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Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.1. Determine general relationships for the second, third, and four central moments (variance = σ2, skewness = S, and kurtosis = K) of the random variable u in terms of its first four ordinary moments:
€
u ,
€
u2 ,
€
u3 , and
€
u4 . Solution 12.1. a) The variance = σ2 is the second central moment, so:
€
σ 2 =1N
u(n) − u ( )2n=1
N
∑ =1N
u2(n) − 2u(n)u + u 2( )n=1
N
∑ = u2 − 2u u + u 2 = u2 − u 2 .
The skewness = S is the third central moment, so
€
S =1N
u(n) − u ( )3
n=1
N
∑ =1N
u3(n) − 3u2(n)u + 3u(n)u 2 − u 3( )n=1
N
∑
= u3 − 3u2u + 3u u 2 − u 3 = u3 − 3u2u + 2u 3.
The kurtosis = K is the fourth central moment, so
€
K =1N
u(n) − u ( )4
n=1
N
∑ =1N
u4 (n) − 4u3(n)u + 6u2(n)u 2 − 4u(n)u 3 + u 4( )n=1
N
∑
= u4 − 4u3u + 6u2u 2 − 4u u 3 + u 4 = u4 − 4u3u + 6u2u 2 − 3u 4 .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.2. Calculate the mean, mean square, variance, and rms value of the periodic time series
€
u(t) = U + U0 cos ωt( ), where
€
U , U0 and ω are positive real constants. Solution 12.2. The given time series is periodic so time averaging over one period will yield the desired results.
Average:
€
u = 12π ω( )
U + U0 cos(ωt)( )0
2π ω
∫ dt =1
2π ω( )U t +
U0
ωsin(ωt)
%
& ' (
) * 0
2π ω
= U .
Mean Square:
€
u2 =1
2π ω( )U + U0 cos(ωt)( )2
0
2π ω
∫ dt
€
=1
2π ω( )U 2 + 2U0U cos(ωt) + U0
2 cos2(ωt)( )0
2π ω
∫ dt
=1
2π ω( )U 2t +
2U0U ω
sin(ωt) +U0
2
2t +
sin(2ωt)2ω
%
& '
(
) *
+
, -
.
/ 0
0
2π ω
= U 2 +U0
2
2.
Variance:
€
u − u ( )2 =1
2π ω( )U + U0 cos(ωt) −U ( )2
0
2π ω
∫ dt = U02 cos2(ωt)
0
2π ω
∫ dt =U02
2.
rms value:
€
u2 = U 2 +U02
2.
And for completeness,
Standard deviation:
€
u − u ( )2 =U0
2
The rms value and the standard deviation are not equal unless
€
u = U = 0 .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.3. Show that the autocorrelation function
€
u(t)u(t + τ) of a periodic series u = Ucos(ωt) is itself periodic. Solution 12.3. The given time series is periodic so time averaging over one period will yield the desired results.
€
u(t)u(t + τ) =U 2
2π ω( )cos(ωt)cos ω(t + τ )( )
0
2π ω
∫ dt
=U 2
2π ω( )cos(ωt) cos(ωt)cos(ωτ) − sin(ωt)sin(ωτ)[ ]
0
2π ω
∫ dt
=U 2
2π ω( )cos(ωτ ) cos2(ωt)
0
2π ω
∫ dt − U 2
2π ω( )sin(ωτ ) cos(ωt)
0
2π ω
∫ sin(ωt)dt
=U 2
2π ω( )cos(ωτ ) 1
22πω
'
( )
*
+ , − 0 =
U 2
2cos(ωτ)
And, since cos(ωt) is periodic, then
€
u(t)u(t + τ) is periodic.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.4. Calculate the zero-lag cross-correlation
€
u(t)v(t) between two periodic series u(t) = cos ωt and v(t) = cos(ωt + φ) by performing at time average over one period = 2π/ω. For values of φ = 0, π/4, and π/2, plot the scatter diagrams of u vs v at different times, as in Figure 12.8. Note that the plot is a straight line if φ = 0, an ellipse if φ = π/4, and a circle if φ = π/2; the straight line, as well as the axes of the ellipse, are inclined at 45° to the uv-axes. Argue that the straight line signifies a perfect correlation, the ellipse a partial correlation, and the circle a zero correlation. Solution 12.4. The given time series is periodic so time averaging over one period will yield the desired results.
€
u(t)v(t) =U 2
2π ω( )cos(ωt)cos ωt + φ( )
0
2π ω
∫ dt
=U 2
2π ω( )cos(ωt) cos(ωt)cosφ − sin(ωt)sinφ[ ]
0
2π ω
∫ dt
=U 2
2π ω( )cosφ cos2(ωt)
0
2π ω
∫ dt − U 2
2π ω( )sinφ cos(ωt)
0
2π ω
∫ sin(ωt)dt
=U 2
2π ω( )cosφ 1
22πω
'
( )
*
+ , − 0 =
U 2
2cosφ
The scatter diagrams are obtained by placing sample points from different times in a two-dimensional (u,v)-coordinate plane. The locus of sample points is obtained by eliminating t from using the equations for u and v:
€
v = cos(ωt + φ) = cosωt cosφ − sinωt sinφ = ucosφ − 1− u2 sinφ . Use the two ends of this extended equality to find:
€
v − ucosφ = − 1− u2 sinφ , or
€
v 2 − 2uv cosφ + u2 cos2 φ = (1− u2)sin2 φ . Further simplify:
€
v 2 − 2uv cosφ + u2 = sin2 φ . This quadratic relationship can be cast in a standard form by switching to sum, v + u, and difference, v – u, coordinates:
€
1− cosφ2
$
% &
'
( ) (v + u)2 +
1+ cosφ2
$
% &
'
( ) (v − u)2 = sin2 φ =1− cos2 φ ,
which implies:
€
(v + u)2
2(1+ cosφ)+
(v − u)2
2(1− cosφ)=1 when φ ≠ 0 or π,
v = u when φ = 0, and v = –u when φ = π. The first possibility is the equation for an ellipse having major and minor axes rotated 45° from the u and v axes. The other two possibilities are just straight lines. When φ = 0 then v = u so the resulting distribution of sample points is a straight line with unity slope.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
In this case, each value of u is linked to a single and equal value of v. This is perfect correlation between u and v.
When φ = π/4, then the locus of possible points becomes
€
(v + u)2
2 + 2+(v − u)2
2 − 2=1, which is:
In this case, each value of u is linked to a two values of v. Here positive v is more likely with positive u, and negative v is more likely with negative u. Thus, this situation corresponds to partial correlation between u and v. When φ = π/2, then the locus of possible points becomes, which is a circle:
In this case, each value of u is linked to a two values of v. Here positive v is equally likely with positive or negative u, and negative v is equally likely with positive or negative u. Thus, this situation corresponds to no correlation between u and v.
u
v
u
v
u
v
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.5. If u(t) is a stationary random signal, show that u(t) and
€
du(t) dt are uncorrelated. Solution 12.5. a) For a stationary signal u(t), use the time-average definition of the correlation and evaluate the integral:
€
limΔt→∞
1Δt
u(t) dudtdt
−Δt 2
+Δt 2
∫ = limT→∞
12Δt
du2
dtdt =
−Δt 2
+Δt 2
∫ limΔt→∞
12Δt
u2(Δt /2) − u2(−Δt /2)[ ] = 0 .
The final equality occurs because u2 is stationary and remains finite at ±∞ while the divisor of the [,]-brackets goes to infinity.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.6. Let R(τ) and S(ω) be a Fourier transform pair. Show that S(ω) is real and symmetric if R(τ) is real and symmetric. Solution 12.6. Start with:
€
S(ω) =12π
e−iωτR(τ )dτ−∞
+∞
∫ ,
and decompose into real and imaginary parts:
€
S(ω) =12π
cosωτ − isinωτ( )R(τ )dτ−∞
+∞
∫ .
Since sinωτ is odd and R(τ) is even in τ, the integral over the imaginary part is zero. Thus,
€
S(ω) =12π
cos(ωτ)R(τ)dτ =−∞
+∞
∫ 12π
cos(−ωτ )R(τ )−∞
+∞
∫ dτ = S(−ω),
which clearly shows that: (1) S(ω) is real because it can be computed from the integral of the two real functions R(τ) and cos(ωτ), and (2) S(ω) is symmetric about ω = 0 because S(ω) = S(–ω).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.7. Compute the power spectrum, integral time scale, and Taylor time scale when
€
R11(τ) = u12 exp −ατ 2( )cos(ωoτ ) , assuming that α and ωo are real positive constants.
Solution 12.7.
€
Se (ω) =u12
2πe−ατ
2
cos(ωoτ)exp −iωτ{ }−∞
+∞
∫ dτ =u12
4πe−ατ
2
ei(ωo−ω )τ + ei(−ωo−ω )τ( )dτ−∞
+∞
∫
The exponents of the two terms in the integrand are:
€
−ατ 2 − i(ω ∓ωo)τ = −α τ 2 +i(ω ∓ωo)
ατ +
(ω ∓ωo)2
4α 2 −(ω ∓ωo)
2
4α 2
&
' (
)
* +
= −α τ +i(ω ∓ωo)
2α&
' (
)
* +
2
−(ω ∓ωo)
2
4α,
where the top sign belongs with the first term. Let
€
β = α τ +i(ω ∓ωo)2α
&
' (
)
* + be the new integration
variable:
€
Se (ω) =u12
4π αexp −
(ω −ωo)2
4α& ' (
) * +
e−β2
dβ−∞
+∞
∫ +u12
4π αexp −
(ω +ωo)2
4α& ' (
) * +
e−β2
dβ−∞
+∞
∫ .
The integral in both terms is
€
π , so the energy spectrum is:
€
Se (ω) =u12
4 παexp −
(ω −ωo)2
4α& ' (
) * +
+ exp −(ω +ωo)
2
4α& ' (
) * +
,
- .
/
0 1 .
The integral time scale can be determined from (12.18) and (12.20) evaluated at ω = 0:
€
Λ t =1# u 2
R11(τ)dτ =0
∞
∫ 2π2u1
2Se (0) =
π
u12
u12
4 πα2exp −
ωo2
4α+ , -
. / 0
=π
2 αexp −
ωo2
4α+ , -
. / 0
.
The Taylor time scale defined in (12.19) involves the second derivative of R11 at τ = 0, which is tedious to determine directly. However the given R11 is relatively easy to expand for τ << 1:
€
R11(τ) = u12 exp −ατ 2( )cos(ωoτ ) = u1
2 1−ατ 2 + ...( ) 1− 12 (ωoτ )
2 + ...( ) = u12 1− α + 1
2ωo2[ ]τ 2 + ...( ) ,
and this is much easier to differentiate:
€
d2R11(τ )dτ 2
#
$ %
&
' ( τ= 0
= −u12 2α +ωo
2[ ].
So,
€
λt = −2R11(0)d2R11(τ)dτ 2
%
& '
(
) * τ= 0
+
, -
.
/ 0
1 2
=−2u1
2
−u12 2α +ωo
2[ ]= α + 1
2ωo2[ ]
−1 2.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.8. There are two formulae for the energy spectrum Se(ω) of the stationary zero-mean signal
€
u(t):
€
Se (ω) =12π
R11(τ)exp −iωτ{ }−∞
+∞
∫ dτ and
€
Se (ω) = limT→∞
12πT
u(t)exp −iωt{ }−T 2
+T 2
∫ dt2
.
Prove these two are identical without requiring the existence of the Fourier transform of
€
u(t). Solution 12.8. Start with the second formula and use t´ and t as the integration variables.
€
Se (ω) = limT →∞
12πT
u(t)exp −iωt{ }−T 2
+T 2
∫ dt2
= limT →∞
12πT
u( ( t )u(t)exp −iω( ( t − t){ }−T 2
+T 2
∫ d ( t −T 2
+T 2
∫ dt .
Alter the t´ integration by introducing a new integration variable τ = t´ – t. The region of integration is a square centered at t´ = t = 0.
The t integration goes from t– to t+. For τ < 0 (the case corresponding to the dashed line),
t– = –T/2 – τ, and t+ = T/2, so t+ – t– = T + τ = T – |τ|. For τ > 0,
t– = –T/2 , and t+ = T/2 – τ, so t+ – t– = T – τ = T – |τ|. The τ integration goes from –T to +T. Thus, the arguments in the integrand and the limits of integration are altered:
€
Se (ω) = limT→∞
12πT
u(t + τ)u(t)e−iωτt= t−
t= t+
∫ dττ=−T
τ= +T
∫ dt = limT→∞
12πT
u(t + τ )u(t)t= t−
t= t+
∫ dt)
* +
,
- .
τ=−T
τ= +T
∫ e− iωτdτ .
From the figure, the difference t+ – t– = T – |τ|. Therefore,
€
Se (ω) = limT→∞
12π
T − τ( )T
1T − τ
u(t + τ)u(t)t= t−
t= t+
∫ dt)
* +
,
- .
τ=−T
τ= +T
∫ e−iωτdτ ,
so contents of the large parentheses is R11(τ) for T >> τ when u(t) is uncorrelated with itself at large time lags, τ >> Λt. This leaves:
t´ =
t +
!
t´
t
t+t–
!
–T
!
t´ =
t –
T
t´ =
t +
T
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
Se (ω) = limT→∞
12π
1−τ
T(
) *
+
, - R11(τ)
τ=−T
τ= +T
∫ e− iωτdτ =12π
R11(τ )−∞
+∞
∫ e−iωτdτ ,
where the final equality holds when T >> τ and
€
R11(τ)→ 0 for τ >> Λt.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.9. Derive the formula for the temporal Taylor microscale λt by expanding the definition of the temporal correlation function (12.17) into a two term Taylor series, and determining the time shift, τ = λt, where this two term expansion equals zero. Solution 12.9. Expand the temporal autocorrelation:
€
R11(τ) = R11(0) + τdR11(τ)dτ
#
$ %
&
' ( τ= 0
+τ 2
2d2R11(τ )dτ 2
#
$ %
&
' ( τ= 0
+ ...
The autocorrelation function is even, so the terms involving odd derivatives are zero. Thus, the appropriate two-term expansion is:
€
R11(τ) = R11(0) +τ 2
2d2R11(τ )dτ 2
#
$ %
&
' ( τ= 0
+ ...
