What is the maximum speed of water flow in the intake pipe?

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The discussion centers on calculating the maximum speed of water flow in an intake pipe located 11.5 meters below a reservoir surface, using Bernoulli's principle. Participants express confusion over the limited data provided for the calculation. They reference Bernoulli's equation to relate pressure, velocity, and height, and discuss the importance of ignoring certain variables to simplify the equation. The conversation includes attempts to substitute known values, such as atmospheric pressure and water density, into the equation to solve for velocity. Ultimately, the focus is on applying Bernoulli's principle correctly to determine the maximum flow speed in the intake pipe.
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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:
 
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Bernoulli's Equation

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

Well you know the density of the water and atmospheric pressure.
 
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 ?
 
Ignore the speed of the water at the surface. (Assume the surface area is much greater than the cross-sectional area of the pipe.)
 
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...
 
It should be:

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