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Power of Water Pump (from continuity?)

  1. Dec 19, 2006 #1
    1. The problem statement, all variables and given/known data

    Water is pumped steadily out of a flooded basement at a speed of 5.5 m/s through a uniform hose of radius 1.2 cm. The hose passes out through a window 2.9 m above the waterline. What is the power of the pump?

    2. Relevant equations

    [tex]R_m=\rho Av[/tex]

    where P is power, W is work, and K is kinetic engery:
    [tex]P=\frac{W}{\Delta t}[/tex]
    [tex]W=\Delta K=\frac{1}{2}mv^2-\frac{1}{2}mv_0^2[/tex]

    3. The attempt at a solution

    [tex]R_v=Av=\pi r^2v=(\pi)(.012 m)^2(5.5 m/s)=0.002488 m^3/s[/tex]

    [tex]R_m=\rho Av=(1000 kg/m^3)(0.002488m^3/s)=2.488kg/s[/tex]

    [tex]Power=\frac{W}{\Delta t}=\frac{\Delta K}{\Delta t}=(\frac{1}{2}m{v_f}^2-\frac{1}{2}m{v_i}^2)/(\Delta t)[/tex]

    Take delta time to be 1s, initial velocity (water sitting in basement) to be 0m/s, final velocity (water in pipe) to be 5.5m/s, mass dealt with in one second is 2.488kg.
    NOTE: I think my error lies here; something was assigned a totally evil value. Probably my choices of velocity.

    But going with the values I picked, I got [tex]P=(\frac{1}{2})(2.488 kg/s)(5.5 m/s)^2=37.63W[/tex]

    This be wrong. I be sad.
  2. jcsd
  3. Dec 19, 2006 #2


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    As a first thought, use Bernoulli to calculate the pressure at the pump knowing the velocities and elevation changes. Use the pressure to calculate power via
    [tex]P = pQ[/tex] where

    [tex]P[/tex]= Power
    [tex]p[/tex]= pressure
    [tex]Q[/tex] = volumetric flow rate
  4. Dec 19, 2006 #3

    Andrew Mason

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    You seem to be overlooking the [itex]\rho gh [/itex] term.

  5. Dec 19, 2006 #4


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    The [tex]\rho g h[/tex] is taken care of in Bernoulli in the [tex]\gamma \Delta z[/tex] terms.
  6. Dec 19, 2006 #5
    I don't think I know that much Bernoulli... the only Bernoulli equation that we've covered in class so far is[tex]p+\frac{1}{2}\rho v^2+\rho gy=constant[/tex]. And the problem claims that it can be solved without even that. Sigh.
  7. Dec 19, 2006 #6


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    That's just a different form of it. That is correct. Like I mentioned, you can calculate the pressure at the pump, i.e. P1 from Bernoulli. You know the constant velocity and density, so the two terms with velocity will drop out.

    Once you get the pressure at the pump from the Bernoulli equation, use the second equation I mentioned to calculate the power.

    A quck run through gets about 71 W. Is that close to your answer?
    Last edited: Dec 19, 2006
  8. Dec 20, 2006 #7

    Andrew Mason

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    The power is the kinetic + potential energy added to the water per unit time. You have figured out the kinetic energy/time at 37.6 W. What is the potential energy added per unit time?

  9. Dec 20, 2006 #8
    A simple equation for power, P, is:

    P = Q * ((gamma)force density) * h

    I get 70.7 W

    Of course this doesn't account for any parasitic losses.
  10. Dec 20, 2006 #9
    No matter what I do, I keep getting Power in the hundreds of Watts.

    Here's the latest work (and thank you for putting up with me!)

    Where the 1's are the pipe at the waterline, and the 2's are the top of the pipe:
    [tex]p_1+\frac{1}{2}\rho {v_1}^2+\rho gy_i=p_2+\frac{1}{2}\rho {v_2}^2+\rho gy_2[/tex]

    Since v is constant along the pipe, the 0.5*rho*v^2 terms cancel.
    Let the waterline be 0m.

    [tex]p_1=p_2+\rho gy_2=10^5 Pa + (1000kg/m^3)(9.81m/s^2)(2.9m) = 128449 Pa[/tex]

    OK. Now to find the force that the pipe exerts.

    [tex]p=\frac{F}{A}[/tex] so [tex]F=pA=(p)(\pi r^2)=(128449 Pa)(\pi)(.012 m)^2=58.11N[/tex]

    Lastly (*gasp*) [tex]Power=Fv=(58.11 N)(5.5 m/s)=319.6 W[/tex]

    And I lose yet again...
    Last edited: Dec 20, 2006
  11. Dec 21, 2006 #10

    Andrew Mason

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    One kg of stationary water in the basement is lifted 2.9 m. and moved to a speed of 5.5 m/s. This is an increase of mgh + .5mv^2 = 1*9.8*2.9 + .5*1*5.5^2 = 28 + 15 = 43 Joules. Since the flow rate is 2.5 kg/sec., the power is 43*2.5 = 107.5 Joules/second.

    Last edited: Dec 21, 2006
  12. Dec 21, 2006 #11


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    You're close. Bernoulli will reduce to [tex]p_1-p_2 = \rho g(y_2-y_1)[/tex]

    The [tex]p_1-p_2[/tex] term is equal to the pressure that the pump is putting out since the pipe is exiting to atmospheric pressure.

    Your next step is where you went wrong. That is not the way to calculate pump power. The power from the pump is a function of pressure and flow rate. Look at the equation that has been given to you twice:

    [tex]P = p Q[/tex]
  13. Dec 21, 2006 #12
    Yessir. Thank you, sir. Sorry for disregarding you, sir.
    Now that the final's over, I no longer have any compunctions about using equations not covered in class or the textbook :wink:
  14. Dec 21, 2006 #13


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    I agree with Andrew's calculation:

    power = (GPE created per second) + (KE created per second)
    = 108.39... Joules per second

    (+ losses due to friction, etc)
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