# CHAPTER 3.8: PROTECTION OF ALTERNATORS AND AC MOTORS¶

## Example 3.8.1, Page number 624¶

In :
#Variable declaration
V = 6600.0           #Alternator Voltage(V)
P = 2000.0*10**3     #Rating of alternator(W)
PF = 0.8             #Power factor of alternator
X = 12.5             #Alternator reactance(%)
I = 200.0            #Current protection(A)
per = 10.0           #Percentage of winding unprotected(%)

#Calculation
I_fl = P/(3**0.5*V*PF)          #Full load current of alternator(A)
x = X*V/(3**0.5*100*I_fl)       #Reactance per phase of alternator(ohm)
x_per = per/100*x               #Reactance of 10% of the winding(ohm)
NA = V/(3**0.5*per)             #Voltage induced in winding(V)
r = ((NA/I)**2-x_per**2)**0.5   #Neutral earthing reactance(ohm)

#Result
print('Neutral earthing reactance, r = %.2f ohm' %r)

Neutral earthing reactance, r = 1.89 ohm


## Example 3.8.2, Page number 624-625¶

In :
#Variable declaration
MVA = 20.0            #Generator rating(MVA)
V = 11.0*10**3        #Generator voltage(V)
ratio_CT = 1200.0/5   #Ratio of current transformer
I_min_op = 0.75       #Minimum operating current of relay(A)
R = 6.0               #Neutral point earthing resistance(ohm)

#Calculation
I_max_fault = ratio_CT*I_min_op       #Maximum fault current to operate relay(A)
x = I_max_fault*3**0.5*100*R/V        #Unprotected portion for R = 6 ohm(%)
R_1 = 3.0                             #Neutral point earthing resistance(ohm)
x_1 = I_max_fault*3**0.5*100*R_1/V    #Unprotected portion for R = 3 ohm(%)
R_3 = 12.0                            #Neutral point earthing resistance(ohm)
x_3 = I_max_fault*3**0.5*100*R_3/V    #Unprotected portion for R = 12 ohm(%)

#Result
print('Unprotected portion of each phase of the stator winding against earth fault, x = %.f percent' %x)
print('Effect of varying neutral earthing resistance keeping relay operating current the same')
print(' (i)   R = 3 ohms')
print('       Unprotected portion = %.1f percent' %x_1)
print('       Protected portion = %.1f percent' %(100-x_1))
print(' (ii)  R = 6 ohms')
print('       Unprotected portion = %.f percent' %x)
print('       Protected portion = %.f percent' %(100-x))
print(' (iii) R = 12 ohms')
print('       Unprotected portion = %.f percent' %x_3)
print('       Protected portion = %.f percent' %(100-x_3))

Unprotected portion of each phase of the stator winding against earth fault, x = 17 percent
Effect of varying neutral earthing resistance keeping relay operating current the same
(i)   R = 3 ohms
Unprotected portion = 8.5 percent
Protected portion = 91.5 percent
(ii)  R = 6 ohms
Unprotected portion = 17 percent
Protected portion = 83 percent
(iii) R = 12 ohms
Unprotected portion = 34 percent
Protected portion = 66 percent


## Example 3.8.3, Page number 625¶

In :
#Variable declaration
kVA = 5000.0      #Alternator rating(kVA)
V = 6600.0        #Alternator voltage(V)
X = 2.0           #Synchronous reactance per phase(ohm)
R = 0.5           #Resistance(ohm)
ofb = 30.0        #Out-of-balance current(%)
R_n = 6.5         #Resistance of resistor earthed to star point(ohm)

#Calculation
I_fl = kVA*1000/(3**0.5*V)                 #Full load current(A)
I_ofb = ofb/100*I_fl                       #Out-of-balance current(A)
x = R_n/((V/(3**0.5*100*I_ofb))-(R/100))   #Portion of winding unprotected(%)

#Result
print('Portion of alternator winding unprotected, x = %.1f percent' %x)

Portion of alternator winding unprotected, x = 22.8 percent


## Example 3.8.4, Page number 625¶

In :
#Variable declaration
I_min = 0.15          #Minimum pick up current of relay(A)
slope = 12.0          #Slope(%)
CT_ratio = 400.0/5    #CT ratio
I_1 = 360.0           #Current(A)
I_2 = 300.0           #Current(A)

