Chapter 15, Problem 45P

A village maintains a large tank with an 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 tank 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 exerted on the stopper by the nozzle.

Answer to Problem 45P

The friction force exerted on the stopper by the nozzle is $27.9\text{ N}$.

Explanation of Solution

Take point 1 to be at the free surface of the water in the tank and point 2 to be inside the nozzle. Since the top of the tank is open to the atmosphere the pressure at point 1 will be equal to atmospheric pressure. Since the tank has large area compared to the nozzle, the speed of water at point 1 will be very small compared to that at point 2 according to continuity equation and can be considered to be zero. Also assume the position of the nozzle to be $y_2=0\text{ m}$.

Write the Bernoulli’s equation.

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

Here, $P_1$ is the pressure at the point 1, $\rho$ is the density of the liquid, $v_1$ is the speed of the liquid at point 1, $g$ 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, $v_2$ is the speed of the liquid at point 2 and $y_2$ is the height of the point 2 above the reference position.

Substitute $P_0$ for $P_1$ , $0\text{ m/s}$ for $v_1$ , $0\text{ m/s}$ for $v_2$ and $0\text{ m}$ for $y_2$ in equation (I) and rearrange it.

  $\begin{array}{c}{P}_0+\frac{1}{2}\rho \left(0\text{ m/s}\right)+\rho g{y}_1=P_2+\frac{1}{2}\rho \left(0\text{ m/s}\right)+\rho g\left(0\text{ m}\right)\\ P_0+\rho g{y}_1=P_2\\ P_2-P_0=\rho g{y}_1\end{array}$        (II)

The diagram of the different forces on the system is shown below.

In the diagram $f$ is the frictional force.

The sum of the forces in the horizontal direction must be zero for the stopper.

Write the equilibrium condition for the stopper.

  $\Sigma F_x=0$        (III)

Here, $\Sigma F_x$ is the sum of the forces in the horizontal direction.

Write the expression for $\Sigma F_x$.

  $\Sigma F_x=F_{\text{water}}-F_{\text{air}}-f$

Here, $F_{\text{water}}$ is the force exerted by water and $F_{\text{air}}$ is the force exerted by air.

Put the above equation in equation (III) and rewrite it for $f$.

  $\begin{array}{c}{F}_{\text{water}}-F_{\text{air}}-f=0\\ f=F_{\text{water}}-F_{\text{air}}\end{array}$        (IV)

Write the equation for pressure.

  $P=\frac{F}{A}$

Here, $P$ is the pressure and $A$ is the area over which the force acts.

Rewrite the above equation for $F$.

  $F=PA$        (V)

Use equation (V) to write the expression for $F_{\text{water}}$.

  $F_{\text{water}}=P_{2}A$

Here, $A$ is the area of the nozzle.

Use equation (V) to write the expression for $F_{\text{air}}$.

  $F_{\text{air}}=P_{0}A$

Put the above two equations in equation (IV).

  $\begin{array}{c}f=P_{2}A-P_{0}A\\ =\left(P_2-P_0\right)A\end{array}$        (VI)

Write the equation for $A$.

  $A=\frac{\pi d_n^2}{4}$        (VII)

Here, $d_n$ is the diameter of the nozzle.

Put equations (II) and (VII) in equation (VI).

  $f=\rho g{y}_1\frac{\pi d^{{}^2}}{4}$        (VIII)

Conclusion:

The value of $P_0$ is $1.013\times {10}^5\text{ Pa}$, The value of $g$ is $9.80{\text{ m/s}}^2$ and the density of water is $1000{\text{ kg/m}}^3$.

Substitute $1000{\text{ kg/m}}^3$ for $\rho$ , $9.80{\text{ m/s}}^2$ for $g$ , $7.50\text{ m}$ for $y_1$ and $2.20\text{ cm}$ for $d_n$ in equation (VIII) to find $f$.

  $\begin{array}{c}f=\frac{\left(1000{\text{ kg/m}}^3\right)\left(9.80{\text{ m/s}}^2\right)\left(7.50\text{ m}\right)\pi {\left(2.20\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}^2}{4}\\ =27.9\text{ N}\end{array}$

Therefore, the friction force exerted on the stopper by the nozzle is $27.9\text{ N}$.

To determine

The mass of water that flows from the nozzle in $2.00\text{ h}$.

Answer to Problem 45P

The mass of water that flows from the nozzle in $2.00\text{ h}$ is $3.32\times {10}^4\text{ kg}$.

Explanation of Solution

When the stopper is removed, point 2 is also at the atmospheric pressure.

Substitute $P_0$ for $P_1$ , $P_0$ for $P_2$ , $0\text{ m/s}$ for $v_1$ , and $0\text{ m}$ for $y_2$ in equation (I) and rearrange it for $v_2$.

