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Calculate pumping power

  1. Jul 11, 2017 #1
    1. The problem statement, all variables and given/known data
    CCcyOXN.png
    (question one is irrelevant for my problem).

    What is the pumping power? From energy balance I got a negative value - but it should be positive since the mechanical work from pump is given into system.

    2. Relevant equations
    The input+output velocity:
    $$v=\frac{\dot{V}}{A}$$
    The friction coefficient:
    $$f=\frac{\left |\Delta p \right |}{\frac{L}{D}\frac{1}{2}\rho v^{2}}$$
    Moody diagram (not shown here).
    The energy balance of open system:
    $$\dot{W}=\dot{m}\cdot \left (h_{out}-h_{in}+\frac{v^{2}_{out}-v^{2}_{in}}{2}+g\cdot (z_{out}-z_{in}) \right )$$
    The specific enthalpy:
    $$h=u+\frac{p}{\rho }$$

    3. The attempt at a solution
    The velocity of input and output is ##v=5\,m\cdot s^{-1}##.
    From calculated Reynolds number and relative roughness I found friction coefficient ##f=0.02## in Moody diagram (that is a correct value).
    The pressure loss due to friction was calculated from the given equation: ##p_{F,out}-p_{F,in}=-5MPa##.
    The hydrostatic pressure loss is ##p_{H,out}-p_{H,in}=\rho \cdot g\cdot (z_{out}-z_{in})=-177kPa##.
    The total pressure loss is ##p_{out}-p_{in}=p_{F,out}-p_{F,in}+p_{H,out}-p_{H,in}=-5000000-177000=-5177kPa##.
    The enthalpy change between outlet and inlet is:
    ##h_{out}-h_{in}=\frac{p_{out}-p_{in}}{\rho }=\frac{-5177000}{1000}=-5177\,J\cdot kg^{-1}##.
    The change of potential energy between outlet and inlet is:
    ##e_{p,out}-e_{p,in}=g\cdot (z_{out}-z_{in})=9.81\cdot 18=174.58\, J\cdot kg^{-1}##
    The resulting power of the pump is:
    ##\dot{W}=1000\cdot 1\cdot (-5177+174.58)=-5MW##
     
  2. jcsd
  3. Jul 11, 2017 #2
    You should be using the mechanical energy balance equation, not the thermal energy balance. The mechanical energy balance is a modified version of the Bernoulli equation that includes frictional energy dissipation.
     
  4. Jul 12, 2017 #3
    But I don't know why I cannot use the energy balance. There are no restrictions to use energy balance. I thought a shaft work in thermodynamics is defined ##dW_{shaft}=\vec{dF}_{ext}\cdot \vec{dl}##, i.e. ##W_{shaft}>0## for pressure forces acting on the system.
     
  5. Jul 12, 2017 #4
    Your problem is with the enthalpy change. Also, isn't zout higher than zin?
     
  6. Jul 12, 2017 #5
    The enthalpy change ##h_{out}-h_{in}## should be negative since ##p_{out}-p_{in}## is negative:
    ##h_{out}-h_{in}=\frac{p_{out}-p_{in}}{\rho }=\frac{-5177000}{1000}=-5177\,J\cdot kg^{-1}##.

    It is, which means that hydrostatic pressure difference ##p_{H,out}-p_{H,in}## is negative and potential energy difference ##e_{P,out}-e_{P,in}## is positive.
     
  7. Jul 12, 2017 #6
    If you have a fluid flowing through an insulated pipe, and there is a pressure drop, according to the open system version of the first law, what is the enthalpy change?
     
  8. Jul 12, 2017 #7
    The enthalpy change is:
    $$h_{out}-h_{in}=u_{out}-u_{in}+\frac{p_{out}-p_{in}}{\rho }$$

    Since all power given to the system increases the mechanical energy of a fluid, the change of internal energy should be zero (there is no change of temperature nor compression of a fluid).
     
  9. Jul 12, 2017 #8
    I don't understand. What is your answer?
     
  10. Jul 12, 2017 #9
    The first law for open system is:
    $$\dot{U}_{out}+\dot{E}_{k,out}+\dot{E}_{p,out}-\dot{U}_{in}-\dot{E}_{k,in}-\dot{E}_{p,in}=\dot{Q}+\dot{W}_{shaft}+p_{in}\cdot \dot{V}_{in}-p_{out}\cdot \dot{V}_{out}$$

    That is:
    $$\dot{H}_{out}+\dot{E}_{k,out}+\dot{E}_{p,out}-\dot{H}_{in}-\dot{E}_{k,in}-\dot{E}_{p,in}=\dot{Q}+\dot{W}_{shaft}$$
     
  11. Jul 12, 2017 #10
    So you are saying that ##\Delta h=0## because ##\Delta u=0## and ##\Delta p/\rho=0## in my pipe example?
     
  12. Jul 12, 2017 #11
    No ##p_{out}-p_{in}## is nonzero which means that ##\Delta h\neq 0##.
     
  13. Jul 12, 2017 #12
    That is not correct.
     
  14. Jul 12, 2017 #13
    I don't understand now. You stated that "there is a pressure drop". That should have impact on enthalpy. Can you explain?
     
  15. Jul 12, 2017 #14
    Delta h is equal to zero. And, as you said, delta p/ rho is not equal to zero. So what does that tell you?
     
  16. Jul 12, 2017 #15
    It tells us that change of internal energy is nonzero. But how is it possible?
     
  17. Jul 12, 2017 #16
    It is due to viscous frictional heating. For a pressure drop of 50 psi, how much of a temperature rise does that amount to, assuming the fluid is water?
     
  18. Jul 12, 2017 #17
    It will be much dissipated heat :) Also the situation with nonzero pressure drop but same temperature on inlet and outlet and adiabatic conditions - it does not make sense (the shaft work cannot be determined from energy balance)?
     
  19. Jul 12, 2017 #18
    I calculate a temperature rise of only 0.1 C. So it is nearly isothermal, but not quite. So on the thermal energy balance, the mechanical energy dissipation is practically negligible, but in the mechanical energy balance, the thermal term would play a huge role. This is why you can't use the version with h to solve a flow problem.
     
  20. Jul 12, 2017 #19
    If I neglect the friction with pipe wall, I get the shaft work:
    ##\dot{W}=(p_{in}-p_{out})\cdot A\cdot v=5177000\cdot 0.2\cdot 5=5177000\,W##
     
  21. Jul 12, 2017 #20
    See the most recent thread started by MexChemE. It is very relevant to all this.
     
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