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Thermodynamics- determining mechanical power

  1. Sep 8, 2013 #1

    MacLaddy

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    1. The problem statement, all variables and given/known data

    Hello all,

    I've worked through the problem below, but I feel my answer is incorrect. I would appreciate it if someone could check my work.

    Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20kW of shaft power. The free surface of the upper reservoir is 45m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m^3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.


    2. Relevant equations

    [itex]e_m = ke =(\frac{P}{\rho})(\frac{1}{2}\vec{V}^2)(gz)[/itex]

    [itex]\dot{E}=\dot{m}e_m[/itex]

    [itex]\dot{m}=\rho\dot{V}[/itex]

    [itex]\eta=\frac{energy-out}{energy-in}[/itex]


    3. The attempt at a solution

    [itex]e_m = ke =(\frac{P}{\rho})(\frac{1}{2}\vec{V}^2)(gz)=gz=(9.81\frac{m}{s^2})(45m)=441.45\frac{m^2}{s^2}=441.45\frac{J}{kg}[/itex]

    [itex]\dot{m}=\rho\dot{V}=(1 \frac{kg}{m^3})(0.03\frac{m^3}{s})=0.03\frac{kg}{s}[/itex]

    [itex]\dot{E}=\dot{m}e_m=(0.03\frac{kg}{s})(441.45\frac{J}{kg})=13.24\frac{J}{s}=13.24W = 13.24*10^{-3}kW[/itex]

    [itex]\eta=\frac{13.24*10^{-3}kW}{20kW}=.66*10^{-3}=0.066[/itex]%

    [itex]20kW-20\eta= 19.99 kW[/itex]


    So 19.99 kW out of 20 are wasted to heat in this problem? doesn't seem correct.
     
  2. jcsd
  3. Sep 8, 2013 #2

    SteamKing

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    You have the density of water as 1 kg / m^3. This is some really, really light water.
     
  4. Sep 8, 2013 #3

    MacLaddy

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    Ha, how did I miss that? Thanks.

    How's the rest of my method look?
     
  5. Sep 8, 2013 #4

    Ygggdrasil

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    You have the wrong value for the density of water. [edit: nevermind, Steam King beat me to it]
     
  6. Sep 8, 2013 #5

    MacLaddy

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    Going through it with the proper density, I come up with 6.76 kW lost to heat. Sound about right?
     
    Last edited: Sep 8, 2013
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