# Chapter 15 : Circuit Breakers¶

## Example 15.1, Page No 486¶

In [1]:
import math
#initialisation of variables
i=5.0
L=5*(10**6)
C=0.01

#Calculations
e=i*math.sqrt(L/C)

#Results
print("The voltage appearing across the pole of C.B.=%.2f V " %e)
R=0.5*math.sqrt(L/C)
print("The value of resistance to be used across contacts, R=%.2f ohms" %R)

The voltage appearing across the pole of C.B.=111803.40 V
The value of resistance to be used across contacts, R=11180.34 ohms


## Example 15.2, Page No 487¶

In [2]:
import math
#initialisation of variables
Vnl=132*math.sqrt(2)/math.sqrt(3)    #peak value of peak to neutral voltage(kV)
Vr1=Vnl*.95  #recovery voltage (kV)

#Calculations
Vr=102.4*.916   # active recovery voltage(kV)
Vrmax=2.0*Vr
fn=16.0*(10**3)
t=1.0/(2*fn)
RRRV=Vrmax*(10**-6)/t

#Results
print("Rate of rise of restriking voltage, RRRV = %.0f kV/micro-sec " %RRRV)

Rate of rise of restriking voltage, RRRV = 6 kV/micro-sec


## Example 15.3, Page No 487¶

In [3]:
import math
#initialisation of variables
Vm=132*math.sqrt(2)/math.sqrt(3)
K1=0.9
K2=1.5
K=K1*K2
sinq=0.92

#Calculations
Vr=K*Vm*sinq
fn=16*(10**3)
RRRV=2*Vr*(10**-6)*fn*2

#Results
print("Average rate of rise of restriking voltage,RRRV=%.3f kV/micro-sec" %RRRV)

Average rate of rise of restriking voltage,RRRV=8.567 kV/micro-sec


## Example 15.4 Page No 504¶

In [4]:
import math
#initialisation of variables
In=1500.0

#Calculations
Ib=2000/(math.sqrt(3.0)*33.0)
Im=2.55*Ib
Is=Ib

#Results
print("rated normal current=%.0f amps" %In)
print("Breaking current=%.2f KA" %Ib)
print("Making current =%.2f kA" %Im)
print("Short time rating  for 3 sec=%.2f kA " %Is)

rated normal current=1500 amps
Breaking current=34.99 KA
Making current =89.23 kA
Short time rating  for 3 sec=34.99 kA


## Example 15.5, Page No 504¶

In [5]:
import math
#initialisation of variables
MVA=10.0
Is=MVA*1000/(math.sqrt(3)*13.8)
print("(i)sustained short circuit current in the breaker =%.0f amps" %Is)
MVA1=100.0
Isc=MVA1*1000.0/(math.sqrt(3)*13.8)

#Calculations
print("(ii)initial symmetrical r.m.s current in the breaker r.m.s=%.0f amps" %Isc)
Im=math.sqrt(2)*Isc
print("(iii)maximum possible d.c component of the short circuit current in the breaker =%.0f amps" %Im)
Im2=1.6*Isc
print("(iv)momentary current rating of the breaker=%.0f amps" %Im2)
Ib=1.2*Isc
print("(v)the current to be interrupted by the breaker =%.0f amps" %Ib)
KVA=math.sqrt(3)*13.8*5016

#Results
print("(vi)the interupting =%.0f KVA"%KVA)
#Answers don't match due to difference in rounding off of digits

(i)sustained short circuit current in the breaker =418 amps
(ii)initial symmetrical r.m.s current in the breaker r.m.s=4184 amps
(iii)maximum possible d.c component of the short circuit current in the breaker =5917 amps
(iv)momentary current rating of the breaker=6694 amps
(v)the current to be interrupted by the breaker =5020 amps
(vi)the interupting =119894 KVA