CHAPTER 14: TRANSFORMERS

Example 14.1, Page number 487

In [1]:
#Variable declaration
N_1 = 500.0     #Number of turns in the hv side i.e primary
N_2 = 100.0     #Number of turns in the lv side i.e secondary
I_2 = 12.0      #Load current i.e secondary(A)

#Calculation
#For step-down transformer
alpha = N_1/N_2    #Transformation ratio
I_1 = I_2/alpha    #Load component of primary current(A)
#For step-up transformer
N1 = 100.0        #Number of turns in the lv side i.e primary
N2 = 500.0        #Number of turns in the hv side i.e secondary
alpha_c = N1/N2   #Transformation ratio

#Result
print('Case(a): Transformation ratio , α = %.f ' %alpha)
print('Case(b): Load component of primary current , I_1 = %.1f A' %I_1)
print('Case(c): Transformation ratio if transformer is used as step-up , α = %.1f ' %alpha_c)
Case(a): Transformation ratio , α = 5 
Case(b): Load component of primary current , I_1 = 2.4 A
Case(c): Transformation ratio if transformer is used as step-up , α = 0.2 

Example 14.2, Page number 488

In [1]:
#Variable declaration
V_h = 2300.0       #Primary voltage i.e hv(V)
V_l = 115.0        #Secondary voltage i.e lv(V)
f = 60.0           #Frequency(Hz)
S = 4.6            #Rating of the step-down transformer(kVA)
V_per_turn = 2.5   #Induced EMF per turn(V/turn)

#Calculation
#Case(a)
N_h = V_h/V_per_turn       #Number of high-side turns
N_l = V_l/V_per_turn       #Number of low-side turns
N_1 = N_h
N_2 = N_l
#Case(b)
V_1 = V_h
V_2 = V_l
S_1 = S
S_2 = S
I_1 = S_1*1000/V_1         #Rated primary current(A)
I_2 = S_2*1000/V_2         #Rated secondary current(A)
I_h = I_1
I_l = I_2
#Case(c)
alpha_stepdown_c = N_1/N_2 #Step-down transformation ratio
alpha_stepup_c = N_l/N_h   #Step-up transformation ratio
#Case(d)
alpha_stepdown_d = I_2/I_1 #Step-down transformation ratio
alpha_stepup_d = I_h/I_l   #Step-up transformation ratio

#Result
print('Case(a): Number of high-side turns , N_h = %.f t = N_1 ' %N_h)
print('         Number of low-side turns , N_l = %.f t = N_2 ' %N_l)
print('Case(b): Rated primary current , I_h = I_1 = %.f A' %I_h)
print('         Rated secondary current , I_l = I_2 = %.f A' %I_l)
print('Case(c): Step-down transformation ratio , α = %.f ' %alpha_stepdown_c);
print('         Step-up transformation ratio , α = %.2f ' %alpha_stepup_c);
print('Case(d): Step-down transformation ratio , α = %.f ' %alpha_stepdown_d);
print('         Step-up transformation ratio , α = %.2f ' %alpha_stepup_d);
Case(a): Number of high-side turns , N_h = 920 t = N_1 
         Number of low-side turns , N_l = 46 t = N_2 
Case(b): Rated primary current , I_h = I_1 = 2 A
         Rated secondary current , I_l = I_2 = 40 A
Case(c): Step-down transformation ratio , α = 20 
         Step-up transformation ratio , α = 0.05 
Case(d): Step-down transformation ratio , α = 20 
         Step-up transformation ratio , α = 0.05 

Example 14.3, Page number 489

In [1]:
#Variable declaration
N_1 = 500.0     #Number of primary turns
N_2 = 25.0      #Number of secondary turns
Z_L = 8.0       #Impedance(ohm)
V_1 = 10.0      #Output voltage of the amplifier(V)

#Calculation
alpha = N_1/N_2         #Step-down transformation ratio
Z_1 = alpha**2*Z_L      #Impedance reflected to the transformer primary at the output of the amplifier(ohm)
I_1 = V_1/Z_1*1000      #Matching transformer primary current(mA)

#Result
print('Case(a): Impedance reflected to the transformer primary at the output of the amplifier , Z_1 = %.f Ω' %Z_1)
print('Case(b): Matching transformer primary current , I_1 = %.3f mA' %I_1)
Case(a): Impedance reflected to the transformer primary at the output of the amplifier , Z_1 = 3200 Ω
Case(b): Matching transformer primary current , I_1 = 3.125 mA

Example 14.4, Page number 490

In [1]:
#Variable declaration
N_1 = 600.0      #Number of primary turns 
N_2 = 150.0      #Some number of secondary turns 
N_3 = 300.0      #Some number of secondary turns 
Z_2 = 30.0       #Resistive load across N_2(ohm)
Z_3 = 15.0       #Resistive load across N_3(ohm)
V_p = 16.0       #Primary applied voltage(V)
cos_theta = 1.0  #Unity PF

#Calculation
#Case(a)
Z_2_prime = Z_2*(N_1/N_2)**2   #Impedance reflected to the primary by load Z_2(ohm)
#Case(b)
Z_3_prime = Z_3*(N_1/N_3)**2   #Impedance reflected to the primary by load Z_3(ohm)
#Case(c)
Z_1 = (Z_2_prime*Z_3_prime)/(Z_2_prime+Z_3_prime) #Total impedance reflected to primary(ohm)
#Case(d)
I_1 = V_p/Z_1                  #Total current drawn from the supply(A)
#Case(e)
P_t = V_p*I_1*cos_theta        #Total power drawn from the supply at unity PF(W)
#Case(f)
R_2 = Z_2
V_2 = V_p*(N_2/N_1)            #Voltage across load Z_2(V)
P_2 = V_2**2/R_2               #Power dissipated in load Z_2(W)
#Case(g)
R_3 = Z_3
V_3 = V_p*(N_3/N_1)            #Voltage across Z_3 in volt
P_3 = V_3**2 / R_3             #Power dissipated in load Z_3(W)
#Case(h)
P_total = P_2+P_3              #Total power dissipated in both loads(W)

#Result
print('Case(a): Impedance reflected to the primary by load Z_2 = %.f Ω' %Z_2_prime)
print('Case(b): Impedance reflected to the primary by load Z_3 = %.f Ω' %Z_3_prime)
print('Case(c): Total impedance reflected to the primary , Z_1 = %.1f Ω' %Z_1)
print('Case(d): Total current drawn from the supply , I_1 = %.1f A' %I_1)
print('Case(e): Total power drawn from the supply at unity power factor , P_t = %.1f W' %P_t)
print('Case(f): Voltage across load Z_2 , V_2 = %.f V' %V_2)
print('         Power dissipated in load Z_2 , P_2 = %.2f W' %P_2)
print('Case(g): Voltage across load Z_3 , V_3 = %.f V' %V_3)
print('         Power dissipated in load Z_3 , P_3 = %.2f W' %P_3)
print('Case(h): Total power dissipated in both loads , P_t = %.1f W ' %P_total)
Case(a): Impedance reflected to the primary by load Z_2 = 480 Ω
Case(b): Impedance reflected to the primary by load Z_3 = 60 Ω
Case(c): Total impedance reflected to the primary , Z_1 = 53.3 Ω
Case(d): Total current drawn from the supply , I_1 = 0.3 A
Case(e): Total power drawn from the supply at unity power factor , P_t = 4.8 W
Case(f): Voltage across load Z_2 , V_2 = 4 V
         Power dissipated in load Z_2 , P_2 = 0.53 W
Case(g): Voltage across load Z_3 , V_3 = 8 V
         Power dissipated in load Z_3 , P_3 = 4.27 W
Case(h): Total power dissipated in both loads , P_t = 4.8 W 

Example 14.5, Page number 491

In [1]:
#Variable declaration
P = 100.0       #Power rating of the single channel power amplifier(W)
Z_p = 3200.0    #Output impedance ohm of the single channel power amplifier(ohm)
N_p = 1500.0    #Number of primary turns in a tapped impedance-matching transformer
Z_L1 = 8.0      #Amplifier output using a tapped impedance-matching transformer(ohm)
Z_L2 = 4.0      #Amplifier output using a tapped impedance-matching transformer(ohm)

#Calculation
#Case(a)
alpha = (Z_p/Z_L1)**0.5      #Transformation ratio
N_2 = N_p/alpha              #Total number of secondary turns to match 8 ohm speaker
#Case(b)
alpha_b = (Z_p/Z_L2)**0.5    #Transformation ratio
N_1 = N_p/alpha_b            #Number of primary turns to match 4 ohm speaker
#Case(c)
turns_difference = N_2-N_1   #Difference in secondary and primary turns
alpha_c = (N_p/turns_difference)  #Transformation ratio
Z_L = Z_p/alpha_c**2         #Impedance that must be connected(ohm)

#Result
print('Case(a): Total number of secondary turns to match 8 Ω impedance speaker , N_2 = %.f t' %N_2)
print('Case(b): Number of turns to match 4 Ω impedance speaker , N_1 = %.f t' %N_1)
print('Case(c): Impedance that must be connected , Z_L = %.2f Ω' %Z_L)
Case(a): Total number of secondary turns to match 8 Ω impedance speaker , N_2 = 75 t
Case(b): Number of turns to match 4 Ω impedance speaker , N_1 = 53 t
Case(c): Impedance that must be connected , Z_L = 0.69 Ω

Example 14.6, Page number 492

In [1]:
#Variable declaration
P = 100.0        #Power rating of the single channel power amplifier(W)
Z_L1 = 8.0       #Amplifier output using a tapped impedance-matching transformer(ohm)
Z_L2 = 4.0       #Amplifier output using a tapped impedance-matching transformer(ohm)
P_servo = 10.0   #Power rating of the servo motor(W)
Z_servo = 0.7    #Impedance of the servo motor(ohm)

#Calculation
root_Z_AB = 8**0.5-4**0.5
Z_AB = (root_Z_AB)**2

#Result
print('Impedance between terminals A and B , Z_AB = %.2f Ω' %Z_AB)
Impedance between terminals A and B , Z_AB = 0.69 Ω

Example 14.7, Page number 494

In [1]:
import math
import cmath

#Variable declaration
V_s = 10.0*cmath.exp(1j*0*math.pi/180)    #Supply voltage of the source(V)
R_s = 1000.0                              #Resistance of the source(ohm)
R_L = 10.0                                #Load resistance(ohm)

#Calculation
#Case(a)
alpha = (R_s/R_L)**0.5       #Transformation ratio of the matching transformer for MPT
#Case(b)
V = abs(V_s)                 #Considering only absolute value of Supply voltage for easy simplification
V_1 = V/2                    #Terminal voltage of the source at MPT(V)
#Case(c)
V_2 = V_1/alpha              #Terminal voltage across the load at MPT(V)
#Case(d)
Z_L = R_L
I_2 = V_2/Z_L                #Secondary load current(A). Method 1
I2 = V/(2*alpha*R_L)         #Secondary load current(A). Method 2
#Case(e)
I_1 = I_2/alpha              #Primary load current drawn from the source(A). Method 1
I1 = V/(2*R_s)               #Primary load current drawn from the source(A). Method 2
#Case(f)
P_L = (I_2)**2*R_L           #Maximum power dissipated in the load(W)
#Case(g)
P_s = (I_1)**2*R_s           #Power dissipated internally within the source(W)
#Case(h)
theta = 0
P_T1 = abs(V)*I_1*math.cos(theta) #Total power supplied by the source(W). Method 1
P_T2 = P_L+P_s               #Total power supplied by the source(W). Method 2
#Case(i)
P_T = P_T1
n = P_L/P_T*100              #Power transfer efficiency(%)

#Result
print('Case(a): Transformation ratio of the matching transformer for MPT , α = %.f ' %alpha)
print('Case(b): Terminal voltage of the source at MPT , V_1 = %.1f V' %V_1)
print('Case(c): Terminal voltage across the load at MPT , V_2 = %.1f V' %V_2)
print('Case(d): Secondary load current , I_2 = %.f mA (Method 1)' %(1000*I_2))
print('         Secondary load current , I_2 = %.f mA (Method 2)' %(1000*I2))
print('Case(e): Primary load current drawn from the source , I_1 = %.f mA (Method 1)' %(1000*I_1))
print('         Primary load current drawn from the source , I_1 = %.f mA (Method 2)' %(1000*I1))
print('Case(f): Maximum power dissipated in the load , P_L = %.f mW' %(1000*P_L))
print('Case(g): Power dissipated internally within the source , P_s = %.f mW' %(1000*P_s))
print('Case(h): Total power supplied by the source , P_T = %.f mW (Method 1)' %(1000*P_T1))
print('         Total power supplied by the source , P_T = %.f mW (Method 2)' %(1000*P_T2))
print('Case(i): Power transfer efficiency , η = %.f percent' %n)
Case(a): Transformation ratio of the matching transformer for MPT , α = 10 
Case(b): Terminal voltage of the source at MPT , V_1 = 5.0 V
Case(c): Terminal voltage across the load at MPT , V_2 = 0.5 V
Case(d): Secondary load current , I_2 = 50 mA (Method 1)
         Secondary load current , I_2 = 50 mA (Method 2)
Case(e): Primary load current drawn from the source , I_1 = 5 mA (Method 1)
         Primary load current drawn from the source , I_1 = 5 mA (Method 2)
Case(f): Maximum power dissipated in the load , P_L = 25 mW
Case(g): Power dissipated internally within the source , P_s = 25 mW
Case(h): Total power supplied by the source , P_T = 50 mW (Method 1)
         Total power supplied by the source , P_T = 50 mW (Method 2)
Case(i): Power transfer efficiency , η = 50 percent