#Calculation
i_1 = I_1/CT_ratio                         #Current(A)
i_2 = I_2/CT_ratio                         #Current(A)
percentage = (i_1-i_2)/((i_1+i_2)/2)*100   #Percentage(%)

#Result
if(percentage>slope):
print('Relay would trip the circuit breaker, since the point lie on +ve torque regime')
else:
print('Relay would not trip the circuit breaker, since the point do not lie on +ve torque regime')

Relay would trip the circuit breaker, since the point lie on +ve torque regime


## Example 3.8.5, Page number 625-626¶

In :
#Variable declaration
MVA = 50.0           #Alternator rating(MVA)
V = 33.0*10**3       #Alternator voltage(V)
CT_ratio = 2000.0/5  #CT ratio
R = 7.5              #Resistor earthed generator neutral(ohm)
I = 0.5              #Current above which pick up current(A)

#Calculation
I_min = CT_ratio*I           #Minimum current required to operate relay(A)
x = I_min*R/(V/3**0.5)*100   #Winding unprotected during normal operation(%)

#Result
print('Winding of each phase unprotected against earth when machine operates at nominal voltage, x = %.2f percent' %x)

Winding of each phase unprotected against earth when machine operates at nominal voltage, x = 7.87 percent


## Example 3.8.6, Page number 626¶

In :
#Variable declaration
MVA = 50.0        #Alternator rating(MVA)
kV = 11.0         #Alternator voltage(kV)
X = 2.0           #Synchronous reactance per phase(ohm)
R = 0.7           #Resistance per phase(ohm)
R_n = 5.0         #Resistance through which alternator is earthed(ohm)
ofb = 25.0        #Out-of-balance current(%)

#Calculation
I_fl = MVA*1000/(3**0.5*kV)                      #Full load current(A)
I_ofb = ofb/100*I_fl                             #Out-of-balance current(A)
x = R_n/((kV*1000/(3**0.5*100*I_ofb))-(R/100))   #Portion of winding unprotected(%)

#Result
print('Portion of winding unprotected, x = %.f percent' %x)

Portion of winding unprotected, x = 56 percent


## Example 3.8.7, Page number 626-627¶

In :
#Variable declaration
MVA = 5.0         #Alternator rating(MVA)
kV = 11.0         #Alternator voltage(kV)
X = 2.0           #Reactance per phase(ohm)
ofb = 35.0        #Out-of-balance current(%)
R_n = 5.0         #Resistance through which star point is earthed(ohm)

#Calculation
I_fl = MVA*1000/(3**0.5*kV)              #Full load current(A)
I_ofb = ofb/100*I_fl                     #Out-of-balance current(A)
x = I_ofb*R_n*100/(kV*1000/3**0.5)       #Portion of winding unprotected(%)
protected = 100.0-x                      #Winding that is protected against earth faults(%)

#Result
print('Percentage of winding that is protected against earth faults = %.2f percent' %protected)

Percentage of winding that is protected against earth faults = 92.77 percent


## Example 3.8.8, Page number 627¶

In :
#Variable declaration
kV = 11.0         #Alternator voltage(kV)
P = 100.0         #Alternator maximum rating(MW)
PF = 0.8          #Power factor
X = 0.1           #Reactance of alternator(pu)
i = 500.0         #Current(A)
per = 10.0        #Windings unprotected(%)

#Calculation
I = P*1000/(3**0.5*kV*PF)      #Rated current of alternator(A)
a = i/I                        #Relay setting
I_n = a*I*100/per              #Current through neutral(A)
R = kV*1000/(3**0.5*I_n)       #Magnitude of neutral earthing resistance(ohm)

#Result
print('Magnitude of neutral earthing resistance, R = %.2f ohm' %R)
print('\nNOTE: ERROR: Unit of resistance is not mentioned in textbook solution')

Magnitude of neutral earthing resistance, R = 1.27 ohm

NOTE: ERROR: Unit of resistance is not mentioned in textbook solution