Chapter 5, Problem 49P

To determine

(a)

To determine the maximum height that the water will reach in the tank.

Explanation of Solution

The diameter of the tank is $D_T$ and the mass flow rate is ${\dot{m}}_{in}$.

The following figure shows the schematic diagram of water tank system.

  Figure-(1)

Write the expression for Bernoulli theorem between the point $1$ and point $2$ by setting bottom of the tank as datum.

   $\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2$    ...... (I)

Here, the pressure at point $1$ is $P_1$, the pressure at point $2$ is $P_2$, the velocity at point $1$ is $V_1$, the velocity at point $2$ is $V_2$, the height at point $1$ is $z_2$, the height at point $2$ is $z_2$, the density is $\rho$ and gravitational acceleration is $g$.

Substitute $0$ for $V_1$, $0$ for $z_1$, $P_{atm}$ for $P_1$ and $P_{atm}$ for $P_2$ in Equation (1).

   $\begin{array}{l}\frac{P_{atm}}{\rho g}+\frac{0}{2g}+z=\frac{P_{atm}}{\rho g}+\frac{V_2^2}{2g}+0\\ \frac{V_2^2}{2g}=z\\ V_2=\sqrt{2gz}\end{array}$    ...... (II)

Write the expression for the mass flow rate at the point $2$ at orifice.

   $m_{orifice}=\rho A_{orifice}{V}_2$    ...... (III)

Here, the area of the orifice is $A_{orifice}$ and the velocity at point $2$ is $V_2$.

Write the expression for the area of the orifice.

   $A_{orifice}=\frac{\pi D_o^2}{4}$

Here, the diameter of the orifice is $D_o$.

Substitute $\frac{\pi D_o^2}{4}$ for $A_{orifice}$ in Equation (II).

   $m_{orifice}=\rho \left(\frac{\pi D_o^2}{4}\right)V_2$    ...... (IV)

Square on both sides in Equation (IV).

   ${\left(m_{orifice}\right)}^2={\rho }^2{\left(\frac{\pi D_o^2}{4}\right)}^2{\left(V_2\right)}^2$    ...... (V)

Substitute $\sqrt{2gz}$ for $V_2$ in Equation (V).

   $\begin{array}{l}{\left(m_{orifice}\right)}^2={\rho }^2{\left(\frac{\pi D_o^2}{4}\right)}^2{\left(\sqrt{2gz}\right)}^2\\ z=\frac{{\left(m_{orifice}\right)}^2}{{\rho }^2{\left(\frac{\pi D_o^2}{4}\right)}^{2}2g}\\ z=\frac{1}{2g}\left(\frac{4\left(m_{orifice}\right)}{\rho \pi D_o^2}\right)\end{array}$    ...... (VI)

Substitute $z_{\max }$ for $z$ and $m_{in}$ for $m_{orifice}$ in Equation (VI).

   $z_{\max }=\frac{1}{2g}\left(\frac{4\left(m_{in}\right)}{\rho \pi D_o^2}\right)$

The maximum height reached by the water of the tank is $\frac{1}{2g}\left(\frac{4\left(m_{in}\right)}{\rho \pi D_o^2}\right)$.

To determine

(b)

The relation between water height as a function of time.

Explanation of Solution

Write the expression for the principle of conservation of mass for the control volume undergoing a change during period $dt$.

   $d{m}_{cv}=n_{in}dt+m_{orifice}dt$    ...... (VII)

Here, the mass flow rate of orifice is $m_{orifice}$.

Write the expression for the mass flow rate of control volume.

   $d{m}_{cv}=\rho \left(\frac{\pi D_T^2}{4}\right)dz$

Substitute $\rho \left(\frac{\pi D_T^2}{4}\right)dz$ for $d{m}_{cv}$ and $\rho \left(\frac{\pi D_o^2}{4}\right)\sqrt{2gz}$ for $m_{orifice}$ in Equation (VII).

   $\begin{array}{l}\rho \left(\frac{\pi D_T^2}{4}\right)dz=m_{in}dt+\rho \left(\frac{\pi D_o^2}{4}\right)\sqrt{2gz}dt\\ dt=\frac{\rho \left(\frac{\pi D_T^2}{4}\right)}{m_{in}-\rho \left(\frac{\pi D_o^2}{4}\right)\sqrt{2gz}}dz\\ dt=\frac{1}{m_{in}}\left(\frac{\rho \left(\frac{\pi D_T^2}{4}\right)}{1-\frac{\rho \left(\frac{\pi D_o^2}{4}\right)}{m_{in}}\sqrt{2gz}}\right)dz\\ dt=\frac{\rho \left(\frac{\pi D_T^2}{4}\right)}{m_{in}}\left(\frac{1}{1-\left(\frac{\rho \left(\frac{\pi D_o^2}{4}\right)}{m_{in}}\sqrt{2g}\right)z^{1/2}}\right)dz\end{array}$    ...... (VIII)

Substitute $b$ for $\left(\frac{\rho \left(\frac{\pi D_o^2}{4}\right)}{m_{in}}\sqrt{2g}\right)$ and $a$ for $\frac{\rho \left(\frac{\pi D_T^2}{4}\right)}{m_{in}}$ in Equation (VIII).

