Modern Engineering Thermodynamics

16.7 Reynolds Transport Theorem 671
dA in
(X, V x , J x ) in
dV
Open system boundary of
volume V and surface area A
σ x
CG
Equivalent closed (fixed
mass) system boundary
V system
dA out
(X, V x , J x ) out
FIGURE 16.19
The vector quantities of Eqs. (16.24), (16.25), and (16.26) and the difference between the open and equivalent closed systems.
In a one-dimensional analysis, ρ, x, J x , and V do not vary across their flow streams, and consequently, they do not
depend upon the cross-sectional area A. In the following equations, we denote the magnitude of a vector quantity
by using the same quantity symbol in italic rather than boldface type. For example, the magnitudes of V, J x ,and
dA are V, J x , and dA, respectively. Then, in a one-dimensional flow, the following simplifications occur
Z
Z Z
J.dA = −ðJ x
dAÞ in
+ ðJ x
and
Z
A
A
= −∑
in
A
J x A +∑ J x A
out
dAÞ out
A
Z
Z
ρxðV .dAÞ = −ðρxV dAÞ in
+ ðρxV dAÞ out
A
A
= −∑
in
= −∑
in
ρVAx +∑ ρVAx
out
_mx+∑
out
_mx
(16.28)
(16.29)
In Eqs. (16.28) and (16.29), we use the fact that the inflow area vector always points in a direction opposite to
the inflow velocity and flux, while the outflow area vector always points in the same direction as the outflow
velocity and flux vectors (see Figure 16.19), so that
and
whereas
and
ðJ x
.dAÞ in
= −ðJ x dAÞ in
ðV .dAÞ in
= −ðVdAÞ in
ðJ x
.dAÞ out =+ðJ x dAÞ out
ðV .dAÞ out
=+ðVdAÞ out
For our one-dimensional analysis, we now require that the system volume V not be a function of time; then, we
can write
Z
∂ðρxÞ
dV = d Z
ρxdV
v ∂t dt V