Flow through pipes in series or flow through a compound pipe.

It can be defined as the pipes which are connected end to end to form a pipeline and are of different length and different diameters.

$L_1, L_2, L_3$ = lengths of pipe 1, 2 and 3

$d_1, d_2, d_3$ = diameters of pipes 1, 2, 3 respectively

$V_1, V_2, V_3$ = velocity of flow through pipes 1, 2, 3

$f_1, f_2, f_3$ = coefficient of friction for pipes 1, 2, 3

H = difference of water level in the two tanks

$\therefore Q=A_1V_1=A_2V_2=A_3V_3$

Difference in liquid surface levels is equal to the sum of total head loss in the pipes.

$\therefore H=\dfrac {0.5V_1^2}{2g}+\dfrac {4f_1L_1V_1^2}{d_1\times 2g}+\dfrac {0.5V_2^2}{2g}+\dfrac {4f_2L_2V_2^2}{d_2\times 2g}+\dfrac {(V_2-V_3)^2}{2g}+\dfrac {4f_3L_3V_3^2}{d_3\times 2g}+\dfrac {V_3^2}{2g}.....................(1)$

If minor losses are neglected, then the above equation becomes as,

$\therefore H=\dfrac {4f_1L_1V_1^2}{d_1\times 2g}+\dfrac {4f_2L_2V_2^2}{d_2\times 2g}+\dfrac {4f_3L_3V_3^2}{d_3\times 2g}.....................(2)$

If coefficient of friction is same for all pipes,

$f_1=f_2=f_3$

$\therefore H=\dfrac {4fL_1V_1^2}{d_1\times 2g}+\dfrac {4fL_2V_2^2}{d_2\times 2g}+\dfrac {4fL_3V_3^2}{d_3\times 2g}$

$\therefore H=\dfrac {4f}{2g}\left[\dfrac {L_1V_1^2}{d_1}+\dfrac {L_2V_2^2}{d_2}+\dfrac {L_3V_3^2}{d_3}\right]$

Numericals

Q1) The difference in water surface level in two tanks, which are connected by three pipes in series of length 300 m, 170 m and 210 m and of diameter 300 mm, 200 mm and 400 mm respectively is 12 m. Determine the rate of flow of water if coefficient of friction are 0.005,0.0052 and 0.0048 respectively.

Considering

(i) minor losses also

(ii) neglecting minor losses

Solution:- Given:

Difference of water level, H=12m

Length of pipe 1, $L_1=300m$

Diameter, $d_1=300mm=0.3m$

Length of pipe 2, $L_2=170m$

Diamter, $d_2=200mm=0.2m$

Length of pipe 3, $L_3=210m$

Diamter, $d_3=400mm=0.4m$

Case-1: Considering minor losses

From continuity, we have

$A_1V_1=A_2V_2=A_3V_3$

$\therefore V_2=\dfrac {A_1V_1}{A_2}=\dfrac{\dfrac\pi 4(d_1)^2}{\dfrac\pi 4(d_2)^2}V_1$

$V_2=\dfrac {d_1^2}{d_2^2}V_1$

$V_2=\left(\dfrac{0.3}{0.2}\right)^2\times V_1$

$V_2=2.25V_1..........................(1)$

Now,

$V_3=\dfrac {A_1V_1}{A_3}=\dfrac{\dfrac\pi 4(d_1)^2}{\dfrac\pi 4(d_3)^2}V_1$

$V_3=\dfrac {d_1^2}{d_3^2}V_1$

$V_3=\left(\dfrac{0.3}{0.4}\right)^2\times V_1$

$V_3=0.5625V_1..........................(2)$

Using the equation of flow through pipes, we have

$\therefore H=\dfrac {0.5V_1^2}{2g}+\dfrac {4f_1L_1V_1^2}{d_1\times 2g}+\dfrac {0.5V_2^2}{2g}+\dfrac {4f_2L_2V_2^2}{d_2\times 2g}+\dfrac {(V_2-V_3)^2}{2g}+\dfrac {4f_3L_3V_3^2}{d_3\times 2g}+\dfrac {V_3^2}{2g}$

