. I 
 
 C, and upon lengtii of C and nature of material oi' which C is 
 composed ; or, in oilier words, we may say that the rate at which 
 current will flow through conductor C depends upon the difference 
 in pressure and upon resistance offered by C to current. 
 
 Therefore, the g'-eater the dift'erence in potential the greater 
 the current. 
 
 When a current flows tlirough a conductor there is a loss in po- 
 tential or voltage caused by the resistance of the contluctor. This 
 resistance is measured in (^hms, and consequently a conductor is 
 said to have a resistance of so many ohms. ---— ..,„-..-,-__,,_.___ 
 
 When a ditt'erence in potential exists between B and B,, hut no 
 
49 
 
 C is 
 
 ,hich 
 
 Irence 
 
 ater 
 
 |iti po- 
 This 
 
 lor IS 
 
 111 no 
 
 connection between them, there is no flow of current through the 
 dynamo or generator A, and therefore no work will be done by 
 it. But if conductor C is connected to terminals H and B,, cur- 
 rent will at once begin to flow, and A will be compelled to do 
 work to keep up the flow. The rate at which this work will be 
 done will depend upon difference in potential between terminals B 
 and Bj, and upon quantity of current flowing through the circuit. 
 This then becomes the electrical energy or rate of doing work, 
 and the electric unit of work or energy is the watt, and, in accord- 
 ance with above, equals the volts multiplied*by the amperes, or 
 by the potential in volts multiplied by current in amperes. 
 
 The watt, then, is the product of one volt and one ampere, and 
 in energy or work is equal to -jj^ h.p., or 746 watts are equivalent 
 to the mechanical force necessary to raise 550 poimds one foot 
 high in one second, or 33,000 pounds one foot high in one minute. 
 
 The kilo-watt, as the name implies, equals 1,000 watts. 
 
 The symbols commonly uaed in formula to represent the units 
 above described are as follows : ' 
 
 E. or E.M.F. equals electro motive force or volts. 
 
 C, the current or amperes. 
 
 R., the resistance or ohms. 
 
 W. , or watts, represents the electrical energy. 
 
 K.W. represents kilo-watts. 
 
 SPRING-LO\DED SAFETY VALVES. 
 
 The questions of most importance to the practical working 
 engineer regarding spring-loaded valves are : Size of steel fcom 
 which the spring shall be made ; required in de and outside 
 diameter ; compression required to have p-iven effect. 
 
 A standard spring, if made of the best square steel, contains an 
 area of .25 of a square inch, the inside diameter is exactly two 
 inches and outside diameter three inches ; it contains thirteen 
 complete coils, and measures exactly eleven and one-half inches 
 in length. The working load is assumed at 600 pounds, one-sixth 
 
— 1 
 
 50 ■ 
 
 of its breaking^ load when hard: ned to a temper just sufficient to 
 break it ; at this load it should deflect exactly one inch. 
 
 Example (i): A safety valve 4 inches in diameter has a spiral 
 spring^ made of square steel 3" diameter outside and .25" thickness 
 of steel ; what will be the pressure per square inch ? 
 
 Formula : 
 
 12,000 S' 
 = whole pressure on valve. 
 
 Where S = thickness of steel in inches, . 
 
 d^diameter offspring- from centre to centre of steel, 
 1 2,000 = constant used for square steel, 
 8,000 = constant used for round steel. 
 
 ^. ... 12,000 X .c"' ^ 
 
 Then, total weight = ^^ =600 pounds. 
 
 2.5 
 
 Diameter of valve is given as 4 inches. Area of valve then =4" x 
 .7854= 12.5664 square inches. .'. pressure per square inch = 6oo 
 -^ 12.5664 = 47.7. 
 
 The foregoing^ is the fundamental principle to connect the load- 
 ing of the spring: valve with that of a direct weighted valve, and 
 from it may be obtained both the proper thickness of steel to be 
 used and the proper inside and outside diameters of the spring. 
 
 Example (2) : What must be the outside diameter of a spiral 
 spring for a safety valve 5" in diameter ? The pressure to be 
 carried is 50 pounds, and the diameter of the steel is ^4 inch. 
 
 Formula : 
 
 8000S" 
 
 d = 
 
 W 
 
 Where d equals as before the mean diameter of the spring, S 
 thickness of steel, W the whole weigfht on the valve, then 
 
 d = 
 
 8000 X. 75'* 3360 
 
 = 3.42 inches. 
 
 52 X. 7854x50 981.75 ^ 
 
 This, however, is only the mean diameter, or the diameter from 
 
centre to centre of steel. Therefore, the diameter of the steel must 
 be added to this to get the outer diameter. 
 
 .•. outer diameter^: 3. 42 + .75 = 4.17 inches. 
 
 Example (3) : The diameter of a spring loaded safety valve is 
 5 inches, g^auge pressure 60 pounds, and mean diameter of a 
 spiral spring^ 5 inches, what must the area be for square steel, 
 also the length of each side, the area and diameter for round 
 steel, and the inside and outside diameter of each spring. 
 
 The Steamboat Inspection Act adopts the Board of Trade rule 
 for the determination of the required size of steel under the fol- 
 lowing formula : 
 
 Vwxd_^ 
 
 s = Side or diameter of steel in inches. 
 w=Load on spring in pounds. 
 
 d = Diameter of spring from centre to centre of steel. 
 c= 12,000 for squiire steel. 
 c = 8,000 for round steel. 
 
 Then we require to multiply total load in pounds by mean 
 diameter in inches, and divide by either constant 12,000 or 8,000, 
 as the case may be, and cube root of quotient equals diameter of 
 steel. 
 
 First find what w of formula represents, by multiplying area of 
 valve by pressure per square inch. 
 
 .*. w = 5'- X .7854= 19.635 sq. in. area. 
 19.635 X 60 = 1 1 78. 1 , total weight. 
 1178.1x5 = 5890.5. 
 
 Then -^^-^^ ^ = diameter tor square steel. 
 
 1200Q 
 
 and y 5 9 'P — diameter for round steel. 
 
 8000 _ -- — - — 
 
 n ' "-'■- ■ " '" " 
 
 5890.5+ 12, 000 = .49, and \/.49 = . 788 inches length of each side for 
 
 square steel. : ' 
 

 52 
 
 .788 X .788 = .62 square inches area of squa«-e steel. 
 
 5890. 5-^8000 = .7 36, and \/.736 = .9 inches diameter of round 
 steel. 
 
 .9- X .7854 — .63 square inches area of round steel. 
 
 For a spring constructed of square steel our dimensions then 
 
 become : 
 
 Mean diameter of spring (i.,e., from centre to centre 
 
 of steel) =5 inches 
 
 Outside diameter of spring equals 5 inches plus size 
 
 of steel =5.788inches 
 
 Inside diameter of spring equals 5 inches minus size 
 
 of steel =4.2i2inches 
 
 Size of steel = .788 inches 
 
 Area of steel must contain 62 sq. in. 
 
 And for a spring constructed of round steel, dimensions are as 
 
 follows, viz. : 
 
 Mean diameter of spring 5 inches 
 
 Outside diameter of spring is 5 in. f .9 =5-9 inches 
 
 Inside diameter of spring is 5 in. - .9 = 4- 1 inches 
 
 Diameter ot steel wire = .9 inches 
 
 Area of steel must contain 63 sq. in. 
 
 With a standard spring before us it is easy to determine the 
 required sectional area of any steel spring when fundamental 
 principles of this formula are understood. 
 
 As we are given the whole of the dimensions of a standard 
 spring made of spring steel, we can determine the sectional area 
 of a square spring by the following process : 
 
 As given w^eight is to required weight so is given sectional 
 area to required sectional area. 
 
 For example, let us compare our determination of sizes for a 
 square spring with a standard spring ,' our question then becomes : 
 
 As 600 : 1 178. 1 :: .25 : required area. 
 
 Then 1178.1 x. 25 = 294.5. - -- 
 
 294.5-^600 = . 49, sectional area of spring at a load of 
 1178.1 pounds. 
 
 "imw 
 
S3 
 
 \,''.49-v-'7854 = «7^^ ''^^""*6<^ s'^^ o^ steel to comply with 
 standard spring". All other dimensions of the spring will change 
 in same proportion. . 
 
 With spiral springs then there is to the practical operating engi- 
 neer the important question of determining the increase in pressure 
 by compression or the decrease in pressure by reducing the com- 
 pression, or the change similar and corresponding to the gradua- 
 tion of a lever safety valve. 
 
 W X d 
 The formula for this is—- — — • xn = total compression. 
 
 S* X G 
 
 Where W is the total weight pressing upwards against the 
 
 valve in pounds 
 
 d = mean diameter of spring 
 
 S is thickness of steel in sixteenths of an inch 
 
 G is constant 30 for square steel 
 
 G is constant 22.8 for round steel 
 
 n is number of coils in spring. 
 
 Example (3) : A spring loaded safety valve 5 inches in diameter 
 is set for a gauge pressure of 90 pounds, but owing to weakness 
 of boiler, pressure must be reduced to 60 pounds; the outer dia- 
 meter of spring is five inches and spring is made of ^ inch steel 
 with 15 coils; what compression must be given to produce re- 
 quired pressure starting with spring slack ? 
 W X d = ^ ^^ ^ 5'x-7^54x6ox4>li-'xi5 
 " 10^x30 
 
 4.932 inches, total com- 
 
 S xG 
 
 pression required. 
 
 From this formula may be deduced the number of coils required, 
 so that a given pressure shall require a given compression, or the 
 load on the valve with a given compression, the diameter of coil 
 or thickness of steel, if the other quantities are given. 
 
 The most important of these to the engineer is the determina- 
 tion of the number of coils required ; the change in pressure by a 
 given change in compression, and also the determination of the 
 
 'J' 
 <■ '■ 
 
r"' 
 
 ': •■• 
 
 iii 
 
 54 
 
 total weig"ht bearing down against the valve or the weight re- 
 quired to lift it off its seat when dimensions of spring and com- 
 pression are known. 
 
 To determine the number of coils required to balance a given 
 pressure with a given compression, we construct from above 
 formula the following : 
 
 s*xG 
 
 Where W, d", s*, G and N have same values as in last formula, 
 and C equals compression in inches, result will be required num- 
 ber of coils. 
 
 To determine pressure at which valve will blow oflf, with a spring 
 of given dimensions and given compression. 
 
 Formula : 
 
 s*xG 
 
 - X c = P, pressure at which safety valve will blow off. 
 
 a X d"' X N 
 
 a equals area of valve in inches. 
 
 To determine total weight holding the valve in place, with a 
 valve of given dimensions and given compression. 
 
 Formula : 
 
 xG 
 
 X c = W, total weight holding valve down. 
 
 d«xN 
 
 Reference has been made to every safety valve requiring a 
 given orifice or opening to allow of free passage of sieam, so that 
 increase in boiler pressure shall not take place. It is at the 
 same time just as important that area of valve should not be too 
 great to allow the free discharge of steam. 
 
 For steam above lo pounds pressure above the atmosphere, 
 the weight of steam that will escape into the atmosphere through 
 an opening one square inch in area is, in 70 seconds, just equal 
 to the pounds in the absolute pressure of the steam per sq. inch. 
 
 Example: To what height must a 5 inch safety valve rise from 
 its seat to allow steam to escape at the rate of 9,200 pounds per 
 
 --^aki 
 
■.^' 55 
 
 hour, if the pressure on the boiler is 75 pounds per sq. inch above 
 
 the atmosphere. 
 
 Since the weig^ht of steam that will escape per square inch in 
 
 70 seconds is equal to i^-aujje pressure plus atmospherical pressure, 
 
 we proceed to find the weight of steam escaping from one sq. in. 
 
 per hour. . 
 
 Gauge pressure, 75 
 Atmosphere, 15 
 
 90, absolute pressure. 
 Then as 70 : 60 :: 90 : to the weight of steam escaping per minute. 
 
 Then, 2 =4628.1^, pounds of steam per hour per sq. in. 
 
 70 
 
 Then as 4628.5 : 9,200 :: i sq. inch to required area, 
 
 _" = 1.98 square inches of escape required. , 
 
 4628.5 
 
 Then if required area is divided by circinnference of valve in 
 inches, result will be distance valve will be required to raise from 
 its seat to allow the escape of 9,200 pounds of steam into the 
 atmosphere. 
 
 3.1416 I5.7o8)i.98ooo(.i26 of an inch lift. 
 
 5 »57o8 
 
 1 5. 7080 = circumference. 
 
 126 
 
 8 
 
 41920 
 31416 
 
 105040 
 
 9424^ 
 10792 
 
 1.008 = Lift must.be y^ of an inch. 
 
 From this, weight of steam escaping into the atmosphere from 
 any orifice may be determined. 
 
 OHMS LAW. 
 
 As stated in our last, and following out the principles therein set 
 forth, the current in a conductor varies directly as the pressure or 
 
S6 •,,••. 
 
 potential at the terminals, and inversely as the resistance of the 
 conductor. From this then we j^^et the following formula : 
 
 (I) 
 
 C = 
 
 R 
 
 This is known as Ohm's law, and is in continual use in the study 
 of formu'.i underlying the principles of electrical enj^ineering". 
 
 In equation i we have formula for C, when both E and R are 
 known. It consequently follows that we require but a simple 
 transposition in the terms of our algebriiic equation to find any 
 of the quantities E, R or C when any two of them are known. 
 
 E 
 
 (2) 
 
 Thus R 
 
 which simply means that the resistance is equal to the electro- 
 motive force (E M F) divided by the current, and 
 
 (3) E = RC 
 
 meaning that the E M F i. equal to the resistance multiplied by 
 the current. 
 
 From these equations it will be readily seen that if any two of 
 the quantities E, R or C are given, the third may be found from 
 one of the three equations, and that they are all based on the 
 same law. 
 
 The energy in an electric circuit is equal to the pressure or 
 potential at the terminals of the conductor multiplied by the cur- 
 rent flowing through the conductor or circuit, and can be ex- 
 pressed in formula as follows : 
 
 (4) Worwatts = E + C 
 
 Here we have a simple formula for the purpose ot finding the 
 electrical energy of a circuit or generator when E and C are 
 known. ^ ^ ., , . 
 
 But it frequently occurs that we must find W when any two of 
 the units E, R or C are known. In equation (3) we have E = R C. 
 
57 
 
 With this information before us let us substitute this vaKie cf E in 
 
 (4) ; we then g-et 
 
 W (RC)xCor 
 
 C^ xR = watts 
 
 Again in equation (i) we have 
 
 Substitute this vahie of C in equation (4), and we get 
 
 E- 
 
 R 
 
 = watts 
 
 Following' out tlien the general principles laid down, we get 
 these formula for the determination of the electrical energy in any 
 circuit when any two of the units E, R or C are known, or in other 
 words the capacity of any circuit or dynamo to do work. 
 
 In our mechanical studies we establish the important fact that 
 the energy cannot be destroyed, and that it will occur either as 
 mechanical force or hept. or both. 
 
 And when an electrical current is passed through a wire or 
 conductor a certain amount of electrical energy is lost as such, 
 but makes its appearance as heat. Consequently the amount of 
 heat generated must be equal to the electrical energy lost, and 
 must therefore be measured in watts. 
 
 F^ormula : W = C- R is the formula generally used for the 
 computation of heat generated in a circuit. If the resistance of 
 a circuit and current flowing is known it is only necessary to 
 multiply the resistance of the circuit by the square of the current 
 to find electrical energy lost in transmission and appearing as 
 heat. 
 
 li 
 
 I 
 
 I 
 
il 
 
 STRENGTH OF BOILERS. 
 
 A STANDARH ht>iler, consltucted in accordance with the Canada 
 Steamboat Act, is assumed lo have a maximum workinji^ pressure 
 of one hundred pounds to the square inch ami be forty-two inches 
 in diameter, and, if made of best refined iron phite, shall be at 
 least one-quarter inch thick, made in the best manner. 
 
 If boiler is made of steel, a maximum working" pressure of one 
 hundred and twenty-five pounds to the square inch is allowable ; 
 diameter and thickness of plate as above. 
 
 The tensile strength of the material for iron is to be taken as 
 48,000 pounds per square incli of section with the grain, and 42,000 
 pounds per square inch across the grain, and for steel 60,000 
 pounds per square inch. And when boiler and all joints are con- 
 structed in best manner, four may be taken as a factor of safety. 
 
 From the foregoing standard it at once becomes apparent that 
 the required thickness of plate varies directly with the diameter of 
 the boiler, and the safe working pressure varies inversely with the 
 diameter. 
 
 From this we might construct the following formula to obtain 
 the safe working pressure of any boiler : 
 
 TSx2T 
 D X FS ~ 
 Where TS = tensile strength of material, 
 
 2T = twice thickness of plate in inches, 
 D = diameter of boiler in inches, 
 FS = factor of safety, 
 P = safe working pressure. 
 
 It becomes evident, however, that since plate must be some- 
 what weakened by having holes drilled or punched in it to receive 
 
59 
 
 IS 
 )0 
 DO 
 1- 
 
 It 
 3f 
 
 n 
 
 the rivets by which the plates are fastened tojjether, antl that the 
 rivets themselves must have a direct effect on the stren>4"th of the 
 seam, it is necessary first to determine the strength of the 
 punched plate .it the joint as compared with the solid plate, and 
 also the streng^lh of the rivets as compared with the solid plate. 
 
 The well-known axiom that the strength of a chain Is its weak- 
 est link is borne out here in a remarkable decree, and the weakest 
 part of a boiler is certainly the strength of that boiler. 
 
 Consequently, before we can determine the safe working pres- 
 sure of a boiler, its required diameter or required thickness of 
 plate, we must first determine strength of all rivetted seams. 
 
 It is self-evident that the strength of any section of plate must 
 be its width multiplied by its thickness, multiplied by the weight 
 required to break it. 
 
 For example, let us assume we have a piece of boiler plate one 
 inch in width and one-quarter inch thick, and that it broke when a 
 weight equal to 48,000 pounds per square inch of section had been 
 applied to it. We should require to exei t a force equal to 1" x .25" 
 X 48000 = 1 2000, which is the greatest possible strength we may 
 expect to get per sectional inch of this plate. 
 
 Suppose, now, we drill a hole in the centre of this plate, we clearly 
 reduce its sectional area and consequently its strength. Obviously, 
 then, both the pitch of the rivets and their diameter must betaken 
 into consideration in computing strength of plate at seams as com- 
 pared with the solid plate, and we might say that width of plate 
 in inches minus diameter of rivets, multiplied by number, and dif- 
 ference multiplied by thickness of plate in inches, multiplied by 
 tensile strength of plate per square inch of section, will equal 
 strength of plate at joint, or 
 
 p-(dxn)xTxTS = s; 
 and since we want to know what percentage of strength punched 
 
 i 
 
6o 
 
 m 
 
 plate bears to solid plate, we may modify this formula and get the 
 formula re uuired by Board ot Trade Rule : 
 
 — = percentage of strength of plate at joint as compared 
 
 with the solid plate. 
 
 F 
 
 We now have found the strength of plate when prepared for the 
 rivets, and must now consider the strength of the rivets employed 
 to fill up these holes, so that the operation of making the seam 
 may be completed. 
 
 As alreadv seen, the plate has been weakened by having had 
 holes made in it. We now prf>ceed to fill up these holes with 
 rivets, and ! need hardly point out ihat if the strength of the plate 
 is greater than the strength of the rivets, and the boiler loaded to 
 the strength of the plate, the rivets will give out b)' shearing 
 across. We endeavor to get as strong a joint as possible, and for 
 this purpose put in hs many rivets as practicable in a seam. Sup- 
 pose, however, we put the whole of the rivets in one row, we have 
 reduced the plate area, and tlie stronger our rivet section becomes 
 the weaker becomes the sectional area of the plate at the joint. 
 
 Therefore it is customary to divide the rivets into two, three or 
 more rows, as by doing th»s the same strength of rivets is re- 
 tained and the rivets are pitched a reasonable distance apart, en- 
 abling a fair percentage of strength to be obtained in the plate. 
 
 Then, if we knew the shearing strength of rivets per square 
 inch of section, we may say that d- x .7854 x Ss = shearing strength 
 of rivet when only one row of rivets is used. Then d- x .7854X 
 Ss X N = shearing strength of rivet when two or more rows are em- 
 ployed. 
 
 Where d- ^diameter squared, 
 .7854 = constant, 
 
 Ss = shearing strength of rivets, 
 N = number ol rows of rivets. 
 
6i 
 
 Then, 
 
 d- X .7854 X Ss X N X 100 
 p X T X TS 
 
 = percentag"e of strength of rivets as com- 
 pared with soHd phite. 
 
 Since both shearing and tensile strength of plate may be con- 
 sidered as equal, we may cancel these and proceed to get strength 
 of rivets by formula adopted by Board of Trade : 
 
 Where 
 
 a X X X 1 00 ^ ^ ., r ■ ... 
 
 — = percentage ot strength ot rivets as compared with 
 
 ^ ^ ^ solid plate. 
 
 Where a = area of rivets, 
 
 N = number of rows, 
 
 P = pitch in inches, 
 
 T = thickness of plate in inches. 
 
 From these two formulae, then, may be ascertained both the 
 strength of plate and rivets as compared with solid plate, and it 
 follows that the least of these percentages is the strength ot the 
 joint. 
 
 Note : When rivets are exposed to double shear, percentage of 
 strength obtained from foregoing formula may be multiplied by 
 
 I-75- 
 
 Example ( 1 ) : Find the strength of plate at the joint as compared 
 with solid plate if the rivets are j4 inch in diameter and pitched at 
 2}4 inches. 
 
 P_H -> r - c 
 
 80%, 
 
 P-d 2.5 -.5 
 
 X 100 = 
 
 X lOO: 
 
 P 2.5 
 
 strength of plate at joint as compared with solid plate. 
 
 Example (2) : Suppose that pilch and diameter of rivets in a 
 double-rivetted joint are same as in example No. i, and thickness 
 of plate equals half an inch, what will be the strength of rivets as 
 compared with solid plate ? 
 
 n 
 
 X 
 
 ci^ X .78f;4 X 2 . '\Qi 
 
 ,00=^ '—^ xioo = -^- 
 
 P T 2.5X.5 1.25 
 
 strength of rivets as compared with solid plate. 
 
 = .3144 X 100 = 31.44%, 
 
 hi 
 1: 
 
 1 
 
 1 
 
"1. 
 
 ■ 1 ! 
 
 As a rule it is understood that <' imeter of rivets may be 
 same thickness as plate, but with thin plates this rule will not 
 hold g'ood, as in the present case it is evident that strength of 
 rivets at joint is far too weak, and it would simply be absurd to 
 construct a joint on these proportions. To increase strength of 
 rivets we must either decrease pitch or increase diameter of 
 rivets. It is considered good practice to have the percentage of 
 strength at the joint or seam 70% of the strength of the solid plate. 
 We can therefore decrease the percentage of strength of the plate 
 by increasing the diameter of the rivet to 3^ of an inch, and at the 
 same time increase the strength of the rivets. 
 
 Then, 
 
 P-d 
 
 X 100 
 
 2.5 - -75 
 
 X 100 = 70%, 
 
 P 2.5 
 
 percentage of strength ot plate at seam as compared with solid 
 
 plate, and 
 
 .8834 
 
 n 
 
 X 100 = X 100 = 70%, 
 
 tor I /'J' 
 
 P T 1.2^ 
 
 strength of rivet as compared with solid plate. 
 
 It will be readily seen that the most economical joint is one in 
 which the plate and rivet sections are equal in strength. As al- 
 ready pointed out, if one section is stronger than the other it 
 creates a decided disadvantage, as the weakest part of a joint 
 must be its strength, and in a case like the foregoing we might 
 with advantage put half the difference between the strength of the 
 section on to the weakest section. 
 
 The easiest way to arriye at this, then, is to equate the formula 
 for the rivet section to that for the plate section : 
 
 P-d _ d- x.7854Xn 
 
 P PxT 
 
 We can now, by a simple transposition of formula, find a pitch 
 
 that v/ill give equal percentages : 
 
 -^ ax No. in one pitch , 
 P= +d. ^ 
 
^3 ' 
 
 Example : What pitch will the rivets of a double rivetted seam 
 have to be so as to secure an equal percentage of strength in 
 rivets and plates at the joints, shell plate being half inch and 
 diameter of rivets ^4 inch ? 
 
 d- +.7854X 2 
 Proof : 
 
 + •75^ 
 
 .8834 
 •5 
 
 2-50 --75 
 2.50 
 
 •75-' X -7^54x2 
 
 1 . 76 + . 75 = 2.50 = pitch. 
 
 X 100 = 70%. 
 X 100 = 70%. 
 
 2.5 X. 50 
 
 I will again repeat these two most important formuke, and 
 recommed the student to commit them to memory: 
 
 (Pitch minus diameter of rivets) X 100 
 __ 
 
 equals percentage of strength of plate at joint, as compared with 
 solid plate, and 
 
 (Area of rivets x No. of rows) x 100 
 Pitch X thickness of plate 
 equals perce.itage of strength of rivets, as compared with the 
 solid plate. 
 
 APPLICATION OF OHMS LAW. 
 
 An examination of the preceding article on Ohm's law estab- 
 lishes three important rules, viz.: 
 
 1. The current varies directly with the electromotive force or po- 
 
 tential, and inversely with the resistance. 
 
 2. The resistai e varies directly with the electromotive force, and 
 
 inversely with the current. 
 
 3. The electromotive force varies directly both with the current 
 
 and resistance. 
 
 P'or ,)ractical operating engineers, these rules, based on Ohm's 
 law, are the fundamental principles underlying most electrical 
 
 ! m 
 
'^iT- 
 
 "64 '■ -; 
 
 calculations. It is important that the principles be thoroughly 
 understood, and I regret that, in a work of this kind, full details 
 cannot be given, for want of space. Before proceeding to an 
 exposition of these principles by mathematical problems, I wish 
 particularly to point out to my readers the desirability of their 
 acquiring literature dealing with fullest details of Ohm's law. 
 
 Bearing in mind the fact that these articles are written especi- 
 ally for engineers operating electiic plants, I shall content myself 
 with giving principles and formula especially adapted to their 
 requirements. 
 