Example 14.8, Page number 495

In [1]:
#Variable declaration
V = 20.0         #No-load voltage(V)
R_s = 18.0       #Internal resistance of the power amplifier(ohm)
R_L = 8.0        #Load resistance(ohm)

#Calculation
V_L = (R_L/(R_L+R_s))* V    #Load voltage(V)
P_L = (V_L)**2/R_L          #Power delivered to the speaker when connected directly to the amplifier(W)
alpha = (R_s/R_L)**0.5      #Turns ratio of the transformer to maximize speaker power
V_2 = V/(2*alpha)           #Secondary voltage(V)
P_L2 = (V_2)**2/R_L         #Maximum power delivered to the speaker(W)

#Result
print('Case(a): Power delivered to the speaker when connected directly to the amplifier , P_L = %.2f W' %P_L)
print('Case(b): Turns ratio of the transformer to maximize speaker power , α = %.1f ' %alpha)
print('Case(c): Maximum power delivered to the speaker using matching transformer of part(b) , P_L = %.1f W' %P_L2)
Case(a): Power delivered to the speaker when connected directly to the amplifier , P_L = 4.73 W
Case(b): Turns ratio of the transformer to maximize speaker power , α = 1.5 
Case(c): Maximum power delivered to the speaker using matching transformer of part(b) , P_L = 5.6 W

Example 14.9, Page number 500

In [1]:
#Variable declaration
kVA = 1.0        #Rating of the transformer(kVA)
V_1 = 220.0      #Primary voltage(V)
V_2 = 110.0      #Secondary voltage(V)
f_o = 400.0      #Original frequency(Hz)
f_f = 60.0       #Frequency for which the transformer is to be used(Hz)

#Calculation
#Case(a)
alpha = V_1/V_2        #Transformation ratio
E_h = V_1*(f_f/f_o)    #Maximum rms voltage applied to HV side(V)
E_1 = E_h
E_l = E_1/alpha        #Maximum voltage output of LV side(V)
#Case(b)
V_h = V_1              #High voltage(V)
I_h = kVA*1000/V_h     #High current(A)
Vh = E_h
kVA_new = Vh*I_h/1000  #kVA rating of transformer under reduced frequency

#Result
print('Case(a): Maximum rms voltage that may be applied to high-voltage side , E_h = %.f V' %E_h)
print('         Maximum voltage output of the low-voltage side , E_l = %.1f V' %E_l)
print('Case(b): kVA rating of transformer under conditions of reduced frequency , V_h*I_h = %.2f kVA' %kVA_new)
Case(a): Maximum rms voltage that may be applied to high-voltage side , E_h = 33 V
         Maximum voltage output of the low-voltage side , E_l = 16.5 V
Case(b): kVA rating of transformer under conditions of reduced frequency , V_h*I_h = 0.15 kVA

Example 14.10, Page number 501

In [1]:
#Variable declaration
kVA = 1.0       #kVA rating of the transformer
V_1 = 220.0     #Primary voltage(V)
V_2 = 110.0     #Secondary voltage(V)
f_o = 400.0     #Frequency(Hz)
f_f = 60.0      #Frequency for which the transformer is to be used(Hz)
P_orig = 10.0   #Original iron losses of the transformer(W)

#Calculation 
B = f_o/f_f                  #Flux density
P_iron = P_orig*B**2         #Iron losses(W)

#Result
print('Iron losses if is operated at reduced frequency of 60 Hz , P_iron = %.f W' %P_iron)
Iron losses if is operated at reduced frequency of 60 Hz , P_iron = 444 W

Example 14.11, Page number 504

In [1]:
#Variable declaration
kVA = 500.0     #Rating of the step-down transformer(kVA)
V_1 = 2300.0    #Primary voltage(V)
V_2 = 230.0     #Secondary voltage(V)
f = 60.0        #Frequency(Hz)
r_1 = 0.1       #Primary winding resistance(ohm)
x_1 = 0.3       #Primary winding reactance(ohm)
r_2 = 0.001     #Secondary winding resistance(ohm)
x_2 = 0.003     #Secondary winding reactance(ohm)

#Calculation
alpha = V_1/V_2             #Transformation ratio
#Case(a)
I_2 = kVA*1000/V_2          #Secondary current(A)
I_1 = I_2/alpha             #Primary current(A)
#Case(b)
Z_2 = complex(r_2,x_2)      #Secondary internal impedance(ohm)
Z_2_m = abs(Z_2)            #Magnitude of Z_2(ohm)
Z_1 = complex(r_1,x_1)      #Primary internal impedance(ohm)
Z_1_m = abs(Z_1)            #Magnitude of Z_1(ohm)
#Case(c)
I_2_Z_2 = I_2*Z_2_m         #Secondary internal voltage drop(V)
I_1_Z_1 = I_1*Z_1_m         #Primary internal voltage drop(V)
#Case(d)
E_2 = V_2+I_2_Z_2           #Secondary induced voltage(V)
E_1 = V_1-I_1_Z_1           #Primary induced voltage(V)
#Case(e)
ratio_E = E_1/E_2           #Ratio of primary to secondary induced voltage
ratio_V = V_1/V_2           #Ratio of primary to secondary terminal voltage

#Result
print('Case(a): Secondary current , I_2 = %.f A' %I_2)
print('         Primary current , I_1 = %.1f A' %I_1)
print('Case(b): Secondary internal impedance , Z_2 = %.5f Ω' %Z_2_m)
print('         Primary internal impedance , Z_1 = %.3f Ω' %Z_1_m)
print('Case(c): Secondary internal voltage drop , I_2*Z_2 = %.2f V' %I_2_Z_2)
print('         Primary internal voltage drop , I_1*Z_1 = %.1f V' %I_1_Z_1)
print('Case(d): Secondary induced voltage , E_2 = %.2f V' %E_2)
print('         Primary induced voltage , E_1 = %.1f V' %E_1)
print('Case(e): Ratio of primary to secondary induced voltages , E_1/E_2 = %.2f ' %ratio_E)
print('         Ratio of primary to secondary iterminal voltages , V_1/V_2 = %.1f ' %ratio_V)
Case(a): Secondary current , I_2 = 2174 A
         Primary current , I_1 = 217.4 A
Case(b): Secondary internal impedance , Z_2 = 0.00316 Ω
         Primary internal impedance , Z_1 = 0.316 Ω
Case(c): Secondary internal voltage drop , I_2*Z_2 = 6.87 V
         Primary internal voltage drop , I_1*Z_1 = 68.7 V
Case(d): Secondary induced voltage , E_2 = 236.87 V
         Primary induced voltage , E_1 = 2231.3 V
Case(e): Ratio of primary to secondary induced voltages , E_1/E_2 = 9.42 
         Ratio of primary to secondary iterminal voltages , V_1/V_2 = 10.0 

Example 14.12, Page number 505

In [1]:
#Variable declaration
V_1 = 2300.0    #Primary voltage(V)
V_2 = 230.0     #Secondary voltage(V)
I_2 = 2174.0    #Secondary current(A)
I_1 = 217.4     #Primary current(A)
Z_2 = 0.00316   #Secondary internal impedance(ohm)
Z_1 = 0.316     #Primary internal impedance(ohm)

#Calculation
alpha = V_1/V_2             #Transformation ratio
#Case(a)
Z_L = V_2/I_2               #Load impedance(ohm)
#Case(b)
Z_p = V_1/I_1               #Primary input impedance(ohm)
Zp = alpha**2*Z_L           #Primary input impedance(ohm)

#Result
print('Case(a): Load impedance , Z_L = %.4f Ω' %Z_L)
print('Case(b): Primary input impedance , Z_p = %.2f Ω (Method 1)' %Z_p)
print('         Primary input impedance , Z_p = %.2f Ω (Method 2)' %Zp)
print('Case(c): Impedance of load Z_L = %.4f Ω, is much greater than internal secondary impedance Z_2 = %.5f Ω' %(Z_L,Z_2))
print('         Primary input impedance Z_p = %.2f Ω, is much greater than the internal primary impedance Z_1 = %.3f Ω' %(Z_p,Z_1))
print('Case(d): Z_L must be much greater than Z_2 so that major part of the voltage produced by E_2 is dropped across Z_L')
print('         As Z_L is reduced in proportion to Z_2, the load current increases and more voltage is dropped internally across Z_2')
Case(a): Load impedance , Z_L = 0.1058 Ω
Case(b): Primary input impedance , Z_p = 10.58 Ω (Method 1)
         Primary input impedance , Z_p = 10.58 Ω (Method 2)
Case(c): Impedance of load Z_L = 0.1058 Ω, is much greater than internal secondary impedance Z_2 = 0.00316 Ω
         Primary input impedance Z_p = 10.58 Ω, is much greater than the internal primary impedance Z_1 = 0.316 Ω
Case(d): Z_L must be much greater than Z_2 so that major part of the voltage produced by E_2 is dropped across Z_L
         As Z_L is reduced in proportion to Z_2, the load current increases and more voltage is dropped internally across Z_2

Example 14.13, Page number 505

In [1]:
#Variable declaration
kVA = 500.0     #Rating of the step-down transformer(kVA)
V_1 = 2300.0    #Primary voltage(V)
V_2 = 230.0     #Secondary voltage(V)
f = 60.0        #Frequency(Hz)
r_1 = 0.1       #Primary winding resistance(ohm)
x_1 = 0.3       #Primary winding reactance(ohm)
r_2 = 0.001     #Secondary winding resistance(ohm)
x_2 = 0.003     #Secondary winding reactance(ohm)
Z_L = 0.1058    #Load impedance(ohm)

#Calculation
alpha = V_1/V_2             #Transformation ratio
R_c1 = r_1+alpha**2*r_2     #Equivalent internal resistance referred to the primary side(ohm)
X_c1 = x_1+alpha**2*x_2     #Equivalent internal reactance referred to the primary side(ohm)
Z_c1 = complex(R_c1,X_c1)   #Equivalent internal impedance referred to the primary side(ohm)
Z_c1_m = abs(Z_c1)          #Magnitude of Z_e1(ohm)
Z_L_prime = alpha**2*Z_L    #Equivalent secondary load impedance referred to the primary side(ohm)
R_L = Z_L                   #Load resistance(ohm)
X_L = 0                     #Load reactance(ohm)
I_1 = V_1/complex((R_c1+alpha**2*R_L),(X_c1+alpha**2*X_L))  #Primary load current(A)

#Result
print('Case(a): Equivalent internal resistance referred to the primary side , R_c1 = %.1f Ω' %R_c1)
print('Case(b): Equivalent internal reactance referred to the primary side , X_c1 = %.1f Ω' %X_c1)
print('Case(c): Equivalent internal impedance referred to the primary side , Z_c1 = %.3f Ω' %Z_c1_m)
print('Case(d): Equivalent secondary load impedance referred to the primary side , α^2*Z_L = %.2f Ω' %Z_L_prime)
print('Case(e): Primary load current , I_1 = %.f A' %abs(I_1))
Case(a): Equivalent internal resistance referred to the primary side , R_c1 = 0.2 Ω
Case(b): Equivalent internal reactance referred to the primary side , X_c1 = 0.6 Ω
Case(c): Equivalent internal impedance referred to the primary side , Z_c1 = 0.632 Ω
Case(d): Equivalent secondary load impedance referred to the primary side , α^2*Z_L = 10.58 Ω
Case(e): Primary load current , I_1 = 213 A

Example 14.14, Page number 508

In [1]:
#Variable declaration
kVA = 500.0        #Rating of the step-down transformer(kVA)
V_1 = 2300.0       #Primary voltage(V)
V_2 = 230.0        #Secondary voltage(V)
R_e2 = 2.0         #Equivalent resistance referred to the primary side(mΩ)
X_e2 = 6.0         #Equivalent reactance referred to the primary side(mΩ)
cos_theta2 = 1.0   #Unity PF

#Calculation
I_2 = kVA/V_2              #Rated secondary current(kA)
R_e2_drop = I_2*R_e2       #Full-load equivalent resistance voltage drop(V)
X_e2_drop = I_2*X_e2       #Full-load equivalent reactance voltage drop(V) 
sin_theta2 = (1-cos_theta2**2)**0.5
E_2 = complex((V_2*cos_theta2+I_2*R_e2),(V_2*sin_theta2+I_2*X_e2)) #Induced voltage(V)
E_2_m = abs(E_2)           #Magnitude of E_2(V)
VR = (E_2_m-V_2)/V_2*100   #Percent voltage regulation at unity PF(%)

#Result
print('Case(a): Rated secondary current , I_2 = %.3f kA' %I_2)
print('Case(b): Full-load equivalent resistance voltage drop , I_2*R_e2 = %.2f V' %R_e2_drop)
print('Case(c): Full-load equivalent reactance voltage drop , I_2*X_e2 = %.2f V' %X_e2_drop)
print('Case(d): Induced voltage when the transformer is delivering rated current at unity PF , E_2 = %.2f V' %E_2_m)
print('Case(e): Voltage regulation at unity PF , VR = %.2f percent' %VR)
Case(a): Rated secondary current , I_2 = 2.174 kA
Case(b): Full-load equivalent resistance voltage drop , I_2*R_e2 = 4.35 V
Case(c): Full-load equivalent reactance voltage drop , I_2*X_e2 = 13.04 V
Case(d): Induced voltage when the transformer is delivering rated current at unity PF , E_2 = 234.71 V
Case(e): Voltage regulation at unity PF , VR = 2.05 percent

Example 14.15, Page number 508

In [1]:
#Variable declaration
kVA = 500.0        #Rating of the step-down transformer(kVA)
V_1 = 2300.0       #Primary voltage(V)
V_2 = 230.0        #Secondary voltage(V)
R_e2 = 2.0         #Equivalent resistance referred to the primary side(mΩ)
X_e2 = 6.0         #Equivalent reactance referred to the primary side(mΩ)
cos_theta2 = 0.8   #Lagging PF