Substituting the values of $V_2$ and $V_3$,

$12.0=\dfrac {0.5V_1^2}{2g}+\dfrac {4\times0.005\times300\times V_1^2}{0.3\times 2g}+\dfrac {0.5(2.25\times V_1)^2}{2g}+\dfrac {4\times0.0052\times 170\times (2.25\times V_1)^2}{0.2\times 2g}+\dfrac {(2.25V_1-0.562V_1)^2}{2g}+\dfrac {4\times0.0048\times210\times(0.5625\times V_1)^2}{0.4\times 2g}+\dfrac {(0.5625V_1)^2}{2g}$

$12.0=\dfrac {V_1^2}{2g}[0.5+20.0+2.53+89.505+2.847+3.189+0.316]$

$V_1=\sqrt{\dfrac {12\times2\times9.81}{118.887}}$

Therefore, $V_1=1.407m/s$

Therefore, rate of flow,

$Q=A\times V=A_1\times V_1$

$Q=\dfrac \pi 4\times (d_1)^2\times V_1$

$Q=\dfrac \pi 4\times (0.3)^2\times 1.407$

$Q=0.09945m^3/s$

Case-2: neglecting minor losses:-

$H=\dfrac {4f_1L_1V_1^2}{d_1\times 2g}+\dfrac {4f_2L_2V_2^2}{d_2\times 2g}+\dfrac {4f_3L_3V_3^2}{d_3\times 2g}$

$12.0=\dfrac {V_1^2}{2g}\left[\dfrac {4\times 0.005\times 300}{0.3}+\dfrac {4\times0.0052\times170\times(2.25)^2}{0.2}+\dfrac {4\times0.0048\times 210\times (0.5625)^2}{0.4}\right]$

$12.0=\dfrac {V_1^2}{2g}[20+89.505+3.189]$

$V_1=\sqrt{\dfrac {12\times2\times9.81}{112.694}}$

Therefore, $V_1=1.445m/s$

Therefore, rate of flow,

$Q=A\times V=A_1\times V_1$

$Q=\dfrac \pi 4\times (d_1)^2\times V_1$

$Q=\dfrac \pi 4\times (0.3)^2\times 1.445$

$Q=0.1021m^3/s$

or Q=102.1 lit/s

Equivalent pipes:-

• It can be defined as a pipe which has its loss of head and discharges equal to the loss of head and discharge of a compound pipe consisting of various are several types of different length and diameters.

• The equivalent pipe of uniform diameter it is known as the equivalent size of pipe.

• Its length is equal to the sum of lengths of the compound by consisting of different pipes.

Here, $L_1=$ Length of Pipe 1, $d_1=$ diameter of pipe 1

$L_2=$ Length of Pipe 2, $d_2=$ diameter of pipe 2

$L_3=$ Length of Pipe 3, $d_1=$ diameter of pipe 3

H = Total head loss

L = length of equivalent pipe

d = diameter of equivalent pipe

Then, $L=L_1+L_2+L_3$

Total head loss in compound pipe, (neglecting minor losses)

$H=\dfrac {4f_1L_1V_1^2}{d_1\times 2g}+\dfrac {4f_2L_2V_2^2}{d_2\times 2g}+\dfrac {4f_3L_3V_3^2}{d_3\times 2g}..................(1)$

Assuming $f_1=f_2=f_3$

Discharge, $Q=A_1V_1=A_2V_2=A_3V_3$

$Q=\dfrac\pi 4 (d_1^2\times V_1)=\dfrac\pi 4 (d_2^2\times V_2)=\dfrac\pi 4 (d_3^2\times V_3)$

or, $V_1=\dfrac {4Q}{\pi d_1^2}, V_2=\dfrac {4Q}{\pi d_2^2}, V_3=\dfrac {4Q}{\pi d_3^2}$

