Chapter 6
Flow in Closed Conduits
6.1 Flow regimes
Water
Water
Valve
Valve
Dye
Dye
(a) Re < 2000
(b) Re > 4000
Figure 6.1-1 Reynolds’ experiment.
Laminar or well-ordered type of flow exists when adjacent fluid layers slide smoothly over
one another. Mixing between layers occurs only on a molecular level. Turbulent flow exists
when packets of fluid particles are transferred between layers, giving the flow a fluctuating
nature. Osborn Reynolds first described the existence of laminar and turbulent flow
quantitatively through his classic experiment in 1883. As shown in Figure 6.1-1, water was
allowed to flow through a transparent pipe at a rate controlled by a valve. Reynolds
introduced a dye having the same specific gravity as water into the flow to observe what was
happening. He found that at low flow rates the dye pattern was regular and formed a single
line of color as show in Figure 6.1-1(a). The pressure drop was also found to directly
proportional to the flow rate. As the flow rate was increased a point was reach where the dye
trace was seen to be unstable and it broke up after a short distance. At still higher flow rates
the dye almost immediately dispersed throughout the pipe cross section. The relationship
between pressure drop and flow rate now became almost quadratic instead of linear.
The stable flow observed initially was called laminar flow. The unstable flow pattern,
characterized by high degree of mixing between the fluid elements, was called turbulent
flow. There is a transition region in between laminar and turbulent where the flow is unstable
but not thoroughly mixed. Laminar flow in a tube persists up to a point where the value of
the Reynolds number is about 2000. Reynolds number is defined as
NRe =
DV 

=
V 2
V / D
The Reynolds number is a ratio of the inertial momentum flux (V2) in the flow direction to
the viscous shear stress or viscous momentum flux in the transverse (V/D) direction.
Turbulent flow occurs when Reynolds number is greater than about 4000. Viscous forces are
a manifestation of intermolecular attractive forces that stabilize the flow. Therefore stable
laminar flow should occur at low Reynolds numbers where viscous forces dominate.
6-1
6.2 Generalized Mechanical Energy Balance Equation
For laminar flow of a fluid in a cylindrical tube of radius R and length L, the HaganPoiseuille equation provides a relationship between volumetric flow rate and pressure drop
across the tube as follows.
Q=
R 3
R 3 Po  PL R R 4 Po  PL 
w =
=
2L
8L
4
4
P2
V2
P1
V1
z1
z2
Figure 6.2-1 A general piping system.
For a general piping system shown in Figure 6.2-1, we need the generalized relationship,
equation (6.2-1), that can account for the effect of pressure drop on incompressible fluid
flow, changes in elevation, tube cross section, changes in fluid velocity, sudden contractions
or expansions, and friction loss through pipe and fittings such as valves and flow meters.
P1

+ gz1 +
1V12
2
+ wp =
P2

+ gz2 +
 2V22
2
+ ef
(6.2-1)
Each term in this equation has units of energy per unit fluid mass flow rate or (length/time)2.
P = pressure
 = fluid density
g = acceleration of gravity
z = elevation relative to a reference surface
V = average fluid velocity
 = kinetic energy correction factor
 = 2 for laminar flow
 = 1 for turbulent flow
wp = work done per unit mass flow rate
 = pump efficiency ( < 1)
ef = friction loss due to piping and fitting
6-2
The friction loss is given by the following equation
ef = 4  f i
i
where
fi =
w
1
V 2
2
Li Vi 2
+
Di 2

j
V j2
2
Kfitting,j
(6.2-2)
= friction factor in tube segment i with length Li and diameter Di.
Vi = average velocity within tube segment i.
Kfitting = friction loss factor or loss coefficient for pipe fittings, some typical values are
given in Table 6.2-1. The velocity Vj in the summation is for the fluid just
downstream of the contraction, expansion, or fitting.
Table 6.2-1 Friction loss factor for various pipe fittings.
Fitting
Kfitting
Globe valve, wide open
7.5
Angle valve, wide open
3.8
Gate valve, wide open
0.15
A1
A2
Gate valve, half open
4.4
Standard 90o elbow
0.7
o
Standard 45 elbow
0.35
Sudden contraction
Tee, through side outlet
1.5
Tee, straight through
0.4
Sudden contraction

A 
0.4 1  2 
(turbulent flow)
A1 

2
A1
A2
Sudden expansion
 A2




1
(turbulent flow)
A

 1

Sudden expansion
The friction factor for laminar flow (NRe =
f=
VD
< 2000) is given by

