Chapter 9: Single Phase Transformer

Example 1: pg 305

In [1]:
# Exa 9.1
#pg 305
#calculate the number of turns, load current
# Given data
V1 = 3000.;# in V
V2 = 300.;# in V
N2 = 86.;# in Turns
Rating = 60.*10**3;# in VA
#calculations
K = V2/V1;
#Transformer ratio,  N2/N1 = K;
N1 = N2/K;# in turns
I2 = Rating/V2;# in A
I1 = Rating/V1;# in A
#results
print "The numbers of primary turns is",N1
print "The secondary full load current in A is",I2
print "The primary full load current in A is",I1
The numbers of primary turns is 860.0
The secondary full load current in A is 200.0
The primary full load current in A is 20.0

Example 2: pg 306

In [2]:
# Exa 9.2
#pg 306
#calculate the max flux density
# Given data
E1 = 3000.;# in V
E2 = 200.;# in V
f = 50.;# in Hz
a = 150.;# in cm**2
N2 = 80.;# turns
#calculations
#Formula E2 = 4.44*phi_m*f*N2;
phi_m = E2/(4.44*f*N2);# in Wb
Bm = phi_m/(a*10**-4);# in Wb/m**2
#results
print "The maximum flux density in Wb/m^2 is",round(Bm,2)
The maximum flux density in Wb/m^2 is 0.75

Example 3: pg 306

In [3]:
# Exa 9.3
#pg 306
#calculate the load current, emf and core flux
# Given data
N1 = 500.;
N2 = 40.;
E1 = 3000.;# in V
f = 50.;# in Hz
Rating = 25*10**3;# in VA
#calculations
K = N2/N1;
I1 = Rating/E1;# in A
print "The primary full load current in A is",round(I1,2)
I2 = I1/K;# in A
print "The secondary full load current in A is",round(I2,1)
# K = E2/E1;
E2 = K*E1;# in V
print "The secondary emf in V is",round(E2)
# e.m.f equation of the transformer, E1 = 4.44*phi_m*f*N1;
phi_m = E1/(4.44*f*N1);# in Wb
phi_m = phi_m*10**3;# in mWb
print "The maximum core flux in mWb is",round(phi_m)
The primary full load current in A is 8.33
The secondary full load current in A is 104.2
The secondary emf in V is 240.0
The maximum core flux in mWb is 27.0

Example 4: pg 307

In [4]:
# Exa 9.4
#pg 307
#calculate the current
# Given data
from math import acos,cos,sin
Rating = 25.;# in KVA
f = 50.;# in Hz
Io = 15.;# in A
Wo = 350.;# in W
Vo = 230.;# in V
#calculations
# No load power factor
phi_o = acos(Wo/(Vo*Io));
# active component of current 
Ic = Io*cos(phi_o);# in A
print "The active component of current in A is",round(Ic,3)
# magnetizing component of current 
Im = Io*sin(phi_o);# in A
print "The magnetizing component of current in A is",round(Im,3)
The active component of current in A is 1.522
The magnetizing component of current in A is 14.923

Example 5: pg 307

In [5]:
# Exa 9.5
#pg 307
#calculate the resistance and reactance
# Given data
V1 = 2200.;# in V
V2 = 110.;# in V
R1 = 1.75;# in ohm
R2 = 0.0045;# in ohm
X1 = 2.6;# in ohm
X2 = 0.0075;# in ohm
#calculations
K = V2/V1;
#R1e = R1+R_2 = R1 + (R2/(K**2));
R1e = R1 + (R2/(K**2));# in ohm
print "Equivalent resistance reffered to primary  in ohm is",R1e
# R2e = R2+R_1 = R2+((K**2)*R1);
R2e = R2+((K**2)*R1);# in ohm
print "Equivalent resistance reffered to secondary  in ohm is",R2e
#X1e = X1+X_2 = X1+(X2/(K**2));
X1e = X1+(X2/(K**2));# in ohm
print "Equivalent reactance reffered to primary  in ohm is",X1e
# X2e = X2+X_1 = X2 + ((K**2)*X1);
X2e = X2 + ((K**2)*X1);# in ohm
print "Equivalent reactance reffered to secondary  in ohm is",X2e
Z1e= R1e+1j*X1e;# in ohm
Z2e= R2e+1j*X2e;# in ohm
print "Equivalent impedance reffered to primary in ohm is : ",round(abs(Z1e),4)
print "Equivalent impedance reffered to secondary in ohm is : ",round(abs(Z2e),5)
Equivalent resistance reffered to primary  in ohm is 3.55
Equivalent resistance reffered to secondary  in ohm is 0.008875
Equivalent reactance reffered to primary  in ohm is 5.6
Equivalent reactance reffered to secondary  in ohm is 0.014
Equivalent impedance reffered to primary in ohm is :  6.6304
Equivalent impedance reffered to secondary in ohm is :  0.01658

