Page: Flow Through Pipes
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Loss of Energy in Pipes:-

When a fluid flows through a pipe, the fluid experiences some resistance because of which some energy of the fluid is lost. Major Energy Losses Minor Energy Losses
Mainly due to friction. Can be calculated using: (a) Darcy-Weisbach Formula (b) Chezy's Formula Mainly due to: (a) Sudden expansion (b) Sudden Contraction (c) Bend in pipe (d) Pipe fitting

Loss of Energy or Head due to friction:-

(1) Darcy-Weisbach Equation:-

$hf=\dfrac{4.f.L.V^2}{d\times 2g}$

where hf = loss of head due to friction

f = co-efficient of friction

=$\dfrac{16}{R_e}$ for $R_e\lt2000$ (Viscous Flow)

=$\dfrac {0.079}{R_e^{\dfrac 14}}$ for $R_e$ varying from 4000 to 10$^6$

V = Mean Density

d = diameter of pipe

L = Length of pipe

(2) Chezy's Formula or equation: -

$v=c\sqrt{mi}$

### Numericals:-

Q1) Find the head lost due to friction in a pipe of diameter 300mm and length 50m through which water is flowing at a velocity of 3m/s using (i) Darcy weisbach (ii) Chezy's Formula (c=60)

(Take V=0.01 stoke (for water))

Solution:

Given: Diameter of the pipe, d = 300mm (divide by '1000' to convert it from 'mm' to 'm')

So, d=0.3 m

Length, L= 50m

Velocity, v=3m/s

Kinematic Viscosity, V=0.01stoke

$V=0.01 cm^2/s$

$V=0.01\times 10^{-4}m^2/s$

$\therefore$ By Darcy's Formula,

$hf=\dfrac {4\times f\times L\times V^2}{d\times 2g}$

All the values are given except 'f'

'f' = coefficient of friction which is the function of Reynolds number '$R_e$'

To find: 'f'

$f=\dfrac{0.079}{R_e^{\dfrac 14}}$

$\therefore R_e=\dfrac{V\times d}{v}$

$=\dfrac {3.0\times 0.30}{0.01\times 10^{-4}}$

$R_e=9\times 10^5$

$\therefore f=\dfrac {0.079}{(9\times 10^5)^{\dfrac14}}$

$f=2.56\times10^{-3}$

or $f=0.00256$

Therefore, Head lost, $hf=\dfrac{4\times0.00256\times 50\times 3^2}{0.3\times 2.0\times 9.81}$

$hf=782.87\times 10^{-3}$

$hf=0.7828m$

Now by using Chezy's formula:-

$V=c\sqrt{mi}$

where, $c=60, m=\dfrac d4=\dfrac {0.30}4=0.075m$

$\therefore 3=60\sqrt{0.075\times i}.................(i)$

or, $\therefore, i=\left(\dfrac {3}{60}\right)^2\times \dfrac 1{0.075}...................(ii)$

By equating, we get

$i=0..333$

But, $i=\dfrac {hf}L$

$\therefore, 0.333=\dfrac {hf}{50}$

$\therefore hf=1.665m$

Q2) Find the diameter of a pipe of length 2000m when the rate of flow of water through the pipe is 200 litres/s and the head lost due to friction is 4m. Take the value of c=50 in Chezy's Formula

Solution:-

Given:

L=200m

$Q=200lit/s=\dfrac {200}{1000}=0.2m^3/s$

c=50

Step 1: Velocity of flow

$V=\dfrac {\text{Discharge}}{\text{Area}}$

$V=\dfrac {Q}{\dfrac \pi 4d^2}$

$V=\dfrac{0.2\times 4}{\pi d^2}...............................(i)$

Step 2: Hydraulic mean depth

$m=\dfrac d4.........................(ii)$

Loss of head per unit length

$i=\dfrac {hf}L=\dfrac 4{2000}$

$i=0.002.....................(iii)$

Now substituting the value of V, m, iand c from the equation (i), (ii), (iii), we get