16
N Re
(6.2-3)
The friction factor for turbulent flow (Re > 4000) can be estimated by
f = { 1.737 ln[0.269


2.185
14

ln (0.269
+
)]}-2
D
D
N Re
N Re
(6.2-4)
In this equation  is the surface pipe roughness and D is the inside pipe diameter.
Representative values for surface roughness are given in Table 6.2-2.
6-3
Table 6.2-2 Surface roughness
Surface
 (ft)
 (mm)
Concrete
0.001-0.01
0.3-3.0
Cast iron
0.00085
0.25
Wrought iron
0.00015
0.045
Galvanized iron
0.0005
0.15
Commercial steel
0.00015
0.046
Drawn tubing
0.000005
0.0015
Equation (6.2-5) developed by Churchill1 adequately predicts the Fanning fiction factor over
the entire range of Reynolds number including a reasonable estimate for the transition region
between laminar and turbulent flow.
 8 12

1
 
f = 2 

( A  B) 3 / 2 
 N Re 

1 / 12
(6.2-5)



1

In this equation A = 2.457 ln 
0.9
(
7
/
N
)

0
.
27

/
D
Re



16
 37,530 

and B = 
 N Re 
16
If the fluid flows through a noncircular duct, then the equivalent diameter, Deq, can be used in
equations (6.2-2, 3, 4, 5). The equivalent diameter is defined as
Deq = 4rH = 4
where
Across
Pwet
rH = hydraulic radius
Across = cross sectional area of the flow
Pwet = wetted perimeter of the duct
Do Di
Figure 6.2-2 Flow through an annular tube.
For the flow through an annular tube, the equivalent diameter is given as
Deq = 4
1
 ( Do2  Di2 ) / 4
= Do  Di
 ( Do  Di )
Churchill SW, Chem. Eng., Nov. 7, 1977, p. 91
6-4
Example 6.2-1. ---------------------------------------------------------------------------------Water is pumped from the upper reservoir to the lower reservoir through the piping system
shown. Determine the power required for the pump if the water flow rate is 60 kg/s. The
fittings from pipe D1 to pipe D2 and from pipe D2 to pipe D3 can be considered to be standard
90o elbows. Data:
h1 = 10 m, h2 = 3 m, L1 = 50 m, L2 = 300 m, L3 = 2 m, D1 = 0.2 m, D2 = 0.5 m, D3 = 0.03 m,
water viscosity = 1 cP = 10-3 kg/ms,  = 1000 kg/m3. The pipe roughness is 0.05 mm. The
pump efficiency is 75%.
(1)
h1
D1, L1
h2
(2)
D3, L3
Globe valve
D2, L2
Solution -----------------------------------------------------------------------------------------Applying the mechanical energy balance between (1) and (2) we have
P1

+ gz1 +
1V12
2
+ wp =
P2

+ gz2 +
 2V22
2
+ ef
Let the reference level be at (2), the end of pipe 3, the energy equation becomes
Patm

+ g(h1 + L1  L3) + 0 + wp =
g(h1 + L1  L3) + wp = gh2 +
D(m)
.2
.5
.03
A(m2)
3.1410-2
1.9610-1
7.0710-4
V(m/s)
1.91
0.306
84.9
 3V32
2
Patm  gh2

 3V32
2
+ ef
+ ef
NRe
3.82105
1.53105
2.55106
6-5
+0+
/D
2.5010-4
1.0010-4
0.0017
f
0.00406
0.00431
0.00600
ef = 4  f i
i
4  fi
i

j

j
V j2
2
V j2
2
Li Vi 2
+
Di 2

j
V j2
2
Kfitting,j
LiVi 2
300  0.306 2
50  1.912
2  84.9 2
= 2 10-3[4.06
+ 4.31
+ 6
]
2 Di
0.2
0.2
0.2
= 5.77 103 m2/s2
Kfitting,j = 0.51.9120.4
sudden contraction, Kfitting = 0.4
+ 0.50.30620.7
standard 90o elbow, Kfitting = 0.7
+ 0.50.30627.5
open globe valve, Kfitting = 7.5
+ 0.584.920.7
standard 90o elbow, Kfitting = 0.7
Kfitting,j = 2.52 103 m2/s2
ef = 5.77103 + 2.52103 = 8.29103 m2/s2
Therefore
g(h1 + L1  L3) + wp = gh2 +
 3V32
2
+ ef
9.81(10 + 50  2) + 0.75wp = 9.813 +
84.9 2
+ 8.29103
2
wp = 1.51104 m2/s2
The power required for the pump is
 wp = 601.51104 = 9.08105 W = 1220 hp
W p = m
Note: 1 hp = 746 W
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6-6