Chapter 14, Problem 14.44P

A village maintains a large tank with ail open top, containing water for emergencies. The water can drain from the tank through a hose of diameter 6.60 cm. The hose ends with a nozzle of diameter 2.20 cm. A rubber stopper is inserted into the nozzle. The water level in the lank is kept 7.50 m above the nozzle. (a) Calculate the friction force exerted on the stopper by the nozzle. (b) The stopper is removed. What mass of water flows from the nozzle in 2.00 h? (c) Calculate the gauge pressure of the flowing water in the hose just behind the nozzle.

To determine

The friction force which is exerted on the stopper by the nozzle.

Answer to Problem 14.44P

The friction force which is exerted on the stopper by the nozzle is $27.96\text{N}$.

Explanation of Solution

Given info: The diameter of a hose is $6.60\text{cm}$, the diameter of the nozzle is $2.20\text{cm}$, the water level in the tank above the nozzle is $7.50\text{m}$ and the time when water flows from the nozzle is $2.00\text{h}$.

Write the equation for Bernoulli’s equation.

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

Here,

$P_1$ is the pressure at the free surface of the water.

$\rho$ is the density of the water.

$g$ is the acceleration due to gravity.

$v_1$ is the velocity of water at the free surface of water.

$y_1$ is the distance between the water level and the nozzle.

$P_2$ is the pressure at the nozzle.

$v_2$ is the velocity at the nozzle.

$y_2$ is the distance at the nozzle.

The velocity at the free surface $v_1$ and the velocity $v_2$ is zero. The distance inside the nozzle $y_2$ is zero.

Substitute $0$ for $v_1$, $0$ for $v_2$ and $0$ for $y_2$ in above equation.

$\begin{array}{c}{P}_{\text{1}}+\frac{1}{2}\rho {\left(0\right)}^2+\rho g{y}_1=P_2+\frac{1}{2}\rho {\left(0\right)}^2+\rho g\left(0\right)\\ P_2-P_1=\rho g{y}_1\end{array}$

Write the equation for forces in equilibrium.

$\begin{array}{c}{\displaystyle \sum F_{\text{x}}}=0\\ P_{2}A-P_{1}A-f_{\text{k}}=0\\ f_{\text{k}}=\left(P_2-P_1\right)A\end{array}$

Here,

$f_{\text{k}}$ is the friction force.

$A$ is the area of the nozzle.

Substitute $\rho g{y}_1$ for $\left(P_2-P_1\right)$ in above equation to find $f_{\text{k}}$.

$f_{\text{k}}=\left(\rho g{y}_1\right)A$

Write the formula for area of nozzle.

$A=\frac{\pi }{4}{d}_{\text{N}}^2$

Here,

$d_{\text{N}}$ is the diameter of nozzle.

Substitute $\frac{\pi }{4}{d}_{\text{N}}^2$ for $A$ in above equation to find $f_{\text{k}}$.

$f_{\text{k}}=\left(\rho g{y}_1\right)\left(\frac{\pi }{4}{d}_{\text{N}}^2\right)$

The density of water is $1000\text{kg}/{\text{m}}^3$.

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$, $9.81\text{m}/{\text{s}}^2$ for $g$, $7.50\text{m}$ for $y_{\text{1}}$ and $2.20\text{cm}$ for $d_{\text{N}}$ in above equation to find $f_{\text{k}}$.

$\begin{array}{c}{f}_{\text{k}}=\frac{\pi }{4}{\left\{2.20\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right\}}^2\left(1000\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(7.50\text{m}\right)\\ =27.96\text{N}\end{array}$

Conclusion:

Therefore, the friction force which is exerted on the stopper by the nozzle is $27.96\text{N}$.

To determine

The mass of water flows from the nozzle.

Answer to Problem 14.44P

The mass of water flows from the nozzle is $3.32\times {10}^4\text{kg}$.

Explanation of Solution

Given info: The diameter of a hose is $6.60\text{cm}$, the diameter of the nozzle is $2.20\text{cm}$, the water level in the tank above the nozzle is $7.50\text{m}$ and the time when water flows from the nozzle is $2.00\text{h}$.

Write the equation of mass flow rate.

