good fo civilengineering student

1. PRESENTER NAME:
VAIBHAV PATEL (131040106039)
VINAY PATEL (131040106040)
DIGVIJAYSINH PUVAR (131040106045)
RAVI SUTHAR (131040106055)
Subject : Applied Fluid Mechanics
B.E. Civil Engineering (6th sem (2016))
Guidance by
Prof. ankit patel
GUJRAT POWER ENGG. AND
RESEARCH INSTITUTE

2. Flow through pipe in
series and parallel

3. 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.

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

5.  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 2
1 1 1 2 2 2 1 1 2 2
1 2
4 4 ( )
. . 0.5
2 2 2 2 2
f l v f l v v v v v
d g d g g g g

    

6.  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 2
1 2 1
1 2 0.5
2 2 2
a bv vp p v
z z
w g w g g
      
1 2
1 2
0
totalloss
v v
h z z H
 
  
totallossh

7. 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 

9.  Neglecting all minor losses.
Head loss in pipe 1 = Head loss in pipe 2
If f1=f2 then
2 2
1 1 1 2 2 2
1 2
2 2
1 1 1 2 2 2
1 2
4 4
* *
2 2
f l v f l v
d g d g
f l v f l v
d d


2 2
1 1 2 2
1 2
l v l v
d d


10. OPENENDEDPROBLEM
Flow 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..

11. 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
2
1
1 2
13
f
fl Q
h
d

2
2
2 2
23
f
fl Q
h
d


12. 2 2 2
1
2 4 4
1 1 1
1 4
0.5 0.5 0.5*0.0826 0.0413
2 2
entry
u Q Q Q
h
g g d d d
 
    
 
2 2 2
1
4 4
2 2
1.0 1.0*0.0826 0.0413
2
exit
u Q Q
h
g d d
  
2 22
1 2 21 2
2 2
1 2
1 1
( ) 4 1 1
0.0826
2 2
anp
d du u Q
h Q
g 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

13. 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

14.  Apply Bernoulli to each pipe separately. For pipe 1:
2
1
1
4*0.008*100
10 1.0
0.05 2*9.81
1.731 /
u
u m s
 
  
 

2
1
1 2
1
4
0.5 1.0
2
vfl
z z
d g
 
    
 
2 2 2 2 2
1 1 1
1
4
0.5 1.0
2 2 2 2 2
a a b b
a b
p u p u u flu u
z z
g g g g g gd g 
       

15. 2
32
2 2 0.0190 /
4
d
Q u m s

 
For pipe 2:
2
1
1
4*0.008*100
10 1.0
0.05 2*9.81
1.731 /
u
u m s
 
  
 

2 2 2 2 2
2 2 2
2
4
0.5 1.0
2 2 2 2 2
a a b b
a b
p u p u u flu u
z z
g 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

16. 2
2 2
4
4 0.02852
D
Q Au u
Q
u
D D

 
 

2
32
2 2 0.0190 /
4
d
Q u m s

 

17.  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 2
1 1 1
1
4
0.5 1.0
2 2 2 2 2
a a b b
a b
p u p u u flu u
z z
g g g g g gd g 
       
2
2
2
2
4
0.5 1.0
2
4*0.008*100
10 1.0
2*9.81
3.2
196.2 1.0
a b
ufl
z z
D g
u
D
u
D
 
    
 
 
  
 
 
  
 

18.  The velocity can be obtained from Q
2
2 2
2
2
5
4
4 0.02852
3.2 0.02852
196.2 1.0
0 241212 1.5 3.2
D
Q Au u
Q
u
D D
D D
D D

 
 

  
   
  
  
which must be solved iteratively
D=0.1058m