In [18]:

```
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)[0]
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,"%"
```

In [1]:

```
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)[0]
#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%.'''
```

In [19]:

```
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)[0]
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%.'''
```

In [38]:

```
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)[0]
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"
```

In [2]:

```
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"
```

In [12]:

```
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"
```

In [22]:

```
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)[1]
#Result:
print "The most economical conductor size is",round(a1,2),"cm**2"
```

In [14]:

```
from __future__ import division
from sympy import *
#Variable declaration:
P = 5 #line load(MW)]
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)[1]
#Result:
print "The most economical size of conductor is",round(a1,2),"cm**2"
```

In [23]:

```
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)[1]
#Result:
print "The most economic size of conductor is",round(a1*10000,2),"cm**2"
```

In [24]:

```
from __future__ import division
from sympy import *
#Variable declaration:
l = 1 #line length(km)
V = 110 #supply voltage(kV)
#Working hours(hrs) Load(MW) Power factor
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)
#The load currents at various loads are :
#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)[1]
#Result:
print "THe most economical size of the conductor is",round(a1,2),"cm**2"
```