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chattkis3
Nov7-04, 07:27 PM
A pump and its horizontal intake pipe are located 11.5 m beneath the surface of a reservoir. The speed of the water in the intake pipe causes the pressure there to decrease, in accord with Bernoulli's principle. Assuming nonviscous flow, what is the maximum speed with which water can flow through the intake pipe?

--Im getting confused because they give you so few numbers to work with. Any ideas? Thanks, --

:smile:

Pyrrhus
Nov7-04, 07:37 PM
Bernoulli's Equation

P + \frac{1}{2} \rho v^2 + \rho gh = constant

Well you know the density of the water and atmospheric pressure.

chattkis3
Nov7-04, 07:58 PM
Ok so this is what I have: (P1) + ((1/2)density*v1^2) + (density*g*h1) = (P2) + ((1/2)density*v2^2) + (density*g*h2)

So for this question I can ignore the left side of this longer version of Bernoulli's? Do I substitute in a value of 1 or 0 so that I can solve for v ?

Doc Al
Nov7-04, 08:20 PM
Ignore the speed of the water at the surface. (Assume the surface area is much greater than the cross-sectional area of the pipe.)

chattkis3
Nov7-04, 08:41 PM
Ok am I on the right track with this? :

1.013 x 10^5 Pa = (1.013E5Pa + (1000 kg/m3*9.8m/s2*11.5m)) + .5(1000 kg/m3)* v^2 + (1000 kg/m3 *9.8m/s2 *11.5m)

Im trying to plug in what I know with Bernoulli's...

Pyrrhus
Nov8-04, 10:00 AM
It should be:

P_{o} + \rho g h = P_{pipe} + \frac{1}{2} \rho v^2