None
Hence the normal losses and design for a choking coil will be the same as those of a transformer of half its rating. The power factor of the choking coil is the ratio of the losses to the volt-ampere input, and will be a minimum when the losses are a minimum, which is for that load which makes the copper and iron losses alike. The design is worked out in the same way as for a transformer, except that the whole space goes to one winding and the gap has to be inserted. If half of this be placed at the centre of each coil, its correct length can be very accurately estimated beforehand, and the stray flux will be comparatively small. If it be outside the coils, the gap will have to be longer than calculated, while the stray flux will be very considerable and may cause trouble with eddies. Application to Auto-Transformers. — An auto-transformer is not in reality a special kind of transformer, but a special way of using one. The two windings are connected in series with the supply across the ends and the load between one end and the joint, or vice versa. If the parts into which the total voltage be divided be Vj and V2, the latter being the load voltage, while the currents in the two parts are Ij and I2, the load for which it is rated (assuming that the voltage is being reduced) will be Hence the rating as an auto-transformer is twice its ordinary rating, and this increase becomes very great when the supply and loadvoltages do not differ much. To design an auto-transformer, it is therefore best to treat it as an ordinary transformer with the rating altered to suit. The normal losses and size are the same as for an ordinary rating, and the nominal efficiency will be higher, for it will be reckoned on the larger effective output. The voltage drop and the magnetizing current will also be reduced in the same ratio; the former because a drop of V^ is necessarily accompanied by a rise of V^ when the total is kept constant, and the magnetizing current is reduced by the fact that it goes through both windings instead of one only. Table 1012 gives the values for the ratio (V^ + ^^j^y. The rating is unaltered if the positions of the load and supply are interchanged, but in that case the magnetizing current is the same as with an ordinary transformer. TABLE 1012. — Comparison of Auto- and Ordinary Transformer Ratings. Supply voltage Load voltage Auto rating V, + Va 4 : 1 3 : 1 2 : 1 3 : 2 V, 1-833 1-50 2 3 4 : 3 1 5 : 4 6:5 10 : 9 1 : 1 6 6 10 00 Application to Multi-Voltage Transformers.— When more than two windings, or sections of one winding, are required, the aggregate volt-amperage should be taken as the sum of the greatest volt-amperages for each section, whether these occur simultaneously or not. The greatest input to the primary will of course be determined by the greatest simultaneous loads for the other windings. The normal losses may be taken the same as those in a transformer rated at half the grand aggregate. The procedure is then exactly as before, but the copper section should be divided between the windings in the proportion which their output or input bears to the aggregate, the corresponding fraction being taken instead of ^ in the equations for the number of turns and section of wire. If the maximum loads do not occur simultaneously, the average temperature rise will be somewhat less than is permissible, but it is unsafe to take advantage of this fact by reducing the dimensions much, for fear of overheating the loaded parts. Preliminary Design for a Choking Coil of the Rectangular Coil Core Type to Absorb i K.V.A. at 100 Volts, 10 Amperes, and 100 Cycles per  second. Loss-length for 20 itiils Stalloy (from curve) Ll = 2-40 mils. Resistivity of copper (hot , A.C ) (assumed) PC = 0-90 X 10~« ohm-inches. Space-factor for iron (estimated) ^ = 0-85. „ „ copper » o-C = 0-45. „ geom. mean >j Voiac = 0-619. Specific cost ratio r n = 2ac/<n = 106. Fundamental length — L^ IVLl_ Vai<rc V PiPc _ 1000 volt-amps, x 2*40 mils 0-619 X 12-5 watts = 0-310 inches. Fundamental surface So-Lo^ = 0-096 inches^. „ volume Vo-Lo« = 0-0298 inchest Conventional cooling surface of iron — Sci = (Sci/So)So « 964x0-096 inches' ^= 92-5inche82. Conventional cooling surface of copper — Sec = (Soc/So)So = 1500 X 0-096 j> = 144 „ Conventional dissipation Pi/Sci intensity for iron — = 12-5 watts -T- 92-5 » = 0*135 watts per inch'^. Conventional dissipation Pc/Soc intensity for copper = 12-5 watts ^144 >i = 0-087 244 TRANSFORMERS. Width of iron space — lis =(lis/U))Lo = 3-40 X 0-310 Distance between yokes — Lc8 = (Lcs/Lo)Lo = 10-60 x 0310 Depth of copper space — Ics = (lcs/Lo)Lo = 2-36 x 0-310 Depth of iron space — Lis-(Li8/Lo)Lo - 7-50x0-310 inches =1'05 inches. „ =3*28 „ „ -0-73 „ „ =2-32 ,. Length of iron — Li = (4li8 + 2Lcs + 4lcs) - (4-20 + 6-56 + 2-92) inches = 1 3-68 inches. Length of mean turn — Lc =(2li8 + 3lcs+2Li8) = (2-10 + 219 + 4-64) = 8-93 Section of iron space — Sis = lisLis = I "05 ins. x 2-32 inches = 2-44 inches^. Section of copper space — Sc = 2Lcslc8 = 6-56 „ xO-73 „ = 480 „ Section of iron — Si = <fiSis = 0-85x2-44 inche82 = 207 „ Section of copper — Sc = crcScs = 0-45x4-80 „ = 216 „ Volume of iron — Vi -LiSi Volume of copper — Vc = LcSc = 13-68 ins. x 2-07 inches = 28-3 inches. = 19-3 „ = 8-93 „ x2-16 Mass of iron — Mi = DiVi Mass of copper — Mc-DcVc 0-282 lb. inch^ 0-322 lb. inch^ x28'3 inches8= 8-0 lbs. 19-3 Mass of standard iron of equal total cost — Ms = 1-75Mi + 3Mc =(14-0 + 186) lbs. = 6-2 „ = 32-6 „ = 14-8kg8. Turns in winding — N^lJPcScM^ 1 f L^-S watts X 2-16 ins.P^ ^34^^^^ I ( pcLc J 10 amps. I 090 x 10"*' ohm-ins. x 893 ins. J = 2 X 92  -7 X 10"' ins. = 855 amps, per inch. DESIGN OF TRANSFORMERS. 245 Flux density (maximum) — =. 100 volts 4/4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4N4 „ current I = 10 amperes. Number of turns N = 184 turns. coils 2. „ turns per coil 92. „ layers per coil 5. „ turns per layer 19, 19, 19, 19, and 16. Conductor, diameter, bare 10S.W.G. 128 mils. „ „ covered 142 „ „ section Sw = 12-87 xl0-»inches2. „ total section of copper Sc = NSw = 2-37 inches^. „ current density I = I/Sw = 778 amps, per inch^. Depth of winding Ics =0*71 inches. Axial length of coil, allowing one extra diameter for spiral 2*84 „ Perimeter of coil section, after taping Lpc *=7'26 „ Former, dimensions of section „ perimeter of section Allowance for comers Length of mean turn ,, wire ^ A ^ 2^ inches. 7-50 213 Lc =9-63 Lw = NLc =1770 Cooling surface of coils Volume of copper Mass of copper Scc = 2LpcLc= 14-52'' x 9-63"= 140 inchesS. Vc = LcSc = 22-8 inchest Mc = (0-322 lb. per inch8)Vc = 7-34 lbs. 246 TRANSFORMERS. Eflfective resistance, hot — R = 0-90 X 10-« ohm-ins. x Lw/Sw = 0124 ohm. Copper loss Pc = RI^ = 1 2 4 watts = 1 -24 %. Dissipation intensity for coils Pc/Scc ^ 0*088 watts per inch^. Estimated temperature rise — <EC = (240'' C. per watt per in.2) x Pc/8cc = 21' C. Width of core lis =1 inch. Distance between yokes Lcs =* 3 inches. „ „ cores 2lcs = 2 „ Length of iron — Li = (4li8 + 2 Lcs + 2 X 2 lcs) = (4 + 6 + 4) ins. = 14 inches. „ flux Lf = (31i8 + 2Lcs + 2x21c8) = (3 + 6 + 4) „ =13 „ Thickness of sheets 20 mils. Number of sheets in each core, r— n-shaped 100. Iron depth, net 2.00 inches. „ gross 1.25 = 235 „ „ section, net 1.25 = 2.00 inches. Cooling surface of iron — Sci = 2.15 + 1.25 = 28 inches x (1 + 2.35) inches = 93.8 inches. Volume of iron 1.25 = 28 inches x (1 + 2.35) inches = 93.8 inches. Mass of iron 1.25 = 28 inches x (1 + 2.35) inches = 93.8 inches. Mass of standard iron of equal total cost — Ms = 1.75 + 3.25 = (13.8 + 22.0) inches = 16.2 kgs. Relative cost mass — = Ms X (1.50 C.P.S.) = 35.8 lbs. x (1.50 C.P.S.) = 0.716 grams. Flux density (maximum) — d8-= Zr. = —1.50-6.6 M.V.S. per inch. ^ •4ffNSi 4.44 X 100 per sec. X 184 X 2.00 in 8.2 o. per m,a = 9,500 lines per cm.* Iron loss per unit volume (from curves for Stalloy) — Pi/Vi = 0.45 watts per inch. total Pi = (Pi/Vi) Vi = 12.6 watts = 1.26 %. Dissipation intensity for iron Pi/Sci = 0.35 watts per inch. Estimated temperature rise for iron — %i = (240) C. per watt in. = 32' C. DESIGK OF TRANSMISSION. Magnetizing field (maximum) required for iron (from curve) = ^ = 6'7 amp.-turns per inch. Excitation (R.M.S.) required for iron Xi = -j^HLy= 62 amp.-tums. „ „ due to current Xm=NI = 1840 „ „ „ to be used by gaps Xq = Xm - Xi - 1778 „ Length of air-gap required — I.-_ J^JJ^A^G _ 45 mils X M.V.S. per in. 1778 amp.-turns _ 154 mils ^ " aidii "" 100'iip.-turns ^ 085 x 0-613 M.V.S. per in.2 " = 2x77 mils. Power-factor cos ♦ = (?!