Put this value in equation (1), we get

$H=\dfrac {4fL_1\times\left(\dfrac {4Q}{\pi d_1^2}\right)^2}{d_1\times 2g}+\dfrac {4fL_2\times\left(\dfrac {4Q}{\pi d_2^2}\right)^2}{d_2\times 2g}+\dfrac {4fL_3\times\left(\dfrac {4Q}{\pi d_3^2}\right)^2}{d_3\times 2g}$

$H=\dfrac {4\times 16fQ^2}{\pi^2\times 2g}\left[\dfrac {L_1}{d_1^5}+\dfrac {L_2}{d_2^5}+\dfrac {L_3}{d_3^5}\right]..................(1)$

Head loss in equivalent pipe,

$H=\dfrac {4fLV^2}{d\times 2g}$

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

$\therefore H=\dfrac {4fL\left(\dfrac {4Q}{\pi d}\right)^2}{d\times 2g}$

$H=\dfrac {4\times 16fQ^2}{\pi^2\times 2g}\left[\dfrac {L}{d^5}\right].................(2)$

Equating equations(1) and (2), we get

$\dfrac {4\times 16fQ^2}{\pi^2\times 2g}\left[\dfrac {L_1}{d_1^5}+\dfrac {L_2}{d_2^5}+\dfrac {L_3}{d_3^5}\right]=\dfrac {4\times 16fQ^2}{\pi^2\times 2g}\left[\dfrac {L}{d^5}\right]$

Cancelling the like terms, we get

$\left[\dfrac {L_1}{d_1^5}+\dfrac {L_2}{d_2^5}+\dfrac {L_3}{d_3^5}\right]=\left[\dfrac {L}{d^5}\right]$

Flow through parallel pipes:-

When the main pipe is divided into two or more branches and which again join together downstream and forms a single pipe, then such branch pipes are said to be connected in parallel.

The discharge increases through the main pipe if the pipes are connected in parallel.

$\therefore Q=Q_1+Q_2.....................(1)$

In this, the loss of head for each branch is the same.

$\dfrac {4f_1L_1V_1^2}{d_1\times 2g}=\dfrac {4f_2L_2V_2^2}{d_2\times 2g}$

Numericals

Q1) A pipeline of 0.6 m diameter is 1.5 km long. To increase the discharge, another line of CM diameter is introduced parallel to the first in the second half of the length. Neglecting minor losses, find the increase in discharge If 4f= 0.04. The head at the inlet is 300 mm.

*Solution: * Given:

Diameter of pipe line, D=0.6m

Length, L=1.5km=1.5$\times$1000=1500m

4f=0.004

f=0.01

Head at inlet, h=300mm = 0.3m

Head at outlet = atmospheric head=0

Therefore, Head loss, hf=0.3m

Length of another parallel pipe, $L_1=\dfrac {1500}2=750m$

Diameter of another pipe, $d_1=0.6m$

Case 1: Discharge for a single pipe of length 1500m and diameter 0.6m

Head loss due to friction,

$hf=\dfrac {4fLV_1^2}{d\times 2g}$

$\therefore, 0.3=\dfrac {4\times0.01\times1500\times V_1^2}{0.6\times 2g}$

$\therefore v_1=\sqrt{\dfrac{0.6\times2\times 9.81\times 0.3}{4\times 0.01\times 1500}}$

$V_1=0.2426m/s$

$\therefore, Q_1=V_1\times A$

$Q_1=0.2426\times \dfrac \pi 4(0.6)^2$

$Q_1=0.0685m^3/s$

Case 2: When an additional pipe of length 750m and diameter 0.6m is connected in parallel with the last half length of the pipe.

