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26 November, 17:49

The electric output of a power plant is 669 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows at a rate of 1.17x 108 L/Hr. The water enters the plant at 23.6°C and exits at 29.8°C. (a) What is the power plant's total thermal power? (b) What is the efficiency of the power plant? (a) MWT (Megawatt thermal) O Answer part (a) (b) Answer part (b)

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  1. 26 November, 19:53
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    a) 1511 MW

    b) 44%

    Explanation:

    The thermal power will be the electric power plus the heat taken away by the cooling water.

    Qt = P + Qc

    The heat taken away by the water will be:

    Qc = G * Cp * (t1 - t0)

    The Cp of water is 4180 J / (kg K)

    The density of water is 1 kg/L

    Then

    G = 1.17 * 10^8 L/h * 1 kg/L * 1/3600 h/s = 32500 kg/s

    Now we calculate Qc

    Qc = 32500 * 4180 * (29.8 - 23.6) = 842*10^6 W = 842 MW

    The total thermal power then is

    Qt = 669 + 842 = 1511 MW

    The efficiency is

    η = P / Qt

    η = 669 / 1511 = 44%
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