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Homework Help: Thermodynamics (Entropy Generation and Heat Engine)

  1. Nov 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A hot gas stream at 600K and 200 kPa is cooled at constant pressure to 300K in a pipe by direct thermal contact with the atmosphere. The mass flow rate of the stream is 0,1 kg/s and the atmospheric temperature and pressure are 300K and 100 kPa. Modeling the gas as an ideal gas with constant specific heat determine;
    a)heat transfer rate to the atmosphere
    b)entropy generation rate associated with the cooling process
    c)if the hot gas stream is used to produce mechanical power by operating a heat engine between the stream and the atmosphere determine the maximum shaft power produced
    d)demonstrate that the mechanical power produced in part c is proportional to the rate of entropy generation in part b

    2. Relevant equations

    3. The attempt at a solution
    Since we don't know what the gas is, i write down c representing Cp (constant pressure problem)
    dT = T2 - T1 (change in temp)
    dS = S2- S1 (change in entropy)

    a) Q=mcdt = 0,1c.(600K-300K) = 30c (heat transfer rate to the atmosphere)

    b) dS = Sgen + (Q/T) so Sgen = dS - (Q/T)
    dS = mcln(T2/T1) = 0,1cln(600/300) = 0.07c
    Sgen = 0.07c - (-30c/300K) = 0.17c (entropy generation)

    What did i miss? I am confused because i didn't make use of pressure throughout my calculations.

    c) max efficiency n = 1 - (Tcold/Thot) = 1 - (300/600) = 0.5
    30c.0,5=15c (maximum shaft work)

    d) i couldn't come up with something on this.
  2. jcsd
  3. Nov 28, 2015 #2
    You didn't miss anything. But you got the temperatures switched in your calculation of ΔS for the gas. It should be -0.07c.
    It seems to me that this part is not correct because the gas temperature is not 600K throughout the pipe. This needs to be taken into account.
  4. Nov 29, 2015 #3
    Thank you for the help.

    As the pipe temperature not being uniform, I've accepted gas stream inlet as a thermal reservoir and calculated from that. But, yes this might not be true.
  5. Nov 29, 2015 #4
    You need to do an integration over the increments of heat leaving the pipe while applying the efficiency equation to each increment. When you do that, you will find that the maximum work is just the cold reservoir temperature (300 K) times Sgen.

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