ME 543
Homework Solutions
29 February 2012
Fluid Turbulence
Problem 6.5
Given that the following isotropic relationship holds:
1 r
g = f + r , then
2 r
1 f
1 2
rg (r) = rf + r2
=
(r f ) , so then
2 r
2 r
1 2
1
rg (r)dr =
(r f )dr = r2 f (r) .
2 0 r
2
0
0
ME 543
Homework Solutions
22 February 2012
Fluid Turbulence
Problem 5.42
For grid turbulence, assuming a power law decay, the turbulent kinetic energy k (t) and the
energy dissipation rate (t) decay as
t
t0
t
t0
k (t) = k0
(t) =
where k0 = k (t0 ) and
k
=
ME 543
Homework Solutions
15 February 2012
Fluid Turbulence
Problem 5.3
For an axisymmetric jet, an approximation to the self-similar velocity prole is
U
= f ( ) = (1 + a 2 )2 , where = r/(x x0 ) .
U0
Note that
r1/2 (x) = S (x x0 ) , or
dr1/2 (x)
=S,
dx
ME 543
Homework Solutions
8 February 2012
Fluid Turbulence
Problem 12.28
The Lagrangian velocity autocorrelation function is given by:
(s) = exp(|s|/TL ) ,
(1)
where TL is the Lagrangian velocity integral time scale. From Equation (12.158) in the text,
a
ME 543
Homework Solutions
1 February 2012
Fluid Turbulence
Problem 1. Consider the equation for a passive scalar in an incompressible, turbulent ow:
2
=D 2 .
+ Uj
t
xj
xk
(1)
Using the Reynolds decompositions = + and Ui = Ui + ui , plugging this into
Equ
ME 543
Homework Solutions
18 January 2012
Fluid Turbulence
Problem 3.27. Assume that u(t) is statistically stationary with zero mean. Then
u(t)u(t) = u(t)
du(t)
1d2
1d 2
=
u (t) =
u (t) = 0 ,
dt
2 dt
2 dt
since u2 (t) is independent of time. Therefore u(t