Example 6: pg 308

In [6]:
# Exa 9.6
#pg 308
#calculate the impedance and copper loss
from math import sqrt
# Given data
V1 = 2200.;# in V
V2 = 440.;# in V
R1 = 0.3;# in ohm
R2 = 0.01;# in ohm
X1 = 1.1;# in ohm
X2 = 0.035;# in ohm
#calculations
K = V2/V1;
Rating = 100;# in KVA
I1 = (Rating*10**3)/V1;# in A
I2 = (Rating*10**3)/V2;# in A
R1e = R1 + (R2/(K**2));# in ohm
X1e = X1+(X2/(K**2));# in ohm
Z1e = sqrt( (R1e**2) + (X1e**2) );# in ohm
# Total copper loss
totalcopperloss = (I1**2)*R1e;# in W
#results
print "The equivalent impedance of the transformer reffered to primary in ohm is",round(Z1e,2)
print "The total copper loss in W is",round(totalcopperloss,2)
The equivalent impedance of the transformer reffered to primary in ohm is 2.05
The total copper loss in W is 1136.36

Example 7: pg 309

In [7]:
# Exa 9.7
#pg 309
#calculate the efficiency
from math import cos,acos
# Given data
Rating = 150000.;# in VA
phi= acos(0.8);# in radians
Pcu = 1600.;# in W
Pi = 1400.;# in W
n = 1/4.;
#calculations
# Total loss of 25% load
totalloss = Pi + (n**2)*Pcu;# in W
# efficiency of transformer of 25% load
Eta = n*Rating*cos(phi)/(n*Rating*cos(phi)+Pi+n**2*Pcu)*100;# in %
#results
print "The efficiency in percent is",round(Eta,2)
The efficiency in percent is 95.24

Example 8: pg 309

In [8]:
# Exa 9.8
#pg 309
#calculate the efficiency
# Given data
from math import acos
Rating = 25.;# in KVA
V1 = 2000.;# in V
V2 = 200.;# in V
Pi = 350.;# in W
Pi = Pi * 10**-3;# in kW
Pcu = 400.;# in W
Pcu = Pcu * 10**-3;# in kW
#calculations
phi= acos(1);# in radians
output = Rating;
losses = Pi+Pcu;
Eta = (output/(output + losses))*100;# %Eta in %
print "The efficiency of full load power in percent is",round(Eta,2)
# For half load
output = Rating/2;# in kW
h = 1.;
Pcu = Pcu*((h/2)**2);# in kW
losses = Pi+Pcu;
# efficiency of half load power 
Eta = (output/(output+losses))*100;# in %
print "The efficiency of half load power in percent is",round(Eta,2)
The efficiency of full load power in percent is 97.09
The efficiency of half load power in percent is 96.53

Example 9: pg 310

In [9]:
# Exa 9.9
#pg 310
#calculate the efficiency
from math import acos,cos,sqrt
# Given data
Rating = 250*10**3;# in VA
Pi = 1.8;# in kW
Pi = Pi * 10**3;# in W
Pcu_f1 = 2000;# in W
phi= acos(0.8);# in radians
Eta = ((Rating*cos(phi))/((Rating*cos(phi))+Pi+Pcu_f1))*100;# %Eta in %
print "The efficiency at full load in percent is",round(Eta,2)
# The maximum efficiency 
Eta_max = Rating * sqrt(Pi/Pcu_f1 );# in VA
Eta_max = Eta_max *10**-3;# in kVA
print "The maximum efficiency in kVA is",round(Eta_max,2)
Eta_max = Eta_max *10**3;# in VA
Pcu = Pi;# in W
Eta_max1 = ((Eta_max*cos(phi))/((Eta_max*cos(phi)) + Pi+Pcu ))*100;# in %
print "The maximum efficiency in percent is",round(Eta_max1,3)
The efficiency at full load in percent is 98.14
The maximum efficiency in kVA is 237.17
The maximum efficiency in percent is 98.138

