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

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The maximum speed of water flow in an intake pipe located 11.5 meters beneath the surface of a reservoir can be calculated using Bernoulli's Equation. The equation states that the sum of pressure energy, kinetic energy, and potential energy remains constant. By substituting known values such as atmospheric pressure (1.013 x 10^5 Pa), water density (1000 kg/m³), and gravitational acceleration (9.8 m/s²), one can derive the maximum velocity of water flow through the pipe. The discussion emphasizes the importance of ignoring the speed of water at the surface due to the larger surface area compared to the pipe's cross-section.

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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
 
Last edited:

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