CHAPTER 3: TRANSFORMER AND PER UNIT SYSTEM¶

Example 3.1, Page number 90-91¶

In [1]:
#Variable declaration
V_1 = 2200.0     #Primary voltage of transformer(V)
V_2 = 220.0      #Secondary voltage of transformer(V)
N_2 = 56.0       #Number of turns in the secondary coil of transformer
kVA = 25.0       #Rating of transformer(kVA)

#Calculation
a = V_1/V_2           #Turns ratio
#For case(i)
N_1 = a*N_2           #Number of primary turns
#For case(ii)
I_1 = kVA*10**3/V_1   #Primary full load current(A)
#For case(iii)
I_2 = kVA*10**3/V_2   #Secondary full load current(A)

#Result
print('(i)   Number of primary turns , N_1 = %.f ' %N_1)
print('(ii)  Primary full load current , I_2 = %.2f A' %I_1)
print('(iii) Secondary full load current , I_2 = %.1f A' %I_2)

(i)   Number of primary turns , N_1 = 560
(ii)  Primary full load current , I_2 = 11.36 A
(iii) Secondary full load current , I_2 = 113.6 A


Example 3.2, Page number 91¶

In [1]:
#Variable declaration
V_1 = 220.0     #Voltage(V)
N_1 = 150.0     #Number of turns in primary side
N_2 = 300.0     #Number of turns in secondary side
f = 50.0        #Frequency(Hz)

#Calculation
#For case(i)
a = N_1/N_2                       #Turns ratio
#For case(ii)
phi_m = V_1/(4.44*f*N_1)*10**3    #Mutual flux(mWb)

#Result
print('(i)  Turns ratio , a = %.1f ' %a)
print('(ii) Mutual flux in the core , Φ_m = %.2f mWb' %phi_m)

(i)  Turns ratio , a = 0.5
(ii) Mutual flux in the core , Φ_m = 6.61 mWb


Example 3.3, Page number 92¶

In [1]:
import math

#Variable declaration
V_1 = 2200.0    #Primary voltage(V)
V_2 = 220.0     #Secondary voltage(V)
P_0 = 350.0     #Power absorbed(W)

#Calculation
cos_phi_0 = P_0/(V_1*I_0)        #No-load power factor
I_w = I_0*cos_phi_0              #Iron loss component of current(A)
phi_0 = math.acos(cos_phi_0)     #Power factor angle
I_m = I_0*math.sin(phi_0)        #Magnetizing component of current(A)

#Result
print('Iron loss component of current , I_w = %.2f A' %I_w)
print('Magnetizing component of current , I_m = %.2f A' %I_m)

Iron loss component of current , I_w = 0.16 A
Magnetizing component of current , I_m = 0.47 A


Example 3.4, Page number 94¶

In [1]:
#Variable declaration
N_1 = 450.0     #Number of turns in the primary side
N_2 = 45.0      #Number of turns in the secondary side
V_1 = 15.0      #Primary coil voltage of transformer(V)

#Calculation
#For case(i)
a = N_1/N_2       #Turns ratio
#For case(ii)
Z_1 = a**2*Z_L    #Load impedance referred to primary(ohm)
#For case(iii)
I_1 = V_1/Z_1     #Primary current(A)

#Result
print('(i)   Turns ratio , a = %.f ' %a)
print('(ii)  Load impedance referred to primary , Z_1 = %.f ohm' %Z_1)
print('(iii) Primary current , I_1 = %.2f A' %I_1)

(i)   Turns ratio , a = 10
(ii)  Load impedance referred to primary , Z_1 = 300 ohm
(iii) Primary current , I_1 = 0.05 A


Example 3.5, Page number 96-97¶

In [1]:
import math

#Variable declaration
V_1 = 400.0        #Primary voltage of transformer(V)
V_2 = 100.0        #Secondary voltage of transformer(V)
I_2 = 100.0        #Current drawn by load(A)
cos_phi_0 = 0.3    #Power factor lagging from the supply
cos_phi_2 = 0.6    #Power factor lagging from the secondary

#Calculation
phi_0 = math.acos(cos_phi_0)    #Power factor angle(radians)
phi_0_deg = phi_0*180/math.pi   #Power factor angle(degree)
phi_2 = math.acos(cos_phi_2)    #Power factor angle(radians)
phi_2_deg = phi_2*180/math.pi   #Power factor angle(degree)
phi_1_deg = phi_1*180/math.pi   #Angle(degree)
a = V_1/V_2                     #Turns ratio
I_2_ = I_2/a                    #Secondary current equivalent to the primary(A)
I_1 = ((I_2_**2)+(I_0**2)+(2*I_2_*I_0*math.cos(phi_1)))**0.5   #Primary current(A)

