# Chapter 22: Protection of Alternators and Transformers¶

### Example 22.1, Page Number: 529¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
MVA = 10                         #MVA rating of alternator
V = 6.6                          #voltage rating of alternator(V)
X = 10                           #per phase reactance of lternator(%)
Iop = 175                         #operating current(A)

#Calculation:
Vph = V*1000/3**0.5                   #phase voltage(kV)
I = round(MVA*10**6/(3**0.5*V*1000))      #full load current(A)

#Let the reactance per phase be x ohms.
r,x = symbols('r x')           #r = earthing resistance required to leave 10% of
#the winding unprotected
x1 = solve(3**0.5*x*I/(6.6*1000)*100-10,x)
X1 = x1*0.1                     #Reactance of 10% winding
E = Vph*0.1                    #E.M.F. induced in 10% winding
Zf = (X1**2+r**2)**0.5
Ief = E/Zf                     #Earth-fault current due to 10% winding

#When this fault current becomes 175 A, the relay will trip
r1 = solve(Ief-175,r)               #A

#Result:
print "Required value of earth resistance is",round(r1,3),"ohm"

Required value of earth resistance is 2.177 ohm


### Example 22.2, Page Number: 530¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
MVA = 10                         #MVA rating of alternator
V = 6.6                          #voltage rating of alternator(V)
CR = 1000/5                      #current ratio of CT
Rn = 7.5                         #resistance of star-point to earth(ohm)
Iop = 0.5                        #operating current of the relay(A)

#Calculation:
#Let x % of the winding be unprotected.
x = symbols('x')
Vph = V*1000/3**0.5                   #phase voltage(kV)
If = 1000/5*Iop              #minimum fault current which will operate the relay(A)
E = Vph*x/100                 #E.M.F. induced in x% winding(V)
Ief = E/Rn                      #Earth fault current which x% winding will cause(A)
#This current must be equal to 100 A.
x1 = solve(Ief-If,x)

#Result:
print "Percentage of unprotected winding is",round(x1,2),"%"

Percentage of unprotected winding is 19.68 %


### Example 22.3, Page Number: 530¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
MVA = 10                         #MVA rating of alternator
V = 6.6                          #voltage rating of alternator(V)
CR = 1000/5                      #current ratio of CT
Rn = 6                         #resistance of star-point to earth(ohm)
Iop = 0.75                        #operating current of the relay(A)

#Calculation:
#Let x % of the winding be unprotected.
x = symbols('x')
Vph = V*1000/3**0.5                   #phase voltage(kV)
If = 1000/5*Iop              #minimum fault current which will operate the relay(A)
E = Vph*x/100                 #E.M.F. induced in x% winding(V)
Ief = E/Rn                      #Earth fault current which x% winding will cause(A)
#This current must be equal to 100 A.
x1 = solve(Ief-If,x)

#(ii) Let r2 = the minimum earthing resistance required to
#provide protection for 90% of stator winding.
#Then, 10% winding would be unprotected
x2 = 10                                 #%
r2 = Vph*x2/If*0.01                     #ohm

#Result:
print "(i) The percentage of each of the stator windings is",round(x1,1),"%"
print "(ii)The minimum resistance to provide protection for 90% of"
print "    the stator winding is",round(r2,2),"ohm"

(i) The percentage of each of the stator windings is 23.6 %
(ii)The minimum resistance to provide protection for 90% of
the stator winding is 2.54 ohm


### Example 22.4, Page Number: 531¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
MVA = 10                         #MVA rating of alternator
V = 6.6                          #voltage rating of alternator(V)
CR = 1000/5                      #current ratio of CT
s = 20                          #earth-fault setting(%)
Iop = 0.75                        #operating current of the relay(A)

#Calculation:
#Since 85% winding is to be protected, 15% would be unprotected
r = symbols('r')            #earthing resistance reqd. to leave 15% of winding unprotected(ohm)
x = 15                                 #%
Ifl = MVA*10**6/(3**0.5*V*1000)       #Full load current(A)
IF = s*Ifl/100                    #Minimum fault current which will operate the relay
Vu = x/100*V*1000/3**0.5             #Voltage induced in 15% of winding(kV)
Ief = Vu/r                        #Earth fault current which 15% winding will cause(A)
#This current must be equal to IF.
r1 = solve(Ief-IF,r)              #ohm

#Result:
print "The value of earthing resistor is",round(r1,2),"ohm"

The value of earthing resistor is 3.27 ohm


### Example 22.5, Page Number: 538¶

In :
from __future__ import division

#Variable declaration:
r1 = 220/11000                        #voltage ratio of transformer
r2 = 600/5                 #current ratio of protective transformer on 220V side

#Calculation:
#Suppose that line current on 220 V side is 600 A

Ipd = 5                  #Phase current of delta connected CTs on 220V side(A)
Ild = 3**0.5*Ipd         #Line current of delta connected CTs on 220 V side(A)

#This Ild will flow through the pilot wires.
Ips = 5*3**0.5           #Phase current of star connected CTs on 11,000 V side(A)

#Now, using this relation: Primary apparent power = Secondary apparent power
I = 3**0.5*220*600/(3**0.5*11000)           #A
r3 = I/Ips                     #Turn-ratio of CTs on 11000 V side

#Result:
print "Turn-ratio of CTs on 11000 V side is (",round(r3,3),": 1 )"

Turn-ratio of CTs on 11000 V side is ( 1.386 : 1 )


### Example 22.6, Page Number: 538¶

In :
from __future__ import division

#Variable declaration:
r1 = 0.4/11                        #line voltage(in kV) ratio of transformer
r2 = 500/5                   #current ratio of protective transformer

#Calculation:
#Suppose the line current on 400 V side is 500 A.
Ipd = 5                     #Phase current of delta connected CTs on 400 V side(A)
Ild = Ipd*3**0.5            #Line current of delta connected CTs on 400 V side(A)
#This Ild will flow through the pilot wires.
Ips = 5*3**0.5             #Phase current of star-connected CTs on 11000 V side(A)

#Primary apparent power = Secondary apparent power
I = 3**0.5*400*500/(3**0.5*11000)             #A
r3 = I/Ips

#Result:
print " The ratio of the protective transformers on 11kV side is",round(r3,3),"i.e, 10.5:5"

 The ratio of the protective transformers on 11kV side is 2.099 i.e, 10.5:5