import math
## Example 8.1
print('Example 8.1');
print('Page No. 222');
## given
V = 240;## Voltage in Volts
I = 8;## Current in Amps
##By ohm's law-> V = I*R
R = V/I;## In ohms
print'%s %.2f %s '%('The resistance of the given circuit is ',R,' ohms')
import math
import numpy
from numpy.linalg import inv
## Example 8.2
print('Example 8.2');
print('Page No. 223');
## given
V1 = 100.;## In Volts
V2 = 50.;## In Volts
R1 = 8.;## Resistance in ohm
R2 = 5.;## Resistance in ohm
R3 = 10.;## Resistance in ohm
R4 = 50.;## Resistance in ohm
##By refering figure 8.3, and applying kirchoff's current law and kirchoff's voltage law in the given circuit diagram, we get following equations:
## I1 = I2 + I3
##V1 - R1*I1 - V2 - R3*I3 = 0
##V2 - R4*I3 + R3*I3 - R2*I2 = 0
A = ([[1, -1, -1],[8, 0, 10],[0 ,55, -10]])
b = ([0,50,50]);
x =numpy.dot(inv(A),b)
print'%s %.2f %s '%('The currents in I1 is ',x[0],' A ')
print'%s %.2f %s '%('The currents in I2 is ',x[1],' A ')
print'%s %.2f %s '%('The currents in I3 is ',x[2],' A ')
import math
## Example 8.3
print('Example 8.3');
print('Page No. 226');
## given
R = 6;## Resistance in ohm
Xc = 16;## Capacitive resistance in ohm
Xl = 24;## Inductive resistance in ohm
Z = ((R**2) + (Xc - Xl)**2)**0.5;## Impedance in ohm
P_f = R/Z;## Power factor = cos(x) = 0.6
x = math.cos(0.6)*57.3;
y = math.sqrt(1 - P_f**2);##y = sin (x)
V = 200;## in Volts(sin wave voltage = ((200*2**1.5)*sinwt)
I = V/Z;## Current in Amperes
P = I**2 * R;## in W
Q = V * I * y;## in VAR
S = V * I;## in VA
print'%s %.2f %s'% ('The actual power is ',P,' W ')
print'%s %.2f %s'% ('The reactive power is ',Q,' VAR ')
print'%s %.2f %s'% ('The apparent power is ',S,' VA ')
## Example 8.4
print ('Example 8.4');
print ('Page No. 232');
##given
pump_1 = 100*10**3;## Required pump in W
T_1 = 8;## Pump Operating time of each day
Inc_op = 0.5;## Increased output per cent
pump_ex = 50*10**3;## Extra pump requried in W
## This question doesnot contain any calculation part.
print('there is no computational part in the problem')
## Example 8.5
print ('Example 8.5');
print ('Page No. 232');
##given
P = 600*10**3;## Power demand of pump in W
T = 8.;## Operating time in hour per day
red = 1.00/10**3.;## off-peak reduction in Pound per 10**3 W month
M_save = P*red;## Monthly saving Pound per month
A_save = M_save*12.;## Annual saving in Pound per year
print'%s %.2f %s'% ('Annual saving is ',A_save,' Pound per year')
import math
import numpy
from numpy import floor
## Example 8.6
print ('Example 8.6');
print('Page No. 234');
##given
T_lamp = 12.*10**3.;## Output for the tungsten filament lamp in lm per 10^3 W
F_tube = 63.*10**3.;## Output for the fluorescent tubes in lm per 10^3 W
Save = F_tube - T_lamp;## in W
print'%s %.2f %s'% ('Energy saving is ',Save,' lm per 10^3 W ')
Save_pcent = (Save/F_tube)*100.;
print'%s %.2f %s'% ('Energy saving per cent is ',floor(Save_pcent),'')
import math
## Example 8.7
print('Example 8.7');
print ('Page No. 235');
##given
N = 40.;## Number of lamps
T1 = 15.;## Operating time in h per day
P = 500.;## POwer from the lamps in W
T2 = 300.;## Total operating time in days per year
C = 2.5/10**3;## Electricity cost in p per Wh
An_Cost = N*P*T1*T2*C*10**-2;## In euro
print'%s %.2f %s'% ('The Annual Cost is ',An_Cost,' Euro ')
##Improvement in light by installing glassfibre skylights
T3 = 5;## Extra Time for natural lighting in h per day
New_An_Cost = N*P*(T1-T3)*T2*C*10**-2;## In euro
print'%s %.2f %s'% ('The New Annual Cost is ',New_An_Cost,' Euro ')
Save = An_Cost - New_An_Cost;## in euro
print'%s %.2f %s'% ('The annual saving for a pay-back period of 2.5 years is ',Save,'')
## Example 8.8
print ('Example 8.8');
print ('Page No. 236');