+ Pc)/IV = (12-6 + 12*4) watt/1000 V.A. = 0-0250. The desired gap is fixed by inserting wooden distance-pieces of the correct thickness between the two parts of the core, and it can be adjusted by trial to give the right inductance exactly, if that is necessary, by inserting press- pahn packing-pieces. It is most convenient to stamp the sheets in the form of a I — I to save labour in assembling, and those constituting each half of the core should be securely bound together by cord or tape. They must be thoroughly impregnated if silent running be required. The two halves may be clamped together by tie-bars, or the case may be designed as shown, so as to act as a clamp as well. The coils can be secured by wooden wedges between them and the cores and yokes. Preliminary Design for an Auto-Transformer of the Rectangular Coil Core Type to give 2 K. V.A. at 210/105 Volts, 9*8/19 Amperes, and 50 Cycles per Second, using Stalloy Iron. Loss allowed in iron (given) Pi = 20 watts = 097 per cent. „ „ copper „ Pc =40 „ -1-94 „ „ geom. mean \/PiPc=28'3„ =1*37 „ Output of secondary winding (estimated) IgVg = 970 volt-amperes. Input to primary winding „ IjVi =1030 „ Aggregate volt-amperage „ (Ii^i + IjV^) = 2000 „ Terminal P.D. of primary winding (given) Vj = 105 volts. „ secondary „ „ V^ = 105 „ Current in primary winding (estimated) I^ = 9*80 amperes. „ secondary „ (given) l^ =9*24 „ 248 TRANSFORMERS. Frequency (given) f Loss-length for 20 mils Stalloy (from curve) Ll Resistivity of copper (hot A.C.) (assumed) pc = 50 cycles per second. -• 3-20 mils. = 0-90 X 10"^ ohm-inches. Space-factor for iron „ „ copper „ geom. mean Specific cost ratio (estimated) crc m = 2ac/cFi ' = 0-85. ^0-40. = 0-583. = 0-94. Fundamental length — Lo = (iZLiiHb ^'2^ -^^« = 0-388 inches. 'J^c VPiPc ^'^83 X 28-3 watts Fundamental surface So — Lo^ „ volume Vo - Lo^ 0151 inches^. 0-0584 inches^. Conventional cooling surface of iron — Sci=(Sci/So)So = 950x0-151 Conventional cooling surface of copper — Sec = (8cc/So)So = 1510 x 0-151 Conventional dissipation intensity for iron — Pi/8ci =20 watts -^ 143 Conventional dissipation intensity for copper — Pe/Scc . = 40 watts -r 228 inches^ =143 inches^. „ =228 „ „ = 0- 1 40 watts per inch^. „ =0175 Width of iron space — lis =(li8/U)Lo Distance between yokes — Lea =(Lcs/Lo)Lo Depth of copper space — Icjs =(lcs/Lo)Lo Depth of iron space — Lis -(LWU)Lo 3-33 X 0-388 10-75 X 0-388 2-40 X 0-388 7-37 X 0-388 inches =1*29 inches. „ =4-17 „ „ =0-93 „ „ =2-86 „ Length of iron — Li =(4lis + 2Lcs + 4lcs) = Length of mean turn — Lc =(2lis + 3lcs + 2Lis) = (5-16 4- 8-34 + 3-72) inches : (2-58 4-2-79 + 5-72) „ 17-22 inches. : 11-09 „ Section of iron space — Sis =lisLis Section of copper space — Scs =2Lcslcs Section of iron — Si = «riSis Section of copper — So = cJ'cScs 1-29 ins. X 2-86 :8-34 „ xO-93 : 0-85x3-69 = 0-40x7-75 inches = 3*69 inches^. „ =7-75 „ inches'-^ = 3-14 „ ,. =3-10 ., DESIGN OP TRANSFORMERS. 249 Volume of iron — Vi =LiSi Volume of copper — Vc =LcSc = 17-22 ins. x 3'14 inches^ = 54*1 inches^. Mass of iron — Mi = DiVi Mass of copper- Mc = DcVc 0-282 lb. inch 0-322 lb. inch Mass of standard iron of equal total cost — Ms = 1-75 Mi + 3Mc =(26-6 + 333) lbs. „ = 1M „ = 59-9 „ =27-2kgs. Turns in primary winding — M _ 1 jPcScl*^ 1 J * 2110 pcLc'19-6 amps. 10 40 watts X 3-10 ins. 2 90 X 10" ohm-ins. x Turns in secondary winding — N = ^-i PcScM^__l__^ I 40 watts X 3-10 ins.g i*=.190 2x90 Turns in secondary winding — N = ^-i PcScM^__l__^ I 40 watts X 3-10 ins.g i*=.190 2x95 „ Section of wire for primary — Swi = Sc/2Ni = 310 in8.2 + 360 Section of wire for secondary — Swi = Sc/2N, = 3-10 „ + 380 = 8-62 X 10-8 inches2. = 816x10-3 „ Flux density (maximum) — Current density (R.M.S.) — 1 = 1/2/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/3 amps. + 8/62 x 10-ins.2 = 1140 amps, per in.2 2 = 1/3/3 = 2/ „ coils Mass of copper, etc. = (0322 lbs. per inch) Vc. 287 29-1 936 0-235 204 21-5 694 4 50 16 6 30 438 inches. inches. lbs. Effective resistance, hot — R = 0-90x10-0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0. Number of sheets in each core (L-shaped) = 135. Iron depth, net = 2.70 inches. gross loss = 3.18 inches. Section, net = 3.71 inches. Cooling surface of iron = 2.11 inches + 1.38 inches. Sci = 2.11 inches + 1.38 inches. Volume of iron 1.25 inches = 1.75 inches. Mass of iron 1.25 inches = 1.75 inches. Mass of iron 1.25 inches = 1.75 inches. Relative cost mass = 1.75 inches = 1.75 inches. K = 0.98 grams. Flux density (maximum) = 10.5 volts. K = 0.98 volts = 0.75 volts = 0.75 volts. Iron loss per unit volume (from curves for Stalloy) = 1.00 watts per inch. Total 1.00 watts = 0.99 watts = 0.99 watts. Dissolve the given weight of the iron in the given area and the weight of the iron in the given area. Importance of iron for iron, per inch = 0.17 watts per inch. Estimated temperature rise for iron = 0.240 watts per inch. Magnetising field (maximum) required for iron, from curve, = 0.28 amps per inch. Excitation (R.M.S.) required for iron, per inch = 0.12 watts per inch. Exciting current, idle component, per inch = 0.426 amps. „ working component two = PiC^i + ^2) — ^'^^7 „ „ total— Io = (1io^+1wo^)* = (0-182 + 0010)* amps. = 0-436 = 4-45 per cent. Equivalent leakage thickness between coils (^5^ - 0*390) ins. = 0*172 inches. „ of winding depths = 0*260 „ „ total Itl = 0*432 „ Total equivalent resistance, referred to secondary — Rt2 = Pc/l2^ = 0-438 ohms. „ leakage reactance, referred to secondary — £rL2« = 2irjLiafN22(Lc/2Lcs)lTL ^0*^10-« henry ^ 50 ^ Ig^,^ ^ ^ ^.