#Calculation
I_2 = kVA/V_2              #Rated secondary current(kA)
R_e2_drop = I_2*R_e2       #Full-load equivalent resistance voltage drop(V)
X_e2_drop = I_2*X_e2       #Full-load equivalent reactance voltage drop(V) 
sin_theta2 = (1-cos_theta2**2)**0.5
E_2 = complex((V_2*cos_theta2+I_2*R_e2),(V_2*sin_theta2+I_2*X_e2)) #Induced voltage(V)
E_2_m = abs(E_2)           #Magnitude of E_2(V)
VR = (E_2_m-V_2)/V_2*100   #Percent voltage regulation(%)

#Result
print('Case(d): Induced voltage when the transformer is delivering rated current at 0.8 PF lagging , E_2 = %.2f V' %E_2_m)
print('Case(e): Voltage regulation at 0.8 PF lagging , VR = %.2f percent' %VR)
Case(d): Induced voltage when the transformer is delivering rated current at 0.8 PF lagging , E_2 = 241.43 V
Case(e): Voltage regulation at 0.8 PF lagging , VR = 4.97 percent

Example 14.16, Page number 508

In [1]:
#Variable declaration
kVA = 500.0        #Rating of the step-down transformer(kVA)
V_1 = 2300.0       #Primary voltage(V)
V_2 = 230.0        #Secondary voltage(V)
R_e2 = 2.0         #Equivalent resistance referred to the primary side(mΩ)
X_e2 = 6.0         #Equivalent reactance referred to the primary side(mΩ)
cos_theta2 = 0.6   #Leading PF

#Calculation
I_2 = kVA/V_2              #Rated secondary current(kA)
R_e2_drop = I_2*R_e2       #Full-load equivalent resistance voltage drop(V)
X_e2_drop = I_2*X_e2       #Full-load equivalent reactance voltage drop(V) 
sin_theta2 = (1-cos_theta2**2)**0.5
E_2 = complex((V_2*cos_theta2+I_2*R_e2),(V_2*sin_theta2-I_2*X_e2)) #Induced voltage(V)
E_2_m = abs(E_2)           #Magnitude of E_2(V)
VR = (E_2_m-V_2)/V_2*100   #Percent voltage regulation(%)

#Result
print('Case(d): Induced voltage when the transformer is delivering rated current at 0.6 PF leading , E_2 = %.1f V' %E_2_m)
print('Case(e): Voltage regulation at 0.6 PF leading , VR = %.2f percent' %VR)
Case(d): Induced voltage when the transformer is delivering rated current at 0.6 PF leading , E_2 = 222.5 V
Case(e): Voltage regulation at 0.6 PF leading , VR = -3.28 percent

Example 14.17, Page number 511

In [1]:
import math

#Variable declaration
kVA = 20.0    #kVA rating of the step-down transformer
S = 20000.0   #Power rating of the step-down transformer in VA
V_1 = 2300.0  #Primary voltage(V)
V_2 = 230.0   #Secondary voltage(V)
P1 = 250.0    #Wattmeter reading(W)
I1 = 8.7      #Input current(A)
V1 = 50.0     #Input voltage(V)

#Calculation
alpha = V_1/V_2               #Transformation ratio
#Case(a)
Z_e1 = V1/I1                  #Equivalent impedance w.r.t HV side(ohm)
R_e1 = P1/I1**2               #Equivalent resistance w.r.t HV side(ohm)
theta = math.acos(R_e1/Z_e1)   #PF angle(radians)
theta_a = theta*180/math.pi   #PF angle(degree)
X_e1 = Z_e1*math.sin(theta)   #Equivalent reactance w.r.t HV side(ohm)
#Case(b)
Z_e2 = Z_e1/alpha**2          #Equivalent impedance w.r.t LV side(ohm)
R_e2 = R_e1/alpha**2          #Equivalent resistance w.r.t LV side(ohm)
X_e2 = Z_e2*math.sin(theta)   #Equivalent reactance w.r.t LV side(ohm)
#Case(c)
I_2 = S/V_2                   #Rated secondary load current(A)
R_e2_drop = I_2*R_e2          #Full-load equivalent resistance voltage drop(V)
X_e2_drop = I_2*X_e2          #Full-load equivalent reactance voltage drop(V)
cos_theta2 = 1.0              #Unity PF
sin_theta2 = (1-cos_theta2**2)**0.5
E_2 = complex((V_2*cos_theta2+I_2*R_e2),(V_2*sin_theta2+I_2*X_e2)) #Induced voltage(V)
E_2_m = abs(E_2)              #Magnitude of E_2(V)
VR_unity_PF = (E_2_m-V_2)/V_2*100   #Transformer voltage regulation(%)
#Case(d)
cos_theta_2 = 0.7             #Lagging PF
sin_theta_2 = (1-(cos_theta_2)**2)**0.5
E2 = complex((V_2*cos_theta_2+I_2*R_e2),(V_2*sin_theta_2+I_2*X_e2))  #Induced voltage(V)
E2_m = abs(E2)                #Magnitude of E2(V)
VR_lag_PF = (E2_m-V_2)/V_2*100      #Transformer voltage regulation(%)

#Result
print('Case(a): Equivalent impedance referred to HV side , Z_e1 = %.2f Ω' %Z_e1)
print('         Equivalent resistance referred to HV side , R_e1 = %.1f Ω' %R_e1)
print('         Equivalent reactance referred to HV side , X_e1 = %.2f Ω' %X_e1)
print('Case(b): Equivalent impedance referred to LV side , Z_e2 = %.1f mΩ' %(1000*Z_e2))
print('         Equivalent resistance referred to LV side , R_e2 = %.f mΩ' %(1000*R_e2))
print('         Equivalent reactance referred to LV side , X_e2 = %.1f mΩ' %(1000*X_e2))
print('Case(c): Transformer voltage regulation , VR = %.2f percent' %VR_unity_PF)
print('Case(c): Transformer voltage regulation , VR = %.2f percent' %VR_lag_PF)
Case(a): Equivalent impedance referred to HV side , Z_e1 = 5.75 Ω
         Equivalent resistance referred to HV side , R_e1 = 3.3 Ω
         Equivalent reactance referred to HV side , X_e1 = 4.70 Ω
Case(b): Equivalent impedance referred to LV side , Z_e2 = 57.5 mΩ
         Equivalent resistance referred to LV side , R_e2 = 33 mΩ
         Equivalent reactance referred to LV side , X_e2 = 47.0 mΩ
Case(c): Transformer voltage regulation , VR = 1.26 percent
Case(c): Transformer voltage regulation , VR = 2.14 percent

Example 14.18, Page number 511

In [1]:
#Variable declaration
V1_rated = 2300.0  #Primary voltage(V)
V_sc = 50.0        #Short circuit test voltage(V)

#Calculation
Pc_sc_Pc = (V_sc/V1_rated)**2    #Fraction of rated core loss

#Result
print('Fraction of P_c measured by wattmeter , P_c(sc) = %.6f*P_c ' %Pc_sc_Pc)
Fraction of P_c measured by wattmeter , P_c(sc) = 0.000473*P_c 

Example 14.19, Page number 512

In [1]:
import math

#Variable declaration
kVA = 20.0    #kVA rating of the step-down transformer
S = 20000.0   #Power rating of the step-down transformer in VA
V_1 = 2300.0  #Primary voltage(V)
V_2 = 230.0   #Secondary voltage(V)
P1 = 250.0    #Wattmeter reading(W)
I1 = 8.7      #Input current(A)
V1 = 50.0     #Input voltage(V)

#Calculation
#Case(a)
I1_Ze1 = V1                         #HV impedance drop(V)
#Case(b)
theta = math.acos(P1/(V1*I1))       #PF angle(radian)
theta_a = theta*180/math.pi         #PF angle(degree)
I1_Re1 = I1_Ze1*math.cos(theta)     #HV side resistance volt drop(V)
#Case(c)
I1_Xe1 = I1_Ze1*math.sin(theta)     #HV side resistance volt drop(V)
#Case(d)
cos_theta1 = 1.0                     #Unity PF
sin_theta1 = (1-cos_theta1**2)**0.5
E1 = complex((V_1*cos_theta1+I1_Re1),(V_1*sin_theta1+I1_Xe1))  #Induced voltage(V)
E1_m = abs(E1)                      #Magnitude of E1(V)
VR = (E1_m-V_1)/V_1*100             #Transformer voltage regulation at unity PF(%)
#Case(e)
cos_theta_1 = 0.7                    #Lagging PF
sin_theta_1 = (1-cos_theta_1**2)**0.5
E_1 = complex((V_1*cos_theta_1+I1_Re1),(V_1*sin_theta_1+I1_Xe1))  #Induced voltage(V)
E_1_m = abs(E_1)                    #Magnitude of E_1(V)
VR_1 = (E_1_m-V_1)/V_1*100          #Transformer voltage regulation at 0.7 PF lagging(%)

#Result
print('Case(a): High voltage impedance drop , I_1*Z_e1 = %.f V' %I1_Ze1)
print('Case(b): HV side resistance volt drop , I_1*R_e1 = %.2f V' %I1_Re1)
print('Case(c): HV side reactance volt drop , I_1*X_e1 = %.2f V' %I1_Xe1)
print('Case(d): Transformer voltage regulation at unity PF , VR = %.2f percent' %VR)
print('Case(e): Transformer voltage regulation at 0.7 PF lagging , VR = %.2f percent' %VR_1)
Case(a): High voltage impedance drop , I_1*Z_e1 = 50 V
Case(b): HV side resistance volt drop , I_1*R_e1 = 28.74 V
Case(c): HV side reactance volt drop , I_1*X_e1 = 40.92 V
Case(d): Transformer voltage regulation at unity PF , VR = 1.27 percent
Case(e): Transformer voltage regulation at 0.7 PF lagging , VR = 2.15 percent

Example 14.20, Page number 515

In [1]:
#Variable declaration
kVA = 500.0       #kVA rating of the step-down transformer
V_1 = 2300.0      #Primary voltage(V)
V_2 = 208.0       #Secondary voltage(V)
f = 60.0          #Frequency(Hz)
P_sc = 8200.0     #Wattmeter reading in SC test(W)
I_sc = 217.4      #Short circuit current(A)
V_sc = 95.0       #Short circuit voltage(V)
P_oc = 1800.0     #Wattmeter reading in OC test(W)
I_oc = 85.0       #Open circuit current(A)
V_oc = 208.0      #Open circuit voltage(V)

#Calculation
alpha = V_1/V_2        #Transformation ratio
#Case(a)
P = P_sc               #Wattmeter reading(W)
I1 = I_sc              #Short circuit current(A)
R_e1 = P/(I1)**2       #Equivalent resistance w.r.t HV side(ohm)
R_e2 = R_e1/alpha**2   #Equivalent resistance referred to LV side(ohm)
#Case(b)
r_2 = R_e2/2           #Resistance of low-voltage side(ohm)
#Case(c)
I_m = I_oc             #Open circuit current(A)
P_cu = I_m**2*r_2      #Transformer copper loss of the LV side winding during OC-test(W)
#Case(d)
P_c = P_oc-P_cu        #Transformer core loss(W)

#Result
print('Case(a): Equivalent resistance to low voltage side , R_e2 = %.3f mΩ' %(1000*R_e2))
print('Case(b): Resistance of low voltage side , r_2 =  %.2f mΩ' %(1000*r_2))
print('Case(c): Transformer copper loss of the LV side winding during OC-test , I_m^2*r_2 = %.2f W' %P_cu)
print('Case(d): Transformer core loss when rated voltage is applied , P_c = %.1f W' %P_c)
print('Case(e): Yes.The error is approximately 5/1800 = 0.278 percent, which is within error produced by the instruments used in test');
print('         We may assume that the core loss is %d W' %P_oc)
Case(a): Equivalent resistance to low voltage side , R_e2 = 1.419 mΩ
Case(b): Resistance of low voltage side , r_2 =  0.71 mΩ
Case(c): Transformer copper loss of the LV side winding during OC-test , I_m^2*r_2 = 5.13 W
Case(d): Transformer core loss when rated voltage is applied , P_c = 1794.9 W
Case(e): Yes.The error is approximately 5/1800 = 0.278 percent, which is within error produced by the instruments used in test
         We may assume that the core loss is 1800 W

Example 14.21, Page number 516

In [1]:
#Variable declaration
V_sc = 50.0           #Short circuit voltage(V)
V_1 = 2300.0          #Rated primary voltage(V)
P_c = 1.8             #Core losses(kW)
P_k = 1.8             #Fixed losses(kW)
P_cu_rated = 8.2      #Rated copper loss(kW)
kVA = 500.0           #Power rating(kVA)
PF = 1.0              #Power factor
P_o = kVA*PF          #Full-load output at unity PF(kW)

