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Need help with Heat Engines

  1. Nov 6, 2004 #1
    A 20% efficient real engine is used to speed up a train from rest to 5 m/s. It is known that an ideal (Carnot) engine using the same cold and hot reservoirs would accelerate the same train from rest to a speed of 6.5 m/s using the same amount of fuel. The engines use air at 300 K as a cold reservoir. Find the temperature of the steam serving as the hot reservoir.

    I don't know how the velocity comes into play in order to get the temperature of the steam serving as the hot reservoir. I know that the formula for efficiency is
    [tex] e = \frac {W_{eng}} {|Q_{h}|} [/tex].
    Any help would be great! thx! :)
     
  2. jcsd
  3. Nov 7, 2004 #2
    can anyone help ?
     
  4. Nov 7, 2004 #3
    I think I got it maybe...
    Since efficiency of a Carnot engine is
    [tex] e = 1 - \frac {T_{c}} {T_{h}} [/tex]
    and e = .2.
    So, we can reorder the formula to get,
    [tex] T_{h} = \frac {T_{c}} {1-e} [/tex]
    What I am confused about is then why do they give us the initial and final velocity of each engine?
     
  5. Nov 7, 2004 #4

    Clausius2

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    Science Advisor
    Gold Member

    You're wrong and have a misconception. The efficiency of a Carnot engine is not 0.2. Your problem states:

    "A 20% efficient real engine is used to speed up a train from .."

    A real engine is not a Carnot engine. That phrase also not mean the efficiency of the engine is the 20% of the carnot efficiency.

    Let' see:

    [tex] v_1=5m/s[/tex]
    [tex] v_2=6.5m/s[/tex]
    [tex] m=[/tex]mass of the train
    [tex] m_f=[/tex]mass of the fuel employed in combustion.
    [tex] L_i=[/tex] Caloric power or combustion enthalpy.

    From this phrase: "It is known that an ideal (Carnot) engine using the same cold and hot reservoirs would accelerate the same train from rest to a speed of 6.5 m/s using the same amount of fuel. The engines use air at 300 K as a cold reservoir." one can write:

    [tex]\eta_{carnot}=\frac{W}{Q_h}=1-\frac{T_c}{T_h}=\frac{1/2 m v_2^2}{m_f L_i}[/tex] because you need to produce a work equal to the kinetic energy needed, and the combustion heat is the fuel mass multiplied by the combustion enthalpy. So that we have the first equation:

    [tex]1-\frac{T_c}{T_h}=\frac{1/2 m v_2^2}{m_f L_i} [/tex] (1);

    From this phrase: "A 20% efficient real engine is used to speed up a train from rest to 5 m/s." one can write again:

    [tex]\eta_{real engine}=0.2=\frac{W}{Q_h}=\frac{1/2 m v_1^2}{m_f L_i}[/tex] (2)

    Dividing (1) and (2) we have:

    [tex]\frac{1-\frac{T_c}{T_h}}{\eta_{real engine}}=\Big(\frac{v_2}{v_1}\Big)^2[/tex]

    and solve for T_h.

    You should know that going straightforward to the solution is not always the best way for solving a problem. They might be more difficult than we could think.
     
  6. Nov 7, 2004 #5
    thx for the help!
    I did think about the problem first, but thats all I could think of. yea...I need to study more or something...I'm not fully understanding the material...
    anyways, thx for the help again! :)
     
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