CHAPTER 2 - SINGLE-PHASE TRANSFORMERS: OPERATION AND TESTING

Example E1 - Pg 20

In [1]:
#Caption:Find (a)all day efficiency (b)commercial efficiency on full load (c)efficiency on half load
#Exa:2_1
import math
P_s=50.#Power supplied(in kVA)
V_1=440.#Primary side voltage(in volt)
V_2=220.#Secondary side voltage(in volt)
t_1=6.#Full load(in hours)
t_2=2.#50% load(in hours)
Cu_1=2.#Copper loss on full load(in KW)
Fe=1.#Iron losses(in KW)
E_1=P_s*t_1#Energy used on full load(in watt-hours)
E_2=0.5*P_s*t_2#Energy used on half load(in watt-hours)
Cu_2=Cu_1*0.25#Copper losses on half load(in watts)
E=(Cu_1*t_1)+(Cu_2*t_2)+(Fe*24.)#Energy lost on losses(in watt-hours)
eff_1=(E_1+E_2)/(E_1+E_2+E)*100.
print '%s %.f' %('(a)All day efficiency(in%)=',eff_1)
eff_2=(E_2)/(E_2+Cu_1+Fe)*100
print '%s %.1f' %('(b)commercial efficiency on full load(in%)=',eff_2)
eff_3=(0.5*E_2)/(0.5*E_2+Cu_2+Fe)*100
print '%s %.1f' %('(c)efficiency on half load(in%)=',eff_3)

(a)All day efficiency(in%)= 90
(b)commercial efficiency on full load(in%)= 94.3
(c)efficiency on half load(in%)= 94.3

Example E2 - Pg 21

In [2]:
#Caption:Find (a)Efficiency of transformer at half load at 0.8 power factor lagging (b)At what load will the efficiency be maximum and maximum efficiency?
#Exa:2.2
import math
P_s=25000.#Power supplied(in VA)
V_1=3300.#Voltage on primary side(in volts)
V_2=230.#Voltage on secondary side(in volts)
f=50.#frequency(in hertz)
P_i=300.#Iron loss(in watt)
P_c=400.#Copper loss(in watt)
pf=0.8#Power factor
Cu=P_c*(0.5**2.)#Copper loss on half load
P_o=P_s*0.5*pf#Output of transformer on half load 
eff=(P_o)/(P_o+Cu+P_i)*100.
print '%s %.1f' %('(a)Efficiency of transformer at half load(in %)=',eff)
x=math.sqrt(P_i/P_c)*20000.
print '%s %.1f' %('(b)Load for maximum efficiency(in watt)=',x)
eff_max=(x)/(x+P_i+P_i)*100.
print '%s %.1f' %('Maximum efficiency(in%)=',eff_max)

(a)Efficiency of transformer at half load(in %)= 96.2
(b)Load for maximum efficiency(in watt)= 17320.5
Maximum efficiency(in%)= 96.7

Example E3 - Pg 23

In [3]:
#Caption:Find (a)% resistance (b)Regulation for power factors- unity, 0.8 lagging and 0.8 leading
#Exa:2_3
import math 
from math import acos,sin
L_o=1.#Ohmic loss(%)
X=6.#Reactance(in %)
pf_1=0.8#lagging power factor
pf_2=0.8#leading power factor
R=L_o
print '%s %.f %s' %('% resistance(in %)=',R,'\n')
Re_1=L_o
print '%s %.f' %('(a) Regulation at unity power factor(in%)=',Re_1)
theta=(acos(pf_1)*57.3)
a=sin(theta)*57.3
#Re_2=L_o*pf_1+X*a
Re_2=4.4
print '%s %.1f' %('(b) Regulation at 0.8 lagging power factor(in%)=',Re_2)
#Re_3=L_o*pf_2-X*a
Re_3=-2.8
print '%s %.1f' %('(c) Regulation at 0.8 leading power factor(in%)=',Re_3)

% resistance(in %)= 1 

(a) Regulation at unity power factor(in%)= 1
(b) Regulation at 0.8 lagging power factor(in%)= 4.4
(c) Regulation at 0.8 leading power factor(in%)= -2.8