Truncate the expansion, evaluate it at t = λt, and set it equal to zero to find:
€
R11(λt ) = 0 = R11(0) +λt2
2d2R11(τ)dτ 2
$
% &
'
( ) τ= 0
+ ...
Now solve for λt to reach:
€
λt2 = −2R11(0)
d2R11(τ )dτ 2
%
& '
(
) * τ= 0
= −2 1R11(0)
d2R11(τ )dτ 2
%
& '
(
) * τ= 0
= −2 d2r11(τ )dτ 2
%
& '
(
) * τ= 0
,
which duplicates (12.19).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.10. When x, r, and k1 all lie in the stream-wise direction, the wave number spectrum
€
S11(k1) of the stream-wise velocity fluctuation
€
u1(x) defined by (12.45) can be interpreted as a distribution function for energy across stream-wise wave number k1. Show that the energy-weighted mean-square value of the stream-wise wave number is:
€
k12 ≡
1u2
k12S11(k1)dk1 = −
1u2−∞
+∞
∫ d2
dr2R11(r)
&
' (
)
* + r= 0
, and that
€
λ f = 2 k12 .
Solution 12.10. The relationship between the stream-wise autocorrelation function and the stream-wise spatial power spectrum is the Fourier inverse of (12.45):
€
R11(r) = S11(k1)e+ ik1rdk1
−∞
+∞
∫ .
Differentiate this twice with respect to r, and evaluate at r = 0:
€
d2R11(r)dr2
"
# $
%
& ' r= 0
= S11(k1)d2
dr2e+ ik1r
"
# $
%
& ' dk1
−∞
+∞
∫"
# $
%
& ' r= 0
= −k12S11(k1)e
+ ik1rdk1−∞
+∞
∫"
# $
%
& ' r= 0
= −k12S11(k1)dk1
−∞
+∞
∫ .
Divide by the mean-square fluctuation
€
" u 2 and multiply by –1 to get the desired form:
€
−1u2
d2R11(r)dr2
#
$ %
&
' ( r= 0
=1u2
k12S11(k1)dk1
−∞
+∞
∫ = k12
For, the specified geometry, R11 is the longitudinal autocorrelation function, so from (12.39)
€
−1u2
d2R11(r)dr2
#
$ %
&
' ( r= 0
=d2 f (r)dr2
#
$ %
&
' ( r= 0
=2λ f2 ,
and this can be combined with the mean-square wave number result to find
€
λ f = 2 k12 .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.11. In many situations, measurements are only possible of one velocity component at one point in a turbulent flow, but the flow has a non-zero mean velocity and moves past the measurement point. Thus, the experimenter obtains a time history of
€
u1(t) at fixed point. In order to estimate spatial velocity gradients, Taylor’s frozen-turbulence hypothesis can be
invoked to estimate a spatial gradient from a time derivative:
€
∂u1∂x1
≈ −1U1
∂u1∂t
where the “1”-axis
must be aligned with the direction of the average flow, i.e. Ui = (U1, 0, 0). Show that this approximate relationship is true when
€
uiui U1 <<1,
€
p ~ ρu12 , and Re is high enough to neglect
the influence of viscosity. Solution 12.11. The place to start is with the “1”-component of the Navier-Stokes equation for
an incompressible fluid:
€
∂u1∂t
+ u1∂u1∂x1
+ u2∂u1∂x2
+ u3∂u1∂x3
= −1ρ∂p∂x1
+ ν∂ 2u1∂x1
2 +∂ 2u1∂x2
2 +∂ 2u1∂x3
2
&
' (
)
* + .
Rearrange this equation to isolate the terms with
€
∂u1 ∂x1 and
€
∂u1 ∂t .
€
u1∂u1∂x1
+ = −∂u1∂t
− u2∂u1∂x2
− u3∂u1∂x3
−1ρ∂p∂x1
+ ν∂ 2u1∂x1
2 +∂ 2u1∂x2
2 +∂ 2u1∂x3
2
&
' (
)
* +
Set
€
u1 = U1 + " u 1 in the two terms, and move parts of them to the other side of the equation:
€
∂ # u 1∂t
+ U1∂ # u 1∂x1
+ = − # u 1∂u1∂x1
− u2∂u1∂x2
− u3∂u1∂x3
−∂U1
∂t+ U1
∂U1
∂x1+1ρ∂p∂x1
&
' (
)
* + + ν
∂ 2u1∂x1
2 +∂ 2u1∂x2
2 +∂ 2u1∂x3
2
&
' (
)
* +
Divide by U1, and note that when Ui = (U1, 0, 0), then
€
u2 = " u 2 and
€
u3 = " u 3 :
€
∂ # u 1∂x1
+1
U1
∂ # u 1∂t
= −# u 1
U1
∂u1∂x1
−# u 2
U1
∂u1∂x2
−# u 3
U1
∂u1∂x3
−1
U1
∂U1
∂t+ U1
∂U1
∂x1+1ρ∂p∂x1
&
' (
)
* + +
νU1
∂ 2u1∂x1
2 +∂ 2u1∂x2
2 +∂ 2u1∂x3
2
&
' (
)
* + .
When Re is high the terms multiplied by ν will be small and can be dropped. Now decompose the pressure into average and fluctuating parts:
€
∂ # u 1∂x1
+1
U1
∂ # u 1∂t
= −# u 1
U1
∂u1∂x1
−# u 2
U1
∂u1∂x2
−# u 3
U1
∂u1∂x3
−1
U1
∂U1
∂t+ U1
∂U1
∂x1+1ρ∂p ∂x1
+1ρ∂ # p ∂x1
&
' (
)
* +
The first three terms inside the parentheses can be simplified using the “1”-direction RANS
equation for this situation:
€
∂U1
∂t+ U1
∂U1
∂x1= −
1ρ∂p ∂x1
−∂ % u 1 % u j∂x j
,
€
∂ # u 1∂x1
+1
U1
∂ # u 1∂t
= −# u 1
U1
∂u1∂x1
−# u 2
U1
∂u1∂x2
−# u 3
U1
∂u1∂x3
−1
U1
−∂ # u 1 # u j∂x j
+1ρ∂ # p ∂x1
&
' ( (
)
* + + .
When
€
" p ~ ρ " u 12 , then every term on the right side contains a factor like
€
" u i " u i U1 , either inside or outside of a differentiation, thus
€
∂u1∂x1
+1U1
∂u1∂t
≅ 0 when
€
" u i " u i U1→ 0
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.12. a) Starting from (12.33), derive (12.34) via an appropriate process of Reynolds decomposition and ensemble averaging. b) Determine an equation for the scalar fluctuation energy =
€
12 " Y 2 , one half the scalar variance.
c) When the scalar variance goes to zero, the fluid is well mixed. Identify the term in the equation from part b) that dissipates scalar fluctuation energy. Solution 12.12. a) First simplify (12.33) for constant density flow by dividing by ρm:
€
∂∂t
ρm˜ Y ( ) +
∂∂x j
ρm˜ Y ̃ u j( ) =
∂∂x j
ρmκm∂∂x j
˜ Y %
& ' '
(
) * * →
∂ ˜ Y ∂t
+∂∂x j
˜ Y ̃ u j( ) =∂∂x j
κm∂ ˜ Y ∂x j
%
& ' '
(
) * * .
Insert the Reynolds decomposition for the fluid velocity and the mass fraction,
€
˜ u j = U j + u j and
€
˜ Y = Y + " Y , into the constant-density passive-scalar conservation equation:
€
∂∂t
Y + # Y ( ) +∂∂x j
(Y + # Y )(U j + u j )( ) =∂∂x j
κm∂∂x j
Y + # Y ( )%
& ' '
(
) * * . (&)
Average the equation, and drop terms that involve first-order averages of the fluctuating components,
€
" Y = 0 = u j ,
€
∂∂t
Y + # Y ( ) +∂∂x j
(Y + # Y )(U j + u j )( ) =∂∂x j
κm∂∂x j
Y + # Y ( )%
& ' '
(
) * * ,
€
∂∂t
Y + # Y ( ) +∂∂x j
Y U j + Y u j + # Y U j + # Y u j( ) =∂∂x j
κm∂∂x j
Y + # Y ( )%
& ' '
(
) * * ,
€
∂Y ∂t
+∂∂x j
Y U j + # Y u j( ) =∂∂x j
κm∂Y ∂x j
%
& ' '
(
) * * , or (†)
€
∂Y ∂t
+ U j∂Y ∂x j
=∂∂x j
κm∂Y ∂x j
− % Y u j
&
' ( (
)
* + + .
where ∂Uj/∂xj = 0 has been used to reach the final form which is identical to (12.34). b) Generate an equation for the scalar fluctuation, by subtracting (†) from (&) to produce:
€
∂ # Y ∂t
+∂∂x j
# Y U j + Y u j + # Y u j − # Y u j( ) =∂∂x j
κm∂ # Y ∂x j
&
' ( (
)
* + + , or
€
∂ # Y ∂t
+ U j∂ # Y ∂x j
+ u j∂Y ∂x j
+ u j∂ # Y ∂x j
−∂∂x j
# Y u j( ) =∂∂x j
κm∂ # Y ∂x j
&
' ( (
)
* + + ,
where ∂Uj/∂xj = 0 and ∂uj/∂xj = 0 have been used to reach the second version. Multiply the lower equation by Y´ to find:
€
" Y ∂" Y
∂t+ U j " Y ∂
" Y ∂x j
+ u j " Y ∂Y ∂x j
+ u j " Y ∂" Y
∂x j
− " Y ∂∂x j
" Y u j( ) = " Y ∂∂x j
κm∂ " Y ∂x j
&
' ( (
)
* + + or
€
∂∂t
12 # Y 2( ) + U j
∂∂x j
12 # Y 2( ) + u j # Y ∂Y
∂x j
+ u j∂∂x j
12 # Y 2( ) − # Y ∂
∂x j
# Y u j( ) = # Y ∂∂x j
κm∂ # Y ∂x j
&
' ( (
)
* + + ,
and average the result to find:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
∂∂t
12 # Y 2( ) + U j
∂∂x j
12 # Y 2( ) + u j # Y ∂Y
∂x j
+ u j∂∂x j
12 # Y 2( ) − # Y ∂
∂x j
# Y u j( ) = # Y ∂∂x j
κm∂ # Y ∂x j
&
' ( (
)
* + + .
The final two terms on the left are zero because
€
" Y = u j = 0, and the term on the right can be rearranged to produce:
€
∂∂t
12 # Y 2( ) + U j
∂∂x j
12 # Y 2( ) + u j # Y ∂Y
∂x j
= # Y ∂∂x j
κm∂ # Y ∂x j
%
& ' '
(
) * * =
∂∂x j
κm∂∂x j
12 # Y 2( )
%
& ' '
(
) * * −κm
∂ # Y ∂x j
∂ # Y ∂x j
.
The terms in this equation for the scalar-fluctuation energy are interpreted as follows:
€
∂∂t
12 # Y 2( ) + U j
∂∂x j
12 # Y 2( ) = −u j # Y ∂Y
∂x j
+∂∂x j
κm∂∂x j
12 # Y 2( )
&
' ( (
)
* + + −κm
∂ # Y ∂x j
∂ # Y ∂x j
.
Time rate of change Production diffusive transport dissipation following the mean flow c) The final term in the last equation dissipates scalar fluctuations. It is always negative and it is largest where the scalar gradients are largest.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.13. Measurements in an atmosphere at 20 °C show an rms vertical velocity of wrms = 1 m/s and an rms temperature fluctuation of Trms = 0.1°C. If the correlation coefficient is 0.5, calculate the heat flux ρcpw !T . Solution 12.13. The heat flux will be:
€
ρCp w # T = ρCpwrmsTrmsw # T
wrmsTrms
$
% &
'
( ) = (1.2kgm−3)(1004m2s−2K−1)(1ms−1)(0.1°C)(0.5)
= 60.24Wm−2
where the factor in large parentheses is the correlation coefficient between vertical velocity fluctuations and temperature fluctuations.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.14. a) Compute the divergence of the constant density Navier-Stokes momentum
equation
€
∂ui∂t
+ u j∂ui∂x j
= −1ρ∂p∂xi
+ ν∂ 2ui∂x j∂x j
to determine a Poisson equation for the pressure.
b) If the equation
€
∂ 2G∂x j∂x j
= δ(x j − ˜ x j ) has solution:
€
G(x j , ˜ x j ) =−1
4π (x j − ˜ x j )2
, then use the
result from part a) to show that: P(x j ) =ρ
4π1
(x j − !x j )2
!x∫ ∂ 2
∂ !x j∂ !xiUiU j +uiuj( )d3 !x in a turbulent
flow.
Solution 12.14. a) Start with
€
∂ui∂t
+ u j∂ui∂x j
= −1ρ∂p∂xi
+ ν∂ 2ui∂x j∂x j
and take the divergence (i.e.
apply
€
∂∂xi
to both sides of the equation). This produces:
€
∂∂t∂ui∂xi
+∂u j
∂xi
∂ui∂x j
+ u j∂∂x j
∂ui∂xi
= −1ρ
∂ 2p∂xi∂xi
+ ν∂ 2
∂x j∂x j
∂ui∂xi
The first, third, and final terms are zero in an incompressible flow, leaving
€
∂ 2p∂xi∂xi
= −ρ∂u j
∂xi
∂ui∂x j
.
b) This is merely an application of a Green’s function solution in an unbounded domain. Start
with
€
∂ 2G∂x j∂x j
= δ(x j − ˜ x j ) and multiply by
€
−ρ∂u j
∂˜ x i
∂ui
∂˜ x j and integrate over all
€
˜ x -space:
−ρ∂uj∂ !xi
∂ui∂ !x jall !x∫ ∂ 2G
∂x j∂x jd3 !x = −ρ δ(x j − !x j )
∂uj∂ !xi
∂ui∂ !x j
d3 !xall !x∫ = −ρ
∂uj∂xi
∂ui∂x j
,
where the final equality is achieved through the sifting property of the Dirac δ-function. Inside the first integral, the differentiations on the x-coordinates can be brought outside the integral because the integral involves the
€
˜ x -coordinates. Thus:
−∂ 2
∂x j∂x jρ∂uj∂ !xi
∂ui∂ !x jall !x
∫ Gd3 !x = −ρ∂uj∂xi
∂ui∂x j
A comparison of this equation with the result part a) then requires:
p = −ρ∂uj∂ !xi
∂ui∂ !x jall !x∫ Gd3 !x = ρ
4π∂uj∂ !xi
∂ui∂ !x jall !x
∫ 1(x j − !x j )
2d3 !x
Here, we note that:
€
∂∂x j
uiu j = u j∂∂x j
ui + ui∂∂x j
u j = u j∂∂x j
ui
for incompressible flow. Thus
€
∂ 2
∂xi∂x j
uiu j =∂u j
∂xi
∂ui∂x j
+ u j∂∂x j
∂ui∂xi
=∂u j
∂xi
∂ui∂x j
,
so that the pressure equation becomes:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
p =ρ
4π1
(x j − ˜ x j )2all ˜ x ∫ ∂2
∂˜ x i∂˜ x juiu j( )d3 ˜ x
Now insert the Reynolds decomposition,
€
u j = U j + " u j , for the velocity factors inside the integral:
€
p =ρ
4π1
(x j − ˜ x j )2all ˜ x ∫ ∂2
∂˜ x i∂˜ x jUiU j + Uiu j + uiU j + ' u i ' u j( )d3 ˜ x .
