Modern Engineering Thermodynamics

672 CHAPTER 16: Compressible Fluid Flow
CRITICAL THINKING
The Reynolds transport theorem is a very powerful mathematical relation often used in advanced engineering courses. Can
you use it to visualize the difference between a Lagrangian (closed system of fixed mass) and a Eulerian (open system of
variable mass) fluid momentum rate balance analysis where X = mV?
and, from the definitions of ρ and x inside the differential volume dV, we get
so that
ρxdV = ðdm=dVÞðdX=dmÞdV = dX
Z
d
=
dt VρxdV d Z
dX =
dX
dt V dt sys
(16.30)
where X sys is the value of dX integrated over the system volume V. Substituting Eqs. (16.28), (16.29), and
(16.30) into Eq. (16.27) gives the complete one-dimensional open system rate balance equation:
Z
ðdX/dtÞ sys
= ∑ J x AÞ −∑ J x A + ∑ _mx−∑ _mx + σ x dV
(16.31)
⎵
in out
in out
V
Net rate
of change
of X inside
the open
system
⎵
Net “conduction”
ði:e:, non-mass
flowÞ transport
rate of X into
the open sytem
⎵
Net mass flow
transport rate
of X into the
open system
⎵
Net production
rate of X inside
the boundary of
the open system
If X is a conserved property like mass, energy, or momentum, then σ x = 0 and the last term on the right side of
Eq. (16.31) vanishes. If X is not conserved, like entropy, then a formula must be found for the variation of σ x
inside V so that the integration of σ x over V can be carried out as indicated (this is what was done for the
entropy production terms discussed in Chapter 7).
EXAMPLE 16.9
Show that Eq. (16.31) reduces to the standard one-dimensional
a. Mass rate balance, Eq. (6.19).
b. Energy rate balance, Eqs. (6.4) and (6.5).
c. Entropy rate balance, Eq. (9.6).
Solution
a. For mass, X = m and x = X/m = m/m = 1. Mass is conserved, so σ x = σ 1 = 0, and since there are no conduction (i.e., nonmass
flow) mechanisms that move mass across a system boundary, J x = J 1 = 0. Then, Eq. (16.31) reduces to
ðdm/dtÞ sys
= ∑ _m −∑ _m
in out
which is identical to the one-dimensional mass rate balance introduced in Eq. (6.19) of Chapter 6.
b. For energy, X = E and x = E/m = e. Energy is conserved, so σ X = σ e = 0, and the non-mass flow energy fluxes are
identified in Chapter 4 as heat and work, soJ e = _q − _w . Then, Eq. (16.31) reduces to
where
ðdE=dtÞ sys
= _Q − _W +∑
in
_me −∑
out
_me
_Q = ∑
in
_qA−∑ _qA and _W = ∑
out
out
_WA−∑
in
_WA
are the net heat and work transport rates of energy, and e = u + V 2 /ð2g c Þ + mgZ/g c is the flow stream specific energy.
These results are identical to those for the one-dimensional energy rate balance originally presented in Chapter 6 in
Eqs. (6.4) and (6.5).