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)[0]
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)[1] #A
#Result:
print "Required value of earth resistance is",round(r1,3),"ohm"
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)[0]
#Result:
print "Percentage of unprotected winding is",round(x1,2),"%"
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)[0]
#(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"
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)[0] #ohm
#Result:
print "The value of earthing resistor is",round(r1,2),"ohm"
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 )"
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"