# Chapter 4 : Performance of Lines¶

## Example 4.1, Page No 65¶

In [1]:
import math
#initialisation of variables
R=0.496		# resistance
X=1.536
Vr=2000.0

#Calculations
Z=(10*2*2/(11*11)) + complex(30)*2*2/(11*11)
Zt=(.04+(1.3*2*2/(11*11))) + complex(0.125,(4.5*2*2/(11*11)))#Transformer impedence
Il=250*1000.0/2000	# line current(amps.)
Pl=Il*Il*R			#line loss(kW)
Po=250*0.8			# output(kW)
cosr=0.8			# power factor
sinr=0.6
n=200*100.0/(200+7.7)
Vs=(Vr*cosr+Il*R)+complex(Vr*sinr+Il*X)
V=math.sqrt((1662**2)+ (1392**2))

#Results
print("efficiency= %.1f percent " %n)
print("Sending end voltage,|Vs|=%.0f volts" %V)

efficiency= 96.3 percent
Sending end voltage,|Vs|=2168 volts


## Example 4.2, Page No 66¶

In [2]:
import math
#initialisation of variables
Vln=400/math.sqrt(3.0)		# Line to neutral voltage(V)
Z=complex(7,11)					#Impedence per phase
Il=231/Z						# line current(amp.)
I=abs(231/Z)
Pi=3*I*I*7
Po=3*I*I*6

#Calculations
print("power input =%.0f watts\n" %Pi)		#Answers don't match due to difference in rounding off of digits
print("power output=%.0f watts\n" %Po)	#Answers don't match due to difference in rounding off of digits
Ze=complex(2,3)		# equivalent impedence(ohm)
Zp=complex(3,5)		# impedence per phase
il=231/Zp			#Line current(amps.)
IL=abs(il)
pi=3*IL*IL*3
po=3*IL*IL*2

#Results
print("power input=%.1f watts" %pi)			#Answers don't match due to difference in rounding off of digits
print("power output = %.0f watts " %po)		#Answers don't match due to difference in rounding off of digits

when load is star connected
power input =6592 watts

power output=5650 watts

power input=14125.0 watts
power output = 9417 watts


## Example 4.3, Page No 66¶

In [3]:
import math
#initialisation of variables
a=100/.5
Xl=2*(10**-7)*math.log(100/.5)		#inductance(H/meter)
XL=20*(1000)*Xl						# inductance of 20 km length
R=6.65								# resistance(ohm)
Rc=20*1000/(58.0*90)				# resistance of copper(ohm)
I=10*1000/(33*.8*math.sqrt(3))		# the current(amps.)

#Calculations
Pl=3*I*I*Rc/(10**6)					#loss (MW)
n=10.0/(10+Pl)
print("Efficiency=%.4f percent " %n)
Vr=19052
cosr=.8				#power factor
sinr=.6
Vs=abs(((Vr*cosr+I*Rc) +complex(Vr*sinr+ I*R)))

#Results
print("Vs =%.0f volts\n" %Vs)	#Answer don't match due to difference in rounding off of digits
Reg=(Vs-Vr)*100/Vr
print("Regulation =%.2f percent" %Reg)

Efficiency=0.9479 percent
Vs =28965 volts

Regulation =52.03 percent


## Example 4.4 Page No 67¶

In [4]:
import math
#initialisation of variables
IR=(400)/((math.sqrt(3)*complex(6.3,9)))

#Calculations
In=abs((IR +IY +IB))		#Neutral current
print("Neutral current =%.2f amps\n" %In)
VR=abs(IR*complex(6,9))
VY=abs(IY*(8))
VB=abs(IB*complex(6,-8))

#Results
print("Voltage across Phase R =%.1f volts \n" %VR)
print("Voltage across Phase Y =%.2f volts \n" %VY)
print("Voltage across Phase B =%.0f volts \n" %VB)

Neutral current =45.18 amps

Voltage across Phase R =227.4 volts

Voltage across Phase Y =222.65 volts

Voltage across Phase B =227 volts



## Example 4.5, Page No 73¶

In [5]:
import math
#initialisation of variables
R=100*.1		#Resistance of line (ohm)
Xl=2*(10**-7)*100*1000*math.log(200/.75)	#inductance of line
X2=Xl*314									#inductive reactance
C=2*(math.pi*100)*8.854*(10**-12)*100*1000*(10**6)/(math.log(200/.75))	# capacitance per phase (micro farad)

