Chapter 12, Problem 22P

(a)

To determine

The maximum height the water will reach in the tank.

Explanation of Solution

Given:

The diameter of the tank is $D_T$.

The mass flow rate is ${\dot{m}}_{\text{in}\text{.}}$.

The diameter of the orifice is $D_o$.

Neglect the frictional losses, since the orifice has rounded entrance.

Calculation:

Consider the point 1 as the free surface of the tank and the point 2 as the exit of the orifice.

Consider the reference line is at the orifice. Therefore, $z_2=0$.

The pressure at the points 1 and 2 is atmospheric. Therefore, $P_1=P_2=P_{atm}$.

The velocity at the tank is very small compared to the orifice. Therefore, $V_1\cong 0$

Write the Bernoulli’s equation between the points 1 and 2 as follows;

  $\begin{array}{l}\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\ \frac{P_{atm}}{\rho g}+\frac{0^2}{2g}+z_1=\frac{P_{atm}}{\rho g}+\frac{V_2^2}{2g}+0\\ z_1=\frac{V_2^2}{2g}\\ V_2=\sqrt{2g{z}_1}\end{array}$        (I)

Write the equation for the mass flow rate through the orifice as follows;

  $\begin{array}{c}{\dot{m}}_{out}=\rho {\dot{V}}_{out}\\ =\rho A_{orifice}{V}_2\\ =\rho \left(\frac{\pi D_o^2}{4}\right)\sqrt{2gz}\\ z=\frac{1}{2g}{\left(\frac{4{\dot{m}}_{out}}{\rho \pi D_o^2}\right)}^2\end{array}$        (II)

When the flow rate is incoming, ${\dot{m}}_{\text{out}}={\dot{m}}_{\text{in}};z=h_{\max }$;

Substitute ${\dot{m}}_{\text{out}}={\dot{m}}_{\text{in}}$ and $z=h_{\max }$ in Equation (II).

$h_{\max }=\frac{1}{2g}{\left(\frac{4{\dot{m}}_{in}}{\rho \pi D_o^2}\right)}^2$

Thus, the maximum height the water will reach in the tank is $\underset{\_}{h_{\max }=\frac{1}{2g}{\left(\frac{4{\dot{m}}_{in}}{\rho \pi D_o^2}\right)}^2}$.

(b)

To determine

The relationship for the water height as a function of time.

Explanation of Solution

The amount of water flows through the orifice at a differential time is;

  $\begin{array}{c}d{m}_{out}={\dot{m}}_{out}dt\\ =\rho \times \frac{\pi D_o^2}{4}\times \sqrt{2gz}dt\end{array}$

The increase in water in the tank at the differential time is;

  $\begin{array}{c}d{m}_{\text{tank}}=\rho A_{\tan k}dz\\ =\rho \times \frac{\pi D_T^2}{4}\times dz\end{array}$

The amount of water enters during the differential time is;

$d{m}_{in}={\dot{m}}_{in}dt$

Write the conservation of mass equation;

  $\begin{array}{l}d{m}_{\text{tank}}={\dot{m}}_{in}dt-{\dot{m}}_{out}dt\\ \rho \times \frac{\pi D_T^2}{4}\times dz={\dot{m}}_{in}dt-\rho \times \frac{\pi D_o^2}{4}\times \sqrt{2gz}dt\\ \rho \times \frac{\pi D_T^2}{4}\times dz=\left({\dot{m}}_{in}-\rho \times \frac{\pi D_o^2}{4}\times \sqrt{2gz}\right)dt\\ \frac{\rho \times \frac{\pi D_T^2}{4}\times dz}{{\dot{m}}_{in}-\rho \times \frac{\pi D_o^2}{4}\times \sqrt{2gz}}=dt\end{array}$

Integrate the above equation and apply the boundary conditions $\left(t=0\text{to}t;z=0\text{to}z\right)$.

  $\begin{array}{l}{\displaystyle \underset{z=0}{\overset{z=z}{\int }}\frac{\rho \times \frac{\pi D_T^2}{4}\times dz}{{\dot{m}}_{in}-\rho \times \frac{\pi D_o^2}{4}\times \sqrt{2gz}}}={\displaystyle \underset{t=0}{\overset{t=t}{\int }}dt}\\ \frac{\frac{\rho \pi D_T^2}{2}}{{\left(\frac{\rho \pi D_o^2\sqrt{2g}}{4}\right)}^2}\left(\frac{\rho \pi D_o^2\sqrt{2gz}}{4}-{\dot{m}}_{in}\ln \frac{{\dot{m}}_{in}-\frac{\rho \pi D_o^2\sqrt{2gz}}{4}}{{\dot{m}}_{in}}\right)=t\end{array}$

Thus, the relationship for the water height as a function of time is $\underset{\_}{\frac{\frac{\rho \pi D_T^2}{2}}{{\left(\frac{\rho \pi D_o^2\sqrt{2g}}{4}\right)}^2}\left(\frac{\rho \pi D_o^2\sqrt{2gz}}{4}-{\dot{m}}_{in}\ln \frac{{\dot{m}}_{in}-\frac{\rho \pi D_o^2\sqrt{2gz}}{4}}{{\dot{m}}_{in}}\right)=t}$.