Chapter 5
5.4b One-Dimensional Flow in a Tube.
Aw

dx
V
dV x
+
x
z
P+
Vx
dP
dx  dz
P
x
g
m
Figure 5.4-5 Momentum balance on the CV of Adx.
We will consider the steady state plug flow in a tube with cross-sectional area A as illustrated
in Figure 5.4-5. Applying the x-momentum balance on the control volume Adx gives
d
(m V x )cv = m Vx  m (Vx  dVx ) +
dt
F
x
F
x
=0
on CV
 m dVx = 0
(5.4-2)
on CV
The forces acting on the control volume (fluid) result from pressure (dFP), gravity (dFg), wall
drag (dFw), and possible external “shaft” work (Fext).
F
x
= dFP + dFg + dFw + Fext
on CV
Each of the force in the above expression is given by
dFP = A[P  (P + dP)] =  AdP
dFg =  Adxgcos  =  Adzg (since dxcos  = dz)
dFw =  wdAw =  wWpdx (Wp = perimeter of the wall)
Fext = 
W
dx
5-25
(5.4-3)
In these expressions, w is the stress exerted by the fluid on the wall and W is the shaft work
performed by the fluid. Substituting the expressions for the forces from Eq. (5.4-3) into the
momentum balance equation (5.4-3) gives
 AdP  Adzg  wWpdx 
W
 m dVx = 0
dx
Dividing the equation by ( A), we obtain
dP
+ gdz +

 wW p
VA
W
dx +
+
dV = 0
Adx
A
A
In this equation V = Vx. Let w =
W
= work done per unit mass of fluid, the equation
Adx
becomes
dP
+ gdz +

 wW p
dx + w + VdV = 0
A
Integrating this expression from the inlet to the outlet gives

Po
dP

Pi
where w =
+ g(zo  zi) +

L
0
 wW p
1
dx + w + (oVo2  iVi2) = 0
2
A
(5.4-4)
 w , and  = kinetic energy correction factor,  = 2 for laminar flow,  = 1 for
turbulent flow. Comparing equation (5.4-4) with the energy equation (5.3-9)

Po
dP
Pi

+ g(zo  zi) +
1
(oVo2  iVi2) + ef + w = 0
2
(5.3-9)
shows that they are identical, provided
ef =

L
0
 wW p
dx
A
For steady flow in a uniform conduit
ef =
 wW p L  w  4 L 
A
  , where Dh = 4
=
  Dh 
Wp
A
In this expression Dh is called the hydraulic diameter for a conduit of any cross-sectional
shape. For a circular tube Dh is the same as the tube diameter D
5-26
D 2
A
=4
=D
4D
Wp
Dh = 4
For this special case of one-dimensional fully developed flow in a straight uniform pipe, the
energy and momentum equations provide the same results for the relationship between
temperature, pressure, density, velocity, and dissipated energy (ef) across the system. From
the energy balance, ef represents the lost energy associated with irreversible effects. From the
momentum balance, ef represents the work required or friction loss to overcome the frictional
force at the wall. In general, the momentum balance gives additional information relative to
the forces exerted on and/or by the fluid in the system through the boundaries. This
information cannot be obtained by the energy equation.
5.4c Definition of the Loss Coefficient and the Friction Factor
The energy equation can be made dimensionless by dividing it by any term within the
equation.