Example 10: pg 311

In [10]:
# Exa 9.10
#pg 311
#calculate the iron loss, full load copper loss
# Given data
from math import acos,cos
phi= acos(1);# in radians
Pout = 500.;# in kW
Pout = Pout*10**3;# in W
Eta = 90.;# in %
n=1/2.;
#calculations
# For full load, Eta= Pout*100/(Pout+Pi+Pcu_f1) or Pi+Pcu_f1= (Pout*100-Eta*Pout)/Eta                                       (i)
# For half load, Eta= n*Pout*100/(n*Pout+Pi+n**2*Pcu_f1) or Pi+n**2*Pcu_f1= (n*Pout*100-n*Eta*Pout)/Eta    (ii)
# From eq(i) and (ii)
Pcu_fl= ((n*Pout*100-n*Eta*Pout)/Eta-(Pout*100-Eta*Pout)/Eta)/(n**2-1)
Pi=(Pout*100-Eta*Pout)/Eta-Pcu_fl
#results
print "The iron loss in W is : ",round(Pi,3)
print "The full load copper loss in watt",round(Pcu_fl,2)
The iron loss in W is :  18518.519
The full load copper loss in watt 37037.04

Example 11: pg 311

In [11]:
# Exa 9.11
#pg 311
#calculate the flux, iron loass and load current
# Given data
from math import acos,sin,cos
Io = 10;# in A
phi_o= acos(0.25);# in radians
V1 = 400.;# in V
f = 50.;# in Hz
N1 =500.;
Im = Io*sin(phi_o);# in A
print "The magnetizing component of no load current in A is",round(Im,2)
Pi = V1*Io*cos(phi_o);# in W
print "The iron loss in W is",Pi
E1 = V1;# in V
#E1 v= 4.44*f*phi_m*N1;
phi_m = E1/(4.44*f*N1);# in Wb
phi_m=phi_m*10**3;# in mWb
print "The maximum value of flux in mWb is",round(phi_m,2)
The magnetizing component of no load current in A is 9.68
The iron loss in W is 1000.0
The maximum value of flux in mWb is 3.6

Example 12: pg 312

In [12]:
# Exa 9.12
#pg 312
#calculate the equivalent resistance
# Given data
from math import sqrt
Rating = 30.*10**3;# in VA
V1 = 2000.;# in V
V2 = 200.;# in V
f = 50.;# in Hz
R1 = 3.5;# in ohm
X1 = 4.5;# in ohm
R2 = 0.015;# in  ohm
X2 = 0.02;# in ohm
#calculations
K = V2/V1;
R1e = R1 + (R2/(K**2));# in ohm
print "The equivalent resistance to primary side in ohm is",R1e
X1e = X1 + (X2/(K**2));# in ohm
print "The equivalent reactance to primary side in ohm is",X1e
Z1e = sqrt( (R1e**2) + (X1e**2) );# in ohm
print "The equivalent impedance to primary side in ohm is",round(Z1e,1)
I1 = Rating/V1;# in A
# Total copper loss in transformer
Pcu_total = (I1**2)*R1e;# in W
print "Total copper loss in W is",Pcu_total
The equivalent resistance to primary side in ohm is 5.0
The equivalent reactance to primary side in ohm is 6.5
The equivalent impedance to primary side in ohm is 8.2
Total copper loss in W is 1125.0

Example 13: pg 313

In [13]:
# Exa 9.13
#pg 313
#calculate the full load secondary voltage
from math import cos,sin,acos
# Given data
Rating = 10;# in KVA
phi= acos(0.8)
V1 = 2000.;# in V
V2 = 400.;# in V
R1 = 5.5;# in ohm
X1 = 12;# in ohm
R2 = 0.2;# in ohm
X2 = 0.45;# in ohm
K = V2/V1;
#R1e = R1 + R_2 = R1 + (R2/(K**2));
R1e = R1 + (R2/(K**2));# in ohm
#X1e = X1 + X_ = X1 + (X2/(K**2));
X1e = X1 + (X2/(K**2));# in ohm
I2 = (Rating*10**3)/V2;# in A
R2e = (K**2)*R1e;# in ohm
X2e = (K**2)*X1e;# in ohm
Vdrop = I2 * ( (R2e*cos(phi)) + (X2e*sin(phi)) );# voltage drop in V
#E2 = V2 +Vd;
E2 = V2;# in V
# The full load secondary voltage 
V2 = E2-Vdrop;# in V
#results
print "The full load secondary voltage in V is",V2
The full load secondary voltage in V is 377.65

Example 14: pg 313

In [14]:
# Exa 9.14
#pg 313
#calculate the efficiency
from math import cos,acos,sqrt
# Given data
Rating = 40*10**3;# in VA
Pi = 400.;# in W
Pcu_f1  = 800.;# in W
#calculations
phi= acos(0.9);# in radians
Eta_f1 = ((Rating*cos(phi))/( (Rating*cos(phi)) + Pi + Pcu_f1 ))*100;# in %
print "Full load efficiency in percent is",round(Eta_f1,2)
# percentage of the full load
Eta_max = Rating*sqrt( Pi/Pcu_f1);# in KVA
Eta_max = Eta_max/Rating*100;# in %
print "The percentage of the full load in percent is",round(Eta_max,2)
Full load efficiency in percent is 96.77
The percentage of the full load in percent is 70.71