## This question doesnot contain any calculation part.
##By refering figure 8.7 which shows Poer factor-load curve for a motor with a capacitor and one without a capacitor.
print('there is no computational part in the problem')
import math
## Example 8.9
print('Example 8.9');
print('Page No. 238');
## This question doesnot contain any calculation part.
##given
l = 500.*10**3.;## Load in VA
P_F = 0.6;## Power Factor
Req_P_F = 0.9;## Required power factor
##Refer to figure 8.8
BC = 2.5;## units
C_rt = 250*10**3;## in VAR (obtained from figure 8.8)
print'%s %.2f %s'%('The required condenser rating is',C_rt,'')
import math
import numpy
from numpy.linalg import inv
## Example 8.10
print ('Example 8.10');
print ('Page No. 240');
P = 100.;## Power in 10^3 W
C = 5.;## Charge in Euro per 10^3 per month
PF = ([1.0, 0.9 ,0.8, 0.7, 0.6, 0.5]);## Power factor
leng=len(PF)
for i in range (0,leng):
VA = (PF[i]/P)*C
a=VA
b=PF[i]
print'%s %.2f %s %.2f %s'% ('Charge per month for power factor ',b,' is' ,a,'Euro')
import math
## Example 8.11
print('Example 8.11');
print ('Page No. 240');
## This question doesnot contain any calculation part.
##given
P_F_1 = 0.7;## Initial power factor
P_F_2 = 0.95;## Final power factor
##Refer Figure 8.10
red_I = 26;##reduction in current in per cent
print'%s %.2f %s'% ('The reduction in current is ',red_I,' per cent ')
P_F_3 = 1.0;## Increased power factor
## From figure 8.10
Save = 4.;## per cent
print'%s %.2f %s'% ('Increase in power factor from 0.95-1.0 only increases saving further by a ',Save,' per cent')
## Example 8.12
print ('Example 8.12');
print ('Page No. 240');
## This question doesnot contain any calculation part.
##given
C = 10000.;## Installation cost of capacitors in Pound
P_F_1 = 0.84;## Initial power factor
P_F_2 = 0.97;## Final power factor
##Refer Figure 8.10
red_dem = 14.;##reduction in maximum demand in per cent
T = 9.;## pay-back time in months
print'%s %.2f %s'% ('The reduction in maximum demand is',red_dem,' per cent')
print'%s %.2f %s'% ('The pay-back time was',T,' months')
## This question does not contain any calculation part.
import math
import numpy
from numpy import floor
## Example 8.13
print ('Example 8.13');
print ('Page No. 244');
##given
T1 = 21.;## in degree celcius
t1 = 8.;## time in h per day
c = 3.5;## cost in p per unit
C1 = 38.;## Total cost in Pound per 10^3 W
T2 = 16.;## in degree celcius
t2 = 8.;## time in h per day
C2 = 27.;## Total cost in Pound per 10^3 W
Save = C1 - C2;## Saving in Pound per 10^3 W
Save_deg = Save/(T1 - T2);## Total Saving in Pound per 10^3 W for each degree drop
Save_per = (Save_deg/C1)*100.;## Saving in percent
print'%s %.2f %s'% ('For each degree drop, an energy saving of ',floor(Save_per),' per cent is achieved')
import math
## Example 8.14
print ('Example 8.14');
print ('Page No. 245');
