Posted by **Ashley** on Friday, November 9, 2012 at 7:04pm.

Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.70 105 Pa and the pipe radius is 2.50 cm. At the higher point located at y = 2.50 m, the pressure is 1.30 105 Pa and the pipe radius is 1.60 cm. (a) Find the speed of flow in the lower section.

(b) Find the speed of flow in the upper section.

(c) Find the volume flow rate through the pipe.

- Physics -
**drwls**, Friday, November 9, 2012 at 7:24pm
(a) P + (1/2)*rho*V^2 + (rgo)*g*y = constant. That's Bernoulli's law, neglecting fluid frictional pressure drops, for incompressible flow. Elevation is y = 2.50 m on the high side, and is zero on the low side.

You know the pressure on both sides, and that

V(high) = V(low)*[(2.5)/(1.6)]^2

= 2.44 V(low)

Now you can solve for V(low).

(b) I already told you the relation between V(high) and V(low). It comes from the continuity equation

(c)(Volume flow rate) = pi*R^2*V

You can used data from either the high or low side.

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