$V=c\sqrt{mi}$

$\dfrac{0.2\times 4}{\pi d^2}=50\sqrt{\dfrac d4\times 0.002}$

$\dfrac{0.2\times 4}{\pi d^2\times 50}=\sqrt{\dfrac d4\times 0.002}$

$\dfrac{0.00509}{ d^2}=\sqrt{\dfrac d4\times 0.002}$

Squaring on both sides, we get

$\dfrac{(0.00509)^2}{ d^4}=\dfrac d4\times 0.002$

$\dfrac{2.59\times 10^{-5}}{ d^4}=\dfrac d4\times 0.002$

Rearranging, we get

$d^5=\dfrac {4\times (2.59\times 10^{-5})}{0.002}$

$d^5=\dfrac {4\times (0.0000259)}{0.002}$

$\therefore d^5=0.0518$

$\therefore d=\sqrt{0.0518}$

$d=0.553m$

Q3) A crude oil of kinematic viscosity 0.4 stoke is flowing through a pipe of diameter 300 mm at the rate of 300 litres per se. Find the head lost due to friction for a length of 50 m of the pipe.

Solution:

Given:

$V=0.4 stoke=0.4cm^2/s=0.4\times 10^{-4}m^2/s$

$d=300mm=0.3m^3/s$

L=50m

Step 1: Velocity of flow

$V=\dfrac {\text{Discharge}}{\text{Area}}$

$V=\dfrac {Q}{\dfrac \pi 4d^2}$

$V=\dfrac{0.3\times 4}{\pi (0.3)^2}$

$V=4.24m/s$

Step 2: Reynold's Number

$R_e=\dfrac {V\times d}v$

$R_e=\dfrac {4.24\times 0.3}{0.4\times 10^{-4}}$

$R_e=3.18\times 10^4$

As the value of $R_e$ lies between 4000 and 10000, the value of 'f' is given by

$f=\dfrac{0.079}{R_e^{\dfrac 14}}$

$f=\dfrac {0.079}{(3.18\times 10^4)^{\dfrac 14}}$

$f=5.91\times10^{-3}$

or, f=0.00591

Now by Darcy's Formula

$hf=\dfrac {4\times f\times L\times V^2}{d\times 2g}$

$hf=\dfrac{4\times 0.00591\times 50\times 4.24^2}{0.3\times 2\times 9.81}$

$\therefore$ Head lost due to friction $hf=3.61m$

Q4) An oil of sp.gr. 0.7 is flowing through a pipe of diameter 300 mm at the rate of 500 litres/s. Find the head lost due to friction and power required to maintain the floor for a length of 1000 m. Take v=0.29 stokes.

Solution: Given:-

s=0.7

Diameter of pipe, $d=300mm=0.3m$

Discharge, $Q=500lit/s=0.5m^3/s$

Length, $L=1000m$

Step 1: Velocity

$V=\dfrac Q{\text{area}}$

$V=\dfrac {0.5}{\dfrac \pi 4\times d^2}=\dfrac {0.5\times 4}{\pi\times 0.3^2}=7.073m/s$

Step2: Reynold's Number

$R_e=\dfrac {V\times d}v$

$R_e=\dfrac {7.073\times 0.3}{0.29\times 10^{-4}}$

$R_e=7.316\times 10^4$

Step 3: Coefficient of friction,

$f=\dfrac {0.079}{R_e^{\dfrac 14}}$

$f=\dfrac {0.079}{(7.316\times 10^4)^{\dfrac 14}}$

$f=0.0048$

$hf=\dfrac {4\times f\times L\times V^2}{d\times 2g}$

$hf=\dfrac {4\times 0.0048\times 1000\times 7.073^2}{0.3\times 2\times 9.81}$

$f=163.18m$

Step 5: Power required

$=\dfrac {\rho. g.Q.hf}{1000}$

Density of oil, $\rho=07\times 1000=700kg/m^3$

Power required $=\dfrac {700\times 9.81\times 0.5\times 16318}{1000}$

Hence power required = 560.28kW

page fluid mechanics 2 fm2 • 821 views
 modified 8 months ago by Sanket Shingote ♦♦ 290 written 9 months ago by Syedahina Mohi • 10