# Chapter 14 : Protective Relayes¶

## Example 14.1, Page No 366¶

In :
import math
#initialisation of variables
If=4000.0	# fault current
I=5*1.25	# operating current of relay

#Calculations
CT=400.0/5	# CT ratio
PSM=If/(I*CT)	# plug setting multiplier

#Results
print("PSM =%.2f" %PSM)
print("operating time for PSM=8 is 3.2sec.")
print("actual operating time = 1.92 sec.")

PSM =8.00
operating time for PSM=8 is 3.2sec.
actual operating time = 1.92 sec.


## Example 14.2, Page No 369¶

In :
import math
#initialisation of variables
Xc=Xl-X

#Calculations
C=1000000.0/(314.0*Xc)

#Results
#Answers don't match due to difference in rounding off of digits
print("X= %.2f" %X)
print("Xc= %.2f" %Xc)

X= 321.39
Xc= 178.61


## Example 14.3, Page No 384¶

In :
import math
#initialisation of variables
Isec1=4000.0/40# secondary current(amps)
PSM=100.0/5# PSM if 100% setting is used
Isec2=4000.0/40
PSM2=100.0/6.25#PSM if setting used is 125%
TMSb=0.72/2.5

#Calculations
PSM1=5000.0/(6.25*40)
to=2.2
tb=to*TMSb
PSMa=5000/(6.25*80)
TMS=1.138/3
PSMa1=6000/(6.25*80)
ta=(2.6*.379)

#Results
print("Actual operating time of realy at b=%.3f sec" %tb)
print("Actual operating time of realy at a=%.3f sec " %ta)

Actual operating time of realy at b=0.634 sec
Actual operating time of realy at a=0.985 sec


## Example 14.4 Page No 399¶

In :
import math
#initialisation of variables
Vph=6600/(math.sqrt(3))
Ifull=5000/(math.sqrt(3)*6.6)

#Calculations
Ib=Ifull*.25
x=Ib*800.0/Vph

#Results
print("Percent of the winding remains unprotected = %.2f " %x)

Percent of the winding remains unprotected = 22.96


## Example 14.5, Page No 399¶

In :
import math
#initialisation of variables
Iph=10000.0/math.sqrt(3)	# phase voltage of alternator(V)
x=1.8*100*10*1000.0/(5*Iph)

#Calculations
print("(i) percent winding which remains unprotected=%.2f " %x)
Ip=Iph*.2
R=1.8*1000.0/(5*Ip)

#Results
print("(ii)minimum  value of earthing resistance required to protect 80 percent of winding =%.4f ohms" %R)

(i) percent winding which remains unprotected=62.35
(ii)minimum  value of earthing resistance required to protect 80 percent of winding =0.3118 ohms


## Example 14.6, Page No 400¶

In :
import math
#initialisation of variables
Ic=360-320    # the difference current (amp)
Io=40*5/400.0

#Calculations
Avg=(360+320)/2    # average sum of two currents
Iavg=340*5/400.0
Ioc=.1*Iavg +0.2

#Results
print("operating current=%.3f amp. " %Ioc)
print("since current through  operating coil is %.3f amp.  " %Io)
print("therefore Relay will not operate ")

operating current=0.625 amp.
since current through  operating coil is 0.500 amp.
therefore Relay will not operate


## Example 14.7 Page No 403¶

In :
import math
#initialisation of variables
Il=400*6.6/33.0		# line current on star side of PT(amps)

#Calculations
Ic=5/math.sqrt(3.0)		# current in CT secondary

#Results
print(" The CT ratio on HT will be %d : %.3f" %(Il,Ic))

 The CT ratio on HT will be 80 : 2.887


## Example 14.8, Page No 404¶

In :
import math
#initialisation of variables
Il=10000.0/((math.sqrt(3.0))*132)
ILV=10000/((math.sqrt(3.0))*6.6)

#Calculations
a=5.0/math.sqrt(3.0)

#Results
print("Ratio of CT on LV side is %.3f : %.3f" %(ILV,a))
print("Ratio of CT on HT side is %.3f : %d" %(Il,5))

Ratio of CT on LV side is 874.773 : 2.887
Ratio of CT on HT side is 43.739 : 5


## Example 14.9 Page No 404¶

In :
import math
#initialisation of variables
Vs=110.0
I=1.0

#Calculations
R2=Vs/(complex(3, -math.sqrt(3))*I)
c=abs(R2)
print("R2=%.2f ohms" %c)
R1=2*c
d=abs(R1)
C=(10**6)/(0.866*d*314)
print("R1=%.2f ohms " %R1)
Vt=d*complex(-0.5,-0.866) + complex(c,-55 )

#Results
print(" Voltage across the terminals of the relay will be (V)= {0:.5f}+{1:.5f}i" .format(Vt.real, Vt.imag))

R2=31.75 ohms
R1=63.51 ohms
Voltage across the terminals of the relay will be (V)= 0.00000+-109.99839i


## Example 14.10 Page No 272¶

In :
import math
#initialisation of variables
Ic=5*0.25		# operating current(amp)
Vsec=5.0/1.25	# secondary voltage(V)
Bm=1.4
f=50
N=50

#Calculations
V=15*Vsec
A=60/(4.44*Bm*f*N)

#Results
print("The knee point must be slightly higher than =%.3f V " %V)
print("Area of cross section=%.6f m_2 " %A)

The knee point must be slightly higher than =60.000 V
Area of cross section=0.003861 m_2


## Example 14.11 Page No 273¶

In :
import math
#initialisation of variables

#Calculations
o_p=5*5*(.1+.1) +5

#Results
print(" VA output of CT =%.0f VA\n " %o_p)

 VA output of CT =10 VA