$Q_1 = $ discharge in 1st parallel pipe

$Q_2 = $ discharge in 2nd parallel pipe

$\therefore, Q=Q_1+Q_2$

But as the length and diameter of each pipe is same

$Q_1=Q_2=\dfrac Q2$

Consider flow through pipe ABC or ABD

head lost due to friction through AB

$=\dfrac{4\times f\times 750\times V_2^2}{0.6\times 2\times 9.81}$

$V_2=\dfrac Q{\dfrac \pi 4(0.6)^2}$

$V_2=\dfrac {4Q}{\pi\times 0.36}$

Head lost due to friction through AB

$\therefore hf=\dfrac {4\times 0.01\times 750}{0.6\times 2\times 9.81}\times \left[\dfrac {4Q}{\pi\times 0.36}\right]^2$

$hf=31.87 Q^2$

Head loss due to friction through BC

$=\dfrac{4\times f\times L_1\times V_2^2}{d_1\times 2\times g}$

$\therefore hf=\dfrac {4\times 0.01\times 750}{0.6\times 2\times 9.81}\times \left[\dfrac {Q}{2\times \pi\times (0.6)^2}\right]^2$

As $V_1=\left[\dfrac {Q}{2\times \pi\times (0.6)^2}\right]^2$

$\therefore hf=\dfrac {4\times 0.01\times 750}{0.6\times 2\times 9.81}\times \left[\dfrac {16Q^2}{4\times \pi^2\times (0.36)}\right]$

$hf=7.969Q^2$

Therefore, Head loss through ABC = Head lost through AB+ head lost through BC

$0.3=31.87Q^2+7.969Q^2$

$0.3=39.839Q^2$

$\therefore Q=\sqrt {\dfrac {0.3}{39.839}}$

$Q_2=0.0867m^3/s$

Increase in discharge = $Q_2-Q_1$

$=0.867-0.0685$

$=0.0182m^3/s$

Q2) For a town water supply, I mean pipeline of diameter 0.4 m is required. As pipes more than 0.35 m diameters are not readily available, two parallel pipes of same diameter were used for water supply. If the total discharge in the parallel pipes is as same as the single main pipe. Find the diameter of the parallel pipe. Assume the coefficient of friction same for all pipes.

Solution: Given: -

Diameter of single main pipe line, d=0.4m

Let the length of single pipe = L

Coefficient of friction = f

$hf=\dfrac{4fLV^2}{d\times 2g}$

$hf=\dfrac{4fLV^2}{0.4\times 2g}.................(1)$

This is loss of head due to friction in single pipe.

In case, of parallel pipes, as the diameter and length are the same, hence discharge in each part will be half the discharge of single main pipe. Velocity will be the same as well.

$V_* = $ Velocity in each parallel pipe

$d_*=$ diameter of each parallel pipe

$\therefore hf=\dfrac{4fLV_*^2}{d_*\times 2g}...................(2)$

Equating (1) and (2), we get

$\dfrac{4fLV^2}{d\times 2g}=\dfrac{4fLV_*^2}{d_*\times 2g}$

Cancelling like terms, we get

$\dfrac{V^2}{0.4}=\dfrac{V_*^2}{d_*}....................(3)$

or, $\dfrac{V^2}{V_*^2}=\dfrac{0.4}{d_*}$

From Continuity equation,

Total flow in single main = sum of flow in two parallel pipes

$V\times\dfrac \pi4(0.4)^2=2\times V_*\times\dfrac\pi 4\times (d_*)^2$

$\dfrac V{V_*}=\dfrac {2\times\dfrac \pi 4d_*^2}{\dfrac \pi 4(0.4)^2}$

$\dfrac V{V_*}=\dfrac {2d_*^2}{0.16}$

Squaring both sides, we get

$\dfrac {V^2}{V_*^2}=\dfrac {4d_*^4}{0.0256}........................(4)$

Comparing equation (3) and (4), we get

$\dfrac {0.4}{d_*}=\dfrac {4d_*^4}{0.0256}$

$d_*^5=\dfrac {0.4\times 0.0256}4$

$d_*^5=(0.00256)$

$d_*=0.303 \ m=30.3 \ cm$