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I am tasked with finding the radius of a hole at the bottom of a tank to obtain the desired outflow rate when the tank is full.

The tank is a cylinder with dimensions Height: 4.8m and Radius: 1.79m. It holds roughly 48316.6897 liters of water.

I have already determined $C_d$ (The coefficient of discharge) to be 0.6372.

The outflow rate I desire is 3500 liters per hour.

I know that $\frac{dh}{dt}= \frac{a_0C_d\sqrt{2gH}}{A}$. where $a_0$ is the radius i need to obtain, and $A$ being the surface area.

From chain rule I can get $\frac{dh}{dt}$ in terms of radius and $\frac{dv}{dt}$(The desired outflow rate)

$\frac{dh}{dt} = \frac{\Delta{V}}{\pi r^2} $

Now I let them equal each other

$\frac{\Delta{V}}{\pi r^2} = \frac{a_0 C_d\sqrt{2gH}}{A}$

This gives me the equation $$a0=\frac{\Delta{V}}{C_d\sqrt{2gH}} $$

Now, substitute in what we know

$3500$ Litres per hour is $\frac{35}{36000} \frac{m^3}{s} $

$$ a0 = \frac{\frac{35}{36000}}{0.6372\sqrt{2\cdot 9.8 \cdot 4.8}} $$

According to this $a_0$ is $0.157304491$ millimeters in radius.

I am doubting my calculations because $3500$ liters going past a hole of radius $0.157304491$ millimeters every hour seems absurd.

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  • $\begingroup$ 3500 liters = 3.5 m3. That being said, 1 hour = 3600 seconds, so the errors cancel out. $\endgroup$ – Wasabi Apr 20 '17 at 15:23
  • $\begingroup$ 3500 liters is not 35 m^3. its 3.5m^3 $\endgroup$ – Fennekin Apr 21 '17 at 3:43
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According to this a0 is 0.157304491 millimetres in radius.

I am doubting my calculations because 3500 litres going past a hole of radius 0.157304491 millimetres every hour seems absurd.

This is area. You need diameter. Multiply it by 4 then divide it by pi then take root of the result. That will be diameter of hole

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I think the problem is that there's something wrong with your first differential equation $$ \frac{dh}{dt}=\frac{a_0C_d\sqrt{2gH}}{A} $$ the lefthand side has the unit $ \frac{m}{s} $, while the righthand side leads to the units $\frac{1}{s}$. If you only need to know the outflow rate or needed radius for the instant where the tank is full, I would use Bernoullis principle: $$ H=\frac{v^2}{2g} \qquad \mbox{and}\qquad v=\frac{Q_t}{\pi r^2} $$ $$ C_d=\frac{Q_{r}}{{Q_t}} $$, where $t =$ theoretical and $r=$ real. When solving for $r$ you get: $$ r = \left( \frac{\left(\frac{Q_r}{C_d}\right)^2}{2\pi^2Hg} \right)^{1/4} \qquad\qquad \star $$ If I put in the given values it returns $$ r=6.88mm $$

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To expand on previous answers, you're solving Bernoulli's equation which describes the conservation of mechanical fluid energy.

$$P_1+\frac{\rho V_1^2}{2}+\rho g h_1 = P_2+\frac{\rho V_2^2}{2}+\rho g h_2$$

Assuming atmospheric pressure at the exit is 0, fluid in the tank has zero velocity, $h$ is the depth of the tank at the outlet and change in potential energy is neglected, the formula simplifies to: $$P_1 = \frac{\rho V_2^2}{2}$$ where $P_1$ is equal to $\rho g h$ and simplifies to: $$g h=\frac{V_2^2}{2}$$ You can solve for the exit velocity and you already know your flow rate: $$V_2 = \sqrt{2gh}{}$$ $$Q_{actual}=C_d A V_2=C_d\frac{\pi D^2}{4}\sqrt{2gh}$$ Now you can solve for the diameter of the outlet: $$D=\sqrt{\frac{4Q_{actual}}{C_D \pi\sqrt{2gh}}}$$

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