(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A pump lifts a liquid of density [itex]\rho[/itex] to a height h and accelerates it from rest to a final velocity v.

What power P does the pump deliver to the liquid, if the liquid is being pumped through

a pipe with a cross sectional area A?

1. Av[itex]\rho[/itex] + [itex]\rho[/itex]gh

2. Not enough information is provided.

3. Av[itex]\rho[/itex](gh + 1/2 v^{2})

4. 1/2 Av^{2}[itex]\rho[/itex]gh

5. Av[itex]\rho[/itex](1/2 v^{2}) - gh

6. Av[itex]\rho[/itex](gh - 1/2 v^{2})

7. 1/2 Av^{3}[itex]\rho[/itex]

8. Av[itex]\rho[/itex]

9. Zero, of course; a liquid will naturally flow in that way without any pumping.

10. Av[itex]\rho[/itex]g h

2. Relevant equations

Bernoulli's principle: P_{1}+ 1/2 [itex]\rho[/itex]v_{1}^{2}+ [itex]\rho[/itex]gh_{1}= P_{2}+ 1/2 [itex]\rho[/itex]v_{2}^{2}+ [itex]\rho[/itex]gh_{2}

P = W/t = [itex]\Delta[/itex]E/t = F dot v

W = [itex]\int[/itex]F dh

v^{2}= 2ah

h = 1/2 at^{2}

3. The attempt at a solution

I'm actually not sure if kinematics can be applied to a fluid. My attempts have so far eliminated answers 1 and 10, though I came up with 10 using the integral of the force F = PA = [itex]\rho[/itex]ghA. Integration gives W = [itex]\rho[/itex]gh^{2}A, and since h/t = v, P = Av[itex]\rho[/itex]gh.

I think what I forgot to take into account was the acceleration-- the pump not only did work to lift the liquid, but also accelerated it. To try to incorporate a change in kinetic energy into the work term, I used

E_{2}= P_{2}+ 1/2 [itex]\rho[/itex]v^{2}+ [itex]\rho[/itex]gh

E_{1}= P_{1}

[itex]\Delta[/itex]E = P_{2}- P_{1}+ 1/2 [itex]\rho[/itex]v^{2}+ [itex]\rho[/itex]gh = [itex]\rho[/itex]gh + 1/2 [itex]\rho[/itex]v^{2}+ [itex]\rho[/itex]gh

Playing around with kinematics I got a = v^{2}/(2h) and t = 2h/v, but all of that together isn't leading me to an answer choice given.

Any advice is much appreciated!

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# Homework Help: Power of a fluid pump

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