fluid_mechanics

5.3 First Law of Thermodynamics—The Energy Equation 225
Section (1) Control volume Section (2) Pipe
R
r
u max
u max
u 1 = u max 1 -
R
_ r 2
u
[ ( ) ] 2 = u max 1 -
[ ( R
_ r )
2
]
F I G U R E 5.6
developed pipe flow.
Simple, fully
p
Work is transferred
by rotating shafts,
normal stresses,
and tangential
stresses.
τ
^
n
δ W• tangential stress = 0
V
Note that the value of W # normal stress for particles on the wetted inside surface of the pipe is zero because
V nˆ is zero there. Thus, W # normal stress can be nonzero only where fluid enters and leaves the
control volume. Although only a simple pipe flow was considered, Eq. 5.62 is quite general and
the control volume used in this example can serve as a general model for other cases.
Work transfer can also occur at the control surface because of tangential stress forces. Rotating
shaft work is transferred by tangential stresses in the shaft material. For a fluid particle, shear
stress force power, dW # tangential stress, can be evaluated as the dot product of tangential stress force,
dF tangential stress , and the fluid particle velocity, V. That is,
dW # tangential stress dF tangential stress V
For the control volume of Fig. 5.6, the fluid particle velocity is zero everywhere on the wetted inside
surface of the pipe. Thus, no tangential stress work is transferred across that portion of the
control surface. Furthermore, if we select the control surface so that it is perpendicular to the fluid
particle velocity, then the tangential stress force is also perpendicular to the velocity. Therefore,
the tangential stress work transfer is zero on that part of the control surface. This is illustrated in
the figure in the margin. Thus, in general, we select control volumes like the one of Fig. 5.6 and
consider fluid tangential stress power transfer to be negligibly small.
Using the information we have developed about power, we can express the first law of thermodynamics
for the contents of a control volume by combining Eqs. 5.59, 5.60, and 5.62 to obtain
0
0t er dV cs
erV nˆ dA Q # net W # shaft cs
pV nˆ dA
cv in net in
(5.63)
When the equation for total stored energy 1Eq. 5.562 is considered with Eq. 5.63, we obtain the
energy equation:
0
0t er dV cs
aǔ p
cv
r V 2
2 gzb rV nˆ dA Q# net W # shaft
in net in
(5.64)
5.3.2 Application of the Energy Equation
In Eq. 5.64, the term 00t cv
er dV represents the time rate of change of the total stored energy,
e, of the contents of the control volume. This term is zero when the flow is steady. This term is
also zero in the mean when the flow is steady in the mean 1cyclical2.
In Eq. 5.64, the integrand of
aǔ p
cs
r V 2
gzb rV nˆ dA
2
can be nonzero only where fluid crosses the control surface 1V nˆ 02. Otherwise, V nˆ is zero
and the integrand is zero for that portion of the control surface. If the properties within parentheses,
ǔ, pr, V 2 2, and gz, are all assumed to be uniformly distributed over the flow cross-sectional
areas involved, the integration becomes simple and gives
aǔ p
cs
r V 2
2 gzb rV nˆ dA a aǔ p
flow
r V 2
gzb m#
2
out
aflow
aǔ p (5.65)
r V 2
gzb m#
2
in