Chapter 12, Problem 8TP

To determine

The speed of water flowing out of the pipe.

Answer to Problem 8TP

  $7.57\text{m}/\text{s}$

Explanation of Solution

Given:

Diameter of pipe =3.5cm

Force= $320\text{N}$

Rate of flowing water = 2.5m/s

Pumped water level = 21m

Formula used:

According to the Bernoulli's principle,

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

Where $P_1$ - applied pressure by the pump, $\rho$ -fluid density, $h_1$ -pipe lower end height, $v_1$ -lower end water speed, $v_2$ -upper end water speed,g-gravity acceleration, $P_2$ -upper end atmospheric pressure.

Calculation:

The left hand side of the equation is evaluated initially.

The cross-sectional area of the pipe is calculated as

  $A=\frac{\pi }{4}{d}^2$

Where, A -Area of the pipe cross section, d -Diameter of the pipe cross section.

Substitute the diameter value 3.5 cm in the above equation.

  $\begin{array}{l}A=\frac{\pi }{4}{(}^{3.5}\\ A=\frac{\pi }{4}{\left((3.5\text{cm)}\left(\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1cm}}\right)\right)}^2\\ A=9.621\times {10}^{-4}{\text{m}}^2\end{array}$

The plunger pressure of fluid due to force is calculated as

  $P_1=\frac{F}{A}$

Where, F -plunger force, P -plunger pressure, A -area of cross-section of pipe

Substitute the above found values in equation

  $\begin{array}{l}{P}_1=\frac{\text{320N}}{\text{9}{\text{.621×10}}^{\text{-4}}{\text{m}}^{\text{2}}}\\ P_1=332601.3{\text{5Nm}}^{\text{-2}}\end{array}$

The lower end water is assumed to be zero.Hence

  $h_1=0\text{m}$

Substitute the above found values in equation:

  $\begin{array}{l}{P}_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=\text{332601}\text{.35}\text{N}/{\text{m}}^{\text{2}}\text{+}\frac{\text{1}}{\text{2}}\text{(1000}\text{kg}/{\text{m}}^{\text{3}}\text{)(2}\text{.5}\text{m}/\text{s}{\text{)}}^{\text{2}}\text{+(1000}\text{kg}/{\text{m}}^{\text{3}}\text{)(9}\text{.8}\text{m}/{\text{s}}^{\text{2}}\text{)(0m)}\\ =332601.35\text{N}/{\text{m}}^{\text{2}}\text{+(500×6}\text{.25)}\text{kg}/{\text{m}}^{\text{3}}\text{-}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\\ =335726.35\text{N}/{\text{m}}^{\text{2}}\end{array}$

Now the right hand side of the equation is evaluated.

The upper end atmospheric pressure is 1 atm

Hence,

  $P_2=1\text{atm}=101300\text{N}/{\text{m}}^{\text{2}}$

Substitute the values

  $\begin{array}{l}{P}_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2=10130\text{0}\text{N}/{\text{m}}^{\text{2}}\text{+}\frac{\text{1}}{\text{2}}\text{(1000}\text{kg}/{\text{m}}^{\text{3}}{\text{)v}}_{\text{2}}^{\text{2}}\text{+(1000}\text{kg}/{\text{m}}^{\text{3}}\text{)(9}\text{.8}\text{m}/{\text{s}}^{\text{2}}\text{)(21m)}\\ =10130\text{0}\text{N}/{\text{m}}^{\text{2}}+(500v_2^2)+(1000\times 9.8\times 2\text{1)}\text{kg}/{\text{m}}^{\text{3}}\text{.}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\text{.m}\\ =101300\text{N}/{\text{m}}^{\text{2}}+(500v_2^2)+20580\text{0}\text{kg}/{\text{m}}^{\text{3}}\text{.}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\text{.m}\\ =307100\text{N}/{\text{m}}^{\text{2}}+(500v_2^2)\end{array}$

Since both sides of the equation is equal, therefore

  $\begin{array}{l}\text{335726}\text{.35}\text{N}/{\text{m}}^{\text{2}}\text{=307100}\text{N}/{\text{m}}^{\text{2}}+(500v_2^2)\\ (500v_2^2)=335726.3\text{5}\text{N}/{\text{m}}^{\text{2}}\text{-307100}\text{N}/{\text{m}}^{\text{2}}\\ v_2^2=\frac{28626.35}{500}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\\ v_2^2=57.25{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\end{array}$

Take square root on both sides $v_2=7.57\text{m}/\text{s}$

Conclusion:

Hence, the speed of water flows out pipe is $7.57\text{m}/\text{s}$