#Calculation
#Case(a)
LF1 = 1.0/4                     #Load fraction
LF2 = 1.0/2                     #Load fraction
LF3 = 3.0/4                     #Load fraction
LF4 = 5.0/4                     #Load fraction
P_cu_fl = 8.2                   #Equivalent copper loss at full-load slip(kW)
P_cu_LF1 = LF1**2*P_cu_fl       #Equivalent copper loss at 1/4 rated load(kW)
P_cu_LF2 = LF2**2*P_cu_fl       #Equivalent copper loss at 1/2 rated load(kW)
P_cu_LF3 = LF3**2*P_cu_fl       #Equivalent copper loss at 3/4 rated load(kW)
P_cu_LF4 = LF4**2*P_cu_fl ;     #Equivalent copper loss at 5/4 rated load(kW)
P_L_fl = P_c+P_cu_fl            #Total losses at rated load(kW)
P_L_1 = P_c+P_cu_LF1            #Total losses at 1/4 rated load(kW)
P_L_2 = P_c+P_cu_LF2            #Total losses at 1/2 rated load(kW)
P_L_3 = P_c+P_cu_LF3            #Total losses at 3/4 rated load(kW)
P_L_4 = P_c+P_cu_LF4            #Total losses at 5/4 rated load(kW)
P_o_fl = P_o                    #Total output at rated load(kW)
P_o_1 = P_o*LF1                 #Total output at 1/4 rated load(kW)
P_o_2 = P_o*LF2                 #Total output at 1/2 rated load(kW)
P_o_3 = P_o*LF3                 #Total output at 3/4 rated load(kW)
P_o_4 = P_o*LF4                 #Total output at 5/4 rated load(kW)
P_in_fl = P_L_fl+P_o_fl         #Total input at rated load(kW)
P_in_1 = P_L_1+P_o_1            #Total input at 1/4 rated load(kW)
P_in_2 = P_L_2+P_o_2            #Total input at 1/2 rated load(kW)
P_in_3 = P_L_3+P_o_3            #Total input at 3/4 rated load(kW)
P_in_4 = P_L_4+P_o_4            #Total input at 5/4 rated load(kW)
n_fl = (P_o_fl/P_in_fl)*100     #Efficiency at rated load(%)
n_1 = (P_o_1/P_in_1)*100        #Efficiency at 1/4 rated load(%)
n_2 = (P_o_2/P_in_2)*100        #Efficiency at 1/2 rated load(%)
n_3 = (P_o_3/P_in_3)*100        #Efficiency at 3/4 rated load(%)
n_4 = (P_o_4/P_in_4)*100        #Efficiency at 5/4 rated load(%)
#Case(b)
PF_b = 0.8                      #PF lagging
Po_fl = P_o*PF_b                #Total output at 1rated load(kW)
Po_1 = P_o*LF1*PF_b             #Total output at 1/4 rated load(kW)
Po_2 = P_o*LF2*PF_b             #Total output at 1/2 rated load(kW)
Po_3 = P_o*LF3*PF_b             #Total output at 3/4 rated load(kW)
Po_4 = P_o*LF4*PF_b             #Total output at 5/4 rated load(kW)
Pin_fl = P_L_fl+Po_fl           #Total input at rated load(kW)
Pin_1 = P_L_1+Po_1              #Total input at 1/4 rated load(kW)
Pin_2 = P_L_2+Po_2              #Total input at 1/2 rated load(kW)
Pin_3 = P_L_3+Po_3              #Total input at 3/4 rated load(kW)
Pin_4 = P_L_4+Po_4              #Total input at 5/4 rated load(kW)
nfl = (Po_fl/Pin_fl)*100        #Efficiency at rated load(%)
n1 = (Po_1/Pin_1)*100           #Efficiency at 1/4 rated load(%)
n2 = (Po_2/Pin_2)*100           #Efficiency at 1/2 rated load(%)
n3 = (Po_3/Pin_3)*100           #Efficiency at 3/4 rated load(%)
n4 = (Po_4/Pin_4)*100           #Efficiency at 5/4 rated load(%)
#Case(c)
R_e2 = 1.417*10**-3             #Equivalent resistance referred to LV side(ohm)
Pc = 1800.0                     #Core losses(W)
I_2 = (Pc/R_e2)**0.5            #Load current for maximum efficiency invariant of LF(A)
#Case(d)
V = 208.0                       #Voltage rating(V)
I_2_rated = kVA*1000/V          #Rated secondary current(A)
LF_max = I_2/I_2_rated          #Load fraction for maximum efficiency
#Case(e)
cos_theta = 1.0                 #Unity PF
V_2 = V                         #Secondary voltage(V)
n_max_e = (V_2*I_2*cos_theta)/((V_2*I_2*cos_theta)+(Pc+I_2**2*R_e2))*100    #Maximum efficiency(%)
#Case(f)
cos_theta2 = 0.8                #PF lagging
n_max_f = (V_2*I_2*cos_theta2)/((V_2*I_2*cos_theta2)+(Pc+I_2**2*R_e2))*100  #Maximum efficiency(%)

#Result
print('Case(a): Tabulation at unity PF:')
print('__________________________________________________________________________________________________________')
print('L.F \t Core loss \t Copper loss \tTotal loss \t Total Output \t Total Input \t Efficiency')
print('    \t   (kW)    \t    (kW)     \t  P_L (kW) \t    P_o(kW)   \t P_L+P_o(kW) \t P_o/P_in(percent)')
print('__________________________________________________________________________________________________________')
print('%.2f  \t   %.1f \t\t   %.3f       \t %.3f \t\t   %.1f \t    %.2f \t %.2f' %(LF1,P_c,P_cu_LF1,P_L_1,P_o_1,P_in_1,n_1))
print('%.2f  \t   %.1f \t\t   %.3f       \t %.2f \t\t   %.1f \t    %.2f \t %.2f' %(LF2,P_c,P_cu_LF2,P_L_2,P_o_2,P_in_2,n_2))
print('%.2f  \t   %.1f \t\t   %.3f       \t %.2f \t\t   %.1f \t    %.2f \t %.2f' %(LF3,P_c,P_cu_LF3,P_L_3,P_o_3,P_in_3,n_3))
print('1  \t   %.1f \t\t   %.3f       \t %.2f  \t\t   %.1f \t    %.1f \t %.2f' %(P_c,P_cu_fl,P_L_fl,P_o_fl,P_in_fl,n_fl))
print('%.2f  \t   %.1f \t\t   %.3f  \t %.1f  \t\t   %.1f \t    %.1f \t %.2f' %(LF4,P_c,P_cu_LF4,P_L_4,P_o_4,P_in_4,n_4))
print('__________________________________________________________________________________________________________')
print('\nCase(b): Tabulation at 0.8 PF lagging:')
print('__________________________________________________________________________________________________________')
print('L.F \t Core loss \t Copper loss \tTotal loss \t Total Output \t Total Input \t Efficiency')
print('\t   (kW)    \t    (kW)     \t  P_L (kW) \t    P_o(kW)   \t P_L+P_o(kW)\t P_o/P_in(percent)')
print('__________________________________________________________________________________________________________')
print('%.2f  \t   %.1f \t\t   %.3f       \t %.3f \t\t   %.1f \t    %.2f \t %.2f' %(LF1,P_c,P_cu_LF1,P_L_1,Po_1,Pin_1,n1))
print('%.2f  \t   %.1f \t\t   %.2f       \t %.2f \t\t   %.1f \t    %.2f \t %.2f' %(LF2,P_c,P_cu_LF2,P_L_2,Po_2,Pin_2,n2))
print('%.2f  \t   %.1f \t\t   %.2f       \t %.2f \t\t   %.1f \t    %.2f \t %.2f' %(LF3,P_c,P_cu_LF3,P_L_3,Po_3,Pin_3,n3))
print('1  \t   %.1f \t\t   %.1f \t\t %.2f \t\t   %.1f \t    %.f \t\t %.2f' %(P_c,P_cu_fl,P_L_fl,Po_fl,Pin_fl,nfl))
print('%.2f  \t   %.1f \t\t   %.1f  \t %.1f \t\t   %.1f \t    %.1f \t %.2f' %(LF4,P_c,P_cu_LF4,P_L_4,Po_4,Pin_4,n4))
print('__________________________________________________________________________________________________________\n')
print('Case(c): Load current at which maximum efficiency occurs regardless of load PF , I_2 = %.1f A' %I_2)
print('Case(d): Load fraction at which maximum efficiency occurs = %.3f (approximately half rated load)' %LF_max)
print('Case(e): Maximum efficiency at unity PF , η_max = %.2f percent' %n_max_e)
print('Case(f): Maximum efficiency at 0.8 PF lagging , η_max = %.2f percent' %n_max_f)
Case(a): Tabulation at unity PF:
__________________________________________________________________________________________________________
L.F 	 Core loss 	 Copper loss 	Total loss 	 Total Output 	 Total Input 	 Efficiency
    	   (kW)    	    (kW)     	  P_L (kW) 	    P_o(kW)   	 P_L+P_o(kW) 	 P_o/P_in(percent)
__________________________________________________________________________________________________________
0.25  	   1.8 		   0.512       	 2.312 		   125.0 	    127.31 	 98.18
0.50  	   1.8 		   2.050       	 3.85 		   250.0 	    253.85 	 98.48
0.75  	   1.8 		   4.612       	 6.41 		   375.0 	    381.41 	 98.32
1  	   1.8 		   8.200       	 10.00  		   500.0 	    510.0 	 98.04
1.25  	   1.8 		   12.812  	 14.6  		   625.0 	    639.6 	 97.72
__________________________________________________________________________________________________________

Case(b): Tabulation at 0.8 PF lagging:
__________________________________________________________________________________________________________
L.F 	 Core loss 	 Copper loss 	Total loss 	 Total Output 	 Total Input 	 Efficiency
	   (kW)    	    (kW)     	  P_L (kW) 	    P_o(kW)   	 P_L+P_o(kW)	 P_o/P_in(percent)
__________________________________________________________________________________________________________
0.25  	   1.8 		   0.512       	 2.312 		   100.0 	    102.31 	 97.74
0.50  	   1.8 		   2.05       	 3.85 		   200.0 	    203.85 	 98.11
0.75  	   1.8 		   4.61       	 6.41 		   300.0 	    306.41 	 97.91
1  	   1.8 		   8.2 		 10.00 		   400.0 	    410 		 97.56
1.25  	   1.8 		   12.8  	 14.6 		   500.0 	    514.6 	 97.16
__________________________________________________________________________________________________________

Case(c): Load current at which maximum efficiency occurs regardless of load PF , I_2 = 1127.1 A
Case(d): Load fraction at which maximum efficiency occurs = 0.469 (approximately half rated load)
Case(e): Maximum efficiency at unity PF , η_max = 98.49 percent
Case(f): Maximum efficiency at 0.8 PF lagging , η_max = 98.12 percent

Example 14.22, Page number 520

In [1]:
import math
import cmath

#Variable declaration
V_1 = 2300.0        #Primary voltage(V)
V_2 = 230.0         #Secondary voltage(V)
P = 20.0            #Power rating of the transformer(kVA)
P_sc = 250.0        #Power measured(W)
V_sc = 50.0         #Short circuit voltage(V)
I_sc = 8.7          #Short circuit current(A)

#Calculation
#Case(a)
V_1b = V_1              #Base voltage(V)
Z_eq_pu = V_sc/V_1
beta = math.acos(P_sc/(V_sc*I_sc))    #Angle(radian)
beta_a = beta*180/math.pi             #Angle(degree)
Zeq_pu = Z_eq_pu*cmath.exp(1j*beta)   #Equivalent pu impedance
#Case(b)
PF_b = 1.0                            #Unity PF
theta_b = math.acos(PF_b)*180/math.pi #PF angle(degree)
V_1_pu = 1.0*cmath.exp(1j*0*math.pi/180)+1.0*cmath.exp(1j*theta_b*math.pi/180)*Zeq_pu  #pu voltage
V_1_pu_m = abs(V_1_pu)                #Magnitude of V_1_pu
#Case(c)
theta = math.acos(0.7)*180/math.pi    #Power factor angle(degrees)
V1_pu = 1.0*cmath.exp(1j*0*math.pi/180)+1.0*cmath.exp(1j*(-theta)*math.pi/180)*Zeq_pu  #pu voltage
V1_pu_m = abs(V1_pu)                  #Magnitude of V1_pu
#Case(d)
VR_unity_PF = (V_1_pu_m-1.0)*100      #Voltage regulation at unity PF(%)
#Case(e)
VR_lag_PF = (V1_pu_m-1.0)*100         #Voltage regulation at 0.7 lagging PF(%)

#Result
print('Case(a): Z_eq(pu)∠β = %.5f∠%.f° p.u' %(abs(Z_eq_pu),beta_a))
print('Case(b): V_1(pu) at unity PF , |V_1(pu)| = %.4f ' %V_1_pu_m)
print('Case(c): V_1(pu) at 0.7 PF lagging , |V_1(pu)| = %.5f ' %V1_pu_m)
print('Case(d): Voltage regulation at unity PF , VR = %.2f percent' %VR_unity_PF)
print('Case(e): Voltage regulation at 0.7 PF lagging , VR = %.2f percent' %VR_lag_PF)
print('Case(f): VRs as found by p.u method are essentially the same as those found in Exs.14-17 and 14-19 using the same data, for the same transformer but with much less effort')
Case(a): Z_eq(pu)∠β = 0.02174∠55° p.u
Case(b): V_1(pu) at unity PF , |V_1(pu)| = 1.0127 
Case(c): V_1(pu) at 0.7 PF lagging , |V_1(pu)| = 1.02146 
Case(d): Voltage regulation at unity PF , VR = 1.27 percent
Case(e): Voltage regulation at 0.7 PF lagging , VR = 2.15 percent
Case(f): VRs as found by p.u method are essentially the same as those found in Exs.14-17 and 14-19 using the same data, for the same transformer but with much less effort

Example 14.23, Page number 521

In [1]:
#Variable declaration
V_1 = 2300.0          #Primary voltage(V)
V_2 = 230.0           #Secondary voltage(V)
S = 500.0             #Power rating of the transformer(kVA)
f= 60.0               #Frequency(Hz)
V_oc = 208.0          #Open circuit voltage(V)
I_oc = 85.0           #Open circuit current(A)
P_oc = 1800.0         #Power measured(W)
V_sc = 95.0           #Short circuit voltage(V)
I_sc = 217.4          #Short circuit current(A)
P_sc = 8200.0         #Power measured(W)

