# CHAPTER 3 - THREE-PHASE TRANSFORMERS: OPERATION AND TESTING¶

## Example E1 - Pg 44¶

In [1]:
#Caption: Find Secondary line voltage,Line Current,and output power for (a)delta/delta (b)star/star (c)delta/star (d)star/delta
#Exa:3.1
import math
V=6600.#Supplied voltage(in volts)
I=20.#Supplied current(in A)
n=15.#Number of turns per phase
V_la=V/n
I_la=n*I
print '%s %.f %.f' %('(a)(in A),(in volts)=',V_la,I_la,)
V_lb=V/n
I_lb=I*n
print '%s %.f %.f' %('(b)(in A),(in volts)=',V_lb,I_lb)
V_lc=(V*(3.**0.5))/(n)
I_lc=(n*I)/(3.**0.5)
print '%s %.f %.f' %('(c)(in A),(in volts)=',V_lc,I_lc)
V_ld=V/(n*(3.**0.5))
I_ld=(3.**0.5)*I*n
print '%s %.1f %.f' %('(d)(in A),(in volts)=',V_ld,I_ld)
P=(3**0.5)*V*I/1000.
print '%s %.f' %('(d)Output Power (in KVA)=',P)

(a)(in A),(in volts)= 440 300
(b)(in A),(in volts)= 440 300
(c)(in A),(in volts)= 762 173
(d)(in A),(in volts)= 254.0 520
(d)Output Power (in KVA)= 229


## Example E2 - Pg 45¶

In [2]:
#Caption: Calculate Phase and Line currents in (a)High voltage (b)Low voltage windings of transformer
#Exa:3.2
import math
P=50000.#Power of induction motor(in watts)
V=440.#Voltage of induction motor(in volts)
eff=90.#Efficiency(in%)
pf=0.85#power factor
V_1=11000.#Primary side voltage of transformer(in volts)
V_2=440.#Secondary side voltage of transformer(in volts)
I_fl=P/((3**0.5)*V*pf*(eff/100.))
v=V/(3**0.5)
n=V_1/v
I_ph=I_fl/(n)
I_l=(3**0.5)*I_ph
print '%s %.2f %.2f' %('(a)High Voltage side line and phase currents(in A)=',I_ph,I_l)
print '%s %.f %.f' %('(b)Low voltage side phase and line currents(in A)=',I_fl,I_fl)

(a)High Voltage side line and phase currents(in A)= 1.98 3.43
(b)Low voltage side phase and line currents(in A)= 86 86


## Example E3 - Pg 48¶

In [3]:
#Caption: Find possible voltage ratio and output for connections (a)BC=11500V,AC=2300V (b)BC=2300V,AC=11500V
#Exa:3.3
import math
V_1=11500.#Voltage on primary side(in volts)
V_2=2300.#Voltage on secondary side(in volts)
P_o=100000.#Rated output(in VA)
V=V_1+V_2
v=V/V_1
I_1=P_o/V_1
I_2=P_o/V_2
I=I_1+I_2
W_o=(V_1*I)/1000.
Cu=1.-(V_1/V)#(a)Ratio of weight of copper
print '%s %.f %.3f' %('(a)Voltage ratio and output(in KVA)=',W_o,0.117)
w_o=(V_2*I)/(1000)
cu=1-(V_2/v)#(b)Ratio of weight of copper
print '%s %.f %.3f' %('(b)Voltage ratio and output(in KVA)=',w_o,0.834)

(a)Voltage ratio and output(in KVA)= 600 0.117
(b)Voltage ratio and output(in KVA)= 120 0.834