Chapter 14, Problem 80AP

(a)

To determine

The speed with which the water leaves the faucet.

Answer to Problem 80AP

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

Explanation of Solution

Given that the diameter of the faucet tap is $2.00\text{cm}$, the diameter of the main pipe is $6.00\text{cm}$, the height of the faucet pipe above main pipe is $2.00\text{m}$, the capacity of the tank is $25.0\text{L}$, and the time taken to fill the tank is $30.0\text{s}$.

The flow rate of water through the pipe is always a constant. Write the expression for the volume flow rate of the water.

  $\text{flow rate}=\frac{\Delta V}{\Delta t}$                                                                                                          (I)

Here, $\Delta V$ is the volume of water transferred, and $\Delta t$ is the time duration.

Write the expression for the flow rate in terms of area of the pipe.

  $\text{flow rate}=Av$                                                                                                          (II)

Here, $A$ is the cross sectional area of the pipe, and $v$ is the speed of flow.

Equate the right hand sides of equations (I) and (II).

  $\frac{\Delta V}{\Delta t}=Av$                                                                                                                 (III)

Write the expression for the cross sectional area of the faucet tap in terms of its diameter.

  $A=\frac{\pi d^2}{4}$                                                                                                                 (IV)

Here, $d$ is the diameter of the faucet tap.

Use equation (IV) in (III) and solve for $v$

  $\begin{array}{c}\frac{\Delta V}{\Delta t}=\left(\frac{\pi d^2}{4}\right)v\\ v=\frac{4\Delta V}{\pi d^2\Delta t}\end{array}$                                                                                                        (V)

Conclusion:

Substitute $2.00\text{cm}$ for $d$, $25.0\text{L}$ for $\Delta V$, and $30.0\text{s}$ for $t$ in equation (V) to find $v$.

  $\begin{array}{c}v=\frac{4\left(25.0\text{L}\right)}{\pi {\left(2.00\text{cm}\right)}^2\left(30.0\text{s}\right)}\\ =\frac{4\left(25.0\text{L}\times \frac{1{\text{m}}^3}{1000\text{L}}\right)}{\pi {\left(2.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\left(30.0\text{s}\right)}\\ =2.65\text{m/s}\end{array}$

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

(b)

To determine

The gauge pressure in the main pipe.

Answer to Problem 80AP

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

Explanation of Solution

Given that the diameter of the faucet tap is $2.00\text{cm}$, the diameter of the main pipe is $6.00\text{cm}$, the height of the faucet pipe above main pipe is $2.00\text{m}$, the capacity of the tank is $25.0\text{L}$, and the time taken to fill the tank is $30.0\text{s}$. It is obtained that speed with which the water leaves the faucet is $2.65\text{m/s}$.

Consider two points 1 and 2 such that point 1 is in the entrance pipe and point 2 is in the faucet tap.

Write the continuity equation for the flow of water through points 1 and 2.

  $A_1{v}_1=A_2{v}_2$                                                                                                               (VI)

Here, $A_1$ is the cross sectional area of the main pipe, $v_1$ is the speed of flow through main pipe, $A_2$ is the cross sectional area of the faucet tap, $v_2$ is the speed of flow through faucet tap.

Rewrite equation (VI) in terms of the diameters of the pipes and solve for $v_1$.

  $\begin{array}{c}\frac{\pi d_1^2}{4}{v}_1=\frac{\pi d_2^2}{4}{v}_2\\ d_1^2{v}_1=d_2^2{v}_2\\ v_1={\left(\frac{d_2}{d_1}\right)}^2{v}_2\end{array}$                                                                                                  (VII)

Write Bernaulli’s equation for the give flow.

  $P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)+\rho g\left(y_2-y_1\right)$                                                                    (VIII)

Here, $P_1$ is the pressure at point 1, $P_2$ is the pressure at point 2, $\rho$ is the density of water, and $\left(y_2-y_1\right)$ represents the vertical separation between the two points.

Since, $P_1-P_2$ represents the gauge pressure in the main pipe, equation (VIII) can be modified as,

  $P_{\text{gauge}}=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)+\rho g\left(y_2-y_1\right)$                                                                        (IX)

Conclusion:

Substitute $6.00\text{cm}$ for $d_1$, $2.00\text{cm}$ for $d_2$, and $2.65\text{m/s}$ for $v_2$ in equation (VII) to find $v_1$.

  $\begin{array}{c}{v}_1={\left(\frac{2.00\text{cm}}{6.00\text{cm}}\right)}^2\left(2.65\text{m/s}\right)\\ =0.295\text{m/s}\end{array}$

Substitute $0.295\text{m/s}$ for $v_1$, $2.65\text{m/s}$ for $v_2$, $1000{\text{kg/m}}^3$ for $\rho$, $9.80{\text{m/s}}^2$ for $g$, and $2.00\text{m}$ for $\left(y_2-y_1\right)$ in equation (IX) to find $P_{\text{gauge}}$.

  $\begin{array}{c}{P}_{\text{gauge}}=\frac{1}{2}\left(1000{\text{kg/m}}^3\right)\left({\left(2.65\text{m/s}\right)}^2-{\left(0.295\text{m/s}\right)}^2\right)+\left(1000{\text{kg/m}}^3\right)\left(9.80{\text{m/s}}^2\right)\left(2.00\text{m}\right)\\ =2.31\times {10}^4\text{Pa}\end{array}$

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