Effect of Turbulent Fluctuations on Mean Flow: Reynolds-Averaging
In a turbulent flow, we can define the mean, steady flow as:
T
1
u ( x, y , z ) = lim ∫ u ( x, y , z , t )dt
T →∞ T
0
This allows us to split the flow properties into a mean and a fluctuating part:
u ( x, y , z , t ) = u ( x, y , z ) + u′( x, y , z, t )
v ( x, y , z , t ) = v ( x, y , z ) + v′( x, y , z, t )
w( x, y , z, t ) = w( x, y , z ) + w′( x, y , z, t )
p ( x, y , z , t ) = p ( x, y , z ) + p′( x, y , z, t )
mean
part
turbulent
fluctuating
part
Note: the mean of u ′ is zero:
u − u = u′
1
T →∞ T
lim
T
0
T
T
1
u′dt
T →∞ T ∫
0
∫ {u − u }dt = lim
1
lim ∫ udt − u = u′
T 0
T →∞
u
⇒
− u = u′
uN
=0
⇒
u′ = 0 ⇐ mean of fluctuations is zero.
Now, we will develop equations which govern the mean flow and try to develop
some insight into how the fluctuations alter the mean flow equations.
Let’s start with incompressible flow and look at the x − momentum:
x − momentum:
 ∂ 2u ∂ 2u ∂ 2u 
1 ∂p
∂u
∂u
∂u
∂u
+u +v
+w
=−
+ν  2 + 2 + 2 
ρ ∂x
∂t
∂x
∂y
∂z
∂y
∂z 
 ∂x
Let’s look at the “averaging” of this equation in time to develop an equation for
the mean flow.
Effect of Turbulent Fluctuations on Mean Flow: Reynolds-Averaging
⇒
1
T →∞ T
lim
T
∫ [ x − mom]dt or
0
x
−
mom
time-average
of x-momentum
The first term is:
1 ∂u
1 ∂ (u + u′)
lim ∫ dt = lim ∫
dt
T →∞ T
T →∞ T
∂t
∂t
0
0
T
But,
T
∂u
= 0 thus:
∂t
T
T
∂u′
1 ∂u
1 ∂u′
=
dt
dt =
lim
∫
∫
T →∞ T
T →∞ T
∂t
∂t
T 0→∂t∞
0
lim
Just as u′ = 0 , we will assume
∂u ′
= 0.
∂t
End result:
∂u
=0
∂t
Let’s skip over to the pressure term and look at its average:
∂p ∂p ∂p′
∂p ∂p ∂p′
=
+
⇒
=
+
∂x ∂x ∂x
∂x ∂x ∂x
∂p ∂p ∂p′
=
+
∂x ∂x ∂x
∂p ∂p ∂p′
=
+
∂x ∂x ∂x
But, p ′ = 0 thus:
∂p ∂p
=
∂x ∂x
Similarly,
∂ 2u ∂ 2u ∂ 2u ∂ 2u ∂ 2u ∂ 2u
+
+
=
+
+
∂x 2 ∂y 2 ∂z 2 ∂x 2 ∂y 2 ∂z 2
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2
Effect of Turbulent Fluctuations on Mean Flow: Reynolds-Averaging
Combining these into x − mom , we now have:
 ∂ 2u ∂ 2u ∂ 2u 
1 ∂p
∂u
∂u
∂u
u +v +w
=−
+ν  2 + 2 + 2 
∂x
∂y
∂z
∂y
∂z 
ρ ∂x
 ∂x
must still work this out
Let’s work out the last term:
u
∂u
∂u
∂u
∂u
∂u
∂u
+w
+v
=u
+w
+v
∂z
∂y
∂x
∂z
∂y
∂x
That’s the easy part. Now, consider u
Question:
What does u
a)
u
∂u
∂x
b)
u
∂u′
∂u
+ u′
∂x
∂x
c)
u
∂u
∂u ′
+ u′
∂x
∂x
d)
none of the above
16.100 2002
∂u
:
∂x
∂u
equal in terms of u & u′ only?
∂x
3