Page: Flow through branched pipe
0

When three or more reservoirs are connected by means of pipes, and which has one or more functions, this type of system is known as the branching pipe system.

Basic equations used for solving such problems:- 1) continuity equation 2) Darcy weisbach equation 3) Bernoulli's equation Numericals

Q1) Three reservoir A, B, C are connected by a pipe system. Find the discharge into or from the reservoirs Bull and Cif the rate of flow from reservoir A is 60 litres per second. Find the height of water level in the reservoir C. f = 0.06 for all pipes.

Solution: Given:-

Length of pipe AD, $L_1=1200m$

Diameter of pipe AD, $d_1=30cm=0.30cm$

Discharge through 'AD', $Q_1=60lit/s=0.06m^3/s$

height of water level in A from reference line, $z_A=40m$

For pipe DB length,

$L_2=600m$, diameter, $d_2=20cm=0.20cm$, $z_B=38$

For pipe DC

Length, $L_3=800m$, diameter $d_3=30cm=0.30m$,

Step 1:

Applying Bernoulli's eqaution to point E and D

$z_A=z_D+\dfrac {P_D}{\rho g}+hf_1$

where $hf_1=\dfrac{4fL_1V_1^2}{d_1\times 2g}$

$V_1=\dfrac {Q_1}{A}=\dfrac {0.06}{\dfrac \pi4(0.3)^2}=0.848m/sec$

$\therefore hf_1=\dfrac{4\times 0.006\times 1200\times (0.848)^2}{0.3\times 2\times 9.81}=3.518m$

$\therefore z_A=z_D+\dfrac {P_D}{\rho g}+3.518$

$40=z_D+\dfrac {P_D}{\rho g}+3.518$ $z_D+\dfrac {P_D}{\rho g}=40-3.518=36.482m$

Hence piezometric head at D=36.482

But $z_B=38m$

Hence water flows from 'B' to 'D'.

Step 2:

Applying Bernoulli's equation to point 'B' and 'D'

$z_B=z_D+\dfrac {P_D}{\rho g}+hf_2$

$38=36.482+hf_2$

$\therefore hf_2=38-36.482=1.518m$

But $hf_2=\dfrac{4fL_2V_2^2}{d_2\times 2g}$

$\therefore, 1.518=\dfrac{4\times 0.006\times 600\times V_2^2}{0.2\times 2\times 9.81}$

$V_2=\sqrt{\dfrac {0.2\times2\times 9.81\times 1.518}{4\times 0.006\times 600}}$

$V_2=0.643m/s$

$Q_2=V_2\times \dfrac \pi4(d_2)^2$

$Q_2=0.643\times\dfrac \pi4\times (0.2)^2$

$Q_2=0.0202m^3/s=20.2lit/s$

Step 3:

Applying Bernoulli's equation to 'D' and 'C'

$z_D+\dfrac {P_D}{\rho g}=z_C+hf_3$

$\therefore 36.482=z_C+\dfrac{4fL_3V_3^2}{d_3\times 2g}$

where, $V_3=\dfrac {Q_3}{\dfrac \pi 4d_3^2}$

but from continuity $Q_1+Q_2=Q_3$

$\therefore Q_3=0.06+0.0202=0.0802m^3/s$

$\therefore V_3=\dfrac {Q_3}{\dfrac \pi 4(0.3)^2}$

$\therefore V_3=\dfrac {0.0802}{\dfrac \pi 4(0.3)^2}=1.134m/s$

$\therefore 36.482=z_C+\dfrac {4\times0.006\times 800\times 1.134^2}{0.3\times 2\times 9.81}$

$36.482=z_C+4.194$

$z_C=36.482-4.194=32.288m$

page fluid mechanics 2 fm2 • 72 views
 modified 7 months ago  • written 7 months ago by Syedahina Mohi • 0