Chapter 14, Problem 46AP

Review. In a water pistol, a piston drives water through a large tube of area A₁ into a smaller tube of area A₂ as shown in Figure P14.46. The radius of the large tube is 1.00 cm and that of the small tube is 1.00 mm. The smaller tube is 3.00 cm above the larger tube. (a) If the pistol is fired horizontally at a height of 1.50 m, determine the time interval required for the water to travel from the nozzle to the ground. Neglect air resistance and assume atmospheric pressure is 1.00 atm. (b) If the desired range of the stream is 8.00 m, with what speed v₂ must the stream leave the nozzle? (c) At what speed v₁ must the plunger be moved to achieve the desired range? (d) What is the pressure at the nozzle? (e) Find the pressure needed in the larger tube. (f) Calculate the force that must be exerted on the trigger to achieve the desired range. (The force that must be exerted is due to pressure over and above atmospheric pressure.)

Figure P14.46

To determine

The time required for the water to travel from the nozzle to the ground.

Answer to Problem 46AP

The time required for the water to travel from the nozzle to the ground is $0.553\text{s}$.

Explanation of Solution

The radius of the large tube is $1.00\text{cm}$ and radius of the small tube is $1.00\text{mm}$. The smaller tube is $3.00\text{cm}$ above the larger tube, the pistol is fired at a height $1.50\text{m}$ with the horizontal.

Formula to calculate the time interval is,

    $t=\sqrt{\frac{2h}{g}}$

Here, $h$ is the height of the pistol above the horizontal and $g$ is the acceleration due to gravity.

Substitute $1.50\text{m}$ for $h$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $t$.

    $\begin{array}{c}t=\sqrt{\frac{2\times 1.50\text{m}}{9.81\text{m}/{\text{s}}^{\text{2}}}}\\ =0.553\text{s}\end{array}$

Conclusion:

Therefore, the time required for the water to travel from the nozzle to the ground is $0.553\text{s}$.

To determine

The speed of the stream to leave the nozzle if the range of the stream is $8.00\text{m}$.

Answer to Problem 46AP

The speed of the stream to leave the nozzle is $14.46\text{m}/\text{s}$ if the range of the stream is $8.00\text{m}$.

Explanation of Solution

Formula to calculate the speed of the stream to leave the nozzle is,

    $v_2=\frac{d}{t}$

Here, $d$ is the range of the stream and $v_2$ is the speed of the stream to leave the nozzle.

Substitute $8.00\text{m}$ for $d$ and $0.553\text{s}$ for $t$ to find $v_2$.

    $\begin{array}{c}{v}_2=\frac{8.00\text{m}}{0.553\text{s}}\\ =14.46\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the stream to leave the nozzle is $14.46\text{m}/\text{s}$ if the range of the stream is $8.00\text{m}$.

To determine

The speed of the plunger is moved to achieve the range of $8.00\text{m}$.

Answer to Problem 46AP

The speed of the plunger is $0.1446\text{m}/\text{s}$ to achieve the range of $8.00\text{m}$.

Explanation of Solution

By continuity equation at the plunger and exit point of the nozzle is,

    $A_1{v}_1=A_1{v}_2$        (I)

Here, $A_1$ is the area of the large tube, $A_2$ is the area of the smaller tube and $v_1$ is the speed of the plunger.

Formula to calculate the area of the large tube is,

    $A_1=\pi r_1{}^2$

Here, $r_1$ is the radius of the large tube.

Formula to calculate the area of the small tube is,

    $A_2=\pi r_2{}^2$

Here, $r_2$ is the radius of the small tube.

Substitute $\pi r_1{}^2$ for $A_1$ and $\pi r_2{}^2$ for $A_2$ in equation (I).

`    $\begin{array}{c}\pi r_1{}^2\times v_1=\pi r_2{}^2\times v_2\\ v_1=\frac{r_2{}^2\times v_2}{r_1{}^2}\end{array}$

Substitute $1.00\text{cm}$ for $r_1$, $1.00\text{mm}$ for $r_2$ and $14.46\text{m}/\text{s}$ for $v_2$ to find $v_1$.

    $\begin{array}{c}{v}_1=\frac{{\left(1.00\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2\times 14.46\text{m}/\text{s}}{{\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2}\\ =0.1446\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the plunger is $0.1446\text{m}/\text{s}$ to achieve the range of $8.00\text{m}$.

To determine

The pressure at the nozzle.

Answer to Problem 46AP

The pressure at the nozzle is $101325\text{Pa}$.

Explanation of Solution

The pressure at the nozzle is equal to the atmospheric pressure.

The atmospheric pressure is equal to the $101325\text{Pa}$, hence the pressure at the muzzle is equal to the $101325\text{Pa}$.

Conclusion:

Therefore, the pressure at the nozzle is $101325\text{Pa}$.

To determine

The pressure needed in the large tube.

Answer to Problem 46AP

The pressure needed in the large tube is $205860\text{Pa}$.

Explanation of Solution

Apply the Bernoulli’s equation at point $1$ and point $2$.

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

Here, $P_1$ is the pressure in the large tube, $y_1$ is the elevation at point $1$, $P_2$ is the pressure at the exit point of the nozzle, $y_2$ is the elevation at point $2$, $\rho$ is the density of the water and $g$ is the acceleration due to gravity.

Substitute $0$ for $y_1$, $0.1446\text{m}/\text{s}$ for $v_1$, $101325\text{Pa}$ for $P_2$, $14.46\text{m}/\text{s}$ for $v_2$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0$ for $y_2$ to find $P_1$.

    $\begin{array}{c}{P}_1+\frac{1}{2}1000\text{kg}/{\text{m}}^{\text{3}}\times {\left(0.1446\text{m}/\text{s}\right)}^2+\rho g\left(0\right)=101325\text{Pa}+\frac{1}{2}1000\text{kg}/{\text{m}}^{\text{3}}\times {\left(14.46\text{m}/\text{s}\right)}^2+\rho g\left(0\right)\\ P_1=101325\text{Pa}+\frac{1}{2}1000\text{kg}/{\text{m}}^{\text{3}}\left({\left(14.46\text{m}/\text{s}\right)}^2-{\left(0.1446\text{m}/\text{s}\right)}^2\right)\\ =205860\text{Pa}\end{array}$

Conclusion:

Therefore, the pressure needed in the large tube is $205860\text{Pa}$.

To determine

The force exerted on the trigger to achieve the range of $8.00\text{m}$.

Answer to Problem 46AP

The force exerted on the trigger to achieve the range of $8.00\text{m}$ is $35.84\text{N}$.

Explanation of Solution

Formula to calculate the force exerted on the trigger is,

    $F=\left(P_1-P_2\right)\times A_1$

Here, $F$ is the force exerted on the trigger.

Substitute $\pi r_1{}^2$ for $A_1$ in the above expression.

    $F=\left(P_1-P_2\right)\times \pi r_1{}^2$

Substitute $1.00\text{cm}$ for $r_1$, $101325\text{Pa}$ for $P_2$, $205860\text{Pa}$ for $P_1$ to find $F$.

    $\begin{array}{c}F=\left(205860\text{Pa}-101325\text{Pa}\right)\times \pi {\left(1.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^2\\ =35.84\text{N}\end{array}$

Conclusion:

Therefore, the force exerted on the trigger to achieve the range of $8.00\text{m}$ is $35.84\text{N}$.