Example 15: pg 314

In [15]:
#pg 314
# Exa 9.15
#calculate the full load efficiency
from math import cos,acos
# Given data
Rating = 8*10**3;# in VA
phi= acos(0.8);# in radians
V1 = 400.;# in V
V2 = 100.;# in V
f = 50.;# in Hz
Pi = 60.;# in W
Wo = Pi;# in W
Pcu = 100.;# in W
#results
# The full load efficiency 
Eta_f1 = ((Rating*cos(phi))/((Rating*cos(phi)) + Pi + Pcu))*100;# in %
#results
print "The full load efficiency in percent is",round(Eta_f1,2)
The full load efficiency in percent is 97.56

Example 16: pg 314

In [16]:
#pg 314
# Exa 9.16
# Given data
from math import acos,cos
Rating = 10*10**3;# in VA
phi= acos(0.8);# in radians
V1 = 500.;# in V
V2 = 250.;# in V
Pi = 200.;# in W
Pcu = 300.;# in W
Isc = 30.;# in A
#calculations
I1 = Rating/V1;# in A
# Pcu/(Pcu(f1)) = (Isc**2)/(I1**2);
Pcu_f1 = Pcu * ((I1**2)/(Isc**2));# in W
# The efficiency at full load
Eta_f1 = Rating*cos(phi)/(Rating*cos(phi) + Pi + Pcu_f1)*100;# in %
#results
print "The full load efficiency in percent is",Eta_f1
The full load efficiency in percent is 96.0

Example 17: pg 315

In [17]:
#pg 315
# Exa 9.17
#calculate the max efficiency
# Given data
from math import acos,cos,sqrt
Rating = 20*10**3;# in VA
phi= acos(0.8);# in radians
V1 = 2000.;# in V
V2 = 200.;# in V
Pi = 120.;# in W
Pcu = 300.;# in W
#calculations
Eta_max = Rating*(sqrt( Pi/Pcu ));# in VA
Pcu = Pi;# in W
# The maximum efficiency of transformer 
Eta_max = ((Eta_max*cos(phi))/( Eta_max*cos(phi) + (2*Pi) ))*100;# in %
#results
print "The maximum efficiency of transformer in percent is",round(Eta_max,3)
The maximum efficiency of transformer in percent is 97.683

Example 18: pg 315

In [18]:
#pg 315
# Exa 9.18
#calculate the resistances
# Given data
Turnratio = 5;
R1 = 0.5;# in ohm
R2 = 0.021;# in ohm
X1 = 3.2;# in ohm
X2 = 0.12;# in ohm
Rc = 350.;# in ohm
Xm = 98.;# in ohm
N1 = 5.;
N2 = 1.;
#calculations
K = N2/N1;
# Evaluation of the equivalent parameters referred to secondary side 
R2e = R2 + ((K**2)*R1);# in ohm
print "The equivalent parameters referred to secondary side are : "
print "The value of  R_2e  is : ",R2e," ohm"
X2e = X2 + ((K**2)*X1);# in ohm
print "The value of  X_2e  is : ",X2e," ohm"
R_c = (K**2)*Rc;# in ohm
print "The value of  R''c  is : ",R_c," ohm"
X_m = (K**2)*Xm;# in ohm
print "The value of  X''m  is : ",X_m," ohm"
The equivalent parameters referred to secondary side are : 
The value of  R_2e  is :  0.041  ohm
The value of  X_2e  is :  0.248  ohm
The value of  R''c  is :  14.0  ohm
The value of  X''m  is :  3.92  ohm