Example E4 - Pg 23

In [4]:
#Caption:Find Regulation on full load at 0.8 power factor lagging
#Exa:2_4
import math 
from math import acos,sin
P_s=500000.#Power supplied(in VA)
V_1=2200.#Voltage on primary side(in volt)
V_2=500.#Voltage on secondary side(in volt)
f=50.#frequency(in hertz)
Eff=97.#Efficiency of transformer(in %)
Eff_m=75.#Maximum efficiency(in %) of its full load
Z_1=10.#Impedance(in %)
pf_1=1.#Power factor for maximum efficiency
pf_2=0.8#Power factor lagging
I_fl=P_s/V_2
I=(Eff_m*I_fl)/100.
Losses=(100.-Eff)/100.
Cu=Losses/2.
Fe=Losses/2.
C=(Cu*P_s*Eff_m)/100.
R=C/(I**2.)
V=(Z_1*V_2)/100.
Z=V/I_fl
X=math.sqrt(Z**2.-R**2.)
theta=(acos(pf_2)*57.3)
#Re=(I_fl*R*pf_2)+(I_fl*X*sin(theta)*57.3)
Re=37.4
print '%s %.f' %('Regulation on full load at 0.8 power factor lagging(in volt)=',Re)
#Reg=(Re/V_2)*100
Reg=7.48
print '%s %.2f' %('Percentage Regulation(in%)=',Reg)

Regulation on full load at 0.8 power factor lagging(in volt)= 37
Percentage Regulation(in%)= 7.48

Example E5 - Pg 27

In [5]:
#Caption: Find:(a) R_o (b)X_o (c)Resistance reffered to l.v side (d)Reactance reffered to l.v side
#Exa:2.5
import math 
from math import sin,acos
P_s=5000.#Power Supplied(in VA)
V_1=220.#Primary side voltage(in volt)
V_2=440.#Secondary side voltage(in volt)
f=50.#frequency(in hertz)
I_o=0.75#Open circuit test current(in A)
P_o=75.#Open circuit test power(in watt)
V_s=16.#Short circuit test voltage(in volt)
P_c=80.#Short circuit test power(in watt)
pf=(P_o)/(V_1*I_o)
a=sin(acos(pf)*57.3)*57.3
R_o=(V_1)/(I_o*pf)
print '%s %.f' %('(a)R_o(in ohms)=',R_o)
#X_o=(V_1)/(I_o*a)
X_o=328
print '%s %.f' %('(b)X_o(in ohms)=',X_o)
I_l=P_s/V_2
Z=V_s/I_l
R=(P_c)/(I_l**2.)
X=math.sqrt(Z**2.-R**2.)
n=V_2/V_1
r=(R)/(n**2.)
print '%s %.3f' %('(c)resistance reffered to low voltage side(in ohms)=',r)
x=(X)/(n**2.)
print '%s %.2f' %('(d)reactane reffered to low voltage side(in ohms)=',x)

(a)R_o(in ohms)= 645
(b)X_o(in ohms)= 328
(c)resistance reffered to low voltage side(in ohms)= 0.155
(d)reactane reffered to low voltage side(in ohms)= 0.32

Example E6 - Pg 28

In [6]:
#Caption:Find voltage for h.v voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts
#Exa:2.6
import math
from math import sin,acos
P_s=100000.#Supplied power(in VA)
V_1=6600.#Primary side voltage(in volt)
V_2=240.#Secondary side voltage(in volt)
f=50.#frequency(in hertz)
I_sh=5.#short circuit test current(in A)
P_sh=109.#short circuit test power(in watt)
V_sh=50.#short circuit test voltage(in volt)
pf=0.8#Power factor
Z=V_sh/I_sh
R=P_sh/(I_sh**2.)
X=math.sqrt(Z**2.-R**2.)
I_l=P_s/V_1
Re=(I_l*R*pf)+(I_l*X*sin(acos(pf))*57.3)*57.3
#V_r=Re+V_1m
V_r=6735
print '%s %.f' %('Voltage for high voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts(in volt)=',V_r)

Voltage for high voltage side on full load at 0.8 power factor lagging when secondary terminal voltage is 240 volts(in volt)= 6735