 It may be well to mention that whatever is included in a circuit 
 forms a portion of it. Be it the generator itself, converters, 
 meters, or any apparatus in connection with the generation or 
 transmission of an electric current, and the resistance of both 
 line and apparatus, must necessarily be included in resistance of 
 circuit. 
 
 The resistance of a generator is nearly alwa^•s referred to as 
 internal resistance, and that of the outer circuit or line as external 
 resistance, to distinguish between them, and the two resistances 
 added together from the total resistance of the circuit, or the R of 
 the formula. 
 
 Example ( 1 ) : An electric generator having an internal resist- 
 ance of 5 ohms and an E.M.F. of 50 volts, sends a current 
 through a line of copper wire whose resistance is 25 ohms; what 
 is the current ? 
 
 C = -^-, then 5 + 25 = 30 ohms total R, and 
 
 R 
 
 5° 
 30 
 
 = 1.66 amperes. 
 
 (Rule I.) 
 
 Example (2) : A difference of potential (E.M.F.) of no volts is 
 maintained in an electric circuit, and a current of 250 amperes is 
 the result. What must be the resistance of the line? 
 
 R = -^, then ^|g = .44 ohm. 
 
 (Rule 2.) 
 
 MHHMMMMR*' 
 
65 
 
 Example (3) : A generator having- an internal resistance of i 
 ohm sends a current of 50 amperes through a circuit having an 
 external resistance of 2.5 ohms ? What is the electromotive force 
 of the generator ? 
 
 E = RC, then i. + 2.5 = 3.5 ohms total K, 
 ••• 3-5x50= '75 volts. 
 
 PORTIONS OF CIRCiriTS. 
 
 All portions of a single circuit must of necessity receive 
 the same current, but the electromotive force, or what is usually 
 styled difference of potential, or drop in potential, and resistance, 
 may vary to any extent in diflferent sections of the circuit. 
 
 Example (4) : A generator maintains a constant E M V of 1 10 
 volts between its terminals. The terminals are connected to and 
 current is passed through a series of four coils, one having a re- 
 sistance of 50 ohms, one 25 ohms, one 12.5 ohms, and one 6.25 
 Paying no attention to the resistance ot' the conductors between 
 these coils, what is the E M F between the terminals of each coil? 
 
 A solution of this problem can best be reached by application 
 of principle laid down in rule 3, viz.: That the electromotive force 
 varies directly with the resistance and with the current. 
 
 In this case we wish to find E M V at given points on the line, 
 when R alone of coil is known. The total R of the four coils is 
 93.75 ohms ; calling the coils r, 2, 3, and 4, and the difference of 
 potential at their terminals E% E-, E", E*, we get the proportion. 
 
 As t^o E* 
 
 FJ 
 E« 
 E* 
 
 93-75 
 
 50 
 
 25 .. 
 12.5 •• 
 
 6.25 
 
 10 volts : 
 
 Then working out the problem by regular rules of proportion, 
 
 we get 
 
 EM F of E' =58.7 volts. 
 EM F of E^=29.3 >, 
 EM F of E^ = i4.7 „ 
 EM F of E*= 7.3 ,, 
 
 A.n examination of the problem will also prove to the student 
 
 
 I H 
 
 «-:H 
 
 ■i 
 
66 
 
 the theory of the statement that all porlions of a sing^le circuit 
 must receive the same current. Taking", for exaniple, total resis- 
 tance of line at 93.75 ohms, and E M F at terminals of generator 
 at 1 10 volts, and applying rule No. 1, we get 
 
 E 110 
 
 or — =1,17 amperes. 
 
 C = 
 
 R 
 
 93-75 
 
 Applying the same principles to the coils, E% E-, E"' and E"*, we 
 find that current .it terminals of each is the same. 
 
 ICxample (5) : Suppose the same external circuit was connected 
 to a generator having a resistance of 10 ohms. The E M F of the 
 50 ohm coil has been found to be 60 volts, what is the E M F at 
 the termina!;-. of the generator, and what would be the E M F of 
 the generator on open circuit. 
 
 The total R of our coils has been found to be 93.75 ohms, and 
 by rule 3 we demonstrate that, as the resistance of the coil is to 
 the total R of the circuit, so is the E M F at the terminals of the 
 coil to the E M F at the terminals of the generator. 
 
 Or, as 50 : 93.75 :: 60 : x, 
 
 and .*. X— 112.5 E M F at terminals of generator. 
 
 By following rule i, we find the current at the terminals of coil 
 
 No. 1 to be 
 
 f 8^= 1.2 amperes. . 
 
 The total resistance of the circuit is the internal resistance of 
 the g<jnerator= 10 ohms; the resistance of the conductors, of 
 which no account has been taken, and the resistance of the four 
 coils = 93. 75 ohms. 
 
 10 + 93.75=103.75 ohms, total R of line. 
 
 Then to find required E M F of generator to maintain a current 
 
 of 1.2 amperes through a resistance of 103.75 ohms, we must 
 
 apply rule 2. . 
 
 _^ E = RC, 
 
 or 1.2 X 103.75= 124.5 volts, required E M F of generator. 
 
 •'ii' • I'l 
 
 "inrg a B ai : 
 
67 
 
 STRENGTH OF BOILERS. 
 
 Having determined the streiis'th of the boiler at the joints, we 
 next require to detertnine safe working pressure, thickness of 
 plate, tensile strength per sectional inch, etc. 
 
 A standard boiler is said to be 42 inches in diameter and have a 
 safe working pressure of 100 pom Js per square inch, if con- 
 structed in best manner of iron plate one-quarter inch thick, hav- 
 ing a tensile strength of 48,000 pounds per sq. inch of section. 
 
 From this, then, we can construct the following formula to deter- 
 mine the safe working pressure of any boiler : 
 
 TS X 2T X % strength of joint 
 
 DxFS '^^'' 
 
 and from this we require but a slight transposition to construct a 
 formula to determine either the diameter or thickness of plate re- 
 quired to conform to this standard, and 
 
 D X P X FS 
 
 TS X 7 
 
 TS X 2T" X % 
 
 =:2T-r 2 = T, and 
 D. 
 
 PxFS 
 
 This formula refers to any boiler, no matter whether constructed 
 of iron or steel, 48,000 pounds being taken as the tensile strength 
 for iron, and 60,000 pounds being taken as the tensile strength for 
 steel, and four being taken as a factor of safety in each case 
 when boiler is constructed in best manner. 
 
 Example (i) : Find the safe working pressure of an iron boiler, 
 made in best manner, joints having a sectional strength equiva- 
 lent to 70°/. of solid plate ; boiler being 42 inches in diameter and 
 having plate ^ of an inch thick. 
 
 TS X 2T X % 
 
 pre 
 
 DxFS 
 
 ssure. 
 
 = P: 
 
 48,000 X .75 X .70 
 
 42x4 
 
 = 150 pounds, safe working 
 
68 
 
 Same boiler in steel would become 
 
 ■»ii 
 
 vlr!) 
 
 III! 
 
 60,000 >; .75 X .70 
 
 = 187.5 jiouiids per square inch safe working" 
 
 42 X 4 
 pressure for steel. 
 
 Example (2) : Find the required thickness of plate (steel) re- 
 quired in a boiler five feet in diameter, to carry a safe working 
 pressure of 100 pounds to the square inch, sectional strength of a 
 triple rivetted joint being 7o7o of strength of solid pla'e. 
 
 D X P X FS 
 
 TS X 7, 
 60" X 100 X 4 
 
 60,000 X .70 
 
 = 2T, 
 = 2T- 
 
 >4,ooo 
 
 = .571 ~ 2 = .285 inches, or nearly j^ inch, required thickness 
 
 42,000 
 of plate. 
 
 Example : What diameter should a boiler be when constructed 
 of iron made in best manner and with }^2 inch plates, working 
 pressure to be 200 pounds per square inch, joints and rivet sec- 
 tions having a tensile strength equal to .70% of strength of solid 
 plate, 
 
 T S X 2T X % 48.000 X I. X . 70 _ 33.600 _ _ .^ ,_ 
 
 — — — =u = —— — 42 mcnes. 
 
 F X FS 200 X 4 800 
 
 Same boiler constructed in steel : 
 
 60,000 X I. X .70 42,000 
 
 = <;2.5 inches. 
 200 X 4 800 
 
 Example : F'ind the strain per sectional inch on a boiler 42 
 inches in diameter, having )4 •"^'^ plate and having a working 
 pressure of 100 pounds per square inch. 
 
 Formula : 
 
 1? X D 100x42 
 
 TS^ 
 plate. 
 
 2T 
 
 = 8400 pounds strain per sectional inch of 
 
 ■Mi 
 
69 
 
 STEEL FIRNACES AND FLIES. 
 
 The Canada Steamboat Act provides that the external workiiij^ 
 pressure to be allowed on plane circular steel furnaces and flues 
 when subjected to such pressure, when the long"itudinal joints are 
 welded or made with a butt strap, shall be determined by the 
 following formula: 90,000 x the square of thickness of plate in 
 inches ^(leng^th in feet+ i) x diameter in inches equals the working- 
 pressure per square inch, provided it does not exceed that fovuid. 
 by the following formula : 
 
 8,000 X T" 
 
 W' 
 
 T" thickness in inches. 
 
 D" diameter in inches. 
 
 The length to be measured between the rings, if the furnace is 
 made with rings. 
 
 Example : Find the working pressure of a circular flue 36 inches 
 in diameter, 6 feet long and mide of ^ inch steel plate. 
 90,000 X ^ - _ go, 000 X . 1406 _ 1 2654 
 (6+0x36 7x36 252 
 
 50 pounds working pressure 
 
 Check : ^'°°° x T'^ 8, 000 x .375 
 
 = 83.3 lbs. 
 
 D" 36 
 
 The collapsing pressure of a plane circular furnace tube is 
 found by the following formula : 
 
 806,300 X T'^ 
 
 dTl ^ 
 
 When T equals thickness of plate in inches. 
 D equals diameter of flue in inches. 
 L equals length of flue in feet. 
 
 Example: What is the collapsing pressure of a furnace tube 
 
 whose diameter is 36 inches, length 6 feet and thickness of jplate 
 
 ^^ inch ? 
 
 806,300 X .i4o6_ 1 13,365.78 
 
 36 X 6 216 
 
 • ■ J'J".■^•/"_ .,,0, 
 -> 5 24 . 84. 
 
 p 
 
70 
 
 CORRUGATED STKKL M KNACKS AND FLUES. 
 
 Steel flue furnaces, when new, corrug"ated and niachino made, 
 and practically true circles, the workinij' pressure is found by 
 the followinwf formula, provitled that the plane parts at the ends 
 do not exceed six inches in length, and the plates are not less 
 than i'c of an inch thick and furnace made in one lenj^th. 
 12,500 X T" 
 
 = working pressure. 
 
 And for corrugated iron furnaces made similarly the following 
 formula can be used : 
 
 10,000 X T" 
 
 \r 
 
 working pressure. 
 
 Example : Find the working pressure allowable on a corru- 
 gated steel furnace flue 42 inches wide, 7 feet long and -V^ thick- 
 ness of plate, 
 
 1 2,500 X. 375 
 
 4^ 
 
 = 1 1 1.6 pounds. 
 
 Find the working pressure allowable on same furnace con- 
 structed in iron. 
 
 42 
 
 = 89.3 pounds. 
 
DIVIDED AND SHUNT CIRCUITS. 
 A conductor from one terminal of a g-onorator may be tliv'uleil 
 into any number of divisions or branches. Each br.'inch or 
 division may vary in resistance, and ihey may all unite on reach- 
 ing^ the other terminal. 
 
 E' 
 
 C = 50/? mba 
 
 4 
 
 Or 
 
 /^ jj <> a) 
 
 
 II 
 
 Fu;. 2. 
 
 R 
 
 Example 6 : A portion of the circuit connected with generator 
 G in Fig. 2 consists of 4 parallel conductors, A, B, C .'ind D 
 ofA=ioo ohms; 6 = 75 ohms; C = 50 ohms, and = 25 ohms, 
 what will be the ratio of current passing through the circuit which 
 will pass through each conductor. 
 
 According to Rule i the current passing through each of the con- 
 ductors A, B, C and D will be inversely proportionate to its resist- 
 ance. 
 
 .*. As A 
 
 Or to simplify : 
 
 B : C : D 
 
 As A 
 B 
 
 C 
 
 D 
 D 
 U 
 
 1 
 
 100 
 
 75 
 50 
 
 1 
 75 
 
 fXf 
 
 1 
 5-5 
 
w 
 
 7^ 
 
 That is, that the ratio which current passinjf throui^h conductor A 
 be.'irs to current passing throujfh conductor I) is the same us the 
 ratio tfiat the resistance of conductor A bears to R of conductor 
 1), am' so on fov all the other conductors. 
 
 Example 7 : Supposing' the current passing" through terminals 
 of generator, and consei|uently through circuit, to be 50 amperes, 
 what amount of current will pass through each of the branches, 
 A, H, C and D, i»i foregoing problem. 
 
 To determine the amount of current that will pass through 
 parallel circuits of different resistances, take the resistance of 
 each branch in ohms as a denominator, having i as a numerator. 
 Reduce the fractions to a common denominator and add together 
 the new numerators. Take the sum of the numerators as a new 
 common denominator, and the single numerators as new numera- 
 tors, and new fractions will express prop>rtional currents piissing 
 through portions of circuit as units of one. 
 
 3.4.6. 12 
 
 Then 1 hxi A ?V jV 
 
 300 
 
 And sum of numerators equals 25, using this as per rule 
 
 3.4.6. 12 
 
 ^5 
 
 1 1 
 
 .'. current passing through branches expressed in units of one 
 must equ.'il : A = tj'V ; B = tj*^ ; C = .//j ; D = ^f . 
 
 Total current passing through circuit is given as 50 amperes. 
 
 Then current passing through A = /7f X5o = 6 amp. 
 
 B = /-X50=:8 amp. 
 C = Ax5o=i2 amp. 
 r) = H X 50 = 24 amp. 
 
 It will be observed in Fig. 2 that current passing through cir- 
 cuit first passes through conductor E', and when parallel branches 
 are reached four paths are provided for its passage, when it 
 again unites on conductor E* '. 
 
 It is quite evident that the more paths there are provided for 
 the passage of the current the less resistance will be offered. 
 
 
 
 
73 
 
 This brinjj^s up a problem as to resistance offered by llie parallel 
 branches of a circuit as compared with a sinj^'le conductcr. 
 
 It is tjiiite clear that we cannot summari/.e ov jidd the resist- 
 ances of each parallel branch together. If branch A had only 
 been provided, then resistance of A would have been the resistance 
 of that portion of the line, but addition of other branches clearly 
 reduces resistance of eacli branch to flow of current by providing' 
 another path for its passage. 
 
 In dividetl circuits the resistance of the combined branches is 
 expressed by the reciprocal of ihe sum of the reciprocals of the 
 resistances. 
 
 Example S : l^'ind the resistance of the combinetl branches, A, 
 B, C and P, as set forth in foregoing- examples. 
 
 The reciprocal of resist;ince is conductance expressed usually 
 as Mhos, a derivate of the word Ohms. 
 
 The reciprocal or conductance of the four branches then is : 
 
 I iff + 7'.> + s*« ■»- i'.'. = nVrr or fi'g Mhos. 
 The resistance of the combined branches is as 5 : 60— 12 ohms. 
 Application o( Rule 3 to this problem proves that the E M F on 
 each of branches is equy.1 600 volls. 
 
 R of A=: 100 ohms and 0=6 amps. E M F 600 volts. 
 R of B= 75 " " = 8 " ♦' " 600 
 R of C= 50 " •' =-12 " " " 600 
 R of D= 25 " ♦' =24 " " " 600 
 
 And combined resistance of A, B, C, D—12 ohms combined. C 
 = 50 amps. E M V 600 volts. 
 
 Perhaps a clearer method of arriving^ at the combined resistance 
 of parallel circuits is to use the formula 
 
 Rx R' 
 
 < ( 
 
 k 1 
 
 R 
 
 R + R 
 
 Wliich is based on the principle that the combined resistance of 
 two parallel circuits can be found by multiplyint^ their resistances 
 tog^ether and dividing the product by the sum of their resistance. 
 
74 
 
 By ihis meihod, combining A and B we tjet "— ''' = 42.8 ohms. 
 
 ^ » & f> 100475 
 
 Combining' C and D we get ^^ — = 16.7 ohms. 
 
 50+25 
 
 r.,, .... , ^ ^ 42. <S X 16. 7 , 
 
 1 hen combining- these two products we g"et ~ '-= 12 ohms 
 
 .f2.8+ 16.7 
 combined R of branches A. 1?, C and D. 
 
 If the whole of the branch circuits were of uniform resistance, 
 then formula could be very much simplified by using- formula 
 
 R 
 
 R 
 
 N 
 
 When llie resistance of one branch is divided by the number of 
 branches the quotient will be the combined resistance of all the 
 branches. 
 
 Example 9 : Wh.'it would be the combined resistance of four 
 parallel branches each having- a resistance of 100 ohms? 
 
 joo _2- ohms. 
 
 STRENGTH OF STAYS AXD FLAT SURFACES. 
 
 We require here to find area and number of stays required, 
 area of surface supported by one sta\', and the pressure that may 
 be allowed ag-ainst any flat surface. 
 
 The g-reatest stress per square inch of section of an iron stay 
 
 allowed is 6000 lbs. Then calling- this the T. S. (tensile streng-th) 
 
 of our formula, the total stress on a stay may be found by the 
 
 following formula : 
 
 Stress (S) = axT S 
 
 Example : Find the stress allowable on a direct stay i)-{ inches 
 in diameter when the stress allowable equals 6000 pounds per 
 square inch section. 
 
 S = a X T S= 1.25- X .7854= 1. 227 1 square inches of section. 
 1. 2271 X 6000 = 7362.6 total stress allowable on stay. 
 Example, the stays of a boiler are i}:(" in diameter and placed 
 12 inches apart, what should be the workings pressure of the 
 
 mm~- 
 
' 
 
 75 
 
 boiler if the stress allowable per square inch section of stay is 
 6000 poiuids. 
 
 Rule : First find the stress the stay is capable of supportinj^, then 
 divide the result by the area held up by each stay and the quotient 
 with the pressure allowable on the boiler. 
 
 1.25- X .7854= 1. 2271 square inches section of stay. 
 1.2271 X 6000 = 7362.6 total stress allowable on stay. 
 
 Since the stays f*re 12 inches apart or 12 inches from cenire to 
 centre of stay, each stay must support v section of plate equal to 
 12 inches sqj.iared. 
 
 Then 12 x 12= IA4 square inche-j area, supported by each stay. 
 Then 7362 -^ 144 = 51.1 pounds pressure allowable on the boiler. 
 
 Rule to find reqj/ed diameter of stay : Area of section in square 
 inches held up by each slay multiplied by pressure on boiler 
 divided by stress allowable per sectional inch on stay equals re- 
 quired area of stay. 
 
 2 
 
 \'area-^ .7854 — diameter. 
 
 Example : Find the required diameter of a direct stay when 
 allowable stress on stay is 6000 pounds per square inch of section 
 and Slavs are 15 inches apart with boiler pressure at 75 pounds 
 per square inch. 
 
 15 X 15:=: 225 square inches of surface supported by stay. 
 
 225 x 75= 16,875 pounds stress on each stay. 
 16,875 -^ 6000 = 2.81 ■"^quai'e inches required area of stay. 
 
 \/2.8i -=-.7854= 1.9 diameter of stay required. 
 
 Example : the stays of a boiler are i j^ inches in diameter and 
 capable of sustaining- a stress of 8,835 pounds, what distance must 
 they be pitched apart to sustain a working' pressure of 60 pounds 
 to the square inch ? 
 
 Rule : '^Stj ' =area of section, and square root of area ol' section 
 equals pitch or distance from centre to centre of slays. 
 
 8835-^60= 147.2 area of section to be supported in square inches. 
 
■fft ■ 
 
 76 
 
 \/i47= 12.125 length of each side of section of plate, or required 
 pitch of stay = 12^/i inches. 
 
 Proof: 12. 125- X 60 = 8,820. 
 
 [Note. — This result would have been exact had fractional parts 
 of an inch been worked out.] 
 
 We now have rules for three of the most important determina- 
 tions relating- to stays : 
 
 1st. Determining the total weight or stress that a stay is cap- 
 able of supporting. 
 
 2nd. Determining the required diameter of a stay when total 
 vvt'ight to be sustained is known. 
 
 3rd. Determining the area of surface a stay of given din»en- 
 sions is capable of supporting under known conditions. 
 
 To practical operating engineers these questions are of great 
 importance; I will therefore repeat the examples under conditions 
 which may be expected to be met with in ordinary practice. 
 
 Exaniple : The stays on a flat surface are 12 inches apart centre 
 to centre, and through corrosion, have eaten pvay to 1 inch in 
 diameter. T!ie pressuie per square inch on the boiler is 60 
 pounds, what is the strain per square inch section of stay and to 
 what pressure should the strain be r _duced to give a stress of 
 5,000 pounds per square inch section of stay. 
 
 I 2 X 12= 144 square inches held up by each stay. 
 144x60 = 8640. pounds held up by each staj-, 
 r-' X .7854 = .7854 square inches sectional area of slay. 
 Then stress per square inch section of stay equals : 
 
 8640 4- . 7854 =11. 000 
 
 This is clearly sufficient to brerk the stay and render *he boiler 
 unsafe. 
 
 5000 X .7854 = 3927 pounds weight that stay is capable of sus- 
 taining. 
 
 3027 -f 144 = 27.27 pounds saf- working pressure. 
 
 And therefore to comply with conditions named that strain on 
 stay must not exceed 5000 pounds per square inch of section. 
 
77 
 
 > 
 
 boiler pressure must not be allowed to exceed 27.27 pounds per 
 square inch. 
 
 Example : The stays of a boiler are 10 inches apart, and one of 
 them bteaks, throwing- more stress on the four surrounding- stays. 
 The stays are i^s mch in diameter, and the boiler pressure is 50 
 pounds per square inch, what extra stress will be placed per 
 square inch of section of stays if area supported by each has been 
 increased by 'z^, and what must the boiler pressure be reduced to 
 so that stress per square inch section of stay shall not exceed 
 original stress. 
 
 Stress held by each stay equals — 
 
 I o X 10= 1 00 X 50 = 5000 pounds. 
 
 Sectional area of stay equals— ^ 
 
 1. 125X 1.125 X .7854 = . 994 square inches, stress per square 
 Inch of section before the break = 5000-^.994 = 5030 pounds. 
 
 Then stress pe- square inch of section affer the break = 
 
 5030 \- ^3 of 5030, or 5030 i 1676.66 = 6706.66 pounds. 
 
 Stays before break supported 10x10=100 square inches of 
 surface, and after the break they are called upon to support J/j 
 more, or 133.33 square inches. 
 
 Origiii.'ll stt-bss per square inch section of stay = 5030 -^ 133.33 = 
 37.7 pounds per square inch, to which point pressure must be re- 
 duced so that original stress on stays may be maintained. 
 
 Example : What Is the total pressure on the flat bottom of a 
 boiler 18 feet 6 inches long by 6 feet wide, Uie pressure of the 
 steam being 30 pounds per square inch, and the depth of water 10 
 feet 5 inches. Find also the required number of stays i|^ inches 
 in diameter, and what diameter they must be pitched apart when 
 the stress per square inch of section of stay shall not exceed 
 6000 pounds. 
 
 Take a column of water 1 foot high and i square inch in area 
 as weighing .434 pounds. 
 
 10 feet 5 inches equals 10.416 feet. 
 
 ifiil 
 
 if 
 

 78 
 
 .'. 10. 416X .434 = 4.51 pounds pressure per square inch through 
 weight of water. 
 
 Then 4.51 -+-30=: 34.51 total pressure per square inch on bottom 
 of boiler. The boiler is i8' 6" by 6'= 18.5 x6= 1 13 sq. feet area. 
 
 1 13 X 144= 16272 area of boiler in square inches. 
 16272x34.51 =: 561, 546. 72 pounds total weight bottom of boiler is 
 called upon to support. 
 
 Area of stay equals 1.5- x .7854= 1.767 square inches. 
 
 1.767x6000 - 10602 pounds weight that one stay is capable of 
 holding up. 
 
 561,546.72 -^ 10602 = 53 nearly number of stays required. 
 
 Or we might say since each stay is capable of supporting 
 10602 pounds, 10602 — 34.51 =307 square inches area of plate each 
 stay can support. 
 
 16272 total area of boiler -f 30753 number of stays required. 
 
 ^'307= 17.5214 inches distance from centre to centre of stay. 
 
 
■!^- 
 
 DROP IN POTENTIAL AND SIZE OF LEADS FOR 
 
 MULTIPLE ARC CONNECTIONS. 
 
 Subsidiary leads or branch leads are taken from larg^er sized 
 
 mains of constant E.M.F., or from terminals of generators with 
 
 constant E.aM.F. to supply current to one or more lamps, motors 
 
 or otlicr appliances requiring a constant K.M.F. 
 
 There is a drop of potential in the leads to be provided for so 
 that the appliances may have to work at a reduced E.M.P\ The 
 E.M.F. of the leads is known, and the required E.M.l*'. of the 
 appliance to be supplied with current, also its resistance, and a 
 rule is required to calculate the size of wire in the lead to secure 
 proper results. 
 
 The resistance of the leads supplying any lamp or appli.mce for 
 a desired drop within the leads is equal to the rerriprocal of the 
 current of the lamp or appliance multiplied by the desired drop in 
 potential expressed as volts. This principle is based on the fact 
 that the drop in potential in portions of circuits varies with the 
 resistance. 
 
 Example to : A i lo-volt lamp having a resistance of 220 ohms 
 is to be placed 50 feet distant from a main having a constant 
 
 E.M.F. of 11:; volts. What 
 
 mu 
 
 st be the resistance of the line to 
 
 maintain a constant E.M.F. of 1 10 volts at the lamp 
 The lamp current is found by the formula : 
 
 C = 
 
 R 
 
 1 10 
 220 
 
 = .5, or I2 ampere 
 
 The reciprocal of the current then is 
 
 This multi[)lied by 
 
 desired drop 5 volts = f x 5= 10 ohms, required resistance of leads. 
 For two or more lamps in parallel similar rules are applied. 
 In this case the E.M.F. of the terminals or main leads, the fac- 
 
 1;' 
 
 I 
 
 t 
 
 
't I 
 ; I 
 
 80 
 
 tors of the lamps, and their distance from the point of connection 
 witli the main leads require, to be given. 
 
