Tutorial problem solution of Thermal Power station (Principles of power system Vk -Mehta)

1. Tutorial solution
Principles of power system by VK
Mehta
Solution#1
Heat produced by fuel by 0.75kg Coal= 0.75x Kcal
Heat equivalent of 1KWh=860Kcal
η=electrical output in heat units/heat of combustion
0.15=860/0.75x
X=7644.44 Kcal/Kg
Solution#2
η overall= ηthermal* ηelectrcial
= 0.30*0.80=0.24
Heat produced/Hour =H= 75MW*860/0.24
=268750*106
Kcal
Coal consumption per hour = H/Calorific value
= (268750*106
Kcal)/ (6400 Kcal/Kg)
=41.992188 tons= 42tons
Solution #3
Units generated per Day:
Maximum energy that can produced according to load factor:
=65,000 kW*0.40*24
=624,000 KWh
As coal consumption 1KWh >> 0.5Kg
So, for 624,000kWh this can be 312,000Kg or 312tons
Efficiency=electrical output in heat units/heat produced by fuel

2. Overall Efficiency= (624,000*860) / (312,000*15000)
= 0.11466= 11.466%
Solution #4
Maximum energy that can be produced in day:
=plant Capacity * hour
= 60,000KW*24h
= 1440,000 KWh
Calorific Values: 6950Kcal/Kg
Electrical output in heat units = 1440, 000*860
= 1238400000 Kcal
Coal consumption per day:
=heat produced by fuel /calorific value
= 1238400000/6950
=178187.0504 Kg = 178.187 tons
Specific fuel consumption:
=178187.0504Kg/1440000kWh
=0.123Kg/kWh
Solution#6
Calorific value???
1Kg/kWh efficiency=15%
0.15=860/1x
X=5733.33kcal/kg
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