Power Plant Engineering

Solution. Total hours = (24 hr) × (30 days) × (11 months) = 7920 hrs
Total energy generated = MW × hr
= 100 × 7920 = 7920 × 100 mW hr = 792 × 10
3
kW hr
Total cost of energy sold
= kW hr × Rs. × kW hr = 792 × 10
3
× 2.5 Rs.
= Rs. 1980000.00.
Example 4. Calculate annual requirement of lignite (fuel) for a thermal power plant rated 2000
mW under following conditions.
Plant rating 2000 mW,
Annual load factor = 
×
Energy Delivered
Rated power
Total hours in year
 = 0.5
Plant efficiency = 
Electrical power output
Thermal power input
 = 0.25
Utility factor of fuel = 
Useful energy in fuel
Available energy in fuel
 = 0.7
Available energy density in coal = 14 mJ/kg.
Solution. Total energy delivered in year
(mW hr)
e
= Rated power × Hours × Load factor hours in year
= 365 × 24 = 8760 hrs
(mW hr)
e
= 2000 × 8760 × 0.5 = 8760000

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NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION
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Thermal energy input = 
Electrical energy output
Plant efficiency
= 
8760000
0.25
= 35040000 (mW hr)
t
Thermal energy input to plant for one year = 35040000 (mW hr)
t
Lignite input to plant for one year
= 
(mW hr)
(Useful mW/kg) in lignite
t
Useful energy in fuel per kg = Available energy in fuel × Utility factor
= (14 mJ/kg) × 0.7 = 9.8 mJ/kg
Lignite input per year = 35040000 (mW hr)
t
Conversion factor:
1 mW hr = 1 mW × 3600 s = 3600 mW.s = 3600 mJ
Lignite required per year = 35040000 × 
3600
9.8
= 1.287 × 10
10
kg = 12870 × 10
6
kg
Since (1 MT = 1000 kg) = 12870 × 10