#Result
print('Primary current , I_1 = %.1f A' %I_1)

Primary current , I_1 = 25.4 A


Example 3.6, Page number 101¶

In [1]:
#Variable declaration
V_1 = 2000.0    #Primary voltage of transformer(V)
V_2 = 400.0     #Secondary voltage of transformer(V)
kVA = 200.0     #Rating of transformer(kVA)
R_1 = 3.0       #Primary resistance(ohm)
X_1 = 12.0      #Primary reactance(ohm)
R_2 = 0.3       #Secondary resistance(ohm)
X_2 = 0.1       #Secondary reactance(ohm)

#Calculation
a = V_1/V_2                         #Turns ratio
R_01 = R_1+(a**2*R_2)               #Total resistance referred to primary(ohm)
X_01 = X_1+(a**2*X_2)               #Total reactance referred to primary(ohm)
Z_01 = ((R_01**2)+(X_01**2))**0.5   #Equivalent impedance referred to primary(ohm)
R_02 = R_2+(R_1/a**2)               #Total resistance referred to secondary side(ohm)
X_02 = X_2+(X_1/a**2)               #Total reactance referred to secondary side(ohm)
Z_02 = ((R_02**2)+(X_02**2))**0.5   #Equivalent impedance referred to secondary side(ohm)

#Result
print('Equivalent impedance referred to primary , Z_01 = %.1f ohm' %Z_01)
print('Equivalent impedance referred to secondary , Z_02 = %.2f ohm' %Z_02)

Equivalent impedance referred to primary , Z_01 = 17.9 ohm
Equivalent impedance referred to secondary , Z_02 = 0.72 ohm


Example 3.7, Page number 103-104¶

In [1]:
import math

#Variable declaration
V_1 = 200.0           #Primary voltage(V)
V_2 = 400.0           #Secondary voltage(V)
R_1 = 0.3             #Primary resistance(ohm)
X_1 = 0.6             #Primary reactance(ohm)
R_2 = 0.8             #Secondary resistance(ohm)
X_2 = 1.6             #Secondary reactance(ohm)
I_2 = 10.0            #Secondary supply current(A)
cos_phi_2 = 0.8       #Power factor lagging

#Calculation
a = V_1/V_2               #Turns ratio
R_02 = R_2+(R_1/a**2)     #Total resistance referred to secondary(ohm)
X_02 = X_2+(X_1/a**2)     #Total reactance referred to primary(ohm)
sin_phi_2 = math.sin(math.acos(cos_phi_2))
V_reg = (abs(E_2)-V_2)/V_2*100                                    #Voltage regulation(percent)

#Result
print('Voltage regulation = %.f percent' %V_reg)

Voltage regulation = 10 percent


Example 3.8, Page number 105-106¶

In [1]:
#Variable declaration
P_i = 1.0     #Iron loss of transformer(kW)
P_cu = 2.0    #Full load copper loss of transformer(kW)
kVA = 200.0   #Rating of transformer(kVA)
pf_1 = 1.0    #Power factor
pf_2 = 0.95   #Power factor

#Calculation
P_cu1 = (3.0/4)**2*P_cu    #Copper loss at 3/4 full load(kW)
P_cu2 = (1.0/2)**2*P_cu    #Copper loss at 1/2 full load(kW)
P_01 = (3.0/4)*kVA*pf_1    #Output power at 3/4 full load and unity pf(kW)
P_in1 = P_01+P_i+P_cu1     #Input power at 3/4 full load and unity pf(kW)
n_1 = (P_01/P_in1)*100     #Efficiency at 3/4 full load and unity pf(percent)
P_02 = (1.0/2)*kVA*pf_2    #Output power at 1/2 full load and 0.95 pf(kW)
P_in2 = P_02+P_i+P_cu2     #Input power at 1/2 full load and 0.95 pf(kW)
n_2 = (P_02/P_in2)*100     #Efficiency at 1/2 full load and 0.95 pf(percent)

#Result
print('Efficiency at 3/4 full load and unity power factor , η_1 = %.2f percent' %n_1)
print('Efficiency at 1/2 full load and 0.95 power factor , η_2 = %.2f percent' %n_2)

Efficiency at 3/4 full load and unity power factor , η_1 = 98.60 percent
Efficiency at 1/2 full load and 0.95 power factor , η_2 = 98.45 percent