#Calculations
print("Using Nominal-T method\n")
Ir=20*1000.0/(math.sqrt(3)*66*.8)
Vr=66*1000/math.sqrt(3)
Vc=complex((38104*.8+ Ir*5),(38104*.6+ Ir*17.55))	# voltage across condenser
Ic=complex(314*(Vc)*.9954*(10**-6))
ise=Ir+Ic
Is=abs(Ir+Ic)
Vs=abs(Vc + (ise*complex(5,17.53)))
VR=abs(Vs*complex(-3199)/complex(5,-3181))# no load recieving end voltage
Reg=(VR-Vr)*100.0/Vr
Pl=3*(Ir*Ir*5 + Is*Is*5)/1000000
n=20*100/(20+Pl)
print("percent regulation=%.1f " %Reg)
print("percent efficiency=%.1f \n" %n)
print("Using Nominal-pi method\n")
Ir1=218.68*complex(.8,-.6)
Ic1=complex(314*.4977*(10**-6)*Vr)
Il=Ir1+Ic1
vs1=Vr+Il*complex(10,35.1)
Vs1=abs(vs1)
Vr1=Vs1*complex(-6398)/complex(10,-6363)
VR1=abs(Vr1)			# no load recieving end voltage
Reg2=(VR1-Vr)*100/Vr
IL=abs(Ir1+Ic1)
Loss=3*IL*IL*10
n=20*100/21.388

#Results
print("percent regulation=%.2f " %Reg2)
print("percent efficiency=%.1f " %n)

Using Nominal-T method

percent regulation=18.2
percent efficiency=93.0

Using Nominal-pi method

percent regulation=18.23
percent efficiency=93.5


## Example 4.6, Page No 78¶

In [6]:
import math
import cmath
#initialisation of variables
R=0.2
L=1.3
C=0.01*(10**-6)

#Calculations
z=complex(R,(L*314.0*(10**-3)))		# serie impedence
Zc=cmath.sqrt(z/y)		# characterstic impedence
Y=cmath.sqrt(y*z)
Vr=132*1000/math.sqrt(3.0)
Ir=0
Vin=(Vr + Ir*Zc)/2			# incident voltage to neutral at the recieving end

#Results
print("Vr =%.3f volts \n" %Vr)		#Answer don't match due to difference in rounding off of digits
print("(i)The incident voltage to neutral at the recieving end {0:.5f}+{1:.5f}i".format(Vin.real, Vin.imag))	#Answer don't match due to difference in rounding off of digits
Vin2=(Vr - Ir*Zc)/2						# The reflected voltage to neutral at the recieving end
print("(ii)The reflected voltage to neutral at the recieving end{0:.5f}+{1:.5f}i".format(Vin2.real, Vin2.imag))		#Answer don't match due to difference inrounding off of digits
Vrp=Vr*cmath.exp(.2714*120*(10**-3))*cmath.exp(complex(1.169*120*(10**-3))/1000.0)#Taking Vrp=Vr+
Vrm=Vr*cmath.exp(-0.0325)*cmath.exp(complex(-.140))/1000			#Taking Vrm=Vr-
v1=Vrm/2				               # reflected voltage to neutral at 120 km from the recieving end
phase_v1=math.degrees(math.atan(v1.imag/v1.real))
v2=Vrp/2			                   #incident voltage to neutral at 120 km from the recieving end
phase_v2=math.degrees(math.atan(v2.imag/v2.real))#Phase angle of v2
print("(iii) reflected voltage to neutral at 120 km from the recieving end =%.2f at angle of %.2f" %(abs(v1),phase_v1))
print("incident voltage to neutral at 120 km from the recieving end = %.2f at angle of %.2f" %(abs(v2),phase_v2))

Vr =76210.236 volts

(i)The incident voltage to neutral at the recieving end 38105.11777+0.00000i
(ii)The reflected voltage to neutral at the recieving end38105.11777+0.00000i
(iii) reflected voltage to neutral at 120 km from the recieving end =32.07 at angle of 0.00
incident voltage to neutral at 120 km from the recieving end = 39372.08 at angle of 0.00


## Example 4.7 Page No 79¶

In [7]:
import math
#initialisation of variables
Ir=40.0*1000/(math.sqrt(3)*132*.8)
Vr=132.0*1000/math.sqrt(3)

#Calculations
vs=Vsp+ Vsm
Vs=abs(vs)
ise=(Vsp-Vsm)/Zc
Is=abs(ise)
n=40*100/P

#Results
print("efficiency=%.1f" %n)		#Answer don't match due to difference in rounding off of digits

efficiency=92.3


## Example 4.8, Page No 80¶

In [8]:
import math
import cmath
#initialisation of variables
yl=complex(0.2714,1.169)*120*(10**-3)
Ir=40*1000/(math.sqrt(3)*132*.8)
A=cmath.cosh(yl)

