# Chapter 7: Supply Systems¶

### Example 7.1, Page Number: 145¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
V1 = 200                            #initial volt of system(V)
V2 = 400                            #final volt of the system(V)

#Calculation:
I1,I2,R1,R2 = symbols('I1 I2 R1 R2')
I2 = V1*I1/V2
W1 = 2*I1**2*2*R1                        #Power loss in 200 V system
W2 = 2*I2**2*2*R2                        #Power loss in 400 V system
# As power loss in the two cases is the same,
R22 = solve(W1/W2-1,R2)
a = R22/R1                               #ratio of cables' cross-section
v = 1/a                                  #ratio of cables' volumes
s = (1-v)*100

#Result:
print "% Saving in feeder copper is ",s,"%"

% Saving in feeder copper is  75 %


### Example 7.2, Page Number: 146¶

In :
from __future__ import division
from sympy import *
import math

#Variable declaration:
#P = power upplied,      W = power loss,
#I = current,            V = voltage,     R = resistance

R,V = symbols('R V')
P1,I1,W1 = symbols('P1,I1,W1')    #for 2-wire system
P2,I2,W2 = symbols('P2,I2,W2')     #for 3-wire system

#Calculation:
#Two-wire d.c. system:
P1 = V*I1
W1 = 2*I1**2*R
L1 = W1/P1*100                    #% power loss

#3-phase, 3-wire a.c. system:
P2 = math.sqrt(3)*V*I2
W2 = 3*I2**2*R
L2 = W2/P2*100                     #% power loss

I2 = solve(L1-L2,I2)
#r2 = P2/P1
r2 = math.sqrt(3)*I2*V/(I1*V)
P1 = 2*P2

#Result:
print '''Additional power which can be supplied at
unity p.f. by 3-phase, 3-wire a.c. system = 100%.'''

Additional power which can be supplied at
unity p.f. by 3-phase, 3-wire a.c. system = 100%.


### Example 7.3, Page Number: 146¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
#Let R be the resistance per conductor in each case

R,V = symbols('R V')
P1,I1,W1 = symbols('P1,I1,W1')    #for 3-wire dc system
P2,I2,W2 = symbols('P2,I2,W2')     #for 4-wire ac system

#Calculation:

#3-wire d.c. system.:
#At balanced loads, implying no power loss in the neutral wire
P1 = 2*V*I1
W1 = 2*I1**2*R
L1 = W1/P1*100                    #% power loss

#3-phase, 4-wire a.c. system.:
#At balanced loads, implying no power loss in the neutral wire
P2 = 3*V*I2
W2 = 3*I2**2*R
L2 = W2/P2*100                     #% power loss

I2 = solve(L1-L2,I2)
r = P2/P1
#solving r, we get
P2 = 1.5*P1

#Result:
print '''Extra power that can be supplied at unity power
factor by 3-phase, 4-wire a.c. system = 50%.'''

Extra power that can be supplied at unity power
factor by 3-phase, 4-wire a.c. system = 50%.


### Example 7.4, Page Number: 148¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
R,V,phy = symbols('R V phy')       #phy is power factor angle
#V is voltage between the conductors
#R is resistance per conductor
P1,I1,W1 = symbols('P1,I1,W1')    #for Single phase 2-wire system
P2,I2,W2 = symbols('P2,I2,W2')     #for 3-phase, 3-wire a.c. system

#Calculation:
#Single phase 2-wire system
P1 = 2*V*I1*math.cos(phy)
W1 = 2*I1**2*R
L1 = W1/P1*100                    #% power loss

#3-phase, 3-wire a.c. system.
P2 = math.sqrt(3)*V*I2*math.cos(phy)
W2 = 3*I2**2*R
L2 = W2/P2*100                     #% power loss

I22 = solve(L1-L2,I2)
r = P1/(math.sqrt(3)*V*I22*math.cos(phy))
#As the power supplied by single phase, 2-wire is 200 kW,
P1 = 200

#Result:
print "Power supplied by 3-phase, 3-wire a.c. system is",P1*r,"kW"

Power supplied by 3-phase, 3-wire a.c. system is 400 kW


### Example 7.5, Page Number: 149¶

In :
from __future__ import division
import math

#Variable declaration:
MVA = 5                   #power transmitted to load
n = 0.9                   #efficiency
pf = 0.8                  #power factor
V = 33                    #receiving end voltage(V)
l = 50*10**3             #length of line(m)
s = 2.85*10**-8           #specific resistance of aluminium(ohm-m)

#Calculation:
P = MVA*pf*10**6                #power transmitted(W)
W = 10*P/100                          #line loss(W)

#(i) Single phase, 2-wire system:
I1 = MVA*10**3/V                        #Apparent power(A)
a1 = 2*I1**2*s*l/W                      #area of cross-section(m**2)
v1= 2*a1*l                              #volume of conductor required(m**3)

#(ii) 3-phase, 3-wire system:
I2 = MVA*10**3/(math.sqrt(3)*V)                        #Apparent power(A)
a2 = 3*I2**2*s*l/W                      #area of cross-section(m**2)
v2 = 3*a2*l                              #volume of conductor required(m**3)

