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Reversible Carnot Cycle and Heat Engine

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A borehole produces 120kg/s of high pressure water at 165°C. A powerplant is to be designed to extract heat from the water, reducing the temp to 147°C. The plant which will initially be assumed to be reversible, will reject heat at 17°C to a river. Take 147°C as the highest temp available and assume the river is so large that there is negligible change in temp. Specific Heat Capacity of high pressure water is 4.2kj/kgK.

    Work out:
    a) Rate of heat input available to the plant.
    b) Rate at which heat is rejected
    c) Maximum power out from the plant and its efficiency.


    2. Relevant equations

    QL/QH = TL/TH


    3. The attempt at a solution

    I am not sure where or how to begin. Any help pointing me in the right direction for each question would be much appreciated.
     
  2. jcsd
  3. Jan 4, 2012 #2
    I am not sure your question makes sense. What do you mean by "to reject heat?" Are you sure that 17 C is not the temperature of the river? If the process reduces the temperature from 165 C to 147 C, then isn't 165 the highest temperature, and not 147? I would not know where to start either. It does not make sense.
     
  4. Jan 6, 2012 #3
    For part a): Use this equation to calculate the heat lost from the water (the heat input to the plant): Q(rate) = m(rate of mass flow) * specific heat * temperature change

    for b): Since the plant is assumed to be reversible, use the equation: Qh(rate)/QL(rate) = Th/TL

    for c): The maximum power output results from a reversible cycle (or a process), which is the case here. Now since the efficiency of a reversible cycle is a function of temperature ONLY, use the equation: η(rev) = (Th-TL)/(Th)
    After you find the maximum efficiency, use the equation: η = [Wnet,output(rate)]/Qh

    I hope that answers your question.
     
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