Take the ensemble average of this equation:
€
p = P =ρ
4π1
(x j − ˜ x j )2all ˜ x ∫ ∂ 2
∂˜ x i∂˜ x jUiU j + ' u i ' u j( )d3 ˜ x .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.15. Starting with the RANS momentum equation (12.30), derive the equation for the kinetic energy of the average flow field (12.46). Solution 12.15. Start with the definition:
€
E ≡ 12UiUi , and the RANS equation for Ui:
€
∂Ui
∂t+ U j
∂Ui
∂x j
= −1ρ∂p ∂xi
− g 1−α T −T0( )[ ]δi3 + ν∂Ui
∂x j∂x j
−∂∂x j
( u i ( u j .
Take the dot product of this equation with Ui:
€
Ui∂Ui
∂t+ UiU j
∂Ui
∂x j
= −1ρ
Ui∂p ∂xi
− gUi 1−α T −T0( )[ ]δi3 + νUi∂Ui
∂x j∂x j
−Ui∂∂x j
( u i ( u j
Recognize
€
E where it occurs:
€
∂E ∂t
+ U j∂E ∂x j
= −1ρ
Ui∂p ∂xi
+ νUi∂Ui
∂x j∂x j
−Ui∂∂x j
& u i & u j .
Now get to work on the remaining terms. The pressure term is fairly easy to get into shape:
€
−1ρ
Ui∂p ∂xi
= −1ρ∂∂xi
p Ui( ) +1ρ
p ∂Ui
∂xi
= −∂∂xi
p Ui
ρ
%
& '
(
) * = −
∂∂x j
p U j
ρ
%
& '
(
) * ,
where the final equality comes from replacing the summed-over index "i" with "j". The body force term is also easily managed:
€
−gUi 1−α T −T0( )[ ]δi3 = −gU3 1−α T −T0( )[ ] = −gU3ρ ρ0
.
The viscous term requires more effort:
€
νUi∂Ui
∂x j∂x j
= νUi∂∂x j
∂Ui
∂x j
$
% & &
'
( ) ) = νUi
∂∂x j
∂Ui
∂x j
+∂U j
∂xi
−∂U j
∂xi
$
% & &
'
( ) ) = 2νUi
∂∂x j
S ij −νUi∂∂x j
∂U j
∂xi
€
= 2νUi∂∂x j
S ij −νUi∂∂xi
∂U j
∂x j
%
& ' '
(
) * * = 2νUi
∂∂x j
S ij = 2ν ∂∂x j
UiS ij( ) − 2νS ij∂Ui
∂x j
= 2ν ∂∂x j
UiS ij( ) − 2νS ij S ij − Ω ij( ) = 2ν ∂∂x j
UiS ij( ) − 2νS ijS ij
where
€
∂U j
∂xi
=12∂U j
∂xi
+∂Ui
∂x j
#
$ % %
&
' ( ( +12∂U j
∂xi
−∂Ui
∂x j
#
$ % %
&
' ( ( ≡ S ij + R ij with
€
S ij and
€
R ij equal to the symmetric
and anti-symmetric parts of the mean velocity gradient tensor, respectively. Because of their symmetry, their tensor product is zero:
€
S ijR ij = 0. The final form for the viscous term occurs because the indices 'i' and 'j' can be exchanged since both are summed over and
€
S ij = S ji. The
Reynolds stress term produces:
€
−Ui∂∂x j
$ u i $ u j = −∂∂x j
Ui $ u i $ u j( ) + $ u i $ u j∂Ui
∂x j
Now reassemble the
€
E equation:
€
∂E ∂t
+ U j∂E ∂x j
= −∂∂x j
p U j
ρ
%
& '
(
) * + 2ν
∂∂x j
UiS ij( ) − 2νS ijS ij −∂∂x j
Ui , u i , u j( ) + , u i , u j∂Ui
∂x j
− gU3ρ ρ0
Rearrange the terms to achieve the final form:
€
∂E ∂t
+ U j∂E ∂x j
=∂∂x j
−p U j
ρ+ 2νUiS ij −Ui & u i & u j
'
( )
*
+ , = + & u i & u j
∂Ui
∂x j
− 2νS ijS ij − gU3ρ ρ0
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
The terms inside the (,)-parenthesis transfer mean kinetic energy from one location in the flow to another. The first term on the right side is the turbulence production term; it is typically negative in this equation. The second term on the right side is the mean-flow kinetic energy dissipation rate. The final term represents kinetic to potential energy conversion.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.16. Derive the RANS transport equation for the Reynolds stress correlation (12.35) via the following steps. a) By subtracting (12.30) from (4.86), show that the instantaneous momentum equation for the fluctuating turbulent velocity ui is:
€
∂ui
∂t+ uk
∂Ui
∂xk
+ Uk∂ui
∂xk
+ uk∂ui
∂xk
= −1ρ0
∂p∂xi
+ ν∂ 2ui
∂xk2 + gα ' T δi3 +
∂∂xk
uiuk .
b) Show that:
€
uiDu j
Dt+ u j
DuiDt
=∂∂t
uiu j( ) +Uk∂∂xk
uiu j( ) +∂∂xk
uiu juk( ).
c) Combine and simplify the results of parts a) and b) to reach (12.35) Solution 12.16. a) Start from (4.86):
€
∂˜ u i∂t
+∂∂x j
˜ u j ˜ u i( ) = −1ρ0
∂˜ p ∂xi
− g 1−α ˜ T −T0( )[ ]δi3 + ν∂ 2 ˜ u i∂x j
2 ,
and insert the Reynolds decompositions (12.24) to find:
€
∂∂t
Ui + ui( ) +∂∂x j
(U j + u j )(Ui + ui)( ) = −1ρ0
∂∂xi
(P + p) − g 1−α T + & T −T0( )[ ]δi3 + ν∂ 2
∂x j2 Ui + ui( ),
and subtract (12.30),
€
∂Ui
∂t+ U j
∂Ui
∂x j
= −1ρ0
∂P∂xi
− g 1−α T −T0( )[ ]δi3 +1ρ0
∂∂x j
∂Ui
∂x j
− ρ0uiu j
'
( ) )
*
+ , , ,
to reach:
€
∂ui
∂t+
∂∂x j
u jUi + U jui + u jui( ) = −1ρ0
∂p∂xi
+ gα & T δi3 + ν∂ 2ui
∂x j2 +
∂∂x j
uiu j .
Now use the fact that ∂Uj/∂xj = 0 and ∂uj/∂xj = 0, and switch the summed-over index from "j" to "k".
€
∂ui
∂t+ uk
∂Ui
∂xk
+ Uk∂ui
∂xk
+ uk∂ui
∂xk
= −1ρ0
∂p∂xi
+ gα & T δi3 + ν∂ 2ui
∂xk2 +
∂∂xk
uiuk .
b) Use "k" as a subscript to denote the dot product within the fluid particle acceleration,
€
DDt
=∂∂t
+ ˜ u k∂∂xk
,
so that:
€
ui
Du j
Dt+ u j
Dui
Dt= ui
∂u j
∂t+ ui ˜ u k
∂u j
∂xk
+ u j∂ui
∂t+ u j ˜ u k
∂ui
∂xk
=∂∂t
uiu j( ) + ˜ u k∂∂xk
uiu j( )
€
=∂∂t
uiu j( ) + Uk + uk( ) ∂∂xk
uiu j( ) =∂∂t
uiu j( ) +Uk∂∂xk
uiu j( ) + uk∂∂xk
uiu j( )
=∂∂t
uiu j( ) +Uk∂∂xk
uiu j( ) +∂∂xk
uiu juk( )
where the final equality follows because ∂uk/∂xk = 0. c) The result of part a) can be written:
€
Dui
Dt+ uk
∂Ui
∂xk
= −1ρ0
∂p∂xi
+ gα & T δi3 + ν∂ 2ui
∂xk2 +
∂∂xk
uiuk .
Use this twice to develop the following two equations:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
u jDui
Dt+ u juk
∂Ui
∂xk
= −u j
ρ0
∂p∂xi
+ u jgα & T δi3 + νu j∂ 2ui
∂xk2 + u j
∂∂xk
uiuk , and
€
ui
Du j
Dt+ uiuk
∂U j
∂xk
= −ui
ρ0
∂p∂x j
+ uigα & T δ j3 + νui
∂ 2u j
∂xk2 + ui
∂∂xk
u juk .
Add these and average using the part b) result to find:
€
∂∂t
uiu j( ) + Uk∂∂xk
uiu j( ) +∂∂xk
uiu juk( ) + u juk∂Ui
∂xk
+ uiuk
∂U j
∂xk
= −u j
ρ0
∂p∂xi
−ui
ρ0
∂p∂x j
+ νu j∂ 2ui
∂xk2 + νui
∂ 2u j
∂xk2 + gα ' T u jδi3 + gα ' T uiδ j 3.
Together the viscous terms allow some simplification:
€
u j∂ 2ui∂xk
2 + ui∂ 2u j
∂xk2 =
∂∂xk
u j∂ui∂xk
#
$ %
&
' ( −
∂ui∂xk
∂u j
∂xk+∂∂xk
ui∂u j
∂xk
#
$ %
&
' ( −
∂u j
∂xk
∂ui∂xk
=∂ 2uiu j
∂xk2 − 2∂ui
∂xk
∂u j
∂xk.
Thus, the final Reynolds-stress transport equation is:
€
∂∂t
uiu j( ) + Uk∂∂xk
uiu j( ) +∂∂xk
uiu juk( ) = −u juk∂Ui
∂xk
− uiuk
∂U j
∂xk
= −1ρ0
u j∂p∂xi
+ ui∂p∂x j
%
& ' '
(
) * * + ν
∂ 2 uiu j
∂xk2 + 2ν ∂ui
∂xk
∂u j
∂xk
+ gα - T u jδi3 + - T uiδ j 3( ),
and this matches (12.35).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.17. In two dimensions, the RANS equations for constant-viscosity constant-density turbulent boundary-layer flow are:
∂U∂x
+∂V∂y
= 0 , U ∂U∂x
+V ∂U∂y
≅ −1ρ∂∂x
P + ρu2( )+ ∂∂y
ν∂U∂y
−uv$
%&
'
() , and 0 ≅ − ∂
∂yP + ρv2( ) ,
where x & y are the streamwise and wall-normal coordinates, U & V are the average streamwise and wall-normal velocity components, u & v are the streamwise and wall normal velocity fluctuations, P is the average pressure, and an overbar denotes a time average. a) Assume that the fluid velocity Ue(x) above the turbulent boundary layer is steady and not turbulent so that the average pressure, Pe, at the upper edge of the boundary layer can be determined from the simple Bernoulli equation: Pe +
12 ρUe
2 = const. Use this assumption, the given Bernoulli equation, and the wall-normal momentum equation to show that:
−1ρ∂P∂x
=UedUe
dx+∂v2
∂x.
b) Use the part a) result, the continuity equation, and the streamwise momentum equation to derive the turbulent-flow von Karman boundary-layer momentum-integral equation:
τ wρ=ddx
Ue2θ( )+Ueδ
* dUe
dx+ddx
v2 −u2( )dy0
∞
∫ ,
where: δ* = 1− UUe
"
#$
%
&'dy
0
∞
∫ , and θ =UUe
1− UUe
"
#$
%
&'dy
0
∞
∫ . In practice, the final term is typically small
enough to ignore, but the efforts here should include it. Solution 12.17. a) Integrate the wall-normal momentum equation in the y-direction to find: P + ρ !v 2 = f (x) = Pe(x) , where f(x) is a function of integration. The equality f(x) = Pe(x) follows when P is evaluated at the edge of the boundary layer where v´ = 0. Differentiate this equation in the x-direction and use the Bernoulli equation for Pe(x):
∂∂x
P + ρ "v 2( ) = ∂Pe∂x = −ρUedUe
dx, or − 1
ρ∂P∂x
=UedUe
dx+∂ #v 2
∂x.
b) Multiply the continuity equation by U, add this to the horizontal momentum equation, and insert the part a) result for ∂P/∂x to reach:
∂U 2
∂x+∂UV∂y
−UedUe
dx−∂∂x
#v 2 − #u 2( ) = ∂∂y
ν∂U∂y
− #u #v$
%&
'
() .
Integrate this equation in the vertical direction from y = 0 to y = ∞. Here, ∫(∂UV/∂y)dy = UV with UV = UeVe as y→∞ and UV = 0 on y = 0. Plus, ∫(∂/∂y)(ν∂U/∂y – !u !v )dy = ν∂U/∂y – !u !v with ∂U/∂y = !u !v = 0 as y→∞ , and ν∂U/∂y = τw/ρ & !u !v = 0 on y = 0. Therefore, the horizontal
momentum equation becomes: ∂U 2
∂x−Ue
dUe
dx−∂∂x
#v 2 − #u 2( )$
%&
'
()
0
∞
∫ dy+UeVe = −τ wρ
. (†)
The continuity equation for the average flow implies: Ve = − ∂U ∂x( )0
∞
∫ dy , so
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
UeVe = −Ue∂U∂x0
∞
∫ dy = − ddx
Ue U0
∞
∫ dy%
&'
(
)*+
dUe
dxU
0
∞
∫ dy .