$m=\rho \left(\frac{\pi }{4}{d}_{\text{N}}^2\right)v_{2}t$ (1)

Here,

$m$ is the mass of water which flows from the nozzle.

$t$ is the time.

Write the equation of velocity at the nozzle from Bernoulli’s equation at the point of hose and nozzle.

$v_2=\sqrt{2g{y}_1}$

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$ and $7.50\text{m}$ for $y_{\text{1}}$ in above equation to find $v_2$.

$\begin{array}{c}{v}_2=\sqrt{2\left(9.81\text{m}/{\text{s}}^2\right)\left(7.50\text{m}\right)}\\ =12.1\text{m}/\text{s}\end{array}$

Thus, the velocity at the nozzle is $12.1\text{m}/\text{s}$.

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$, $2.20\text{cm}$ for $d_{\text{N}}$, $12.1\text{m}/\text{s}$ for $v_2$ and $2.00\text{h}$ for $t$ in equation (1) to find $m$.

$\begin{array}{l}m=\left[\begin{array}{l}\left(1000\text{kg}/{\text{m}}^3\right)\left[\frac{\pi }{4}{\left\{2.20\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right\}}^2\right]\\ \left(12.1\text{m}/\text{s}\right)\left\{2.00\text{h}\left(\frac{3600\text{s}}{1\text{h}}\right)\right\}\end{array}\right]\\ m=3.32\times {10}^4\text{kg}\end{array}$

Conclusion:

Therefore, the mass of water flows from the nozzle is $3.32\times {10}^4\text{kg}$.

To determine

The gauge pressure of the flowing water in the hose just behind the nozzle.

Answer to Problem 14.44P

The gauge pressure of the flowing water in the hose just behind the nozzle is $7.2\times {10}^4\text{Pa}$.

Explanation of Solution

Given info: The diameter of a hose is $6.60\text{cm}$, the diameter of the nozzle is $2.20\text{cm}$, the water level in the tank above the nozzle is $7.50\text{m}$ and the time when water flows from the nozzle is $2.00\text{h}$.

Write the equation for Bernoulli’s equation in the hose and the point of nozzle.

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

The level is same at the hose and the nozzle.

Substitute $0$ for $y_1$ and $0$ for $y_2$ in above equation.

$\begin{array}{l}{P}_{\text{1}}+\frac{1}{2}\rho v_1^2+\rho g\left(0\right)=P_2+\frac{1}{2}\rho v_2^2+\rho g\left(0\right)\\ P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)\end{array}$ (2)

Write the equation of continuity equation.

$\begin{array}{c}\left(\frac{\pi }{4}{d}_{\text{H}}^2\right)v_1=\left(\frac{\pi }{4}{d}_{\text{N}}^2\right)v_2\\ v_1=v_2{\left(\frac{d_{\text{N}}}{d_{\text{H}}}\right)}^2\end{array}$

Here,

$d_{\text{H}}$ is the diameter of the hose.

Substitute $12.1\text{m}/\text{s}$ for $v_2$, $2.20\text{cm}$ for $d_{\text{N}}$ and $6.60\text{cm}$ for $d_{\text{H}}$ in above equation.

$\begin{array}{c}{v}_1=\left(12.1\text{m}/\text{s}\right){\left(\frac{\left(2.20\text{cm}\right)}{\left(6.60\text{cm}\right)}\right)}^2\\ =1.35\text{m}/\text{s}\end{array}$

Thus, the value of velocity $v_1$ is $1.35\text{m}/\text{s}$.

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$, $1.35\text{m}/\text{s}$ for $v_1$ and $12.1\text{m}/\text{s}$ for $v_2$ in equation (2) to find $m$.

$\begin{array}{c}{P}_1-P_2=\frac{1}{2}\left(1000\text{kg}/{\text{m}}^3\right)\left\{{\left(12.1\text{m}/\text{s}\right)}^2-{\left(1.35\text{m}/\text{s}\right)}^2\right\}\\ =7.2\times {10}^4\text{Pa}\end{array}$

Conclusion:

Therefore, the gauge pressure of the flowing water in the hose just behind the nozzle is $7.2\times {10}^4\text{Pa}$.