#Result:
print "(i) Single phase, 2-wire system, Vol. of conductor required is",round(v1,2),"m^3"
print "(ii)3-phase, 3-wire system, Vol. of conductor required",round(v2,2),"m^3"

(i) Single phase, 2-wire system, Vol. of conductor required is 16.36 m^3
(ii)3-phase, 3-wire system, Vol. of conductor required 12.27 m^3


### Example 7.6, Page Number: 149¶

In :
from __future__ import division

#Variable declaration:
V1 = 11*10**3                   #Supply voltage(V)
pf = 0.8
R1 = 0.15                      #Total resistance in ac system(ohm)
R2 = 0.05                      #Total resistance in dc system(ohm)
r1 = 0.15                     #voltage drop in ac system(%)
r2 = 0.25                     #voltage drop in dc system(%)

#Calculation:
#Single phase system:
I1 = r1*V1/R1                   #line current(A)
P1 = V1*I1*pf/1000              #power received by consumer(kW)

#D.C. two-wire system:
P2 = P1
V2 = (P2*10**3*R2/r2)**0.5             #supply voltage(V)

#Result:
print "The supply voltage is",V2,"V"

The supply voltage is 4400.0 V


### Example 7.7, Page Number: 153¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
I = 200                      #current(A)
l = 1                      #length of cable(km)
c1 = 5                         #cost per unit(paise/kWh)
r1 = 10                        #interrest & depreciation(%)
ro = 1.73*10**-6              #resistivity(ohm-m)

#Calculation:
#If 'a' is the area os cross-section of the cnductor,
#cost of cable including installation = (20*a + 20)    Rs./m.

a = symbols('a')
R = ro*l*10**5/a
E = 2*I**2*R*8760/1000           #energy lost per annum
C = c1/100*E                  #cost of energy lost(Rs)

#capital cost for 1 km length of the cable is = Rs. 20,000*a.

V = r1/100*20000*a             #Variable annual charge(Rs)
a1 = solve(C-V,a)

#Result:
print "The most economical conductor size is",round(a1,2),"cm**2"

The most economical conductor size is 1.74 cm**2


### Example 7.8, Page Number: 153¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
V = 33                          #kV
pf = 0.8                        #power factor
c1 = 4                          #paise/kWh
r1 = 10                         #interest and depreciation(%)
ro = 10**-6                     #ohm-cm

#Calculation:
a = symbols('a')                #size of conductor(cm**2)
R = ro*l*10**5/a                #ohm
I = P*10**3/(math.sqrt(3)*V*pf)            #current (A)
E = 3*I**2*R*8760/1000           #kWh
C = c1/100*E                        #Annual cost of energy lost(R)
#The capital cost (variable) of the cable = 25000*a Rs/km
V = r1/100*25000*a                 #Rs
a1 = solve(C-V,a)

#Result:
print "The most economical size of conductor is",round(a1,2),"cm**2"

The most economical size of conductor is 0.71 cm**2


### Example 7.9, Page Number: 154¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
I = 250                         #current through the feeder(A)
c1 = 5                          #Rs/kg
r1 = 10                         #interest and depreciation(%)
c2 = 5                          #paise/kWh
d = 8.93                        #density of wire(gm/cm**3)
ro = 1.73*10**-8                #speciafic resistance(ohm-m)

#Calculation:
l = 1                           #length of wire(1 m say)
a = symbols('a')                #conductor size(m**2)
R = ro*l/a                      #ohm
E = 2*I**2*R*8760/1000          #kWh
C = c2/100*E                   #annual cost os energy lost(Rs)
m = 2*a*1*d*10**3                     #Mass of 1 metre feeder
CC = c1*m                    #capital cost(Rs)
V = r1*CC/100                    #Variable Annual charge(Rs)
a1 = solve(C-V)

#Result:
print "The most economic size of conductor is",round(a1*10000,2),"cm**2"

The most economic size of conductor is 3.26 cm**2


### Example 7.10, Page Number: 154¶

In :
from __future__ import division
from sympy import *

#Variable declaration:
l = 1                           #line length(km)
V = 110                         #supply voltage(kV)

t1 = 6;               P1 = 20 ;            pf1 = 0.8
t2 = 12;              P2 = 5 ;             pf2 = 0.8
t3 = 6;                P3 = 6 ;             pf3 = 0.8

c1 = 6                  #
n = 365                 #no.of days used
r1 = 10                 #interest and dep-reciation(%)

#Calculation:
a = symbols("a")                #cross-section of conductor(cm**2)
R = 0.176/a                    #Resistance per km of each conductor(ohm)
#At 20 MW,
I1 = P1*1000/(math.sqrt(3)*V*pf1)                #A
I2 = P2*1000/(math.sqrt(3)*V*pf2)                #A
I3 = P3*1000/(math.sqrt(3)*V*pf3)                #A

E1= 3*R*1/1000*(I1**2*t1+I2**2*t2+I3**2*t3)      #Energy loss per day in 3-phase line(kWh)
E = E1*n                        #energy lost per annum(kWh)
C = c1*E/100                   #Annual cost of energy(Rs)
v = r1*6000*a/100                  #variable annual charge(Rs/kWh)
a1 = solve(C-v,a)

#Result:
print "THe most economical size of the conductor is",round(a1,2),"cm**2"

THe most economical size of the conductor is 1.56 cm**2