Combine this with (†), rearrange the integrals on the left side, and pull ∂/∂x outside the integrals (where it changes to d/dx):
ddx
U 2 −UUe − "v 2 + "u 2( )0
∞
∫ +dUe
dxU −Ue( )
0
∞
∫ dy = −τ wρ
.
Change the sign of all the terms and form δ* = 1− UUe
"
#$
%
&'dy
0
∞
∫ & θ =UUe
1− UUe
"
#$
%
&'dy
0
∞
∫ from the
remaining integrals to reach: τ wρ=ddx
Ue2θ( )+ d
dx!v 2 − !u 2( )dy
0
∞
∫ +Ueδ* dUe
dx.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.18. Staring from (12.38) and (12.40), set r = re1 and use
€
R11 = u2 f (r) , and
€
R22 = u2g(r) , to show that
€
F(r) = u2 f (r) − g(r)( )r−2 and
€
G(r) = u2g(r) . Solution 12.18. For homogeneous isotropic turbulence, the spatial correlation function is given by (12.40):
€
Rij = F(r)rirj +G(r)δij . When r = (r1, r2, r3) = (r, 0, 0), the longitudinal correlation given by (12.38) is:
€
u2 f (r) = R11 = F(r)r1r1 +G(r) = F(r)r2 +G(r). where
€
u2 is the velocity variance (it is independent of direction), and r1 = r in this case. When r = (r1, r2, r3) = (r, 0, 0), the lateral correlation given by (12.38) is:
€
u2g(r) = R22 = F(r)r2r2 +G(r) = 0 +G(r) . because r2 = r in this case. This equation implies:
€
G(r) = u2g(r) , so the longitudinal correlation result becomes:
€
u2 f (r) = F(r)r2 + u2g(r). Solve this for F(r) to find:
€
F(r) =u2
r2f (r) − g(r)( ) .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.19. a) Starting from Rij from (12.39), compute
€
∂Rij ∂rj for incompressible flow. b) For homogeneous-isotropic turbulence use the result of part a) to show that the longitudinal,
€
f (r), and transverse,
€
g(r), correlation functions are related by
€
g(r) = f (r) + r 2( ) df (r) dr( ). c) Use part b), and the integral length scale and Taylor microscale definitions to find
€
2Λg = Λ f and
€
2λg = λ f .
Solution 12.19. a)
€
∂∂rj
Rij = ui(x)∂∂rj
u j (x + r) = 0 because the fluctuating velocity field is
incompressible.
b)
€
∂∂rj
Rij = 0 = u2 ∂∂rj
f (r) − g(r)( )rirjr2
+ g(r)δij% & '
( ) *
. At this point with
€
r = r12 + r2
2 + r32 , so the
differentiations & summations can get completely out of hand unless they are completed in a systematic manner. The following relationships are needed:
€
∂∂rj
f (r) − g(r)( ) = $ f − $ g ( )rj
r,
€
∂g(r)∂rj
= # g rj
r,
€
∂∂rj
1r2#
$ %
&
' ( = −
2r3rjr
= −2rjr4
, and
€
∂ri∂rj
= δij ,
where a prime denotes differentiation of the function with respect to its argument. First rewrite the main equation in terms of a dot product so that the operations and the various terms are more obvious:
€
0 =∂∂r1
∂∂r2
∂∂r3
# $ %
& ' (
f (r) − g(r)( ) r12
r2+ g(r) f (r) − g(r)( ) r1r2
r2f (r) − g(r)( ) r1r3
r2
f (r) − g(r)( ) r2r1r2
f (r) − g(r)( ) r22
r2+ g(r) f (r) − g(r)( ) r2r3
r2
f (r) − g(r)( ) r3r1r2
f (r) − g(r)( ) r3r2r2
f (r) − g(r)( ) r32
r2+ g(r)
#
$
* * *
%
* * *
&
'
* * *
(
* * *
.
The outcome will be a column vector, but only the first entry of this vector is needed to determine an equation relating f and g because all three directions are equivalent in homogenous isotropic turbulence. Using the equations above, the first entry of the column vector will be:
€
0 = " f − " g ( ) r1r
r12
r2 + " f − " g ( ) r2
rr2r1r2 + " f − " g ( ) r3
rr3r1r2 − 2( f − g)r1
2 r1r4 − 2( f − g)r1r2
r2
r4
− 2( f − g)r1r3r3
r4 + ( f − g) 2r1r2 + ( f − g) r1
r2 + ( f − g) r1r2 + " g r1
r
Combine like terms:
€
0 = " f − " g ( ) r1r− 2( f − g) r1
r2+ 4( f − g) r1
r2+ " g r1
r= " f r1
r+ 2( f − g) r1
r2
Continue combining, and divide out the common factor of r1/r to reach a single equation involving f and g:
€
0 = " f +2r( f − g) , or
€
g = f +r2
" f .
Therefore,
€
f − g = −r2dfdr
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
Rij (r) = " u 2 −r2
df (r)dr
rirj
r2+ f (r) +
r2
df (r)dr
$
% &
'
( ) δij
+ , -
. / 0
= " u 2 f (r)δij +r2
df (r)dr
δij −rirj
r2$
% &
'
( )
+ , -
. / 0
.
c)
€
Λg = g(r)dr0
∞
∫ = f +r2
% f &
' (
)
* + dr
0
∞
∫ = f +12
ddr(rf ) − f
2&
' (
)
* + dr
0
∞
∫ =12
rf[ ]0∞
+12
f (r)dr0
∞
∫ = 2Λ f
For the Taylor length scale, compute two derivatives of g:
€
" g = " f +12
" f +r2
" " f , so
€
" " g =32
" " f +12
" " f +r2
" " " f = 2 " " f +r2
" " " f . Now evaluate at r = 0 to find
€
" " g (0) = 2 " " f (0) because f
and g are even functions
€
" " " f (0) = 0 . Finally, use the definition of the Taylor scales and the current results,
€
λ f2 = −2 d2 f
dr2$
% &
'
( )
r= 0
=2* * f (0)
and
€
λg2 = −2 d2g
dr2$
% &
'
( )
r= 0
=2* * g (0)
=2
2 * * f (0)=12λ f2 , to find:
€
λg =12λ f .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.20. In homogeneous turbulence:
€
Rij (rb − ra ) = ui(x + ra )u j (x + rb ) = Rij (r) , where
€
r = rb − ra . a) Show that
€
∂ui(x) ∂xk( ) ∂u j (x) ∂xl( ) = − ∂ 2Rij ∂rk∂rl( )r= 0 . b) If the flow is incompressible and isotropic, show that
€
− ∂u1(x) ∂x1( )2 = − 12 ∂u1(x) ∂x2( )2 = +2 ∂u1(x) ∂x2( ) ∂u2(x) ∂x1( ) = u2 d2 f dr2( )r= 0
[Hint: expand f(r) about r = 0 before taking any derivatives.] Solution 12.20. a) Start with
€
Rij (rb − ra ) = ui(x + ra )u j (x + rb ) and take the divergence with respect to the r-variables.
€
∂∂rb,k
Rij (rb − ra ) =∂∂rk
Rij (r) = ui(x + ra )∂∂rb,k
u j (x + rb ) = ui(x + ra )∂∂xk
u j (x + rb )
€
∂ 2
∂ra,l∂rb,kRij (rb − ra ) = −
∂ 2
∂rl∂rkRij (r) =
∂∂ra,l
ui(x + ra )∂∂rb,k
u j (x + rb ) =∂∂xl
ui(x + ra )∂∂xk
u j (x + rb )
Now take the limit as the r-variables go to zero:
€
−∂ 2
∂rl∂rkRij (0) =
∂∂xl
ui(x)∂∂xk
u j (x) =∂ui∂xl
∂u j
∂xk.
b) First choose i = j = k = l = 1, and evaluate the formula from part a) using the results of Exercise 12.18 part b):
€
∂u1∂x1
#
$ %
&
' (
2
= −∂ 2
∂ 2r1R11(0) = −u2 ∂
2
∂ 2r1f (r) +
r2df (r)dr
1− r12
r2#
$ %
&
' (
* + ,
- . /
.
Use the hint and expand f(r) about r = 0:
€
f (r) =1+r2
2d2 f (0)
dr2+ ...=1+
r2
2" " f (0) + ...,
€
∂u1∂x1
#
$ %
&
' (
2
= −u2 ∂2
∂ 2r11+
r2
2* * f (0) +
r2
2* * f (0) 1− r1
2
r2#
$ %
&
' ( + ...
+ , -
. / 0
.
Simplify before starting the differentiation.
€
∂u1∂x1
#
$ %
&
' (
2
= −u2 ∂2
∂ 2r11+ r2 * * f (0) − r1
2
2* * f (0) + ...
+ , -
. / 0
= − * u 2 ∂∂r1
2r * * f (0) r1r− r1 * * f (0) + ...
+ , -
. / 0
€
∂u1∂x1
#
$ %
&
' (
2
= −u2 ∂∂r1
2r1 * * f (0) − r1 * * f (0) + ...{ } = −u2 * * f (0) .
Second choose i = j = 1 and k = l = 2, and use the same expansion
€
∂u1∂x2
#
$ %
&
' (
2
= −u2 ∂2
∂ 2r21+ r2 * * f (0) − r1
2
2* * f (0) + ...
+ , -
. / 0
= −u2 ∂∂r2
2r * * f (0) r2r
+ ...+ , -
. / 0
= −2u2 * * f (0)
Third choose, i = k = 1 and j = l = 2, but this time the expansion is different because δij = 0.
€
∂u1∂x2
#
$ %
&
' ( ∂u2∂x1
#
$ %
&
' ( = −
∂ 2
∂r1∂r2R12(0) = −u2 ∂ 2
∂r1∂r2
r2
df (r)dr
−r1r2r2
#
$ %
&
' (
* + ,
- . /
= u2 ∂ 2
∂r1∂r2
r1r22
0 0 f (0) + ...* + ,
- . /
=u2
20 0 f (0)
Putting these three cases together produces the desired result:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
− ∂u1(x) ∂x1( )2 = − 12 ∂u1(x) ∂x2( )2 = +2 ∂u1(x) ∂x2( ) ∂u2(x) ∂x1( ) = u2 d2 f dr2( )r= 0 .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.21. The turbulent kinetic energy equation contains a pressure-velocity correlation,
€
K j = p(x)u j (x + r). In homogeneous isotropic turbulent flow, the most general form of this correlation is:
€
K j = K(r)rj . If the flow is also incompressible, show that K(r) must be zero. Solution 12.21. Compute the divergence of Kj:
€
∂∂rj
K j = p(x) ∂∂rj
u j (x + r) = 0 .
because the flow is incompressible. Now insert the homogeneous-isotropic form of Kj.
€
0 =∂∂rj
K(r)rj( ) =dKdr
rjrrj + K
∂rj∂rj
=dKdr
r + 3K .
Use the first and last parts of this equality to find:
€
1KdKdr
= −3r
.
Integrate this equation:
€
ln(K) = −3ln(r) + const , and then exponentiate to reach:
€
K =constr3
.
However, const must be zero because K must remain finite as r → 0. Thus, K(r) = 0 in incompressible homogeneous-isotropic turbulence.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.22. The velocity potential for two-dimensional water waves of small amplitude ξo on a deep pool can be written:
€
φ(x1,x2,t) =ωξoke+kx2 cos ωt − kx1( )
where x1 and x2 are the horizontal and vertical coordinates with x2 = 0 defining the average free surface. Here, ω is the temporal radian frequency of the waves and k is the wave number. a) Compute the two-dimensional velocity field:
€
u = (u1,u2) = ∂φ ∂x1 ,∂φ ∂x2( ). b) Show that this velocity field is a solution of the two-dimensional continuity and Navier-Stokes equations for incompressible fluid flow. c) Compute the strain rate tensor
€
Sij = 12 ∂ui ∂x j + ∂u j ∂xi( ) .
d) Although this flow is not turbulent, it must still satisfy the turbulent kinetic energy equation that contains an energy dissipation term. Denote the kinematic viscosity by ν, and compute the kinetic energy dissipation rate in this flow: ε = 2νSijSij where the over bar implies a time average over one wave period = 2π/ω. Only time averages of even powers of the trig-functions are non-zero, for example:
€
cos2 ωt − kx( ) = sin2 ωt − kx( ) =1 2 while
€
cos ωt − kx( ) = sin ωt − kx( ) = 0 . e) The original potential represents a lossless flow and does not include any viscous effects. Explain how this situation can occur when the kinetic-energy dissipation rate is not zero.