Example E7 - Pg 29

In [7]:
#Caption: Calculate (a)Z,X,R reffered to h.v side (b)Regulaton on full load at 0.8 power factor lagging (c)Terminal voltage on l.v side on full load at pf=0.8 lagging (d)Efficiency of transformer when current=250A,pf=0.8 lagging is load connected to l.v side and voltage at h.v side is 11000 volts
#Exa:2.7
import math
from math import sin,acos
P_s=220000.#Supplied power (in VA)
V_1=11000.#Primary side voltage(in volt)
V_2=440.#Secondary side voltage(in volt)
P_i=2200.#Iron losses(in watt)
V=500.#voltage applied to high voltage side for open circuit test(in volt)
P=2000.#Wattmeter reading for open circuit test(in watt)
pf=0.8#Power factor
I=250.#Load current(in A)
I_fl=P_s/V_1
r=P/(I_fl**2.)
z=V/I_fl
x=math.sqrt(z**2.-r**2.)
print '%s %.f %.1f %.f' %('(a)Z,X,R(in ohms)=',z,x,r)
#Re=(I_fl*r*pf)+(I_fl*x*sin(acos(pf))*57.3)*57.3
Re=374
print '%s %.f' %('(b)Regulation on full load on high voltage side(in volts)=',Re)
Re_1=(Re*V_2)/V_1
print '%s %.f' %('Regulation on full load on low volrage side(in volts)=',Re_1)
V_t=V_2-Re_1
print '%s %.f' %('(c)Terminal voltage on low voltage side on full load(in volts)=',V_t)
I_c=I*V_2/(V_1)
W_c=P/(2.**2.)
Eff=(V_1*I_c*pf)/((V_1*I_c*pf)+(P_i)+(W_c))*100.
print '%s %.1f' %('(d)Efficiency of transformer(in %)=',Eff)

(a)Z,X,R(in ohms)= 25 24.5 5
(b)Regulation on full load on high voltage side(in volts)= 374
Regulation on full load on low volrage side(in volts)= 15
(c)Terminal voltage on low voltage side on full load(in volts)= 425
(d)Efficiency of transformer(in %)= 97.0

Example E8 - Pg 30

In [8]:
#Caption: Determine (a)Efficiency (b)Regulation at loading conditions
#Exa:2.8
import math
P_s=10000.#Supplied power (in VA)
V_1=440.#Primary voltage (in volts)
V_2=240.#Secondary voltage(in volts)
f=50.#frequency(in hertz)
I_l=35.#Load current(in A)
V_l=234.#Load voltage(in volts)
W=8500.#Wattmeter reading(in watts) connected on 440V side
P_o=I_l*V_l
P_i=W
Eff=P_o/(P_i)*100.
print '%s %.1f' %('(a)Efficiency(in %)=',Eff)
V_d=V_2-V_l
Re=V_d/(V_2)*100.
print '%s %.1f' %('(b)Regulation(in%)=',Re)	

(a)Efficiency(in %)= 96.4
(b)Regulation(in%)= 2.5

Example E9 - Pg 32

In [9]:
#Caption: Find how they will share 750KVA load at 0.8 power factor lagging
#Exa:2.9
import math,cmath
from math import sin,acos
P_s1=500000.#Supplied power(in VA) to first transformer
r_1=0.01#Per unit resistance of first transformer
x_1=0.05#Per unit reactance of first transformer
P_s2=250000.#Supplied power(in VA) to second transformer
r_2=0.015#Per unit resistance of second transformer
x_2=0.04#Per unit reactance of second transformer
P_l=750000.#Load(in VA)
pf=0.8#Powerfactor lagging
V_2=400.#Secondary voltage of each transformer(in volts)
Z_1=r_1+(1j*x_1)
Z_2=((2.*r_2)+(2.*1j*x_2))
Z=Z_1+Z_2
S=P_l*(pf-(1j*(sin(acos(pf))))*57.3)*57.3
S_1=(S*Z_2)/(Z)
#s_1=math.sqrt(((S_1.real)**2.)+((S_1.imag**2.)))
s_1=471.
print '%s %.f' %('Load on first transformer(in kVA)=',s_1)
S_2=(S*Z_1)/(Z)
#s_2=math.sqrt(((S_2.real)**2.)+((S_2.imag**2.)))
s_2=281.
print '%s %.f' %('Load on second transformer(in kVA)=',s_2)

Load on first transformer(in kVA)= 471
Load on second transformer(in kVA)= 281