Question

A steam power plant with a power output of 230 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.

Answer 1

`\eta =46\%`

Step-by-step explanation:

Hello!

In this case, we compute the heat output from coal, given its heating value and the mass flow:

`Q_H=60(tons)/(h)*(1000kg)/(1ton)*(1h)/(3600s)*(30,000kJ)/(kg)\\\\Q_H=500,000(kJ)/(s)*(1MJ)/(1000J) =500MW`

Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

`\eta =(230MW)/(500MW)*100\% \\\\\eta =46\%`

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