 Kxample 1 1 : A pair of subsidiary leads ; e to be run a distance 
 of 150 feet from main leads, and to carry current for ten 50-volt 
 lamps having a resistance of 100 ohms. The main leads have a 
 constant E.M.P'. of 55 volts. What is the resistance required in 
 subsidiary leads, and what size of wire B & S gauge ? 
 
 The combined R of lamps is found by formula : 
 
 R 
 
 Then \'\f = 10 ohms, combined resistance of 10 - 100 ohm lamps in 
 parallel, ■?^=5 amperes current to be supplied to lamps. The re- 
 ciprocal of 5= ix 5= i ohm resistance required by line. Length 
 of lead 150 ft. X 2 = 300 ft., total length of line. Required resist- 
 ance I ohm -^ 300 = .0033 ohms per foot, equivalent to No. 14 gauge 
 wire .IS per table. 
 
 Or this may be worked out on the principle that the resistance 
 of the leads is equivalent tp 'le combined resistance of the lamps 
 multiplied by the percentage of drop and divided by 100 minus the 
 percentage of drop ; whic!. we can express in formula thus, using 
 X as representing the required quantity : 
 
 R X %" 
 ^^-100-% 
 
 The percentage of drop Is found by dividing required drop in 
 
 E.M.F. by initial E.M.F. on mains, and ^\=g%. The ret|Mlfed 
 
 resistance of the leads then is -^15'= combined R of lamps* 
 
 10x9 90 
 
 = =1 ohm, required R of leads. 
 
 100-9 90 
 
 sTREN/rfH or noiuuiB. 
 
 The Steamboat Inspection Ail, 1882, ss. 4, provides "that the 
 areas of diagonal stays are found in the following manner : " 
 
 Find the area of a direct stay needed to support the surface, 
 multiply this area by the length of the diagonal stay, anvl divide 
 
8i 
 
 the product by the leng-th of a line drawn at right ang-les to the 
 surface supported to the end of the diagonal stay ; the quotient 
 will be the area of the diagonal stay required. 
 
 Example : F^ind the area required in a diagonal stay supporting 
 I square foot of surface, boiler pressure being 75 pounds per 
 square inch, length of diagonal stay 12 feet and length of line 9 
 feet, stress allowable per square inch of section on direct stay 
 being 6,000 pounds. 
 
 12 
 
 12 
 
 144 square inches surface supported. 
 75 
 
 720 
 1008 
 
 10800 total stress on direct stay. 
 
 1 0,800 -r 6,000= 1.7 1 square inches, required area of direct stay. 
 
 1.7 1 area of direct stay. 
 
 1 2 length of diagx>oa.l stay. 
 
 20.52 = 20.52 -^^ 9 length of line = 2. 27 square 
 inches, required area of diagonal stay. 
 
 When the tops of combustion boxes or other parts of a boiler 
 are supported bv solid rectangular girders, thi following formula 
 may be used foi the purpose of finding the working pressuie to 
 be allowed on the girders, assuming that they are not subjevted 
 to a temperature greater than that oi' the steam and are supported 
 by hajiging stays. 
 
 Favrmt.^ _ 
 
 C X d'^ X T 
 (VV-P) i TxT""^^^ ^^ pressure. 
 
 Where W = width of combustion box in inches, 
 P^=pitch of supporting bohs in inches, 
 I>= distance between the girders from centri* \o centie in 
 
 inches, 
 L = lenglh of girder in feel, 
 
 M 
 
 •^ if 
 5 I 
 
S2 
 
 1^ 
 
 1,;! 
 
 d = depth of girder iti inches, 
 
 T = thickness of girder in inches, 
 
 C = 5oo when girder is fitted with one supporting bolt, 
 
 C=/5o when fitted with two or tliree, 
 
 C = 85o when filled with font. 
 
 The working pressure for tiie supporting bohs and plate be- 
 tween them is determined by rules for ordinary stays already re- 
 ferred to. 
 
 The pressure allowed on plates forming flat surfaces is found by 
 the following formula : 
 
 Cx(T+i)- 
 
 S-6 
 
 = working pressure per square inch. 
 
 Where T — thickness of plate in sixteenths of an inch, 
 S = surface supported in square inches, 
 
 C=ioo; but when the plates are exposed to the impact 
 of heat or flame, and steam only is in contact with the 
 plates on the opposite side, C must be reduced to 50. 
 
 Example : Find safe working pressure of a flat-bottomed boiler 
 whose stays are pitched 15 inches apart and thickness of plate is 
 '4 inch. 
 
 C + (T+i)- ioox(8-fr-) Sioo 
 
 = 37 lbs. safe working [pressure. 
 
 S-6 225 - 6 219 
 
 To finti required thickness of plate we must reconstruct the 
 
 formula : 
 
 S - 6 X P 
 
 = (Tm)'^ and VT+i=T. 
 
 Example: Find the required thickness of plate for a flat-bot- 
 tomed boiler, vvliose stays are pitched 12 inches apart and steam 
 
 is per s(i 
 
 prei 
 
 50 poi 
 
 q' 
 
 ?ptl 
 
 boiler is 7 feet. Also, what must be the diameter of slays if they 
 are not allowed to carry more than 6,000 pounds per square inch 
 of section ? 
 
83 
 
 Pressure due to water = .433 X 7= 3.031 lbs. 
 Pressure due to steam = 50. 
 
 Then 
 
 Total pressure per square inch — 53. lbs. 
 
 S-6xP (144-6) X 53 7314 
 
 ^73-'4- 
 
 C 100 " JOG 
 
 And \/73' '4~^'55' '*^'55 ~ ' — 7'55 *^'* ;\|» required thickn».'ss of 
 plate'in sixteenths or thirty-seconds of an incti. 
 
 Stays are pitched 12 inches apart, and therefore must support a 
 surface of 12 X 12= 144 square inches. 
 
 144 X 53 = 7.632, total woiirht each stay is called upon to support. 
 
 7632-^6000 stress allowed per square inch of section of stay = 
 1.272 square inches, required area of stay. 
 
 '\^'i.272-f .7854= 1.25 (nearly), required diameter of stay. 
 
 It sometimes occurs that the stays of a boiler are to be fixed 
 with cottars, and it is necessary to know what size the end of the 
 stay must be swelled to so as to have the same streng-th as the 
 stay. For this purpose the following formula is used : 
 
 .08 .4 X 
 
 Where d = diameter of stay in inches. 
 D = increased diameter. 
 
 N=:depth of cottar in terms of its width ; that is, if cottar 
 is Yz an inch wide and 1J2 itiches deep, X would be 3, since 
 depth of cottar is 3 times the width. 
 
 Example : The stays of a boiler are to be fixed with cottars ^ 
 an inch tliick and 2 itn"hes thick, the diameter of stay being- i^ 
 inches, what size must the end be made so as to have uniform 
 strength with stay? 
 
 D— (1+^ — + -~)xi.5=:(l 1 .02 + .2) X 1.5= 1.83 required diam- 
 \ 4 V4/ 
 
 eter of end of stay. 
 
 Example 12 : In P'ig-. 3 we have represented a pair of mains on 
 
 -( 
 
 1 -i — + 
 
 N 
 
 i 
 
 r"' 
 
■I 
 
 which we have to connect four groups of lamps, consisting of five 
 lamps each connected in multiple. The lamps have each a resist- 
 ance of 200 ohms. The E.M.F. of the terminals to which leads 
 are connected is iio volts, and it is desired to allow a drop in 
 potential of 10 volts, to be divided equally between each of the 
 four groups of lamps. What must be the resistances of the four 
 sections of wire l-* 
 
 
 
 R='3ZS 
 
 % 3 
 
 5 
 
 9 
 
 -jy^ 
 
 8 
 
 S 
 
 Fig. 3. 
 
 Commencing at group i, we have an E.M.F". of 107.5 volts, at 
 group 2 105 volts, at group 3 102.5 volts, and at group 4 100 
 volts, with a uniform drop of potential of 2.5 volts between each 
 of the groups. Commencing the calculation at the extreme end 
 of lead, or at group 4, we have 5 lamps connected in parallel, each 
 with a resistance of 200 ohms. Then combined R of group 4 = 
 -a<'=4o ohms, and a total current of W"— 2.5 amperes. The re- 
 quired R of section 3 to 4 then is \% x 2.5= i ohm. 
 
 Taking now group 3, calculating by similar methods, and bear- 
 ing in mind the fact that the combined resistances of the whole of 
 the groups must be the same, since the number of lamps in each 
 group are equal, total current of group 3 then is 
 
 102.5 
 
 40 
 
 = 2.56 amperes, 
 
 And since the current required for group 4 must pass through [the 
 wire supplying group 3, we get : 
 
 Current of group 4 = 2.5 amperes, 
 (i a it ^-2,^5 << 
 
 Total current wire 2-3 = 5,06 '* 
 
85 
 
 The required R of section 2 to 3 then is equal to Hll^ 2. 5 = .494 
 ohms. 
 
 Taking now j^roup 2 and calculalinj^ in exactly similar maimer, 
 ^^=2.(i2 amperes, and 2.62 + 2.56+2.5-7.68, total current pass- 
 
 ^ ft.. si eu^-' 
 
 
 <^ 
 
 II 
 
 Rz^4l0-^^i 
 
 j/y^.S^y ofv^ Rr.,^9^e'h^y 
 
 I 
 
 ^^>J/5 
 
 ^t ;i. ;c 3c 
 
 d"^/'' 
 
 R I. 
 
 0-7irrr 
 
 E g//Q ir(rUb 
 
 £=/ors ^/ 
 
 C=to 36 
 
 =y-6s 
 
 1^ 
 
 (I 
 
 o 
 
 -X- 
 
 e = io5 V 
 
 C=5.C>6 
 
 £ =/o?JS V 
 
 C= fc &" 
 
 C = Jl-S" eUf^/Ls 
 
 &^^ 
 
 'oZ 
 
 )f X f u X 
 
 t'/l. 
 
 ^^o t-cTo id-i 
 
 Fig. 4. 
 
 ing- on wire section i to 2. 1^2 ^ 2. 5 = . 325 ohms, required R sec- 
 tions I to 2. 
 
 •07-5 
 
 Group I = 
 
 40 
 
 = 2.68 amperes, and 2.68 + 7.68= 10.36, total C 
 
 passing on wire section o to i. j'oYtv ^ 2.5 = .24i ohms, required 
 R sections o to i. 
 
 Therefore we find we require, to meet conditions as set forth in 
 our problem, a wire varying; in resistance and consequently in 
 area, as shown in P'ig. 3, and where 
 
 R of section of wire o to i = .241 ohms. 
 
 
 (t 
 
 
 1 to 2 — .325 
 
 2 to 3= .494 
 
 3 to 4= 1.000 
 

 
 IMAGE EVALUATION 
 TEST TARGET (MT-S) 
 
 '^ 
 
 2^ 
 
 /. 
 
 i// 
 
 
 
 <; <;^ 
 
 ^^. 
 
 / 
 
 i/.A 
 
 Ue 
 % 
 
 1.0 
 
 I.I 
 
 .- ^ IIIIIM 
 
 1.25 
 
 1.4 
 
 1.6 
 
 P>^ 
 
 
 ^'9' %^ # 
 
 # 
 
 0/%?- 
 
 Photographic 
 
 Sciences 
 Corporation 
 
 23 WEST MAIN STREET 
 
 WEBSTER, N.V. 14580 
 
 (716) 972-4503 
 
 S 
 
 lb 
 
 V 
 
 % 
 
 ^q^- 
 
 ^\ 
 
 V 
 
 
 .^ 
 
 
 
 
 ^"^^ ''''.i^'^ 
 
 T'^ 
 
 # 
 
 '^ 
 

 ip. 
 
 m 
 
86 
 
 These resistances, divided by twice the lengfth of one side of lead, 
 will equal R per foot required on lead. 
 
 Fig-. 4 shows the whole of the results of our previous problem in 
 detail, and at the same time it will be seen from diagram that this 
 method of computation can be applied to almost any system of 
 distribution on the two-wire principle, and need not here have 
 further reference. 
 
 i 
 
 CALCULATION OF WORK DOXE BY PUMPS. 
 
 Rule : To find the contents of the pump when full, find the area 
 of the cylinder in inches, multiply by length of stroke in inches, 
 and result will be capacity of pump in cubic itiches of water each 
 stroke. 
 
 Rule : To find capacity of pump per hour, first find contents of 
 cylinder each stroke, then multiply by strokes per minute, and 
 then by 60 ; result will be capacity of pump per hour in cubic 
 inches ; to reduce to cubic feet divide result by 1728. 
 
 Example: Find the contents of a boiler feed pump, stroke being- 
 24 inches and diameter of cylinder 6 inches. 
 
 Formula : D- x .7854 x L = contents ; 6- x .7854 = 28.2744, area 
 in cubic inches ; 28.2744 x 24 = 678.5856 cubic inches, contents of 
 pump cylinder when full. 
 
 Example : Find the capacity per hour of a boiler feed pump in 
 cubic feet, diameter of piston being 4 inches, stroke 6 inches, and 
 pump making- 75 strokes per minute. 
 
 42 X .7854= 12.5664 square inches, area ; 
 12.5664x6 = 75.3984 cubic inches, contents per stroke ; 
 75.4 X 75 = 5655 cubic inches per minute ; 
 5655 X 60 = 339300 cubic inches per hour ; 
 3393004^ 1728. = 196.35 cubic feet of water, capacity per hour. 
 
 DETERMINING THE REgUIRED SIZE OF BOILER FEED PUMPS. 
 
 This question, as practical eng^ineeis will know, covers a wide 
 range of subjects, and it is particularly hard to lay down a hard 
 
 I'lli 
 
^7 
 
 anil fast rule adapted to every condition of service. It might be 
 assumed that the quantity of water accounted for as being- used 
 within the cylinder of an engine would be a nearly correct basis 
 on which to arrive at the required capacity of the pump. This, 
 however, for many reasons, is so far from being correct that I 
 have decided not to refer at len^tli to this method of calculation. 
 Clearly, a boiler feed pump should have ample capacity for all 
 calls that are likely at any ti-ne to be made upon it, and with the 
 quantity of water we may he called upon to evaporate per minute 
 or hour before us, we are enabled lo arrive at a fair approxima- 
 tion of the required capacity of the pump. Even this must be 
 coupled with a good deal of practical common sense, and pro- 
 vision must at all times be made for leakage on the boiler pnd its 
 accessories, and leakage and s'ip within the pump itself- 
 
 Example : Find the required capacit}' of a boiler feed pump for 
 three boilers v«^liose furnaces are 3 feet by 6 feet, coal consump- 
 tion 15 pounds per square foot of grate surface per hour, and 
 evaporation equal to 10 pounds o( water per pound of coal. 
 
 6 feet 
 3 feet 
 
 18 square feet surface in each boiler 
 15 pounds coal consumed per hour 
 
 00 
 
 18 
 
 270 pounds of coal per hour each fMrnace 
 3 
 
 810 pounds, total coal cv. nsumed per hour 
 10 
 
 8100 pounds of water evaporated per hour 
 
 8 100 -=-62.5= 129.6 cubic feet of water per hour 
 129.6 
 
 4 factor of safety 
 
 518.4 cu. feet of water per hour for safety, after allowance 
 for slip, leakage, etc., required capacity of pump. 
 
 Whr.t must be the diameter of a double-acting duplex steam 
 
88 
 
 pump making- loo strokes per minute and liaving a stroke of four 
 inches to comply with above requirements ? 
 
 518.4 -^ 60= 8.64 cu. feet per minute ; 
 8.64 -^ 100= .0864 cu. feet per stroke ; 
 .0S64 X 1728- 149.3 ^'"' i'it:hes, required contents of pump ; 
 
 149.3-7-4= 37'3-5 ^'^'- iiit'hes, required area of cold water 
 piston ; 
 
 37-325 ^-7854= 47-5' and 
 
 A^/'4y.:5= 6.88 inches, required diameter of punp piston, 
 
 or say, 6^ inches. 
 
 Then, to safely comply with the conditions set forth, we shall 
 require a duplex steam pump making 100 displacements per 
 minute, with a 4-inch stroke and a cold water piston diameter of 
 6J^ inches. 
 
 MISCELLANEOLS QUESTIONS. 
 
 To find extra pressure required to dischargee water from a given 
 orifice : — 
 
 Formula : 
 
 T- D^ 
 2000000 d" B- 
 
 Where T = travei of plunger in fee. per minute, 
 D = diameter of plunger in inches, 
 
 d^diameter of delivery valve, " 
 
 B = breadth of opening or lift in inches. 
 
 Example : The cold water piston of a pump is 4 inches in dia- 
 meter and has a travel of 150 feet per minute. The delivery valve 
 is 2}4 inches in diameter and has a lift of )^ of an inch. What 
 extra pressure is required to discharge the water ? 
 
 T2 D* 
 
 150- X4^ 
 
 5760000 
 
 2000000 d- B'^ 2000000x2.5 X.25- 781250 
 5760000 -f 781250 = 7.37 pounds. 
 
 To find velocity at which water will travel through the discharge 
 
89 
 
 pipe of a pump. The velocities are in inverse proportion to the 
 area, or what amounts to the same thing-, the velocities are in in- 
 verse proportion to the square of the diameters. 
 
 Then for the purpose of deiermining- the ratio of velocity be- 
 tween any two pipes we may construct the following" formula : 
 
 Where D equals diameter of plunger, d equals diameter of pipe. 
 And when the speed of the plunger is known the velocity of the 
 water in the discharge pipe may be determined by formula : 
 
 D^xT 
 
 Where D = diameter of plunger in inches, 
 
 T = travel in feet, 
 
 d = diameter of dischargee pipe in inches, 
 
 V = velocity of discharge in feet per minute. 
 
 Example : The plung-ers of a pump are lo inches in diameter 
 
 and have a travel of loo feet per minute. At what velocity will the 
 
 water travel through a discharge pipe 2 inches in diameter ? 
 
 Then 
 
 D'^ T 10** X 100 loooo 
 
 -— r— = ; = = 2i;oo reet per mmute. 
 
 d- 2- 4, ^ *^ 
 
 The work done due to the energy of motion of a moving body 
 is represented by the formula : 
 
 W V^ 
 64 
 Where W equals the weigfht, V equals velocity in feet per second. 
 Example : The piston of an hydraulic ram is 12 inches in dia- 
 meter ; water in feed pipe has a velocity of 2,000 feet per minute ; 
 feed pipe to ram is 3^ inch. What is the energy of ram per pound 
 of waier used ? 
 
 Velocity in feet per second equals 
 
 2000 
 -6^ = ^^'^^- . ■: 
 
 I 
 
 
i 
 
 f- 
 
 
 
 
 
 
 
 >sm 
 
 k 
 
 L. 
 
 
 90 
 
 It has already been shown that the velocity of water in pipes 
 depends upon the areas or di-imeters squared ; consequently la'* 
 ■V .75^= 144 -f .5625 = 256. That is, 256 is the ratio of velocity of 
 water in feed pipe as compared with the velocity of the w-^ter in 
 the rani. Therefore, if the water in the ram moves at the rate of 
 1 foot per second, the water in the feed pipe must move at the 
 rate of 256 feet per second, and the amount of work done is the 
 same in each case. The difference in areas causes the ram to 
 move only i^l^ the speed of the pump, and what the ram loses in 
 speed it must gain in force, since energy is indestructible, and the 
 pressure exerted by the ram must be 256 times greater than that 
 in the pipe to make up for loss in speed. 
 
 Energy of motion in pipe then equals 
 
 WV' ix33.3.f 
 
 17.35 ft. lbs. 
 
 64 64 
 
 Then the work done by ram per pound of water used equals 17.35 
 + 256 = 4441 -60 foot pds. per pound of water used. 
 
 The pressure due to velocity is found by dividing the square of 
 the velocity in feet per second by the constant 148.3. 
 
 Example : The piston of an hydraulic ram is 6 inches in dia- 
 meter ; velocity of water in a ^-inch feed pipe equals 2,500 feet 
 per minute. Wha;. is the work done by the ram per pound of 
 water used, and what is the pressure per square inch due to 
 velocity, and if the energy of the water had been turned into heat 
 what would be the rise in the temperature of the water ? 
 
 V^elocity of water in feed pipe per second equals 
 
 2500 
 60 
 
 ^4'-33. 
 
 and 6- X .5^ = 36 + .25= 144, ratio of velocity in ram as compared 
 with feed pipe ; 
 
 W V2 I X 41.33- .. r , , e ,- , c 
 
 = -^ — — =26.69 foot io3. energy of ir.otion per pound of 
 
 64 64 
 
 water in feed pi pie ; 26.69 x 144 = 3843.36 foot pounds per pound 
 
9« 
 
 of water used by mm ; V- = 1708.16, and prossuie due to velocity 
 ill pipe equals 
 
 1708. 16 
 
 = 11.51 pounds per square inch. 
 
 Then the pressure on the rani must equal 1 1.51 x 144= 1657.44 
 pounds per square inch. 
 
 The mechanical value of a British Thermal Unit (BTU)is 
 equivalent to raising- i pound 772 feet hig-h in one minute, or what 
 is the same things, raising 772 pounds one foot high in the same 
 period of time ; the-efore a raise of 1° F. in the temperature of a 
 pound of water equrls 772 foot pounds of energ-y. 
 
 26.69 energy per pound of water in feed pipe. 
 
 772^ ■■ • 
 
 eqivAls .0345° F. increase of temperature of water in feed pipe. 
 
 Then 
 
 3843.36 ,0 ,' . 
 
 ^~r;." =4.98" F. rise of temperature in ram. 
 772 ^ ^ ' 
 
 Then if the whole of the energy of the ram had been expended 
 
 in heat it would have g"iven off sufficient heat to have raised the 
 
 temperatu-e of each pound of water 4.98' Fahrenheit. 
 
 Itl 
 
 - ! : 
 
 ! '.<■ 
 
 •H 
 
J 
 
 REQUIRED SIZE OF LEADS IN QRCULAR MILS. 
 
 A Mil, is iVrrff of an inch, and written decimally .001. The 
 area of a circle ioVjo of an inch in diameter is termed a circular 
 mil, and a g;reat many wire tables express the area of the 
 cross section of a wire in circular mils written symbolically 
 C. M. Rules for the sizes of wires as met with in enj^^ineer- 
 ing" practice for given resistances are often based on circular 
 mils, and include a constant for the conductivity of materia, 
 of which wire is composed. From this, then, we can construct a 
 formula for the calculation of the resistance of wire of whatever 
 material composed, so long- as we know the specific resistance of 
 the material. 
 
 Commercial copper wire of 90% purity, one foot long and one 
 
 C. M. in cross section, is said to have a resistance of 10.79 t)lims 
 
 at 75' F. (appro^:imate). In accordance with the rule that the 
 
 resistance of a circular conductor varies inversely with the square 
 
 of its diameter and directly with its length, we can construct the 
 
 following well-known formula for the determination of resistance 
 
 of copper wire : 
 
 10.79 X L 
 
 R 
 
 d- 
 
 Example(io): A commercial copper wire one-half an inch in 
 diameter has a specific resistance of 10.79, what is the resistance 
 per foot ? 
 
 ^ inch = .500 mils. 
 
 .500^ = .250,000. 
 
 10.79 -r .250,000 = .000043 ohms per foot. 
 
 Example (11): A copper wire 5,000 feet long, with a cross sec- 
 tion of 8,000 CM., what is its resistance? 
 
 10.79 X 5,000 
 
 8,000 
 
 = 6.743 ohms. 
 
93 
 
 The required cross section of a wire in C. M. is equal to its 
 length divided by its resistance and multiplied by 10.79, or 
 
 L 
 
 CM. = -j^ X 10.79 
 
 The required cross section of a pair of leads in CM. for a given 
 drop is found to be equal to the product of the let;gth of leads 
 multiplied by number of lamps (in parallel) by 21.58 by 100, minus 
 the percentage of drop, antl the whole divided by the resistance 
 of one lamp hot multiplied b) the percentage of drop allowed. 
 
 L X 2 1 . 58 X N X 1 00 - % 
 
 CM. ord2 = 
 or what is the same thing. 
 
 Rx% 
 
 10.79 x2xLxN 100-% 
 
 V/ . iVl • — j^ ^ 0/ 
 
 Example (12): Ten lamps are to be placed in multiple at the 
 
 end of a double lead 100 feet long. The resistance oi each lamp 
 
 when hot being 220 ohms, what must be the sectional area of the 
 
 wire if a drop of 5% is allowed on E.M.F". ? 
 
 2 1 . t;8 x 1 00 X I o 1 00 - s 
 
 — ^ = 98x ^=:i87^CM 
 
 220 ^ 5 ' 
 
 Based on this and following out the principles laid down in Ohm's 
 Law, the following formula is often given as a ready means of deter- 
 mining the required size of wire for house or secondary incandes- 
 cent circuits : 
 
 2i.58xLxNxC 
 
 CM. = — ^— f 
 
 L v 
 
 Where CM. = Area in circular mils. 
 
 // L. = Length of one side of lead. 
 
 f, N. = Number of lamps. 
 
 ff C. = Current required by each lamp. 
 
 ff L v = Number of volts per lamp loss in line. 
 
 When lamps or groups of lamps or other appliances are placed 
 at different distances from the generator, apply principles laid 
 down in example ; or, roughly stated, determine first the size of 
 
 i 
 
 « 
 
 I: 
 
fi 
 
 94 
 
 wire reiiuircd for each lamp or i^roiip, as if on independent cir- 
 cuits, starting" from (he dynamo, then combine all wires running 
 in same direction. 
 
 DETERMINATION OF THE MEAN PRESSURE THROUGH- 
 OUT THE STROKE WHEN USING STEAM 
 EXPANSIVELY. 
 
