Calculating Energy Efficiency in a Cogeneration Power Plant

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SUMMARY

The discussion centers on calculating energy efficiency in a cogeneration power plant operating at a steady state. The plant generates 80 MW of electricity, with a thermal efficiency of 40%. The energy input rate, denoted as \dot{Q}_{in}, can be determined using the relationship between work output and heat transfer. The calculations reveal that \dot{Q}_{in} is 200 MW, while the energy discarded to the environment is 120 MW.

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Homework Statement


A cogeneration power plant is operating in a thermodynamic cycle at steady state. The plant provides electricity to a community at a rate of 80 MW. The energy discharged from the power plant by heat transfer is denoted by \dot{Q}_{out}. Of this, 70 MW is provided to the community for water heating and the remainder is discarded to the environment without use. The electricity is valued at $0.08 kW\cdoth. If the cycle thermal efficiency is 40%, determine the (a) rate energy is added by heat transfer, \dot{Q}_{in} in MW, (b) rate energy is discarded to the environment, in MW.


Homework Equations


W_{cycle} = Q_{in} - Q_{out}
\dot{Q}=\dot{W}
\eta = \frac{W_{cycle}}{Q_{in}}=\frac{Q_{in}-Q_{out}}{Q_{in}}


The Attempt at a Solution


I haven't the slightest idea how to even start this. There aren't even any worked examples in the book. I only know that the rate of heat transfer has to be equal to 80 MW (the power supplied by the plant).
 
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Welcome to PF, treynolds147! :smile:

I think it's not the rate of heat transfer that is 80 MW.
It's the generated useful power \dot W_{cycle} that is 80 MW.
With an \eta of 40%, you should be able to find \dot Q_{in}.
 

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