Water steadily enters a tank of diameter DT at a flow rate Q. A small hole in th

Question

Water steadily enters a tank of diameter DT at a flow rate Q. A small hole in the bottom of the tank (with diameter DO) allows water to escape. We will show in class that the water velocity through the hole is given by , where H is the depth of water in the tank. If the tank is initially empty, derive an expression for the maximum height (Hmax) that the water will reach in the tank. Check your answer using your intuition: is Hmax proportional to, inversely proportional to, or independent of Dr, Do and Q? Does this make sense? Once this maximum height is reached, the flow coming into the tank (i.e. Q) is turned off. If the hole diameter is one-fifth of the tank diameter, and the maximum height is 50 cm, how quickly will the water surface fall as soon as the incoming flow is turned off?

Explanation / Answer

dV/dt = Q - D02*(2gH)1/2 V = DT2H dH/dt = (Q/DT2) -(D0/DT)2*(2gH)1/2 H is maximized when dH/dt = 0 0 = (Q/DT2) -(D0/DT)2*(2gH)1/2 H = (Q/D02)2 / 2g b) Q = 0 dH/dt = (Q/DT2) -(D0/DT)2*(2gH)1/2 dH/dt = -(D0/DT)2*(2gH)1/2 H = 50 cm = 0.5 m D0/DT = 1/5 dH/dt = -(1/5)2*(2*9.8 m/s2 *0.5m)1/2 = -0.125219807 m/s ~ 12.5 cm/s

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