^32.^^^^.225 ^^ mch sec. 8*/ 5 Total equivalent impedance, referred to secondary — ZT2='(RT2^ + JBrL22«2)* = (0*192 + 0051)« ohms = 0*493 „ Internal lag, cos 4» = Rrg/ZTg = 0*889. 4» = 27'. Voltage drop at unity power-factor = iRT2l2 zero (lagging) „ = 1*92 1 „ 0*899 „ „ = JZt2I2 = 2*02 volts = 1*92 % = 104 „ = 0*99 „ = 2*28 „ =2*18 „ Efficiency at Unity Power-Factor. Load  the efficiency is considerably higher than the limit for the rating. DESIGN OF TRANSFORMERS. 253 Preliminary Design for an Oil-Insulated Transformer of the Hex- agonal Ring Type with Rectangular Coils to give 5 K.V.A. at 2000/200 Volts, 2*6/25 Amperes, and 80 Cycles per Second, using Stalloy Iron. Loss allowed in iron „ „ copper „ geom. mean (given) Pi = 60 watts =M6% Pc =120 „ =2-32,, VPiP"c = 84-9 „ =l-64„ Output of secondary winding (given) IgVg = 5000 volt-amperes. Input to primary winding (estimated) I^Vj = 5180 „ Aggregate volt-amperage „ (I^Vj + IgVg) = 10, 1 80 „ Terminal P.D. of primary winding (given) „ secondary „ „ Current in primary winding (estimated) „ secondary „ (given) = 2000 volts. = 200 ,. Ij =2-59 amperes. I, =25-0 „ Frequency (given) f Loss-length for 20 mils Stalloy (from curve) Ll Resistivity of copper (hot, A.C.) (assumed) pc = 80 cycles per second. = 2-75 mils. = 0-90 X 10-« ohm-inches. Space-factor for iron „ „ copper „ geom. mean. Specific cost ratio (estimated) <ri =0-70. <rc =0-33. /s/oi<rc =0*481. m = 2<rc/<ri = 0-943. Fundamental length — .(JjVkT^V^^ 10,180 volt-amps, x 2-75 mils^^.^gg ,^^^^ J^ s/P^ 0-481 X 84*9 watts Fundamental surface So = Lo^ = 0470 inches^. „ volume Vo = Lo^ = 0-322 inches*. Conventional cooling surface of iron — Sci = (Sci/So)So = 738 X 0-470 Conventional cooling surface of copper — Sec = (8cc/So)8o = 2034 x 0-470 Conventional dissipation intensity for iron — Pi/Sci = 60 watts -r 347 Conventional dissipation intensity for copper- Pc/8cc = 1 20 watts -J- 955 inches2 = 347 inches^. :955 = 0-173 watts per inch^. = 0126 254 TRANSFORMERS. Width of iron space — Depth of copper space — Lis = (1.8.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0.0. x6-64 = 127 inches, = 127 inches, Mass of standard of equal cost — Ms = 1-75 mill + 3.5 = (91-0 + 122-4) , = 40-8, lbs = 213, lbs = 96-5 kgs. Turns in primary winding — 1.2.2.3.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5.5 Section of wire for secondary — Sw2 = Sc/2N2 = 6-64 „ -=-272 = 24-4xl0-« „ DESIGN OF TRANSFORMERS. Flux density (maximum) — 2000 volts 255 2 = 0-566 M.V.S. per inch 4ffNiSi 4-44 X 80 per sec. x 1314 x 7*59 ins. = 8780C.G.S. Current density (R.M.S.) — I = 1/Swi = 2-59 amps, -r- 2-53 x 10-^ ins.2 = 1 020 amps, per inch. Design for an Oil-Insulated Transformer of the Hexagonal Ring Type with Rectangular Coils to give 5 K.V.A. at 2000/200 Volts, 2-6/25 Amperes, and 80 Cycles per Second, using Stalloy Iron. (Plate 6.) Primary. 2000 Secondary. 200 Total. Units. Rated voltage V ... volts. „ current 1 2-59 250 126 amperes. Number of turns N 1224 turns „ coils -6 6 6 „ turns per coil 204 21 „ layers per coil 17 2 „ turns per layer 20-4 17 56 11-10 7 176 Conductor, diameter, bare < ... S.W.G. mills. Logs. Covering Section 8w 2-463 24-33 10-8ins.2 Total section of copper 8c^ NSw 3-02 307 609 inches. Per inch. Depth of winding 1cs 1-156 1-55 inches. Axial length of coil, allowing one extra diameter for spiral 0-34 1-43 209 2-28 Perimeter of coil section, after taping Lpc 4-4 5-2 Former dimensions of section H^H 5x2| inches. Perimeter of section 19-5 15-0 Allowance for corners 23 1-2 ... Length of mean turn Lc 21-8 16-2 190 ... Allowance for comers 23 1-2 ... Length of wire Lc 21-8 16-2 190 ... Cooling surface of coils Sec = 6LpcL(^ 6x96 inches. Volume of copper Vc = LcSc 65-8 49-7 115-5 inches. Mass of copper Mc = (0*322 lbs. per inch)Vt' 21-2 160 37-2 lbs. 256 TRANSFORMERS. Effective resistance, hot — R = 0-90 X 10-^ ohm-ins. x Lw/8w Copper loss Pc = RI^ Dissipation intensity for coils Pc/Scx; Estimated temperature rise — 'rc = (200' a per watt/in.*) x Pc/8cc Primary. 9-74 65-3 0114 23 Secondary. 0-0755 47-2 0-094 19 Total. 0-180 112-5 0-104 21 Units. ohms. watts. watts per inch. 