유체역학 3판(한티미디어)chapter05

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유체역학 3판(한티미디어)chapter05에 대한 보고서 자료입니다.

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5.3 m& m& m .
in out t
- = element
¶
¶
r q q f r
¶
¶
v rd r d v r q q f
r
r ( ) sin - é r + ( vr )dr (r dr)d (r dr) sin d
ë ê
ù
û ú
+ +
+ + æè ç
öø ÷
- é +
ë ê
ù
û ú
+ æè ç
öø ÷
r q f r
¶
¶q
vqdr r q r q q q f
dr
d v v d dr r
dr
d
2 2
sin ( ) sin
+ + æè ç
öø ÷
- +
é
ë ê
ù
û ú
+ æè ç
öø ÷
r q r
¶
¶f
r f q f f f v dr r dr d v v d dr r dr d
2 2
( )
= + æè ç
öø ÷
é
ë ê
ù
û ú
¶
¶
r q q f
t
r
dr
drd d
2
2
sin
Because some areas are not rectangular, we used an average length (r + dr / 2).
Now, subtract some terms and divide by rdqdfdr:
- - -
+
-
+
r q r q
¶
¶
r q
¶
¶q
r q q v v
r
v r dr
r
v
r dr
r r r r sin sin ( )sin ( ) ( ) sin
2 2
-
+
=
+ æè ç
öø ÷
¶
¶f
r
¶r
¶
q f ( v ) sin
r
dr
r t
r dr
r
2 2
2
Since dr is infinitesimal (r + dr)2 / r = r and (r + dr / 2) / r = 1. Divide by r sinq
and there results
¶r
¶
¶
¶
r
¶
¶q
r
q
¶
¶f
r r q f t r
v
r
v
r
v
r
v r r + ( ) + ( ) + + =
sin
1 1 ( ) 2 0
5.4 For a steady flow
¶r
¶t
= 0. Then, with v = w = 0 Eq. 5.2.2 yields
¶
¶
r r
r
x
u du
dx
u d
dx
( ) = 0 or + = 0.
Partial derivatives are not used since there is only one independent variable.
5.5 Since the flow is incompressible
D
Dt
r
= 0. This gives
3 2 3
ˆ 1 ˆ 200 1 ˆ 200 ˆ r cos2 r sin2
p p
p i i i i
r r r r r q q
¶ ¶ r r
q q
¶ ¶q
\\Ñ = + = é - ù - êë úû
v
or
u
x
w
z
¶r
¶
¶r
¶
+ = 0.
Also,
v v
Ñ×V = 0, or
¶
¶
¶
¶
u
x
w
z
+ = 0.

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유체,   역학,   유체역학,   유체역학 3판(한티미디어)chapter05