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27 Nov 2019

If a coal burning electrical generating plant burns 2 tons of coal to generate 6000 kWh of electricity, calculate the efficiency of the plant as the ratio of electricity output to fuel energy input.

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Nestor Rutherford
Nestor RutherfordLv2
8 Aug 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

A typical coal-fired power plant burns 290 metric tons of coalevery hour to generate 730 MW of electricity. 1 metric ton=1000 kg.The density of coal is 1500 kg/m^3 and its heat of combustion is 28MJ/kg. Assume that all heat is transferred from the fuel to theboiler and that all the work done in spinning the turbine istransformed into electrical energy.

a.) Suppose the coal is piled up in a 13.0 m 14.0 m room. How tallmust the pile be to operate the plant for one day? (in m)

b.) What is the power plant's thermal efficiency? (in %)

A typical coal-fired power plant burns 250 metric tons of coalevery hour to generate 800 MW of electricity. 1 metric ton= 1000kg. The density of coal is 1500 {\rm kg}/{\rm m}^{3} and its heatof combustion is 28 \;\rm MJ/kg. Assume that all heat istransferred from the fuel to the boiler and that all the work donein spinning the turbine is transformed into electricalenergy.
Suppose the coal is piled up in a 8.0 m \times 15 m room. How tallmust the pile be to operate the plant for one day?
What is the power plant's thermal efficiency?

A typical coal-fired power plant burns 320 metric tons of coalevery hour to generate 760 MW of electricity. 1 metric ton= 1000kg. The density of coal is 1500 kg/m^3 and its heat of combustionis 28 MJ/kg. Assume that all heat is transferred from the fuel tothe boiler and that all the work done in spinning the turbine istransformed into electrical energy.

What is the power plant's thermal efficiency?

What I know:
e= W/Qh
1 Watt = 1 Joule/sec

How I've tried to solve it:
320 metric tons * 1000 kg = 320000kg / 28 MJ/kg = 11428.6 MJ? I'mnot sure if I've done it to get the right units.
I know I need both the above number and the 760 MWatts to be inJoules per second.
When I divide the two resulting numbers I get 6.7% efficiency whichisn't correct. Can someone explain where I'm going wrong?

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