Chapter 15, Problem 69P

(a)

To determine

The speed at which the water leaves the faucet.

Answer to Problem 69P

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

Explanation of Solution

Write the equation for the volume flow rate.

  $Q=\frac{\Delta V}{\Delta t}$        (I)

Here, $Q$ is the volume flow rate, $\Delta V$ is the volume of liquid flowed in time interval $\Delta t$ .

Write the equation for the volume flow rate in terms of the speed of the water.

  $Q=Av$

Here, $A$ is the area through which the liquid moves and $v$ is the speed of the liquid.

Rewrite the above equation for $v$ .

  $v=\frac{Q}{A}$        (II)

Write the equation for $A$ .

  $A=\frac{\pi d^2}{4}$

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

Put the above equation in equation (II).

  $\begin{array}{c}v=\frac{Q}{\frac{\pi d^2}{4}}\\ =\frac{4Q}{\pi d^2}\end{array}$        (III)

Conclusion:

Substitute $25\text{ L}$ for $\Delta V$ and $30.0\text{ s}$ for $\Delta t$ in equation (I) to find $Q$ .

  $\begin{array}{c}Q=\frac{25.0\text{ L}\frac{1000{\text{ cm}}^3}{1.0\text{ L}}}{30.0\text{ s}}\\ =833{\text{ cm}}^3\text{/s}\end{array}$

Substitute $833{\text{ cm}}^3\text{/s}$ for $Q$ and $2.00\text{ cm}$ for $d$ in equation (III) to find $v$ .

  $\begin{array}{c}v=\frac{4\left(833{\text{ cm}}^3\text{/s}\right)}{\pi {\left(2.00\text{ cm}\right)}^2}\\ =265\text{ cm/s}\frac{1\text{ m}}{100\text{ cm}}\\ =2.65\text{ m/s}\end{array}$

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

(b)

To determine

The gauge pressure in the $6\text{ cm}$ main pipe.

Answer to Problem 69P

The gauge pressure in the $6\text{ cm}$ main pipe is $2.31\times {10}^4\text{ Pa}$.

Explanation of Solution

Take point 1 to be in the entrance pipe and point 2 to be at the faucet pipe.

Write the continuity equation of fluids.

  $A_1{v}_1=A_2{v}_2$

Here, $A_1$ is the area of the entrance pipe, $v_1$ is the speed of the water at point 1, $A_2$ is the area of the faucet pipe and $v_2$ is the speed of the water at point 2.

Rewrite the above equation for $v_1$.

  $v_1=\frac{A_2{v}_2}{A_1}$        (IV)

Write the equation for $A_2$ .

  $A_2=\frac{\pi d_2^2}{4}$

Here, $d_2$ is the diameter of the faucet tap.

Write the equation for $A_1$ .

  $A_1=\frac{\pi d_1^2}{4}$

Here, $d_1$ is the diameter of the entrance pipe.

Put the above two equations in equation (IV).

  $\begin{array}{c}{v}_1=\frac{\left(\frac{\pi d_2^2}{4}\right)v_2}{\left(\frac{\pi d_1^2}{4}\right)}\\ ={\left(\frac{d_2}{d_1}\right)}^2{v}_2\end{array}$        (V)

Write the Bernoulli’s equation.

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

Here, $P_1$ is the pressure at the point 1, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity, $y_1$ is the height of the point 1 above the reference position, $P_2$ is the pressure at the point 2 and $y_2$ is the height of the point 2 above the reference position.

Rearrange the above equation.

  $\begin{array}{c}{P}_1-P_2=\left(\frac{1}{2}\rho v_2^2-\frac{1}{2}\rho v_1^2\right)+\left(\rho g{y}_2-\rho g{y}_1\right)\\ =\frac{1}{2}\rho \left(v_2^2-v_1^2\right)+\rho g\left(y_2-y_1\right)\end{array}$        (VI)

Write the equation for gauge pressure.

  $P_{\text{gauge}}=P_1-P_2$

Here, $P_{\text{gauge}}$ is the gauge pressure.

Put the above equation in equation (VI).

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

Conclusion:

The density of water is $1000{\text{ kg/m}}^3$ , value of $g$ is $9.80{\text{ m/s}}^2$ and the value of $v_2$ is found to be $2.65\text{ m/s}$ in part (a).

Substitute $2.00\text{ cm}$ for $d_2$ , $6.00\text{ cm}$ for $d_1$ and $2.65\text{ m/s}$ for $v_2$ in equation (V) 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 $1000{\text{ kg/m}}^3$ for $\rho$ , $2.65\text{ m/s}$ for $v_2$ , $0.295\text{ m/s}$ for $v_1$ , $9.80{\text{ m/s}}^2$ for $g$ and $2.00\text{ m}$ for $y_2-y_1$ in equation (VII) 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 $6\text{ cm}$ main pipe is $2.31\times {10}^4\text{ Pa}$.