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Homework Help: Problem on the second law of thermodynamics

  1. Feb 14, 2005 #1
    A locomotive steam engine opperating with a total train mass of 99.9x10^6 kg climbs a hill of height 1000m in 10 minutes. If the steam engine produces steam at a temperature of 390k and exhausts steam at 300k, and if the engine were at the theoretically best efficiency possible, what would be the rate, in mass per unit time, at which the same is exhausted be?

    I figured out the efficiency, and the work done by the try... I get the heat in, then use that to get the heat exhausted (I get 4.22x10^11 J for reference).. but then i'm stuck. It's more not really understanding the question than a physics then.. I just don't really know what to do with this, almost like I can't figure out how I should manipulate the units :yuck:
  2. jcsd
  3. Feb 14, 2005 #2


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    I didn't understand what the problem's asking...You can find:heat consumed,work done and heat exhausted.Now,what do they mean by "the same is exhausted"...?And mass of what...?

  4. Feb 14, 2005 #3


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    The problem requests you determine the Mass/Time ([tex]\breve{M}[/tex]) flow rate of the working fluid (steam) through the turbine (in at 390 degK and exhausted at 300 degK), which produces the work. If [tex]\breve{M}[/tex] is the problem "unknown", how did you calculate all the above mentioned quantities ("heat in", "heat exhausted", etc.) without [tex]\breve{M}[/tex]?? Or to phrase it differently, if you have indeed computed [tex]\breve{M}[/tex], then you've completed the problem, and you should be happy!

    Best way to proceed here is to show your work, both symbolic equations and subsequent substitutions, and in particular, how you computed [tex]\breve{M}[/tex], which is the problem requirement.

    (Hint: For the theoretical Rankine Cycle ("Steam Engine"), the Turbine segment is considered Isentropic and Adiabatic. {Heat Capacity of Steam}={2x10^3 Joules/Kg/degK} )

    d - Didn't see your msg. Maybe we're asking the same thing!!
    Last edited: Feb 14, 2005
  5. Feb 14, 2005 #4
    You see, that's is the same exact problem I'm having. Like I said, it's not as much as a problem with the physics as with the wording. I have no clue what the problem is asking.

    Interesting, the answer in the back of the book is "270 kg/s"... I have no idea where they got that from. The question I posted is word for word, so I'm afraid I can't really clarify further.
  6. Feb 14, 2005 #5

    Andrew Mason

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    I believe the maximum possible theoretical efficiency of a steam engine is the Carnot cycle:
    [tex]\eta = \frac{T_H - T_C}{T_H}[/tex]

    which, for these temperatures is 23.1%.

    The problem first asks you to find the power (work/time) that the steam engine produces. Then you have to find the amount of work per unit mass of water that the steam engine produces (this is just the heat energy/kg of steam that is converted to work - which is 23.1% of the amount of heat needed to change the temperature of 1 kg of water from 300C to 390K) Once you have those two numbers you can answer the question.

  7. Feb 16, 2005 #6
    Can you clarify the part after finding the power? It seems like one big run on sentence. Is it the amount of (work per unit mass of water) that the engine produces? Or the amount of work per unit (mass of water that the engine produces)? And is which of those is the heat energy/kg of steam converted to work? And is it the (heat energy/kg of steam converted) that is 23.1% of the amount of heat needed to change the temperature 90K? Or is it the heat energy/(kg of steam that is converted to work which is 23.1% of the amount of heat needed to change the temperature 90K)? Thanks.
    Last edited: Feb 16, 2005
  8. Feb 16, 2005 #7

    Andrew Mason

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    No. That makes absolutely no sense.

    The energy input is that required to heat the water from 300-390 K. The energy output is the work, which is 23.1% of the input energy.

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