Thermodynamics - efficiency question

  1. The burning of gasoline in a car releases about 3.0 x 10^4 kcal/gal. If a car averages 41 km/gal when driving 90 km/h, which requires 25 hp, what is the efficiency of the engine under those conditions?

    I think I have an answer, but I'm not sure if it's correct.

    1 horsepower = 746 watts

    25 hp = 18650 watts

    1 kcal = 4186 joules (watts / sec)

    e = W / Qh = 1 - (Ql / Qh)

    1 gal / 41 km at 91 km / h means that that 2.26 gallons will be used in one hour.

    Since no temperatues are given I don't think that the 1 - Ql / Qh is needed...

    Converting 30000 kcal / gal to watts I get 34883 joules/sec. Since 2.26 gal are used, I multiplied the above by 2.26 to get 78836.

    Dividing 18650 watts (from hp) by 78836 I got an efficiency of 23.46%.
     
  2. jcsd
  3. Andrew Mason

    Andrew Mason 6,801
    Science Advisor
    Homework Helper

    You don't need to find Qc but it is just:[tex]Qc = Qh - W[/tex]. You don't need temperatures to find this.

    You should state your answer algebraically so you and others can follow the physical reasoning. ie:

    [tex]\eta = W/Q_h = (dW/dt)/(dQ_h/dt)[/tex]

    Your answer is almost right. I get 2.195 gal/hr (90/41) not 2.26.

    [tex]dQ_h/dt = 2.195 * gal/hr = 3 x 10^4 * 2.195 * 4.186 KJ/hr = 2.76 x 10^8 J/hr = 7.66 x 10^4 J/sec[/tex]

    [tex]dW/dt = Power = 25 * 746 J/sec = 1.87 x 10^4 J/sec[/tex]

    So:

    [itex]\eta[/itex] = 1.87/7.66 = 24.4%

    AM
     
    Last edited: May 14, 2006
  4. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    I have not rigorously checked your answer, however the efficency appears reasonable for a car, just beware of rounding too early in your calculations, this could induce significant errors. For example you obtained 78836 watts for the input power of the engine, however I obtained [itex]78798\frac{2}{3}[/itex].

    Just a small point that I sould point out is that here you said;
    I'm sure this is just a typo, but this should be joules = watts * sec. Power is work done (energy) divided by time, therefore it follows that energy is the product of power and time.

    ~H

    Sorry AM, didn't see your post. I sould learn to type faster :wink:
     
  5. Thank you.
     
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