Chapter 9, Problem 83P

(a)

To determine

The speed with which the water leave the showerhead.

Answer to Problem 83P

The speed with which the water leave the showerhead is $26\text{m/s}$.

Explanation of Solution

Find the equation for the speed with which the water leave the showerhead.

$\begin{array}{c}{P}_1+\rho g{y}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{y}_2+\frac{1}{2}\rho v_2^2\\ \frac{1}{2}\rho \left(v_1^2-v_2^2\right)=P_2-P_1+\rho g\left(y_2-y_1\right)\\ v_1^2-\frac{A_1^2}{A_2^2}{v}_1^2=\frac{2P_{\text{gauge}}}{\rho }-2gh\\ v_1=\sqrt{\frac{\frac{2P_{\text{gauge}}}{\rho }-2gh}{1-{\left[\frac{N{r}_1^2}{r_2^2}\right]}^2}}\end{array}$ (I)

Here, $P_2-P_1=P_{\text{gauge}}$ is the gauge pressure, $y_2-y_1=h$ is the height of the showehead, $\rho$ is the density of water, $g$ is the acceleration due to gravity, $r_2$ is the radius of the output pipe, $r_1$ is the radius of the showehead, $A_2$ is the area of the output pipe, $A_1$ is the area of the showehead and $N$ is the number of holes.

Conclusion:

Substitute $6.3\text{mm}$ for $r_2$, $0.33\text{mm}$ for $r_1$, $36$ for $N$, $410\text{kPa}$ for $P_{\text{gauge}}$, $6.7\text{m}$ for $h$, $1.00\times {10}^3{\text{kg/m}}^3$ for $\rho$ and $9.80{\text{m/s}}^2$ for $g$ in equation (I) to find $v_1$.

$\begin{array}{c}{v}_1=\sqrt{\frac{\frac{2\left(\left(410\text{kPa}\right)\left(\frac{{10}^3\text{Pa}}{1\text{kPa}}\right)\right)}{1.00\times {10}^3{\text{kg/m}}^3}-2\left(9.80{\text{m/s}}^2\right)\left(6.7\text{m}\right)}{1-{36}^2{\left[\frac{0.33\text{mm}}{6.3\text{mm}}\right]}^4}}\\ =\sqrt{\frac{\frac{2\left(410\times {10}^3\text{Pa}\right)}{1.00\times {10}^3{\text{kg/m}}^3}-2\left(9.80{\text{m/s}}^2\right)\left(6.7\text{m}\right)}{1-{36}^2{\left[\frac{0.33\text{mm}}{6.3\text{mm}}\right]}^4}}\\ =26\text{m/s}\end{array}$

Thus, the speed with which the water leave the showerhead is $26\text{m/s}$.

(b)

To determine

The speed with which the water moves through the output pipe of the pump.

Answer to Problem 83P

The speed with which the water moves through the output pipe of the pump is $2.6\text{m/s}$.

Explanation of Solution

Find the equation for the speed with which the water moves through the output pipe of the pump.

$\begin{array}{c}{v}_2=\frac{A_1}{A_2}{v}_1\\ =\frac{N{r}_1^2}{r_2^2}{v}_1\end{array}$ (II)

Here, $v_2$ is the speed with which the water moves through the output pipe of the pump.

Conclusion:

Substitute $6.3\text{mm}$ for $r_2$, $0.33\text{mm}$ for $r_1$, $26\text{m/s}$ for $v_1$ and $36$ for $N$ in equation (II) to find $v_2$.

$\begin{array}{c}{v}_2=36{\left(\frac{0.33\text{mm}}{6.3\text{mm}}\right)}^2\left(26\text{m/s}\right)\\ =2.6\text{m/s}\end{array}$

Thus, the speed with which the water moves through the output pipe of the pump is $2.6\text{m/s}$.