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a steam power plant with a power output of 170 mw consumes coal at a rate of 60 tons/h. if the heating value of the coal is 30,000 kj/kg, determine the overall efficiency of this plant.

Sagot :

The overall efficiency of this steam power plant which consumes coal at a rate of 60 tons/h and provides a power output of 170 MW is 34 %

Win = m Q

Win = Power input

m = Consumption rate

Q = Heating value

Q = 30000 kj / kg

m = 60 tons / h = 60 * ( 1000 / 3600 )

m = 16.67 kg / s

Win = 16.67 * 30000

Win = 500000 KW

Win = 500 MW

η = Wout / Win * 100

η = Efficiency

Wout = Power output

Wout = 170 MW

η = 170 / 500 * 100

η = 0.34 * 100

η = 34 %

Therefore, the overall efficiency of this plant is 34 %

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