Solution 12.22. a) Using the prescribed potential
€
φ(x1,x2,t) =ωξoke+kx2 cos ωt − kx1( ) , the velocity
field is obtained by differentiation:
€
ui =∂φ∂x1, ∂φ∂x2
$
% &
'
( ) =ωξoe
+kx2 sin ωt − kx1( ),cos ωt − kx1( )( ) .
b) For the continuity equation:
€
∂u1∂x1
+∂u2∂x2
=ωξoe+kx2 cos ωt − kx1( ) ⋅ −k + k[ ] = 0
The NS-momentum equation for incompressible flow,
€
∂ui∂t
+ u j∂ui∂x j
= −1ρ∂p∂xi
− gδi2 + ν∂ 2ui∂x j∂x j
,
can be rewritten in mixed notation:
€
∂∂xi
∂φ∂t
+pρ
+12∇φ
2+ gx2
&
' (
)
* + = −ω ×u+ ν∇ ×ω using vector
identities where
€
ω =∇ ×u is the vorticity. For the above potential flow,
€
ω =∇ ×u = 0, so the modified left side of the NS-momentum equation is all that’s left. Thus,
€
∂φ∂t
+pρ
+12∇φ
2+ gx2 = fn(t)
but the temporal “function” of integration can be combined with φ or set to zero without loss of generality. When this is done, all that’s left is a Bernoulli equation that determines p from φ. Therefore, the potential provided in this problem does produce a solution of the full NS equations provided that fluctuating stresses occur at the water surface. For example, the fluctuating surface pressure can be obtained from the above Bernoulli equation evaluated on the water surface:
€
0 =∂φ∂t
+pρ
+12∇φ
2&
' (
)
* + x2= 0
+ gξ = −ω 2ξoksin ωt − kx1( ) +
psρ
+ω 2ξo
2
2+ gξo sin ωt − kx1( )
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
where ps is the surface pressure. The usual dispersion relationship for deep water waves,
€
ω 2 = gk , allows the first and final terms to cancel, leaving
€
psρ
= −ω 2ξo
2
2. Thus, the normal stress
on the surface will be:
€
σ 22 = −ps + 2µ∂u2∂x2
= −ps + 2µωkξo cos(ωt − kx1) , and the shear stress on
the surface will be:
€
σ12 = µ∂u1∂x2
+∂u2∂x1
$
% &
'
( ) = 2µωkξo sin(ωt − kx) . The work rate per unit span of the
flow will be:
€
σ12u1 +σ 22u2 = 2µω 2kξo2 − psωξo cos(ωt − kx). Thus the average rate of surface-
stress work input will be:
€
σ12u1 +σ 22u2 = 2µω 2kξo2.
c) Use the results of part a) and compute all four derivatives:
€
∂u1∂x1
= −kωξoe+kx2 cos ωt − kx1( )
€
∂u1∂x2
= kωξoe+kx2 sin ωt − kx1( )
€
∂u2∂x1
= kωξoe+kx2 sin ωt − kx1( )
€
∂u2∂x2
= kωξoe+kx2 cos ωt − kx1( )
Now assemble the 2 × 2 matrix for
€
Sij = 12 ∂ui ∂x j + ∂u j ∂xi( ) :
€
Sij =ωξoe+kx2
−k cos ωt − kx1( ) k sin ωt − kx1( )k sin ωt − kx1( ) k cos ωt − kx1( )
% & '
( ) *
= −k 2φ1 −tan ωt − kx1( )
−tan ωt − kx1( ) −1
% & '
( ) *
where φ is the velocity potential defined in the problem statement. d)
€
2νSijSij = 2νk 4φ 2 1+ tan2 ωt − kx1( ) + tan2 ωt − kx1( ) +1( )
€
= 4νk 4φ 2 1+ tan2 ωt − kx1( )( ) = 4νk 2ω 2ξo2e+2kx2 = 2νSijSij = ε ≠ 0.
e) The turbulent kinetic energy equation contains an energy transfer term on the other side of the equation that includes the viscosity:
€
∂∂x j
2νuiSij( ) = 2ν ∂∂x j
−k 2φωξoe+kx2 sin ωt − kx1( ) cos ωt − kx1( ){ }
1 −tan ωt − kx1( )−tan ωt − kx1( ) −1
( ) *
+ , -
.
/ 0 0
1
2 3 3
€
= −2ν ∂∂x j
k 3φ 20
−tan2 ωt − kx1( ) −1' ( )
* + ,
-
. /
0
1 2 = 2ν
∂∂x1
∂∂x2
' ( )
* + ,
0kω 2ξo
2e+2kx2
' ( )
* + ,
€
= 2ν 2k 2ω 2ξo2e+2kx2( ) = 4νk 2ω 2ξo
2e+2kx2 = ε ! Thus, the kinetic energy that is dissipated is transferred from somewhere else. The most likely source is the work input from the surface stresses (see part b). The total rate of energy dissipation per unit span can be obtained by integrating ε through the depth:
€
ρεdx2 = 4µk 2ω 2ξo2 e+2kx2
−∞
0
∫−∞
0
∫ =4µk 2ω 2ξo
2
2k= 2µkω 2ξo
2 .
This is precisely the rate at which work (per unit depth) is done by the surface forces. Thus, the energy flow for ideal water waves can be understood using the turbulent kinetic energy equation.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.23. A mass of 10 kg of water is stirred by a mixer. After one hour of stirring, the temperature of the water rises by 1.0 °C. What is the power output of the mixer in watts? What is the size η of the dissipating eddies? Solution 12.23. Start from a thermodynamic description of the power delivered to the water.
Power =
€
mCpΔTΔt
=(10kg)(4200m2s−2K−1)(1K)
3600s=11.67W
The kinetic energy dissipation rate per unit mass is:
€
Powermass
= ε =11.67W10kg
=1.167m2s−3 .
Thus, the Kolmogorov scale is:
€
η = ν 3 ε( )1 4
= (1.0 ×10−6m2s−1)3 1.167m2s−3( )1 4
= 3.04 ×10−5m. However, the actual dissipation scale is larger by about a factor of 30, so the dissipation length scale is approximately one millimeter.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.24. In locally isotropic turbulence, A.N. Kolmogorov determined that the wave number spectrum can be represented by
€
S11(k) ν 5ε ( )1 4
=Φ kν 3 4 ε 1 4( ) in the inertial-subrange and dissipation-range of turbulent scales, where Φ is an undetermined function. a) Determine the equivalent form for the temporal spectrum
€
Se (ω) in term of the average kinetic energy dissipation rate
€
ε , the fluid’s kinematic viscosity ν, and the temporal frequency ω. b) Simplify the results of part a) for the inertial range of scales where ν is dropped from the dimensional analysis. c) To obtain the results for parts a) and b) an implicit assumption has been made that leads to the neglect of an important parameter. Add the missing parameter and redo the dimensional analysis of part a). d) Use the missing parameter and ω to develop an equivalent wave number. Insist that your result for Se only depend on this equivalent wave number and
€
ε to recover the minus-five-thirds law. Solution 12.24. a) First layout the units of the various quantities using square brackets to denote “units of”.
€
Se (ω)[ ] =length2
time2⋅
1frequency
=length2
time,
€
ε [ ] =length2
time3,
€
ν[ ] =length2
time, and
€
ω[ ] =1time
.
Four parameters and two independent units means there should be two dimensionless groups. By inspection these are:
€
Π1 =Se (ω)ν
and
€
Π2 =ωνε
, so
€
Se (ω)ν
= fn ω νε
%
& '
(
) *
b) If ν must drop out of the dimensional analysis then
€
fn( ) = const ⋅ ( )−2 , so:
€
Se (ω)ν
= const ω νε
%
& '
(
) *
−2
, or
€
Se (ω) = const ε ω 2
$
% &
'
( )
c) The implicit assumption is that the flow has zero average velocity, or that the observer moves at the average velocity. Thus, for a stationary observer, the average flow speed U must also be a parameter. When this is added to the parameters used for part a), there will be another dimensionless group,
€
Π3 =U νω , leading to:
€
Se (ω)ν
= fn ω νε , Uων
%
& '
(
) * (&)
d) Wave number has units of (length)–1, so ω/U has the right units to be a surrogate wave number. If Se can only depend on ω through ω/U, then its normalization changes:
€
Se (ω)−∞
+∞
∫ dω = & u 2 = Se ω U( )−∞
+∞
∫ d ω U( ) .
This normalization relationship, requires
€
Se (ω) =Se ω U( )
U, so (&) becomes:
€
Se (ω U)νU
= fn ω νε , Uων
%
& '
(
) *
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
To ensure that ω and U always occur as a ratio, the two dimensionless arguments in (&) can be
used to create a dimensionless surrogate wave number:
€
ωνε
%
& '
(
) *
1 2Uων
%
& '
(
) * −1
=ωUν 3 4
ε 1 4. In addition,
the divisor of
€
Se (ω U) must not contain U or ω alone. Again using the dimensionless arguments in (&) it can be converted to a dimensionally-equivalent form:
€
1νU
⇒1νU
Uων
%
& '
(
) * ω
νε
%
& '
(
) *
1 2
=1
ν 3 2ω1 2ω1 2ν1 4
ε 1 4=
1ν 5 4ε 1 4
Thus, the dimensionless law becomes:
€
Se ω U( )ν 5 4ε 1 4
= fn ωUν 3 4
ε 1 4%
& '
(
) * .
When ν is eliminated from this expression, Kolmogorov’s –5/3 power law is recovered:
€
Se ω U( )ν 5 4ε 1 4
= const ωUν 3 4
ε 1 4%
& '
(
) *
−5 3
→
€
Se ω U( ) = const ⋅ ε 2 3 ⋅ ωU%
& '
(
) * −5 3
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.25. Estimates for the importance of anisotropy in a turbulent flow can be developed by assuming that fluid velocities and spatial derivatives of the average-flow (or RANS) equation are scaled by the average velocity difference ΔU that drives the largest eddies in the flow having a size L, and that the fluctuating velocities and spatial derivatives in the turbulent kinetic energy (TKE) equation are scaled by the kinematic viscosity ν and the Kolmogorov scales η and uK [see (12.50)]. Thus, the scaling for a mean velocity gradient is:
€
∂Ui ∂x j ~ ΔU L , while the mean-
square turbulent velocity gradient scales as:
€
∂ui ∂x j( )2~ uK η( )2 = ν 2 η4 , where the “~” sign
means “scales as”. Use these scaling ideas in parts a) and d) below. a) The total energy dissipation rate in a turbulent flow is
€
2νS ijS ij + 2ν # S ij # S ij , where
€
S ij =12∂Ui
∂x j
+∂U j
∂xi
#
$ % %
&
' ( ( and
€
" S ij =12∂ui
∂x j
+∂u j
∂xi
$
% & &
'
( ) ) . Determine how the ratio
€
" S ij " S ijS ijS ij
depends on the
outer-scale Reynolds number:
€
ReL = ΔU ⋅ L ν . b) Is average-flow or fluctuating-flow energy dissipation more important? c) Show that the turbulent kinetic energy dissipation rate,
€
ε = 2ν $ S ij $ S ij can be written:
€
ε = ν∂ui∂x j
∂ui∂x j
+∂ 2
∂xi∂x j
uiu j
%
& '
(
) * .
d) For homogeneous isotropic turbulence, the second term in the result of part c) is zero but it is
non-zero in a turbulent shear flow. Therefore, estimate how
€
∂ 2
∂xi∂x j
uiu j∂ui∂x j
∂ui∂x j
depends on
ReL in turbulent shear flow as means of assessing how much impact anisotropy has on the turbulent kinetic energy dissipation rate. e) Is an isotropic model for the turbulent dissipation appropriate at high ReL in a turbulent shear flow? Solution 12.25. a) Use the scaling ideas in the problem statement and the relationship
€
η∝L ⋅ Re−3 4 to find:
€
" S ij " S ijS ijS ij
~ν η2( )
2
ΔU L( )2=ν 2L2
(ΔU)2L4
L41η4
=ν 2
(ΔU)2L2L4
η4= ReL
−2 ReL3 4( )
4= ReL
b) Thus since ReL >> 1 in turbulent flow, the fluctuating-flow energy dissipation is more important. c) Start with the definition of the dissipation rate and the strain rate tensor:
€
ε = 2ν $ S ij $ S ij =ν2∂ui
∂x j
+∂u j
∂xi
&
' ( (
)
* + +
2
=ν2
∂ui
∂x j
&
' ( (
)
* + +
2
+ 2 ∂ui
∂x j
∂u j
∂xi
+∂u j
∂xi
&
' (
)
* +
2&
'
( (
)
*
+ +
The first and last terms under the final overbar are the same, so
€
ε = ν∂ui∂x j
%
& ' '
(
) * *
2
+ ν∂ui∂x j
∂u j
∂xi.
The second term can be manipulated using the requirement of fluctuating flow incompressibility:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
ν∂ui
∂x j
∂u j
∂xi
= ν∂∂x j
ui
∂u j
∂xi
$
% &
'
( ) − + u i
∂ 2u j
∂x j∂xi
= ν∂∂x j
ui
∂u j
∂xi
$
% &
'
( )
€
= ν∂ 2
∂x j∂xiuiu j( ) − ∂
∂x j
u j∂ui∂xi
%
& '
(
) * =
∂ 2
∂x j∂xiuiu j( ) =
∂ 2
∂x j∂xiuiu j .
Thus,
€
ε = ν∂ui∂x j
%
& ' '
(
) * *
2
+∂ 2
∂xi∂x j
uiu j
+
,
- -
.
/
0 0 .
d) Use the simple scaling ideas for the ratio to find:
€
∂ 2
∂xi∂x j
uiu j
∂ui ∂x j( )2 ~
ΔU L( )2
ν η2( )2=ΔULν
&
' (
)
* + 2η4
L4= ReL
2 ReL−3 / 4( )
4= ReL
−1 .
e) This implies that anisotropy will not be important at high Reynolds number. Therefore, an isotropic dissipation model should be suitable for engineering purposes.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.26. Determine the self-preserving form of the average stream-wise velocity Uz(z,R) of a round turbulent jet using cylindrical coordinates where z increases along the jet axis and R is the radial coordinate. Ignore gravity in your work. Denote the density of the nominally-quiescent reservoir fluid by ρ. a) Place a stationary cylindrical control volume around the jet's cone of turbulence so that circular control surfaces slice all the way through the jet flow at its origin and at a distance z downstream where the fluid density is ρ. Assuming that the fluid outside the jet is nearly stationary so that pressure does not vary in the axial direction and that the fluid entrained into the volume has negligible x-direction momentum, show
J0 ≡ ρ0U02
0
d /2∫ 2πRdR = ρUz
2 (z,R)2π0
D/2∫ RdR ,
where J0 is the jet's momentum flux, ρ0 is the density of the jet fluid, and U0 is the jet exit velocity. b) Simplify the exact mean-flow equations ∂Uz
∂z+1R∂∂R
RUR( ) = 0 , and Uz∂Uz
∂z+UR
∂Uz
∂R= −
1ρ∂P∂z
+νR∂∂R
R∂Uz
∂R"
#$
%
&'−1R∂∂R
RuzuR( )− ∂∂z
Ruz2( ) ,
when ∂P/∂z ≈ 0, the jet is slender enough for the boundary layer approximation ∂/∂R >> ∂/∂z to be valid, and the flow is at high Reynolds number so that the viscous terms are negligible. c) Eliminate the average radial velocity from the simplified equations to find:
Uz∂Uz
∂z−1R
R ∂Uz
∂zdR
0
R
∫#$%
&'(
∂Uz
∂R= −
1R∂∂R
RuzuR( )
where R is just an integration variable. d) Assume a similarity form: Uz (z,R) =UCL (z) f (ξ ) , −uzuR =Ψ(z)g(ξ ) , where ξ = R δ(z) and f and g are undetermined functions, use the results of parts a) and c), and choose constant values appropriately to find Uz (z,R) = const. J0 ρ( )1 2 z−1 f R z( ) . e) Determine a formula for the volume flux in the jet. Will the jet fluid from the nozzle be diluted with increasing z?
