# CHAPTER 3.6: CIRCUIT BREAKER¶

## Example 3.6.1, Page number 545¶

In :
import math

#Variable declaration
f = 50.0          #Generator frequency(Hz)
kV = 7.5          #emf to neutral rms voltage(kV)
X = 4.0           #Reactance of generator & connected system(ohm)
C = 0.01*10**-6   #Distributed capacitance(F)

#Calculation
#Case(a)
v = 2**0.5*kV                           #Active recovery voltage i.e phase to neutral(kV)
V_max_restrike = v*2                    #Maximum restriking voltage i.e phase to neutral(kV)
#Case(b)
L = X/(2.0*math.pi*f)                   #Inductance(H)
f_n = 1/(2.0*math.pi*(L*C)**0.5*1000)   #Frequency of transient oscillation(kHZ)
#Case(c)
t = 1.0/(2.0*f_n*1000)                  #Time(sec)
avg_rate = V_max_restrike/t             #Average rate of rise of voltage upto first peak of oscillation(kV/s)

#Result
print('Case(a): Maximum re-striking voltage(phase-to-neutral) = %.1f kV' %V_max_restrike)
print('Case(b): Frequency of transient oscillation, f_n = %.1f kHz' %f_n)
print('Case(c): Average rate of rise of voltage upto first peak of oscillation = %.f kV/s' %avg_rate)
print('\nNOTE: Changes in the obtained answer from that of textbook is due to more approximation in the textbook')

Case(a): Maximum re-striking voltage(phase-to-neutral) = 21.2 kV
Case(b): Frequency of transient oscillation, f_n = 14.1 kHz
Case(c): Average rate of rise of voltage upto first peak of oscillation = 598413 kV/s

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


## Example 3.6.3, Page number 545-546¶

In :
import math

#Variable declaration
kV = 132.0       #Voltage(kV)
pf = 0.3         #Power factor of the fault
K3 = 0.95        #Recovery voltage was 0.95 of full line value
f_n = 16000.0    #Natural frequency of the restriking transient(Hz)

#Calculation
kV_phase = kV/3**0.5                    #System voltage(kV)
sin_phi = math.sin(math.acos(pf))       #Sinφ
K2 = 1.0
v = K2*K3*kV/3**0.5*2**0.5*sin_phi      #Active recovery voltage(kV)
V_max_restrike = 2*v                    #Maximum restriking voltage(kV)
t = 1.0/(2.0*f_n)                       #Time(sec)
RRRV = V_max_restrike/(t*10**6)         #Rate of rise of restriking voltage(kV/µ-sec)

#Result
print('Rate of rise of restriking voltage, R.R.R.V = %.2f kV/µ-sec' %RRRV)

Rate of rise of restriking voltage, R.R.R.V = 6.25 kV/µ-sec


## Example 3.6.5, Page number 565¶

In :
#Variable declaration
kV = 132.0        #Voltage(kV)
C = 0.01*10**-6   #Phase to ground capacitance(F)
L = 6.0           #Inductance(H)
i = 5.0           #Magnetizing current(A)

#Calculation
V_pros = i*(L/C)**0.5/1000    #Prospective value of voltage(kV)
R = 1.0/2*(L/C)**0.5/1000     #Resistance to be used across the contacts to eliminate the restriking voltage(k-ohm)

#Result
print('Voltage across the pole of a CB = %.1f kV' %V_pros)
print('Resistance to be used across the contacts to eliminate the restriking voltage, R = %.2f k-ohm' %R)
print('\nNOTE: ERROR: Unit of final answer R is k-ohm, not ohm as in the textbook solution')

Voltage across the pole of a CB = 122.5 kV
Resistance to be used across the contacts to eliminate the restriking voltage, R = 12.25 k-ohm

NOTE: ERROR: Unit of final answer R is k-ohm, not ohm as in the textbook solution


## Example 3.6.6, Page number 567¶

In :
#Variable declaration
I = 1200.0     #Rated normal current(A)
MVA = 1500.0   #Rated MVA
kV = 33.0      #Voltage(kV)

#Calculation
I_breaking = MVA/(3**0.5*kV)   #Rated symmetrical breaking current(kA)
I_making = I_breaking*2.55     #Rated making current(kA)
I_short = I_breaking           #Short-time rating(kA)

#Result
print('Rated normal current = %.f A' %I)
print('Breaking current = %.2f kA' %I_breaking)
print('Making current = %.f kA' %I_making)
print('Short-time rating = %.2f kA for 3 secs' %I_short)

Rated normal current = 1200 A
Breaking current = 26.24 kA
Making current = 67 kA
Short-time rating = 26.24 kA for 3 secs


## Example 3.6.8, Page number 569¶

In :
#Variable declaration
kVA = 7500.0    #Rated kVA
X_st = 9.0      #Sub-transient reactance(%)
X_t = 15.0      #Transient reactance(%)
X_d = 100.0     #Direct-axis reactance(%)
kV = 13.8       #Voltage(kV). Assumption

#Calculation
kVA_base = 7500.0                       #Base kVA
kVA_sc_sustained = kVA_base/X_d*100     #Sustained S.C kVA
I_sc_sustained = kVA_base/(3**0.5*kV)   #Sustained S.C current(A). rms
I_st = kVA*100/(X_st*3**0.5*kV)         #Initial symmetrical rms current in the breaker(A)
I_max_dc = 2**0.5*I_st                  #Maximum possible dc component of the short-circuit(A)
I_moment = 1.6*I_st                     #Momentary current rating of the breaker(A)
I_interrupt = 1.1*I_st                  #Current to be interrupted by the breaker(A)
I_kVA = 3**0.5*I_interrupt*kV           #Interrupting kVA

#Result
print('Case(a): Sustained short circuit in the breaker = %.1f A (rms)' %I_sc_sustained)
print('Case(b): Initial symmetrical rms current in the breaker = %.f A' %I_st)
print('Case(c): Maximum possible dc component of the short-circuit in the breaker = %.f A' %I_max_dc)
print('Case(d): Momentary current rating of the breaker = %.f A (rms)' %I_moment)
print('Case(e): Current to be interrupted by the breaker = %.f A (rms)' %I_interrupt)
print('Case(f): Interrupting kVA = %.f kVA' %I_kVA)

Case(a): Sustained short circuit in the breaker = 313.8 A (rms)
Case(b): Initial symmetrical rms current in the breaker = 3486 A
Case(c): Maximum possible dc component of the short-circuit in the breaker = 4931 A
Case(d): Momentary current rating of the breaker = 5578 A (rms)
Case(e): Current to be interrupted by the breaker = 3835 A (rms)
Case(f): Interrupting kVA = 91667 kVA