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

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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