Chapter 10, Problem 83GP

(a)

To determine

To Find: The velocity of the water.

Answer to Problem 83GP

The velocity of water coming out from each sprinkler head is $9.13\text{ m/s}$ .

Explanation of Solution

Given:

The diameter of the pipe, $d=1.9\text{ cm = }0.019\text{ m}$

Angle, $\theta =35°$

Radius, $r=8\text{ m = }\Delta x$

Formula used:

By using the equation of continuity

  $v_{0_x}=v_0\times \cos \theta \text{ and }{v}_{0_y}=v_0\times \sin \theta$

In x-direction, there is no acceleration,

  $\text{So, }x(t)=x_0+v_{0_x}\times t$

In the Y direction, there is an acceleration due to gravity.

  $\text{So, }y(t)=y_0+v_{0_y}\times t+\frac{1}{2}\times a\times t^2$

Calculation:

Here, d = diameter of the pipe

  $v_0$ = velocity

  $v_{0_y}$ = velocity of y

  $v_{0_x}$ = velocity of x

Substitute the given values in the above equation,

  $\begin{array}{l}x-x_0=\Delta x=v_{0_x}\times t\\ r=v_{0_x}\times t,t=\frac{r}{v_{0_x}}\\ 0=v_{0_y}\times t-\frac{1}{2}\times g\times t^2\\ v_{0_y}\times t=\frac{1}{2}\times g\times t^2\\ v_{0_y}=\frac{1}{2}\times g\times t=\frac{1}{2}\times g\times \frac{r}{v_{0_x}}\\ v_0\times \sin \theta =\frac{1}{2}\times g\times \frac{r}{v_0\times \cos \theta }\\ v_0^2=\frac{9.8\times 8}{\sin (2\theta )}=\frac{78.4}{0.9396}=83.4\\ v_0=9.13\text{ m/s}\end{array}$

Conclusion:

Thus, the velocity of water coming out from each sprinkler head is $9.13\text{ m/s}$ .

(b)

To determine

To Find: The rate of flow of water.

Answer to Problem 83GP

The rate of flow of water in liters/second, when each head diameter is 3 mm is $0.258\text{ lit/sec}$ .

Explanation of Solution

Given:

the diameter of each head= $d=3\text{ mm = }0.003\text{ m}$

output velocity $v_{\text{output}}=9.13\text{ m/s}$

Formula used:

By using the continuity equation, we can get the rate of flow in liter per second

  $\begin{array}{l}{A}_{\text{input}}\times v_{\text{input}}=A_{\text{output}}\times v_{\text{output}}\\ Q\text{ = }\frac{\text{Volume}}{\text{Time}}\text{ = }4\times \pi \times r^2\times v_{\text{output}}=4\times \pi \times \frac{d^2}{4}\times v_{\text{output}}\end{array}$

Calculation:

Substitute the given values in the above equation,

  $\begin{array}{l}{A}_{\text{input}}\times v_{\text{input}}=A_{\text{output}}\times v_{\text{output}}\\ Q\text{ = }\frac{\text{Volume}}{\text{Time}}\text{ = }4\times \pi \times r^2\times v_{\text{output}}=4\times \pi \times \frac{d^2}{4}\times v_{\text{output}}\\ Q=4\times \pi \times {(}^{0.003}\times 9.13=2.58\times {10}^{-4}{\text{ m}}^{\text{3}}\text{/ sec}\\ Q=2.58\times {10}^{-1}\text{ lit}\text{/ sec = }0.258\text{ litre / second}\end{array}$

Conclusion:

Thus, the rate of flow of water in litre/second, when each head diameter is 3 mm is $0.258\text{ lit/sec}$

(c)

To determine

To Find: The speed of water.

Answer to Problem 83GP

The speed of water flowing inside the 1.9 cm diameter pipe is $0.91\text{ m/s}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Rate of flow }Q=0.258{\text{ m}}^3\text{/s = 4}\times {\text{A}}_{\text{out}}\times {\text{V}}_{\text{out}}\\ \text{Diameter of pipe }d=1.9\text{ cm = }0.019\text{ cm}\end{array}$

Formula used:

The total energy can be found from the KE at the bottom of motion by the following formula,

  $\begin{array}{l}{\text{A}}_{\text{input}}\times {\text{v}}_{\text{input}}\text{= 4}\times {\text{A}}_{\text{output}}\times {\text{v}}_{\text{output}}\\ {\text{A}}_{\text{input}}\text{ = }\frac{\text{π}}{\text{4}}\times {\text{d}}^{\text{2}}\end{array}$

Calculation:

Substitute the given values in the above equation,

  $v_{input}=\frac{4\times A_{output}\times v_{output}}{A_{input}}=\frac{0.258}{\frac{\pi }{4}\times {0.019}^2}=0.91\text{ m/s}$

Conclusion:

Thus, the speed of water flowing inside the 1.9 cm diameter pipe is $0.91\text{ m/s}$ .