Electrical Machine Design

Legend: : Formula | : Limits, Range | : Assumption, etc.

Item 2. Total Number of turns, T
T_L.T. = E_L.T./Vt_formula.
T_H.T. = a T_L.T..
Where: a = ratio of transformation.
Subject for correction, should the copper and iron losses differ greatly from the guaranteed losses.

Item 3. Number of Coils, C
C_L.T., customarily assumed to be 4 coils., or divisible by 4 so that the coils maybe efficiently grouped for series or parallel connection.
C_H.T., customarily assumed to be 2 coils, that is one for each limb.

Item 4. Turns per Coil, T/C
T/C = Turns (item 02) / Coils (item 03)..

Item 5. Full-Load Current, amp, I
I_L.T. = KVA/E_L.T.
I_H.T. = a I_L.T..
Where: a = ratio of transformation.
For primary current, the exciting current is usually neglected during initial design.

Space Factor, sf
Use: sf = 10 / (30 + kv) ,
where kv = the kilovolts at the high tension terminlas for transformers between 50- and 200-kva output.
Should be 20 per cent larger than that calculated for transformers of 1,000 KVA or more and 20 per cent smaller to transformers having the ratings up to about 7½ kva output.

Item 6. Current Density, Amp per square inch, D
Use: D = I/CS.
For standard distribution transformer : 900 - 1,600 amp per sq. in.,
for Central General Stations, 1,500 - 2,100 amp per sq. in. and
for large, oil-insulated, water-cooled, or oil forced-circulated transformers: 1,800 - 2,600 amp per sq. in.
For secondary winding, initial value is assumed to be 900 amp per sq. in. For correction when the cross sectional area of each winding is finalized.

Note: Solve the Dimensions of the Window, Item 24 to Item 32, before proceeding further.

Item 7. Cross section of each conductor, square inch, CS
Use: CS_L.T. = I_L.T. / D_L.T.
Sizes of conductor are limited by the dimensions of the window. For correction when the dimension of the window is finalized.
For high tension, trial CS_H.T. = CS_L.T. / a
where: a = ratio of transformation.

Item 8. Dimension of conductors, t x w
Use: w_L.T = H - 2x dcc)/ (T/layer_L.T.).
Double-cotton covering [dcc] is assumed as ¼ inch.
Use: t_L.T = CS_L.T. / w_L.T..
For w_H.T. select a suitable wire from the table and take note of the Cross Sectional Area in sq. in., it should be near the calculated value in item 7 above.

Item 9. Number of turns per layer per coil, T/layer/coil
Use: T/layer/coil_L.T. = T/coils, since it is imperative that there is only one layer per coil in the secondary to give ample space for the primary winding, which usually occupy more space in the winding layer.
Use: T/layer/coil_H.T. = H_layer x T/inch..
Where the T/inch is the approximate number of turns per inch dcc from the wire table.

Item 10. Number of layers per coil, _layers/coil
For layers/coil_H.T. = ( T/coil_H.T. ) / (T/layers/coil_H.T.)
For layers/coil_L.T. usually assumed as 1 layer per coil.

Item 11. Taps
Assume no taps.

Item 12. Volts per coil, V/coil
V/coil = E / coils..
V/coil rarely exceed 5,000.

Item 13. Volts per layers, V/layers_H.T. only
V/layers= V/coil_H.T. x [(coils x Turns/layer/coil) / (Turns/coil).
The potential difference between layers must be between 200 to 300 volts.

Item 14. Length of winding layer, H_layer
Use: H_layer = H - L_ins.
For secondary winding, L_ins is usually assumed as ¼ inch at each end.
L_ins for primary winding is usually assumed as 3/8 inch at each end.

Item 15. Length of layer including insulation, L_layer+ins
L_{layer + ins} = H
The height of the windows, H limits this value.

Item 16. Insulation between layers, inch, L_ins
Assume a 0.01 inch at each end of the layer for clearance and insulation for the primary winding.

Width of the Opening, D
For the computation of the space availability, D > t + s + g + p, within the opening, the following must be calculated:

Material
Radial thickness, in,

t = 0.125

Cotton tape binding

0.018

Varnished cambric
0.010

Pressboard
0.300

Clearance
0.067

s = 2(t + 0.020) + 0.015

Inner L.T. coil (one layer) incl. cotton on wire
t + 0.020

Parchment paper (0.010) b/w coils; allow
0.015

Outer L.T. coil (one layer)
t + 0.020

g = 0.020 + 0.020 +0.018 +0.005 + note

Pressboard around outer L.T. coil
0.020

Spacers of treated wood to form ducts
note

Pressboard
0.020

2 layers cotton tape
0.018

Parchment paper for H.T.
0.005

p = (layers/coils) x D + [(layers/coil)-1] x 0.010 + 0.080 + 2% x S

Layers/coil X Diameter of H.T. coil
(layers/coils) x D

Two 0.005 paper b/w layers
[(layers/coil)-1] x 0.010

Cotton tape, half lapped
0.018

Allowance for bulging
2% of S

Note For spacers: use at least 3/16 to 1/4 inch for oil ducts plus the insulation using 0.2 + 0.035 kv, where kv is the rms value of the kilovolts.

Space Available on D
Use: D/2 which should be greater than the total of t + s + g + p, above.

Item 17. Insulation on wire, _t
Assume double cotton-covered insulation on wire for both secondary and primary windings.
Assume a ¼ inch at each end of the layer for clearance and insulation for the secondary winding.

Item 18. Length per turn, feet, m.l.t.
m.l.t._L.T. = [2 ( L + S ) + p(2t + s)]/12.
m.l.t._H.T. = [2 ( L + S ) + p(2t + 2s + p)]/12.

Item 19. Total length, all turns in series, feet, WL
WL = T x m.l.t. ft.

Item 20. Weight in copper, lb, W_Cu
W_{Cu L.T.} = 0.32 (t x w x m.l.t. x 12) lb.
W_{Cu H.T.} = Wt/lb/1K-ft x m.l.t._H.T..
where the weight per lb. per 1,000 ft., Wt/lb/1K-ft is read from the wire table.

Item 21. Resistance at 75°C-all coils in series, R_75°C
R_{75°C H.T.} = R/1K-ft_60°C x (234.5 + 70)/ (234.5 +60) x m.l.t./12
R_{75°C L.T.} = R_{75°C H.T.} x (CS _H.T. / CS _L.T.) x (m.l.t./1000) ohm.
Where the resistance per 1,000 ft at 60°C, R/1K-ft_60°C is read from the wire table.

Item 22. IR drop, volts, IR
IR = I x R_75°C

Item 23. Full-load Copper loss, W_Cu
W_Cu = I²R = I²R_L.T. + I²R_H.T..

Note: When Total Copper Loss is reasonable, within the guarantee, then proceed to Item 33 for the calculation of efficiences.

Items 24 - 32