#Calculation
#Case(a)
S_b = S ;             #Base voltage(kVA)
Psc = 8.2             #Power measured in kW during SC-test
P_Cu_pu = Psc/S_b     #Per unit value of P_Cu at rated load
#Case(b)
Poc = 1.8             #Power measured during OC-test(kW)
P_CL_pu = Poc/S_b     #Per unit value of P_CL at rated load
#Case(c)
PF = 1.0              #Unity Power factor
n_pu = PF/(PF+P_CL_pu+P_Cu_pu)*100                   #Efficiency at rated load,unity PF(%)
#Case(d)
PF_d = 0.8                                           #Lagging Power factor
n_pu_d = PF_d/(PF_d+P_CL_pu+P_Cu_pu)*100             #Efficiency at rated load,0.8 lagging PF
#Case(e)
LF = (P_CL_pu/P_Cu_pu)**0.5                          #Load fraction producing maximum efficiency
#Case(f)
n_pu_max = (LF*PF)/((LF*PF)+2*(P_CL_pu))*100         #Maximum efficiency at unity PF load
#Case(f)
n_pu_max_g = (LF*PF_d)/((LF*PF_d)+2*(P_CL_pu))*100   #Maximum efficiency at 0.8 lagging PF load


#Result
print('Case(a): Per unit copper loss at rated load , P_Cu(pu) = %.4f p.u = R_eq(pu)' %P_Cu_pu)
print('Case(b): Per unit core loss at rated load , P_CL(pu) = %.4f p.u' %P_CL_pu)
print('Case(c): Efficiency at rated load, unity PF , η_pu = %.2f percent' %n_pu)
print('Case(d): Efficiency at rated load, 0.8 PF lagging , η_pu = %.2f percent' %n_pu_d)
print('Case(e): Load fraction producing maximum efficiency , L.F = %.3f ' %LF)
print('Case(f): Maximum efficiency at unity PF load , η_pu = %.2f percent' %n_pu_max)
print('Case(g): Maximum efficiency at 0.8 PF lagging , η_pu = %.2f percent' %n_pu_max_g)
print('Case(h): All efficiency values are identical to those computed in solution to Ex.14-21')
print('Case(i): Per-unit method is much simpler and less subject to error than conventional method')
Case(a): Per unit copper loss at rated load , P_Cu(pu) = 0.0164 p.u = R_eq(pu)
Case(b): Per unit core loss at rated load , P_CL(pu) = 0.0036 p.u
Case(c): Efficiency at rated load, unity PF , η_pu = 98.04 percent
Case(d): Efficiency at rated load, 0.8 PF lagging , η_pu = 97.56 percent
Case(e): Load fraction producing maximum efficiency , L.F = 0.469 
Case(f): Maximum efficiency at unity PF load , η_pu = 98.49 percent
Case(g): Maximum efficiency at 0.8 PF lagging , η_pu = 98.12 percent
Case(h): All efficiency values are identical to those computed in solution to Ex.14-21
Case(i): Per-unit method is much simpler and less subject to error than conventional method

Example 14.24, Page number 521

In [1]:
#Variable declaration
V_1 = 2300.0          #Primary voltage(V)
V_2 = 230.0           #Secondary voltage(V)
S = 500.0             #Power rating of the transformer(kVA)
f= 60.0               #Frequency(Hz)
V_oc = 208.0          #Open circuit voltage(V)
I_oc = 85.0           #Open circuit current(A)
P_oc = 1800.0         #Power measured(W)
V_sc = 95.0           #Short circuit voltage(V)
I_sc = 217.4          #Short circuit current(A)
P_sc = 8200.0         #Power measured(W)

#Calculation
S_b = S ;             #Base voltage(kVA)
Psc = 8.2             #Power measured in kW during SC-test
P_Cu_pu = Psc/S_b     #Per unit value of P_Cu at rated load
Poc = 1.8             #Power measured during OC-test(kW)
P_CL_pu = Poc/S_b     #Per unit value of P_CL at rated load
#Case(a)
LF1 = 3.0/4           #Load fraction of rated load 
PF1 = 1.0             #Unity Power factor
n_pu_LF1 = (LF1*PF1)/((LF1*PF1)+P_CL_pu+(LF1)**2*P_Cu_pu)*100     #Efficiency at rated load,unity PF(%)
#Case(b)
LF2 = 1.0/4           #Load fraction of rated load
PF2 = 0.8             #Lagging PF
n_pu_LF2 = (LF2*PF2)/((LF2*PF2)+P_CL_pu+(LF2)**2*P_Cu_pu)*100     #Efficiency at 1/4 rated load,0.8 lagging PF(%)
#Case(c)
LF3 = 5.0/4           #Load fraction of rated load
PF3 = 0.8             #Leading PF
n_pu_LF3 = (LF3*PF3)/((LF3*PF3)+P_CL_pu+(LF3)**2*P_Cu_pu)*100     #Efficiency at r1/4 rated load,0.8 leading PF(%)


#Result
print('Case(a): p.u efficiency at 3/4 load unity PF , η_pu = %.2f percent' %n_pu_LF1)
print('Case(b): p.u efficiency at 1/4 load 0.8 PF lagging , η_pu = %.2f percent' %n_pu_LF2)
print('Case(c): p.u efficiency at 5/4 load 0.8 PF leading , η_pu = %.2f percent' %n_pu_LF3)
Case(a): p.u efficiency at 3/4 load unity PF , η_pu = 98.32 percent
Case(b): p.u efficiency at 1/4 load 0.8 PF lagging , η_pu = 97.74 percent
Case(c): p.u efficiency at 5/4 load 0.8 PF leading , η_pu = 97.16 percent

Example 14.25, Page number 522

In [1]:
import math
import cmath

#Variable declaration
kVA_1 = 500.0                     #Power rating of the transformer 1(kVA)
R_1_pu = 0.01                     #Per-unit value of resistance of the transformer 1
X_1_pu = 0.05                     #Per-unit value of reactance of the transformer 1
Z_1_pu = complex(R_1_pu,X_1_pu)   #Per-unit value of impedance of the transformer 1
PF = 0.8                          #Lagging PF
V_2 = 400.0                       #Secondary voltage(V)
S_load = 750.0                    #Increased system load(kVA)
kVA_2 = 250.0                     #Power rating of the transformer 2(kVA)
R_pu_2 = 0.015                    #Per-unit value of resistance of the transformer 2
X_pu_2 = 0.04                     #Per-unit value of reactance of the transformer 2

#Calculation
Z_pu_1 = complex(R_pu_2,X_pu_2)   #New transformer p.u. impedance
#Case(a)
V_b1 = 400.0                      #Base voltage(V)
V_b2 = 400.0                      #Base voltage(V)
Z_pu_2 = (kVA_1/kVA_2)*(V_b1/V_b2)**2*(Z_pu_1)        #New transformer p.u impedance
Z_2_pu = Z_pu_2                                       #New transformer p.u impedance
#Case(b)
cos_theta = PF                                        #Power factor
sin_theta = (1-(cos_theta)**2)**0.5
S_t_conjugate = (kVA_1+kVA_2)*complex(cos_theta,sin_theta)    #kVA of total load
#Case(c)
S_2_conjugate = S_t_conjugate*(Z_1_pu/(Z_1_pu+Z_2_pu))        #Portion of load carried by the smaller transformer(kVA)
S_2_conjugate_m = abs(S_2_conjugate)                          #Magnitude of S_2_conjugate(kVA)
S_2_conjugate_a = cmath.phase(S_2_conjugate)*180/math.pi      #Phase angle of S_2_conjugate(degree)
#Case(d)
S_1_conjugate = S_t_conjugate*(Z_2_pu/(Z_1_pu+Z_2_pu))        #Portion of load carried by the original transformer(kVA)
S_1_conjugate_m = abs(S_1_conjugate)                          #Magnitude of S_1_conjugate(kVA)
S_1_conjugate_a = cmath.phase(S_1_conjugate)*180/math.pi      #Phase angle of S_1_conjugate(degree)
#Case(e)
S_1 = S_1_conjugate_m
S_b1 = kVA_1                                                  #Base power of trancsformer 1(kVA)
LF1 = (S_1/S_b1)*100                                          #Load factor of the original transformer(%)
#Case(f)
S_2 = S_2_conjugate_m
S_b2 = kVA_2                                                  #Base power of trancsformer 1(kVA)
LF2 = (S_2/S_b2)*100                                          #Load factor of the new transformer(%)

#Result
print('Case(a): New transformer p.u impedance , Z_p.u.2 = (%.2f+j%.2f) p.u' %(Z_pu_2.real,Z_pu_2.imag))
print('Case(b): kVA of total load , S*_t = (%.f+j%.f) kVA ' %(S_t_conjugate.real,S_t_conjugate.imag))
print('Case(c): Portion of load now carried by the smaller transformer , S*_2 = (%.2f+j%.2f) kVA = %.1f∠%.1f° kVA' %(S_2_conjugate.real,S_2_conjugate.imag,S_2_conjugate_m,S_2_conjugate_a))
print('Case(d): Portion of load now carried by the original transformer , S*_1 = (%.2f+j%.1f) kVA = %.1f∠%.1f° kVA' %(S_1_conjugate.real,S_1_conjugate.imag,S_1_conjugate_m,S_1_conjugate_a))
print('Case(e): Load factor of the original transformer , L.F_1 = %.1f percent' %LF1)
print('Case(f): Load factor of the new transformer , L.F_w = %.1f percent' %LF2)
print('Case(g): Yes. Reduce no-load voltage of the new transformer to some value below that of its present value so that its share of the load is reduced')
Case(a): New transformer p.u impedance , Z_p.u.2 = (0.03+j0.08) p.u
Case(b): kVA of total load , S*_t = (600+j450) kVA 
Case(c): Portion of load now carried by the smaller transformer , S*_2 = (206.76+j190.54) kVA = 281.2∠42.7° kVA
Case(d): Portion of load now carried by the original transformer , S*_1 = (393.24+j259.5) kVA = 471.1∠33.4° kVA
Case(e): Load factor of the original transformer , L.F_1 = 94.2 percent
Case(f): Load factor of the new transformer , L.F_w = 112.5 percent
Case(g): Yes. Reduce no-load voltage of the new transformer to some value below that of its present value so that its share of the load is reduced

Example 14.26, Page number 523

In [1]:
import math
import cmath

#Variable declaration
kVA_1 = 500.0                     #Power rating of the transformer 1(kVA)
R_1_pu = 0.01                     #Per-unit value of resistance of the transformer 1
X_1_pu = 0.05                     #Per-unit value of reactance of the transformer 1
Z_1_pu = complex(R_1_pu,X_1_pu)   #Per-unit value of impedance of the transformer 1
PF = 0.8                          #Lagging PF
V = 400.0                         #Secondary voltage(V)
S_load = 750.0                    #Increased system load(kVA)
kVA_2 = 250.0                     #Power rating of the transformer 2(kVA)
R_pu_2 = 0.015                    #Per-unit value of resistance of the transformer 2
X_pu_2 = 0.04                     #Per-unit value of reactance of the transformer 2

#Calculation
Z_pu_1 = complex(R_1_pu,X_1_pu)                   #New transformer p.u. impedance
Z_pu1 = complex(R_pu_2,X_pu_2)                    #New transformer p.u. impedance
#Case(a)
V_b = V                                           #Base voltage(V)
#Case(b)
S_b = kVA_1                                       #Base power(kVA)
I_b = S_b/V_b*1000                                #Base current(A)
#Case(c)
Zb = V_b/I_b                                      #Base impedance(ohm). Method 1
Z_b = V**2/(S_b*1000)                             #Base impedance(ohm). Method 2
#Case(d)
Z_1 = Z_b*Z_pu_1*1000                             #Actual impedance of larger transformer(mΩ)
Z_1_m = abs(Z_1)                                  #Magnitude of Z_1(Ω)
Z_1_a = cmath.phase(Z_1)*180/math.pi              #Phase angle of Z_1(degree)
#Case(e)
V_b1 = V_b                                        #Base voltage(V)
V_b2 = V_b                                        #Base voltage(V)
Z_pu_2 = (kVA_1/kVA_2)*(V_b1/V_b2)**2*(Z_pu1)    #New transformer p.u impedance
Z_2_pu = Z_pu_2
Z_2 = Z_b*Z_2_pu*1000                             #Actual impedance of smaller transformer(mΩ)
Z_2_m = abs(Z_2)                                  #Magnitude of Z_2(Ω)
Z_2_a = cmath.phase(Z_2)*180/math.pi              #Phase angle of Z_2(degree)
#Case(f)
cos_theta = 0.8                                   #Lagging power factor
sin_theta = (1-(cos_theta)**2)**0.5
S_T = (kVA_1+kVA_2)*complex(cos_theta,-sin_theta) #kVA of total load
I_T = S_T*1000/V_b                                #Total current(A)
I_1 = I_T*(Z_2/(Z_1+Z_2))                         #Actual current delivered by larger transformer(A)
I_1_m = abs(I_1)                                  #Magnitude of I_1(A)
I_1_a = cmath.phase(I_1)*180/math.pi              #Phase angle of I_1(degree)
#Case(g)
I_2 = I_T*(Z_1/(Z_1+Z_2))                         #Actual current delivered by smaller transformer(A)
I_2_m = abs(I_2)                                  #Magnitude of I_2(A)
I_2_a = cmath.phase(I_2)*180/math.pi              #Phase angle of I_2(degree)
#Case(h)
Z1 = Z_1/1000                                     #Actual impedance of larger transformer(Ω)
E_1 = I_1*Z1+V_b                                  #No-load voltage of larger transformer(V)
E_1_m = abs(E_1)                                  #Magnitude of E_1(V)
E_1_a = cmath.phase(E_1)*180/math.pi              #Phase angle of E_1(degree)
#Case(i)
Z2 = Z_2/1000                                     #Actual impedance of smaller transformer(Ω)
E_2 = I_2*Z2+V_b                                  #No-load voltage of smaller transformer(V)
E_2_m = abs(E_2)                                  #Magnitude of E_2(V)
E_2_a = cmath.phase(E_2)*180/math.pi              #Phase angle of E_2(degree)

