Chapter 9, Problem 39P

To determine

Find the average velocity of the water leaving the tank.

Answer to Problem 39P

The average velocity of the leaving water in $\frac{\text{in}}{\text{s}}$, $\frac{\text{ft}}{\text{s}}$, and $\frac{\text{m}}{\text{s}}$ are $\underset{\_}{33.48\frac{\text{in}}{\text{s}}}$, $2.79\frac{\text{ft}}{\text{s}}$ and $0.85\frac{\text{m}}{\text{s}}$ respectively.

Explanation of Solution

Given data:

Refer to the given Figure Problem 9.38 in the textbook, which shows the tank is filled by water using pipes 1 and 2.

The water level increases in the rate of 0.1 in/s.

The diameter of the tank is $\left(d_{\text{tank}}\right)$ 6 in.

Formula used:

Write the expression for the volume flow rate.

$Q=V_{\text{average}}{A}_c$

Here,

$A_c$ is cross-sectional area, and

$V_{\text{average}}$ is average velocity.

Calculation:

Consider that the constant water density.

From the given figure, write the volume flow rate expression for inlet and outlet volumetric flow rate expressions.

$Q_1+Q_2-Q_3=A_{\text{tank}}\frac{\text{change}\text{in}\text{the}\text{height}\text{of}\text{water}}{\text{time}}$

Re-arrange the equation,

$\begin{array}{l}{Q}_1+Q_2-A_{\text{tank}}\frac{\text{change}\text{in}\text{the}\text{height}\text{of}\text{water}}{\text{time}}=Q_3\\ V_1{A}_1+V_2{A}_2-A_{\text{tank}}\frac{\text{change}\text{in}\text{the}\text{height}\text{of}\text{water}}{\text{time}}=V_3{A}_3\\ \left[\begin{array}{l}{V}_1\left(\frac{\pi }{4}\right){\left(d_1\right)}^2+V_2\left(\frac{\pi }{4}\right){\left(d_2\right)}^2-\\ \left(\frac{\pi }{4}\right){\left(d_{\text{tank}}\right)}^2\left(0.1\frac{\text{in}}{\text{s}}\right)=V_3\left(\frac{\pi }{4}\right){\left(d_3\right)}^2=Q_3\end{array}\right]\left[\begin{array}{l}\therefore A=\frac{\pi }{4}{d}^2,and\\ \frac{\left[\begin{array}{l}\text{change}\text{in}\text{the}\\ \text{height}\text{of}\text{water}\end{array}\right]}{\text{time}}=0.1\frac{\text{in}}{\text{s}}\end{array}\right]\end{array}$

Substitute $2\frac{\text{ft}}{\text{s}}$ for $V_1$, $1.5\frac{\text{ft}}{\text{s}}$ for $V_2$, 1 in. for $d_1$, 1.75 in. for $d_2$, 1.5 in. for $d_3$, and 6 in. for $d_{\text{tank}}$,

$\begin{array}{l}\left(2\frac{\text{ft}}{\text{s}}\right)\left(\frac{\pi }{4}\right){\left(1\text{in}\text{.}\right)}^2+\left(1.5\frac{\text{ft}}{\text{s}}\right)\left(\frac{\pi }{4}\right){\left(1.75\text{in}\text{.}\right)}^2-\left(\frac{\pi }{4}\right){\left(6\text{in}\text{.}\right)}^2\left(0.1\frac{\text{in}\text{.}}{\text{s}}\right)=\left(V_3\right)\left(\frac{\pi }{4}\right){\left(1.5\text{in}\text{.}\right)}^2=Q_3\\ \left\{\left[\begin{array}{l}\left(2\frac{\text{ft}}{\text{s}}\right)\left(\frac{\pi }{4}\right){\left(\frac{1\text{ft}}{12}\right)}^2+\left(1.5\frac{\text{ft}}{\text{s}}\right)\left(\frac{\pi }{4}\right){\left(\frac{1.75\text{ ft}}{12}\right)}^2-\\ \left(\frac{\pi }{4}\right){\left(\frac{6\text{ft}}{12}\right)}^2\left(\frac{0.1}{12}\frac{\text{ft}}{\text{s}}\right)=\left(V_3\right)\left(\frac{\pi }{4}\right){\left(\frac{1.5\text{ft}}{12}\right)}^2=Q_3\end{array}\right]\right\}\left[\therefore 1\text{in}\text{.}=\frac{1}{12}\text{ft}\right]\\ 0.0109\frac{{\text{ft}}^{\text{3}}}{\text{s}}+0.02505\frac{{\text{ft}}^{\text{3}}}{\text{s}}-0.001636\frac{{\text{ft}}^{\text{3}}}{\text{s}}=0.01227{\text{ft}}^{\text{2}}\left(V_3\right)=Q_3\\ 0.0343\frac{{\text{ft}}^{\text{3}}}{\text{s}}=0.01227{\text{ft}}^{\text{2}}\left(V_3\right)=Q_3\end{array}$

The values of $Q_3$ and $V_3$ are,

$Q_3=0.0343\frac{{\text{ft}}^{\text{3}}}{\text{s}}$,

And

$V_3=\frac{0.0343\frac{{\text{ft}}^{\text{3}}}{\text{s}}}{0.01227{\text{ft}}^{\text{2}}}$

$V_3=2.79\frac{\text{ft}}{\text{s}}$ (1)

Convert the unit of value in equation (1) from $\frac{\text{ft}}{\text{s}}$ to $\frac{\text{in}}{\text{s}}$.

$\begin{array}{c}{V}_3=2.79\times 12\frac{\text{in}}{\text{s}}\left[\therefore 1\text{ft}=12\text{in}\text{.}\right]\\ =33.48\frac{\text{in}}{\text{s}}\end{array}$

Convert the unit of value in equation (1) from $\frac{\text{ft}}{\text{s}}$ to $\frac{\text{m}}{\text{s}}$.

$\begin{array}{c}{V}_3=2.79\times 0.3048\frac{\text{m}}{\text{s}}\left[\therefore 1\text{ft}=0.3048\frac{\text{m}}{\text{s}}\right]\\ =0.85\frac{\text{m}}{\text{s}}\end{array}$

Conclusion:

Hence, the average velocity of the leaving water in $\frac{\text{in}}{\text{s}}$, $\frac{\text{ft}}{\text{s}}$, and $\frac{\text{m}}{\text{s}}$ are $\underset{\_}{33.48\frac{\text{in}}{\text{s}}}$, $2.79\frac{\text{ft}}{\text{s}}$ and $0.85\frac{\text{m}}{\text{s}}$ respectively.