Chapter 8, Problem 36P

To determine

Find the average velocity of the water leaving the tank.

Answer to Problem 36P

The average velocity of the leaving water in the tank is $\underset{\_}{2.79\frac{\text{ft}}{\text{s}}}$.

Explanation of Solution

Given data:

Refer to the Figure Problem 8.36 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.

$Q=V_{\text{average}}{A}_c$ (1)

Here,

$A_c$ is cross-sectional area, and

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

Calculation:

Consider that the constant water density.

From the Figure Problem 8.35 in the textbook, 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,

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

Refer to the Figure Problem 8.35 in the textbook, 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 $d_{\text{tank}}$,

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

Reduce the equation as follows,

$\begin{array}{l}0.0109\frac{{\text{ft}}^{\text{3}}}{\text{s}}+0.02505\frac{{\text{ft}}^{\text{3}}}{\text{s}}-0.001636\frac{{\text{ft}}^{\text{3}}}{\text{s}}=0.01227{\text{ft}}^{\text{2}}\left(V_3\right)=Q_3\\ 0.0343\frac{{\text{ft}}^{\text{3}}}{\text{s}}=0.01227{\text{ft}}^{\text{2}}\left(V_3\right)=Q_3\end{array}$

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

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

And

$\begin{array}{c}{V}_3=\frac{0.0343\frac{{\text{ft}}^{\text{3}}}{\text{s}}}{0.01227{\text{ft}}^{\text{2}}}\\ =2.79\frac{\text{ft}}{\text{s}}\end{array}$

Conclusion:

Hence, the average velocity of the leaving water in the tank is $\underset{\_}{2.79\frac{\text{ft}}{\text{s}}}$.