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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