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  1. The maximum efficiency of a heat engine operating between reservoirs at TH = 505K and TC = 323 K is:
    Efficiency = 1 – TC/TH = 0.36
    We can determine the minimum heat rejected, QC, using the 1st Law for a closed system, where the system is the heat engine.
    QH – QC = W or QC = QH – W
    So, we need to determine QH. We can use the efficiency to accomplish this since the efficiency of a heat engine is defined as:
    Efficiency = W / QH
    So: QH = W / Efficiency = 233 MW
    Therefore, QC = 233 MW – 84 MW = 149 MW
    QC = 149 MJ/sec * 24 hours * 3600 sec/hour
    QC = 1.288×10^7 MJ

    Enjoy !

Thermodynamics & Heat Transfer question… Steam efficiency etc..?

A power plant taps steam superheated by geothermal energy to 505 K (the temperature of the hot reservoir) and uses the steam to do work in truning the turbine of an eletric generator. The steam is then converted back into water in a condenser at 323 K (the temperature of the cold reservoir), after which the water is pumped back down into the ground where it is heated again. The output of the generator is 84MW. Calculate:
The maximum efficiency at which this plant can operate.

The minimum amount of rejected heat (in MJ) that must be removed from the condenser every 24 hours.

Would appresiate if you showed working with the answer so i can work though it myself, but just the answer is fine. Cheers :)