Power Plant Engineering

= Rs. 205 × 10
5
.
Ans.
Example 2. The maximum (peak) load on a thermal power plant of 60 mW capacity is 50 mW at
an annual load factor of 50%. The loads having maximum demands of 25 mW, 20 mW, 8 mW and, 5 mW
are connected to the power station.
Determine: (a) Average load on power station (b) Energy generated per year (c) Demand factor
(d) Diversity factor.
Solution.
(a) Load factor = Average load/Maximum demand
Average load = 0.5 × 50 = 25 mW
(b) E = Energy generated per year
= Average load × 8760
= 219 × 10
6
kWh.
(c) Demand factor = Maximum demand/Connected load
= 50/(25 + 20 + 8 + 5) = 0.86
(d) Diversity factor = 
1
2
M
M
where
M
l
= Sum of individual maximum demands = 25 + 20 + 8 + 5 = 58 mW

---

136
POWER PLANT ENGINEERING
M
2
= Simultaneous maximum demand = 50 mW
Diversity factor = 
58
50
= 1.16.
Ans.
Example 3. In a steam power plant the capital cost of power generation equipment is Rs. 25 ×
10
5
. The useful life of the plant is 30 years and salvage value of the plant to Rs. 1 × 10
5
. Determine by
sinking fund method the amount to be saved annually for replacement if the rate of annual compound
interest is 6%.
Solution.
P = Capital cost = Rs. 20 × 10
5
S = Salvage value = Rs. 1 × 10
5
n = Useful life = 30 years
r = Compound interest
A = Amount to be saved per year for replacement
A = 
[(P – S) ]
{(1
)
1
n
r
r
+
−
= 
5
5
30
[(20 10
1 10 )0.06]
{(1
0.06)
1}
×
− ×
+
−
= Rs. 24,000.