Chapter 9, Problem 49P

(a)

To determine

The speed of water in the hose.

Answer to Problem 49P

The speed of water in the hose is $\underset{\_}{v_2=39.1\text{cm/s}}$.

Explanation of Solution

Write the continuity equation

$\frac{\Delta V}{\Delta t}=A_1{v}_1=A_2{v}_2$ (I)

Here, $\frac{\Delta V}{\Delta t}$ is the volume flow rate, $A_1$ and $A_2$ is inner and outer area, $v_1$ and $v_2$ are velocities

Write continuity equation in case hose and obtain expression of $v_2$ using equation (I)

$\begin{array}{c}{A}_1{v}_1=A_2{v}_2\\ v_2=\frac{A_1}{A_2}{v}_1\end{array}$ (II)

Replace $A_1$ and $A_2$ by $\pi r_1{}^2$, and $\pi r_2{}^2$ respectively in equation (II)

$\begin{array}{c}{v}_2=\frac{\pi r_1{}^2}{\pi r_2{}^2}{v}_1\\ =\frac{r_1{}^2}{r_2{}^2}{v}_1\end{array}$ (III)

Conclusion:

Substitute $1.00\text{mm}$ for $r_1$, $8.00\text{mm}$ for $r_2$, and $25.0\text{m/s }$ for $v_1$ in equation (III)

$\begin{array}{c}{v}_2=\frac{{\left(1.00\text{mm}\right)}^2}{{\left(8.00\text{mm}\right)}^2}\left(25.0\text{m/s }\right)\\ =39.1\text{cm/s}\end{array}$

Therefore, the speed of water in the hose is $\underset{\_}{v_2=39.1\text{cm/s}}$.

(b)

To determine

The volume flow rate.

Answer to Problem 49P

The volume flow rate is $\underset{\_}{\frac{\Delta V}{\Delta t}=78.5{\text{cm}}^3/\text{s}}$.

Explanation of Solution

From equation (I) in part (A) the expression for volume flow rate is

$\begin{array}{c}\frac{\Delta V}{\Delta t}=A_1{v}_1\\ =\pi r_1{}^2{v}_1\end{array}$ (IV)

Conclusion:

Substitute, $25\text{m/s}$ for $v$, $1.0\text{mm}$ for $r$ in equation (IV)

$\begin{array}{c}\frac{\Delta V}{\Delta t}=3.14{\left(1.0\text{mm}\times \frac{{10}^{-1}\text{cm}}{1\text{mm}}\right)}^2\left(25\text{m/s}\times \frac{{10}^2\text{cm}}{1\text{m}}\right)\\ =78.5{\text{cm}}^3/\text{s}\end{array}$

Therefore, the volume flow rate is $\underset{\_}{\frac{\Delta V}{\Delta t}=78.5{\text{cm}}^3/\text{s}}$.

(c)

To determine

The mass flow rate.

Answer to Problem 49P

The mass flow rate $\underset{\_}{\frac{\Delta m}{\Delta t}=78.5\text{g/s}}$.

Explanation of Solution

Write the expression for mass flow rate

$\frac{\Delta m}{\Delta t}=\rho A_1{v}_1$ (VI)

Here, $\rho$ is the density, $A_1$ is the area, $v_1$ is the velocity

Conclusion:

Substitute, $1.0\text{g}/{\text{cm}}^3$ for $\rho$, $78.5{\text{cm}}^3/\text{s}$ for $A_1{v}_1$, in equation (VI)

$\begin{array}{c}\frac{\Delta m}{\Delta t}=\left(1.0\text{g}/{\text{cm}}^3\right)\left(78.5{\text{cm}}^3/\text{s}\right)\\ =78.5\text{g}/\text{s}\end{array}$

Therefore, the mass flow rate $\underset{\_}{\frac{\Delta m}{\Delta t}=78.5\text{g/s}}$.