Chapter 9, Problem 109P

To determine

The velocity of water through main pipe.

Answer to Problem 109P

The velocity of water is $0.116\text{m}/\text{s}$.

Explanation of Solution

Depth pipe from the street level is $0.90\text{m}$, gauge pressure is $52.0\text{kPa}$, radius of main pipe is $5.00\text{cm}$, radius of faucet is $1.20\text{cm}$, height of faucet from the street is $4.20\text{m}$.

Write the equation of continuity at the entrance point and the faucet.

$\left(\pi r_1^2\right)v_1=\left(\pi r_2^2\right)v_2$

Here, $r_1$ is the radius of entrance hole, $r_2$ is radius of faucet, $v_1$ is the speed of water at the entrance hole, and $v_2$ is the velocity of water at the faucet.

Rewrite the above relation in terms of $v_2$.

$v_2={\left(\frac{r_1}{r_2}\right)}^2{v}_1$ (I)

Write the equation for Bernoulli’s theorem at the entrance point and the faucet.

$P_1+\rho g{y}_1+\frac{1}{2}\rho v_1^2=P_{atm}+\rho g{y}_2+\frac{1}{2}\rho v_2^2$

Here,$p_1$ is the pressure at the entrance hole, $\rho$ is the density of water, $y_1$ is depth to entrance hole from the street level, $P_{atm}$ is the atmospheric pressure, $g$ is the acceleration due to gravity, and $y_2$ is the height to faucet from the street level.

Rewrite the above relation in terms of $v_1^2$.

$\begin{array}{c}{v}_1^2=\frac{2\left(P_{atm}-P_1\right)}{\rho }+2g\left(y_2-y_1\right)+{\left({\left(\frac{r_1}{r_2}\right)}^2{v}_1\right)}^2\\ {\left({\left(\frac{r_1}{r_2}\right)}^2{v}_1\right)}^2-v_1^2=\frac{2\left(P_1-P_{atm}\right)}{\rho }-2g\left(y_2-y_1\right)\\ v_1=\sqrt{\frac{\frac{2\left(P_1-P_{atm}\right)}{\rho }-2g\left(y_2-y_1\right)}{{\left(\frac{r_1}{r_2}\right)}^4-1}}\end{array}$

Conclusion:

Substitute $52.0\text{kPa}$ for $\left(P_1-P_{atm}\right)$, $1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $4.20\text{m}$ for $y_2$, $-0.90\text{m}$ for $y_1$, $5.00\text{cm}$ for $r_1$, and $1.20\text{cm}$ for $r_2$ in the above equation to find $v_1$.

$\begin{array}{c}{v}_1=\sqrt{\frac{\left(\frac{2\left(52.0\text{kPa}\left(\frac{{\text{10}}^{\text{3}}\text{Pa}}{\text{1}\text{Pa}}\right)\right)}{1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}}\right)-\left(2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(4.20\text{m}--0.90\text{m}\right)\right)}{{\left(\frac{5.00\text{cm}}{1.20\text{cm}}\right)}^2-1}}\\ =\sqrt{\frac{0.220{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{16.36}}\\ =0.116\text{m}/\text{s}\end{array}$

Therefore, the velocity of water is $0.116\text{m}/\text{s}$.