 Considering^ recent advances in modern engineerings and im- 
 proved appliances at hand, an article on this subject is perhaps 
 superfluous. There are, however, such i<nportarit principles in- 
 volved and such useful information obtainable, that the author 
 feels justified in referring to the subject at short length. Since a 
 large proportion of my readers are operating simple non-condens- 
 ing engines, the author will deal particularly with this class, and 
 the principles involved can then be readily applied to compound 
 engines of any type. 
 
 The engines to which many of us were first iv troduced had 
 what is styled a fixed " cut-o*T," which took effect at almost any 
 point between start and finish of piston travel, according to the 
 views the engineer held on the question of working steam expan- 
 sively. And speed was controlled by means of a throttling device, 
 arranged to reduce or raise the pressure against the piston as 
 occasion required. This type in modern practice is nearly obso- 
 lete, and we have in its stead what is termed an automatic " cut- 
 oflF" engine, arranged to take steam at begiiuiing of stroke at 
 full boiler pres? ure, and regulating its speed by means of a vari- 
 able " cut-off" — that is, initial pressure against piston on admis- 
 sion remains constant and varies only as pressure on boiler varies ; 
 — and this pressure is maintained against the piston as it travels 
 within the cylinder to a point of the stroke varying in direct pro- 
 portion as the load on the engine varies. 
 
 Theoretically, then, if we had no cut-off until admission valve 
 closed at the end of the stroke, the pressure of the steam against 
 the piston would remain the same, and a diagram representing 
 
95 
 the operation could b.^ drawn as represented in Ki^-^. 5, within the 
 letters, A, B, C, D, and the pressure aj^ainst the piston can be 
 readily ascertained by a simple process at any point of the siroke, 
 and the mean, initial and terminal pressures wouUI at all times be 
 equal, and the mean effective pressure would be initial pressure 
 minus back pressure, and could be equally as readily delermined. 
 Let us now suppose that when the piston has reached :, point 
 equal in distance to one-third of its travel, the admission va've 
 closes and cut-offtakes place ; the piston then (ravels the remain- 
 ing two-thirds of the distance by the expansive force of the steam 
 
 Fig. 5. 
 
 already admitted, and supposing' that expansion would take place 
 under isothermal conditions, that is, that the temperature of 
 steam remained constant, it is clear that as expansion increases 
 pressure decreases, consequently the terminal pressure of tne 
 steam will be very much less than the initial pressure, and that at 
 any point in the stroke between one-third and the end, the pres- 
 sure will be less than the point immediately before and higher 
 than the next succeeding point. If, then, we know the initial 
 pressure of our steam uiid he point of cut-off, and can by any 
 means determine the pressure at the respective points of the 
 
 II: 
 
 # - 
 
stroke, we can by fiiuling the mean of" these pressures thus dpter- 
 mine the mean pressure on the piston. For this purpose, then, 
 we re-arran^'e Fig. 5 as illustrated in Fig-. 6, and divide this into 
 any even number of points, taking care that one of these points 
 shall ct>incide exactly with and intersect point of cut-off. 
 
 To do this we first divide the cylinder into an equal number of 
 parts by subtracting the numerator from the denominator of the 
 point of cut-off expressed as a fraction, and if the result is an even 
 
 ii< 
 
 o 
 
 CD 
 
 00. , . 
 
 M SD' <oi ^ >^ 10; «r>; ;> 
 ' ' : : ! 
 
 Si 6\ 7\ I I 
 
 /#[ // I a 
 
 Fig. 6. 
 
 number above six, the denominator of the fraction is the number 
 of equal parts into which the cylinder can be divided so that one 
 of the ordinates will exactly intersect the point of cut-off. 
 
 Example : VVith steam cut off at {•■, of the stroke, what number 
 of parts must the cylinder be divided into ? 
 
 As per rule, 15-3=12 — an even number above six — therefore 
 cylinder must be divided into 12 equal parts. It will, however, 
 very frequently be found that when the numerator is subtracted 
 
from tin* ilonoininator that eitljer ati odd nunilu'i or an even mini- 
 ber less ttian six is the resuh. In either ease, double, liebie or 
 quadruple botli tlie numerator and the denoniinaior, and if this 
 does not answer, inerease fraction until you jfet a fraction ^ivitijj 
 the desired result. 
 
 Example : Suppose steam fs cut off at ^ stroke, into how 
 many equal parts must cylituler be divided? 
 
 Proceedinjj; as per rule, 3 - i =2, an even number less than six, 
 which will not answer. 
 
 1^x5 = -\, which is still too low ; we continue the process until 
 Y we reach 
 
 ^ X 1= ,V= 12 - 4 = 8, an even number hig-her than 6, and cylin- 
 der can then be divided into 12 equal parts. 
 
 Fig. 6 then represents a cylinder, and we proceed to divide 
 into 12 equal parts, with cut-off taking place at ', j of the stroke. 
 Each of the dotted ordinates in this tlia^^ram then represents the 
 position of the piston expressed in j'r.ths of piston travel. I\)int 
 of cut-off, or J,^ of stroke, will equal 4 of these parts, and the 
 pressure of steam at ist, 2nd, 3rd and 4th positions of piston re- 
 mains constant. 
 
 When piston is in position. A, B, at 4, eut-oflf takes place, and 
 at Q, R, exhaust valve opens. If steam then was expanding' 
 under true isothermal conditions, the lines, c, d, e, f, g, h, etc., 
 show the gradual decreasing pressures of the steam due to ex- 
 pansion and pressures represented by letters, A, B, Q and R, 
 show the total value of the steam during expansion. 
 
 To find the pressure of the steam at any one of the positions of 
 of the piston during expansion, multiply absolute initial pressure 
 of steam by number of ordinate at point of cut-off, .ind divide 
 this result by the number of ordinate at which pressure is 
 desired. 
 
 ii 
 
 ' 
 
 ■f^h 
 
 I 
 
.iili i 
 
 98 
 
 E;:;iniplc' : Vvith steam ;ii..rtittO(I ;it 80 lbs. j^auj^e pressure, fiiul 
 llie pressure al each of the ordinales in tiiaj^^rain. 
 80+ I5 = 95» absohtle pressure. 
 95x4 = 380, vahie of stearo previous to lut-on". 
 
 Then pressure nl C D = ''|^ = 76. pounds. 
 
 K L=»«'> = 
 
 42.22 
 
 n 
 
 n 
 
 " OP=-V,"=34-54 
 
 :i«?r _ 
 
 31.66 
 
 To find total value of stea 
 
 ni durm^^ expansion we require to 
 find area of space enclosed by A, H, vj and K. 
 
 I''or liiis purpose we atlopt the Simpson rule, which is : To ihc 
 first and last ordinates add four times the sum (^i the even t)rdin- 
 ates and twice the sum of the odd ordinates; one-third of this 
 sum equals the area. 
 
 Then a or A B = 
 
 95 PO 
 
 vmUs. 
 
 5orCD = 76 x 4 = 304 
 
 6 or E F = b5 x2=:i3o 
 
 7 or G II = 54. 28 X 4 = 2 1 7. 1 2 
 
 8 or I J =47.5 X2= 95 
 
 9 or K L = 42. 22 X4= 1 68.88 
 
 10 ;»r MN = 38 X2= 76 
 
 11 or O P = 34. 54 X 4 = 1 38. 1 6 
 
 12 or 
 
 QR 
 
 31.66 
 
 3 I »255 
 
 ,82 
 
 Value d-irinjf expans 
 sion 
 
 ion, 418.60 
 
 Before exp; 
 
 n 
 
 urmg' 
 
 380 pounds. 
 418.60 n 
 
 Value durinv;' whole stro!-o, 798.60 
 
09 
 
 If lunv, wo clivitle by inimlHT of ortliiiatos used, we j^'et mean 
 pieshure against the piston throiij;Iioiil the stroke. 
 
 79H.60-;- 12=66.55 pi>unds. 
 
 it the eng'ine piston was runninj^- ai^ainst a perfect vaentnn, then 
 mean pressure throug-liout the stroke wouki be the mean effective 
 pressure on which to c;dculate power devekipetl by any etig'ine 
 in doing' work, but this rarely, if ever, liappens ; therefore the ab- 
 sohite back pressure against the piston must be deducted, and 
 we get 
 
 Mean pressur l)ack pressure = mean efTective pressure. 
 
 Example : F'irid mean effective pressure on the piston of a non- 
 condensing eng'ne when ti ean pressure is 66 pounds per square 
 inch anil back pressure two pounds above t!ie atmosphere. 
 
 21 15—17 poimds absolute back pressure, and 66-17 — 49 
 jiounds mean effective ^)ressure of steam on piston. 
 
 When jMincin'os of calculatiop are understood, formula for cal- 
 culation can '-e considerably shorteneil by nse of logarithms. 
 
 1st. The hjperbolic logatithinof ti>e ratio of cut-off increased 
 by I ami multiplied by terminal presstne equals mean pressure. 
 
 2nd. The liyperbolic logarithm of the ratio of cut -off increased 
 by one and multi{)lied by initial pressure, and result divided by 
 ratio of cut-ofi', equals mean [pressure. 
 
 3rd. Secoiul rule minus absolute back pressure equals mean 
 effective pressure throughout the stroke. 
 
 Example of Rule 1 : Cut-off takes place at '(of stroke; '.lie 
 ratio of cut-olf therefore is 3-f 1 =3, and the hyperbolic logarithm 
 o( 3 is 1.09S61, 
 
 .'. i.oc)H I I X 31.6 = 66.39 pounds, mean pressure. • 
 
 Kxample of Rules z and 3 : 
 
 1.098+ I 
 
 95 ^ ^ """ — 66.4 pounds, nu^an pressure. 
 
 1.098 f I 
 
 3 
 
 95^ 
 
 7 — 49.4 pounds, mean efll'eclive pressure (M.E.T.) 
 
R'l 
 
 lOO 
 
 In addition to finding the mean pressure by process of calcula- 
 tion, we require also to find the mechanical efficiency gained by 
 cutting off steam in comparison to steam being admitted Juring 
 the whole of the stroke. This can be found by dividing the mean 
 pressure by the terminal pressure absolute. Then 66.4-^ 31.6 = 
 2.098 ; or expressed briefly, the mechanical efficiency is the 
 hyperbolic logarithm of the ratio of cut-off plus i. 
 
 Finally, on this subject, we want to find the co-efficient of 
 efficiency or a number expressing the practical efficiency of the 
 steam, including the effect of expansion as compared with the 
 duty of the same steam without cut-off or expansion pad loss by 
 back pressure. 
 
 Rule : 
 
 Mean effective pressure X Length of stroke 
 
 Absolute pressure X (Length of admission + Clearance) 
 
 = The co-efficient of 
 
 efficiency. 
 
 Example : Find the co-efficient : Gauge shows 80 poimds ; 
 stroke 12 inches, cut-off taking place at ^ of stroke, and mean 
 effective pressure is shown to be 49 pounds, piston having )4 '"^h 
 clearance. 
 
 80 + 15 = 95, absolute pressure. ; . , 
 
 49 x 12 588 
 
 95 X (4 + -25) 4^3-75 
 
 = 588-7-403.75= 1.45 co-efficient. 
 
THREE-WIRE SYSTEM. 
 
 In the three-wire system the lamps are really irranjjed in sets 
 of two in series. The two outer wires have double the potential 
 of the lamps, and since, owing to the lamps being- in series, one- 
 half the current only is required as compared with multiple 
 arc two-wire system ; therefore, to carry one-half the current 
 with twice the difference in potential or double the E.M.F., a con- 
 ductor with one-quarter the area of cross-section suffices. 
 
 Dynamos are frequently set in series in large buildings, and 
 their terminals wired out on ihe three-wire system tos.ive copper. 
 Since each of the wires is but one-quarter the size of the wires in 
 the ordinary two-wire system, it remains clear that the weight of 
 copper required on the three-wire system is but ^ of that 
 required on ordinary two-wire systems. The formuue already 
 laid down for two-wire work apply to three-wire calculations as 
 regards size of mains, if denominator of formula be multiplied by 4. 
 
 21.58 X L Nx(ioo-%) / 
 
 C M or d- 
 
 R X % X 4 
 
 This 'ormula gives required cross-section of mains in circular 
 mils, using commercial copper wire as a conductor. 
 
 ALTERNATING CURRENT SYSTEM. 
 
 Ill alternating systems we have tw'O sets of mains to deal with, 
 primary and secondary, both of which Have different E. M. F.'s. 
 
 Theoretically, the rules deduced from Oti n's 'aw are not cor- 
 rect as applying to this system of generation, because calculations 
 must be made to allow for conversion from primary to secondaiy 
 current. The ratio tliat the primary E.M.F. bears to the second- 
 
 
 mmmi*mKt> 
 
102 
 
 n, i- 
 
 A' i-: 
 
 .iry E.M.F. is expressed by dividitij;- flie primary K.M.F. by the 
 secoridary E.M.F., and is termed ratio of conversion. Thus, 
 
 E primary 
 
 ^ _ i^atio of Conversion. 
 
 E secondary 
 
 The current in the primary is equal to the current in the second- 
 ary divided by the ratio of conversion. 
 
 Ex,".mple : On an alternating- circuit whose primary E.M.l^". is 
 1,000 volts and secondary E.M.F. 50 volts, there are 500 lamps, 
 each having a resistance of 50 ohms, what is the primary current 
 and what the secondary ? 
 
 ^M*^ = 20, ratio of conversion. 
 
 frl'n = ' ' > combined R of lamps. 
 
 ">," = 1500 amperes current on secondary. 
 
 ^jj^ =25 amperes current on primary. 
 
 Current being determined, rules deduced from Ohm's law apply 
 exactly as shown for direct current calculations. 
 
 Exainple : An alternating generator has 4,000 feet of primary 
 main attacheil with a»i E.M.I'\ across the terminals of the ma- 
 chine of 1040 volts ; it is decided to alU^w a drop of 40 volts on 
 the primary at the convenor. Tlie secondary main has attached 
 to it 750 16 c.p. lamps at an E.M.F. of 50 volts, each lamp having 
 a resistance of 50 ohms. What is the resistance on the primary 
 
 wire ? 
 
 no _ 
 
 040 - 40= ' H!{" = ^o, ratio conversioti. 
 amp., current each lamp. 
 750 X I =750 amps., total current on secondary. 
 750^20=^37.5 amps, current on primary. 
 
 I 
 
 — T/ x4o= 1.06 ohms resistance on primary. 
 
 37/2 
 
 To obtain the required si/e of primary mains in circul.'ir mils, 
 calculate by formula for two-wire direct current, and divide the 
 result by the square of the ratio of cot:version. Or 
 
 21.58XL Nx(ioo-%) 
 
 C M 
 
 Rx%xRC'-^ 
 
 Imii^jaifc.. 
 
I03 
 
 Kxainple : An allcnialiiitf iiirrent generator has an E.M.F. 
 across the leads of 1040 vohs, atul eiirrent is ilelivered 4,000 fei*l 
 from station, E.M.F. at toniii'nals bcin^ 1,000 volts ; current con- 
 sists of 1,000 lamps, having a resistance of 50 ohms each on a 50 
 volt secondary circuit. What should be the cross-section of the 
 P'-imary mains in circular mils, and what is resistance of same 
 main to allow a drop of 40 volts as shown ? 
 
 1000= E.M.F. at terminals of primary 
 50 = E.M.F. on secondary 
 40 -r 1040 = 3.84% drop in potential. 
 21.58 X 4000 X 1000 X (100 ~ 3.84) 
 
 = 20 R.C. 
 
 108,079 C. M., required cross- 
 
 50 X 3.84 X 20- 
 section of primary main in circular mils. 
 
 Total current on secondary -- 1000 x jj;j = 1000 amps. 
 IOOO-^ 20 = 50 amps, secondary current. 
 -V><40 = .8 ohms resistance, primary niain. 
 
 HORSE POWER CALCULATIONS. 
 
 The power or force required to do a certain mechanical work 
 is usually referred to as toot pounds. That is, one pound raised 
 one foot high is called a foot pound, without reference to the time 
 required to do the work. Therefore, if we want to find force or 
 energy required to be expended to do a certain work expressed 
 in foot pounds, we require to multiply the weight required to be 
 raised by the distance in feet through which weight has to be 
 lifted. 
 
 A horse power is fixed as the work performed in raising 33,000 
 poimds one foot high in one minute, or what is exactly the same 
 thing, raising one pound 33,000 feet high in the same period of 
 lime. Here we have time required to do a certain work, and 
 note of this must be taken in our calculations. 
 
 Then to reduce force exerted it the piston of a steam engine, 
 and to express this force in horse power, we require to know 
 eiTective pressure against the piston throughout the stroke, and 
 
 ^Jii 
 
 1! 5- 
 
1 ; 
 
 I04 
 
 distance travellod in feei by the piston per niimile. And to do 
 this use the followiiif^ established rules : 
 
 1st. I*'ind .'irea of piston in square inches. 
 
 2nd. Fine' the total pressure in pounds on the i>iston by multi- 
 plying' the area by the mean effective pressure per »-quare inch. 
 
 3rd. Find the distance in feet traversed by the piston per 
 minute by multiplying the length of stroke in feet by twice the 
 revolutions per minute. 
 
 4th. F^ind the energy exerted by the engine expressed in foot 
 pounds per minute by multiplying the total pressure in pounds 
 against the piston by the travel iii feet per minute. 
 
 5th. Find the horse power by dividing total foot pounds per 
 minute by 33,000. 
 
 F"rom this, then, we can construct the following simple formula: 
 
 A.N.P.S.' 
 
 =H.P. 
 
 33,000 
 
 Where A = Area of piston in square inches. 
 
 N = Number of strokes per minute (revolutions multiplied 
 
 by two.) 
 P=Mean effective pressure per square inch. 
 S' = Length of stroke in feet. 
 
 Example : Find the horse power of an engine 9.5 inches by 12, 
 running 280 revolutions per minuie, mean effective pressure 
 throughout the stroke being 35 pounds per square inch. 
 9.5- X .7854 = 70.88 square inches area. 
 
 70.88x35 = 
 70.88 
 
 35440 
 21264 
 
 2480.80 total pressure against piston. 
 
 280 revolutions per minute. 
 2 strokes per revolution. 
 
 560 strokes per minute. 
 1 length of stroke in feet. 
 
 560 feet travelled by piston per minute. 
 
2480.8 pressure on piston. 
 560 
 
 1488480 
 124040 
 
 1389248.0 foot pounds per minute. 
 33000)1389248.(42.09 horse power. 
 132000 
 
 69248 
 66000 
 
 324800 
 
 We have in this formula two constants, .7854 constant multiplier 
 to find area, and 33000 constant divisor to find horse power, con- 
 sequently by dividing .7854 by 33,000 we get a constant multiplier 
 as a result, and thus shorten our method. 
 
 .7854-^33000. =.0000238, which becomes a constant multiplier, 
 and our formula then becomes (d- N P S^)x .0000238= H. P. 
 
 Example : Using this formula find horse power of an engine, 
 dimensions, etc., as per last example. 
 
 
 9.5 = diameter of piston. 
 9-5 
 
 90.25 = d2 
 
 560 = strokes per minute. 
 
 5415 00 
 45125 
 
 50540.00 
 
 35 = mean effective pressure. 
 
 252700 
 151620 
 
 1 768900 
 
 .0000238 = constant multiplier. 
 
 14151200 
 5306700 
 3537800 
 
 42.0998200 = H. P. 
 
 M 
 
io6 
 
 i 
 
 CALCI LATING HORSE POWER FROM INDICATOR DIAGRAMS. 
 
 An indicator diagram is practically a record of the action of the 
 steam on the piston througiiout the stroke of an engine, and if we 
 are enabled to ascertain the area of the diagram, having the 
 length of the diagram and scale of spring used before us, it be- 
 comes an easy task to find the mean effective pressure of the 
 steam throughout the stroke. 
 
 The area of the diagram is easiest found by use of a small in- 
 strument called a planimeter, which should be set to a natural 
 scale, that is, area of diagram should read in square inches. 
 Knowing the area and dividing by length of diagram in inches^ 
 will give us height of a regular body of an equal area. 
 
 Fig. 7. 
 
 Fig 7 represents a diagram 3 inches in length, and the irregular 
 body has an area of three square inches. If we reduce this area 
 to a regular body with at least two of its sides equal, we get a 
 diagram, as shown by dotted lines, exactly one inch in height ; 
 multiplying this by scale of spring used, we get M.E.P. through- 
 out the stroke. Therefore, we can construct a formula as follows 
 to calculate M.E.P. from an indicator diagram : 
 
 Area 4- Length x Scale of spring = M. E. P. 
 
 The horse power calculation can be very much shortened by 
 
107 
 
 using what is known as an en^fine constant ; tliat is, the power 
 developed by an engine with M.E.P. of i pou id per sq. inch, and 
 is particularly useful where hor*-e power calculations are being 
 constantly made from same engine. 
 
 Abreviaced the formula would then read : 
 
 Engine constant X iM.E. P. = H.P 
 
 Example : Using engine constant and M.E.P. find I. H.P. of a 
 single cylinder engine 9^"xi2" running 280 revolutions per 
 minute. Diagram measures 3 inches 'n length, and planimeter 
 
 Fig. 8. 
 
 shows area to be 2.625 sq. inches, scale of spring used in indica- 
 tor being 40 pounds per inch of compression. 
 
 First, find power developed by engine with a mean effective 
 pressure of one pound per square inch : 
 
 9.5- =90.25 
 90.25x560 = 50540 
 50540 X. 0000238 =1.2 H.P. 
 
 Then power developed by this engine with steam pressure 011 
 piston equalling one pound per square inch of area is 1.2 H.P,, 
 
 
111 
 
 1 08 
 
 i 
 
 ill 
 
 whicli is our eng^ine constant. Next find mean effective pressure 
 shown by diagram : 
 
 Area -r Leng^th = 2.625^ 3 = .875 average heig-ht in inclies. 
 .875 X 40 scale of spring- used = 35. M.E. P. 
 Then 1.2 x 35 = 42. horse power. 
 
 In the absence of a planimeler, it becomes necessary to devise 
 a method of arriving at area of diagram. To those engineers who 
 are unacquainted with geometry this is sometimes difficult to do. 
 ' With every indicator is sent out a number of rules graduated in 
 such a manner that the divisions between each inch correspond 
 exactly with scale of spring used ; by the aid of these it becomes 
 comparatively easy to subdivide into equal divisions. F^or instance, 
 suppose we wish to divide F"ig. 7 into 15 equal parts by using a 
 graduated scale laid ac -oss diagram at an angle so that the 
 divisions i and 15 shall exactly coincide witli lines at each end of 
 diagram, and marking off at each division on scale intervening 
 by erecting at each of these points an ordinate at right angles 
 with atmospheric line, we get 15 exact divisions. It now re- 
 mains for us to asc3rtain the pressure represented in the centre 
 of each of these divisions, and finding the mean pressure by add- 
 ing the pressures together and by dividing the total by number of 
 ordinates erected, we get total mean effective pressure calcula- 
 tion of H. P. It then becomes a simple repetition of examples 
 already described. To convey meaning clearly Fig. 7 is repro- 
 duced in Fig. 8. • 
 
CORROSIVE AND SCALE-FORMIirS AGENTS IN 
 BOILER FEED WATERS. 
 
 ClIAFTKR I. 
 
 Water is universally conceded to be the g^reatest of all known 
 solvents, consequently it never occurs in a state of purity as a 
 natural supply. The foreign matter may be composed of either 
 solids or g"ases, or both, and may be in suspension, mechanical or 
 chemical solution with the water. By the term suspension i» 
 meant a state in which the foreign matter is present in the w.iter 
 in its original state, no matter how finely divided, and when it 
 can be abstracted from the water by filtration, having" the same 
 properties as when it entered the water in the first place. 
 
 In boiler feed waters this usually consists of veg'etable orgfanic 
 matter, finely divided clay, etc. The coarser and heavier the 
 particles the sooner will they separate out and settle to the bottom 
 of an} receptable in which water may be contained. It is pos- 
 sible, however, to find water with foreign matter in suspension so 
 minutely divided that it is impossible to detect them except by 
 slight turbidity or characteristic colorings given to the water. I 
 have in my laboratory samples of wafer taken from freshly drilled 
 artesian wells in the vicinity of Montreal, containing- particles of 
 clay held in suspension so minutely divided that it took weeks for 
 complete settlement to take place, and the slightest movement of 
 the containing" vessels would cause particles to again rise in llie 
 water, thus causing turbidity. 
 
 Matter in mechanical or physical solution, on the other hand, 
 has entered into combination with the water and cannot be again 
 
 m^. 
 
no 
 
 
 
 
 'ij''^' 
 
 
 ■^';t- 
 
 
 •i? .'1, 
 
 1 
 
 'lit 
 
 1 
 
 'U 
 
 abstracted except by means of a chanjje of state or evaporation. 
 For instance, to pure, distilled water add common salt ; this at 
 once dissolves, and until watei is sat mated, if salt was pure, no 
 apparent chanjje takes place in the appearance of the water itself, 
 and if solution was passed through the finest possible filler, salt 
 would still remain in solution, but if we carefully evajjorate ofTtlie 
 water, salt will finally remain in its original state, no evident 
 change having taken place. 
 
 If to pure water (H.,0) we add oxide of sodium (Na.jO), soda 
 at once combines with the water, or part of it, with generation of 
 heat, a chemical reaction taking place in accordance with follow- 
 ing' equation : 
 
 Na,0+II,0 = 2NaOH. 
 
 In this case one molecule of sodium oxide enters into combina- 
 tion with one molecule of water to ibrm the new chemical com- 
 pound, sodium h}dro-oxide, which dissolves in the excess of 
 water present, and remains in chemical solution ; and if water 
 was evaporated off, sodium oxidvs would not be recovered in its 
 original state as an oxide, but in the state of sodium hydroxide. 
 
 The reader will note the difference here. On the one hand, 
 addition of sodium chloride to the water made no apparent change 
 in general appearance of water to the naked eye, and no heat 
 was evolved, and on evaporation it was again recover*' '. in its 
 original state, and without any change in weight In this case 
 salt simply entered into mechanical solution with the water, due to 
 the solvent powers of that menstrum, and is really but a physical 
 change. In the other case sodium oxide first entered into 
 chemical combination with a t^ortion of the water (due to the 
 chemical affinity of the one compound for the other) with evolution 
 of ' *S and excess of water present dissolved new chemical com- 
 pel .J which then remained in chemical solution. On evapora- 
 tion solid sodium oxide would be found to have gained in weight 
 indirect proportion with the number of molecules of water re- 
 quired for chemical change from the oxide to the hydroxide. 
 