Po
dP
Pi
+ g(zo  zi) +

1
(oVo2  iVi2) + ef + w = 0
2
(5.3-9)
If the dividing term is the kinetic energy per unit mass of fluid, we obtain the dimensionless
loss coefficient, Kf
Kf =
ef
V2 /2
A loss coefficient can be defined for any element that causes resistance to flow such as a
length of pipe, a valve, a pipe fitting, a contraction, or an expansion. The total friction loss is
then calculated from the sum of the losses in each element.
ef =
K
fi
i
1 2
Vi
2
The pipe wall stress (w) is the flux of momentum from the fluid to the tube wall. Therefore it
can be made dimensionless by dividing by the flux of momentum V2 carried by the fluid
along the conduct. However a factor of ½ is chosen for the flux of momentum so that it
represents the kinetic energy per unit volume (½V2). The Fanning friction factor f is defined
as
f =
w
V 2 / 2
Mechanical and civil engineers usually use the definition of the Darcy friction factor fD given
by
5-27
fD = 4
w
V 2 / 2
= 4f
The advantage of this definition is that the lost energy will have a form that shows a direct
dependence on the kinetic energy per unit mass. Since
w =
1
fDV2
8
The lost energy in terms of the Darcy friction factor fD is given by
ef =
 wW p L  w  4 L  1
 L 
  = fDV2  
=
  Dh  2
A
 Dh 
Thus, it is important to know which definition is implied when data for friction factors are
used. The Fanning friction factor can be related to the loss coefficient
Kf =
ef
2
V /2
=
2 w
V2 
 4L 
2 fV 2
  = 2
2
 Dh  V
 4L 
4 fL
  =
Dh
 Dh 
Example 5.4-32 ---------------------------------------------------------------------------------Consider the flow in a sudden expansion from a small conduit to a larger one. The
conditions upstream of the expansion (point 1) are known, as well as the areas A1 and A2.
Find the velocity and pressure downstream of the expansion (V2 and P2) and the loss
coefficient, Kf.
A1
V1
P1
A2
V2
P2
x
P1a
Figure 5.4-6 Flow through a sudden expansion.
Solution -----------------------------------------------------------------------------------------Applying the mass balance to the fluid in the shaded area give
A1
A2
The downstream pressure (P2) can be obtained from the energy equation
V2 = V1
2
Darby, R., Chemical Engineering Fluid Mechanics, Marcel Dekker, 2001, p. 124
5-28
P2  P1

+
1
(V22  V12) + ef = 0
2
Since the above equation contains two unknowns, P2 and ef. We need another equation, the
steady-state momentum balance
F
x
 (V2  V1) = 0
m
on CV
The forces acting on the control volume are substituted into the equation to obtain
P1A1 + P1a(A2  A1)  P2A2 + Fwall = A1V1(V2  V1)
In this equation P1a is the pressure force from the solid surface on the left-hand boundary of
the system, and Fwall is the drag force of the wall on the fluid at the horizontal boundary of
the system. Since the fluid pressure cannot change suddenly, P1a  P1. Neglect Fwall because
the horizontal boundary of the system is relatively small, we have
A

(P1  P2)A2 = A1V12  1  1
 A2

From the energy equation
ef =
P1  P2

2
 1 2   A1  
A1 2  A1
1
2
2
+ (V1  V2 ) =
V1 
 1 + V1 1   
  A2  
2
A2
 A2
 2


 A 
A 
1
A
ef = V12  2 1   2 1  1   1 
  A2 
2
A2
 A2 

2
2
2
 1
 = V12 1  A1 

 2
A2 


2

A 
The loss coefficient for the flow through a sudden expansion is defined as Kf = 1  1  ,
A2 

therefore
ef =
1
KfV12
2
The experimental value of the loss coefficient for a sharp 90o contraction is Kf = 0.5(1  2)
D
where  = 1 . For most systems, the loss coefficient cannot be determined accurately from
D2
applying the macroscopic momentum and energy balances. They must be determined from
experimental data.
5-29
Example 5.4-4 ----------------------------------------------------------------------------------
Figure 5.4-7 A 90o horizontal reducing elbow.
Water is flowing through a 90o horizontal reducing elbow as shown. Determine the retaining
forces Fx and Fy required to keep the elbow in place. Neglecting frictional loss, D1 = 0.20 m,
D2 = 0.15 m, P1 = 1.5 bar (gage), V1 = 6.0 m/s. 1 bar = 105 Pa. Water density is 1000 kg/m3.
Solution -----------------------------------------------------------------------------------------From the mass balance
2
A
 .20 
V2 = V1 1 = 6.0 
 = 10.67 m/s
A2
 .15 
The downstream pressure (P2) can be obtained from the energy equation with ef = 0
P2  P1

+
P2 = P1 +
1
(V22  V12) + ef = 0
2
1
(V12  V22) = 1.5105 + 500(62  10.672) = 1.11105 Pa
2
Applying the x-momentum balance on the reducing elbow
Fx + P1A1 + A1V1(V1x  V2x) = 0
Fx =  A1(P1 + V12) =  (.102)( 1.5105 +100062) =  5843 N
The force Fx is actually in the negative x-direction. A similar calculation for the y-momentum
balance gives Fy = 3974 N. [Fy  P2A2 + A1V1(V1y  V2y) = 0]
5-30