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Power Consumption and Entropy Generation

  1. Feb 11, 2016 #1
    1. The problem statement, all variables and given/known data

    2-3-15

    123.jpg

    2. Relevant equations

    P = power. W = work. U = internal energy. S = entropy. t = time. Q = heat. T = temperature. F = force. d = distance.

    P = ΔW/Δt= ΔU/Δt

    ΔS = ΔQ/T

    dm/dt = ρ⋅dV/dt

    W = F ⋅ d

    ΔU = Q - W

    Where m is mass, V is volume, and ρ is the density of the fluid. Water ρ = 1000 kg/m3
    3. The attempt at a solution

    Since Uint = 4.2T
    Uin = (4.2)(300K) = 1260J
    Uout = (4.2)(370K) = 1554J

    P = ΔU/Δt = (1554J - 1260J) = 294 J/sec

    I don't know if this is correct. And I'm not sure how to approach (b), regarding entropy generation.
     
    Last edited: Feb 11, 2016
  2. jcsd
  3. Feb 11, 2016 #2
  4. Feb 11, 2016 #3
    I'm really lost. Any hints? Maybe this should be in engineering. It's for my Mechanical Engineering 205 class...
     
  5. Feb 12, 2016 #4

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    Gold Member

    The problem statement did tell you that the specific internal energy and specific entropy were [itex]u = 4.2 T [/itex] and [itex]s = 4.2 \ln T [/itex] respectively. Let's just concentrate on the specific internal energy for now.

    The "specific" part of specific internal energy tells you that it's quantifying the energy per amount of stuff. This amount of stuff is usually measured in units of kilograms (kg) in problems like these, but don't be surprised in future coursework if it's measured in moles, pounds, tons, etc.

    So here, [itex] u [/itex] isn't in units of energy. But rather it's in units of energy per unit mass. I'm guessing, probably J/kg.

    Also, you haven't figured the change in time into your equations either. You've suddenly converted units of J to J/sec without any rationale.

    Figure out both of those, and I suspect you'll get the right answer. :smile: [Hint: if you're clever, you might be able to figure them both out in a single step.]
     
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