Pertemuan 13 - 14
CLOSED CONDUIT FLOW 1
Reynolds Experiment
•
Reynolds Number
•
•
Laminar flow: Fluid moves in smooth streamlines
Turbulent flow: Violent mixing, fluid velocity at a point varies randomly
with time
Transition to turbulence in a 2 in. pipe is at V=2 ft/s, so most pipe flows
are turbulent
•
Laminar
Bina Nusantara
Turbulent

 2000
Laminar flow
hf V

 4000
Turbulent f low
hf V 2
VD 
Re 
2000  4000 Transition flow
 
Shear Stress in Pipes
•
Steady, uniform flow in a pipe: momentum flux is zero and pressure
distribution across pipe is hydrostatic, equilibrium exists between
pressure, gravity and shear forces
dp
s ) A  W sin    0 (D ) s
ds
dp
dz

sA  As
  0 (D ) s
ds
ds
D
d
p
[
 (  z )]
4
ds

D dh

4 ds
4 L 0
hf 
D
 Fs  0  pA  ( p 
0
0 
0 
h1  h2 
•
•
Bina Nusantara
Since h is constant across the cross-section of the pipe (hydrostatic),
and –dh/ds>0, then the shear stress will be zero at the center (r = 0)
and increase linearly to a maximum at the wall.
Head loss is due to the shear stress.
•
•
Applicable to either laminar or turbulent flow
Now we need a relationship for the shear stress in terms of the Re
and pipe roughness
Darcy-Weisbach Equation
0

V

D
e
ML-1T-2
ML-3
LT-1
ML-1T-1
L
L
 0  F (  , V ,  , D, e)
 4  F ( 1 ,  2 )
Repeating variables :  , V , D
hf 
4L
4L

D 0
V 2 F (Re,
0
e
 1  Re;  2 
; 3 
D
V 2
0
e
 F (Re, )
2
D
V

e
 0  V F (Re, )
D
hf  f
2

D
L V2
D 2g
e 

8 F (Re, D )
L V2
D 2g
Darcy-Weisbach Eq.
Bina Nusantara
e
)
D
f  8 F (Re,
e
)
D
Friction factor
Laminar Flow in Pipes
•
Laminar flow -- Newton’s law of viscosity is valid:
dV
r dh

dy
2 ds
dV

dr
r dh

2  ds
r dh

dr
2  ds
 
dV
dy
dV
dr
dV
V 
r 2 dh
C
4  ds
 r
r 2 dh 
1  
V  0
r
4  ds 
 0

r 2 dh
C 0
4  ds



2



•
  r 2 
V  Vmax 1    
  r0  


Bina Nusantara
Velocity distribution in a pipe (laminar flow) is parabolic with
maximum at center.
Discharge in Laminar Flow
 dh 2 2
( r0  r )
4  ds
 dh 2 2
Q   VdA  0r0 
( r0  r )( 2rdr )
4  ds
V 
 dh ( r 2  r02 ) 2

4  ds
2
Q
r04 dh
8 ds
D 4 dh

128 ds
V 
Q
A
V 
Bina Nusantara
D 2 dh
32  ds
r0
0
Head Loss in Laminar Flow
V 
D 2 dh
32  ds
32 
dh
 V
ds
D 2
32 
dh  V
ds
D 2
32 
h2  h1  V
( s2  s1 )
D 2
h1  h2  h f
hf 
Bina Nusantara
32 LV
D 2
hf 

32 LV
D 2
32 LV V 2 / 2
D 2 V 2 / 2

L
 64(
)( ) V 2 / 2
V D D

64 L
( ) V 2 / 2
Re D
hf  f
L V 2
D 2
f 
64
Re
Nikuradse’s Experiments
•
In general, friction factor
e
)
Function of ReD
and roughness
f  F (Re,
–
•
Laminar region
f 
•
–
Independent
64 of roughness
–
Smooth pipe curve
k
Re 
1/ 4
Rough
Blausius
f 
Turbulent
region
Re
•
–
All curves coincide @ ~Re=2300
Rough pipe zone
•
f 
64
Re
All rough pipe curves flatten out and
become independent of Re
Smooth
Blausius OK for smooth pipe
f 
0.25

5.74 
 e


log 10 
 3.7 D Re 0.9 

2
Laminar
Bina Nusantara
Transition
Turbulent
Moody Diagram
Bina Nusantara
Pipe Entrance
•
Developing flow
–
–
•
Includes boundary layer and core,
viscous effects grow inward from the wall
Fully developed flow
–
Shape of velocity profile is same at all points along pipe
 0.06 Re
Le