Example 19: pg 315

In [19]:
#pg 315
# Exa 9.19
#calculate the resistances
# Given data
from math import acos,cos,sin,sqrt
Rating = 100.*10**3;# in VA
V1 = 11000.;# in V
V2 = 220.;# in V
Wo = 2*10**3;# in W
Vo = 220.;# in V
Io = 45.;# in A
#calculations
phi_o = acos(Wo/(Vo*Io));
I_c = Io*cos(phi_o);# in A
I_m = Io*sin(phi_o);# in A
Ro= V2/I_c;# in ohm
Xo= V2/I_m;# in ohm
Wsc= 3*10**3;# in W
Vsc= 500.;# in V
Isc= 9.09;# in A
R1e= Wsc/Isc**2;# in ohm
Z1e= Vsc/Isc;# in ohm
X1e= sqrt(Z1e**2-R1e**2);# in ohm
K= V2/V1;
R2e= K**2*R1e;# in ohm
X2e= K**2*X1e;# in ohm
Z2e= K**2*Z1e;# in ohm
#results
print "The value of R''o is : ",Ro," ohm"
print "The value of X''o is : ",round(Xo)," ohm"
print "The value of R1e is : ",round(R1e,4)," ohm"
print "The value of Z1e is : ",round(Z1e,4)," ohm"
print "The value of X1e is : ",round(X1e,4)," ohm"
print "The value of R2e is : ",round(R2e,4)," ohm"
print "The value of X2e is : ",round(X2e,4)," ohm"
print "The value of Z2e is : ",round(Z2e,4)," ohm"
The value of R''o is :  24.2  ohm
The value of X''o is :  5.0  ohm
The value of R1e is :  36.3073  ohm
The value of Z1e is :  55.0055  ohm
The value of X1e is :  41.3206  ohm
The value of R2e is :  0.0145  ohm
The value of X2e is :  0.0165  ohm
The value of Z2e is :  0.022  ohm

Example 20: pg 316

In [20]:
#pg 316
# Exa 9.20
#calculate the efficiency
# Given data
from math import acos,cos
V1 = 250.;# in V
V2 = 500.;# in V
Pcu = 100.;# in W
Pi = 80.;# in W
V = V2;# in V
A = 12.;# in A
#calculations
phi= acos(0.85);# in radians
# The efficiency of the transformer 
Eta = ((V*A*cos(phi))/( V*A*cos(phi) + Pi+Pcu ))*100;# in %
print "The efficiency of the transformer in percent is",round(Eta,2)
The efficiency of the transformer in percent is 96.59

Example 21: pg 317

In [21]:
#pg 317
# Exa 9.21
#calculate the iron and copper loss
# Given data
from math import acos,cos,sin
VA = 400.*10**3;# in Mean
Eta_fl = 98.77/100;# in %
phi1= acos(0.8);# in radians
phi2= acos(1);# in radians
Eta_hl = 99.13/100;# in %
n = 1/2.;
#calculations
#For full load,  Eta_f1 = ((VA*cosd(phi1))/( VA*cosd(phi1) + Pi + Pcu_f1 )) or Pi+Pcu_f1 = VA*cosd(phi1)*(1-Eta_fl)/(Eta_f1)                   (i)
#For half load,  Eta_hl = n*VA*cosd(phi2)/(n*VA*cosd(phi2)+Pi+n**2*Pcu_f1) or Pi+n**2*Pcu_f1 = n*VA*cosd(phi2)*( 1-Eta_hl)/Eta_hl    (ii)
# From eq(i) and (ii)
Pcu_fl=(n*VA*cos(phi2)*( 1-Eta_hl)/Eta_hl-VA*cos(phi1)*(1-Eta_fl)/(Eta_fl))/(n**2-1);# in W
Pi=VA*cos(phi1)*(1-Eta_fl)/(Eta_fl)-Pcu_fl;# in W
# The copper loss on half load 
C_loss_half_load=n**2*Pcu_fl;# in W 
#results
print "The iron loss on full load and half load remain same in W which are : ",round(Pi,4)
print "The copper loss on full load in W is : ",round(Pcu_fl,3)
print "The copper loss on half load in W is : ",round(C_loss_half_load,3)
The iron loss on full load and half load remain same in W which are :  1012.0226
The copper loss on full load in W is :  2972.993
The copper loss on half load in W is :  743.248

Example 22: pg 317

In [22]:
#pg 317
# Exa 9.22
#calculate the efficiency
# Given data
from math import cos,acos
VA = 100.*10**3;# in VA
Eta_max = 98.40/100;# in %
Eta_max1 = 90./100;# in %
phi= acos(1);# in radians
#Eta_max = (Eta_max1*VA*cos(phi)/(Eta_max1*VA*cos(phi) + 2*Pi);
Pi = (Eta_max1*VA*cos(phi)/Eta_max - Eta_max1*VA*cos(phi))/2;# in W
Pcu = Pi;# in W
n = 0.9;
# Pcu_fl/Pcu = (VA/(0.9*VA) )**2;
Pcu_fl = Pcu*(VA/(0.9*VA) )**2;# in W
Eta_fl = ( (VA*cos(phi))/( (VA*cos(phi)) + Pi + Pcu_fl ) )*100;# in %
#results
print "The efficiency of a transformer in percent is",round(Eta_fl,4)
The efficiency of a transformer in percent is 98.3913