#Result
print('Case(a): Base voltage , V_b = %.f V' %V_b)
print('Case(b): Base current , I_b = %.2f kA' %(I_b/1000))
print('Case(c): Base impedance , Z_b = %.2f Ω (Method 1)' %Zb)
print('         Base impedance , Z_b = %.2f Ω (Method 2)' %Z_b)
print('Case(d): Actual impedance of larger transformer , Z_1 = %.2f∠%.1f° mΩ' %(Z_1_m,Z_1_a))
print('Case(e): Actual impedance of smaller transformer , Z_2 = %.2f∠%.2f° mΩ' %(Z_2_m,Z_2_a))
print('Case(f): Actual current delivered by larger transformer , I_1 = %.f∠%.2f° A' %(I_1_m,I_1_a))
print('Case(g): Actual current delivered by smaller transformer , I_2 = %.1f∠%.1f° A' %(I_2_m,I_2_a))
print('Case(h): No-load voltage of larger transformer , E_1 = %.2f∠%.2f° V' %(E_1_m,E_1_a))
print('Case(i): No-load voltage of smaller transformer , E_2 = %.2f∠%.2f° V' %(E_2_m,E_2_a))
Case(a): Base voltage , V_b = 400 V
Case(b): Base current , I_b = 1.25 kA
Case(c): Base impedance , Z_b = 0.32 Ω (Method 1)
         Base impedance , Z_b = 0.32 Ω (Method 2)
Case(d): Actual impedance of larger transformer , Z_1 = 16.32∠78.7° mΩ
Case(e): Actual impedance of smaller transformer , Z_2 = 27.34∠69.44° mΩ
Case(f): Actual current delivered by larger transformer , I_1 = 1178∠-40.32° A
Case(g): Actual current delivered by smaller transformer , I_2 = 702.9∠-31.1° A
Case(h): No-load voltage of larger transformer , E_1 = 415.24∠1.65° V
Case(i): No-load voltage of smaller transformer , E_2 = 415.24∠1.65° V

Example 14.27, Page number 525

In [1]:
import math
import cmath

#Variable declaration
P_L = 14400.0           #Load output power(W)
V_L = 120.0             #Load voltage(V)
V_b1 = 120.0            #Base voltage at point 1(V)
V_b2 = 600.0            #Base voltage at point 2(V)
V_b3 = 120.0            #Base voltage at point 3(V)
S_b3 = 14.4             #Base power(kVA)
X_2 = 0.25              #Reactance(p.u)
X_1 = 0.2               #Reactance(p.u)
I_L = 120.0             #Load current(A)

#Calculation
#Case(a)
R_L = P_L/(V_L**2)                          #Resistance of the load(ohm)
#Case(b)
Z_bL = (V_b3**2)/(S_b3*1000)                #Base impedance(ohm)
#Case(c)
Z_L_pu = R_L/Z_bL                           #Per unit load impedance 
#Case(d)
Z_2_pu = complex(0,X_2)                     #Per unit impedance of T2
#Case(e)
Z_1_pu = complex(0,X_1)                     #Per unit impedance of T1
#Case(g)
I_bL = (S_b3*1000)/V_b3                     #Base current in load(A)
#Case(h)
I_L_pu = I_L/I_bL                           #Per unit load current
#Case(i)
V_R_pu = I_L_pu*Z_L_pu                      #Per unit voltage across load 
#Case(j)
I_S_pu = I_L_pu                             #Per unit current of source
Z_T_pu = Z_L_pu+Z_1_pu+Z_2_pu               #Total p.u impedance
V_S_pu = I_S_pu*Z_T_pu                      #Per unit voltage of source
V_S_pu_m = abs(V_S_pu)                      #Magnitude of V_S_pu
V_S_pu_a = cmath.phase(V_S_pu)*180/math.pi  #Phase angle of V_S_pu(degrees)
#Case(k)
V_S = V_S_pu*V_b1                           #Actual voltage across source(V)
V_S_m = abs(V_S)                            #Magnitude of V_S(V)
V_S_a = cmath.phase(V_S)*180/math.pi        #Phase angle of V_S(degrees)
#Case(l)
I_x_pu = I_L_pu                             #p.u current at point x
Z_x_pu = Z_L_pu+Z_2_pu                      #p.u impedance at point x 
V_x_pu = I_x_pu*Z_x_pu                      #p.u voltage at point x
#Case(m)
V_x = V_x_pu*V_b2                           #Actual voltage at point x(V)
V_x_m = abs(V_x)                            #Magnitude of V_x(V)
V_x_a = cmath.phase(V_x)*180/math.pi        #Phase angle of V_x(degrees)


#Result
print('Case(a): Resistance of the load , R_L = %.f Ω' %R_L)
print('Case(b): Base impedance of the load , Z_bL = %.f Ω' %Z_bL)
print('Case(c): Per-unit load impedance , Z_L(pu)  = (%.f+j%.f) p.u' %(Z_L_pu.real,Z_L_pu.imag))
print('Case(d): Per-unit impedance of transformer T2 , Z_2(pu)  = j%.2f p.u' %Z_2_pu.imag)
print('Case(e): Per-unit impedance of transformer T1 , Z_1(pu)  = j%.1f p.u' %Z_2_pu.imag)
print('Case(f): See Fig.14-23b in textbook page no-526')
print('Case(g): Base current in the load ,  I_bL = %.f A (resistive)' %I_bL)
print('Case(h): Per-unit load current , I_L_pu = (%.f+j%.f) p.u' %(I_L_pu.real,I_L_pu.imag))
print('Case(i): Per-unit voltage across load , V_R_pu = (%.f+j%.f) p.u' %(V_R_pu.real,V_R_pu.imag))
print('Case(j): Per-unit voltage of source , V_S_pu = %.3f∠%.2f° pu' %(V_S_pu_m,V_S_pu_a))
print('Case(k): Actual voltage across source , V_S = %.1f∠%.2f° pu' %(V_S_m,V_S_a))
print('Case(l): Per-unit voltage at point x in the second transmission loop , V_x(pu) = (%.f+j%.2f) p.u' %(V_x_pu.real,V_x_pu.imag))
print('Case(m): Voltage at point x , V_x = %.1f∠%.f° V' %(V_x_m,V_x_a))
Case(a): Resistance of the load , R_L = 1 Ω
Case(b): Base impedance of the load , Z_bL = 1 Ω
Case(c): Per-unit load impedance , Z_L(pu)  = (1+j0) p.u
Case(d): Per-unit impedance of transformer T2 , Z_2(pu)  = j0.25 p.u
Case(e): Per-unit impedance of transformer T1 , Z_1(pu)  = j0.2 p.u
Case(f): See Fig.14-23b in textbook page no-526
Case(g): Base current in the load ,  I_bL = 120 A (resistive)
Case(h): Per-unit load current , I_L_pu = (1+j0) p.u
Case(i): Per-unit voltage across load , V_R_pu = (1+j0) p.u
Case(j): Per-unit voltage of source , V_S_pu = 1.097∠24.23° pu
Case(k): Actual voltage across source , V_S = 131.6∠24.23° pu
Case(l): Per-unit voltage at point x in the second transmission loop , V_x(pu) = (1+j0.25) p.u
Case(m): Voltage at point x , V_x = 618.5∠14° V

Example 14.28, Page number 526

In [1]:
import math
import cmath

#Variable declaration
V_1 = 11.0                #Tr.1 voltage(kV)
V_b1 = 11.0               #Base Tr.1 voltage(kV)
S_1 = 50.0                #KVA rating of power for Tr.1
S_2 = 100.0               #KVA rating of power for Tr.2
Z_1_pu = complex(0,0.1)   #Per unit impedance of Tr.1
Z_2_pu = complex(0,0.1)   #Per unit impedance of Tr.2
V_b2 = 55.0               #Base Tr.2 voltage(kV)
S_b = 100.0               #Base power(kVA)
PF = 0.8                  #Power factor of the load
Z_line = complex(0,200)   #Line impedance(ohm)
V_L = 10.0                #Load voltage(kV)
V_Lb3 = 11.0              #Base line voltage at point 3(V)
V_b3 = 11.0               #Line voltage at point 3(V)
P_L = 50.0                #Power rating of each transformer(kW)
cos_theta_L = 0.8         #Lagging PF of load

#Calculation
#Case(a)
Z_T1 = Z_1_pu*(V_1/V_b1)**2*(S_2/S_1)      #p.u impedance of T1
#Case(b)
Z_T2 = Z_2_pu*(V_1/V_b3)**2*(S_2/S_1)      #p.u impedance of T2
#Case(c)
V_b = V_b2                                 #Base voltage(V)
Z_b_line = (V_b**2)/S_b*1000               #Base line impedance(ohm)
Z_line_pu = Z_line/Z_b_line                #p.u impedance of the transmission line
#Case(d)
V_L_pu = V_L/V_Lb3                         #p.u voltage across load
#Case(f)
I_bL = S_b/V_b3                            #Base current in load(A)
#Case(g)
VL = V_b3                                  #Load voltage(kV)
I_L = P_L/(VL*cos_theta_L)                 #Load current(A)
I_L_pu = I_L/I_bL                          #p.u load current
theta = math.acos(cos_theta_L)             #PF angle(radian)
theta_a = theta*180/math.pi                #PF angle(degree)
I_Lpu = I_L_pu*complex(math.cos(theta),-math.sin(theta))   #p.u current in complex form
#Case(h)
Z_series_pu = Z_T1+Z_line_pu+Z_T2          #p.u series impedance of the transmission line
V_S_pu = I_Lpu*Z_series_pu+V_L_pu          #p.u source voltage
V_S_pu_m = abs(V_S_pu)                     #Magnitude of V_S_pu
V_S_pu_a = cmath.phase(V_S_pu)*180/math.pi #Phase angle of V_S_pu(degrees)
#Case(i)
V_S = V_S_pu*V_b1                        #Actual value of source voltage(kV)
V_S_m = abs(V_S)                         #Magnitude of V_S_pu(kV)
V_S_a = cmath.phase(V_S)*180/math.pi     #Phase angle of V_S(degrees)

#Result
print('Case(a): Per-unit impedance of transformer T1 , Z_T1 = j%.1f p.u' %Z_T1.imag)
print('Case(b): Per-unit impedance of transformer T2 , Z_T2 = j%.1f p.u' %Z_T2.imag)
print('Case(c): Per-unit impedance of transmission line , Z(line)_pu = j%.4f p.u' %Z_line_pu.imag)
print('Case(d): Per-unit voltage across load , V_L_pu = (%.3f+j%.f) p.u' %(V_L_pu.real,V_L_pu.imag))
print('Case(e): See Fig.14-24b in textbook Page no-527')
print('Case(f): Base current in load , I_bL = %.3f A' %I_bL)
print('Case(g): Per-unit load current , I_L_pu = (%.1f%.3fj) p.u' %(I_Lpu.real,I_Lpu.imag))
print('Case(h): Per-unit source voltage , V_S_pu = %.3f∠%.2f° pu' %(V_S_pu_m,V_S_pu_a))
print('Case(i): Actual value of source voltage , V_S = %.1f∠%.2f° kV' %(V_S_m,V_S_a))
Case(a): Per-unit impedance of transformer T1 , Z_T1 = j0.2 p.u
Case(b): Per-unit impedance of transformer T2 , Z_T2 = j0.2 p.u
Case(c): Per-unit impedance of transmission line , Z(line)_pu = j0.0066 p.u
Case(d): Per-unit voltage across load , V_L_pu = (0.909+j0) p.u
Case(e): See Fig.14-24b in textbook Page no-527
Case(f): Base current in load , I_bL = 9.091 A
Case(g): Per-unit load current , I_L_pu = (0.5-0.375j) p.u
Case(h): Per-unit source voltage , V_S_pu = 1.081∠10.84° pu
Case(i): Actual value of source voltage , V_S = 11.9∠10.84° kV

Example 14.29, Page number 528

In [1]:
#Variable declaration
Z_pu_1 = 1j*0.1      #p.u impedance 
MVA_2 = 80.0         #MVA rating os system 2
MVA_1 = 100.0        #MVA rating of T1 and T2
V_2 = 30.0           #Voltage(kV)
V_1 = 32.0           #Voltage(kV)
Z_pu_2 = 1j*0.15     #p.u impedance 
V_b1 = 100.0         #Base voltage of T1(kV)
Z_line = 1j*60       #Line impedance(ohm)
MVA_M1 = 20.0        #MVA rating of motor load 1
Z_pu_M1 = 1j*0.15    #p.u impedance of motor load M1
MVA_M2 = 35.0        #MVA rating of motor load 2
Z_pu_M2 = 1j*0.25    #p.u impedance of motor load M2
MVA_M3 = 25.0        #MVA rating of motor load 3
Z_pu_M3 = 1j*0.2     #p.u impedance of motor load M3
V_M = 28.0           #Voltage across motor loads M1,M2,M3(kV)

