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Chapter 3: Problem 106

When the drain plug is pulled, water flows from a hole in the bottom o a large, open cylindrical tank. Show that if viscous effects are negligible and if the flow is assumed to be quasisteady. then it takes 3.41 times longer to empty the entire tank than it does to empty the first half of the tank. Explain why this is so.

Short Answer

Expert verified
The procedure shows that it takes approximately 3.41 times longer to empty the entire tank than to empty the first half. The discrepancy is due to the fact that the speed of the efflux decreases as the fluid level drops, due to a smaller pressure head.

Step by step solution

01

Define the Equations

Start by writing out Torricelli's Law which gives the speed of efflux from a hole under a fluid column as \(v = \sqrt{2gh}\), where \(v\) is the speed, \(g\) is the acceleration due to gravity, and \(h\) is the height of fluid column above the hole. The flow rate from the hole then, is given by \(Q = Av = A\sqrt{2gh}\), where \(A\) is the cross-sectional area of the hole.
02

Form the Equation for Water Level Height

The rate at which the water level drops is given by \(-dh/dt = Q/A_{tank}\), where \(A_{tank}\) is the cross-sectional area of the tank (assumed to be greater than \(A\)). This leads to a differential equation: \(-dh/dt = (A/A_{tank})\sqrt{2gh}\).
03

Solve the Differential Equation

Rearranging and integrating both sides gives: \( \int_{H/2}^H \frac{dh}{\sqrt{h}} = \int_0^t (A/A_{tank}) \sqrt{2g} dt\). Solving both integrals gives \(2(\sqrt{H} - \sqrt{H/2}) = (A/A_{tank}) \sqrt{2g} t.\) Solve for \(t\) to find the time required to drain from \(H/2\) to \(H\).
04

Calculate Time to Drain Entire Tank

Similarly, you can find the time to drain the entire tank by changing the limits of the first integral to \(0\) and \(H\). This gives \(2\sqrt{H} = (A/A_{tank}) \sqrt{2g} t*. \) Solving for \(t*\) gives the time to empty the entire tank.
05

Compare the Two Times

Divide the two times found to show that it takes \(t*/t = \frac{2(\sqrt{H} + \sqrt{H/2})}{2\sqrt{H}} = 1 + \sqrt{1/2} ≈ 3.41\) times longer to empty the entire tank than to empty the first half.

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