  $\begin{array}{c}{P}_0+\frac{1}{2}\rho \left(0\text{ m/s}\right)+\rho g{y}_1=P_0+\frac{1}{2}\rho v_2^2+\rho g\left(0\text{ m}\right)\\ \rho g{y}_1=\frac{1}{2}\rho v_2^2\\ v_2^2=2g{y}_1\\ v_2=\sqrt{2g{y}_1}\end{array}$        (IX)

Write the equation for density of water.

  $\rho =\frac{m_{\text{water}}}{V}$        (X)

Here, $\rho$ is the density of water, $m_{\text{water}}$ is the mass of water and $V$ is the volume of water.

Write the equation for volume of water.

  $V=Ay$        (XI)

Here, $y$ is the distance through which the water falls.

Write the equation for the velocity of water at point 2.

  $v_2=\frac{y}{t}$

Here, $t$ is the time.

Rewrite the above equation for $y$ .

  $y=v_{2}t$

Put equation (IX) in the above equation.

  $y=\sqrt{2g{y}_1}t$        (XII)

Put equations (VII) and (XII) in equation (XI).

  $V=\frac{\pi d_n^2}{4}\sqrt{2g{y}_1}t$

Put the above equation in equation (X) and rewrite it for $m_{\text{water}}$.

  $\begin{array}{c}\rho =\frac{m_{\text{water}}}{\frac{\pi d_n^2}{4}\sqrt{2g{y}_1}t}\\ m_{\text{water}}=\frac{\pi \rho t{d}_n^2}{4}\sqrt{2g{y}_1}\end{array}$        (XIII)

Conclusion:

Substitute $1000{\text{ kg/m}}^3$ for $\rho$ , $2.00\text{ h}$ for $t$ , $2.20\text{ cm}$ for $d_n$ , $9.80{\text{ m/s}}^2$ for $g$ and $7.50\text{ m}$ for $y_1$ and in equation (XIII) to find $m_{\text{water}}$.

  $\begin{array}{c}{m}_{\text{water}}=\frac{\pi \left(1000{\text{ kg/m}}^3\right)\left(2.00\text{ h}\frac{60\times 60\text{ s}}{1.00\text{ h}}\right){\left(2.20\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)}^2}{4}\sqrt{2\left(9.80{\text{ m/s}}^2\right)\left(7.50\text{ m}\right)}\\ =3.32\times {10}^4\text{ kg}\end{array}$

Therefore, the mass of water that flows from the nozzle in $2.00\text{ h}$ 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 45P

The gauge pressure of the flowing water in the hose just behind the nozzle is $7.26\times {10}^4\text{ Pa}$.

Explanation of Solution

Assume point 1 to be in the wide hose and point 2 to be just outside the nozzle.

Write the continuity equation of fluids.

  $A_1{v}_1=A_2{v}_2$        (XIV)

Here, $A_1$ is the area through which the liquid moves at point 1 and $A_2$ is the area through which the liquid moves at point 2.

Write the equation for $A_1$.

  $A_1=\frac{\pi d_h^2}{4}$

Here, $d_h$ is the diameter of the hose.

Write the equation for $A_2$.

  $A_2=\frac{\pi d_n^2}{4}$

Put the above two equations in equation (XIV) and rewrite it for $v_1$.

  $\begin{array}{c}\frac{\pi d_h^2}{4}{v}_1=\frac{\pi d_n^2}{4}{v}_2\\ v_1={\left(\frac{d_n}{d_h}\right)}^2{v}_2\end{array}$        (XV)

Take the vertical position of point 1 and point 2 to be zero.

Substitute $P_0$ for $P_2$ , $0\text{ m}$ for $y_1$ and $0\text{ m}$ for $y_2$ in equation (I) and rearrange it.

  $\begin{array}{c}{P}_1+\frac{1}{2}\rho v_1^2+\rho g\left(0\text{ m}\right)=P_0+\frac{1}{2}\rho v_2^2+\rho g\left(0\text{ m}\right)\\ P_1+\frac{1}{2}\rho v_1^2=P_0+\frac{1}{2}\rho v_2^2\\ P_1-P_0=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)\end{array}$        (XVI)

Conclusion:

Substitute $9.80{\text{ m/s}}^2$ for $g$ and $7.50\text{ m}$ for $y_1$ in equation (IX) to find $v_2$ .

  $\begin{array}{c}{v}_2=\sqrt{2\left(9.80{\text{ m/s}}^2\right)\left(7.50\text{ m}\right)}\\ =12.1\text{ m/s}\end{array}$

Substitute $2.20\text{ cm}$ for $d_n$ , $6.60\text{ cm}$ for $d_h$ and $12.1\text{ m/s}$ for $v_2$ in equation (XV) to find $v_1$ .

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

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

  $\begin{array}{c}{P}_1-P_0=\frac{1}{2}\left(1000{\text{ kg/m}}^3\right)\left[{\left(12.1\text{ m/s}\right)}^2-{\left(1.35\text{ m/s}\right)}^2\right]\\ =7.26\times {10}^4\text{ Pa}\end{array}$

Therefore, the gauge pressure of the flowing water in the hose just behind the nozzle is $7.26\times {10}^4\text{ Pa}$.