Solution 12.26. a) Use the stationary CV shown above, consider only the steady mean flow, and ignore turbulent fluctuations. In this case the CV momentum equation is:
ρUz (U ⋅n)Surface∫ dA = − Pn ⋅ ez dA
Surface∫ ,
since there are no shear stresses on any of the CV boundaries. When the fluid entrained into the volume has negligible z-direction momentum, only the inlet (z = 0) and outlet (z) surfaces
z!
R!
Uz(z,R)!
UCL(z)!
d!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
provide a contribution, and if the pressure does not vary in the z-direction, then the above equation reduces to:
− ρUz2"# $%0 2πRdR+
0
d 2
∫ ρUz2"# $%z 2πRdR
0
D 2
∫ ≅ 0 .
where D is a radial location that is comfortably outside the jet's cone of turbulence. [Here it must be noted that this equation is approximate. The jet entrains reservoir fluid and induces the reservoir fluid to move. Thus, the assumption of negligible pressure gradient is not precisely correct.] Noting that the first integral on the left is the jet's momentum flux and that ρUz
2!" #$0 = ρ0U02 , the above equation simplifies to:
ρ0U02 2πRdR ≡ J0 ≅
0
d 2
∫ ρUz2 2πRdR
0
D 2
∫ ,
where J0 is the jet's momentum flux. b) The mean-flow continuity equation must be retained as is. However, the mean-flow z-direction momentum equation can be simplified.
Uz∂Uz
∂z+UR
∂Uz
∂R= −
1ρ∂P∂z
+νR∂∂R
R∂Uz
∂R"
#$
%
&'−1R∂∂R
RuzuR( )− ∂∂z
Ruz2( ) .
When ∂P/∂z ≈ 0, then
Uz∂Uz
∂z+UR
∂Uz
∂R≅ +
νR∂∂r
R∂Uz
∂R"
#$
%
&'−1R∂∂R
RuzuR( )− ∂∂z
Ruz2( ) .
If the jet is slender enough for the boundary layer approximation ∂/∂R >> ∂/∂z to be valid, then
Uz∂Uz
∂z+UR
∂Uz
∂R≅ +
νR∂∂R
R∂Uz
∂R"
#$
%
&'−1R∂∂R
RuzuR( ) .
For high Reynolds number flow, the viscous terms are negligible, so
Uz∂Uz
∂z+UR
∂Uz
∂R≅ −
1R∂∂R
RuzuR( ) .
c) Use the continuity equation to eliminate UR from the problem: ∂Uz
∂z+1R∂(RUR )∂R
= 0 → UR = −1R
R ∂Uz
∂zdR
0
R
∫
so that the boundary layer RANS equation becomes:
Uz∂Uz
∂z−1R
R ∂Uz
∂zdR
0
R
∫#$%
&'(
∂Uz
∂R= −
1R∂∂R
RuzuR( ) (3*)
where R is an integration variable. d) Use the similarity solution: Uz (z,R) =UCL (z) f (ξ ) , −uzuR =Ψ(z)g(ξ ) , and ξ = R δ(z) , and evaluate each term in (3*), converting R to ξ wherever possible
Uz∂Uz
∂z=UCL f
∂∂z
UCL f( ) =UCL f !UCL f +UCL !f ⋅ −R !δδ 2
$
%&
'
()=UCL !UCL f
2 −UCL2 f !f ξ !δ
δ;
1R
R ∂Uz
∂zdR
0
R
∫"#$
%&'
∂UR
∂R=
δ 2
rRδ∂Uz
∂zd R
δ
(
)*
+
,-
0
R
∫"#$
%&'UCL .f 1
δ=1ξ
ξ∂Uz
∂zd ξ
0
ξ
∫"#$
%&'UCL .f
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
=" U CL
ξfξ dξ
0
ξ
∫ −UCL
ξ" δ δ
" f ξ 2dξ 0
ξ
∫' ( )
* + , UCL " f =
UCL " U CL " f η
fξ dξ 0
ξ
∫ −UCL2 " f η
" δ δ
" f ξ 2dξ 0
ξ
∫
€
=UCL " U CL " f
ξfξ dξ
0
ξ
∫ −UCL2 " f η
" δ δ
fξ 2( )0ξ− 2 fξ dξ
0
ξ
∫(
) *
+
, -
€
=UCL " U CL " f
ξ+ 2UCL
2 " f ξ
" δ δ
%
& '
(
) * fξ dξ 0
ξ
∫ −UCL2 ξf " f " δ
δ; and
1R∂∂r
RuzuR( ) = 1δ R δ( )
∂∂ R δ( )
(R δ)uzuR( ) = 1δξ
∂∂ξ
ξuzuR( ) = − RSCLδξ
∂∂ξ
ξg( ) ,
Now reassemble (3*) and cancel terms.
€
UCL " U CL f 2 −UCL2 f " f ξ " δ
δ−
UCL " U CL " f ξ
+ 2UCL2 " f ξ
" δ δ
&
' (
)
* + fξ dξ 0
ξ
∫ + UCL2 ξf " f " δ
δ=
RSCL
δξ∂∂ξ
ξg( )
€
UCL " U CL f 2 − UCL " U CL " f ξ
+ 2UCL2 " f ξ
" δ δ
&
' (
)
* + fξ dξ 0
ξ
∫ =RSCL
δξ∂∂ξ
ξg( )
Multiply by δ and divide by
€
UCL2 to find:
€
δ # U CL
UCL
$
% &
'
( ) f 2 −
δ # U CL
UCL
+ 2 # δ $
% &
'
( )
# f ξ
fξ dξ 0
ξ
∫ =RSCL
UCL2
$
% &
'
( ) 1ξ∂∂ξ
ξg( ) .
The three functions outside the [,]-brackets only depend on the similarity variable ξ. Thus, the flow will show (the simplest) self-similarity when:
€
δ # U CL
UCL
= C1,
€
δ # U CL
UCL
+ 2 # δ = C2, and
€
RSCLUCL2 = C3, (a,b,c)
where C1, C2, and C3 are constants. Equations (a) and (b) require:
€
2 " δ = C2 −C1, or δ = C2 −C12
"
#$
%
&'(z− zo ) .
It is easiest to choose C2 – C1 = 2, and to assume zo = 0, so that δ = z. Thus (a) implies: !UCL
UCL
=C1z
, or lnUCL =C1 ln z+C4 which is the same as UCL =C5zC1 .
The constraint found in part a) requires:
J0 ≡ 2πρ Uz2 (z,R)
0
D/2∫ RdR = 2πρδ 2 UCL
2 f 2 (ξ )0
∞
∫ Rδd Rδ
$
%&
'
()= 2πρδ 2UCL
2 f 2 (ξ )0
∞
∫ ξdξ .
Substituting in the required similarity forms for δ and UCL leads to:
€
J0 ≡ 2πρx2C5
2x 2C1 f 2(ξ)0
∞
∫ ξdξ .
Here the integral is just a number and
€
2πρC52 is just a product of constants, therefore 2+2C1, the
exponent of x, must be zero, and this implies C1 = –1 and
€
C5 = const. J0 ρ , where "const." is a dimensionless constant. When all this is drawn together, the final form for the self-similar round jet velocity field is:
Uz (R, z) = const. J0 ρ( )1 2 z−1 f R z( ) . e) The volume flux Q in the jet will be:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Q = 2π UzRdR0
D 2
∫ = 2πUCLδ2 fξ dξ0
∞
∫ = 2πconst. J0ρ
#
$%
&
'(
1 2
z−1z2 fξ dξ0
∞
∫ ∝J0ρ
#
$%
&
'(
1 2
z .
Thus, the volume flux in the jet increases linearly with downstream distance, so the jet fluid will be diluted with increasing x.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.27. Consider the turbulent wake far from a two-dimensional body placed perpendicular to the direction of a uniform flow. Using the notation defined in the Figure, the result of Example 12.5 may be written:
<Begin Equation> FD lρUo
2 =θ =U(x, y)Uo
1−U(x, y)Uo
"
#$
%
&'+
v2 −u2
Uo2
(
)**
+
,--dy
−∞
+∞
∫ ,
</End Equation> where θ is the momentum thickness of the wake flow (a constant), and U(x,y) is the average horizontal velocity profile a distance x downstream of the body. a) When ΔU << Uo, find the conditions necessary for a self-similar form for the wake’s velocity deficit, U(x, y) =Uo −ΔU(x) f (ξ ) , to be valid based on the equation above and the steady two-dimensional continuity and boundary-layer RANS equations. Here, ξ = y/δ(x) and δ is the transverse length scale of the wake. b) Determine how ΔU and δ must depend on x in the self-similar region. State your results in appropriate dimensionless form using θ and Uo as appropriate.
Solution 12.27. a) The steady boundary layer RANS equations are:
€
∂U∂x
+∂V∂y
= 0, and
€
U ∂U∂x
+V ∂U∂y
= −∂uv∂y
. First combine the momentum & continuity equations to eliminate V:
€
U ∂U∂x
−∂U∂x
dy 0
y
∫% & '
( ) *
∂U∂y
= −∂uv∂y
where y is an integration variable. Now place the assumed self-similar form for the horizontal velocity,
€
U(x,y) =Uo −ΔU(x) f (ξ) , and an equivalent form for the Reynolds shear stress,
€
" u " v = RSCL (x)g(ξ) , into this equation:
€
Uo −ΔUf( )∂ Uo −ΔUf( )
∂x−
∂ Uo −ΔUf( )∂x
dy 0
y
∫& ' (
) * +
∂ Uo −ΔUf( )∂y
= −∂(RSCLg)
∂y
Evaluate the derivative and work to put as much of the equation as possible in terms of the similarity variable ξ.
€
Uo −ΔUf( ) −Δ $ U f + ΔU $ f yδ 2
$ δ &
' (
)
* + − −Δ $ U f + ΔU $ f y
δ 2$ δ
&
' (
)
* + dy
0
y
∫- . /
0 1 2 −ΔU
$ f δ
&
' (
)
* + = −RSCL
$ g δ
€
−(Uo −ΔUf )Δ $ U f + (Uo −ΔUf )ΔU $ f ξ$ δ δ−ΔUΔ $ U
$ f δ
fdy 0
y
∫ + ΔU( )2$ f δ
$ δ δ
$ f ξ dy 0
y
∫ = −RSCL$ g δ
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Convert the integrations to the similarity variable:
€
−(Uo −ΔUf )Δ $ U f + (Uo −ΔUf )ΔU $ f ξ$ δ δ−ΔUΔ $ U $ f fdξ
0
ξ
∫ + ΔU( )2 $ f $ δ δ
$ f ξ d0
ξ
∫ ξ = −RSCL$ g δ
.
Evaluate the second integral by parts, and cancel the opposite terms to find:
€
−(Uo −ΔUf )Δ $ U f + UoΔU $ f ξ$ δ δ−ΔUΔ $ U $ f fdξ
0
ξ
∫ − ΔU( )2 $ f $ δ δ
fd0
ξ
∫ ξ (
) *
+
, - = −RSCL
$ g δ
.
Rearrange each term and place coefficients that depend only on x in square brackets.
€
− UoΔ $ U [ ] f + UoΔU$ δ δ
&
' ( )
* + ξ $ f + ΔUΔ $ U [ ] f 2 − ΔUΔ $ U + ΔU( )2
$ δ δ
&
' ( )
* + $ f fdξ 0
ξ
∫ =−RSCL
δ
&
' ( )
* + $ g
When
€
ΔU « Uo, only the first two terms on the left matter:
€
UoΔ # U [ ] f + UoΔU# δ δ
%
& ' (
) * ξ # f ≈ −RSCL
δ
%
& ' (
) * # g .
Now divide by
€
ΔU( )2 and multiply by δ:
€
UoΔ # U δΔU( )2
%
& ' '
(
) * *
f +Uo # δ ΔU%
& ' (
) * ξ # f ≈ −RSCL
ΔU( )2%
& ' '
(
) * * # g . This scaling of
the turbulent stresses suggests that similarity will be achieved when the factors in square brackets are constants.
€
UoΔ # U δΔU( )2
= C1 ,
€
Uo " δ ΔU
= C2, and
€
−RSCLΔU( )2
= C3 .
b) When
€
ΔU « Uo, the origin constraint determined from the formula provided in the problem
statement becomes:
€
F D bρUo
2 ≈ΔU(x)δ(x)
Uo
f (ξ )dξ −∞
+∞
∫ . Thus, similarity requires the product
€
ΔUδ to
be independent of x. If
€
ΔUδ = C4, then the first similarity requirement from part a) implies
€
Δ # U ΔU( )3
=C1
UoC4
→
€
1ΔU( )2
∝ x , or
€
ΔU ∝ x−1 2 and
€
δ ∝ x +1 2 .
Here, C4 has units but Uo and the fundamental length scale θ are all that’s needed to put these laws into proper dimensionless form:
€
ΔU Uo = const1 ⋅ x θ( )−1 2 , and
€
δ = const2 ⋅ (θx)+1 2 .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.28. Consider the two-dimensional shear layer that forms between two steady streams with flow speed U2 above and U1 below y = 0, that meet at x = 0, as shown. Assume a self-similar form for the average horizontal velocity:
€
U(x,y) =U1 + (U2 −U1) f (ξ) with
€
ξ = y δ(x). a) What are the boundary conditions on f(ξ) as y → ±∞?
b) If the flow is laminar, use
€
∂U∂x
+∂V∂y
= 0 and
€
U ∂U∂x
+V ∂U∂y
= ν∂ 2U∂y 2
with
€
δ(x) = νx U1 to
obtain single equation for f(ξ). There is no need to solve this equation.
c) If the flow is turbulent, use:
€
∂U∂x
+∂V∂y
= 0 and
€
U ∂U∂x
+V ∂U∂y
= −∂∂y
uv( ) with
€
−uv = U2 −U1( )2g(ξ) to obtain a single equation involving f and g. Determine how δ must depend on x for the flow to be self-similar. d) Does the laminar or the turbulent mixing layer grow more quickly as x increases?