'C. Width of core Radius of inscribed circle Side of hexagon, inner „ outer Length of iron and of flux I18 =2 00 inches. Ics =2-44 „ ;y|lc8 =2-85 „ 2 ;y3(lis+lcs) = 5i LfNLi = 2V3(Ii8 + 21c8) = 23-8 „ Thickness of sheets Number of sheets in each core Iron depth, net „ „ gross— 2 packets 2 J ins. thick + 1 duct J in. wide Lis = „ section, net 81 - Cooling surface of iron — Sci = 23-8 ins. x (4 x 2-00 + 4 x 2 J) ins. = Volume of iron Vi = LjSi Mass of iron Mi = (0-282 lbs. per inch8)Vi = „ standard iron Of equal total cost — Ms = 1-75 Mi + 3Mc-(94+112)lbs. = 206 „ = 93-4kgs. Relative cost mass — = Ms X {^/P^JhMi±}2^2}' X (f/50 C.P.S.)t = 206 lbs. X {^59 X 112-5/5086}8x (l-6)« = 172 x 10'S lbs. = 0-78 grams. 20 miles. 200. 4-00 inches. 4-75 „ 8-00 inches. 405 inches. 190 inches. 53-7 lbs. Flux density (maximum) — 2000 volts 4ffN,Si 4*44 X 80 per sec. x 1224 x 8-00 ins. = 0-575 M.V.S. per inch. = 8900C.G.S. Iron loss per unit volume (from curves for Stalloy) — Pi/Vi „ total Pi = (Pi/Vi)Vi Dissipation intensity for iron Pi/Sci Estimated temperature rise for iron — tm = (200** C. per watt/inch*) x Pi/Sci = 29"* C. 0-31 watts per inch. 59 watts. 0-145 watts per inch. DESIGN OF TRANSFORMERS. 257 Excitation (Et.M.S.) required for iron ^i = —^^HLy =101 amp. -turns. Magnetising field (maximum) required for iron (from curve) — rt = 6*0 amp.-tums per inch. „ 3 lap joints — Xj = 3 X (30 amp.-tums per M. V.S./in.2)(S = 52 required, whole Xm = Xj + Xj = 1 53 Iio =Xm/Ni = 0*125 amperes. Iwo = Pi/Vi =0030 Exciting current, idle component „ „ working component „ „ total— lo = (Iio^ + Iwo^)i - (0-0156 + 0-0009)* amps. = 0128 = 4 '95 per cent. Equivalent leakage thickness between coils „ „ ^ of winding depths total = 0'18 inches. -0-52 „ Itl = 0-70 „ Total equivalent resistance, referred to secondary -7- Rt2 = Pc/V = 0180 ohms. „ „ leakage reactance, referred to secondary — J6rL2«-=i2ir;x.fN22(Lc/LF)lTL 0-20 X 10-« henry 80 , „^« 19*0 ^ ^^ . ^ , ,^ = -, — r X — X 126^ X TTo-H X 0-70 ms. =» 0*142 „ mch sec. 23*8 Total equivalent impedance, referred to secondary — Zt2 = (Rt2^ + LtjV)* = (0-033 + 0-020)* ohms « 0-229 „ Internal lag cos + = RT2/ZT2 = 0786 4» = 38*. Voltage drop at unity power-factor = RT2I2 =4-50 volts = 2*25 percent. „ „ zero (lagging) power-factor -=jBrL2*»l2'=^'^^ » =178 „ „ 0-786 „ „ -=.Zt2I2 =5-73 „ =2-86 „ Efficibncy at Unity Power-Factor. Load .... 1250  Input .... 1818 2587 8873 6172 6486 Totalloss. 501 8-87 8-18 8-32 8-64 per cent. Efficiency. 94-99 96-63 96 82 96-68 9686 1 *' 17 258 TRANSFORMERS. Preliminary Desig:n for an Oil-Insulated Transformer of the Rect- angular Coil Core Type to give lo K.V.A. at 6350/230 Volts, i'63/43-s Amperes, and 50 Cycles per Second, using Stalloy Iron. Loss allowed in iron „ „ copper „ geom. mean \ (given) Pi =160 watts = 1-55 per cent. P^ =160 „ =1-55 „ >/P^c =160 „ =1-55 „ Output of secondary winding Input to primary winding Aggregate volt-amperage (given) I5V2 = 10,000 volt-alnperes. (estimated) \^^ =10,320 „ „ (IiVi + IaV^)^ 20,320 Terminal P.D. of primary winding (given) „ secondary „ „ = 6350 volts. = 230 .. Current in primary winding (estimated) I^ = 1*63 amperes. secondary „ (given) = 43-5 Frequency (given) f Loss-length for 20 mils Stalloy (from curve) Ll Resistivity of copper (hot, A.C.) (assumed) pc = 50 cycles per second. = 0'085 mm. = 23xl0"*ohm-mm. Space-factor for iron „ „ copper „ geom. mean , Specific cost ratio (estimated) <ri =0-80. ac =0-50. n/^ =0-632. m = 2ac/ai=l-25i^ Fundamental length — (I^Vi -I- l2Va )LL . 20,320 volt-amps, x 0-085 mm. .. ^ , , "" V^N/PiP^"" 0-632 xl60 watts " * °'"'" Fundamental surface 80= Lo^= 292 mm.^ „ volume Vo = L-o* = 5000 mm.^ mm.2 = 0-286 metres^. = 0-436 Conventional cooling surface of iron — 8ci = (Sci/So)So = 980x292 Conventional cooling surface of copper — .^^/ 8oc-(8cc/So)So « 1490x292 Conventional dissipation intensity for iron — Pi/Sci =160 watts ^ 0-286 metres^ = 560 watts per metre^. Conventional dissipation intensity for copper — Pc/8co =160 watts -^ 0-436 metres2 = 367 „ „ DESIGN OF TRANSFORMERS. 259 Width of iron space — *J lis =(lis/Lo)Lo = 3-49 X 17-1 mm. = 59 6 mm. Distance between yokes — 1 Lcs = (Lcs + 10-40 X 171 „ = 178 „ Depth of copper space — 1 Lcs = (Lcs + 12-65 X 171 „ = 39-5 „ Depth of iron space — Us - (Lcs + 7-65 X 171 „ = 131 „ Length of iron — Li = (4-18 + 2Lcs + 4Lcs + 4Lcs) = (238 + 356 + 158) mm. = 752 mm. Length of mean turn — Lcs = (Lcs + 12-65 X 131 „ Section of iron space — Sis = 1Lcs = 59-6 mm. x 131 mm. = 7,810 mm.