Fluid Mechanics: Tricky Questions Solved (5)

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Tags Fluid Mechanics

Question

A cylindrical tank of cross-section A=1A = 1 m2^2 contains water up to a height H=5H = 5 m. A small hole of cross-section a=1a = 1 cm2^2 is punched at the bottom. Find the time taken for the tank to empty completely. Take g=10g = 10 m/s2^2.

Solution — Step by Step

By Torricelli’s law, when water height is hh:

v=2ghv = \sqrt{2gh}

Volume conservation: A ∣dh/dt∣=av=a2ghA\, |dh/dt| = a v = a\sqrt{2gh}.

dhdt=−aA2gh\frac{dh}{dt} = -\frac{a}{A}\sqrt{2gh} dhh=−aA2g dt\frac{dh}{\sqrt{h}} = -\frac{a}{A}\sqrt{2g}\, dt ∫H0h−1/2dh=−aA2g T\int_H^0 h^{-1/2} dh = -\frac{a}{A}\sqrt{2g}\, T

LHS: [2h]H0=−2H[2\sqrt{h}]_H^0 = -2\sqrt{H}.

T=Aa2HgT = \frac{A}{a}\sqrt{\frac{2H}{g}}

A/a=1/10−4=104A/a = 1/10^{-4} = 10^4. 2H/g=10/10=1\sqrt{2H/g} = \sqrt{10/10} = 1 s.

T=104×1=104 s≈2.78 hoursT = 10^4 \times 1 = 10^4 \, \text{s} \approx 2.78 \, \text{hours}

Final answer: T≈10000T \approx 10000 s (≈2.78\approx 2.78 h).

Why This Works

Torricelli’s law gives instantaneous efflux speed; the volume balance turns this into a differential equation in hh. Solving it integrates the variable speed over time.

A common shortcut: the time to empty is 2\sqrt{2} times the time to drop to half height, because T∝HT \propto \sqrt{H}. Useful for cross-checking.

Alternative Method

Use energy conservation at each instant — the kinetic energy of the jet equals the potential energy lost. Same Torricelli result, slightly more setup.

Students plug v=2gHv = \sqrt{2gH} once and divide volume by flow rate: T=AH/(a2gH)T = AH/(a\sqrt{2gH}). This gives half the right answer because it assumes constant speed, but speed decreases as water drops.

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