A steam power plant with a power output of 101.393 MW consumes coal at a rate of 55,309.154 kg/h. If the heating value of 23,888.055 kJ/kg, determine the overall efficiency of this plant. in percent (%) with three decimals.

Question

Engineering

Gustavo Andrade

24 June, 12:03

A steam power plant with a power output of 101.393 MW consumes coal at a rate of 55,309.154 kg/h. If the heating value of 23,888.055 kJ/kg, determine the overall efficiency of this plant. in percent (%) with three decimals.

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Jordan · Answer 1

27.626%

Explanation:

First we find the amount of energy the plant consumes

Q = (55309.154) (23888.055) = 1321228113KJ/h

then we convert kj / h to MW

Q=1321228113 / (3600*1000) = 367.008Mw

then we calculate the efficiency

η=101.393/367.008=0.27626=27.626%