Solution 12.28. a) As y → –∞, f(ξ)→ 0; and as y → +∞, f(ξ)→ 1. b) First combine the equations to eliminate V:
€
U ∂U∂x
−∂U∂x
dy −∞
y
∫& ' (
) * +
∂U∂y
= ν∂ 2U∂y 2
.
where y is an integration variable. Now plug in the similarity form for U(x,y) with ΔU = U2 – U1:
€
− U1 + ΔUf( )ΔU $ f y $ δ δ 2
+ ΔU $ f y $ δ δ 2
dy −∞
y
∫( ) *
+ , - ΔU $ f 1
δ= νΔU $ $ f 1
δ 2
Now substitute in
€
δ(x) = νx U1 and convert to ξ in place of y:
€
− U1 + ΔUf( )ΔUξ % f 12x
+ ΔU( )2 % f 12x
ξ % f dξ −∞
ξ
∫( ) *
+ , -
= νΔU % % f U1
νx
Cancel the common factor of
€
ΔU x and integrate by parts inside the {,}-braces:
€
− U1 + ΔUf( )ξ% f 2
+ ΔU % f 2
ξf − fdξ −∞
ξ
∫( ) *
+ , -
= U1 % % f
Simplify and divide by U1 to find:
€
−ξ $ f 2−ΔUU1
$ f 2
fdξ −∞
ξ
∫ = $ $ f .
c) Start with
€
U ∂U∂x
−∂U∂x
dy −∞
y
∫& ' (
) * +
∂U∂y
= ν∂ 2U∂y 2
and put in the similarity forms for U(x,y) and the
Reynolds shear stress:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
− U1 + ΔUf( )ΔU $ f y $ δ δ 2
+ ΔU $ f y $ δ δ 2
dy −∞
y
∫( ) *
+ , - ΔU $ f 1
δ= ΔU( )2 $ g 1
δ.
Convert to ξ in place of y, and rearrange the results:
€
− U1 + ΔUf( )ΔU $ δ δξ $ f + ΔU( )2
$ δ δ
$ f $ f ξdξ−∞
ξ
∫) * +
, - .
= ΔU( )2 $ g 1δ
.
Integrate by parts and simplify:
€
−U1ΔU $ δ δξ $ f − ΔU( )2
$ δ δ
$ f fdξ −∞
ξ
∫ = ΔU( )2 $ g 1δ
.
Divide by
€
ΔU( )2 δ and regroup the terms on the left:
€
−U1
ΔU$ δ ξ $ f +
ΔUU1
$ f fdξ −∞
ξ
∫)
* +
,
- . = $ g .
Here, neither U1 nor ΔU depend on x, so the requirement for self-similarity means that
€
δ ∝ x . d) The turbulent shear layer (
€
δ ∝ x) grows more quickly with increasing x than the laminar shear layer (
€
δ ∝ x 1/2).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.29. Consider an orifice of diameter d that emits an incompressible fluid of density ρo at speed Uo into an infinite half space of fluid with density ρ∞. With gravity acting and ρ∞ > ρo, the orifice fluid rises, mixes with the ambient fluid, and forms a buoyant plume with a diameter D(z) that grows with increasing height above the orifice. Assuming that the plume is turbulent and self-similar in the far-field (z >> d), determine how the plume diameter D, the mean centerline velocity Ucl, and the mean centerline mass fraction of orifice fluid Ycl depend on the vertical coordinate z via the steps suggested below. Ignore the initial momentum of the orifice fluid. Use both dimensional and control-volume analysis as necessary. Ignore streamwise turbulent fluxes to simplify your work. Assume uniform flow from the orifice. a) Place a stationary cylindrical control volume around the plume with circular control surfaces that slice through the plume at its origin and at height z. Use similarity forms for the average vertical velocity Uz (z,R) =Ucl (z) f R z( ) and nozzle fluid mass fraction Y (z,R) = (ρ∞ − ρ) (ρ∞ − ρo ) =Ycl (z)h R z( ) to conserve the flux of nozzle fluid in the plume, and
find: mo = ρoUo dAsource∫ = ρoY (z,R)Uz (z,R)2πRdR0
D 2∫ .
b) Conserve vertical momentum using the same control volume assuming that all entrained fluid enters with negligible vertical momentum, to determine:
− ρoUo2 dA
source∫ + ρ(z,R)Uz2 (z,R)2πRdR
0
D 2∫ = g ρ∞ − ρ(z,R)[ ]dV
volume∫ , where ρ =Y ρo + (1−Y )ρ∞ . c) Ignore the source momentum flux, assume z is large enough so that YCL << 1, and use the results of parts a) and b) to find: Ucl (z) =C1. B ρ∞z3 and (ρ∞ − ρo ) ρ∞( )Ycl (z) =C2 B2 g3ρ∞
2z53 , where C1 and C2 are dimensionless constants, and B = (ρ∞ − ρo )gUo dAsource∫ .
Solution 12.29. Use the similarity forms provided,
Uz (z,R) =Ucl (z) f R z( ) and Y (z,R) = (ρ∞ − ρ) (ρ∞ − ρo ) =Ycl (z)h R z( ) to complete this exercise. Since the z-dependences of two functions are sought, it will be necessary to consider both conservation of mass and vertical momentum. a) Choose a conical control volume with a flat end at a height of z above the orifice that encloses the plume. Ignoring turbulent fluxes, conservation of orifice fluid then implies:
!mo = ρoUoπd 2
4= ρoY (z,R)w(z,R)2πRdR0
D 2∫ .
This equation provides a constraint on the possible z-dependence of the Ycl and Ucl. Substitute in the trial self-similar forms for
€
Y and Uz, and change the integration variable to ξ = R/z.
z!
R!
ρ∞"g!
D(z)"UCL(z)f(R/z)"
d!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
ρoUoπd 2
4= 2πρoYcl (z)Ucl (z) h R z( ) f R z( )RdR
0
D 2∫ = 2πρoYcl (z)Ucl (z)z
2 h(ξ ) f (ξ )ξ dξ0
D 2z∫
Since D is presumed proportional to z, the integral is just a pure number, so we can reduce the above formula to:
€
Uod2 = const ⋅ z2Ycl (z)Ucl (z) (1)
b) Now conserve momentum with the same control volume, and assume that all of the fluid entrained though the conical sides of the control volume either enters horizontally and contributes no vertical momentum or that it enters slowly enough so that its vertical momentum contribution is negligible. In addition, assume the conical control volume boundary to be stress free. Under these conditions, the integral vertical-momentum equation becomes:
− ρoUo2
source∫ dA+ ρ(z,R)wUz
2 (z,R)2πRdR0
D 2∫ = − gρ(z,R)dV
volume∫ − P n ⋅ ez( )surface∫ dA .
c) The first integral on the right results from the gravitational body force, and the second term is the pressure integral. The pressure on the conical control surface will be the hydrostatic pressure: P = Po – ρ∞gz. If we assume that the flow is basically upward within the plume, the radial momentum equation can be used to find ∂P/∂r ≈ 0 so the foregoing pressure equation can be used at any radial location. So, use the static pressure relationship and Gauss' Theorem to transform the pressure integral
€
− P n ⋅ ez( )surface∫ dA = − ∇P ⋅ ezdVvolume∫ = + ρ∞gdVvolume∫ . and then combine it with the body force integral to find:
− ρoUo2
source∫ dA+ ρ(z,R)Uz
2 (z,R)2πRdR0
D 2∫ = g ρ∞ − ρ(z,R)[ ]dV
volume∫ .
When the source momentum is negligible, the first term on the left can be dropped. With this, the assumed self-similarity relations produce:
ρ∞ −Ycl (z)h(R z) ρ∞ − ρo( ){ }Ucl2 (z) f 2 (R z)2πRdR
0
D 2∫ ≅ gYcl (z)h(R z) ρ∞ − ρo( )dV
volume∫ Far above the plume origin a lot of mixing will have taken place and Ycl << 1, so ignore it inside the first integral.
ρ∞Ucl2 (z) f 2 (R z)2πRdR
0
D 2∫ ≅ g ρ∞ − ρo( ) Ycl (ζ ) h(R ζ )2πRdRdζ
0
D 2∫0
z∫
As before, change the variable of integration to ξ = R/z (or ξ = R/ζ in the radial integrations where ζ is the vertical direction integration variable) and divide out the common factor of 2π:
€
ρ∞z2Ucl
2 (z) f 2(ξ)ξdξ0
D 2z∫ = g ρ∞ − ρo( ) ζ 2Ycl (ζ ) h(ξ)ξdξ
0
D 2ζ∫[ ]0
z∫ dζ .
On both sides, the ξ-integrations yield pure numbers, therefore:
€
ρ∞z2Ucl
2 (z) = const ⋅ g ρ∞ − ρo( ) ζ 2Ycl (ζ )0
z∫ dζ .
Now substitute for
€
ζ 2Ycl (ζ ) from (1):
€
ρ∞z2Ucl
2 (z) = const ⋅ g ρ∞ − ρo( )Uod2 1
Ucl (ζ )0
z∫ dζ .
Assume Ucl(z) has a power law form:
€
Ucl = Az−b , integrate and solve for A and b.
€
ρ∞A2z2−2b = const ⋅ g ρ∞ − ρo( )Uod
2 zb+1
A(b +1)
This relationship implies:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
A = const ⋅ g(ρ∞ − ρo)Uod2
ρ∞
&
' (
)
* +
1 3
, and 2 – 2b = b +1 or b = 1/3.
Use the definition of the buoyancy flux,
€
B = (ρ∞ − ρo)gUodA∫ =π4g(ρ∞ − ρo)gUod
2 , and (1) to
evaluate Ycl(z). The final results being:
€
Ucl (z) = C1 B ρ∞z3 , and
€
Ycl (z) = " C 2Uod2 ρ∞ Bz53 or
€
ρ∞ − ρoρ∞
Ycl (z) = C2 B2 g3ρ∞2z53 ,
where the C’s are dimensionless constants. It is also possible to obtain the result for Ucl entirely from dimensional analysis. For high Reynolds number flow far above the orifice, the parameters are: Ucl, B, ρ∞, z (“High Re” implies that µ doesn't matter; “far above” implies severe dilution & complete "memory loss" by the flow, therefore d and
€
˙ m o don't matter). The parameter matrix is: Ucl B ρ∞ z –––––––––––––––––––––– Mass: 0 1 1 0 Length: 1 1 -3 1 Time: -1 -3 0 0 Determine the number of dimensionless groups: 4 parameters - 3 dimensions = 1 group Construct the dimensionless groups:
€
Π1 = B ρ∞Ucl3 z
Write a dimensionless scaling law:
€
Π1 = const , or
€
Ucl = const ⋅ B ρ∞z( )1 3 . The mean centerline mass fraction of orifice fluid can be obtained by conserving orifice fluid, this amounts to finding Ycl(z) from (1) and the relationship for Ucl.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.30. Laminar and turbulent boundary layer skin friction are very different. Consider skin friction correlations from zero-pressure-gradient (ZPG) boundary layer flow over a flat plate placed parallel to the flow.
Laminar boundary layer: Cf =τ 0
12 ρU
2 =0.664Rex
(Blasuis boundary layer)
Turbulent boundary layer: (see correlations in §12.9) Create a table of computed results at Rex = Ux/ν = 104, 105, 106, 107, 108, and 109 for the laminar and turbulent skin friction coefficients, and the friction force acting on 1.0 m2 plate surface in sea-level air at 100 m/s and in water at 20 m/s assuming laminar and turbulent flow. Solution 12.30. Merely calculate and tabulate results for air: ρ = 1.2 kg/m3 & ν = 1.5×10–5 m2/s, and for water: ρ = 103 kg/m3 & ν = 10–6 m2/s. The forces below are in Newtons, and the turbulent skin friction numbers were obtained from White's formula:
€
Cf ≅0.455
ln 0.06Rex( )[ ]2.
log Re Cf laminar Cf turbulent
Force, air, laminar
Force, air, turbulent
Force, water, lam.
Force, water, turb.
4 6.64E-03 1.11E-02 3.98E-01 6.67E-01 1.33E+01 2.22E+01 5 2.10E-03 6.01E-03 1.26E-01 3.61E-01 4.20E+00 1.20E+01 6 6.64E-04 3.76E-03 3.98E-02 2.26E-01 1.33E+00 7.52E+00 7 2.10E-04 2.57E-03 1.26E-02 1.54E-01 4.20E-01 5.14E+00 8 6.64E-05 1.87E-03 3.98E-03 1.12E-01 1.33E-01 3.74E+00 9 2.10E-05 1.42E-03 1.26E-03 8.51E-02 4.20E-02 2.84E+00
The laminar results fall faster with increasing Re, and the turbulent skin friction numbers are always higher than the laminar ones. The forces in water are also much larger than those in air at the same Reynolds number in the same flow regime.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.31. Derive the following logarithmic velocity profile for a smooth wall:
€
U + = 1 κ( ) ln y + + 5.0 by starting from
€
U = u* κ( ) ln y + + const. and matching the profile to the edge of the viscous sublayer assuming the viscous sublayer ends at y = 10.7 v/u*. Solution 12.31. In the viscous sublayer the velocity profile is linear:
€
U(y) = u*2y ν .
At the presumed edge of the viscous sublayer (y = 10.7lν = 10.7ν/u*), the velocity is:
€
U(y =10.7lν ) = u*2 10.7ν u*
ν=10.7u*.
Fitting this condition at the edge of the logarithmic region,
€
U = u* κ( ) ln y + + const., leads to:
€
10.7u* = u* κ( ) ln 10.7( ) + const., or
€
const.=10.7u* − u* κ( ) ln 10.7( ) . Thus, the logarithmic law becomes:
€
Uu*
=1κln y + +
const.u*
=1κln y + +10.7 − 1
κln(10.7) =
1κln y + +10.7 − 5.8 =
1κln y + + 5,
where the final numerical values have been rounded to one significant digit and a value of κ = 0.41 has been assumed.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.32. Derive the log-law for the mean flow profile in a zero-pressure gradient (ZPG) flat plate turbulent boundary layer (TBL) through the following mathematical and dimensional arguments. a) Start with the law of the wall,
€
U u* = f yu* ν( ) or
€
U + = f (y +), for the near-wall region of the
boundary layer, and the defect law for the outer region,
€
Ue −Uu*
= F yδ
$
% & '
( ) . These formulae must
overlap when y+ → +∞ and y/δ → 0. In this matching or overlap region, set U and
€
∂U ∂y from both formula equal to get two equations involving f and F. b) In the limit as y+ → +∞, the kinematic viscosity must drop out of the equation that includes df/dy+. Use this fact, to show that
€
U u* = AI ln yu* ν( ) + BI as y+ → +∞ where AI and BI are constants for the near-wall or inner boundary layer scaling. c) Use the result of part b) to determine
€
F(ξ) = −AI ln ξ( ) − BO where ξ = y/δ, and AI and BO are constants for the wake flow or outer boundary layer scaling. d) It is traditional to set AI = 1/κ, and to keep BI but to drop its subscript. Using these new requirements determine the two functions, fI and FO, in the matching region. Which function explicitly depends on the Reynolds number of the flow?