Example 3.9, Page number 108¶

In [1]:
#Variable declaration
P_i = 350.0      #Iron loss of transformer(W)
P_cu = 650.0     #Full load copper loss of transformer(W)
kVA = 30.0       #Rating of transformer(kVA)
pf = 0.6         #Power factor

#Calculation
#For case(i)
P_tloss = (P_i+P_cu)*10**-3      #Total full load loss(kW)
P_out = kVA*pf                   #Output power at full load(kW)
P_in = P_out+P_tloss             #Input power at full load(kW)
n_1 = (P_out/P_in)*100           #Efficiency at full load(percent)
#For case(ii)
kVA_out = kVA*(P_i/P_cu)**0.5    #Output kVA corresponding to maximum efficiency(kVA)
P_01 = kVA_out*pf                #Output power(kW)
#For case(iii)
P_tloss1 = 2*P_i*10**-3          #Total loss(kW)
P_in1 = P_01+P_tloss1            #Input power(kW)
n_2 = (P_01/P_in1)*100           #Efficiency(percent)

#Result
print('(i)   Full load efficiency , η = %.2f percent' %n_1)
print('(ii)  Output kVA corresponding to maximum efficiency = %.1f kVA' %kVA_out)
print('(iii) Maximum efficiency , η = %.2f percent' %n_2)

(i)   Full load efficiency , η = 94.74 percent
(ii)  Output kVA corresponding to maximum efficiency = 22.0 kVA
(iii) Maximum efficiency , η = 94.97 percent


Example 3.10, Page number 109-110¶

In [1]:
#Variable declaration
kVA = 12.0        #Rating of transformer(kVA)
n = 0.97          #Maximum efficiency at unity pf
pf = 1.0          #Power factor
t_1 = 8.0         #Time(hours)
pf_1 = 0.8        #Lagging power factor
t_2 = 10.0        #Time(hours)
pf_2 = 0.90       #Leading power factor
t_3 = 6.0         #Time at no-load(hours)

#Calculation
P_01 = kVA*pf                         #Output power at full load and unity pf(kW)
P_in1 = P_01/n                        #Input power at full load(kW)
P_tloss = P_in1-P_01                  #Total loss(kW)
P_cu = P_tloss/2                      #Copper loss at 12 kVA(kW)
P_024 = P_1*t_1+P_2*t_2+P_3*t_3       #All-day output power(kWh)
P_i24 = 24*P_cu                       #Iron loss for 24 hours(kWh)
P_cu24 = P_cu*t_1*((P_1/pf_1)/kVA)**2+P_cu*t_2*((P_2/pf_2)/kVA)**2   #Copper loss for 24 hours(kW)
P_in24 = P_024+P_i24+P_cu24           #All day input power(kWh)
n_allday = (P_024/P_in24)*100         #All day efficiency(percent)

#Result
print('All-day efficiency , η_allday = %.1f percent' %n_allday)

All-day efficiency , η_allday = 96.0 percent


Example 3.11, Page number 111¶

In [1]:
import math

#Variable declaration
V_1 = 200.0       #Voltage(V)
f = 50.0          #Frequency(Hz)
I_0 = 0.6         #Current(A)
P_0 = 80.0        #Power(W)

#Calculation
cos_phi_0 = P_0/(V_1*I_0)                    #Power factor
sin_phi_0 = math.sin(math.acos(cos_phi_0))
I_w = I_0*cos_phi_0                          #Working component of no-load current(A)
I_m = I_0*sin_phi_0                          #Magnetizing component of no-load current(A)
R_0 = V_1/I_w                                #No-load circuit resistance(ohm)
X_0 = V_1/I_m                                #No-load circuit reactance(ohm)

#Result
print('No-load circuit resistance , R_0 = %.1f ohm' %R_0)
print('No-load circuit reactance , X_0 = %.1f ohm' %X_0)
print('\nNOTE : Changes in obtained answer from that of textbook is due to more precision')

No-load circuit resistance , R_0 = 500.0 ohm
No-load circuit reactance , X_0 = 447.2 ohm

NOTE : Changes in obtained answer from that of textbook is due to more precision


Example 3.12, Page number 112-113¶

In [1]:
import math

#Variable declaration
kVA = 25.0        #Rating of transformer(kVA)
V1 = 2200.0       #Voltage at low-voltage primary side(V)
V2 = 220.0        #Voltage at low-voltage secondary side(V)
V_1 = 40.0        #Voltage at h.v side(V)
I_1 = 5.0         #Current at high voltage side(A)
P = 150.0         #Power at high voltage side(W)

