89
CHAPTER 5
The Differential Forms
of the
Fundamental Laws
5.1
0
=
 +
⋅
∫
∫
∂ρ
∂
ρ
t
d V
V ndA
c v
c s
. .
. .
$
.
v
Using Gauss’ theorem:
0
=
 +
∇ ⋅

=
+ ∇ ⋅

∫
∫
∫
∂ρ
∂
ρ
∂ρ
∂
ρ
t
d V
V d V
t
V
d V
c v
c v
c v
. .
. .
. .
(
)
(
)
.
v
v
v
v
Since this is true for all arbitrary control volumes (i.e., for all limits of
integration), the integrand must be zero:
∂ρ
∂
ρ
t
V
+ ∇ ⋅
=
v
v
(
)
.
0
This can be written in rectangular coordinates as

=
+
+
∂ρ
∂
∂
∂
ρ
∂
∂
ρ
∂
∂
ρ
t
x
u
y
v
z
w
(
)
(
)
(
).
This is Eq. 5.2.2.
The other forms of the continuity equation follow.
5.2
&
&
.
m
m
m
t
in
out
element

=
∂
∂
(
29
ρ
θ
ρ
∂
∂
ρ
θ
v
rd dz
v
r
v
dr
r
dr d dz
r
r
r
(
)
(
)

+
+
+

+
ρ
ρ
∂
∂θ
ρ
θ
θ
θ
θ
v drdz
v
v
d
drdz
(
)
+
+

+
+
=
+
ρ
θ
ρ
∂
∂
ρ
θ
∂
∂
ρ
θ
v
r
dr
d dr
v
z
v
dz
r
dr
d dr
t
r
dr
d drdz
z
z
z
2
2
2
(
)
.
Subtract terms and divide by
rd drdz
q
:


+


+
=
+
ρ
∂
∂
ρ
∂
∂θ
ρ
∂
∂
ρ
∂
∂
ρ
θ
v
r
r
v
r
dr
r
v
r
z
v
r
dr
r
t
r
dr
r
r
r
z
(
)
(
)
(
)
/
/
.
1
2
2
Since
dr
is an infinitesimal,
(
)/
(
/
)/
.
r
dr
r
r
dr
r
+
=
+
=
1
2
1
and
Hence,
∂ρ
∂
∂
∂
ρ
∂
∂θ
ρ
∂
∂
ρ
ρ
θ
t
r
v
r
v
z
v
r
v
r
z
r
+
+
+
+
=
(
)
(
)
(
)
.
1
1
0
This can be put in various forms.
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