Chapter 3, Problem 3.142P

To determine

(a)

The differential equation for the free surface height during draining with respect to time.

Answer to Problem 3.142P

The differential equation for the free surface height during draining with respect to time is $\frac{dh\left(t\right)}{h\left(t\right)}=\frac{D_2^4}{D_1^4}dt$.

Explanation of Solution

Given information:

The diameter of the cylindrical tank is $D$ . The initial height of the liquid is $h_0$ . The losses are negligible. The stopper is removed at time $t=0$.

The Figure-(1) shows a cylinder of diameter $D$ and height $h_0$.

Figure-(1)

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system for no losses in the system.

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2-\frac{dh\left(t\right)}{dt}\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=-\frac{dh\left(t\right)}{dt}\end{array}$ …… (I)

Here, the pressure at section (1) is $p_1$, the velocity at section (1) is $V_1$, the height of the section (1) from the datum is $z_1$, the pressure at section (2) is $p_2$, the velocity at section (2) is $V_2$, the height of the section (2) from the datum is $z_2$, the density of water is $\rho$, and the acceleration due to gravity is $g$.

Since the pressure at the section (1) and section (2) is atmospheric, so the difference between the pressures is zero.

$p_1-p_2=0$

Substitute $0$ for $p_1-p_2$ in Equation (I).

$\begin{array}{l}\left(\frac{0}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=-\frac{dh\left(t\right)}{dt}\\ -\frac{dh\left(t\right)}{dt}=\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)\end{array}$ …… (II)

Write the expression for the flow rate at section (1).

$Q=A_1{V}_1$ …… (III)

Here, the area at section (1) is $A_1$ and the velocity at section (1) is $V_1$.

Write the expression for the area at section (1).

$A_1=\frac{\pi }{4}{D}_1^2$

Here, the diameter at section (1) is $D_1$.

Substitute $\frac{\pi }{4}{D}_1^2$ for $A_1$ in Equation (III).

$Q=\frac{\pi }{4}{D}_1^2{V}_1$ …… (IV)

Write the expression for the flow rate at section (2).

$Q=A_2{V}_2$ …… (V)

Here, the flow rate is $Q$, the jet area at section (2) is $A_2$ and the velocity at section (2) is $V_2$.

Write the expression for the area at section (2).

$A_2=\frac{\pi }{4}{D}_2^2$

Here, the diameter at section (2) is $D_2$.

Substitute $\frac{\pi }{4}{D}_2^2$ for $A_2$ in Equation (V).

$Q=\frac{\pi }{4}{D}_2^2{V}_2$ …… (VI)

Since the flow rate at both the sections is equal, it satisfies the continuity equation.

Substitute $\frac{\pi }{4}{D}_1^2{V}_1$ for $Q$ in Equation (V).

$\begin{array}{l}\frac{\pi }{4}{D}_1^2{V}_1=\frac{\pi }{4}{D}_2^2{V}_2\\ D_1^2{V}_1=D_2^2{V}_2\\ V_1=\frac{D_2^2}{D_1^2}{V}_2\end{array}$ …… (VII)

Write the expression for velocity at section (2).

$V_2=\sqrt{2gh\left(t\right)}$

Here, the water level is $h$.

Substitute $\sqrt{2gh\left(t\right)}$ for $V_2$ in Equation (VII).

$V_1=\sqrt{2gh\left(t\right)}\left(\frac{D_2^2}{D_1^2}\right)$

Substitute $\sqrt{2gh\left(t\right)}\left(\frac{D_2^2}{D_1^2}\right)$ for $V_1$, $\sqrt{2gh\left(t\right)}$ for $V_2$, $h\left(t\right)$ for $z_1$ and $0$ for $z_2$ in Equation (II).

$\begin{array}{l}-\frac{dh\left(t\right)}{dt}=\left(\frac{{\left\{\sqrt{2gh\left(t\right)}\left(\frac{D_2^2}{D_1^2}\right)\right\}}^2}{2g}-\frac{{\left(\sqrt{2gh\left(t\right)}\right)}^2}{2g}\right)+\left(h\left(t\right)-0\right)\\ -\frac{dh\left(t\right)}{dt}=\frac{2gh\left(t\right)}{2g}\left(\frac{D_2^4}{D_1^4}-1\right)+h\left(t\right)\\ -\frac{dh\left(t\right)}{dt}=h\left(t\right)\left(\left(\frac{D_2^4}{D_1^4}-1\right)+1\right)\\ \frac{dh\left(t\right)}{h\left(t\right)}=-\frac{D_2^4}{D_1^4}dt\end{array}$

The negative sign is due to decrease in the head level in the tank.

Conclusion:

The differential equation for the free surface height during draining with respect to time is $\frac{dh\left(t\right)}{h\left(t\right)}=\frac{D_2^4}{D_1^4}dt$.

To determine

(b)

The expression for the time $t_0$ to drain the entire tank.

Answer to Problem 3.142P

The expression for the time $t_0$ to drain the entire tank is

Explanation of Solution

Given information:

The diameter of the cylindrical tank is $D$ . The initial height of the liquid is $h_0$ . The losses are negligible. The stopper is removed at time $t=0$.

Write the differential equation for the free surface height.

$\frac{dh\left(t\right)}{h\left(t\right)}=-\frac{D_2^4}{D_1^4}dt$

Integrate the above expression on both the sides.

$\begin{array}{l}{\displaystyle \int \frac{dh\left(t\right)}{h\left(t\right)}}=-\frac{D_2^4}{D_1^4}{\displaystyle \int dt}\\ \ln \left\{h\left(t\right)\right\}=-\frac{D_2^4}{D_1^4}t+\text{C}\end{array}$ …… (VIII)

Apply boundary conditions to Equation (VIII).

At $t=0$ ; $h\left(t\right)=h_0$.

Substitute $h_0$ for $h\left(t\right)$ in Equation (VIII).

$\begin{array}{l}\ln \left\{h_0\right\}=\frac{D_2^4}{D_1^4}\times 0+\text{C}\\ \text{C}=\ln h_0\end{array}$

Substitute $\ln h_0$ for $\text{C}$ in Equation (VIII).

$\begin{array}{l}\ln \left\{h_0\right\}=-\frac{D_2^4}{D_1^4}t+\ln h_0\\ \ln \left\{h\left(t\right)\right\}-\ln h_0=-\frac{D_2^4}{D_1^4}t\\ \frac{h\left(t\right)}{h_0}=e^{{\left(\frac{D_2}{D_1}\right)}^{4}t}\\ h\left(t\right)=h_0{e}^{-{\left(\frac{D_2}{D_1}\right)}^{4}t}\end{array}$

Since the tank is empty after time $t_0$.

Conclusion:

The force per unit width of the water is $68248.19\text{N}/\text{m}$.