PRESENTER NAME:VAIBHAV PATEL (131040106039)VINAY PATEL (131040106040)DIGVIJAYSINH PUVAR (131040106045)RAVI SUTHAR (131040106055)

Subject : Applied Fluid MechanicsB.E. Civil Engineering (6th sem (2016))

Guidance by Prof. ankit patel

GUJRAT POWER ENGG. AND RESEARCH INSTITUTE

Flow through pipe in series and parallel

Flow through pipe in series

•When pipes of different diameters are connected end to end to form a pipe line, they are said to be in series. The total loss of energy (or head) will be the sum of the losses in each pipe plus local losses at connections.

An arrangement of such pipe line between two reservoir is shown in fig.

Total head loss for the system H = height difference of reservoirs

hf1 = head loss for 200mm diameter section of pipe hf2 = head loss for 250mm diameter section of pipe h entry = head loss at entry point h enp = head loss at join of the two pipes h exit = head loss at exit point

H = hf1 + hf2 + hentry + henp + hexit

2 2 2 2 21 1 1 2 2 2 1 1 2 2

1 2

4 4 ( ). . 0.52 2 2 2 2

f l v f l v v v v vd g d g g g g

Applying Bernoulli’s equation for point (1) and (2)

As p1=p2

And

Where H is the height difference between the levels of liquid in two tanks.

If the pipes are more than two, then the equation of total loss should modified.

2 2 21 2 1

1 2 0.52 2 2a bv vp p vz z

w g w g g

1 2

1 2

0

totalloss

v vh z z H

totallossh

Flow through pipe in parallel

Many times the flow from one reservoir to another reservoir is increased by connecting number of pipes in parallel as shown in fig.

Assume Q1 and Q2 are the discharges through the pipes 1 and 2

1 2Q Q Q

Neglecting all minor losses.

Head loss in pipe 1 = Head loss in pipe 2

If f1=f2 then

2 21 1 1 2 2 2

1 2

2 21 1 1 2 2 2

1 2

4 4* *2 2

f l v f l vd g d g

f l v f l vd d

2 21 1 2 2

1 2

l v l vd d

OPEN ENDED PROBLEMFlow through pipe in series

Example : Consider the two reservoirs shown in figure 16, connected by a single pipe that changes diameter over its length. The surfaces of the two reservoirs have a difference in level of 9m. The pipe has a diameter of 200mm for the first 15m (from A to C) then a diameter of 250mm for the remaining 45m (from C to B). Find Q..

For both pipes use f = 0.01. Total head loss for the system H = height

difference of reservoirs hf1 = head loss for 200mm diameter section of

pipe hf2 = head loss for 250mm diameter section of

pipe h entry = head loss at entry point h anp = head loss at join of the two pipes h exit = head loss at exit point H = hf1 + hf2 + hentry + henp + hexit =9m

21

1 213

ffl Qhd

2

22 2

23f

fl Qhd

2 2 21

2 4 41 1 1

1 40.5 0.5 0.5*0.0826 0.04132 2entryu Q Q Qhg g d d d

2 2 21

4 42 2

1.0 1.0*0.0826 0.04132exitu Q Qhg d d

2 221 2 21 2

2 21 2

1 1( ) 4 1 10.08262 2anp

d du u Qh Qg g d d

Substitute value in equation

H = hf1 + hf2 + hentry + henp + hexit =9m

and solve for Q, to give Q = 0.158 m3/s

Flow through pipe in parallel Example :Two pipes connect two reservoirs (A and B)

which have a height difference of 10m. Pipe 1 has diameter 50mm and length 100m. Pipe 2 has diameter 100mm and length 100m. Both have entry loss kL = 0.5 and exit loss kL=1.0 and Darcy f of 0.008.

Calculate: a) rate of flow for each pipe b) the diameter D of a pipe 100m long that could replace the two

pipes and provide the same flow

Apply Bernoulli to each pipe separately. For pipe 1:

21

1

4*0.008*10010 1.00.05 2*9.81

1.731 /

u

u m s

21

1 21

40.5 1.02vflz z

d g

2 2 2 2 21 1 1

1

40.5 1.02 2 2 2 2

a a b ba b

p u p u u flu uz zg g g g g gd g

232

2 2 0.0190 /4dQ u m s

For pipe 2:

21

1

4*0.008*10010 1.00.05 2*9.81

1.731 /

u

u m s

2 2 2 2 22 2 2

2

40.5 1.02 2 2 2 2

a a b ba b

p u p u u flu uz zg g g g g gd g

Again pAand pB are atmospheric, and as the reservoir surface move s slowly uA and uB are negligible, so

2

2 2

44 0.02852

DQ Au u

QuD D

232

2 2 0.0190 /4dQ u m s

b) Replacing the pipe, we need Q = Q1 + Q2 = 0.0034 + 0.0190 = 0.0224 m3/s

For this pipe, diameter D, velocity u , and making the same assumptions about entry/exit losses, we have

2 2 2 2 21 1 1

1

40.5 1.02 2 2 2 2

a a b ba b

p u p u u flu uz zg g g g g gd g

22

2

2

40.5 1.02

4*0.008*10010 1.02*9.81

3.2196.2 1.0

a buflz z

D g

uD

uD

The velocity can be obtained from Q 2

2 2

2

2

5

44 0.02852

3.2 0.02852196.2 1.0

0 241212 1.5 3.2

DQ Au u

QuD D

D DD D

which must be solved iteratively

D=0.1058m