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2nd Law of Thermodynamics

  1. Feb 14, 2009 #1
    1. The problem statement, all variables and given/known data
    An electric generating station is designed to have an electric output power 1.20 MW using a turbine with two-thirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at 98°C.

    (a) Find the rate at which the station exhausts energy by heat, as a function of the fuel combustion temperature Th.

    (b) Find the exhaust power for Th = 838°C.


    (c) Find the value of Th for which the exhaust power would be only half as large as in part (b).




    2. Relevant equations
    P=W/t e= 1- tc/th


    3. The attempt at a solution
    ok i really dont know where to start on this one
    any kind of help would be appreciated..thanks!
     
  2. jcsd
  3. Feb 16, 2009 #2

    Andrew Mason

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    First of all, you have to determine how much heat per unit of time the station has to exhaust. If the efficiency if 2/3 of a Carnot engine operating between Th and 98C what is the efficiency? (be careful to convert temps to K).

    How much heat is needed to produce 1.2 MW? So how much heat is left over after you subtract the 1.2 MW output?

    AM
     
  4. Feb 16, 2009 #3
    so does that mean the efficiency would be
    e = 2/3(1- 371/Th) ??

    what do i do with this..it asks for the rate at which it exhausts energy by heat
     
  5. Feb 16, 2009 #4
    Trying to figure this part out aswell. Any additional direction would be appreciated.
     
  6. Feb 16, 2009 #5

    Andrew Mason

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    That looks right.
    You are trying to found Qout.
    Use the definition of efficiency (ie in terms of Work done to Qin) to give you Qin. How is the work done related to Qin and Qout?

    AM
     
  7. Feb 17, 2009 #6
    ok yea i dont understand this at all

    so if 2/3(1-371/Th) is right..wat am i solving for..that just gives me the efficiency
    im so confused!
     
  8. Feb 17, 2009 #7
    Thanks Andrew, that helped me.

    You're solving for Q_out, which is like the energy you put in minus the work that you put out. I was solving for Q_in so that part confused me.

    However, if you can get Q_in then getting Q_out is just one step away.

    If you can do this then the problem is solved. You have the right efficiency, but that efficiency can also equal something in terms of work and Q_in <--- useful.
     
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