^ Section of copper space. — 8c8=2Lcs = 1Lcs =, - 356 „ X 39-5 „ -14,070 „ Section of iron — Si = 0-80 x7810 mm. = 6,248 „ Section of copper — Sc = 0-50 x 14,070 Volume of iron — Vi = 1Lcs = 752 mm. X 6248 Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Vc -Lcs = = 500 „ x7035 „ = 3-52 x 10* „ Volume of copper — Mass of iron — Mi = DiVi - l-^f ^^' X 4-70 x 10« mm.s = 367 kgs. 10* mm. Mass of copper— Mc-DcVc = 8|^:x 3-52x10- „ =313 „ Mass of standard of equal cost — Ms = 1-75Mi + 3Mc =(64-2 + 939) kgs. =158 „ = 349 lbs. Relative cost mas s — = Ms X {s/P^li{lJ, + l,y^)V X (f/50 C.P.S.)« = 158 kgs. X {160/10,160)8 =0-618 grams = l-37xlO-Mb8. Turns in primary winding — N Jl i PcSc ) * 1 (160 watts X 7035 mm.2 | *^qo^o t '^'21, ( pcLc J '"3-26amps. ( 23 x 10" ohm-mm. x 500 mm. ( "" °®' = 2x1515 „ Turns in secondary winding — 2I2 ( pcLc J 870 amps. ( 23 x 10" ohm-mm. x 500 mm. f " ^^* " = 2x57 „ 26o TRANSFORMERS. Section of wire for primary — 8wi = 8c/2Ni = 7035 mm.2-^ 6060 = M6 mm.2 Section of wire for secondary — 8w2 = 8c/2N2 = 7035 „ ^228 -30-8 „ Flux density (maximum) — 6350 volts )T sec. X 3030 x 6248 mm.2 : 15,100 C.G.S. xa. V, 6350 volts IK, iA-8Tra « ®" 4ffN^r4^i4ir50^r-^x 30 30 x 6248 mm > = ^'^^ "" ^^' """• Current density (R.M.S.) — I B Ij/Swi ^ V8w2 = 1 '63 amps. 4-116 mm.* = 1 -40 amps, per mm.* Magnetizing field (maximum) required for iron (from curve) — rt == 1 '20 amp. -turns, per mm. Excitation (R.M.S.) Required for iron X = -, -1*Lf = 0.39 amp.-tums. Xj = 4 X (20 amp.-tums per 10" V.S./mm.2)d8= 121 „ (R.M.S.) required for whole Xm = Xi + Xj «760 „ Exciting current, idle component Xm/Nj = 0.251 amperes. working component Xwo = Pi/Vj = 0.252 „ Total — lo = (1.0 + 1.0) = (0.0630 + 0.0006) amps. - 0.252 = 15.4 per cent. Total equivalent resistance, referred to secondary — Rt2 = Pc/1/2 = 0.0846 ohms. „ leakage reactance, referred to secondary — J6^L2a>^ 2ir;x.fNo2(Lc/2Lc8) x Ics/S ^8^x]lO:^hen2 mm. sec. odd o Total equivalent impedance, referred to secondary — Zts=(Rt2^ + <£tl2^<«>^)*-(7160 + 9280)*x 10-8 ohms = 0*128 „ Internal lag cos 4» = RT2/ZT2 = 0-66 + = 49'. Voltage drop at unity power-factor =.Rt2I2 = 3*68 volts = 1*60 per cent. „ zero (lagging) power-factor •=;jBrL2a)l2 = 4*18 „ =1*82 „ 0*66 „ „ nZtsIj =5-57 „ «2-42 „ Efffcikncy at Unity Power-Facior. Load. 2600 ' 5000 ! 7500 Iron loss. 160 1 160 160 Copper loss 10 , 40 90 Input. 2670 ' 5200 1 7760 ToUlloss. 6-37 8*85 3*28 Efficiency. 93*63 1 96*15 9677 10,000 12,500 wafctB. 160 160 160 , 250 10,820 i 12,910 M 8-10 8*18 per cent 96*90 1 t 96*82 " DESIGN OF TRANSFORMERS. 261 This design has been worked out for the same rating as the one on Plate 7, which also uses alloyed iron, and with which it should be com- pared. It will be noticed that this design is appreciably cheaper, partly owing to a difference in the qualities of the iron, but largely due to the fact that the one in the plate uses square coils instead of rectangular. The next design is also for the same rating, but uses Lohys iron instead of Stalloy. It costs about 75 per cent, more than the one with Stalloy, the equivalent standard masses being 349 and 615 lbs., but its magnetising current is much less, owing to the lower flux density at which it is worked. The saving by using Stalloy, however, is more than enough to permit of an increase of size sufficient to bring the exciting current down to the same figure, when there will be the advantage in its favor of considerably reduced losses. Only the preliminary designs have been worked out for these two; the final designs will not differ greatly from that in Plate 7, except in the actual dimensions. Preliminary Design for an Oil-Insulated Transformer of the Rectangular Coil Core Type to give 10 K.V.A. at 6350/230 Volts, 63/43S Amperes, and 50 Cycles per Second, using Lohys Iron. Loss allowed in iron, copper, and zinc, mean (given) Pi = 160 watts = 1.55 percent. Pc = 160 watts = 1.55 percent. Pc = 160 watts = 1.55 percent. Output of secondary winding, unit to primary, aggregate volt-amperage, (given) IgVg = 10,000 volt-amperes, (estimated) l^V^ = 10,320, (estimated) l^V^ = 10,320, (terminal) P.D. of primary winding, (given) I2, (given) I2 = 6350 volts. V2 = 230 volts. V2 = 230 volts. V2 = 230 volts. Frequency, (given) loss-length for 15 miles, Lohys (from curve) Resistivity of copper (hot, A.C.) (assumed) per 50 cycles per sec. = 0110 mm. = 23 X 10" ohm-mm. 262 TRANSFORMERS. Space factor for iron (estimated) of = 0.80 „ copper „ <rc = 0.50 „ geom. mean „ v^t^c = 0.632 Specific cost ratio „ m = 5.5 Fundamental length — , _ diVi -H laVa)^ _ 20.20 volt-amps, x 0110 mm. Fundamental surface So = Lo = 488 mm. „ volume Vo = U^ = 10,800 mm. Conventional cooling surface of iron — Sci = (8ci/8o)So = 1050 x 488 mm. = 0512 meters. Conventional cooling surface of copper — 8cc = (Scc/8o)8o = 1460x488 „ = 0.712 „ Conventional dissipation intensity for iron — Pi/8ci = 160 watts -^ 0.512 meters = 313 watts per meter. Conventional dissipation intensity for copper — Pc/8oc = 160 „ -0.712 „ = 225 „ Width of iron space — lis = (li8/Lo)Lo = 3.80 x 22.1 mm. = 84.0 mm. Distance between yokes — Lcs=(Lcs/Lo)U = 9.80x22.1 „ = 217 „ Depth of copper space — Ics =(lc8/Lo)Lo = 2.15x22.1 „ = 47.5 „ Depth of iron space — Ics =(lc8/Lo)Lo = 2.15x22.1 „ = 47.5 „ Depth of iron space — Ics =(lc8/Lo)Lo = 2.15x22.1 „ = 47.5 „ Depth of iron space — Ics =(lc8/Lo)Lo = 2.15x22.1 „ = 47.5 „ Depth of iron space — Ics =(lc8/Lo)Lo = 2.15x22.1 „ = 47.5 „ Depth of iron space — Ics =(lc8/Lo)Lo = 2.15x22.1 „ = 47.5 „ Depth of iron space — Ics =(lc8/Lo)Lo = 2.15x22.1 „ = 47.5 „ Depth of iron space — Ics =(lc8/Lo)Lo = 2.15x22.1 „ = 47.5 „ Depth of iron space — Lis=(Li8/10)U = 8-33x22-1 = 184 „ Length of iron — Li = (4lis + 2Lcs + 1Lcs) = (336 + 434 + 190) mm. = 960 mm. Length of mean turn — Lc =.(2lis + 3lcs + 2Lis) =(168 + 143 + 368) „ =679 „ Section of iron space — 8i8 = lisLis = 840 mm. x 184 mm. = 15,450 mm.* Section of copper space — 8c8 = 2Lcslc8 =434 „ X47-5 „ =20,600 „ Section of iron — 8i =<ri8is =0-80x15,460 mm.2= 12,360 „ Section of copper — 8c =acScs =0-50x20,600 „ =10,300 „ Volume of iron — Vi =Li8i =960 mm. X 12,360 mm.* = 11-87 x 10« mm.^ Volume of copper — Vc = LcSc = 679 „ X 10,300 „ = 7-00 x 10« „ DESIGN OF TRANSFORMERS. 263 Mass of iron — Mi = DiVi =. ?7^^ X 11-87 X 10» mm.^ = 925 kgs. 10® mm.8 ° Mass of copper — Mc=DcVc ^ 8-90kg8 y.QQ^^Q, ^gg.g 10** mm.^ Mass of standard of equal cost — Ms =Mi + 3Mc .= (92-5 + 186-9) kgs. =280 „ = 615 lbs. Relative cost mass — =.Ms X {^/PiP^/i(IlVl + IjV^)}' X (f/50 C.RS.)i = 280 kgs. X {160/10,160}8 = 1095 grams = 2-42 xl0-« lbs. Turns in primary winding — M -.± f PcScl_ _1 J 160 watts X 10,300 mm.« ( h ^, ^^ 2x1578 „ Turns in secondary winding — N -- i PcSc l*_ 1 / 160 watts X 10,300 mm.^ l^^i.o 2x1578 „ Turns in secondary winding — N -- i PcSc l*_ 1 / 160 watts X 10,300 mm.^ l^^i.o 2x1578 „ Turns in secondary winding — 8w2 = 8c/2Ni = 10,300 mm.2 + 6312 = 163 mm.2 Section of wire for primary — 8w2 = 8c/2Ni = 10,300 mm.2 + 6312 = 163 mm.2 Section of wire for secondary — 8w2 = 8c/2Ni = 10,300 mm.2 + 6312 = 163 mm.2 Flux density (maximum) — V, 6350 volts ^ ^^ -^_. „ ^ ^7/Ki e = m — KA olTc — To-QfiA 2 = ^'736 ^ 1^ V.S. per mm.^ 4fTNiSi 4*44 X 50 per sec. x 3156 x 12,360  curve) — If = 0.21 amp.-tums per mm. Excitation (R.M.S.) required for iron Xi = -^IfLp = 143 amp.-tums. „ „ 4 lap joints— Xj = 4 X (20 amp.-tums per 10"^ V.S./mm.2)0 = 59 „ (R.M.S.) required, whole Xm = Xj + Xj = 202 Exciting current, idle component Iio = Xm/Ni = 0.0641 amperes. „ working component Iwo = Pil^i — 0.0252 „ „ total — lo = (Iio^ + Iwo^)* = (4100 + 634)* x 10-8 amps. = 0.0689 = 4.23 per cent. 264 TRANSFORMERS. Total equivalent resistance, referred to secondary — Rtj = Pc/Ig^ = 00846 ohm. „ leakage reactance, referred to secondary — CXLjom 2irii.fN2*'^(Lc/2Lcs) X lcs/3 8 X 10-» henry 50,, ^„ 679 475 mm. ^, ^o = ^- X - X 1 182 X j^ X ^ 0-138 mm. sec. 434 3 Total equivalent impedance, referred to secondary — ZT2 = (RT,^ + <CTL2V)i = (0-0072 + 00191)* ohms =0162 Internal lag'cos ♦ = Rt2/Zt2 = 0*52 + = 59\ Voltage drop at unity {tower-factor = IIT2I2 ^ ^'^^ ^^^ ^^ '^0 per cent. „. „ zero (lagging) power-factor =<£tl2*»I2'^^*00 „ =2-61 „ „ 0-52 „ „ ^ZTgl., =7-05 „ =3-07 „ Efficiency at Unity Power-Factor. Load.... 2500, 5000   96-90 , 96-82 Preliminary Design for an Oil-Insulated Transformer of the Rectangular Coil Core Type to give 100 K.V.A. at 6000/2200 Volts, 17/45/5 Amperes, and 50 Cycles per Second, using Stalloy Iron. Loss allowed in iron, copper, mean. (given) = 900 watts = 0-885 per cent. = 900 watts = 0-885 per cent. = 0-885 = 0-885 = 0-885 Output of secondary winding (given) IgVj = 100 kilo-volt-amperes. Unit to primary winding (estimated) I^Vj = 101-8 Aggregate volt-amperage = (I^V^ + IjV2) = 20r8 Terminal P.D. of primary winding (given) V^ = 6000 volts. „ secondary „ „ Vj = 2000 „ Descidence of Transformer. 