#Calculations
Z_1_pu = Z_pu_1*(MVA_2/MVA_1)*(V_2/V_1)**2     #p.u imepedance of T1
Z_2_pu = Z_pu_2*(MVA_2/MVA_1)*(V_2/V_1)**2     #p.u imepedance of T2
V_b_line = V_b1*(V_1/V_2)                      #ase voltage of the long-transmission line(kV)
MVA_b = 80.0                                   #Base MVA rating
V_b = V_b_line
Z_line_pu = Z_line*(MVA_b/(V_b_line)**2)       #p.u impedance of the transmission line
Z_M1_pu = Z_pu_M1*(MVA_2/MVA_M1)*(V_M/V_1)**2  #p.u impedance of motor load M1
Z_M2_pu = Z_pu_M2*(MVA_2/MVA_M2)*(V_M/V_1)**2  #p.u impedance of motor load M2
Z_M3_pu = Z_pu_M3*(MVA_2/MVA_M3)*(V_M/V_1)**2  #p.u impedance of motor load M3

#Result
print('Case(a): Per-unit impedance of T1 , Z_1(pu) = j%.4f p.u' %Z_1_pu.imag)
print('Case(b): Per-unit impedance of T2 , Z_2(pu) = j%.4f p.u' %Z_2_pu.imag)
print('Case(c): Base voltage of the long-transmission line between T1 and T2 , V_b(line) = %.1f kV' %V_b_line)
print('Case(d): Per-unit impedance of the transmission line , Z(line)_pu = j%.4f p.u' %Z_line_pu.imag)
print('Case(e): Per-unit impedance of motor load M1 , Z_M1(pu) = j%.4f p.u' %Z_M1_pu.imag)
print('Case(f): Per-unit impedance of motor load M2 , Z_M2(pu) = j%.4f p.u' %Z_M2_pu.imag)
print('Case(g): Per-unit impedance of motor load M3 , Z_M3(pu) = j%.2f p.u' %Z_M3_pu.imag)
print('Case(h): See Fig.14-25b in textbook page no-529')
Case(a): Per-unit impedance of T1 , Z_1(pu) = j0.0703 p.u
Case(b): Per-unit impedance of T2 , Z_2(pu) = j0.1055 p.u
Case(c): Base voltage of the long-transmission line between T1 and T2 , V_b(line) = 106.7 kV
Case(d): Per-unit impedance of the transmission line , Z(line)_pu = j0.4219 p.u
Case(e): Per-unit impedance of motor load M1 , Z_M1(pu) = j0.4594 p.u
Case(f): Per-unit impedance of motor load M2 , Z_M2(pu) = j0.4375 p.u
Case(g): Per-unit impedance of motor load M3 , Z_M3(pu) = j0.49 p.u
Case(h): See Fig.14-25b in textbook page no-529

Example 14.30, Page number 533

In [1]:
#Variable declaration
V_pa = 1000.0      #Phase voltage(V)
I_1a = 1.0         #Line current in primary(A)
V_2a = 100.0       #Voltage across secondary(V)
Ic_a = 10.0        #Current in lower half of auto-transformer(A)
V_s = 100.0        #Voltage in secondary winding(V)
I_2b = 10.0        #Current in secondary(A)
V_1b = 1000.0      #Voltage across primary(V)
Ic_b = 1.0         #Current in lower half of auto-transformer(A)

#Calculation
S_T1 = (V_pa*I_1a+V_2a*I_1a)/1000          #Total kVA transfer in step-down mode
S_T2 = (V_s*I_2b+V_1b*I_2b)/1000           #Total kVA transfer in step-up mode
S_x_former_c = V_pa*I_1a/1000              #kVA rating of th autotransformer in Fig.14-27a
V_1 = V_pa
S_x_former_d = V_1*Ic_b/1000               #kVA rating of th autotransformer in Fig.14-26b

#Result
print('Case(a): Total kVA transfer in step-down mode , S_T = %.1f kVA transferred' %S_T1)
print('Case(b): Total kVA transfer in step-up mode , S_T = %.1f kVA transferred' %S_T2)
print('Case(c): kVA rating of the autotransformer in Fig.14-27a , S_x-former = %.f kVA' %S_x_former_c)
print('Case(d): kVA rating of the autotransformer in Fig.14-26b , S_x-former = %.f kVA' %S_x_former_d)
print('Case(e): Both transformers have the same kVA rating of 1 kVA since the same autotransformer is used in both parts.')
print('         Both transformers transform a total of 1 KVA. But the step-down transformer in part(a) conducts only 0.1 kVA')
print('         while the step-up transformer in the part(b) conducts 10 kVA from the primary to the secondary')
Case(a): Total kVA transfer in step-down mode , S_T = 1.1 kVA transferred
Case(b): Total kVA transfer in step-up mode , S_T = 11.0 kVA transferred
Case(c): kVA rating of the autotransformer in Fig.14-27a , S_x-former = 1 kVA
Case(d): kVA rating of the autotransformer in Fig.14-26b , S_x-former = 1 kVA
Case(e): Both transformers have the same kVA rating of 1 kVA since the same autotransformer is used in both parts.
         Both transformers transform a total of 1 KVA. But the step-down transformer in part(a) conducts only 0.1 kVA
         while the step-up transformer in the part(b) conducts 10 kVA from the primary to the secondary

Example 14.31, Page number 535

In [1]:
#Variable declaration
S = 500.0         #kVA rating of the distribution transformer
V_1 = 2300.0      #Primary voltage(V)
V_2 = 208.0       #Secondary voltage(V)
f = 60.0          #Frequency(Hz)
P_sc = 8200.0     #Wattmeter reading in SC test(W)
I_sc = 217.4      #Short circuit current(A)
V_sc = 95.0       #Short circuit voltage(V)
P_oc = 1800.0     #Wattmeter reading in OC test(W)
I_oc = 85.0       #Open circuit current(A)
V_oc = 208.0      #Open circuit voltage(V)
LF1 = 0.2         #Load fraction 
LF2 = 0.4         #Load fraction 
LF3 = 0.8         #Load fraction 
LF4 = 1.25        #Load fraction 
PF1 = 0.7         #Power factor
PF2 = 0.8         #Power factor
PF3 = 0.9         #Power factor
PF_fl = 1.0       #Power factor
PF4 = 0.85        #Power factor
t1 = 4.0          #Period of operation(hours)
t2 = 4.0          #Period of operation(hours)
t3 = 6.0          #Period of operation(hours)
t_fl = 6.0        #Period of operation(hours)
t4 = 2.0          #Period of operation(hours)

#Calculation
#Case(a)
t = 24.0                                          #Hours in a day
P_c  = P_oc                                       #Wattmeter reading(W). OC test
W_c = (P_c*t)/1000                                #Core loss over 24 hour period(kWh)
#Case(b)
Psc = P_sc/1000                                   #Wattmeter reading(W). SC test
P_loss_1 = (LF1**2)*Psc                           #Power loss for 20% Load(kW)
P_loss_2 = (LF2**2)*Psc                           #Power loss for 40% Load(kW)
P_loss_3 = (LF3**2)*Psc                           #Power loss for 80% Load(kW)
P_loss_fl = Psc                                   #Power loss for 100% Load(kW)
P_loss_4 = (LF4**2)*Psc                           #Power loss for 125% Load(kW)
energy_loss1 = P_loss_1*t1                        #Enegry loss for 20% Load(kWh)
energy_loss2 = P_loss_2*t2                        #Enegry loss for 40% Load(kWh)
energy_loss3 = P_loss_3*t3                        #Enegry loss for 80% Load(kWh)
energy_loss_fl = P_loss_fl*t_fl                   #Enegry loss for 100% Load(kWh)
energy_loss4 = P_loss_4*t4                        #Enegry loss for 125% Load(kWh)
W_loss_total = energy_loss1+energy_loss2+energy_loss3+energy_loss_fl+energy_loss4 #Total energy loss(kWh)
#Case(c)
P_1 = LF1*S*PF1                                   #Power output for 20% load(kW)
P_2 = LF2*S*PF2                                   #Power output for 40% load(kW)
P_3 = LF3*S*PF3                                   #Power output for 80% load(kW)
P_fl = S*PF_fl                                    #Power output for 100% load(kW)
P_4 = LF4*S*PF4                                   #Power output for 125% load(kW)
Energy_1 = P_1*t1                                 #Energy delivered for 20% load(kWh)
Energy_2 = P_2*t2                                 #Energy delivered for 40% load(kWh)
Energy_3 = P_3*t3                                 #Energy delivered for 80% load(kWh)
Energy_fl = P_fl*t_fl                             #Energy delivered for 100% load(kWh)
Energy_4 = P_4*t4                                 #Energy delivered for 125% load(kWh)
W_out_total = Energy_1+Energy_2+Energy_3+Energy_fl+Energy_4    #Total energy delivered in 24hrs(kwh)
#Case(d)
n = W_out_total/(W_out_total+W_c+W_loss_total)*100 #All-day efficiency(%)

#Result
print('case(a): Core loss over the 24 hour period , W_c = P_c*t = %.1f kWh' %W_c)
print('Case(b): Total energy loss over the 24 hour period = %.2f kWh' %W_loss_total)
print('Case(c): Total energy output over the 24 hour period = %.f kWh' %W_out_total)
print('Case(d): All-day efficiency = %.1f percent' %n)
case(a): Core loss over the 24 hour period , W_c = P_c*t = 43.2 kWh
Case(b): Total energy loss over the 24 hour period = 112.87 kWh
Case(c): Total energy output over the 24 hour period = 7142 kWh
Case(d): All-day efficiency = 97.9 percent

Example 14.32, Page number 540

In [1]:
#Variable declaration
S_1 = 10.0        #VA rating of small transformer
V = 115.0         #Voltage rating of transformer(V)
V_2_1 = 6.3       #Voltage rating of one part of secondary winding(V)
V_2_2 = 5.0       #Voltage rating of other part of secondary winding(V)
Z_2_1 = 0.2       #Impedance of one part of secondary winding(ohm)
Z_2_2 = 0.15      #Impedance of other part of secondary winding(ohm)

#Calculation
#Case(a)
V_2 = V_2_1+V_2_2                  #Voltage across secondary winding(V)
I_2 = S_1/V_2                      #Rated secondary current when the LV secondaries are connected in series-aiding(A)
#Case(b)
I_c = (V_2_1-V_2_2)/(Z_2_1+Z_2_2)  #Circulating current when LV windings are paralled(A)
percent_overload = (I_c/I_2)*100   #Percent overload produced

#Result
print('Case(a): Rated secondary current when the low-voltage secondaries are connected series-aiding , I_2 = %.3f A' %I_2)
print('Case(b): Circulating current when LV windings are paralled , I_c = %.2f A' %I_c)
print('         Percent overload produced = %.f percent' %percent_overload)
Case(a): Rated secondary current when the low-voltage secondaries are connected series-aiding , I_2 = 0.885 A
Case(b): Circulating current when LV windings are paralled , I_c = 3.71 A
         Percent overload produced = 420 percent

Example 14.33, Page number 541

In [1]:
#Variable declaration
S = 20.0         #kVA rating of transformer
N_1 = 230.0      #Number of primary turns
N_2 = 20.0       #Number of secondary turns
V_1 = 230.0      #Primary voltage(V)
V_2 = 20.0       #Secondary voltage(V)
V_sc = 4.5       #Short circuit voltage(V)
I_sc = 87.0      #Short circuit current(A)
P_sc = 250.0     #Power measured(W)

#Calculation
V_h = V_sc                  #Short circuit voltage on HV side(V)
I_h = I_sc                  #Short circuit current on HV side(A)
Z_eh = V_h/I_h              #Equivalent impedance reffered to the high side when coils are series connected(ohm)
Z_el = Z_eh*(N_2/N_1)**2    #Equivalent impedance reffered to the low side when coils are series connected(ohm)
I_2_rated = (S*1000)/V_2    #Rated secondary current when coils are series connected(A)
I_2_sc = S/Z_el             #Secondary current when coils in Fig.14-31a are short-circuited with rated voltage applied to HV side(A)
percent_overload = (I_2_sc/I_2_rated)*100  #Percent overload

#Result
print('Case(a): Equivalent impedance reffered to the high side when coils are series-connected , Z_eh = %.4f ohm' %Z_eh)
print('Case(b): Equivalent impedance reffered to the low side when coils are series connected , Z_el = %.2e ohm' %Z_el)
print('Case(c): Rated secondary current when coils are series connected , I_2(rated) = %.e A' %I_2_rated)
print('Case(d): Secondary current when coils in Fig.14-31a are short-circuited with rated voltage applied to HV side , I_2(sc) = %.1e A' %I_2_sc)
print('         Percent overload produced = %.f percent' %percent_overload)
Case(a): Equivalent impedance reffered to the high side when coils are series-connected , Z_eh = 0.0517 ohm
Case(b): Equivalent impedance reffered to the low side when coils are series connected , Z_el = 3.91e-04 ohm
Case(c): Rated secondary current when coils are series connected , I_2(rated) = 1e+03 A
Case(d): Secondary current when coils in Fig.14-31a are short-circuited with rated voltage applied to HV side , I_2(sc) = 5.1e+04 A
         Percent overload produced = 5114 percent

Example 14.34, Page number 548

In [1]:
#Variable declaration
I_L = 100.0            #Load current(A)
cos_theta = 0.7        #Power factor lagging
S = 60.0               #kVA rating of transformer
V_1 = 2300.0           #Primary voltage(V)
V_2 = 230.0            #Secondary voltage(V)