1/6
D
4.4Re
Laminar flow
Turbulent flow
Pressure
Fully developed
flow region
Entrance length Le
Entrance
pressure drop
Region of linear
pressure drop
Le
Bina Nusantara
x
Entrance Loss in a Pipe
•
In addition to frictional losses, there are minor losses due to
–
–
–
–
•
•
•
Losses generally determined by experiment and then corellated with
pipe flow characteristics
Loss coefficients are generally given as the ratio of head loss to
velocity head
Abrupt inlet, K ~ 0.5
K – loss coefficent
–
–
–
Bina Nusantara
Entrances or exits
Expansions or contractions
Bends, elbows, tees, and other fittings
Valves
K ~ 0.1 for well-rounded inlet (high Re)
K ~ 1.0 abrupt
hL pipe outlet
 abrupt
or
hL 
KK
~ 0.5
2 pipe inlet
V
2g
V2
K
2g
Elbow Loss in a Pipe
•
•
A piping system may have many minor losses which are all correlated to
V2/2g
Sum them up to a total system loss for pipes of the same diameter
•
Where,
hL  h f   hm 
m

V2  L
f
  Km 

2g  D m

hL  Total head loss
h f  Frictional head loss
hm  Minor head loss for fitting m
K m  Minor head loss coefficien t for fitting m
Bina Nusantara
EGL & HGL for Losses in a Pipe
•
•
•
Bina Nusantara
Entrances, bends, and other flow transitions cause the EGL to drop an
amount equal to the head loss produced by the transition.
EGL is steeper at entrance than it is downstream of there where the slope
is equal the frictional head loss in the pipe.
The HGL also drops sharply downstream of an entrance
Ex(10.2)
Given: Liquid in pipe has = 8 kN/m3. Acceleration = 0. D = 1 cm,  = 3x10-3 N-m/s2.
Find: Is fluid stationary, moving up, or moving down? What is the mean velocity?
Solution: Energy eq. from z = 0 to z = 10 m
2
V12 p1
V22 p2
1

 z1  hL   2

 z2
2g 
2g

200,000
110,000
 hL 
 10
8000
8000
90
hL 
 10
8
hL  1.25 m (moving upward)
hL 
32 μLV
γD 2
γD 2
V  hL
32 μL
V  1.25
8000*( 0.01 )2
32*3x10 -3*10
V  1.04 m / s
Bina Nusantara
1
Ex (10.4)
•
•
•
Given: Oil (S = 0.97, m = 10-2 lbf-s/ft2) in 2 in pipe, Q = 0.25 cfs.
Find: Pressure drop per 100 ft of horizontal pipe.
Solution:
Q
0.25

 11.46 ft / s
A  ( 2 / 12) 2 / 4
VD 0.97 * 1 / 94 * 11.46 * ( 2 / 12)
Re 

 360 (laminar)

10  2
32 μLV
32*10-2*100*11.46
Δp 

 91.7 psi/100 ft
D2
( 2 / 12 )2
V 
Bina Nusantara
Ex. (10.8)
Given: Kerosene (S=0.94, =0.048 N-s/m2). Horizontal 5-cm pipe. Q=2x10-3 m3/s.
Find: Pressure drop per 10 m of pipe.
Solution:
V2 p
V2 p
α1 1  1  z1  hL  α2 2  2  z 2
2g
γ
2g
γ
32 μLV
hL 
γD 2
0  0  0.5 
32 μLV
V22
 α2
00
2g
γD 2
α2 2 32 μL
V2 
V  0.5  0
2g
γD 2
2 2
32 * 4 * 10  5 * 10
V2 
V  0.5  0
2g
0.8 * 62.4 * (1 / 32) 2
V22  8.45V  16.1  0
V  1.60 ft / s
0.8 * 1.94 * 1.6 * (0.25 / 12)
Re 
 1293 (laminar)
4 * 10  5
Q  V * A  1.6 *  * (0.25/12) 2 / 4  1.23 * 10  3 cfs
Bina Nusantara
Ex. (10.34)
Given: Glycerin@ 20oC flows commercial steel pipe.
Find: h
2
Solution:   12,300 N / m,   0.62 Ns / m
V2
p
V2
p
α1 1  1  z1  hL  α2 2  2  z 2
2g
γ
2g
γ
p1
p
 z1  hL  2  z 2
γ
γ
p
p
h  1  z1  ( 2  z 2 )  hL
γ
γ
VD VD
0.6 * 0.02
Re 


 23.5 (laminar)


5.1 * 10  4
32 μLV
32( 0.62)(1)( 0.6)
h  h L 

 2.42 m
2
γD
12,300 * (0.02) 2
Bina Nusantara
Ex. (10.43)
Given: Figure
Find: Estimate the elevation required in the upper reservoir to produce a water
discharge of 10 cfs in the system. What is the minimum pressure in the
pipeline and what is the pressure there?
Solution:
V2
p
V2
p
α1 1  1  z1   hL  α2 2  2  z 2
2g
γ
2g
γ
0  0  z1   hL  0  0  z 2
L V 2

 hL   K e  2 K b  K E  f

D  2g

K e  0.5; K b  0.4 (assumed) ; K E  1.0; f
V 
V2 p
V2 p
α1 1  1  z1   hL  αb b  b  zb
2g
γ
2g
γ
L
430
 0.025 *
 10.75
D
1
Q
10