265 Current in primary winding (estimated) „ secondary „ (given) Frequency (given) Loss-length for 20 miles Stalloy (from curve) Resistivity of copper (hot, A.C) (assumed) = 17-0 amperes. = 45-5 „ f Ll PC = 50 cycles per second. = 3-20 mils. = 0-90 X 10-« ohm-inches. Space-factor for iron „ „ copper „ geom. mean Specific cost ratio (estimated) ai = 0-80 ac -0 34 x/^ =0-521 m = 2<rc/'<ri = 0*85 Fundamental length — , (IiVi-hI2V2 )Ll 201-8 K.V.A.X 3-20 mils ' _ . . , Lo = t—^ li^ ^ — = e\ KOI — S7u\ 71 = 1 '38 mches. v/^j[^^/PjP^, 0-521 X 900 watts Fundamental surface 80 = Lo^ = 1 '^^ inches*. „ volume Vo = Lo* = 2 62 inches*. Conventional cooling surface of iron — Sci = (8ci/8o)So = 943x1-90 Conventional cooling surface of copper- Sec = (Scc/So^So = 1520 X 1-90 Conventional dissipation intensity for iron — Pi/8ci = 900 watts -r 1 790 Conventional dissipation intensity for copper- Pc/8cc = 900 watts -r- 2890 inche8'«=1790inches2. = 2890 = 0-502 watts per inch*. = 0-312 Width of iron space — lis =(li8/Lo)Lo Distance between yokes Lcs = (Lc8/U)U Depth of copper space — ics =(lc8/Lo)Lo Depth of iron space — Li8«(Li.s/U)U = 3-28x1-38 = 10-92x1-38 2-45x1-38 7-28x1-38 inches = 4*52 inches. „ =15-08 „ „ = 3*38 „ » =10-05 „ Length of iron — Li = (4lis + 2Lcs + 4lc8) = (18-08 -h 30*16 + 13-52)inches = 61-8 inches. Length of mean turn — Lc-=(2lis + 3lc8 + 2Lis) = (904 + 10-14 -H 20-10) „ =393 „ 266 TRANSFORMERS. Section of iron space — Sis = lisLis = ^'52 ins. x 10*05 inches = 45*5 inches^. Section of copper space — Sc = 2Lcslcs = 30-16 „ x 3-38 „ = 1018 „ Section of iron — Si = <riSis Section of copper — Sc — •'cScs = 0-80 = 0-34 X 45-5 inches = 36 4 X 101-8 „ = 34-6 Volume of iron — Vi = LiSi Volume of copper- Vc=LcSc «6r8ins. x36-4 -39-3 „ x34-6 inches^ = 2250 inches. » =1360 „ Mass of iron — Mi = D,Vi Mass of copper — Mc=DcVc 0-282 lb. inch^ 0-322 lb. inch* [2250 :1360 inches* = 634 lbs. „ =438 „ Mass of standard of equal cost — M8=l-75Mi+3Mc =(1108+1314) lbs. = 2422,, =1100kg8. Turns in primary winding — N = 1PcScl*_ 1 I 900 W 8 2Ii I pcLc i 340 amps. I 090 x 10"' watts X 34-6 ins.* ohm-ins. x 39-3 -— l* = 874 turns. ) ms. j = 2x437 „ ' Turns in secondary winding — N =_^ / PqSc I*^ 1 J 9 00 watts x 34- mounds x 34-6 ins.* ohm-ins. x 39-3 -— l* = 874 turns. ) ms. j = 2x437 „ ' Turns in secondary winding — N =_^ / PqSc I*^ 1 J 9 00 watts x 34- mounds x 34-6 ins.* ohm-ins. x 39-3 -— l* = 874 turns. ) ms. j = 2x437 „ ' Turns in secondary winding — N =_^ / PqSc I*^ 1 J 9 00 watts x 34- mounds x 34-6 ins.* ohm-ins. x 39-3 -— l* = 874 turns. ) ms. j = 2x437 „ ' Turns in secondary winding — N =_ 6 in s. 1 = 396 2I2 t pcLc i 91-0 amps. I 090 x lO"" ohm-ins. 3'9-3 ins: J = 2x163 „ Section of wire for primary — Swi = Sc/2Ni = 34-6 ins. 2-^1748 Section of wire for secondary — Sw2 = Sc/2N.2 = 34-6 ins. 2-^1748 = 19-8 X 10-' inches. = 531x10-' „ Flux density (maximum) — /«•= ^1 = _ 6000 volts 4ffNiSi~4-44 X 50 per sec. x 874 x 36*4 ins.^ Current density (R.M.S.) — f = 1/Swi = 1/Sw2= 17 amps. + 19*8 x lO"* ins.2 = 0-850 M.V.S. per inch. = 13,180 C.G.S. = 859 amps, per inch. DESIGN OF TRANSFORMERS. Design for an Oil-Insulated Transformer of the Rectangular Coil Core Type, to give 100 K.V.A. at 6000/2200 Volts, 17/45"5 Amperes, and 50 Cycles per Second, using Stalloy Iron. (Plate 10.) Primary. Secondary Rated voltage „ current Number of turns „ coils „ turns per coil „ layers per coil „ turns per layer Conductor, dimensions, bare „ „ covered „ section „ total section of copper Sc = „ current density f = 1000 170 2200 45-5 864 8 108 9 12 320 2 160 5 32 :200x 100 360x150 1230x130 390x180 Sw i 20-0 540 NSwi 17-3 17-3 I/Sw 850 843 Depth of winding loss Axial length of coil, allowing one extra space for spiral Perimeter of coil section after taping Lpc Former, dimensions of section „ perimeter of section Allowance for corners Length of mean turn Lc „ wire Lw=NLc Sec Vc = LcSc M7 2-99 0-90 12-9 8-52'27-8 8i X 13J I 5^ X lOf I 44-0'320 I 3-5 2-7 47 5 34-7 I 41,000 j 11 M7 2-99 0-90 12-9 8-52'27-8 8i X 13J I 5^ X lOf I 44-0'320 I 3-5 2-7 47 5 34-7 I 41,000 j 11 M7 2-99 0-90 12-9 8-52'27-8 8i X 13J I 5^ X lOf I 44-0'320 I 3-5 2-7 47 5 34-7 I 41,000 j 11 M7 2-99 0-90 12-9 8-52'27-8 8i X 13J I 5^ X lOf I 44-0'320 I 3-5 2-7 47 5 34-7 I 41,000 j 11 M7 2-99 Cooling surface of coils Volume of copper Mass of copper Mass of copper 3240 I 1930 820 I 600 264! 193 Effective resistance, hot R = 0-90 X 10-* ohm-ins. x Lw/Sw Copper loss Pc = RI^ Dissipation intensity for coils Pc/Scc Estimated temperature rise — RC = (200"^ C. per watt/in.2) Rc/Soc Total. 1-84 634 0165 33 0185 383 0198 40 34-6 847 2-07 411 5170 1420 457 Units. volts, amperes. turns. mills. I 10-8 ins.2 inches. amps, per in.2 inches. inches. inches. lbs. 0-443'ohms. 917 I watts. 0-177 I watts I per in.2 36 I C. 268 TRANSFORMERS. Width of core. lis = 4| inches. 