Chapter 14, Problem 44P

(a)

To determine

The frictional force exerted on the stopper by the nozzle.

Answer to Problem 44P

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

Explanation of Solution

Given that the diameter of the hose is $6.60\text{cm}$, the diameter of the nozzle is $2.20\text{cm}$, the water level is $7.50\text{m}$ above the nozzle.

Consider two points, point 1 at the free surface of the water in the tank and point 2 inside the nozzle.

Apply Bernoulli’s theorem to the given situation.

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

Here, $P_1$ is the pressure at the top of the tank, $\rho$ is the density of water, $v_1$ is the speed of water at the top of the tank, $g$ is the acceleration due to gravity, $y_1$ is the height at which the water at the top of the tank occupies, $P_2$ is the pressure at the nozzle, $v_2$ is the speed of water coming through the nozzle, and $y_2$ is the height at which the nozzle present.

Reduce the expression (I) for $P_2-P_1$.

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

Since the pressure $P_1$ is equal to the atmospheric pressure $P_0$, and the speed of water is zero on both sides, equation (II) can be rewritten as,

  $P_2-P_0=\rho g\left(y_1-y_2\right)$                                                                                              (III)

The total force acting on the stopper must be zero.

  $F_{\text{water}}-F_{\text{air}}-f=0$                                                                                                   (IV)

Here, $F_{\text{water}}$ is the force exerted by the water, $F_{\text{air}}$ is the force exerted by the air outside, and $f$ is the frictional force exerted on the stopper by the nozzle.

Write the expression for the force exerted by water on the stopper.

  $F_{\text{water}}=P_{2}A$                                                                                                               (V)

Here, $A$ is the cross-sectional area of the nozzle.

Write the expression for the force exerted by air on the stopper.

  $F_{\text{air}}=P_{0}A$                                                                                                                (VI)

Use equation (V), and (VI) in (IV) and solve for $f$.

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

Write the expression for the cross sectional area of the nozzle.

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

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

Use equation (III), and (VIII) in (VII).

  $f=\frac{\pi d^2\rho g\left(y_1-y_2\right)}{4}$                                                                                              (IX)

Conclusion:

Substitute $2.20\text{cm}$ for $d$, $1000{\text{kg/m}}^3$ for $\rho$, $7.50\text{m}$ for $\left(y_1-y_2\right)$, and $9.80{\text{m/s}}^2$ for $g$ in equation (IX) to find $f$.

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

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

(b)

To determine

The mass of water flows through the nozzle in $2.00\text{h}$ when the stopper is removed.

Answer to Problem 44P

The mass of water flows through the nozzle in $2.00\text{h}$ when the stopper is removed is $\underset{\_}{3.32\times {10}^4\text{kg}}$.

Explanation of Solution

Equation (II) gives the pressure difference at both ends of the stopper.

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

When the stopper is removed, the pressure on both points become atmospheric pressure $P_0$. Moreover, considering $v_1=0$, equation (II) can be reduced to find $v_2$.

  $v_2=\sqrt{2g\left(y_1-y_2\right)}$                                                                                                    (X)

Write the expression for the mass of water leaving the nozzle.

  $m=\rho V$                                                                                                                   (XI)

Here, $m$ is the mass of water, and $V$ is the volume of the water leaving the nozzle.

Write the expression for the volume of water leaving the nozzle with speed $v_2$ in time $t$.

  $V=A{v}_{2}t$                                                                                                                (XII)

Use equation (XII) and (VIII) in (XI).

  $m=\frac{\pi d^2\rho v_{2}t}{4}$                                                                                                        (XIII)

Conclusion:

Substitute $7.50\text{m}$ for $\left(y_1-y_2\right)$, and $9.80{\text{m/s}}^2$ for $g$ in equation (X) 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 $1000{\text{kg/m}}^3$ for $\rho$, $2.20\text{cm}$ for $d$, $12.1\text{m/s}$ for $v_2$, and $2.00\text{h}$ for $t$ in equation (XIII) to find $m$.

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

Therefore, the mass of water flows through the nozzle in $2.00\text{h}$ when the stopper is removed is $\underset{\_}{3.32\times {10}^4\text{kg}}$.

(c)

To determine

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

Answer to Problem 44P

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

Explanation of Solution

Write the continuity equation for the flow of water through the hose and nozzle.

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

Here, $A_1$ is the cross sectional area of the hose, $v_1$ is the speed of water in the hose, $A_2$ is the cross sectional area of the nozzle, and $v_2$ is the speed of water through the nozzle.

Solve equation (XIV) for $v_1$.

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

Rewrite equation (XV) in terms of the diameters of the hose and nozzle.

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

Rewrite equation (II) for the current situation, such that $P_2=P_0$, $P_1$ is the pressure inside the hose, and $y_2=y_1$.

  $P_1-P_0=\frac{\rho }{2}\left(v_2^2-v_1^2\right)$                                                                                            (XVII)

Since the gauge pressure $P_{\text{gauge}}$ inside the pipe is equal to $P_1-P_0$, the equation (XVII) can be modified as,

  $P_{\text{gauge}}=\frac{\rho }{2}\left(v_2^2-v_1^2\right)$                                                                                             (XVII)

Conclusion:

Substitute $6.60\text{cm}$ for $d_1$, $2.20\text{cm}$ for $d_2$, and $12.1\text{m/s}$ for $v_2$ in equation (XVI) to find $v_1$.

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

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

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

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