- #1

cepheid

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According to Torricelli's theorem, the velocity of a fluid draining from a hole in a tank is v ~= (2gh)

^{1/2}, where h is the depth of water above the hole. Let the hole have an area A

_{0}, and the cylindrical tank have cross-sectional area A

_{b}>> A

_{0}. Derive a formula for the time to drain the tank completely from an initial depth h

_{0}.

My work:

The volume flow rate out of the hole is equal to the rate of change of the volume in the tank:

[tex] A_0 v = \frac{dV}{dt} = A_b \frac{dh}{dt} [/tex]

[tex] v = \sqrt{2gh} = \frac{A_b}{A_0} \frac{dh}{dt} [/tex]

Assuming that we start from t

_{0}= 0, and that the tank is drained after a time T, we can separate variables and integrate:

[tex] \int_0^T {dt} = T = \frac{A_b}{A_0} \int_{h_0}^0 {\frac{dh}{\sqrt{2gh}}} [/tex]

One thing that bothered me was that I never made use of the information that A

_{b}>> A

_{0}. I thought at first maybe I was supposed to make some approximation somewhere based on that. But then I dug out my first year physics text and saw that Torricelli's theorem was derived from Bernoulli's eqn, and that this information regarding the two areas was used in the derivation. So maybe that's the only reason they gave it to us. Still, is everthing else ok?