jkwiens.com: Answer: Fluid leaving a hemispherical vessel

Answer: Fluid leaving a hemispherical vessel

Thursday, 20 of December , 2007 at 11:56 pm

Question: A hemispherical vessel of radius R has a small rounded orifice of area at the bottom. If $A_h \gg A_0$ , how much time does it require to lower the level from to ?

Answer: First, lets assume fluid is incompressible and has a inviscid, steady, barotropic flow. If this is the case, the rate fluid leaves the vessel will be equal to the rate the fluid drops in the vessel. Therefore,

$A_h \frac{\partial h}{\partial t} = A_0 u$

$\therefore \frac{\partial h}{\partial t} = \frac{A_0}{A_h} u$

where $\frac{\partial h}{\partial t}$ is the rate the water level drops and is the rate the water leaves the orifice.

Now, if we use Bernoulli’s equation, we will get:

$\frac{1}{2} (\frac{\partial h}{\partial t})^2 + \frac{P_{atm}}{\rho_0} + gh = \frac{1}{2}u^2 + \frac{P_{atm}}{\rho_0}$

$\frac{1}{2} (\frac{A_0}{A_h})^2 u^2 + gh = \frac{1}{2}u^2$

Since $A_h \gg A_0$ , $\therefore (\frac{A_0}{A_h})^2 \approx 0$ . Using this approximation, we find that

$u = \sqrt{2gh}$

Next, we will need to find .

According to Pythagoras ,

$r = (R^2 - R^2 - h^2 +2hR)^{\frac{1}{2}} = (2hR - h^2)^{\frac{1}{2}}$

Therefore,

$A_h = \pi r^2 = \pi (2hR-h^2)$

Now using the fact that $A_h \frac{\partial h}{\partial t} = A_0 u$ , we have

$\pi(2hR - h^2)\frac{\partial h}{\partial t} = A_0 \sqrt{2gh}$

$\int_{h_1}^{h_2} \pi (2 h^{\frac{1}{2}}R - h^{\frac{3}{2}})dh = \int_0^t A_0 \sqrt{2g}dt$

$\pi(\frac{4}{3}Rh^{\frac{3}{2}} - \frac{2}{5}h^{\frac{5}{2}}) \arrowvert_{h_1}^{h_2} = A_0 \sqrt{2g}t$

$t = \frac{2\pi}{A_0 \sqrt{2g}}(\frac{2}{3}R(h_1^{\frac{3}{2}}-h_2^{\frac{3}{2}}) - \frac{1}{5}(h_1^{\frac{5}{2}} - h_2^{\frac{5}{2}}))$

Category: Answers, Fluid Dynamics

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