#Calculation
#For case(i)
Z_01 = V_1/I_1                #Equivalent impedance referred to primary(ohm)
R_01 = P/I_1**2               #Equivalent resistance referred to primary(ohm)
phi = math.acos(R_01/Z_01)    #Power factor angle(radians)
phi_deg = phi*180/math.pi     #Power factor angle(degree)
X_01 = Z_01*math.sin(phi)     #Equivalent reactance referred to primary(ohm)
#For case(ii)
a = V1/V2                     #Turns ratio
Z_02 = Z_01/a**2              #Equivalent impedance referred to secondary(ohm)
R_02 = R_01/a**2              #Equivalent resistance referred to secondary(ohm)
X_02 = X_01/a**2              #Equivalent reactance referred to secondary(ohm)
#For case(iii)
I_2 = kVA*10**3/V2            #Secondary side current(A)
E_2 = V2+I_2*Z_02             #Secondary induced voltage(V)
VR = (E_2-V2)/V2*100          #Voltage regulation(percent)

#Result
print('Case(i)')
print(' Equivalent resistance referred to primary , R_01 = %.1f ohm' %R_01)
print(' Equivalent reactance referred to primary , X_01 = %.1f ohm' %X_01)
print(' Equivalent impedance referred to primary , Z_01 = %.1f ohm' %Z_01)
print('\nCase(ii)')
print(' Equivalent resistance referred to secondary , R_02 = %.2f ohm' %R_02)
print(' Equivalent reactance referred to secondary , X_02 = %.3f ohm' %X_02)
print(' Equivalent impedance referred to secondary , Z_02 = %.2f ohm' %Z_02)
print('\nCase(iii)')
print(' Voltage regulation , VR = %.1f percent' %VR)

Case(i)
Equivalent resistance referred to primary , R_01 = 6.0 ohm
Equivalent reactance referred to primary , X_01 = 5.3 ohm
Equivalent impedance referred to primary , Z_01 = 8.0 ohm

Case(ii)
Equivalent resistance referred to secondary , R_02 = 0.06 ohm
Equivalent reactance referred to secondary , X_02 = 0.053 ohm
Equivalent impedance referred to secondary , Z_02 = 0.08 ohm

Case(iii)
Voltage regulation , VR = 4.1 percent


Example 3.13, Page number 114-115¶

In [1]:
#Variable declaration
kVA = 120.0              #Rating of autotransformer(kVA)
V1 = 220.0               #Volteage at upper part of coil(V)
V2 = 2200.0              #Volteage at lower part of coil(V)

#Calculation
I_pq = kVA*10**3/V1      #Current of upper winding(A)
I_qr = kVA*10**3/V2      #Current of lower winding(A)
I_1 = I_pq+I_qr          #Current in primary side(A)
V_2 = V1+V2              #Voltage across the secondary side(V)
kVA_1 = I_1*V2/1000      #Rating of autotransformer(kVA)
kVA_2 = I_pq*V_2/1000    #Rating of autotransformer(kVA)

#Result
print('kVA ratings of the autotransformer')
print('\t kVA_1 = %.f kVA' %kVA_1)
print('\t kVA_2 = %.f kVA' %kVA_2)

kVA ratings of the autotransformer
kVA_1 = 1320 kVA
kVA_2 = 1320 kVA


Example 3.14, Page number 119¶

In [1]:
import math
import cmath

#Variable declaration
kVA_1 = 100.0      #Rating of transformer(kVA)
kVA_2 = 200.0      #Rating of transformer(kVA)
E_1 = 500.0        #Secondary induced voltage in 100 kVA transformer(V)
E_2 = 450.0        #Secondary induced voltage in 200 kVA transformer(V)
Z_1 = 0.05         #Impedance of 100 kVA transformer
Z_2 = 0.08         #Impedance of 200 kVA transformer

#Calculation
Z1 = Z_1*E_1/(kVA_1*10**3/E_1)   #Actual impedance of first transformer(ohm)
Z2 = Z_2*E_2/(kVA_1*10**3/E_2)   #Actual impedance of second transformer(ohm)
I_c = (E_1-E_2)/complex(0,Z1+Z2) #Circulating current(A)

#Result
print('Circulating current , I_c = %.2f∠%.f° A' %(abs(I_c),cmath.phase(I_c)*180/math.pi))

Circulating current , I_c = 174.22∠-90° A


Example 3.15, Page number 125¶

In [1]:
#Variable declaration
V_L1 = 11.0     #Supply voltage(kV)
I_P1 = 6.0      #Current drawn by transformer(A)
a = 11.0        #Turns ratio