 12.73 ft / s
A  / 4 * 12
12.732
z1  100  0.5  2 * 0.4  1.0  10.75
 133 ft
2 * 32.2
V2 p
0  0  z1   hL  1 * b  b  zb
2g
γ
pb
V2 
L V 2
 z1  zb  b   K e  K b  f 
γ
2g 
D  2g
300  12.732

 133  110.7  1.0  0.5  0.4  0.025

1  2 * 32.2

 1.35 ft
pb  62.4 * ( 1.53)  0.59 psig
Re 
Bina Nusantara
VD


12.73 * 1
1.14 * 10
5
 9 * 105
Ex. (10.68)
Given: Commercial steel pipe to carry 300 cfs of water at 60oF with a head loss of 1
ft per 1000 ft of pipe. Assume pipe sizes are available in even sizes when the
diameters are expressed in inches (i.e., 10 in, 12 in, etc.).
Find: Diameter.
Solution:
  1.22 x10 5 ft 2 / s; k s  1.5x10 4 ft
Assume f = 0.015
Relative roughness:
Get better estimate of f
Re 
L V2
hf  f
D 2g
1  0.015 *
1
5
VD

Q
1000 (Q /( / 4) D )
D
2g
33,984
k s 1.5x104

 0.00002
D
8.06
2 2

( / 4) D 2
D

Re 
D
D  8.06 ft
Q
( / 4) D
300

( / 4)(8.06)1.22 x10
f=0.010
1
22,656
D5
D  7.43 ft  89 in.
Use a 90 in pipe
Bina Nusantara
5
 3.9 x106
Ex. (10.81)
Given: The pressure at a water main is 300 kPa gage. What size pipe is needed to
carry water from the main at a rate of 0.025 m3/s to a factory that is 140 m
from the main? Assume galvanized-steel pipe is to be used and that the
pressure required at the factory is 60 kPa gage at a point 10 m above the
main connection.
Find: Size of pipe.
Solution:
hf  f
L V2
L (Q /( / 4) D 2 ) 2
 f
D 2g
D
2g
Assume f = 0.020
1/ 5
 fL Q 2 

D  8
 h f  2g 


1/ 5
 0.02 140 (0.025) 2 

 8
2
 14.45


9
.
81


 0.100 m
Relative roughness:
k s 0.15

 0.0015
D 100
Friction factor:
f  0.022
1/ 5
 fL Q 2 

D  8
 h f  2g 


V2 p
V2 p
α1 1  1  z1   hL  α2 2  2  z 2
2g
γ
2g
γ
300,000
60,000
 hf 
 10
9810
9810
h f  14.45 m
Bina Nusantara
1/ 5
 0.022 
D  0.100

 0.020 
Use 12 cm pipe
 0.102 m
Ex. (10.83)
Given: The 10-cm galvanized-steel pipe is 1000 m long and discharges water into the
atmosphere. The pipeline has an open globe valve and 4 threaded elbows;
h1=3 m and h2 = 15 m.
Find: What is the discharge, and what is the pressure at A, the midpoint of the line?
Solution:
V2
p
V2
p
α1 1  1  z1   hL  α2 2  2  z 2
2g
γ
2g
γ
0  0  12  (1  K e  K v  4 K b  f
L V2
)
00
D 2g
V2 p
V2 p
α A A  A  z A   hL  α2 2  2  z 2
2g
γ
2g
γ
D = 10-cm and assume f = 0.025
24 g  (1  0.5  10  4 * 0.9  0.025
1000 2
)V
0.1
24 g
V2 
265.1
V  0.942 m / s
pA
L V2
 15  ( 2 K b  f )
γ
D 2g
pA
500 (0.942) 2
 ( 2 * 0.9  0.025
)
 15  9.6 m
γ
0.1
2g
p A  9810 * ( 9.26)  90.8 kPa
Q  VA  0.942( / 4)( 0.10) 2  0.0074 m 3 / s
Re 
VD


0.942 * 0.1
1.31x10
So f = 0.025
Bina Nusantara
6
 7 x10 4
Near cavitation pressure, not good!
Ex. (10.95)
Given: If the deluge through the system shown is 2 cfs, what horsepower is the pump
supplying to the water? The 4 bends have a radius of 12 in and the 6-in pipe is
smooth.
Find: Horsepower
Solution:
V2
p
V2
p
α1 1  1  z1  h p  α2 2  2  z 2   hL
2g
γ
2g
γ
0  0  30  h p  0  60 
V 
V22
L
(1  0.5  4 K b  f
)
2g
D
Q
2

 10.18 ft / s
A
( / 4)(1 / 2) 2
So f = 0.0135
h p  60  30  1.611(1  0.5  4 * 0.19  0.0135
V22
 1.611 ft
2g
Re 
Bina Nusantara
VD


10.18 * (1 / 2)
1.22 x10  5
 4.17 x10
5
 107.6 ft
Qh p
p
 24.4 hp
550
1700
)
(1 / 2)