Chapter 14, Problem 14.80AP

The water supply of a building is fed through a main pipe 6.00 cm in diameter. A 2.00-cm-diameter faucet tap, located 2.00 m above the main pipe, is observed to fill a 25.0-L container in 30.0 s. (a) What is the speed at which the water leaves the faucet? (b) What is the gauge pressure in the 6-cm main pipe? Assume the faucet is the only “leak” in the building.

To determine

The speed at which the water leaves the faucet.

Answer to Problem 14.80AP

The speed at which the water leaves the faucet is $2.65\text{m}/\text{s}$.

Explanation of Solution

Given info: The diameter of main pipe is $6.00\text{cm}$, the diameter of faucet pipe is $2.00\text{cm}$, the height of the main pipe is $2.00\text{m}$, the volume of the container is $25.0\text{L}$ and the time in which the container is fill is $30.0\text{s}$.

Formula to calculate the flow rate of the water is,

$R=\frac{V}{t}$

Here,

$R$ is the flow rate of the water.

$V$ is the volume of the container.

$t$ is the time in which the container is filled.

Substitute $25.0\text{L}$ for $V$ and $30.0\text{s}$ for $t$ in the above equation.

$\begin{array}{c}R=\frac{25.0\text{L}\times \left(\frac{{10}^3{\text{cm}}^{\text{3}}}{1\text{L}}\right)}{30.0\text{s}}\\ =833{\text{cm}}^{\text{3}}/\text{s}\end{array}$

Thus the flow rate of the water is $833{\text{cm}}^{\text{3}}/\text{s}$

Formula to calculate the area of the faucet is,

$A_2=\pi \frac{d_2{}^2}{4}$

Here,

$A_2$ is the area of the faucet.

$d_2$ is the diameter of the faucet.

Substitute $2.00\text{cm}$ for $d_2$ in the above equation to find the value of $A_2$

$\begin{array}{c}{A}_2=\pi \frac{{\left(2.00\text{cm}\right)}^2}{4}\\ =3.14{\text{cm}}^2\end{array}$

Thus the area of the faucet pipe is $3.14{\text{cm}}^2$.

Formula to calculate the flow rate of the water is,

$R=A_2{v}_2$

Here,

$v_2$ is the speed of the water leaves at the faucet pipe.

Rearrange the above expression for $v_2$.

$v_2=\frac{R}{A_2}$

Substitute $833{\text{cm}}^{\text{3}}/\text{s}$ for $R$ and $3.14{\text{cm}}^2$ for $A_2$ in the above equation to find the value of $v_2$.

$\begin{array}{c}{v}_2=\frac{833{\text{cm}}^{\text{3}}/\text{s}}{3.14{\text{cm}}^2}\\ =265.28\text{cm}/\text{s}\times \left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\\ =2.65\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed at which the water leaves the faucet is $2.65\text{m}/\text{s}$.

To determine

The gauge pressure in the main pipe.

Answer to Problem 14.80AP

The gauge pressure in the main pipe is $2.31\times {10}^4\text{Pa}$.

Explanation of Solution

Given info: The diameter of main pipe is $6.00\text{cm}$, the diameter of faucet pipe is $2.00\text{cm}$, the height of the main pipe is $2.00\text{m}$, the volume of the container is $25.0\text{L}$ and the time in which the container is fill is $30.0\text{s}$.

Formula to calculate the area of main pipe is,

$A_1=\pi \frac{d_1{}^2}{4}$

Here,

$A_1$ is the area of the main pipe.

$d_1$ is the diameter of the main pipe.

Substitute $6.00\text{cm}$ for $d_1$ in the above equation.

$\begin{array}{c}{A}_1=\pi \frac{{\left(6.00\text{cm}\right)}^2}{4}\\ =28.26{\text{cm}}^2\end{array}$

Thus the area of the main pipe is $28.26{\text{cm}}^2$.

Write the continuity equation,

$A_1{v}_1=A_2{v}_2$

Here,

$A_1$ is the area of the main pipe.

$v_1$ is the speed of the water leaves at the main pipe.

$A_2$ is the area of the faucet pipe.

$v_2$ is the speed of the water leaves at the faucet pipe.

Rearrange the above expression for $v_1$

$v_1=\frac{A_2{v}_2}{A_1}$

Substitute $28.26{\text{cm}}^2$ for $A_1$, $3.14{\text{cm}}^2$ for $A_2$ and $2.65\text{m}/\text{s}$ for $v_2$ in the above equation to find the value of $v_1$.

$\begin{array}{c}{v}_1=\frac{\left(3.14{\text{cm}}^2\right)\left(2.65\text{m}/\text{s}\right)}{28.26{\text{cm}}^2}\\ =0.295\text{m}/\text{s}\end{array}$

Thus, the speed of the water leaves at the main pipe is $0.295\text{m}/\text{s}$.

The difference between the heights of the pipe is,

$h=h_2-h_1$

Here,

$h$ is the difference between the heights of the pipe.

$h_2$ is the height of the faucet pipe.

$h_1$ is the height of the main pipe.

Formula to calculate the gauge pressure is,

$P_{\text{g}}=P_1-P_2$

Here,

$P_{\text{g}}$ is the gauge pressure.

$P_2$ is the pressure in faucet pipe.

$P_1$ is the pressure in main pipe.

Write the Bernoulli’s theorem,

$P_1+\frac{1}{2}\rho v_1{}^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2{}^2+\rho g{h}_2$

Here,

$\rho$ is the density of water.

Rearrange the above equation for $P_1-P_2$.

$P_1-P_2=\frac{1}{2}\rho \left(v_2{}^2-v_1{}^2\right)+\rho g\left(h_2-h_1\right)$

Substitute $P_{\text{g}}$ for $P_1-P_2$ and $h$ for $h_2-h_1$ in the above equation.

$P_{\text{g}}=\frac{1}{2}\rho \left(v_2{}^2-v_1{}^2\right)+\rho gh$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $2.65\text{m}/\text{s}$ for $v_2$, $0.295\text{m}/\text{s}$ for $v_1$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $2.0\text{m}$ for $h$ in the above equation to find the value of $P_{\text{g}}$.

$\begin{array}{c}{P}_{\text{g}}=\frac{1}{2}\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left[{\left(2.65\text{m}/\text{s}\right)}^2-{\left(0.295\text{m}/\text{s}\right)}^2\right]+\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(2.0\text{m}\right)\\ =2.31\times {10}^4\text{kg}/{\text{ms}}^{\text{2}}\times \left(\frac{1\text{Pa}}{1\text{kg}/{\text{ms}}^{\text{2}}}\right)\\ =2.31\times {10}^4\text{Pa}\end{array}$

Conclusion:

Therefore, the gauge pressure in the main pipe is $2.31\times {10}^4\text{Pa}$.