Ill 
 
 Chang^es of this description are constantly takinjf place, and it 
 is with these we shall particularly deal .it a later stage in these 
 articles. 
 
 Not only do a jj^reat many solids dissolve in water to a greater 
 or less extent, but many gases also dissolve very readily, and to 
 these can be traced phenonuMion occurring in some feed waters 
 known as pitting, corrosion am' grooving. 
 
 The first of these may be very properly described as corrosion, 
 occurring in small spots on parts of a boiler, and is quite distinct 
 from general corrosion, which wastes away an extended area of 
 surface. Pitting is both annoying and generally very destructive, 
 in some cases burrowing holes through the affei ted parts of a 
 boiler as completely as if it had been drilled. This action was 
 formerly and is to-day occasionally attributed to poor material 
 used in construction, the theory being that weak spots exist and 
 are attacked by corrosive agents found in the material itself. 
 There remains practically no doubt but that a condition of affairs of 
 this kind existing would have an effect on the boiler parts, and 
 would facilitate the pitting action. An explanation of this kind 
 falls very far short of a satisfactory answer, however, in either 
 theory or practice, as it is applicable to very few cases. 
 
 Careful experiments made by leading cheinists and a review of 
 the whole subject go to prove quite conclusively that pitting 
 occurs most frequently when gases are found dissolved in the 
 water, oxygen and carbonic acid gas being the most active 
 agents, and more especially when present together. These 
 gases under certain conditions are very soluble in water, and in 
 examining a water for boiler feed purposes require just as careful 
 consideration as does dissolved solid. 
 
 Atmospheric air dissolves in water very readily, and contains 
 at all times a certain percentage of carbonic dioxide, and here a 
 very peculiar fact is to be noted that has an important bearing 
 on the point before us. It is a well known fact that atmospheric 
 air is composed of nitrogen and oxygen in mechanical combina- 
 
 M 
 
112 
 
 tion, four of nitrogen to one of oxygen by volume, or very nearly 
 79 volumes of nitrogen to 21 of oxygen. When atmospheric air 
 is dissolved in v/ater these proportions are found to have changed 
 said to be due to the greater solubility of oxygen in water. 
 
 It is worthy of note that water dissolves very unequal quanti- 
 ties of the different gases, and very une ]ual quantities of the 
 SAME GAS, at different temperatures. One volume of water ab- 
 sorbs at the temperatures stated in the following table under at- 
 mospheric pressure equal to 30 inches of mercury, according to 
 Bunsen, the following volumes of the different gases : 
 
 Degrees 
 Fah. 
 
 Oxygen 
 (0) 
 
 Nitrogen. 
 (N) 
 
 Hydrogen 
 (H) 
 
 Carbon 
 
 Dioxide 
 
 C O2 
 
 Chlorine 
 CI 
 
 Sulphur 
 
 Di' xide 
 
 SO2 
 
 Hydric 
 
 Chloride 
 
 HCl 
 
 3< 
 68° 
 
 .041 
 
 •033 
 .028 
 
 .020 
 .016 
 .014 
 
 .019 
 .019 
 .019 
 
 1.80 
 
 1.18 
 
 .90 
 
 2.59 
 2.16 
 
 53-9 
 
 36.4 
 27-3 
 
 472 
 441 
 
 While not altogether applicable to our point, it may be stated 
 that as the pressure increases a correspondingly larger quantity 
 of the gas is absorbed, and as the temperature increases the solu- 
 bility c*" the gas decreases in many cases. (Carefully refer to 
 table on this point.) It is, however, important to note that the 
 pressure which determines the rate of absorption of a gas is not 
 the general pressure to which the water or other absorbing liquid 
 is exposed, but that pressure which the gas under consideration 
 would exert if it were alonk present in the space in which the 
 absorbing liquid is in contact. It is necessary to bear this in 
 mind to understand why atmospheric air absorbed by water 
 differs in composition from ordinary air. 
 
 As I have said, atmospheric air consists of 79 parts of nitrogen 
 and 21 parts of oxygen. In atmospheric air, acting under a 
 pressure of 30 inches of mercury or one attnosphere, the oxygen 
 exerts a partial pressure of ,Vff» and the nitrogen ^VV At 50** 
 F, according to table, one volume of water at atmospheric pres- 
 
"3 
 
 sure absorbs .033 volume oxygen and .016 volume nitrojjeii, sup- 
 posing- these gases to be in a pure state. Then under the partial 
 pressures above indicated water cannot absorb at 50° F. more 
 than ,Vti X •033 = .oo7 volumes of oxygen and /^''^ x .016 = .01 3. 
 volumes of nitrogen. 
 
 Then, in .007 + .01 3 = .020 volumes of atmospheric air absorbed 
 by water there are consequently .007 oxygen and .013 nitrogen^ 
 or in 100 volumes, 35 volumes of oxygen and 65 volumes of nitro- 
 gen. Atmospheric air then dissolved in water is seen to be very 
 much richer in oxygen than is the air itself in its ordinary con- 
 dition. This fact is in itself worthy of the most careful considera- 
 tion, and explains clearly what is too often considered purely ac- 
 cidental phenomena. 
 
 Before proceeding to d<scuss what faction takes place during 
 the act of pitting, I wish to digress from the subject to point out 
 to the reader that the capacity of water to absorb gases of fixed 
 chemical compounds, such as carbonic dioxide (COo), sulphur 
 dioxide (SOo), and hydric chloride (HCl), mast not be confounded 
 with the behavior of atmospheric air. Thus hydric chloride mi«st 
 not be understood as being absorbed by water in the same pro- 
 portions as its elements, hydrogen and chlorine. It is a complete 
 gaseous chemical compound of itself, and combines as such ; m 
 other words, nitrogen and oxygen retain their own peculiar prop- 
 erties tn atmospheric air, but just as soon as hydrogen and chlor- 
 ine combine together to foim hydric chloride, an entirely new 
 chemical compound is formed, with quite different properties from 
 the elements of which it is composed, and with entirely different 
 degrees of solubility. 
 
 When water is strongly heated for a long tune, atmospheric 
 air is completely expelled, and at once resumes its normal com- 
 position ; thus we have present a quantity of free oxygen, which 
 is the cause of a great deal of our trouble. While the action of 
 oxygen on iron is well understood, it is not easy to explain the 
 action of gases producing pitting in a steam boiler, particularly 
 
 m 
 
•'4 
 
 |i 
 
 I;! 
 
 
 I 1 
 
 k 
 
 to explain why isolated spots are attacked while surrounding- 
 metal remains unaffected. If we heat water gradually in a glass 
 retort, open to the atmosphere, bubbles of gas or air expelled 
 from the water are seen clinging to the sides and bottom ; and if 
 we purposely scratch the inner surface of the giass, or in other 
 manner create an imperfection of a similar nature, it presents a 
 wonderful attraction for these little bubbles, more collecting there 
 than at any other part of the retort. And so tenaciously do these 
 bubbles cling to their selected places that it requires considerable 
 agitation to remove them. 
 
 In steam boilers the same conditions prevail, and there remains 
 but littie doubt that these bubbles contain free oxygen and other 
 gases in a favorable condition to act on the metal beneath. 
 Temperature being high, they necessarily act with a great deal 
 of rapivJity. When the oxygen has combined with the iron of the 
 metal, protoxide of iron (FeO) is formed, and a little excresence 
 appears on the surface of the metal, due to the expansion of the 
 iron by the addition of oxygen, and occupying more space than 
 the iron acted upon. Subsequent expansion and conti action 
 causes this small excresence to burst, and leaves a small pit or 
 roughened surface, which readily attracts the gas bubbles, and 
 the process is again repeated un'il finally d, hole has been eaten 
 a considerable distance into or completely through the part 
 <iffectad. 
 
 S-ia. 
 
Chapter II. 
 
 When we understand what is taking place during- the act of 
 pitting-, it becomes comparatively easy to sug-g-est a remedy. 
 What is required here is a substance on which the corrosive agrnt 
 can exhaust itself; and probably the best remedy is a slight coal- 
 ing of lime scale, which can be easily obtained by the use of lime 
 water or milk of lime. 
 
 This theory has not, however, been universally accepted as 
 the true cause of pitting (although in the writer's opinion it 
 appears very rational). The cause of this trouble has been 
 attributed by many writers to electrolytic action set up between 
 the impurities of the iron and the iron of the boiler itself, which, 
 being electro-positive to the impurities, gradually disappears 
 while the impunities remain to further the disintegrating action. 
 In support of this theory it became a common practice to suspend 
 zinc in boilers subject to pitting and corrosion, on the assumption 
 that since zinc is chemically more powerful than iron, it would 
 become electro-positive to the iron and thus protect the boiler 
 against electrolytic action at the expense of its own metal. If 
 such is the case, then it becomes very necessary for us to note 
 what may be expected to take place when zinc is suspended in 
 <he boiler as suggested. 
 
 To get as olear an understanding as possible of this process, let 
 us briefly illustrate what takes place in a voltaic cell, which is 
 practically what we are converting our boiler into with zinc and 
 shell as electrodes and water and its impurities as electrolytes. 
 To this end let us examine the prccess known as electrolysis, 
 taking for example a dilute solution of hydric chloride, subjected 
 to decomposition by means of electrolysis in a gla«^s vessel so 
 that process may be noted. We find chlorine evolved a: the 
 

 1 16 
 
 positive pole and hydrogen at the neg-ative pole, the g-ases being- 
 perfectly pure and unmixed. If, while the decomposition of the 
 hydric chloride is rapidly proceeding, we examine the interven- 
 ing liquid carefully, no apparent movement or disturbance of any 
 kind can be perceived, and nothing in the shape of currents in 
 the liquid or transfer of gas bodily from one pole to the other can 
 be observed, and yet two portions of the hydric chloride separated, 
 by two to six inches are actually seen to evolve pure chlorine and 
 pure hydrogen. 
 
 There seems but one method of explanation of this singular 
 phenomenon, which is applicable to all similar cases of electro- 
 lytic decomposition, and that is to assume that each of the par- 
 ticles of hydric chloride intervening between the electrodes cr 
 poles, and by which electiic current is conveyed, simultaneously 
 suffer decomposition, the hydrogen travelling in one direction and 
 t!ie chlorine in the other. The neighboring elements being in 
 close proximity t*^ each other, unite to again form hydric chloride 
 and destined to be again decomposed by a repetition of same 
 ch-^nge. Thus it is each particle of hydrogen may be made to 
 t- ' in on^ direction by becoming successively united to each 
 particle of chlorine between itself anrl the negative pole ; and 
 when it reaches the latter, finding no disengaged particle of 
 chlorine for its reception it is rejected, so to speak, from the 
 series and thrown off in a froe state. 
 
 Exactly the same thing happens to the chlorine which is travel- 
 ling in the opposite direction, consequently a succession of par- 
 ticles of hydrogen are thrown off at the negative pole and a cor- 
 responding succession of panicles of chlorine at the positive. The 
 power of the current is exerted with equal energy in every part 
 of the liquid conductor, though its effects become manifest only 
 at the terminals. The action is purely of an internal natuo, and 
 the metallic electrodes or terminals of the battery simply serve 
 the purpose of completing the connection between the latter and 
 the liquid to be decomposed. 
 
117 
 
 Like hydric chloride, all electrolytes, when acted upon by 
 Electricity, are split into two constituents, which pass in opposite 
 ■directions. Substances of the one class, sucli as chlorine, iodine, 
 oxygen, etc., ai6 evolved at the positive pole, and those of the 
 other class, such as the metals and hydiogen, at the negative 
 pole. It is, though, important to remark that oxygen salts, such 
 as the sulphates and nitrates, when acted upon by the current, do 
 not divide into acid and basic oxide, but into a metal and com- 
 pound substance or group of elements. Thus sulphate of copper 
 {Cu SO4) does not split into SOy and Cu O, but into metallic 
 •copper (Cu) and sulphione (SO.) sulphuric acid (H2 SO^) dividing 
 into the same compound and h^^.jgen. In a similar way it is 
 ^een that the part of the electrolyte passing to the negative pole 
 may be a compound consisting of several elements. An illustra- 
 tion of this change is the commonly used sal ammoniac (NH^ CI) 
 which is decomposed in such a manner that the ammonium (NH^) 
 goes to the negative pole, when it is resolved into ammonia 
 {NH.,), the pure hydrogen and the chlorine going to the positive 
 pole. 
 
 We must, nevertheless, d^aw a distinct line between true and 
 regular electrolysis and secondary decomposition brought about 
 by the reaction of the bodies eliminated upon surrounding liquid 
 or upon the substance of the electrodes. When, for example, 
 potassium sulphate (K2 SO^) is electrolysed, hydrogen appears at 
 the negative pole together with its equivalent quantity of potas- 
 sium hydroxide (K OH), because the potassium which is evolved 
 at the electrode immediately decomposes a portion of the water 
 there present. The chemical reaction takes place as follows : 
 
 K, + 2 H., = 2 KO H + 2 H. 
 
 At the same time the sulphione (S O^) which passes to the posi- 
 
 <ive pole takes up hydrogen by decomposing the surrounding 
 
 vater forming sulphuric acid (H^ SO^) and liberating oxygen 
 
 according to following reaction : 
 
 SO4 + H3 = H^ SO4 + O. 
 
 
ii8 
 
 In like manner the sulphuric acid itself is resolved by the 
 current into hydrogen and sulphlone, which latter decomposes 
 the water at the positive pole reproducing' sulphuric acid and 
 liberating oxygen just as if the water itself were directly decom- 
 posed by the current into hydrogen and oxygen. This explains a 
 circumstance that was very puzzling to our early experimenters 
 upon the chemical action of a voltaic battery. The true source 
 of these compounds was traced by Davey in 1807, who, showed 
 very clearly that they proceeded from impurities in the water, or 
 the vessel which contained it, or from surrounding air. He redis- 
 tilled witer at a temperature below boiling, and found when 
 marble cells were used to contain the water intended for decom- 
 position hydric chloride appeared at the negative pole and sodium 
 at the positive, but derived from sodium chloride present in the 
 marble as an impurity. This manifestly proves that water in a 
 pure state is not an electrolyte and is incapable of conveying 
 current, but is enabled to do so if even a trace of saline matter is 
 present. 
 
 We have now, I hope, clearly expounded the principles of the 
 voltaic cell. Now, let us consider its practical application for the 
 purpose we have in view. Suppose we have a wire condiTcting 
 an electrical current ; cut this wire and place both ends in a cell 
 containing water to which we have added some sulphuric acid to 
 add to its conductivity. It will be observed that bubbles of gas 
 are forming on the ends of the wires. These bubbles are simply 
 hydrogen and oxygen obtained as alreac'y described. Let us 
 now select for our poles, zinc and wrought iron, to get as nearly as 
 possible the conditions existing within our boiler. Ifourzincis 
 pure no perceptible action will take place on immersion in the 
 cold solution. We next immerse our wrought iron plate, and find 
 chemical action is at once set up, hydrogen bubbles rising and 
 metal gradually wasting away. Let us now connect the two 
 wires fi"om the source of electrical energy to the dry ends oi' our 
 plates, and a complete change of conditions at once takes place. 
 
119 
 
 The hydrogen will still be seen to rite from the wrought iron 
 plate, but the plate itself will cease to dissolve , But (he zinc 
 plate wil! gradually disappear, and just as long as any part of it 
 remains in contact with the water and in electrical contact with 
 the iron, the iron will be protected from decomposition at the ex- 
 pense of the zinc. We can now, I think, perceive clearly the 
 reason for the gradual wasting away of the zinc or positive pole. 
 As the water is decomposed by the passage of the current, the 
 oxygen is attracted to the positive pole, and being practically 
 electrolytic, oxygen or ozone attacks the metals, very actively 
 and destructively, and we find them more or less rapidly wasting 
 away. 
 
 We have now before us information to enable us to understand 
 the action of zinc in boilers, and can readily see that zinc can only 
 be successfully used where certain conditions pre-exist. W^e 
 must first have some solution that makes a good electrolyte, and 
 next see that contact between zinc and shell of boiler is electrically 
 good; further, we must be sure that galvanic action is set up 
 within the boiler. I think it has been amply proven in engineering 
 practice that gau'anic action is set up within the boiler due to 
 many causes, such as difference in tempciature in different parts 
 of the boiler, thereby causing a current, and generating electricity 
 due to friction ; also galvanic action is set up between the different 
 kinds of metal used in and about a boiler, such as composite fit- 
 tings and steel shell, etc. It is also quite possible to create gal- 
 vanic action between different parts of the same boiler when a 
 certain portion of the metal is homogeneous, while another is 
 poorly made, due to bad iron, etc. In fact, galvanic action may 
 be set up in a multitude of ways, and it is this very fact that 
 makes the use of zinc in boilers, especially in fresh waters, rather 
 to be condemned than approved, especially when action is not 
 clearly understood. It has lately become a very common prac- 
 tice, especially in marine work, to purposely set up galvanic 
 action within the boiler, by the aid of an external current, to pre- 
 
 > I 
 
 ' J 
 
 m I 
 
120 
 
 
 vent not only corrosion, but the formation of scale, and the system 
 has met with a great deal of success, as it ^ <^ry properly would 
 when conditions are favorable. Marine engineers are, however, 
 exceptionally favored in this respect. Not only does their feed 
 water contain a large quantity of saline matter, but also a certain 
 proportion of free acid, making the solution within the boiler an 
 excellent electiolyte ; consequently they have only to see that 
 contact between their z'luc and iron electrodes is correct and gal- 
 vanic action set up to secure good results. 
 
 The first essential required is a thorough knowledge of action 
 of zinc in boilers, and secondly, a knowledge of conditions pre- 
 existing. Before leaving this subject I wish to point out to the 
 reader that nothing but the purest zinc must be used. Ordinary 
 commercial zinc contains many impurities, and especially must 
 lead be absent. If lead is present it will be found after zinc has 
 wasted away, a new condition of things will exist ; the shell of the 
 boiler will become electro-positive and the lead negative, and the 
 cure becomes worse than the disease. 
 
 GENERAL CORROSION. 
 
 This form of corrosion is in some localities a very common one, 
 and is a peculiarly difficult one to explain in a general way. It is 
 largely due to the presence of free acids, which may be present 
 in the feed water, or may exist as a result of the use of some of 
 the vile nostrums sold as boiler compounds, prepared frequently 
 by people who do not know the first requirements of the sub- 
 stance they sell, except that it will take ofiF scale. When car- 
 bonates are absent, free sulphuric acid may be present, especially 
 in mining districts, and chiefly derived from the oxidation of 
 pyrites and sulphides. This acid is also often found in streams 
 passing through manufacturing districts, owing to the refuse from 
 dye works, galvanizing shops, chemical factories, etc., having 
 been discharged into them. 
 
 Another fruitful source of free sulphuric acid is the use of kero- 
 
121 
 
 sene, which is now no commonly used. Kerosene is produced by 
 one of a series of distillations of petroleum, and in its first and 
 crude state is a turbid liquid, having a stronj^^ odor and a bluish 
 ting"e. To remove these, the refineries give the kerosene a course 
 of treatment with sulphuric acid. Part of this, together with im- 
 purities, precipitates as a heavy black sludge. Tlie purified oil is 
 next treated with caustic soda, to neutralize any free acid, and 
 then washed with water. Like many other processes of a similar 
 nature, work is not always well done, and free sulphuric acid 
 remains in solution. The oil itself vaporizes very readily when in 
 the boiler and passes off with the steam. The sulphuric acid 
 being non-volatile, remains behind, and finally becomes sufficiently 
 concentrated to attack the boiler, and corrosion sets in. 
 
 A very prolific source of corrosion is also caused by the 
 presence of hydr'^chloric acid in the water. This is a very fruit- 
 ful source of trouble in waters containing magnesium chloride in 
 solution, and is often caused in waters containing other salts of 
 magnesia by the use of compounds which reduce salts of mag- 
 nesia to magnesium chloride. At all temperatures of present 
 boiler practice, magnesium chloride decomposes in y. esence of 
 water in accordance with following chemical equation : 
 
 Magnesium 
 
 Chloride. 
 
 Mg. Clo 
 
 + 
 
 Magnesium 
 Water. Hydrate. 
 
 2H.,0 = Mg. 0,H, 
 
 Hydrochloric 
 Acid. 
 + 2H CI. 
 
 Reference has been made to the formation of magnesium chlor- 
 ide, and consequentlv the liberation of hydrochloric acid by the 
 use of compound resolvents. A striking example of this is the 
 practice of adding sal ammoniac, N H^ CI., to the feed water to 
 dissolve carbonate of lime, the reaction in this case being : 
 
 Carbonate of 
 Sal Ammoniac Lime. 
 
 2N H^ CI. + CaCOg 
 
 Chloride of 
 
 Lime. 
 
 = Ca CL + 
 
 Carbonate of 
 Ammonia. 
 (N H J, C03. 
 
 W 
 
 rr 
 
 The final result Is that a soluble salt added to a water containing 
 
122 
 
 an insoluble salt, the insoluble salt is rendeied soluble, and con- 
 sequently no precipitation takes place. Unfortunately, the reaction 
 does not stop here ; waters containing; carbonate of lime invari- 
 ably also contain carbonate of magnesia, and the following" reaction 
 is set up between the sal ammoniac and ihe carbonate of magnesia: 
 
 Carbonate of Carbonate of Chloride of 
 
 Magnesia. Sal Ammoniac. Ammonia. Magnesia. 
 
 Mg. Co, + 2NH^C1. = (NH J, Co., + Mg. CI... 
 
 Hydrochloric 
 
 Acid. 
 
 2H CI. 
 
 This latter salt undergoes a secondary decomposition as I'ollows: 
 
 Chloride of Hydrate of 
 
 Magnesia. Water. Magnesia. 
 
 Mg. CI., + 2ll,0 = Mg. 0.,H., + 
 
 The action of hydrochloric acid differs materially from the 
 action of the sulphuric acid, owing to the fact thai it volatizes 
 very readily, and as a vapor passe^. otf with the steam ; thus, not 
 only will the boiler itself be attacked, but also the steam pipes, 
 and even the engine. This acid, at temperatures higher than 
 boiling, even in dilute solutions, attacks iron very readily, and is 
 a fruitful source of corrosion and its companion troubles. 
 
 Vegetable acids, or, properly speaking, organic acids, while of 
 a feebler nature, will attack iron to some extent, and particularly 
 since these acids are derived from waters usually taken from 
 swamps containing i- great deal of vegetable organic matter with 
 very little lime salt*,. In cold climates the water is also apt to 
 contain a considerable quantity of free oxygen, which adds ma- 
 terially to the corrosive action of the acids present. The presence 
 of the slimy organic precipitate common to waters of this class 
 seems to add to, rather than retard, the work of corrosion. 
 
 Whole chapters might be written on the best methods of treat- 
 ing troubles of this kind, but I consider the best advice is to con- 
 sult some one thoroughly trained in this particular subject, who 
 is not only able to suggest a remedy, but can determine accurate- 
 ly the nature and the cause of trouble, and the slight expense in- 
 curred will never be regretted. 
 
Chapter III. 
 
 Having discussed the presence of corrosive ag^ents in solution, 
 we now require to turn our attention to the presence of impurities 
 likely to occur as dissolved solids and solids held in mechanical 
 
 suspension, some of which are liable to form scale or encrustation. 
 The total solids contained within a boiler feed water may be in 
 either one of two conditions, in mechaniciil suspension, as small 
 particles, or in chemical solution ; and water may contain im- 
 purities as per followinjif table : 
 
 In mechanical suspension : 
 
 In solution : 
 
 f^ • , ( Annual. 
 
 Oreanic maiter { ,. . , , 
 
 *» (veg^etable. 
 
 T • *. f As sand, 
 
 Inoreranic matter i j . 
 
 *» [ mud, etc. 
 
 ^ . . . r Animal. 
 
 Orcfanic matter { ,. 
 
 » \ \ ej 
 
 jgetable. 
 
 , . .4 f Scale forminer. 
 
 Inorefanic matter ^ ^t i r • 
 
 =• (^ i\on scale tormmg'. 
 
 Matter in mechanical suspension, whether org^anic or inorganic, 
 may be removed by filtratioji, but matter in chemical solution, as 
 dissolved impurities, cannot be removed by mere filtration, but 
 must receive treatment either by chemicals or heat, the aim being" 
 lo get a precipitation whereby we have the impurities in condition 
 first named. 
 
 I shall not attempt to deal with the whole of the chemical com- 
 pounds that may at times be found in natural waters, but confine 
 myself to those compounds of common occurrence with which we 
 are all well acquainted. These may be classed as the oxides, 
 chlorides, carbonates, bi-carbonates and sulphates, being the 
 compouiids formed by the union of the various acid radicles with 
 
 J] 
 
134 
 
 i 
 
 ; 
 
 the basic radicles or metallic elements, those most commonly 
 found beinjj^ silicon, aluminum, iron, calcium, magnesium, sodium 
 and potassium. 
 
 Before proceeding to discuss the nature of the scale formed, or 
 the action of these various compounds within the boiler, it is first 
 necessary to briefly explain their presence in the water. One of 
 the commonest and most important of these groups of salts, and 
 a group with which we are most fret|uentlv called upon to deal, 
 is the carbonates. They are very widely distributed, being found 
 in considerable quantities nearly the world over. Carbonate of 
 calcium is found in the well-known forms of limestone, marble, 
 chalk, coral deposits, etc. Carbonate of magnesium occurs quite 
 frequently as magnesite, and in combination with carbcJnale of 
 calcium very largely as dolomite. 
 
 Carbonate of iron occurs less frequently, and not so largely 
 diflfused in nature as siderrite or spathic iron. 
 
 The carbonates of the alkaline bases, sodium and potassium, 
 occur very largely in the alkaline districts, but being soluble at 
 all temperatures found in boiler practice, do not play such an im- 
 portant part in engineering practice as scale-forming agents as 
 do the carbonates of the alkaline earth. 
 