#Calculations
phase_A=math.degrees(math.atan(A.imag/A.real)) #Phase angle of A
Vr=132*1000/math.sqrt(3)
B=Zc*cmath.sinh(yl)
phase_B=math.degrees(math.atan(B.imag/B.real))  #Phase angle of B
Vs=(A*Vr+B*IR)
f=abs(B)
d=abs(Vs)
C=cmath.sinh(yl)/Zc
phase_C=math.degrees(math.atan(C.imag/C.real)) #Phase angle of C
D=cmath.cosh(yl)
phase_D=math.degrees(math.atan(D.imag/D.real))		#Phase angle of D

#Results
print("A=%.2f at an angle of %.2f " %(abs(A),phase_A))
print("B=%.1f at an angle of %.0f " %(abs(B),phase_B))
print("C=%.2f at an angle of %.2f " %(abs(C),phase_C))
print("D=%.2f at an angle of %.2f " %(abs(D),phase_D))

A=0.99 at an angle of 0.26
B=54.6 at an angle of 64
C=0.00 at an angle of -89.92
D=0.99 at an angle of 0.26


## Example 4.9 Page No 81¶

In [9]:
import math
#initialisation of variables
Ir=218.7*complex(0.8,-0.6)
Ic1=complex(314)*0.6*(10**-6)*76200
Il=Ic1+Ir
Vs=76200 + Il*complex(24,48.38)

#Calculations
phase_Vs=math.degrees(math.atan(Vs.imag/Vs.real))
Pl=3*24*abs(Il)*abs(Il)/1000000.0   #The Loss(MW)
n=40*100/(40+Pl)
print("Using Nominal- pi method")
print("Vs=%.0f volts at an angle of %.2f \n" %(abs(Vs),phase_Vs))
print("efficiency=%.2f percent" %n)
print("\nUsing Nominal-T method")
Vc=76200*complex(0.8,0.6)+218.7*complex(12,24.49)
Ic=complex(314)*1.2*(10**-6)*complex(63584,51076.0)
Is=complex(199.46,23.95)
Vs=(Vc + Is*complex(12,24.49))/1000.0
phase_Vs=math.degrees(math.atan(Vs.imag/Vs.real))	#Phase angle of Vs
Pl1=3*12*((200.89**2)+ 218.7**2)/1000000   #The loss(MW)
n1=40*100/(40+Pl1)

#Results
print("Vs=%.2f  at an angle of %.2f " %(abs(Vs),phase_Vs))
print("efficiency=%.2f percent\n" %n1)

Using Nominal- pi method
Vs=87299 volts at an angle of 3.95

efficiency=91.28 percent

Using Nominal-T method
Vs=86.25  at an angle of 40.70
efficiency=92.65 percent



## Example 4.10 Page No 92¶

In [10]:
import math
import cmath
#initialisation of variables
R=0.1557*160
GMD=(3.7*6.475*7.4)**(1.0/3)
Z1=2*(10**-7)*math.log(560/0.978)*160*1000
XL=63.8

#Calculations
C=(10**-9)*2*(10**6)*math.pi*160*1000.0/(36*math.pi*math.log(560/.978))
jwC=complex(314)*1.399*(10**-6)/160.0
Zc=cmath.sqrt(Z/jwC)
y=cmath.sqrt(Z*jwC)
yl=y*160
A=cmath.cosh(yl)
B=Zc*cmath.sinh(yl)
C=cmath.sinh(yl)/Zc
Ir=50000/(math.sqrt(3)*132)
VS=152.34
Vnl=abs(Vs)/abs(A)
reg=(Vnl-76.208)*100/76.208
n=50000*.8*100/abs(Ps)

#Results
print("Vs line to line =%.2f kV\n" %VS)
print("sending end current Is(A){0:.5f}+{1:.5f}i".format(Is.real, Is.imag)) #Answer don't match due to difference in rounding off of digits
print("sending end power=%.0f kW" %Ps)
print("sending end p.f =%.3f" %pf)
print("percent regulation=%.1f " %reg)
print("percent efficency=%.1f " %n)

Vs line to line =152.34 kV

sending end current Is(A)211.28696+-129.31797i
sending end power=55350 kW
sending end p.f =0.829
percent regulation=17.2
percent efficency=72.3