fluid_mechanics

5.2 Newton’s Second Law—The Linear Momentum and Moment-of-Momentum Equations 209
E XAMPLE 5.13
GIVEN Air flows steadily between two cross sections in a long,
straight portion of 4-in. inside diameter pipe as indicated in Fig.
E5.13, where the uniformly distributed temperature and pressure at
each cross section are given. If the average air velocity at section 122
is 1000 fts, we found in Example 5.2 that the average air velocity at
section 112 must be 219 fts. Assume uniform velocity distributions
at sections 112 and 122.
SOLUTION
The control volume of Example 5.2 is appropriate for this problem.
The forces acting on the air between sections 112 and 122 are
identified in Fig. E5.13. The weight of air is considered negligibly
small. The reaction force between the wetted wall of the pipe and
the flowing air, R x , is the frictional force sought. Application of
the axial component of Eq. 5.22 to this control volume yields
The positive x direction is set as being to the right. Furthermore,
for uniform velocity distributions 1one-dimensional flow2, Eq. 1
becomes
1u 1 21m # 12 1u 2 21m # 22 R x p 1 A 1 p 2 A 2
(2)
From conservation of mass 1Eq. 5.122 we get
so that Eq. 2 becomes
cs
urV nˆ dA R x p 1 A 1 p 2 A 2
Solving Eq. 4 for R x , we get
The equation of state gives
r 2 p 2
RT 2
and the equation for area A 2 is
Thus, from Eqs. 3, 6, and 7
m # m # 1 m # 2
m # 1u 2 u 1 2 R x A 2 1 p 1 p 2 2
R x A 2 1 p 1 p 2 2 m # 1u 2 u 1 2
A 2 pD2 2
4
m # a p 2
RT 2
b a pD2 2
4 b u 2
Linear Momentum—Pressure, Change in Speed, and Friction
(1)
(3)
(4)
(5)
(6)
(7)
FIND Determine the frictional force exerted by the pipe wall on
the air flow between sections 112 and 122.
p 1
A 1
F I G U R E E5.13
English Engineering 1EE2 units are often used for this kind of
flow. The gas constant, R, for air in EE units is
Thus, from Eqs. 5 and 8
or
V Control volume
V 2 =
1 1000 ft/s
Section (1)
p 1 = 100 psia
T 1 = 540 °R
Pipe
R x
R 17161ft # lb21slug # °R2
53.31ft # lb21lbm # °R2
32.1741lbmslug2
Hence, m # 118.4 psia21144 in. 2 ft 2 2
353.31ft # lb21lbm # °R24 1453 °R2
p14 in.22
41144 in. 2 ft 2 2 11000 ft s2 9.57 lbms
p14 in.22
R x 1100 psia 18.4 psia2
4
19.57 lbm211000 fts 219 fts2
32.1741lbm ft21lb s 2 2
1025 lb 232 lb
R x 793 lb
p 2 A 2
Section (2)
p 2 = 18.4 psia
T 2 = 453 °R
Flow
(8)
(Ans)
COMMENT For this compressible flow, the pressure difference
drives the motion which results in a frictional force, R x , and
an acceleration of the fluid (i.e., a velocity magnitude increase).
For a similar incompressible pipe flow, a pressure difference results
in fluid motion with a frictional force only (i.e., no change in
velocity magnitude).
y
x
E XAMPLE 5.14
Linear Momentum—Weight, Pressure, Friction,
and Nonuniform Velocity Profile
GIVEN
upward.
Consider the flow of Example 5.4 to be vertically
FIND Develop an expression for the fluid pressure drop that
occurs between sections 112 and 122.