   $dt=a\left(\frac{1}{1-b{z}^{1/2}}\right)dz$    ...... (IX)

Write the general expression for $z$.

   $\begin{array}{l}z=x^2\\ x=z^{1/2}\end{array}$    ...... (X)

Differentiate Equation (X).

   $dz=2xdx$

Substitute $x^2$ for $z$ and $2xdx$ for $dz$ in Equation (X).

   $\begin{array}{c}dt=a\left(\frac{1}{1-b{\left(x^2\right)}^{1/2}}\right)2xdx\\ dt=a\left(\frac{2xdx}{1-bx}\right)\end{array}$    ...... (XI)

Write the expression for the dissolve of fraction.

   $\text{Numerator}=A\times \left(\text{denominator}\right)+B\times \frac{d\left(\text{denominator}\right)}{dt}$

Dissolve the fraction in Equation (X).

   $\begin{array}{l}2x=A\times \left(1-bx\right)+B\times \frac{d}{dx}\times \left(1-bx\right)\\ 2x=A\times \left(1-bx\right)-bB\end{array}$    ...... (XII)

Equate the coefficients of $x$ on the left hand side and right hand side in Equation (XII).

   $\begin{array}{l}2x=-Abx\\ 2x=-Abx\\ A=-\frac{2}{b}\end{array}$

   $\begin{array}{l}Ax-bBx=0\\ Ax=bBx\\ A=bB\end{array}$    ...... (XIII)

Substitute $-\frac{2}{b}$ for $A$ in Equation (XIII).

   $\begin{array}{l}-\frac{2}{b}=bB\\ B=-\frac{2}{b^2}\end{array}$

Substitute $\frac{-2}{b^2}$ for $B$ and $\frac{-2}{b}$ for $A$ in Equation (XII).

   $\begin{array}{l}2x=\frac{-2}{b}\times \left(1-bx\right)+\frac{-2}{b^2}\times \frac{d}{dx}\times \left(1-bx\right)\\ 2x=\frac{-2}{b}\times \left(1-bx\right)-b\frac{-2}{b^2}\end{array}$

Substitute $\left(\frac{-2}{b}\times \left(1-bx\right)-\frac{-2}{b^2}b\right)$ for $2x$ in Equation (XI).

   $\begin{array}{l}dt=a\left(\frac{\left(\frac{-2}{b}\times \left(1-bx\right)-\frac{-2}{b^2}b\right)}{1-b{\left(x^2\right)}^{1/2}}\right)dx\\ dt=\frac{2a}{b}\left(\frac{1}{1-bx}-1\right)dx\end{array}$    ...... (XIV)

Integrate Equation (XII) on both sides.

   $\begin{array}{l}{\displaystyle \int dt}={\displaystyle \int \frac{2a}{b}\left(\frac{1}{1-bx}-1\right)dx}\\ t=\frac{2a}{b}\left(\ln \left(1-bx\right)-x\right)\end{array}$    ...... (XV)

Substitute $z^{1/2}$ for $x$ Equation (XIII).

   $t=\frac{2a}{b}\left(\ln \left(1-b{z}^{1/2}\right)-z^{1/2}\right)$    ...... (XVI)

Substitute $\left(\frac{\rho \left(\frac{\pi D_o^2}{4}\right)}{m_{in}}\sqrt{2g}\right)$ for $b$ and $\frac{\rho \left(\frac{\pi D_T^2}{4}\right)}{m_{in}}$ for $a$ in Equation (XVI).

   $\begin{array}{l}t=\frac{2\frac{\rho \left(\frac{\pi D_T^2}{4}\right)}{m_{in}}}{\left(\frac{\rho \left(\frac{\pi D_o^2}{4}\right)}{m_{in}}\sqrt{2g}\right)}\left(\ln \left(1-\left(\frac{\rho \left(\frac{\pi D_o^2}{4}\right)}{m_{in}}\sqrt{2g}\right)z^{1/2}\right)-z^{1/2}\right)\\ t=\frac{\frac{1}{2}\rho \pi D_T^2}{{\left(\frac{1}{4}\rho \pi D_o^2\sqrt{2g}\right)}^2}\left(\left(\frac{1}{4}\rho \pi D_o^2\sqrt{2g}\right)-m_{in}\ln \frac{m_{in}-\frac{1}{4}\rho \pi D_o^2\sqrt{2g}}{m_{in}}\right)\end{array}$

The relation between the time and the height is $t=\frac{\frac{1}{2}\rho \pi D_T^2}{{\left(\frac{1}{4}\rho \pi D_o^2\sqrt{2g}\right)}^2}\left(\left(\frac{1}{4}\rho \pi D_o^2\sqrt{2g}\right)-m_{in}\ln \frac{m_{in}-\frac{1}{4}\rho \pi D_o^2\sqrt{2g}}{m_{in}}\right)$.