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Thermodynamic prob

  1. Jan 18, 2012 #1
    A heat Engine recieves 6 kg/h gasoline which has a calorific value of 42000 kj/kg. If heat rejected from the engine is 40 kw, calculate
    a- the power done by the engine
    b- the engine effiecincy

    -this wt i got today in my exam and unfortuntely i didnt study it well but i did the following :
    1- 6 kg/h -----> 1/600 kg/s
    then i multiplied it by the 42000 kj/kg
    so i can get kj/s ( which is the power )
    so is that correct ??

    2- i said effieciency = w/Qh = 1 - Qh/Ql

    and Qh = W - Ql = 42000 - 40

    so is that too correct ?

    * i knw i didnt study well :cry: and sorry if i posted it in a wrong section
     
    Last edited: Jan 18, 2012
  2. jcsd
  3. Jan 18, 2012 #2
    Efficiency is 1 - Ql/Qh
     
  4. Jan 18, 2012 #3
    thanks for quick reply
    sure i knw that
    but my question is : wt i wrote is right or wrong ?? bcse i got diff units and no more details
     
  5. Jan 18, 2012 #4
    The output of the engine is Qh-Ql. Multiplying fuel rate of use by caloric content only provides Qh. You must subtract Ql from it to arrive at power output. Ql is heat rejected and it is lost.

    You have the equation for efficiency wrong. You wrote
    "2- i said effieciency = w/Qh = 1 - Qh/Ql "
     
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