#Calculation
#For delta-wye connections
V_dP1 = V_L1                  #Phase voltage at primary side(kV)
V_dP2 = V_dP1*10**3/a         #Phase voltage at secondary side(V)
V_dL2 = 3**0.5*V_dP2          #Line voltage at secondary side(V)
I_dP1 = a/3**0.5              #Phase current in the primary(A)
I_dL2 = a*I_dP1               #Line current in secondary(A)
#For Wye-delta connection
V_wP1 = V_L1*10**3/3**0.5     #Phase voltage at primary side(V)
V_wP2 = V_wP1/a               #Phase voltage at secondary(V)
V_wL2 = V_wP2                 #Line voltage at secondary(V)
I_wP2 = a*I_P1                #Phase current in secondary(A)
I_wL2 = 3**0.5*I_wP2          #Line current in secondary(A)

#Result
print('For delta-wye connection')
print(' (i)  Line voltage at secondary side , V_L2 = %.f V' %V_dL2)
print(' (ii) Line current in the secondary , I_L2 = %.2f A' %I_dL2)
print('\nFor wye-delta connection')
print(' (i)  Line voltage at secondary side , V_L2 = %.2f V' %V_wL2)
print(' (ii) Line current in the secondary , I_L2 = %.2f A' %I_wL2)

For delta-wye connection
(i)  Line voltage at secondary side , V_L2 = 1732 V
(ii) Line current in the secondary , I_L2 = 69.86 A

For wye-delta connection
(i)  Line voltage at secondary side , V_L2 = 577.35 V
(ii) Line current in the secondary , I_L2 = 114.32 A


Example 3.16, Page number 132¶

In [1]:
import math
import cmath

#Variable declaration
V_b = 220.0           #Voltage(V)
f = 50.0              #Frequency(Hz)
S_b = 600.0           #Base value of rating(VA)
R = 3.0               #Resistance(ohm)
X_L = 5.0             #Inductance(ohm)
Z = complex(R,X_L)    #Impedance(ohm)

#Calculation
I_b = S_b/V_b                #Base value of current(A)
Z_b = V_b**2/S_b             #Base impedance(ohm)
R_pu = R/Z_b                 #Per unit value of resistance
X_Lpu = X_L/Z_b              #Per unit value of inductance
Z_pu = complex(R_pu,X_Lpu)   #Per unit of value of impedance
Z_pu_alt = abs(Z)/Z_b        #Per unit of value of impedance-alternative method

#Result
print('Per unit value of resistance , R_pu = %.3f ' %R_pu)
print('Per unit value of inductance , X_Lpu = %.3f ' %X_Lpu)
print('Per unit of value of impedance , Z_pu = %.3f∠%.2f° ' %(abs(Z_pu),cmath.phase(Z_pu)*180/math.pi))
print('Per unit of value of impedance by alternative method , Z_pu = %.3f ' %Z_pu_alt)

Per unit value of resistance , R_pu = 0.037
Per unit value of inductance , X_Lpu = 0.062
Per unit of value of impedance , Z_pu = 0.072∠59.04°
Per unit of value of impedance by alternative method , Z_pu = 0.072


Example 3.17, Page number 132-134¶

In [1]:
import math
import cmath

#Variable declaration
MVA_T1 = 100.0       #Rating of transformer T1(MVA)
V_T1_hv = 220.0      #Voltage of h.v side of T1(kV)
V_T1_lv = 132.0      #Voltage of l.v side of T1(kV)
X_T1 = 0.2           #Impedance of T1(pu)
MVA_T2 = 50.0        #Rating of transformer T2(MVA)
V_T2_hv = 132.0      #Voltage of h.v side of T2(kV)
V_T2_lv = 66.0       #Voltage of l.v side of T2(kV)
X_T2 = 0.05          #Impedance of T2(pu)
X_L = 4.0            #Line impedance(ohm)
P = 50.0             #Power absorbed(MW)
pf = 0.6             #Lagging power factor

#Calculation
S_b = MVA_T1                                  #Base apparent power(MW)
V_b = V_T1_hv                                 #Base voltage(kV)
a = V_T1_hv/V_T1_lv                           #Turns ratio for first transformer
V_bline = V_T1_hv/a                           #Base voltage of line(kV)
Z_bline = V_bline**2/S_b                      #Base impedance of line(ohm)
X_puline = X_L/Z_bline                        #Per unit reactance of line
X_pu_T1 = X_T1*(V_T1_hv/V_b)**2*(S_b/MVA_T1)  #Per unit reactance of first transformer

Grid to bus voltage , V_gb = 176∠11.63° kV