- #1
symphwar
- 9
- 0
Homework Statement
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 v2)
4. 1/2 Av2[itex]\rho[/itex]gh
5. Av[itex]\rho[/itex](1/2 v2) - gh
6. Av[itex]\rho[/itex](gh - 1/2 v2)
7. 1/2 Av3[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
Homework Equations
Bernoulli's principle: P1 + 1/2 [itex]\rho[/itex]v12 + [itex]\rho[/itex]gh1 = P2 + 1/2 [itex]\rho[/itex]v22 + [itex]\rho[/itex]gh2
P = W/t = [itex]\Delta[/itex]E/t = F dot v
W = [itex]\int[/itex]F dh
v2 = 2ah
h = 1/2 at2
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]gh2A, 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
E2 = P2 + 1/2 [itex]\rho[/itex]v2 + [itex]\rho[/itex]gh
E1 = P1
[itex]\Delta[/itex]E = P2 - P1 + 1/2 [itex]\rho[/itex]v2 + [itex]\rho[/itex]gh = [itex]\rho[/itex]gh + 1/2 [itex]\rho[/itex]v2 + [itex]\rho[/itex]gh
Playing around with kinematics I got a = v2/(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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