Question

Determine the rate of flow of water through a pipe of diameter 20 cm and length 50 m when one end of the pipe is connected to a tank and other end is open to the atmosphere. The pipe is horizontal and the height of water in the tank is 4 m above the centre of the pipe. (f = 0.01)

Answer 1

Answer:

85.89  lit/sec

Explanation:

Check problem 11.16(PAGE NO. 484 IN REVISED 9TH EDITION)  in  A TEXTBOOK IN FLUID MECHAMICS AND HYDRAULIC MACHINES  R.K Bansal.

Answer 2

Answer:

The rate of water flow through a pipe is $85.89 \text { litres/s }$.

Explanation:

Pipe diameter, $\quad d=20 \mathrm{~cm}=0.20 \mathrm{~m}$

Pipe length, $L=50 \mathrm{~m}$

Water height, $\quad H=4 \mathrm{~m}$

Co-efficient of friction, $\quad f=.009$

Let the velocity of water in the pipe $=V \mathrm{~m} / \mathrm{s}$.

By applying Bernoulli's equation to the top of the water surface in the reservoir and to the outlet of the pipe, we have [Taking point 1 above and point 2 at the exit of the tube].

$\frac{p_{1}}{\rho g}+\frac{V_{1}^{2}}{2 g}+z_{1}=\frac{p_{2}}{\rho g}+\frac{V_{2}^{2}}{2 g}+z_{2}+\text { all losses }$

Considering datum line passing through the centre of pipe:

$\begin{aligned}0+0+4.0 &=0+\frac{V_{2}^{2}}{2 g}+0+\left(h_{i}+h_{f}\right) \\4.0 &=\frac{V_{2}^{2}}{2 g}+h_{i}+h_{f}\end{aligned}$

But velocity in pipe $=V$,

 $\begin{aligned}\therefore V &=V_{2} \\4.0 &=\frac{V^{2}}{2 g}+h_{i}+h_{f} \ldots . .(i)\end{aligned}$

From equation, $h_{i}=0.5 \frac{V^{2}}{2 g}$ and $h_{f}$ the equation is given as:

$h_{f}=\frac{4 \cdot f \cdot L \cdot V^{2}}{d \times 2 g}$

Substituting these values, we have

$\begin{aligned}4.0 &=\frac{V^{2}}{2 g}+\frac{0.5 V^{2}}{2 g}+\frac{4 \cdot f \cdot L \cdot V^{2}}{d \times 2 g} \\&=\frac{V^{2}}{2 g}\left[1.0+0.5+\frac{4 \times .009 \times 50}{0.2}\right]=\frac{V^{2}}{2 g}[1.0+0.5+9.0] \\&=10.5 \times \frac{V^{2}}{2 g} \\V &=\sqrt{\frac{4 \times 2 \times 9.81}{10.5}}=2.734 \mathrm{~m} / \mathrm{sec}\end{aligned}$

$\quad Rate of flow, Q=A \times V=\frac{\pi}{4} \times(0.2)^{2} \times 2.734=0.08589 \mathrm{~m}^{3} / \mathrm{s}$

$=85.89 \text { litres } / \mathrm{s} .$

Therefore, the rate of flow of water through a pipe is $85.89 \text { litres/s}$.