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Problem with Q

  1. Nov 22, 2004 #1
    Seriously Stuck..

    A coal fired plant generates 600MW of electric power. The plant uses 4.8 x 10^6 kg of coal a day. The heat of combustion of coal is 3.3 x 10^7 J/kg. The steam that drives the turbines is at a temperature of 573K, and the exhaust water is 310K.
    1.) What is the overall efficiency of the plant for generating electric power?
    2.) How much thermal energy was wasted?


    Ok..this is what I THINK I know....Qh= 3.3 x 10^7 J/kg
    Th= 573K Tc= 310K
    mass= 4.8 x 10^6 kg

    Therefore Carnot effeciency is e=Th-Tc/Th
    so Carnot eff. worked out is about 46%

    AND I know that P=W/t = Fd/T = FV
    Now...I get lost...I know I need to find the Work, to find my Qc...but they only gave me power in MW, and I need to find wasted thermal energy?? Any thoughts??
     
    Last edited: Nov 22, 2004
  2. jcsd
  3. Nov 23, 2004 #2

    Clausius2

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    Wrong. Why are you using the Carnot efficiency? Is there Carnot anywhere here?. Which cycle is using the plant? Sure it is not the Carnot cycle. It's the Rankine cycle.

    -Thermal power extracted from the hot focus:

    [tex] \dot Q_h=\dot m_{fuel} h_{fuel} [/tex] where h is the heat of combustion and m is the mass flow of fuel.

    - Electric Power generated=Mechanical Power extracted from the turbine rotor (assuming electric efficiency near 1). So that, the efficiency is given by:

    [tex] \eta=\frac{\dot W_{electric}}{ \dot Q_h}[/tex]

    Due to the first principle:

    [tex] \dot W= \dot Q_h-\dot Q_c[/tex] being the last term the heat power wasted to the cold focus. The question of your problem about wasting thermal energy is bad formulated. It means the heat power wasted. You know, due to the second principle some amount of heat extracted from the fuel has to be wasted to the cold environment.

    What do you think?
     
  4. Nov 23, 2004 #3
    Thanks!...I found the Qh by using the Qh=mh equation...

    I found the Work by saying that there is 600MW of Power a day...so, 6 x 10^8 Watts of Power a day, 3600 times 24 will equal 8.64 x10^4 seconds in a day...

    P=W/T 6 x 10^8W (8.64 x 10^4s) = Work

    And then I have to solve for Qc for thermal energy waste.
    Using the first law Qh= W + Qc

    And after much thought I think I got the right answer...lol.
     
  5. Nov 23, 2004 #4

    Clausius2

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    Of course, I'm not going to check the numerics, but if you want an advice, think of why did you used the Carnot efficiency. It is a misconception you have to be sure you have understood why it is wrong. All of student begineers here seem to run automatically towards Carnot efficiency when you hear the word efficiency. It is not the first time.
     
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