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What is the entropy increase per kilogram

  1. Apr 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Suppose a power plant delivers energy at 9.7E2 MW using steam turbines. The steam goes into the turbines superheated at 665 K and deposits its unused heat in river water at 298 K. Assume that the turbine operates as an ideal Carnot engine.

    a. If the river flow rate is 37 m3/s, estimate the average temperature increase of the river water immediately downstream from the power plant.

    b. What is the entropy increase per kilogram of the downstream river water in J/kg·K?

    2. Relevant equations

    http://www.physics.iastate.edu/getfile.php?FileID=3135
    http://www.physics.iastate.edu/getfile.php?FileID=3170

    3. The attempt at a solution

    I solved for efficiency using e=1-T_L/T_H

    Then used e=W/Q_H to get Q_H in watts

    I could find Q_L with Q_H=Q_L+W but don't know where to go from there.

    Maybe convert the volumetric flow rate to mass flowrate using density and then somehow use Q=mc(delta T)??
     
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Apr 23, 2007 #2

    Andrew Mason

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    You are on the right track. The question asks you to find the amount of heat flow per second and then to assume that the heat is carried away by 37m^3 of water per second. How much heat is dumped into the water each second (in joules) and what temperature increase does that give to 37,000 kg of water?

    For entropy, just use dS = dQ/T

    The question is somewhat confusing because it tells you that the heat flows into the river at 298 K and then it asks you to find the temperature of the water. But you will see that the temperature increase is very small.

    AM
     
  4. Apr 24, 2007 #3
    So keep going with what I was doing, find Q_L, then Q_L=mc(delta T), and solve for delta T where Q_L is in watts and "m" is in kg per second?

    All those efficiency and Q_H=Q_L+W etc. equations can be in either watts or joules right?

    Thanks,
    Brandon
     
  5. Apr 24, 2007 #4

    Andrew Mason

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    [itex]Q = m(c\Delta T)[/itex] . So differentiating with respect to time: [itex]dQ/dt = dm/dt(c\Delta T)[/itex] You are given dQ/dt (megawatts or 10^6 joules /sec) and dm/dt (convert to kg/sec from m^3/sec) so that should enable you to find [itex]\Delta T[/itex].

    Efficiency can be thought of as the work per unit time divided by heat supplied per unit time.

    AM
     
    Last edited: Apr 24, 2007
  6. Apr 24, 2007 #5
    Calculus... I tutor this introductory physics class and the students haven't even had calculus yet. Any idea of how I can explain this to them without calculus.
     
  7. Apr 24, 2007 #6

    Andrew Mason

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    In time [itex]\Delta t[/itex], the heat flow is [itex]\Delta Q = P\Delta t/e[/itex]

    This same heat flows to the water in the same time period. So:

    [tex]\Delta Q = \Delta m c\Delta T = P\Delta t/e[/tex]

    Now [itex]\Delta m = \dot m \Delta t [/itex] (flow rate x time) so:

    [tex]\dot m c\Delta T = P/e[/tex]

    AM
     
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