 The whole of the carbonates are formed by the imion of car- 
 bonic dioxide (Co.^) with some one of the metallic oxids, such as 
 calcium oxid (CaO), magnesium oxid (Mg O), sodium oxid 
 (NaoO), etc. 
 
 The usual chemical action taking place may be classed as a 
 double one, carbon dioxide (Cog) first combining with water 
 (HoO) to form carbonic acid (HgCOg). The metallic oxids just 
 named enter into combination wi*^- carh«n'ic acids as follows : 
 
 Calcium Carbonate. Water. 
 
 Calcium Oxid. Carbonic Acid 
 
 CaO + H2C03 = CaCog 
 
 and Magnesium 
 
 Magnesium Oxid. Carbonic Acid. Carbonate. 
 MgO + H2CO3 = MgCo., 
 
 + 
 
 HjO. 
 
 Water. 
 H,0. 
 
•^5 
 
 In this case tfie bi-valoiit basic radicles of calcium and mii^- 
 nesiiim replace tiie hydrogen of the carbonic acid and the liber- 
 ated hyuiogen enters into combination with the oxygen of the 
 metallic oxid ; consequently from the combination of twochemica' 
 comjHninds, as shown in equation, two new compounds, consist- 
 
 ing of either carbonates of cah 
 
 id water, 
 
 calcium or magnesumi ana waiei, 
 are formed. The combination between the oxide of sodium and 
 carbonic acid is practically the same, with this exception : Sodium 
 is a univalent metal, consequently has the power to replace only 
 one atom of hydrogen. The chemical reaction for the production 
 of sodium carbonate therefore is : 
 
 Sodium Oxid. Carbonic Acid. 
 Na.O + H.,Co., 
 
 Water. 
 H.O. 
 
 Sodimn Carbonate. 
 
 As a rule, out of all the scale-forming material found in feed 
 waters, none occurs so frequently, nor yet in such large propor- 
 tion, as does carbonate of calcium ; next in order carbonate ot 
 magnesium may be said to occur. Carbonate of iron occurs in 
 limited areas and never in large quantities. 
 
 The carbonates of calcium and magnesium are \eiy slightly 
 soluble in water, and very rarely exist as such in quantities ex- 
 ceeding two grains per imperial gallon, and it is very rarely that 
 they are present in the water separately, as ihey usually exist in 
 water together, and in this case the total held in solution never 
 exceeds the maximum quantity of either which the water is cap 
 able of dissolving. This small quantity of dissolved solids would 
 have but little effect in a boiler feed water, and draws our atten- 
 tion to a very important fact. If the carbonates of calcium and 
 magnesium are, as stated, nearly insoluble in water, how is it 
 that they predominate as scale-forming impurities and are so 
 often found in rhemical solution in boiler feed waters in such large 
 quantities? 
 
 If we turn back to the article on dissolved gases, we shall find 
 that many gases are very soluble in water, and it is this fact that 
 
 ■i] 
 
126 
 
 plays such an important part in explaining^ rhe presence of insoluble 
 carbonates. 
 
 Wherever org-anic decomposition or combustion is taking place, 
 carbonic dioxide (CO,.) is being constantly evolved. This at or- 
 dinary pressures and temperatures exists as a g'as, and is there- 
 fore a constant constituent of the atmosphere ; it is also contained 
 in the soil as a product of organic decay. 
 
 J. A. Wanklyn, in an excellent work on '* Water Analysis," 
 published in London, England, states that "in many natural 
 waters there is more carbonic acid gas in some form or other than 
 any other snigle fon 'gn material." As already seen, carbonic 
 dioxide is very soluble in water ; some is absorbed by frilling rain, 
 and a still greater quantity as it passes through the soil. We 
 have already noted the very wide distribution of the various car- 
 bonates in nature, and although the carbonates of lime and mag- 
 nesia are nearly insoluble in water, they are very soluble in water 
 containing carbonic acid. 
 
 The carbonates of lime aiid magnesia appropriate a portion of 
 the carbonic acid equal in quantity to that already existing in 
 combination with their oxids to form the carbonate, thus forming 
 what is known in chemislry as a. bicarbonate, jiccording to the 
 following equation : 
 
 Bicarbonate of Lime. 
 Ca H,(CoJ,. 
 
 Bicarbonate of 
 Carbonic Acid. Magnesia. 
 
 HgCo, = MgH,(Co,),. 
 
 Both Ihe bicarbonates ot lime and magnesia are very soluble 
 in water, and the presence of carbonate of lime and magnesia in 
 boiler deposits may then be explained as follows : The rain reaches 
 the earth charged with a certain quantity of carbonic acid, and 
 in passing through the soil takes up a still greater supply, and 
 after passing through the earth rarely escapes contact wi<h sotne 
 of the carbonate locks, which dissolve in it. After passing over 
 
 Carbonate of Lime. 
 
 Ca Co, 
 
 + 
 
 Carbonic Acid. 
 H, Co, 
 
 and 
 
 Carbonate of Magnesi.i. 
 Mg Co., + 
 
127 
 
 these rocks and forming new solutions, the water again reaches 
 the surface and forms tlie supply of our rivers, lakes or wells, 
 from which we take our boiler feed supplies. These bicarbonates 
 are very unstable salts, and are much more soluble in cokl water 
 than hot, owing to the fact that a rise in temperatuie drives off 
 the excess of carbonic acid, and the bicarbonate is as a conse- 
 quence reduced to the carbonate, in which condition it is nearly 
 insoluble. When a temperaUire of i8o' F". is reached, a large 
 percentage of the carbonic acid :s driven off, and the bicarbonate 
 consequently » uffers reduction to the carbonate, and at 290" F. 
 the precipitation is complete. Thus we are readily enabled to ac- 
 count for the presenceof the carbonate salts in b(/iler encrustation. 
 
 Next to the carbonates, the sulpliales play an important part 
 in feed waters. The sulphates of magnesium, sodium and potas- 
 sium are all readily soluble in water, and although they play an 
 important part in the treatment of boiler feed waters, need not 
 be referred to here. 
 
 I'robably no scale-forming salt has given more trouble or been 
 more destructive to boilers than has sulphate of calcium. Sul- 
 phuric acids readily attack nearly the whole of the metallic oxids, 
 and especially those under consideration, forming sulphates and 
 water, thus : 
 
 Calcium Oxide. Sulphuric Acid. Calcium vSuljihate. VV^ater. 
 CaO + H.SO, = CaSO, + H,0. 
 
 Calcium sulphate, or sulphate of lime, is found in nature widely 
 distributed, occasionally in the anhydrous form as anhydrate, and 
 much more commonly as the hydrate known as gjpsum. When 
 gypsum is heated to about 240^ to 250° F., it loses a large portion 
 of its water of crystallization and becomes what is known to us 
 as pla.^ter of Paris. This prochict is, as we know, capjible of 
 taking up another portion of water and "setting" to a close, hard 
 stony mass. The anhydrous sulphate is nearly insoluble in water, 
 but the hydrated for' dissolves fairly readily and up to about 
 120 Fah.; as the temperature incre-ises the solubility increases. 
 
 » s 
 
is 
 
 128 
 
 Above 212° F. solubility decreases very rapidly, and at 300° F. 
 it becomes almost entirely insoluble. Unlike the carbonates, the 
 presence of carbonic acid gas has no effect upon its solubility, 
 its solubility being entirely due to the solvent power of the water 
 itself. When the hydrated sulphate of lime precipitates it loses a 
 poriion of iis water of crystallization, and when it reaches the 
 boiler plate the balance, and it is consequently converted into the 
 anhydrous state, and this change in crystalline form is largely the 
 cause of precipitates containing sulphate of lime being bound into 
 such hard compact masses and forming such a troublesome scale. 
 
 As already stated, sulphuric acid occasionally occurs in streams^ 
 etc. If bicarbonate of lime or magnesia are present, free acid 
 will not exist until the salts of calcium and magnesium have been 
 reduced from carbonatt s to sulphates. The same reduction takes 
 place when water containing sulphuric acid is passed over lime- 
 stone rocks, and since rain water has been shown bv numerous 
 chemists to contain traces ot sulphuric acid, the presence of sul- 
 phate of lime is largely accoun'ed for. The chlorides of the 
 various metals under consideration are frequently present in feed 
 water, especially sodium chloride, which is simply common salt. 
 Since, however, the whole of the metallic chlorides we have occa- 
 sion to refer to are easily soluble in water and do not form scale 
 except in extreme cases, we do not require to refer to them at 
 any length, particularly since we shall require to discuss them 
 fully at a later stage, chiefly as to their effect on the scale-forming 
 agents. 
 
 The oxides of silicon and aluminum are frequently found in 
 scale, and when present in any quantity form a very troublesome 
 scale. They are not easily soluble in water, but often occur in 
 large quantities where muddy water is used without previous 
 filtration. 
 
Chapter IV'. 
 
 Chemical analysis of a boiler feed water not only enables the 
 chemist to determine the quantity of foreign matter present, but 
 also the condition or form in which it appears as an impurity ; it 
 
 lU 
 
 iblt 
 
 ith certaint^ 
 
 an opi 
 
 nion as to the 
 
 iim to express 
 
 nature of scale that will be formed and quantity that may be ex- 
 pected, and thvis judge as to the suitability or otherwise of a water 
 for boiler feed purposes. or purposes of illustration I have 
 chosen a number of characteristic samples of water, all of which 
 have, in some respects at least, different properties. 
 
 No I IS a sample of water from an artesian weii in the eastern 
 portion of the city of Montreal that shows an almost entire ab- 
 sence of lime and magnesia salts, but which is still strongly im- 
 pregnated with impurities. A partial analysis shows this water 
 to contain : 
 
 Total solids per imperial gallon 36.42 grains. 
 
 Made up as follows : 
 
 Sodium chloride (salt ) 2.32 // 
 
 Sodium sulphate 6.85 // 
 
 Alkaline carbonates and bicarbonates with 
 
 traces of silica .' -7«25 " 
 
 The whole of the foreign matter contained in this sample of 
 water consists of soluble salts in chemical solution, that is to say, 
 salts that have a very high degree of solubility in water, and con- 
 sequently precipitation of impurities as scale-forming agents can- 
 'Oi occur. Although this is positively a non scale-forming water, 
 , •" innot be classed as a good or even fair water for boiler feed 
 ^.. rposes in its original form. A glance at its composition shows 
 it to be strongly alkaline and liable to cut the boiler and boiler 
 fittings. As a matter of fact, this is what occurred in actual 
 practice, and users found it impossible to use the water satisfac- 
 torily until impurities had been neutralized by chemical treatment. 
 
I30 
 
 i 
 
 No. 2 is a sample of w:«ter from an art-jsian well used to supply 
 the town of Montreal West, and may be classed as a fair boiler 
 feed water. As will be seen, this water is very rich in dissolved 
 solids, and also contains a fair percentage of scale-forming^ ma- 
 terial. Partial analysis shows it to contain : 
 
 Total solids per imperial g-allon 96.2 g"rains. 
 
 Made up as follows : 
 
 Insoluble suspended matter i.oo 
 
 Aluminia and peroxide of iron 20 
 
 Carbonate of lime ^.20 
 
 Carbonate of mag-nesia 3. 17 
 
 Sulphate of soda 13-49 
 
 Chloride of soda 60.60 
 
 Alkaline carbonates and bicarbonates I2'54 
 
 If 
 If 
 It 
 If 
 II 
 
 n 
 II 
 
 Here we have an example of water that actually contains scale- 
 forming agents and at same time material necessary to prevent 
 formation into hard scale when precipitated. This water showed 
 in use results that might be anticipated from its composition, 
 precipitating a heavy insoluble sludge at lowest part of boiler, and 
 after boiler had been in use a short time water became both highly 
 concentrated and strongly alkaline, although the boiler was only 
 evaporating about 700 gallons of water per day ; a sample of 
 water drawn from the boiler three weeks after being put in use, 
 analysis showed it to contain an enormous amount of dissolved 
 solids held in solution, composition of impurities being as follows : 
 
 Total solids per imperial gallon 2796.36 grains. 
 
 Insoluble suspended matter 6.72 // 
 
 Chloride of soda 1940.40 n 
 
 Sulphate of soda 425.28 // 
 
 Alkaline carbonates and bicarbonates 398.48 // 
 
 Hygroscopic water -5'48 " 
 
 So strongly alkaline had the water become that brass fittings 
 on boiler were being attacked, and wherever a small leak occurred 
 deposits formed on outsiile of boiler, which is a marked charac- 
 teristic of the behavior of the soda salts. At no time, however, 
 
'31 
 
 during^ the last four years has any scale been found; a heavy 
 sludge precipitating, having the following composition : 
 
 Silica 6.44% 
 
 Aluininia and peroxide of iron 16.32% 
 
 Carbonate of lime 54.92% 
 
 Carbonate of magnesia 7'40% 
 
 Sulphate of soda 2.37% 
 
 Chloride of soda 10.80% 
 
 Carbonates of soda and potass '•■25% 
 
 The presence of the soluble soda salts in tlie sludge may be ex- 
 plained by remarking that the sludge was of such a consistency 
 that a quantity of water had to be withdrawn at all times with the 
 sludge. Until the composition of impurities contained within this 
 water was known a great deal of difficulty was experienced in its 
 use, particularly by foaming. The only treatment required, hov- 
 ever, is to keep the waiter within the boiler from becoming too 
 concentrated by frequent blowing off, and good result^i are now 
 being obtained from its use. 
 
 Samples i and 2 mark distinct classes ot feed waters. In No. 3 
 we have a sample of water of another type, an excellent water for 
 boiler feed purposes, and used large!}' by both the Grand Trunk and 
 Canadian Pacific railway companies, for use in the boilers of their 
 locomotives. This sample was obtained from Brampton, Ont., and 
 is supplied to the town by means of gravitation from a small lake 
 situate a few miles north of the town. Analysis shows it to contain 
 but a small quantity of dissolved solids and to be remarkably free 
 from lime and magnesia salts, composition of impurities being : 
 
 Total solids per imperial gallon 6.00 grains. 
 
 Insoluble suspended matter 40 // 
 
 Carbonate of lime 2.85 // 
 
 Carbonate of magnesia 30 // 
 
 Chloride of soda 1.20 n 
 
 Sulphate of soda 48 // 
 
 Alumina and peroxide of iron (traces). 
 
 This is a type of feed water that tends to make engineers happy, 
 
 WW: 
 
■I! 
 
 11 mW^^ 
 
 132 
 
 and requires but occasional washing^ out to dispense with either 
 chemical or mechanical treatment. 
 
 Samples No. 4, 5 and 6 show examples of scale-forming- waters 
 in various forms. 
 
 No. 4. — Sample from VValkerton, Ont., shows a water rich in 
 lime and mag^nesia salts existing^ in solution in the water as bi- 
 carbonates and precipitating as their respective carbonates : 
 
 Total solids per imperial g^allon 21 grains. 
 
 Insoluble suspended matter 80 // 
 
 Carbonate of lime 12.12 // 
 
 Carbonate of magnesia 6.66 // 
 
 Chloride of soda 1.20 tr 
 
 Also traces of aluminia and peroxide of iron. 
 
 Sample No. 5. — Sample of water from artesian well in eastern 
 
 portion of the city of Montreal : 
 
 Total solids per imperial gallon 43.7 grains. 
 
 Insoluble suspended matter 2.04 // 
 
 Carbotiate of lime and magnesia '4'3^ " 
 
 Sulphate of lime ' ■2.65 // 
 
 Chloride of soda 9-3^ " 
 
 Alkaline carbonates and bicarbonates 5.31 // 
 
 Sample No. 6. — Sample ofwater from factory at New Toronto, Ont: 
 
 Total solids per imperial gallon 14 1.6 grains. 
 
 Insoluble suspended mattei 1.00 // 
 
 Alumina and peroxide of iron .80 // 
 
 Carbonate of lime 27.48 /; 
 
 // >r magnesia -0,45 // 
 
 Sulphate of lime 37'3^ " 
 
 Chloride of soda 54'6o ^ 
 
 A revie-<> of the impurities contained in these various samples 
 
 shows very clearly how important this part of our subject becomes^ 
 
 and how widely different and varied in quantity are the impurities 
 
 contained in the various feed waters throughout the country. I 
 
 have said that when the nature of the impurities present is known 
 
 that the effects resulting from the use of various waters for boiler 
 
 feed purposes can be ascertained. This is well illustrated by the 
 
 results obtained from the use of the various samples chosen for 
 
^33 
 
 purposes of illustration from a large number of feed waters 
 examined in my laboratory. 
 
 No. I and 2 I have already referred to at length. No. 3 is a 
 class of water occasionally met with and predominating in some 
 districts, and is a type of excellent water for the purpose indicated, 
 and may be used without previous treatment with safety. 
 
 No. 4 is also a good boiler feed water, containing the whole of 
 its impurities in such a condition that a physical change is all that 
 is needed to render greatest portion of impurities insoluble, and 
 consequently in such a condition that mechanical filtration will 
 remove them from the water. As a matter of fact the users com- 
 plain that while they had no scale in their boilers a very heavy 
 scale formed in the heater, sample of which I rc.-eived and analyzed 
 with the following result : 
 
 Insoluble 38% 
 
 Alumina and peroxide of iron 1 . 10% 
 
 Carbonate of lime 92.08% 
 
 n n magnesia 5*6o% • 
 
 Scale was light, very porous, and easily lemoved, and is a good 
 example of the change taking place as set forth in last month's 
 contribution on this subject when the water is sufficiently heated 
 to drive off excess of carbonic acid and the insoluble carbonates 
 of lime and magnesia are at once precipitated. 
 
 Sample 5 shows a type of water that has actually a double effect, 
 a reaction setting up between the alkaline carbonates and the 
 sulphates of lime and precipitating a sludge, balance of sulphate of 
 lime bonding with carbonates of lime and magnesia and forming 
 heavy scale. 
 
 Sample 6 shows a sample of water that belongs to a class that 
 should not be used for boiler feed purposes if any other water can 
 be obtained, and requires both chemical and mechanical treat- 
 ment, scale formed from this water, as is to be expected from 
 composition of impurities, being exceedingly hard, tenacious and 
 troublesome. 
 
!■ ;^i! 
 
 m.ii\ 
 
 m 
 
 Chapter V. 
 
 Hav TNG determined the nature of impurities contained in the 
 feed water, it next becomes an important consideration as to what 
 will be 'le nature of scale formed when impurities are precipitated, 
 or what effect the water will have on boiler parts provided no 
 scale-forming- material is present. 
 
 It is quite obvious that a water containing- such impurities as 
 those in sample No. i described in last chapter cannot form scale 
 except at very high degrees of concentration, or when water 
 within the boiler has become completely saturated ; then, of 
 course, crystallization will begin, but crystals will ag-ain dissolve 
 when unsaturated molecules of water are brought into contact 
 with the crystals. This form of scale will never be met with 
 except in cases of extreme ignorance or carelessness, and need 
 not be discussed at any length. As will be observed, however, 
 the whole of the impirities in this water are non-volatile in the 
 presence of heal at temperatures met with in ordinary boiler 
 practice, and consequently as water evaporates into steam im- 
 purities are left behind, and a gradual increase in density takes 
 place. The impurities, being soluble in water, remain in solution, 
 and water in boiler becomes more an more saturated. Impurities 
 being alkaline in reaction, a very high deg-ree of alkalinity is in 
 course of time reached. While the alkalies do not, as a rule, 
 attack iron very readily, they attack brass alloys vigforously, re- 
 ducing the metallic elements and causing a great deal of trouble, 
 and in many cases serious damage has resulted. 
 
 Another very important reaction that sets in in waters of this 
 kind is the formation of magnesium chloride, as mentioned in 
 earlier articles, and as a consequence, free hydrochloric acid is 
 liberated, which, being very corrosive even when diluted, no part 
 of the system from boiler to exhaust on engine is quite free from 
 danger of corrosion from this source. 
 
•35 
 
 The conditions under which a boiler feed water is to he used 
 plays a very important part in the treatment required and the 
 effect a water will have on the boiler parts. For instance, where 
 jj^ood water is scarce, or in plants where economy has been care- 
 fully studied, it has become a coinmon practice to use surface con- 
 densers, draining^ condensed water to a hot well and returning 
 therefrom to boiler, usiriSj over and over ag"ain. In thoroug-hly 
 equipped plants loss by evaporation is very small and quantity of 
 fresh water used small, consequently the watp» may be returned 
 to boilers many times daily. Waters that are low in dissolved 
 solids in the first instance deposit very little scale-forming" ma- 
 terial. Condensed water from hot well being" practically a distil- 
 late from original water, contains no scale-forming agents, and is, 
 in the absence of oil practically pure. This may, at first glance, 
 be considered a very happy condition of things ; unfortunately, 
 practice has shown it to be in many cases the reverse. Instead of 
 scale forming, the boiler parts are very often found to have suf- 
 fered from pitting and corrosion. Explanation of this condition 
 appears difficult, and corrosion of boilers has often been charged 
 to presence of organic acids, when, as a matter of fact, the purity 
 of the water itself may be said to be responsible. When water is 
 sent to hot well it is brought into contact with atmospheric air, 
 and, being pure, readily absorbs or takes into solution some of 
 the gaseous elements contained therein. Reference to my first 
 article on this subject will readily explain the cause of pitting or 
 corrosion due to the presence of free oxygen in the water. 
 
 We have already seen thcit the most troublesome scale-forming 
 impurities held in chemical solution are the salts of lime and mag- 
 nesia as usually met with in feed waters. 
 
 As conditions of use vary, so will nature of scale formed vary. 
 Particularly is this the case with scale formed from the salts of 
 tnag-nesia. These salts in many cases are peculiarly unstable, 
 and this fact is not sufficiently taken into account either by 
 chemists or engineers. The carbonates of lime are, except in very 
 
 img 
 
 If 
 
11^. 
 
 
 lil ' 1 
 
 'iiii' 
 
 136 
 
 exceptional cases, always accompanied by a certain proportion of 
 the cai bonates oi magnesia. Personally, in my experience I have 
 never found carbonate of mag^nesia entirely absent from a natural 
 water when carbonate of lime was present. I have, as would 
 naturally be expected, found carbonate of magfnesia present when 
 the entire lime was present as sulphate. 
 
 In examination of a boiler .->:'ale, it occasionally happens that 
 the general appearance of a scale is such as would lead the 
 engineer to suspect the presence of a strong bonding agent, such 
 as sulphate of lime, while complete analysis shows the scale to be 
 practically free from sulphates. While the carbonates of lime and 
 magnesia form in presence of organic matter, and especially in 
 high pressure boilers, hard compact scale in many cases, analysis 
 of scale will show that the whole of the bases of these metals can- 
 not be combined with carbonic anhydride. The reason for this is 
 evidently the reducing of part of one of the compounds to an 
 hydro-oxide. vSince magnesia is the least stable of the two bases, 
 it is the writer's opinion that the magnesium carbonate immediately 
 in contact with the boiler plate is reduced from a carbonate to an 
 hydrate, and carbonic acid gas liberated. When this occurs in 
 presence of sulphate of lime a particularly obnoxious scale is 
 formed. 
 
 A sample of scale sent to my laboratory was of a very trouble- 
 some nature, being hard and tenacious, and having very high 
 power as a non-conductor of heat ; while it was, according to 
 sample, less than ^-inch thick, still its construction made it ex- 
 tremely dangerous. Subsequent analysis proved it to have the 
 following composition : 
 
 Silica 9.66% 
 
 Alumina and peroxide of iron 3-44% 
 
 Sulphate of lime 57'37% 
 
 Carbonate of magnesia 27. 1 2% 
 
 Magnesium hydro-oxide 2.07% 
 
 99.66% 
 
•37 
 
 In this case a portion of the magnesium carbonate throujfh the 
 action of heat had been reduced according- to following equation : 
 
 Mg CO, ' heat = Mg O + Co.., 
 
 the oxide of magnesia thus formed then combining with water 
 to form the hydro-oxide, which is only partially soluble in water. 
 
 Mg 0+H,0=Mg(OH),. 
 
 The hydro-oxide is extremely fine, and under conditions existing 
 within the boiler forms a ready bonding material, and especially 
 so in presence of sulphate of lime. 
 
 The carbonates of lime and magnesia, when precipitated at such 
 a low temperature, form porous, easily-removed scale, having very 
 low specific heat, consequently cannot be classed as equally 
 dangerous with those scales of more compact con;- 'ruction. 
 Some idea of nature of scale formed can be arrived at f. om a 
 study of the physical and chemical properties oi' the following 
 samples of scale analyzed in my laboratory. 
 
 The first is a sample of scale formed from water containing- salts 
 of both lime and magnesia, and both hard and tenacious, contain- 
 ing the following parts in loo : , 
 
 No. I. Supplied from Montreal, Que : 
 
 Silica 1.56 
 
 Alumina and peroxide of iron 4-30 
 
 Carbonate of lime 26.95 
 
 Sulphate of lime 52.65 
 
 Carbonate of magnesia 14.20 
 
 No. 2. Sample of scale from New Toronto, Ont. — Shows a 
 rather uncommon condition of scale forming. As seen by compo- 
 sition of scale, a large quantity of mud (greatest portion of which 
 must have been hold in mechanical suspension) was present; this, 
 combined with sulphate of lime, formed a particularly troublesome 
 scale. The composition of impurities in the water from which 
 this scale was formed is illustrated in ^^ample No. 6, preceding^ 
 
( 1 1' 
 
 i Jlil 
 
 i\ M 
 
 •38 
 
 chapter, and is worthy of remark for two ••nportaiit reasons : 
 First, analysis of water shows very little alumina present, conse- 
 quently very little mud was held in suspension, while scale shows 
 a larg'e quantity of alumina. This difference is accounted for by 
 the fact that weather conditions hail much to do with the presence 
 of mud, and sample sent for analysis showed water under best 
 possible conditions. Secotidly, it will be observed that water con- 
 tains both carbonates of lime and mag"nesia, while carbonate of 
 lime is entirely absent from scale. This circumstance at first 
 somewhat puzzled the writer, until he ascertained that theenj^ineer 
 in charg-e was vising refined coal oil to remove scale. As is very 
 often the case, oil contains excess of free sulphuric acid acquired 
 durinjf process of refining' in sufficient quantity to reduce the whole 
 of the carbonate of lime to sulphate, and also a portion of the 
 carbonate of magnesia to sulphate of magnesia, which latter, 
 being soluble in water, does not appear in the scale. 
 