## This question doesnot contain any calculation part.
##refer Table 8.6
O_1 = 1750.;
O_2 = 0.;
O_3 = 2.;
O_4 = 150.;
O_5 = 1900.;
O_6 = 0.;
I_1 = 580.;
I_2 = 1658.;
I_3 = 0.5;
I_4 = 40.;
I_5 = 1698.;
I_6 = 11.;
D_1 = 300.;
D_2 = 869.;
D_3 = 0.5;
D_4 = 40.;
D_5 = 900.;
D_6 = 37.;
print(' \t\t\t\t\t\t ENERGY COSTS FOR HEATING STEEL BILLETS')
print('components\t\t\t\t\t\t kmol \t\t\t\t\t\t % Composition by volume' '\n\n\n')
print('\t\t\t\t\t\t\t (1)Oil fired (2)Induction (3)Direct resistance')
print'%s %.2f %s %.2f %s %.2f %s '%('Components(10^3 W/tonne) \t\t ',O_1,' \t\t ',I_1,' \t\t ',D_1,'')
print'%s %.2f %s %.2f %s %.2f %s '%('Fuel(electricity) \t\t ' ,O_2,' \t\t ',I_2,' \t\t ',D_2,'')
print'%s %.2f %s %.2f %s %.2f %s '%('Components(10^3 W/tonne) \t\t ',O_3,' \t\t ',I_3,' \t\t ',D_3,'')
print'%s %.2f %s %.2f %s %.2f %s '%('Components(10^3 W/tonne) \t\t ',O_4,' \t\t ',I_4,' \t\t ',D_4,'')
print'%s %.2f %s %.2f %s %.2f %s '%('Components(10^3 W/tonne) \t\t ',O_5,' \t\t ',I_5,' \t\t ',D_5,'')
print'%s %.2f %s %.2f %s %.2f %s '%('Components(10^3 W/tonne) \t\t ',O_6,' \t\t ',I_6,' \t\t ',D_6,'')
import math
## Example 8.15
print('Example 8.15\n\n');
print('Page No. 247\n\n');
## This question doesnot contain any calculation part.
##refer Table 8.7
El = 35.;## Percentage of electricity produced from primary fuel
En_1 = 50.;## Endothermic gas (m^3)
En_2 = 100.;## Endothermic gas (m^3)
En_3 = 200.;## Endothermic gas (m^3)
G_1 = 97.;## Gas use (10^3 Wh)
G_2 = 194.;## Gas use (10^3 Wh)
G_3 = 386.;## Gas use (10^3 Wh)
El_1 = 24.;## Electricity use (10^3 Wh)
El_2 = 48.;## Electricity use (10^3 Wh)
El_3 = 95.;## Electricity use (10^3 Wh)
P_1 = 69.;## Primary energy (10^3 Wh)
P_2 = 137.;## Primary energy (10^3 Wh)
P_3 = 271.;## Primary energy (10^3 Wh)
Endothermicgas=['50 \t\t', '100 \t\t ','200 \t\t ']
gasuse= ['\t\t 97 \t\t ','\t\t194 \t\t ','\t\t 386 \t\t ']
electricityuse=[' \t\t 24 ',' \t\t 48\t\t ',' \t\t 95 \t']
primaryenergy= [' \t\t 69 ',' 137 ',' \t 271 ']
print('Endothermicgas\t\t gasuse\t\t electricityuse \t\t primaryenergy')
for row in zip(Endothermicgas , gasuse, electricityuse,primaryenergy):
print ' '.join(row)