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Homework Help: Energy Balance on a pump with sea water

  1. Sep 16, 2008 #1
    My main question in order to solve this problem is...should kinetic energy be neglected? Ok, so here's the problem question:

    Sea water with a density of rho = 1025 (kg/m^3) flows through a pump of 0.21 (m^3/s). The pump inlet is 0.25 (m) in diameter. At the inlet the pressure is -0.15 (m of Mercury). The pump outlet is 0.152 (m) in diameter, is 1.8 (m) above the inlet. The outlet pressure is 175 (kPa). If the inlet and exit temperature are equal, how much power does the pump add to the fluid?

    So I've done an energy balance on the system (which would be the pump itself. Which I constructed to look like a U to account for the 1.8m distance). Thus, work is negative in this case because work is being done on the system. That said...I neglect internal energy changes because temperature is constant so I'm left with all things I know except for entering and exiting velocity of the fluid which is in the kinetic energy term for the energy balance. So...should I neglect kinetic energy? Because the temperature is the same at the inlet and the outlet and because it's a pump system??

    Many thanks in advance!
  2. jcsd
  3. Sep 16, 2008 #2


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    Homework Helper

    Welcome to PF!

    Hi biomedchem! Welcome to PF! :smile:

    You have the volume per second …

    and you can calculate the area of each pipe …

    so you can find the velocity through each pipe! :wink:
  4. Sep 16, 2008 #3
    Agreed! Very much agreed! I just don't know if whether or not I should neglect the kinetic energy term? I at first figured that I should...because the other terms are usually much larger than the kinetic energy term so that I could in fact just 'forget about it'. But since power is being supplied to the fluid and the diameter at the inlet varies from the diameter at the outlet then I talked myself out of neglecting the kinetic energy term. But then, am I thinking about this the right way?

    My energy balance looks like this: where 2 denotes the outlet and 1 denotes the inlet.

    -(work rate, Ws) = [ g*(y2) + (((v2)^2)/2) + ((P2)/rho)]*(rho*v2*A2) - [ g*(y1) + (((v1)^2)/2) + ((P1)/rho)]*(rho*v1*A1)

    and I said that because energy is being done on the system (the fluid) so that Ws < 0. Also, we can find v2 and v1 easily, but both (rho*v2*A2) and (rho*v1*A1) end up giving me (rho*0.21) so that I can simplify my energy balance to:

    Ws = (rho*0.21) * [ g*(y2 - y1) + (0.5*[((v2)^2) - ((v1)^2))]) + ((P2 - P1)/rho)]

    y1 is 0, y2 is 1.8...etc. We have everything (including P1...where I understood the negative pressure to be gauge pressure so I calculated the absolute pressure) to calculate the power which is just Ws since power is energy per time. So...without neglecting the kinetic energy term this is how much power is added to the fluid right? Am I going about this right way?

    Again, maaaany thanks!
  5. Sep 16, 2008 #4


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    If the kinetic (or potential) energy per unit mass is small when compared to the enthalpy, one typically ignores the kinetic (or potential) energy.

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