# Chapter 21 : WATER POWER ENGINEERING¶

## Example 21.1 pg : 906¶

In :
import math

#total installed capacity
#load factor
#plant factor
#utilization factor

#Given
c = 10000.;               				#capacity of each generator;
n = 3.;                				#number of generator
l1 = 12000.;            				#initial load on plant
l2 = 26000.;            				#final load on plant

# Calculations and Results
tc = n*c;
print "Total installed capacity = %i kW."%(tc);

avg = (l1+l2)/2;       				#average load
pk = l2;               				#peak load
lf = avg*100/pk;
lf = round(lf*10)/10;
print "load factor = %.2f percent."%(lf);

#take any time duration t hours
pf = avg*100/tc;
pf = round(pf*10)/10;
print "plant factor = %.2f percent."%(pf);

uf = pk*100/tc;
uf = round(uf*10)/10;
print "utilization ratio = %.2f percent."%(uf);

Total installed capacity = 30000 kW.
load factor = 73.10 percent.
plant factor = 63.30 percent.
utilization ratio = 86.70 percent.


## Example 21.2 pg : 906¶

In :
import math

#pondage to be provided

#Given
Q = 40.;         				# minimum flow in river
H = 30.;         				#net head
lf = 0.73;       				#load factor
eita = 0.6;     				#plant efficiency

# Calculations and Results
P = 9.81*Q*H*eita;
pk = P/lf;
pk = round(pk*10)/10;
print "maximum generation capacity of generator = %.2f kW."%(pk);

pp = pk-P;           				#power develop from pondage
Q = pp/(9.81*H*eita);
pr = Q*4*3600/10000;
pr = round(pr*10)/10;
print "Pondage required = %.2fD+4 cubic metre."%(pr);

maximum generation capacity of generator = 9675.60 kW.
Pondage required = 21.30D+4 cubic metre.


## Example 21.3 pg : 907¶

In :
import math

#maximum load factor

#Given
c = 15000.;            				#installed capacity of plant
lf = 0.3;             				#load factor
eita = 0.82;         				#plant efficiency
H = 25;              				#working head

# Calculations and Results
avg = c*lf;        				#average power developed
Q = avg/(9.81*H*eita);
Q = round(Q*100)/100;
print "minimum discharge required in the stream = %.2f cumecs."%(Q);

Q = 32;         				#for second case
P = 9.81*H*Q*eita;
lf = P*100/c;
lf = round(lf*10)/10;
print "maximum load factor = %.2f percent."%(lf);

minimum discharge required in the stream = 22.38 cumecs.
maximum load factor = 42.90 percent.