# µ ρ ```Laminar Boundary Layer Order of Magnitude Analysis
δ (x)
u∞
&micro;∞
ρ∞
x
y
c
∗
Changes in x − direction occur over a distance c
∂
1
1
∂
⇒
~
or, we write
= O( )
∂x C
∂x
C
∗ Changes in y − direction occur over a distance δ
1
1
∂
∂
~
or,
= O( )
⇒
∂y
∂y δ
δ
∗ δ &lt;&lt; C (boundary layer is thin)
ρ u c
∗ Re = ∞ ∞ &gt;&gt; 1 (Reynolds number is large)
&micro;∞
∗ Changes in x − velocity are proportional to u ∞
∆u = O(u ∞ )
∗ Changes in pressure are proportional to ρ ∞ u ∞2
∆p = O( ρ ∞ u ∞2 )
Now, let’s do the order of magnitude analysis:
Conservation of Mass
∂u ∂v
+
=0
∂x ∂y
Now, we introduce our assumed orders. For example:
u
∂u
= O( ∞ )
∂x
c
Laminar Boundary Layer Order of Magnitude Analysis
Since we have not assumed an order for v variations, just leave this as simply
∆v and say,
∂v
∆v
= Ο( )
∂y
δ
∂u ∂v
+
= 0 , they obviously have equal (though opposite) magnitudes and
∂x ∂y
therefore orders. Thus, the changes in v scale as:
Since
∆v = Ο(u ∞
δ
C
)
Since v = 0 at a wall, ∆v = Ο(v) itself and we can say
v
δ
= Ο( )
u∞
c
x − momentum
ρu
∂u
∂u
∂p
∂ 2u
∂ 2u
+ ρv
=− +&micro; 2 +&micro; 2
∂x
∂x
∂y
∂x
∂y
Ο( ρ ∞
&micro; u
&micro; u
u ∞2
u2
u2
) Ο ( ρ ∞ ∞ ) Ο ( ρ ∞ ∞ ) Ο( ∞ 2 ∞ ) Ο ( ∞ 2 ∞ )
c
c
c
δ
c
If we compare the last two terms, clearly
&micro;∞u∞
c
2
&lt;&lt;
&micro;∞u∞
δ2
Since we have assumed δ &lt;&lt; C . Thus, we can assume that &micro;
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∂ 2u
is small.
∂y 2
2
Laminar Boundary Layer Order of Magnitude Analysis
In order for any viscous terms to remain, this requires that &micro;
∂ 2u
be equal in
∂x 2
order to the remaining inviscid terms. That is, e.g
∂ 2u
∂u
) = Ο( &micro; 2 )
∂x
∂y
Ο ( ρu
⇒ Ο(
ρ ∞ u ∞2
c
&micro; ∞u∞
)
δ2
&micro;∞
) = Ο(
δ
⇒
( ) 2 = Ο(
)
c
ρ ∞u∞ c
In other words,
δ
C
1
= Ο(
Re
⇐ Major result!
)
y − momentum
∂v
∂v
∂p
∂ 2v
∂ 2v
ρu + ρv = − + &micro; 2 + &micro; 2
∂x
∂y
∂y
∂x
∂y
ρ ∞ u ∞2
δ
1
Ο( ρ ∞ u 2 ) Ο( ρ ∞ u 2 ) Ο(
) Ο( &micro; ∞ u ∞ 3 ) Ο( &micro; ∞ u ∞ )
δ
δc
c
c
c
2
∞
δ
2
∞
δ
Normalizing all of these orders by
and
δ
C
= Ο(
Ο(
1
Re
1
Re
ρ u C
1
and substituting in for Re = ∞ ∞
2
&micro;∞
ρ ∞u∞ / C
) gives:
) Ο(
Clearly, except for
1
Re
) Ο( Re ) Ο(
1
Re
3
) Ο(
2
1
Re
)
∂p
, all other terms are small as Re increases. Thus, we must
∂y
conclude that:
∂p
is small
∂y
∂p
⇒ 0=
is the y − momentum equation for a boundary layer
∂y
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3
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