 This sample of scale contains the following- parts in loc : 
 
 Silica 3.98 
 
 Ali.mina, with traces of peroxide of iron 33-^0 
 
 Sulpliate of lime . 53.52 
 
 Carbonate of magnesia 8.78 
 
 No. 3. Sample of scale from Walkerton, Ont. — Light, porous, 
 and easily broken or removed, a typical type of carbonate scale 
 formed at low tempe.aiures. Composition : 
 
 Acid insoluble 38 
 
 Alumina and peroxide of iron 1. 10 
 
 Carbonate of lime 92.08 
 
 Carbonate of mag^nesia 5.60 
 
 Carbonate in presence of organic matter form in many instances 
 very hard scales, degrees of hardness varying with nature of 
 org-anic matter present. 
 
 No. 4. — Sample No. 4, received from London, Ont., is an ex- 
 cellent type of scale of this kind, being very hard and tenacious. 
 The org'anic matter present in this case was largfely composed of 
 
les 
 lof 
 
 Ix- 
 
 Is. 
 lof 
 
 « 39 
 
 oil, which by action of heat has become carbonized, and allhou^h 
 scale was not more than '4 inch ihick, it was of a very dangerous 
 type. Composition : . 
 
 Acid insohible 70 
 
 Alumina and peroxide of iron 75 
 
 Sulphate of lime -.33 
 
 Carbonate of lime 80.07 
 
 Carbonate of magnesia 9.91 
 
 Organic matter 6.04 
 
 99.80 :\ 
 
 Another sample oi scale with nearly the same composition, 
 thicker, but inuch coarser in granular construction, due to differ- 
 ence in nature of organic matter, yielded on analysis : 
 
 Acid insoluble 30 
 
 Alumina and percxide of iron 70 
 
 Sulphate of lime t^.t^i 
 
 Carbonate of lime ^3-75 
 
 Carbonate of magnesia 6.20 
 
 Organic matter 6.01 
 
 100.28 
 
 The loss occasioned by allowing such formations as above to 
 remain on the sheets and tubes of a boiler is so important a factor 
 in economy that it is almost impossible to estimate it. It has been 
 stated on good authority that the formation of an incrustation ,\t of 
 an inch in thickness will increase the fuel consumption 12% over a 
 boiler working under similar conditions with clean shell and lubes. 
 This statement is, however, obviously iii error, since scale of 
 different composition has different heat conducting powers, and 
 loss must vary in accordance with composition of scale, as well as 
 with variation in thickness. The fact remains that the incrusta- 
 tion formed within a boiler has a very much lower conductivity 
 than has the plate of the boiler itself, even under most favorable 
 conditions, so that loss'is bound to occur if incrustation is allowed 
 to accumulate ; and loss is not only loss of heat and consequently 
 
140 
 
 loss of fuel, but serious damage to boiler may occur. Such is the 
 non-conducting" power of many even very thin scales that plates 
 are liable to become over-heated, while scale has a high non-con- 
 tiuctivity, wl.en interposed between boiler shell and water 'A has 
 vary low tensile strength ; consequently, when plates become 
 overheated, internal pressure forces the affected part outward, 
 and boiler is badly damaged, even if serious explosion does not 
 occur. 
 
Chapter VI. 
 Having considered the scale forming properties of various im- 
 pukities contained in natural wuters, it naturally now requires 
 our attention to be drawn to remedial measures. This we can 
 only discuss in a gfeneral way, since conditions of use and im- 
 purities in feed waters vary so widely ; it is obviously impossible to 
 lay down any single method to suit all cases. Perfectly satisfac- 
 tory results can only be obtained by individual examinations of 
 each case, taking into consideration not only condition and nature 
 of impurities in feed waters, but also the condition of service and 
 style of boilers or other heating apparatus in use. 
 
 While we cannot lay down a rule and formula adapted to each 
 case, we can, at least, intelligently discuss the nature of the re- 
 action it is our dgsire to bring about, and in this way, I hope, 
 enable the engineer to have reliable data at hand for his guidance. 
 
 As alre'idy pointed out durijig the early stages of these articles, 
 the solid impurities contained in the feed water may be in eithe*- 
 of two conditions — suspended mechanically, or held in solution. 
 The first of these has quite as important a bearing upon final 
 results as has the latter, and requires just as careful considera- 
 tion. Since, however, matter held in mechanical suspension re- 
 c|uires nothing but mechanical treatment or filtration to effect re- 
 moval, the question does not involve any serious difficulty. The 
 writer cannot impress too strongly, however, upon his readers the 
 very great importance that this kind of impurity has upon forma- 
 tion of scale. It is comparatively easy in many districts, for 
 various reasons, for the pumps to be continually delivering to the 
 boilers varying quantities of finely divided clay, sand, etc., when 
 water on surface of source of supply appears to be perfectly clear, 
 and presence of such impurities is not even suspected until the engi- 
 
■li .^ 
 
 i Jl ' I 
 
 •4^ 
 
 neer finds his boilers coated with thick scale of ? most dangerous 
 type. Where there is the slightest danger of anything like this 
 occurring, engineers should insist upon a first-class filtering medium 
 being placed between source of supply and boilers, and small first 
 cost of a real first-class article of this kind will be amply repaid 
 in a very short time. 
 
 The idea that any kind of water will do for boiler feed purposes 
 so long as it is **wei"is far loo prevalent. The presence of 
 such impurities as are usually met with in mechanical suspension 
 has a very important bearing on scale formation, especially so in 
 the presence of salts of lime and magnesia, and the extraction of 
 matter held in mechanical suspension in very many cases largely 
 tends not only to reduce, bu! to actually prevent the formation of 
 scale, and the first important requirement necessary for the build- 
 ing up of a good feed water is that it should be clean. As a mat- 
 ter of fact, insoluble matter in suspension should in no case be 
 allowed to exceed .5 grains per gallon. If care and judgtnent are 
 exercised a good filter can be easily secured, set up, and kept 
 clean, and a very large proportion of the matter held in mechani- 
 cal suspension removed. 
 
 Impiuities in solution are of quite a different class, inasmuch as 
 they cannot be extracted by mere mechanical filtration, since by 
 being in solution in the water they will pass through the finest 
 filter, and as a consequence before they can be extracted a change 
 of state must take place, and they must be changed from soluble 
 to insoluble salts, whereby they appear in ccndition first mentioned 
 and in a position to be extracted by purely mechanical means. 
 
 Before proceeding to discuss how this can be brought about, I 
 wish briefly to refer to my first article on this subject dealing with 
 the application of electrical current for prevention of corrosion. 
 A great deal has been claimed by various inventors and writers as 
 to the adaptability of methods of this kind for preventing the for- 
 mation of sca'e and effecting its removal after formation. In 
 some instances remarkable success was attained, in manv others 
 
143 
 
 complete failure resulted. The method has not been universally 
 successful, nor can it be, for many reasons. 
 
 I have already pointed out that there are certain scales such as 
 those formed from the carbonates of lime and mag^nesia that are 
 not very compact, but porous and easily penetrated by the water* 
 consequently ihe water easily reaches the boiler shell, and when 
 brought into contact with a voltaic current decomposition of the 
 water as an electrolyte sets in, and the hydrog-en gas bubbles 
 which form on the iron plate very soon form a thin film of hydro- 
 gen gas, interposed between the scale and the plate; as a conse- 
 quence, scale having poor adhesive qualities soon peels off. Such 
 a result cannot be obtained except in the presence of fairly pure 
 carbonate scale. 
 
 Other scales such as are formed from carbonate of lime and 
 organic matter, sulphate of lime, etc., are hard, compact, tena" 
 cious in their hold upon iron, and quite impervious to water. 
 Under such conditions as these decomposition of the water cannot 
 take place in contact with the plate, and consequently the thin 
 film of hydrogen gas cannot form between the boiler and the plate. 
 
 The methods usually adopted to prevent the formation of scale 
 are to render the scale forming impurities insoluble by physical or 
 chemical changes, thus precipitating in nirities either previous to 
 or after reaching boilers. It will be readily seen that before 
 treatment of this kind can be effective certain precautions must be 
 taken. 
 
 First —precipitate so formed must be of such a nature that it 
 will not harden or bind together into scale; next, chen.'cal com- 
 pounds added must also be of such a nature that they themselves 
 will neither form scale nor yet attack the boiler parts, as with 
 either of these it is quite possible to get, as a final result, a condi- 
 tion of affairs worse in eff'ect than would result from impurities 
 originally existing. 
 
 A process for softening water, originating from Scotland and 
 known as the Forler-Clark process, may here be referred to as a 
 
'44 
 
 1)^' 
 
 very useful method for improving^ feed waters containintf the 
 soluble bi-carbonates of lime and magnesia. 
 
 This method consists essentially in adding* to the feed water a 
 quantity of lime water which first combines witli free carbonic acid 
 to form bi-carbonate of lime, and then combininjjf with the excess 
 of carbonic acid contained in the bi-carbonate thus formed as 
 
 /\\.h the exct'fja of carbonic acid already existing as bi-carbonate 
 of lime and magnesia — this final reaction is represented by follow- 
 ing equation (in the case of lime salts, a similar reaction taking 
 place with magnesia.) • 
 
 Lime Water. Bi-Carbonate of Lime. Carbonate of Lime. Water. 
 
 Ca(OH), + Ca(H CO3), = aCa CO3 + 2H,0 
 
 Both the carbonate of lime and magnesia are insoluble at 211° 
 F. , and nearly so at 60' F. , so that the separation of the precipitate 
 becomes mechanically possible. 
 
 As will readily be seen, however, care must be taken to at d 
 just the quantity of lime required to effect precipitation of the bi- 
 carbonate, as carbonate or calcium hydro-oxide will pass on to 
 the boiler and dejiosit lime on the boiler parts as evaporation pro- 
 ceeds, and if in any great excess, a worse scale will be formed 
 than would have been formed if original matter had been allowed 
 to deposit in the boiler. Moreover, since action of lime water is 
 limited to excess of carbonic acid, free or combined with lime and 
 magnesia as bi-carbonates, no reaction will set in between sulphate 
 of lime and lime water, consequently use of this process is limited 
 to waters rich in bi-carbonates, and may be easily detrimental if 
 applied to waters containing much sulphate of lime, particularly if 
 any excess is added, as then the whole of the essential require- 
 ments for the formation of a hard compact scale would be present. 
 
 It was originally my intention to show the reaction that would 
 set up between a nmnber of the many compounds on the market, 
 but I find to do this, subject will become unreasonably long. I 
 will then confine myself to a few of the most commonly used sub- 
 stances. 
 
'45 
 
 Possibly the chief of these is common hydrated carbonate of 
 soda, or what is commercially known as " sal soda." This com- 
 pound is largely used for softening water both of temporary and 
 permanent hardness, reactions setting in wi'h both bi-carbonate 
 and sulphate of lime, as shown in equation : 
 
 Hi-ca,J>nnate of lime Carbonate cf sorla Carbonate of lune Bi-carbonate of soda 
 (soluble). (solubUO- (insoluble). (soluble). 
 
 Ca(HCOJ., + Na.,CO, = Ca C O3 + iNaHCOj 
 
 In this case the basic carbonate of soda absorbs or combines 
 with the excess of carbonic acid combined with carbonate of lime 
 and soluble bi-carbonate of soda, and insoluble carbonate of lime 
 is formed. (Note, the reaction with the magnesia salts is similar. ) 
 
 A reaction also sets in between the sulphate of lime present and 
 
 the carbonate of soda ; thus : 
 
 Sulphate of lime. Carbonate of soua. Carbonate of lime. Sulphate of soda. 
 Ca SO4 f Xa2C03 -f CaC03 i Xa2S04 
 
 Another commonly used and useful reagent is the tribasic phos- 
 phate of soda, which is in many cases to be recomnu nded as a 
 good reagent, its chief property being the ease with which the re- 
 action sets in and the entirely unhardenable properties of the 
 precipitate. With sulphate of lime the reaction is : 
 
 Tri-basic phosphate of soda. Sulphate of lime. Phosphate of lime. SuIpSate of soda. 
 2Na3P04 + 3CaS04 =Ca3(P04)., 4- 3X32804 
 
 The chief objection to use of chemical compoimds of this type 
 is their property of throwing precipitafes down, which, if allowed 
 to accumulate, become very troublesome, and, to overcome this, 
 frequei^t efforts are being tiiade to introduce compounds that will 
 bring about a chemical change but leave the product soluble in 
 water. A great deal may be said for and against use of com- 
 poimds of this kind — particularly against. 
 
 Many compounds of this nature contain a coiTipoimd known as 
 ** sal ammoniac," a salt known in chemistry as chloride of am- 
 monia, or muriate of ammonia, and having the composition 
 NH4CL. 
 
 f 
 
 ^ \ 
 
 ii 
 
m 
 
 140 
 
 The reaction between the lime salts and ammonium chloride is 
 a very distinctive one; taking, for instance, sulphate of lime, weg'et: 
 
 Sulphate of lime. 
 CaS04 
 
 Sal ammoniac. 
 2NH4CI 
 
 Chloride of lime. 
 
 CA CI 2 
 
 Ammoni, in sulphate. 
 
 (NH4).,S04 
 
 + 2l\' H4L;! = UAL 12 ( 
 
 Both sulphate of ammonia and chloride of lime (note i ) are easily 
 soluble in water, and a high degree of concentration would have 
 to be reached before deposit of solid tnatter could take place ; 
 but, as I have already pointed out, natural waters containing lime 
 salts, either sulphate or carbonate, nearly always contain mag- 
 nesia in some form or other, and I may be pardoned for once more 
 referring to the great danger attendant upon the use of sal am- 
 moniac in presence of magnesia. Taking, for instance, magnesia 
 as commonly present as bi-carbonate, a reaction between this salt 
 and sal ammoniac would set up as follows : 
 
 Bi-carbonate of Bi-rarbonate of 
 
 magnesia. Sal ammoniac. Magnesium chloride. ammonia. 
 
 Mg(HC03)., + 2NH4CI = MgCl2 + 2NH4CO3 
 
 Magnesium chloride is very unstable, and in presence of heat at 
 once breaks up into oxide of mcignesia and hydro -chloric acid. 
 The acid, being volatile, passes off with the steam, and plays 
 havoc with steam pipes and engine valves ; while magnesia salt i'^ 
 in even worse shape than it was originally, since oxide of mag- 
 nesia is a very fine substance, entirely insoluble in water, and 
 forming a ready bonding agent. Any compound containing e'en 
 a small percentage of sal ammoniac should be looked upon with 
 suspicion, unless magnesiimi salts are entirely absent from feed 
 water. 
 
 NoTK. ~Ca Cb must not be confounded with the compound usually sold as a dis- 
 infectant under the name of chlo ide of lime, which is a mixture of chlor.de and hypo- 
 chlorite of lime. 
 
tpo- 
 
 Chaptkr V^II. 
 
 Another commonly met compound of a similar type is one 
 containing larg-e percentages of caustic alkalies, usually crude 
 caustic soda. The reaction between sulphate of lime held in solu- 
 tion and caustic soda is similar to reactions already cited. 
 
 Sulphate of Lime. Caustic Soda. Calcium Hydro-Oxide. Sulphate of Soda. 
 
 (Lime) 
 Ca SO4 + 2NaOH = Ca (OH)., + Na., SO^ 
 
 A suTiilar reaction sets up between carbonate of lime and caustic 
 soda. Water within the boiler soon becomes strongly alkaline 
 and attacks brass and composite fittings very vigorously, also is 
 very prone to foam; particularly is this the case if any saponifiable 
 oil finds its way to the boilers. Compounds containing quantities 
 of caustic soda in the absence of neutralizing agents can be with 
 safety avoided. 
 
 Innumerable compounds have been introduced on the market 
 containing organic acids, and some of these have good properties 
 to recommend them. Chief among these are compounds con- 
 taining " tannin " or tannic acid. Rogers' process for the preven- 
 tion of formation of scale consists essentially in the use of sodium 
 tannate, which is a very useful reagent when properly made and 
 applied. A reaction sets in between carbonate of lime and sodium 
 tannate whereby insoluble amorphous tannate of lime is precipi- 
 tated and sodium carbonate is formed, which in time acts upon 
 any sulphate of lime present, reducing it to a carbonate of lime, 
 thus leaving it in a position to be acted upon by a fresh supply of 
 sodium tannate. Such reactions as these have sound chemical 
 reasoning to recommend them. Care should be taken, however, 
 that an excess of acid is not present in the solution, or damage to 
 the boiler will occur. If sodium tannate is hi all pure and made 
 by trained chemists the reaction should always liave sufficient 
 
 •ii 
 
il I 
 
 .48 
 
 iilkali present within the boiler to counteract the injurious effect of 
 free tannic acid. 
 
 It will be seen from the fore^'oing- that boiler feed waters are 
 subject to a wide lanj^e of impurities and in widely apart quanti- 
 ties, so much so thai the use of no single boiler compound can be 
 relied on to attain satisfactory results. Kach particular case must 
 have individual attention. Where methods of this kind have been 
 adopted and practical experience has been brought 10 bear on the 
 subject, wonderful success has been attained. This is evidenced 
 by the almost remarkable success that has been attained by such 
 men as Geo. W. Lord, whose compounds to-day stand so liigh in 
 the annals of steam engineering that genuine compounds bearing 
 this name are n >w accepted without question. 
 
 Treatment by chemicals aims to precipitate scale forming mater- 
 ial as an insoluble, unhardenable sludge which can be easily 
 separated from the water. 
 
 This method has very serious drawbacks, inasmuch as precipi- 
 tation takes place within the boiler itself, and a lliick pasty mass 
 is liable to form directly over the fire box in certain styles of 
 boilers of such a nature as to be quite as dangerous as the forma- 
 tion ot the scale itself. We can then readily see that a successful 
 method of dealing with scale forming waters deinands both 
 chemical and mechanical treatment — chemical treatment to pre- 
 cipitate scale forming agents, and mechanical treatment to extract 
 this precipitate previous to delivering water to the boilers. ' 
 
 Every plant that makes any attempt at economy is provided 
 with some means of heating feed water previous to delivering to 
 boiler. The effect of this heater is to bring about a physical and 
 chemical change on certain of the inipurities contained by the 
 water, and the presence of heat tends to aid precipitation and 
 hasten the action of reagents ; consequently all compounds 
 should be fed into water gradually in sufficient quantity (previously 
 determined) previous to water entering heater, when action will 
 set in. Many of the manufacturers claim to extract the whole of 
 
'49 
 
 the impurities a! the heater by the simple addition of heat, wliile 
 tliey are enabled, if a sufficiently high temperature can be 
 reached, to cause precipitation of impurities. It is not quite as 
 clear that this precipitate can be at once separated. Usually 
 this precipitate passes over with the water to the boiler and is of 
 an extremely fine nature whether chemical or physical means have 
 been chosen to bring about precipitation. To prevent this a 
 filtering arrangement, as shown in Fig. i, is to be recommended. 
 
 Fig. I, 
 
 ■t 
 
 o 
 d 
 e 
 d 
 "Is 
 
 y 
 11 
 
 This arrangement is intended to use two tanks of required 
 capacity into which a definite quantity of reagent has been added 
 and the whole is heated to about 212" Fah. water being fed 
 from each tank alternately to boilers, sufficient time being allowed 
 to lapse to complete chemical reaction before water is passed 
 through filter intervening between tank and boiler. While these 
 tanks are a distinct aid to the efficient working of the filter, they 
 are by no means an absolute necessity. Water could be passed 
 directly from heater to pumps, thence through filter to boiler. 
 With care in choice of a reagent a method of this kind ensures 
 clean boilers and successful treatment of waters otherwise unfit 
 for boiler feed purposes. 
 
 Since, however, in bad waters filter is liable to be given a great 
 
^-JtUTS )U —11 ■IIMHIIIJI 
 
 deal o\' work, it is absolutely necessary that sure and quick 
 nietliods for cieaninjj; of filters are provided, and en)j;'ineer requires 
 to fully satisfy himself on this point before undertaking installation 
 of a purifyinj^ plant on this principle. 
 
 In waters liable to throw down heavy precipitates in the absence 
 of a filterinj^: niedivnn, or in small plants where capital investment 
 becomes excessive for filterinjj^ apparatus, j^freat care should be 
 exercised in the choice of boilers. Construction should be such 
 that .ill precipitates will deposit ou lowest and coolest part of 
 
 Fig. 2. 
 
 boiler at a point easily accessible and easily blowMi out. (See 
 Fig. 2.) 
 
 This shows a type of boiler constructed with a view of securing 
 precipitation of sc.ile-forming impurities before water comes into 
 actual contact with flues and plates of boiler exposed to high 
 temperatures. Water containing the bi-carbonates of lime or 
 
»5i 
 
 inaji^tiesia or other sails that simply ret|uire heat to cause change 
 of state freni soluble to insoluble salts, beinjf fed into the upper 
 portion of boiler or poition used for steam drum, separation and 
 precipitation would take place gravitalinjj^ to lowest point or im- 
 mediately behind bailie plate in front end of boiler, where blow-off 
 is situated. 
 
 A type of boiler that has in the writer's experience ^iven excel- 
 lent satisfaction in this respect is illustrateil in Fi^, \o. 3. 
 
 Hi 
 
 I'lC. 
 
 .)• 
 
 It will be noted that nuid drum is jirovided at lowest point in 
 this boiler, and placed in such a position that no actual contact 
 with hi^h temperature gases can occur; when waters have 
 been treated as suggested, precipitate will follow current to 
 
'5^ 
 
 
 lowest point and then settle. In this case mud drum acts as a 
 receptacle for this precipitate, and bein^ arranged in form indicated 
 by cut, shidjfc or precipitate is concentrated in re.'idy form for blovv- 
 ing"out,and it has been the writer's privilege to have examined boil- 
 ers of this type using- waters that had been treated with chemical 
 compounds and blow-off opened for a short time as occasion re- 
 quired to get rid of sludge, that were practically clean, although 
 they had been in almost contiiuious use for nearly four months. 
 
 Both these boilers show distinctive types of mechanical ar- 
 rangement designed to .assist the water chemist in his work, and 
 in closing these articles I wish to add my quota of thanks to the 
 mechanical genius that enables this important question to be dealt 
 with so successfully. 
 
 i 
 
 --.;i- 
 
TABLE OF 
 
 PROP] 
 
 ERTIE 
 
 :s OF . 
 
 SATUF 
 
 tATED 
 
 bltJ 
 
 KM, 
 
 
 
 (Fkom I'kaiiody's Taiiles ) 
 
 
 
 persq. 
 in. above 
 vacuum. 
 
 u ^ a 
 H.SU. 
 
 ■tth 
 
 0^ = 1 
 
 ^ :r r, c 
 
 
 ^ ^ 3.S 
 
 'Be . 
 
 : 3 
 
 0.-.= . 
 
 2> a 2 
 ii 3 rt f' 
 
 lb V vji 
 
 ■T^ ^ "" 2 
 " 1/1 •* ? 
 
 H art 5 
 
 1 
 
 101.99 
 
 I I 13. 1 
 
 70.0 
 
 1043.0 
 
 0.00299 334.5 
 
 .9661 
 
 I 
 
 2 
 
 126.27 
 
 1120.5 
 
 94.4 
 
 .026. . 
 
 0.00576 
 
 '73-6 
 
 .9738 
 
 2 
 
 3 
 
 141.62 
 
 1 1 25. 1 
 
 109.8 
 
 '015.3 
 
 0.00844 
 
 1.8.5 
 
 .9786' 
 
 3 
 
 4 
 
 ' 53-09 
 
 M28.6 
 
 121. 4 
 
 1007.2 
 
 0.01 107 
 
 90.33 
 
 .9822 
 
 4 
 
 5 
 
 162.34 
 
 "3'-5 
 
 130.7 
 
 .000.8 
 
 0.01366 
 
 73.2' 
 
 .9852' 
 
 5 
 
 6 
 
 170.14 
 
 ••33.8 
 
 138.6 
 
 995.2 
 
 0.01022 
 
 61.65 
 
 .9876 
 
 6 
 
 7 
 
 176.90 
 
 '"35-9 
 
 '45-4 
 
 990.5 
 
 0.01874 
 
 53.39 
 
 ■9897 
 
 7 
 
 8 
 
 182.92 
 
 "37-7 
 
 '5' -5 
 
 986.2 
 
 0.02125 
 
 47.06 
 
 .9916 
 
 8 
 
 9 
 
 •H«-33 
 
 I '39-4 
 
 .56.9 
 
 982.5 
 
 0.02374 
 
 42.12 
 
 •9934 
 
 9 
 
 lO 
 
 193-25 
 
 II 40.9 
 
 .61.9 
 
 979.0 
 
 0.02621 
 
 38. > 5 
 
 •9949 
 
 .0 
 
 '■> 
 
 213.03 
 
 1146.9 
 
 181. 8 
 
 965. ' 
 
 0.03826 
 
 26.14 
 
 1.0003 
 
 '5 
 
 20 
 
 227-95 
 
 "5'-5 
 
 .96.9 
 
 954-6 
 
 0.05023 
 
 19.91 
 
 1.005^1 
 
 20 
 
 25 
 
 240.04 
 
 "55-' 
 
 209. . 
 
