Question

Review: In a water pistol, a piston drives water through a large tube of area ${\mathit{A}}_{\mathbf{1}}$ into a smaller tube of area ${\mathit{A}}_{\mathbf{2}}$ as shown in Figure P14.78. 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 ${\mathit{V}}_{\mathbf{2}}$must the stream leave the nozzle? (c) At what speed ${\mathit{V}}_{\mathbf{1}}$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.)

Short answer

(a) The time interval required for the water to travel from the nozzle to the ground is $t=0.553s$.

(b) The speed $V_2$ must the stream leave the nozzle is $V_{nozzle}=14.5m/s$.

(c) The speed $V_1$ must the plunger be moved to achieve the desired range is $V_1=0.145m/s$.

(d) The pressure at the nozzle is $P_2=1.013\times {10}^{5}Pa$.

(e) The pressure needed in the larger tube is $P_1=2.06\times {10}^{5}Pa$ .

(f) The force that must be exerted on the trigger to achieve the desired range is $F_1=33.0N$.

Step-by-step solution

A concept:

The flow rate (volume flux) through a pipe that varies in cross-sectional area is constant; that is equivalent to stating that the product of the cross-sectional area A and the speed $\mathit{v}$ at any point is a constant. This result is expressed in the equation of continuity for fluids:

role="math" localid="1663667478243" ${\mathit{A}}_{\mathbf{1}}{\mathit{V}}_{\mathbf{1}}\mathbf{-}{\mathit{A}}_{\mathbf{2}}{\mathit{V}}_{\mathbf{2}}\mathbf{-}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}$

The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume have the same value at all points along a streamline for an ideal fluid.

This result is summarized in Bernoulli’s equation:

role="math" localid="1663667571358" $\mathit{P}\mathbf{+}\mathit{\rho }\mathit{g}\mathit{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{\rho }{\mathit{v}}^{\mathbf{2}}\mathbf{=}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}$

Here, $\mathit{P}$ is the pressure, $\mathit{\rho }$ is the density, $\mathit{g}$ is the gravity, and $\mathit{y}$ is the distance.

(a) Determine the time interval:

Since the pistol is fired horizontally, the emerging water stream has initial velocity components of $(v_{0x}=v_{nozzle},v_{0y}=0)$. Then, $∆y=v_{0y}t+\frac{1}{2}{a}_y{t}^2$, with $a_y=-g$, gives the time of flight as,

$$\begin{gathered} t=\sqrt{\frac{2(∆y)}{a_y}} \\ =\sqrt{\frac{2(-1.50m)}{9.8m/s^2}} \\ =0.553s \end{gathered}$$

(b): Define the speed V2  that must the stream leave the nozzle:

With $a_x=0$, and $v_{ox}=v_{nozzle}$, the horizontal range of the emergent stream is $∆x=v_{nozzle}t$

Where, t is the time of flight from above.

Thus, the speed of the water emerging from the nozzle is,

$$\begin{gathered} v_{nozzle}=\frac{∆x}{t} \\ =\frac{8.00m}{0.553s} \\ =14.5m/s \end{gathered}$$

(c) Achieve the desired range of velocity:

Consider the given data as below.

The radius,

The radius,

From the equation of continuity,

The speed of the water in the larger cylinder is,

$$\begin{gathered} V_1=\frac{A_2}{A_1}{V}_2 \\ =\frac{A_2}{A_1}{v}_{nozzle} \\ =\frac{{\mathrm{\pi r}}_2^2}{{\mathrm{\pi r}}_1^2}{v}_{nozzle} \\ =\left(\frac{r_2}{r_1}\right)v_{nozzle} \end{gathered}$$

Substitute known values in the above equation.

$$\begin{gathered} V_1={\left(\frac{1.00mm}{10.0mm}\right)}^2\times 14.5m/s \\ =0.145m/s \end{gathered}$$

(d) The pressure at the nozzle:

The pressure at the nozzle is atmospheric pressure, that is

$P_2=1.013\times {10}^{5}Pa$

(e) Find the pressure needed in the larger tube:

With the two cylinders horizontal, $y_1=y_2$and gravity terms from Bernoulli’s equation can be neglected, leaving

$P_1+\frac{1}{2}{\rho }_w{v}_1^2=P_2+\frac{1}{2}{\rho }_w{v}_2^2$

So, the needed pressure in the larger cylinder is

$$\begin{gathered} P_1=P_2\frac{1}{2}{\rho }_w(v_2^2-v_2^1) \\ =(1.013\times {10}^{5}Pa)+\frac{1.00\times {10}^{3}kg/m^3}{2}{(14.5m/s)}^2-{(0.145m/s)}^2 \\ =2.06\times {10}^{5}Pa \end{gathered}$$

(f) The force that must be exerted on the trigger to achieve the desired range:

To create an overpressure of $∆P$in the larger cylinder.

$$\begin{gathered} ∆P=2.06\times {10}^{5}Pa \\ =1.05\times {10}^{5}Pa \end{gathered}$$

The force that must be exerted on the piston is,

$$\begin{gathered} F_1=∆P{A}_1 \\ =∆P\times {\mathrm{\pi r}}_1^2 \\ =(1.05\times {10}^5\mathrm{Pa})(3.14){(1.00\times {10}^{-2}\mathrm{m})}^2=33.0\mathrm{N} \end{gathered}$$