# Physics

Water drives a waterwheel (or turbine) of radius 3.0m. The water enters at a speed of v1 = 7.0 m/s and leaves at a speed of v2= 3.0 m/s.
A) if 150kg of water passes through per second, what is the rate at which the water delivers angular momentum to the waterwheel? [answer: 1800 kg m^2/s^2]

B) what is the torque the water applies to the waterwheel? [answer: 1800 m N]

C)If the water causes the waterwheel to make on revolution every 5.5 s, how much power is delivered to the wheel?

Can someone tell me what equations to use or how to approach this because I have no idea.

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1. A and B are really asking the same question, and the results are the same, since 1 N = 1 kg m/s^2.

Assume that the water enters and leaves the waterwheel tangent to the wheel. The loss of angular momentum of the water, measured about the axis of the wheel, equals the TORQUE imparted to the wheel. The water wheel is probably NOT, in steady state, gaining angular momentum due to the action of the water; the water is instead overcoming friction or an electrical generaltor torque.

Ther torque provided by the water is
(mass flow rate)*(change in V)*R
= 150 kg/s* 4 m/s * 3.0 = 1800 N m

C) Power = Torque*(angular velocity)
= 1800 M n* (2 pi rad/5.5 s) = 2056 W.
They want you to use only two significant figures, so that is why they say the answer is 2100 W.

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posted by drwls
2. GOOD GOD, that was something.

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