#Calculation
#Case(a)
V_L = 230.0                                #Voltage across load(V)
P_T = (3**0.5*V_L*I_L*cos_theta)/1000      #Power consumed by the plant(kW)
kVA_T = P_T/cos_theta                      #Apparent power(kVA)
#Case(b)
kVA = S                                    #kVA rating of transformer
V_p = V_2                                  #Phase voltage(V). delta-connection on load side
I_P2_rated = (kVA*1000)/(3*V_p)            #Rated secondary phase current(A)
I_L2_rated = 3**0.5*I_P2_rated             #Rated secondary line current(A)
#Case(c)
percent_load = I_L/I_L2_rated*100          #Percent load on each transformer   
#Case(d)
V_L_d = 2300.0
I_P1 = (kVA_T*1000)/(3**0.5*V_L_d)         #Primary phase current(A)
I_L1 = I_P1                                #Primary line current(A). Y-connection
#Case(e)
kVA_transformer = kVA/3                    #kVA rating of each transformer

#Result
print('Case(a): Power consumed by the plant , P_T = %.1f kW' %P_T)
print('         Apparent power , kVA_T = %.1f kVA' %kVA_T)
print('Case(b): Rated secondary phase current of transformer bank , I_P2(rated) = %.f A' %I_P2_rated)
print('         Rated secondary line current of transformer bank , I_L2(rated) = %.1f A' %I_L2_rated)
print('Case(c): Percent load on each transformer = %.1f percent' %percent_load)
print('Case(d): Primary phase current drawn by each transformer , I_P1 = %.f A' %I_P1)
print('         Primary line current drawn by each transformer , I_L1 = %.f A' %I_L1)
print('Case(e): kVA rating of each transformer = %.f kVA' %kVA_transformer)
Case(a): Power consumed by the plant , P_T = 27.9 kW
         Apparent power , kVA_T = 39.8 kVA
Case(b): Rated secondary phase current of transformer bank , I_P2(rated) = 87 A
         Rated secondary line current of transformer bank , I_L2(rated) = 150.6 A
Case(c): Percent load on each transformer = 66.4 percent
Case(d): Primary phase current drawn by each transformer , I_P1 = 10 A
         Primary line current drawn by each transformer , I_L1 = 10 A
Case(e): kVA rating of each transformer = 20 kVA

Example 14.35, Page number 548

In [1]:
#Variable declaration
I_L = 100.0            #Load current(A)
cos_theta = 0.7        #Power factor lagging
S = 60.0               #kVA rating of transformer
V_1 = 2300.0           #Primary voltage(V)
V_2 = 230.0            #Secondary voltage(V)

#Calculation
#Case(a)
V_L = 230.0                                #Voltage across load(V)
P_T = (3**0.5*V_L*I_L*cos_theta)/1000      #Power consumed by the plant(kW)
kVA_T = P_T/cos_theta                      #Apparent power(kVA)
#Case(b)
kVA = S                                    #kVA rating of transformer
V_p = V_2                                  #Phase voltage(V)
I_P2_rated = (kVA*1000)/(3*V_p)            #Rated secondary phase current(A)
I_L2_rated = 3**0.5*I_P2_rated             #Rated secondary line current(A)
#Case(c)
percent_load = I_L/I_L2_rated*100          #Percent load on each transformer   
#Case(d)
V_L_d = 2300.0
I_P1 = (kVA_T*1000)/(3**0.5*V_L_d)         #Primary phase current(A)
I_L1 = 3**0.5*I_P1                         #Primary line current(A)
#Case(e)
kVA_transformer = kVA/3                    #kVA rating of each transformer

#Result
print('Case(a): Power consumed by the plant , P_T = %.1f kW' %P_T)
print('         Apparent power , kVA_T = %.1f kVA' %kVA_T)
print('Case(b): Rated secondary phase current of transformer bank , I_P2(rated) = %.f A' %I_P2_rated)
print('         Rated secondary line current of transformer bank , I_L2(rated) = %.1f A' %I_L2_rated)
print('Case(c): Percent load on each transformer = %.1f percent' %percent_load)
print('Case(d): Primary phase current drawn by each transformer , I_P1 = %.f A' %I_P1)
print('         Primary line current drawn by each transformer , I_L1 = %.1f A' %I_L1)
print('         Primary line current drawn by a Δ-Δ bank is √3 times the line current drawn by a Y-Δ bank')
print('Case(e): kVA rating of each transformer = %.f kVA' %kVA_transformer)
Case(a): Power consumed by the plant , P_T = 27.9 kW
         Apparent power , kVA_T = 39.8 kVA
Case(b): Rated secondary phase current of transformer bank , I_P2(rated) = 87 A
         Rated secondary line current of transformer bank , I_L2(rated) = 150.6 A
Case(c): Percent load on each transformer = 66.4 percent
Case(d): Primary phase current drawn by each transformer , I_P1 = 10 A
         Primary line current drawn by each transformer , I_L1 = 17.3 A
         Primary line current drawn by a Δ-Δ bank is √3 times the line current drawn by a Y-Δ bank
Case(e): kVA rating of each transformer = 20 kVA

Example 14.36, Page number 554

In [1]:
import math
import cmath

#Variable declaration
V_L = 33.0                #Line voltage(kV)
f = 60.0                  #Frequency(Hz)
PF1 = 1.0                 #Unity power factor for I_AB
PF2 = 0.7                 #Lagging power factor for I_BC
PF3 = 0.9                 #Leading power factor for I_CA
IAB = 10.0                #Magnitude of I_AB(kA)
IBC = 15.0                #Magnitude of I_BC(kA)
ICA = 12.0                #Magnitude of I_CA(kA)

#Calculation
V_AB = V_L*cmath.exp(1j*0*math.pi/180)                 #Line voltage taken as reference voltage(kV)
V_AB_a = cmath.phase(V_AB)*180/math.pi                 #Voltage angle(degree)
V_BC = V_L*cmath.exp(1j*-120*math.pi/180)              #Line voltage(kV)
V_BC_a = cmath.phase(V_BC)*180/math.pi                 #Voltage angle(degree)
V_CA = V_L*cmath.exp(1j*120*math.pi/180)               #Line voltage(kV)
V_CA_a = cmath.phase(V_CA)*180/math.pi                 #Voltage angle(degree)
theta_1 = math.acos(PF1)*180/math.pi                   #PF1 angle(degree)
theta_2 = math.acos(PF2)*180/math.pi                   #PF2 angle(degree)
theta_3 = math.acos(PF3)*180/math.pi                   #PF3 angle(degree)
I_AB = IAB*cmath.exp(1j*(theta_1)*math.pi/180)         #Current(kA)
I_BC = IBC*cmath.exp(1j*(V_BC_a-theta_2)*math.pi/180)  #Current(kA)
I_CA = ICA*cmath.exp(1j*(V_CA_a+theta_3)*math.pi/180)  #Current(kA)
#Case(a)
I_AC = -I_CA
I_A = I_AB+I_AC                         #Phase current(kA)
I_A_m = abs(I_A)                        #Magnitude of I_A(kA)
I_A_a = cmath.phase(I_A)*180/math.pi    #Phase angle of I_A(degrees)
#Case(b)
I_BA = -I_AB
I_B = I_BC+I_BA                         #Phase current(kA)
I_B_m = abs(I_B)                        #Magnitude of I_B(kA)
I_B_a = cmath.phase(I_B)*180/math.pi    #Phase angle of I_B(degrees)
#Case(c)
I_CB = -I_BC
I_C = I_CA+I_CB                         #Phase current(kA)
I_C_m = abs(I_C)                        #Magnitude of I_C(kA)
I_C_a = cmath.phase(I_C)*180/math.pi    #Phase angle of I_C(degrees)
#Case(d)
phasor_sum = I_A+I_B+I_C                #Phasor sum of line currents

#Result
print('Case(a): Current in line A , I_A = %.2f∠%.2f° kA' %(I_A_m,I_A_a))
print('Case(b): Current in line B , I_B = %.2f∠%.2f° kA' %(I_B_m,I_B_a))
print('Case(c): Current in line C , I_C = %.2f∠%.2f° kA' %(I_C_m,I_C_a))
print('Case(d): Phasor sum of line currents , ΣI_L = (%.f+j%.f) ' %(phasor_sum.real,phasor_sum.imag))
Case(a): Current in line A , I_A = 21.04∠-18.68° kA
Case(b): Current in line B , I_B = 24.81∠-171.34° kA
Case(c): Current in line C , I_C = 11.44∠66.30° kA
Case(d): Phasor sum of line currents , ΣI_L = (-0+j0) 

Example 14.37, Page number 556

In [1]:
#Variable declaration
kVA_1 = 20.0     #kVA rating of transformer 1
kVA_2 = 20.0     #kVA rating of transformer 2
kVA_3 = 20.0     #kVA rating of transformer 3
V_1 = 2300.0     #Primary voltage(V)
V_2 = 230.0      #Secondary voltage(V)
kVA = 40.0       #kVA supplied by the bank
PF = 0.7         #Lagging power factor at which bank supplies kVA

#Calculation
kVA_transformer = kVA/3**0.5                          #kVA load carried by each transformer
percent_ratedload_Tr = kVA_transformer/kVA_1*100      #Percent load carried by each transformer
kVA_V_V = 3**0.5*kVA_1                                #Total kVA rating of the transformer bank in V-V
ratio_banks = kVA_V_V/(kVA_1+kVA_2+kVA_3)*100         #Ratio of V-V bank to Δ-Δ bank transformer ratings
kVA_Tr = kVA/3 
percent_increase_load = kVA_transformer/kVA_Tr*100    #Percent increase in load on each transformer when one transformer is removed

#Result
print('Case(a): kVA load carried by each transformer = %.1f kVA/transformer' %kVA_transformer)
print('Case(b): Percent rated load carried by each transformer = %.1f percent' %percent_ratedload_Tr)
print('Case(c): Total kVA rating of the transformer bank in V-V = %.2f kVA' %kVA_V_V)
print('Case(d): Ratio of V-V bank to Δ-Δ bank transformer ratings = %.1f percent' %ratio_banks)
print('Case(e): Percent increase in load on each transformer when one transformer is removed = %.1f percent' %percent_increase_load)
Case(a): kVA load carried by each transformer = 23.1 kVA/transformer
Case(b): Percent rated load carried by each transformer = 115.5 percent
Case(c): Total kVA rating of the transformer bank in V-V = 34.64 kVA
Case(d): Ratio of V-V bank to Δ-Δ bank transformer ratings = 57.7 percent
Case(e): Percent increase in load on each transformer when one transformer is removed = 173.2 percent

Example 14.38, Page number 562

In [1]:
#Variable declaration
V = 440.0           #Rated voltage of SCIM(V)
hp = 100.0          #Rated power of SCIM(hp)
PF = 0.8            #Power factor
V_1 = 155.0         #Primary voltage of transformer(V)
V_2 = 110.0         #Secondary voltage of transformer(V)
V_a = 110.0         #Armature voltage(V)
V_L = 440.0         #Load voltage(V)
n = 0.98            #Efficiency of the transformer

#Calculation
I_L = 124*1.25                          #Motor line current(A). Appendix A-3
alpha = V_a/V_L                         #Transformation ratio
I_a = (3**0.5/2)*(I_L/(alpha*n))        #Current in the primary of the scott transformers(A)
kVA = (V_a*I_a)/((3**0.5/2)*1000)       #kVA rating of the main and teaser transformers

#Result
print('Case(a): Motor line current , I_L = %.f A' %I_L)
print('Case(b): Transformation ratio , α = N_1/N_2 = V_a/V_L = %.2f ' %alpha)
print('Case(c): Current in the primary of the scott transformers , I_a = %.f A' %I_a)
print('Case(d): kVA rating of the main and teaser transformers , kVA = %.1f kVA' %kVA)
Case(a): Motor line current , I_L = 155 A
Case(b): Transformation ratio , α = N_1/N_2 = V_a/V_L = 0.25 
Case(c): Current in the primary of the scott transformers , I_a = 548 A
Case(d): kVA rating of the main and teaser transformers , kVA = 69.6 kVA

Example 14.39, Page number 570

In [1]:
#Variable declaration
I_L = 1.0       #Load current(kA)
V_m = 750.0     #Peak voltage(kV)

#Calculation
V_L = (V_m)/2**0.5           #Maximum allowable rms voltagethat may be applied to the lines using ac(kV)
S_T_ac = 3**0.5*V_L*I_L      #Total 3-phase apparent power(MVA)
I_rms = I_L                  #rms value of load current(kA)
I_dc =I_rms*2**0.5           #Maximum allowable current that can be delivered by dc transmission(kA)
V_dc = V_m                   #dc voltage(kV)
S_T_dc = V_dc*I_dc           #Total dc apparent power delivered by two lines(MVA)
S_ac_line = S_T_ac/3         #Power per ac line
S_dc_line = S_T_dc/2         #Power per dc line

#Result
print('Case(a): Maximum allowable rms voltage that may be applied to the lines using ac , V_L = %.1f kV' %V_L)
print('Case(b): Total 3-phase ac apparent power delivered by three lines , S_T = %.1f MVA' %S_T_ac)
print('Case(c): Maximum allowable current that can be delivered by dc transmission , I_dc = %.3f kA' %I_dc)
print('Case(d): Total dc apparent power delivered by two lines , S_T = %.1f MVA' %S_T_dc)
print('Case(e): Power per ac line , S/ac line = %.1f MVA/line' %S_ac_line)
print('Case(f): Power per dc line , S/dc line = %.1f MVA/line' %S_dc_line)
Case(a): Maximum allowable rms voltage that may be applied to the lines using ac , V_L = 530.3 kV
Case(b): Total 3-phase ac apparent power delivered by three lines , S_T = 918.6 MVA
Case(c): Maximum allowable current that can be delivered by dc transmission , I_dc = 1.414 kA
Case(d): Total dc apparent power delivered by two lines , S_T = 1060.7 MVA
Case(e): Power per ac line , S/ac line = 306.2 MVA/line
Case(f): Power per dc line , S/dc line = 530.3 MVA/line