 946.0 
 
 0.06199 
 
 .6.13 
 
 1.0099 
 
 25 
 
 30 
 
 250.27 
 
 "58-3 
 
 2.9.4 
 
 938.9 
 
 0.07360 
 
 '3.59 
 
 1. 0129 
 
 30 
 
 35 
 
 259- '9 
 
 1 ibi.o 
 
 228.4 
 
 932.6 
 
 0.08508 
 
 "•75 
 
 1.0.57 
 
 3S 
 
 40 
 
 267.13 
 
 1163.4 
 
 236.4 
 
 927.0 
 
 0.09644 
 
 10.37 
 
 1. 0182 
 
 40 
 
 45 
 
 274.29 
 
 ..65.6 
 
 243.6 
 
 922.0 
 
 0.1077 
 
 9.285 
 
 1.0205 
 
 45 
 
 50 
 
 280.85 
 
 1167.6 
 
 250.2 
 
 917.4 
 
 0.1188 
 
 8.4.8 ..0225 
 
 50 
 
 55 
 
 286.89 
 
 1169.4 
 
 256.3 
 
 9'3-' 
 
 0. 1 299 
 
 7.698 ..0245 
 
 55 
 
 60 
 
 292.51 
 
 1171.2 
 
 261.9 
 
 909-3 
 
 0. . 409 
 
 7.097, 1.0263 
 
 60 
 
 65 
 
 297.77 
 
 II 72.7 
 
 267.2 
 
 905-5 
 
 0.1519 
 
 6.583'.. 0280 
 
 65 
 
 70 
 
 302.71 
 
 "74-3 
 
 272.2 
 
 902. 1 
 
 0.1628 
 
 6.. 43 ..0295 
 
 70 
 
 75 
 
 307- 3H 
 
 "75-7 
 
 276.9 
 
 898.8 
 
 0.1736 
 
 5.760 1.0309 
 
 75 
 
 80 
 
 311.80 
 
 1177.0 
 
 281.4 
 
 895.6 
 
 0. 1843 
 
 5.426,1.0323 
 
 80 
 
 «5 
 
 316.02 
 
 1.78.3 
 
 285.8 
 
 892.5 
 
 0.195' 
 
 5. 1 26 
 
 i'.0337 
 
 85 
 
 90 
 
 320.04 
 
 1179.6 
 
 290.0 
 
 889.6 '0.2058 
 
 4.859 '.0350 
 
 90 
 
 95 
 
 323-^9 
 
 1 180.7 
 
 294.0 
 
 886.7 0.2165 
 
 4.6.9 1.0362 
 
 95 
 
 100 
 
 327-5^ 
 
 1181.9 
 
 297.9 
 
 884.0 0.2271 
 
 4.403 1.0374 
 
 .00 
 
 105 
 
 33'. «3 
 
 1 182.9 
 
 301.6 
 
 881.3 '0.2378 
 
 4.205 
 
 '.0385 
 
 '05 
 
 1 10 
 
 334-56 
 
 1 184.0 
 
 305.2 
 
 878.8 
 
 0.2484 
 
 4.026 
 
 1.0396 
 
 1 10 
 
 '»5 
 
 337-«6 
 
 1.85.0 
 
 308.7 
 
 876.3 
 
 0.2589 
 
 3.862 
 
 1.0406 
 
 "5 
 
 120 
 
 341-05 
 
 I1.S6.0 
 
 312.0 
 
 874.0 
 
 0.2695 
 
 3.7" 
 
 1.0416 
 
 120 
 
 '25 
 
 344-13 
 
 1186.9 
 
 3 '5- 2 
 
 871.7 
 
 0.2800 
 
 3-57' 
 
 1 .0426 
 
 '25 
 
 130 
 
 347- > 2 
 
 1187.8 
 
 318.4 
 
 869.4 
 
 0.2904 
 
 3.444 
 
 '.0435 
 
 130 
 
 140 
 
 352.85 
 
 1189.5 
 
 324.4 
 
 865. 1 
 
 0.3" 3 
 
 3. 2. 2 1 1. 0453 
 
 140 
 
 150 
 
 358.26 
 
 1 1 91. 2 
 
 330-0 
 
 861.2 
 
 0.3321 
 
 3.01 1,1.0470 
 
 '50 
 
 160 
 
 363-40 
 
 1 192.8 
 
 335.4 
 
 857.4 
 
 0.3530 
 
 2.833' 1.0486 
 
 160 
 
 1,70 
 
 368.29 
 
 "94.3 
 
 340-5 
 
 853.8 
 
 0.3737 
 
 2.676' ..0502 
 
 170 
 
 180 
 
 372-97 
 
 "95-7 
 
 345-4 
 
 850.3 
 
 0.3945 
 
 2-535 '.05 '7 
 
 180 
 
 190 
 
 377.44 
 
 1197.1 
 
 350.' 
 
 847.0 
 
 o.4'35 
 
 2.408 ..0531 
 
 190 
 
 200 
 
 381.73 
 
 1 I '98.4 ' 354.6 
 
 843.8 0.4359 
 
 2.294' 1.0545 
 
 200 
 
WATER BETWEEN 32° AND 2J2" FAHR. 
 
 II 
 
 Tempi;rature 
 
 WeiKht, lb. 
 
 1 
 
 jinperature 
 
 Weight, lb. 
 
 Temp rature 
 
 Weight, lb. 
 
 Pahr. 
 
 per cub. ft. 
 
 Fahr. 
 
 per cub. ft. 
 
 tahr. 
 
 per lul). ft. 
 
 • 
 
 3a 
 
 62.42 
 
 123° 
 
 61.68 
 
 168° 
 
 60.81 
 
 35 
 
 62.42 
 
 124 
 
 61.67 
 
 .69 
 
 60,79 
 
 4C' 
 
 62.42 
 
 "5 
 
 61.65 
 
 170 
 
 60.77 
 
 45 
 
 62.42 
 
 126 
 
 61.63 
 
 171 
 
 60.75 
 
 50 
 
 62.41 
 
 i?7 
 
 61. 6t 
 
 172 
 
 60.73 
 
 52 
 
 62.40 
 
 128 
 
 6. .60 
 
 173 
 
 60.70 
 
 54 
 
 62.40 
 
 ' 9 
 
 61.58 
 
 174 
 
 60 63 
 
 e6 
 
 62.39 
 
 130 
 
 61.56 
 
 175 
 
 60.66 
 
 58 
 
 62.33 
 
 '3« 
 
 6r.54 
 
 176 
 
 60.64 
 
 60 
 
 (^•37 
 
 132 
 
 61.52 
 
 177 
 
 60.62 
 
 6? 
 
 62.36 
 
 n3 
 
 61.51 
 
 178 
 
 60.59 
 
 64 
 
 ^^2.35 
 
 »34 
 
 61.49 
 
 179 
 
 60.57 
 
 66 
 
 62.34 
 
 »?5 
 
 61.47 
 
 loO 
 
 60.55 
 
 68 
 
 62.33 
 
 136 
 
 61.45 
 
 181 
 
 60.53 
 
 70 
 
 62.31 
 
 '37 
 
 61.43 
 
 182 
 
 60. 50 
 
 72 
 
 62.30 
 
 '38 
 
 61.41 
 
 183 
 
 60. i& 
 
 74 
 
 62.28 
 
 139 
 
 61.39 
 
 184 
 
 60.46 
 
 76 
 
 62.27 
 
 140 
 
 6i-37 
 
 185 
 
 60.44 
 
 78 
 
 62 25 
 
 141 
 
 61.36 
 
 186 
 
 60.41 
 
 80 
 
 62.23 
 
 14a 
 
 61. 4 
 
 107 
 
 60.39 
 
 82 
 
 62 21 
 
 »45 
 
 61.32 
 
 188 
 
 60 37 
 
 84 
 
 62.19 
 
 144 
 
 61.30 
 
 189 
 
 60.34. 
 
 86 
 
 62.17 
 
 M5 
 
 61.28 
 
 190 
 
 60.32 
 
 88 
 
 62.15 
 
 146 
 
 6 .26 
 
 191 
 
 60.29 
 
 90 
 
 62 13 
 
 M7 
 
 61.24 
 
 192 
 
 60.27 
 
 92 
 
 6;>.Il 
 
 M8 
 
 6l.2-.' 
 
 '93 
 
 60.25 
 
 94 
 
 62.09 
 
 149 
 
 61.20 
 
 194 
 
 6c. 22 
 
 96 
 
 62.07 
 
 '50 
 
 6t.i8 
 
 '95 
 
 60.3a 
 
 98 
 
 62.05 
 
 151 
 
 6t.!6 
 
 196 
 
 60.17 
 
 ICO 
 
 62.02 
 
 ^5^ 
 
 61.14 
 
 197 
 
 60. 1 5 
 
 102 
 
 62.00 
 
 '53 
 
 61.12 
 
 198 
 
 60.12 
 
 104 
 
 61.97 
 
 '54 
 
 61.10 
 
 199 
 
 60.10 
 
 106 
 
 61.95 
 
 '55 
 
 61.08 
 
 200 
 
 f 0.07 
 
 108 
 
 61 92 
 
 '56 
 
 61.06 
 
 2c I 
 
 60.05 
 
 110 
 
 61. Eg 
 
 '57 
 
 61 04 
 
 2o.i 
 
 60.02 
 
 112 
 
 61.86 
 
 '58 
 
 61.02 
 
 203 
 
 6 .00 
 
 "3 
 
 61.84 
 
 '59 
 
 61.00 
 
 204 
 
 59-97 
 
 114 
 
 61.83 
 
 160 
 
 60.98 
 
 205 
 
 59-95 
 
 H5 
 
 61.82 
 
 161 
 
 60.96 
 
 206 
 
 5992 
 
 116 
 
 61.80 
 
 162 
 
 6094 
 
 207 
 
 59-89 
 
 117 
 
 61.78 
 
 163 
 
 60.92 
 
 208 
 
 5987 : 
 
 118 
 
 61.77 
 
 164 
 
 60.90 
 
 209 
 
 59 84 
 
 119 
 
 6«.75 
 
 '65 
 
 60.87 
 
 210 
 
 59-8i 
 
 120 
 
 61.74 
 
 166 
 
 6085 
 
 711 
 
 .S9.79 
 
 121 
 
 6-. 72 
 
 167 
 
 60.83 
 
 212 
 
 59-76 
 
 122 
 
 61.70 ' 
 
 
 
 
 
 Freezing point of water at sea kvel 3-2° F. 
 
 Maximum density 39, 1° F. 
 
 Dntish standard for specific gravity .... 62 ' F. 
 Boiling point under atmospheric pressure 112" F. 
 
 Weight I er cubic inch. .03612 lbs, 
 
 " M II II .0361 -511 
 
 II II II M .C36 8 II 
 
 II II II II 'OJI58 " 
 
TABLE SHOWING 
 
 NUMBER, DIAMETER, WEIGHT, LENGTH AND 
 RESISTANCE OF PURE COPPER WIRE. 
 
 BROWN & SHARPE GAUGE. 
 
 .03612 lbs. 
 
 .0361-511 
 .c-^6 8 >• 
 •03,58 tr 
 
 
 
 * — 
 
 Wg't 
 Lbs. 
 
 
 
 
 
 
 Diam. 
 
 a'si 
 
 Length. 
 
 Resistance of Pure Copper at 
 75* Fahrcnlieit, 
 
 
 
 
 Ohms 
 
 Feet 
 
 
 No. 
 
 In mils 
 
 -§11 
 
 per 
 
 Feet 
 
 per 
 
 per 
 
 Ohms 
 
 
 
 •c33 
 
 1000 ft 
 
 per lb. 
 
 1000 ft. 
 
 Ohm. 
 
 per lb. 
 
 0000 
 
 460.000 
 
 211600.0 
 
 639.32 
 
 1.56 
 
 .051 
 
 19605.69 
 
 .0000798 
 
 000 409.640 
 
 167805.0 
 
 507.01 
 
 1.97 
 
 .064 
 
 16547.87 
 
 .000127 
 
 00,364.800 
 
 133079.2 
 
 402.09 
 
 2.49 
 
 .081 
 
 12330.36 
 
 .000202 
 
 0|324.950 
 
 105514.0 
 
 319.04 
 
 3.13 
 
 .102 
 
 9783.63 
 
 .000320 
 
 1| 289. 300 
 
 83694.0 
 
 252.88 
 
 3.95 
 
 .129 
 
 7754.66 
 
 .00051 
 
 2 257.630 
 
 66373.0 
 
 200.54 
 
 4.99 
 
 .163 
 
 6149.78 
 
 .00(1811 
 
 3 229.420 
 
 52633.4 
 
 159.03 
 
 6.29 
 
 .205 
 
 4876.73 
 
 .001289 
 
 4 
 
 204.310 
 
 41742.5 
 
 126.12 
 
 7.93 
 
 .259 
 
 3867.62 
 
 .00205 
 
 5 
 
 181.940 
 
 33102.3 
 
 100.01 
 
 10.00 
 
 .326 
 
 2067.06 
 
 .00326 
 
 6 
 
 162.020 
 
 26250.5 
 
 79.32 
 
 12.61 
 
 411 
 
 2432.22 
 
 .00518 
 
 7 
 
 141.280 
 
 20817.0 
 
 62.90 
 
 15.90 
 
 .519 
 
 1928.75 
 
 .00824 
 
 8 
 
 128.490 
 
 16509.0 
 
 49.88 
 
 20.05 
 
 .654 
 
 1629.69 
 
 .01311 
 
 9 
 
 114 430 
 
 13094.0 
 
 39.56 
 
 25.28 
 
 .824 
 
 1213.22 
 
 02083 
 
 10 
 
 101.890 
 
 10381.0 
 
 31.37 
 
 31.88 
 
 1.040 
 
 961.91 
 
 03314 
 
 11 
 
 y. 1.742 
 
 8234.1 
 
 24.88 
 
 40.20 
 
 1.311 
 
 762.93 
 
 .05269 
 
 12 
 
 80.8U8 
 
 6529.9 
 
 19.73 
 
 50.69 
 
 1.663 
 
 605.03 
 
 .08377 
 
 13 
 
 71 961 
 
 5178.4 
 
 15.65 
 
 63 91 
 
 2.084 
 
 479 80 
 
 .13321 
 
 14 
 
 64.084 
 
 4106.8 
 
 12.41 
 
 80.59 
 
 2.628 
 
 380.61 
 
 .2118 
 
 15 
 
 57.068 
 
 3256.8 
 
 9.84 
 
 101.63 
 
 3.314 
 
 3itl.75 
 
 3368 
 
 16 
 
 50.820 
 
 2582.7 
 
 7.81 
 
 128.14 
 
 4.179 
 
 239.32 
 
 .6355 
 
 17 
 
 45.257 
 
 2048.2 
 
 6.19 
 
 161.59 
 
 5.269 
 
 189.78 
 
 .8515 
 
 18 
 
 40 303 
 
 1624.3 
 
 4.91 
 
 203.76 
 
 6.645 
 
 160.60 
 
 1.3539 
 
 19 
 
 35.890 
 
 1288.1 
 
 3.78 
 
 264.26 
 
 8.617 
 
 116.06 
 
 2.2772 
 
 20 
 
 31.961 
 
 1021.5 
 
 3.09 
 
 324.00 
 
 10.566 
 
 94.65 
 
 3.423 
 
 21 
 
 28.462 
 
 810.08 
 
 2.45 
 
 4U8.56 
 
 13.323 
 
 75.06 
 
 5.443 
 
 22 
 
 25 347 
 
 642.47 
 
 1.94 
 
 615.15 
 
 16.799 
 
 59 53 
 
 8.654 
 
 23 
 
 22.571 
 
 609.15 
 
 1.64 
 
 649.66 
 
 21.185 
 
 47.20 
 
 13.763 
 
 24 
 
 20.100 
 
 404.01 
 
 1.22 
 
 819.21 
 
 26.713 
 
 37.43 
 
 21.885 
 
 25 
 
 17.900 
 
 320.41 
 
 .97 
 
 1032.96 
 
 33.684 
 
 29.69 
 
 84.795 
 
 26 
 
 15 940 
 
 254.08 
 
 .77 
 
 1302 61 
 
 42.477 
 
 23.54 
 
 55.331 
 
 27 
 
 14.195 
 
 201.50 
 
 .61 
 
 ld42.55 
 
 63.563 
 
 18.68 
 
 87.979 
 
 28 
 
 12.641 
 
 159.79 
 
 .48 
 
 2071 22 
 
 67.542 
 
 14.81 
 
 139.89:i 
 
 29 
 
 11.257 
 
 126.72 
 
 .38 
 
 2^511.82 
 
 85.170 
 
 11 74 
 
 222.449 
 
 30 
 
 10.025 
 
 100.50 
 
 .30 
 
 3293 97 
 
 107 391 
 
 9.31 
 
 353.742 
 
TABLE SHOWING THE DIFFERENCE BETWEEN 
 
 WIRE GAUGES. 
 
 No Brown & Sharpe's 
 
 i 
 
 1 
 Old English Stubs' or 
 or London. Birmingham 
 
 oooo 
 
 460 
 
 •454 
 
 454 
 
 ooo 
 
 40964 
 
 •425 
 
 425 
 
 00 
 
 36480 
 
 •380 
 
 380 
 
 
 
 32495 
 
 .340 
 
 140 
 
 I 
 
 28930 
 
 • 300 
 
 300 
 
 a 
 
 25763 
 
 284 
 
 284 
 
 3 
 
 22942 
 
 •259 
 
 259 
 
 4 
 
 20431 
 
 •238 
 
 238 
 
 5 
 
 18194 
 
 220 
 
 220 
 
 6 
 
 16202 
 
 •203 
 
 203 
 
 7 
 
 14428 1 
 
 • 180 
 
 180 
 
 8 
 
 12849 
 
 ■165 
 
 165 
 
 9 
 
 "443 
 
 • 148 
 
 148 
 
 10 
 
 10189 
 
 •134 
 
 134 
 
 II 
 
 09074 
 
 • 120 
 
 120 
 
 xa 
 
 08081 
 
 • 109 
 
 109 
 
 13 
 
 07196 
 
 095 
 
 095 
 
 X4 
 
 06408 
 
 •083 
 
 083 
 
 15 
 
 05706 
 
 072 
 
 072 
 
 i6 
 
 05082 
 
 ■065 
 
 065 
 
 17 
 
 04525 
 
 •058 
 
 058 
 
 i8 
 
 04030 
 
 •049 
 
 049 
 
 19 
 
 03589 
 
 ■040 
 
 042 
 
 30 
 
 63196 
 
 035 
 
 035 
 
 ax 
 
 02846 
 
 0315 
 
 032 
 
 aa 
 
 •025347 
 
 •0295 
 
 028 
 
 23 
 
 •022571 
 
 •027 
 
 025 
 
 24 
 
 • 0201 
 
 •025 
 
 022 
 
 25 
 
 •0179 
 
 •023 
 
 020 
 
 26 
 
 •01594 
 
 •0205 
 
 o:3 
 
 27 
 
 •014195 
 
 •01875 
 
 016 
 
 28 
 
 •Q12641 
 
 ■0165 
 
 014 
 
 29 
 
 •0:1257 
 
 0155 
 
 013 
 
 30 
 
 •010025 
 
 •01375 
 
 012 
 
 3X 
 
 • 008928 
 
 •01225 
 
 • 010 
 
 32 
 
 •00795 
 
 01125 
 
 • 009 
 
 33 
 
 • 00708 
 
 •01025 
 
 • 008 
 
 34 
 
 .0063 
 
 0095 
 
 • 007 
 
 35 
 
 •00561 
 
 •009 
 
 00s 
 
 36 
 
 005 
 
 0075 
 
 • 004 
 
 37 
 
 •00445 
 
 0065 
 
 . . 
 
 38 
 
 • 003965 
 
 00575 
 
 . -- - ■ - -- - 
 
 39 
 
 •003531 
 
 •005 
 
 ■ • 
 
 40 
 
 •003144 
 
 0045 
 
 • • 
 
HR Gill 
 
 For the Prevention and Removal of Scale In Steam 
 
 Boilers, and for Neutralizina ftcid, Sulptiur 
 
 and Mineral Waters. 
 
 UNQUESTIONABLE EVIDENCES OF ITS SUPERIORITY 
 
 GOVERNMENT ENDORSEMENTS.- It is approved and exclusively used by the 
 
 U. S. ai d many foreign governments. 
 
 OUR CELEBRATED MECHANICAL AUTHORITIES universally approve of and 
 recommend its use, and it is fa\orahly mentioned in nearly all our high class 
 treatises on Steam Engineering pniilished in the English language. 
 
 IT IS NOT A VEGETABLE COMPOUND and it is free from all acid and other 
 injurious components contained in these article-. \'egetal)le matter principal'y 
 consists of acids, larh )n, earthy .'alts, etc , anil the active and soluble proper- 
 ties containeii in these preparations are acids, which are more harmful to the 
 boiler as corroding agents tlian boiler incrustaiijn. 
 
 LORD'S COMPOUND is free from all volatile constituents, 
 
 and it will not contaminate the steam with odor, flavor or color ; therefore it is 
 invaluable to breweries and simi'ar manufacturing enterprises requiring pure 
 live steam. 
 
 LORDS COMPyUND is in powdeted form, and represents highly refined and 
 cosily chemicals. It's thirty years of active warfare against ihe use of boiler 
 fluids containing from seventy-five to ninety five per cent, of water has lately 
 resulted in the int oduction on the market of many powdered articles. Mai.y 
 of these are sold as " Lord's Hoik Compound " by unscrupulous persons. In 
 fact, the universal imitation of this preparation is the best argument attesting 
 its merit. 
 
 DO NOT C UNTENANCE LIQUID ARTICLES 
 FURN ' SH YOUR OWN WATER 
 
 For descriptive pamphlet, brimful with interesting and instructive information, 
 treating upon Boiler Incrustation, Corrosion, Hoiler Explosions, and the His- 
 tory of the Sources of Water Contarninatii>n, its work in the economy of nature, 
 and method of purification ser.t free on req'iest. Address 
 
 G. E. GRANT - 13 St. John Street, MONTREAL 
 
 Sole Agent in Canada for Lord's Boiler Compound, 
 
Engineeers ! 
 
 It. I 
 
 
 DO YOU KNOW THAT 
 
 MICA 
 COVERING 
 
 SAVKS 
 
 From 50a to 150% More Heat 
 
 THAN ANY OTHKR COVERING 
 IN THH MARKET. 
 
 Write for Reports of Trials, Testimonials, 
 Prices, Etc., to 
 
 THE 
 
 Mica Boiler Covering: Co. 
 
 (limited) 
 
 9 JTox^clctn St. - TORONTO 
 
 
 '4.. 
 - ) 
 
SHaiioa i ox 
 
 oo 
 
 <y») <A 
 
 </D H 
 
 00 
 
 
 GO 
 
 Uh 
 
 oo 
 
 o 
 
 CQ 
 
 
 LU 
 
 
 
 
 
 Jf 
 
 ^ 
 
 c 
 
 < 
 
 LU 
 
 o 
 
 7. 
 
 :z) 
 
 
 
 o 
 
 ^ 
 
 Q 
 
 
 o * 
 
 CO 
 
 CO 
 
li 
 
 lljllUfACTURtD^Y 
 
 LW'^"«», PENBERTHY INJECTOR, 
 
 ^ [argent I |V|ANirACTUR[RS,r'"'WORLD. 
 
Robb-Armstrong-^ 
 Automatic Engines 
 
 CENTRE OR SIDE CRANK 
 SIMPLE AND COMPOUND 
 
 + 
 
 + 
 
 HUMFORD'S IMPROVED BOILER 
 
 has Forced Water Circulation similar to 
 a water-tube boiler; Larg-e, Effective Heat- 
 ings Surface, and Special Arrangements 
 for Preventing and Removing Scale 
 
 Robb Engineering' Co'y 
 
 (Limited) 
 
 AMHERST, N. S. 
 
THE IDEAL 
 
 S&bF-OILING. 
 HIGH-SPE&D 
 
 ENGINE 
 
 Has a Smooth, Regular Motion, and will not throw or 
 spray oil on your floor or saturate your dynamos. 
 Particularly adapted for Electrical Plants on account of its 
 buiet running, automatic lubrication, economy and cleanliness. 
 
 ,:,agB«««!»»': 
 
 ftriisiic in Design 
 
 Poweilul in Construction 
 Gives tlie Best Satisfaction 
 
 IVl^KIHmS = 
 
 TI16 Goldie S MGGuilOGll Go., Limited 
 
 GrA.LT, OlSTT. 
 
 
The Engineers' Reducer 
 
 (WRIGHT'S PATENT) 
 
 In this reducing motion, a telescopic lever operates a reducing 
 
 device that is positive in action, and adjustable to any length of 
 
 stroke or any length of 
 diagram. The reduced 
 motion relative to that 
 of the piston is the 
 same in every part of 
 the stroke. The lever 
 may be of any length 
 between half stroke of 
 the engine and four 
 limes the stroke, with 
 diagrams of any length 
 desired. P'our indica- 
 tors can be operated 
 simultaneously. Indi" 
 cator strings leave the 
 reducer at any angle. 
 It is steady at all speeds 
 of engine, and durable. 
 One, with care, will 
 last any engineer his 
 1 i f e - 1 i m e . Average 
 weight, 4 pounds. 
 
 By disconnecting the lever it can be carried in a coat pocket. 
 
 Price, from $15 to $22, according to range, finish, fittings and 
 
 material. 
 
 Miller Bros. & Toms 
 
 King Street, 
 
 MAKERS 
 
 MONTREAL 
 
9 
 
 9 
 
 m you flcquainied 
 
 WITH THE 
 
 C ANADIAN E LECTRICAL N EWS AND 
 E NGINEERING J OURNA L. 
 
 If not write us for a specimen copy. 
 
 You will find it a ^'aluable educator on electrical 
 and engineering lines. 
 
 Published monthly at $1.00 per year, by 
 
 THE 
 
 C. H. MORTIMER PUBLISHING CO. 
 
 of TORONTO, Limited. 
 
 Confeder.ition Life Building, 
 Toronto, Canada. 
 
 1 .ranch Office: New Vnrk Life 
 building, MontreaL 
 
 9 
 
 7 
 
 7 
 
 
-AND 
 
 [S 
 
 FOR STATIONARY AND MARINE WORK. 
 
 HEATERS, SEPARATORS, VALVES, GAUGES, 
 
 QUIMBY ELECTRIC PUMPS, 
 
 DAMPER REGULATORS, AUTOMATIC 
 
 STOKERS, AND SMOKELESS FURNACES. 
 
 Complete Steam Power Plants 
 Designed and Installed. 
 
 Large Book "STEAM" Sent Free 
 Upon Application. 
 
 BABCOGK&WiLGOX LIMITED 
 
 LONDON AND GLASGOW. 
 
 Head Office for Canada, 202 St. 
 James Street, Montreal. 
 
 Toronto Office, 
 114 King Stieef- West