Solution 12.32. a) Set U from both formulae equal:
€
fI yu* ν( ) =
Ue
u*− FO
yδ
%
& ' (
) * .
Set
€
∂U ∂y from both formula equal:
€
u*
ν# f I
yu*
ν
$
% &
'
( ) = −
1δ
# F Oyδ
$
% & '
( ) .
b) In the limit as y+ → +∞, the kinematic viscosity must drop out of the second equation in part
a); thus:
€
" f Iyu*
ν
$
% &
'
( ) → const ⋅ yu*
ν
$
% &
'
( )
−1
as
€
yu*
ν→∞. This means:
€
dfIdy + = AI
1y + , or
€
fI (y+) = AI ln y
+ + BI c) Plug this result into the second equation from part a):
€
u*
ν# f I
yu*
ν
$
% &
'
( ) =
u*
νdfI
dy + = AIu*
νν
yu*=
AI
y= −
1δ
# F Oyδ
$
% & '
( )
Now use the last equality and one integration to find:
€
FOyδ
#
$ % &
' ( = −AI ln
yδ
#
$ % &
' ( − BO
d) Now use the first equation from part a), plus the results of parts b) and c) with AI = 1/κ and BI = B, to find:
€
fIyu*
ν
#
$ %
&
' ( =
1κln yu*
ν
#
$ %
&
' ( + B =
Ue
u*− −
1κln y
δ
#
$ % &
' ( − BO
,
- .
/
0 1 =
Ue
u*− FO
yδ
#
$ % &
' (
Work with the middle equality and introduce some δ’s on the left side:
€
1κln y
δδu*
ν
%
& '
(
) * + B =
Ue
u*+1κln y
δ
%
& ' (
) * + BO →
€
1κln y
δ
$
% & '
( ) +1κln δu
*
ν
$
% &
'
( ) + B =
Ue
u*+1κln y
δ
$
% & '
( ) + BO
Thus,
€
BO =1κln δu
*
ν
%
& '
(
) * + B −
Ue
u*, so
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
€
fIyu*
ν
#
$ %
&
' ( =
1κln yu*
ν
#
$ %
&
' ( + B , and
€
FOyδ
#
$ % &
' ( = −
1κln yu*
ν
#
$ %
&
' ( +1κln δu
*
ν
#
$ %
&
' ( + B −
Ue
u*
in the overlap region. The outer function FO explicitly depends on the Reynolds number through
the term containing
€
δu*
ν.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.33. For zero pressure gradient, the Von Karman boundary layer integral equation simplifies to Cf = 2dθ/dx. Use this fact, and (12.90) to determine Cf and numerically compare this result to Cf obtained from (12.93). Do the results match well for 106 < Rex < 109? What difference does the choice of log-law constants make? Consider (κ,B) pairs representative of the nominal modern values for pipes (0.41, 5.2) and boundary layers (0.38, 4.2). Solution 12.33. Compute the skin friction by differentiating the momentum thickness correlation (12.90):
Cf = 2dθdx
= 2 ddx
0.016x Uexν
!
"#
$
%&−0.15(
)**
+
,--= 0.032 0.85( ) Uex
ν
!
"#
$
%&−0.15
= 0.0272Rex−0.15 .
The evaluation of (12.93) is a bit more involved and requires use of (12.90), and (12.91) rewritten in terms of Reynolds numbers:
Reθ = 0.016Rex0.85 , and Reδ* = Reθexp
7.11κln(Reθ )
!"#
$%&= 0.016Rex
0.85exp 7.11κln(0.016Rex
0.85 )!"#
$%&
.
Thus, (12.93) becomes:
Cf =2.0
κ −1 ln Reδ*( )+3.30"# $%2 =
2.0
κ −1 ln 0.016Rex0.85exp 7.11κ
ln(0.016Rex0.85 )
&'(
)*+
,
-.
/
01+3.30
"
#22
$
%33
2 .
The constant B doesn't matter, but κ does enter the formulation for Cf. For clarity, the comparison should be made graphically.
Given that (12.90) is for a zero-pressure-gradient (ZPG) turbulent boundary layer (TBL), it matches the result from (12.93) to within ±5% or so using the ZPG TBL value of κ. The match between (12.90) and (12.93) using the pipe-flow value of κ is not as good; the difference is 10% or more throughout the range of Rex considered here.
0.00E+00%
5.00E'04%
1.00E'03%
1.50E'03%
2.00E'03%
2.50E'03%
3.00E'03%
3.50E'03%
4.00E'03%
1.00E+06% 1.00E+07% 1.00E+08% 1.00E+09%
(12.90).%
(12.93),%0.38%
(12.93),%0.41%
Cf
Rex
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.34. Prove (12.90) and (12.91) by considering a stationary control volume that resides inside the channel or pipe and has stream-normal control surfaces separated by a distance dx and stream-parallel surfaces that coincide with the wall or walls that confine the flow. Solution 12.34. The two calculations are nearly the same. For (12.90) consider a slab-like control volume that has thickness dx in the stream-wise direction, width B perpendicular to the mean flow, and height h that spans the inside of the channel. For steady fully developed flow, conservation of axial momentum implies:
− ρu2 (y)"# $%x Bdy0
h
∫ + ρu2 (y)"# $%x+dx Bdy0
h
∫ = p(x)Bh− p(x + dx)Bh− 2τ wBdx ,
where τw is the wall shear stress, and the factor of 2 in the final term arises because of the shear stress on the upper and lower walls of the channel. Here the flux terms on the left side are equal & opposite, and the pressure at x+dx may be expanded in a Taylor series to find:
0 = p(x)Bh− p(x)+ dp dx( )dx( )Bh− 2τ wBdx , or dp dx( )dxBh = −2τ wBdx . Divide by Bhdx to reach (12.90):
dp dx = −2τ w h . For (12.91) consider a disk-like cylindrical control volume that has thickness dx and spans the inside of the pipe of diameter d. For steady fully developed flow, conservation of axial momentum implies:
− ρu2 (R)"# $%xpipe area∫ dA+ ρu2 (R)"# $%x+dx
pipe area∫ dA = p(x)π d
4
2
− p(x + dx)π d4
2
−πdτ wdx ,
where τo is the wall shear stress. Here again, the flux terms on the left side are equal & opposite, and the pressure at x+dx may be expanded in a Taylor series to find:
0 = p(x)π d4
2
− p(x)+ dp dx( )dx( )π d42
−πdτ wdx , or dp dx( )dxπ d4
2
= −πdτ wdx .
Divide by (πd2/4)dx to reach (12.91): dp dx = −4τ w d .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.35. The log-law occurs in turbulent channel, pipe, or boundary layer flows and should be absent in laminar flows in the same geometries. The extent of the log-law is governed by Reτ = δ+ = δ/lν = δu*/ν, where δ is the channel half-height (h/2), pipe radius (d/2), or full boundary-layer thickness, as appropriate for each flow geometry. a) For laminar channel flow, show that Reτ = (3 / 2)Reh , and compute Reτ at an approximate transition Reynolds number of Reh ~ 3,000. b) For laminar pipe flow, show that Reτ = 2Red , and compute Reτ at an approximate transition Reynolds number of Red ~ 4,000. c) For the Blasius boundary layer, show that Reτ ≅ 2.9Rex
1 4 , and compute Reτ at a transition Reynolds number of Rex ~ 106. d) If mean profile measurements are made in a wall-bounded turbulent flow at Reτ ~ 102, do you expect the profiles to display the log-law? Why or why not? e) Repeat part d) when Reτ > 103. f) The log-law constants (κ and B) are determined from fitting (12.88) to experimental data. At which Reτ are κ and B most likely to be accurately determined: 102, 103, or 104? Solution 12.35. a) For laminar channel flow, the pressure gradient, channel height, and viscosity determine the wall shear stress and the average velocity:
dpdx
= −2τ wh
and Uave = −h2
12µdpdx
, so that 12µUave
h2= −
dpdx
=2τ wh
=2ρu*
2
h.
where τ w = ρu*2 . Use the two ends of the final extended equality to find:
ρu*2
h 2⋅ρ h 2( )3
µ 2=ρ2u*
2 h 2( )2
µ 2= Reτ
2 =12µUave
h2⋅ρ h 2( )3
µ 2=32Reh .
Thus, Reτ = (3 / 2)Reh which produces Reτ = 67 at Reh = 3,000. b) For laminar pipe flow, the pressure gradient, diameter, and viscosity determine the wall shear stress and the average velocity:
dpdx
= −4τ wd
and Uave = −d 2
32µdpdx
, so that 32µUave
d 2= −
dpdx
=4τ wd
=4ρu*
2
d.
Use the two ends of the final extended equality to find: 2ρu*
2
d 2⋅ρ d 2( )3
2µ 2=ρ2u*
2 d 2( )2
µ 2= Reτ
2 =32µUave
d 2⋅ρ d 2( )3
2µ 2= 2Red .
Thus, Reτ = 2Red which produces Reτ = 89 at Red = 4,000. c) For laminar boundary layer flow, the flow speed U, downstream distance x, and viscosity determine the wall shear stress and the boundary layer thickness:
τ w =0.332ρU 2
Rex1 2 and δ ~ 5.0 νx
U= 5.0 x
Rex1 2 .
Use the first relationship to find:
u*2 =0.332U 2
Rex1 2 which implies: u*
2 δ2
ν 2= Reτ
2 =0.332U 2
Rex1 2
δ 2
ν 2.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Now use the relationship for the overall-boundary layer thickness δ to eliminate δ on the right side.
Reτ2 =0.332U 2
Rex1 2
1ν 2
5.0 xRex
1 2
!
"#
$
%&
2
= 0.332(5.0)2 Rex2
Rex3 2 = 8.3Rex
1 2
Thus, taking a square root leads to: Reτ = 2.9Rex1 4 which produces Reτ = 92 at Rex =106.
d) At Reτ ~ 102, the flow is likely transitionally turbulent so there will not be an extensive log-law region, if one appears at all. At such a low Reynolds number, the outer flow is unlikely to be fully independent of the viscosity. e) At Reτ > 103, the outer flow should be fully turbulent so a distinct log-law region is expected. f) Fitting is always better when there is more data to fit, and the extent of the log-law increases with increasing Reynolds number. Thus, data from the highest Reynolds number, Reτ = 104, should produce the most accurate log-law constants.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.36. A horizontal smooth pipe 20 cm in diameter carries water at a temperature of 20 °C. The drop of pressure is dp/dx = –8 N/m2 per meter. Assuming turbulent flow, verify that the thickness of the viscous sublayer is ≈ 0.25 mm. [Hint: Use dp/dx as given by (12.97) to find τw = 0.4 N/m2, and therefore u* =0.02 m/s.] Solution 12.36. From (12.96), ∂p/∂x = –4τw/d, therefore:
τ w = −d4∂ p∂x
= −(0.2m)4
(−8Pa /m) = 0.4Pa , so u* =τ wρ=
0.4Pa103kgm−3 = 0.02ms
−1 .
From Section 12.9, the viscous sublayer thickness is:
€
5lν = 5 ν u*( ) = 5(1.0 ×10−6m2s−1 0.02ms−1) = 2.5 ×10−4m = 0.25mm .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.37. The cross-section averaged flow speed Uav in a round pipe of radius a may be written:
Uav ≡volume flux
area=
1πa2 U(y)2πr dr
0
a
∫ =2a2 U(y)(a− y)dy
0
a
∫ ,
where r is the radial distance from the pipe's centerline, and y = a – r is the distance inward from the pipe's wall. Turbulent pipe flow has very little wake, and the viscous sublayer is very thin at high Reynolds number; therefore assume the log-law profile, U(y) = u* κ( ) ln yu* ν( )+B , holds throughout the pipe to find
Uav ≅ u* 1 κ( ) ln au* ν( )+B−3 2κ#$ %& . Now use the definitions: Cf = τ w
12 ρUav
2 , Red = 2Uava ν , f = 4Cf = the Darcy friction factor, κ = 0.41, B = 5.0, and switch to base-10 logarithms to reach (12.105). Solution 12.37. Start with given relationships and integrate:
€
Uav =2a2 U(y)(a − y)dy
0
a
∫ ≅2a2
u*
κln yu*
ν
'
( )
*
+ , + u*B
-
. /
0
1 2 (a − y)dy
0
a
∫
=2νa2κ
u*aν
ln yu*
ν
'
( )
*
+ , −
u*yν
ln yu*
ν
'
( )
*
+ ,
-
. /
0
1 2 dy
0
a
∫ +2u*Ba2 a − y[ ]dy
0
a
∫
=2ν 2
a2κu*
au*
νln β( ) −β ln β( )
-
. / 0
1 2 dβ
0
au* ν
∫ +2u*Ba2 ay − y
2
2-
. /
0
1 2
0
a
=2ν 2
a2κu*
au*
νβ lnβ −β( ) − β
2
2ln β( ) +
β 2
4-
. /
0
1 2
0
au* ν
+2u*Ba2 a2 −
a2
2-
. /
0
1 2
=2ν 2
a2κu*
au*
νau*
νln au*
ν−au*
ν
'
( )
*
+ , −
12au*
ν
'
( )
*
+ ,
2
ln au*
ν
'
( )
*
+ , +
14au*
ν
'
( )
*
+ ,
2-
. /
0
1 2 + u*B
=2ν 2
a2κu*
12au*
ν
'
( )
*
+ ,
2
ln au*
ν
'
( )
*
+ , −
34au*
ν
'
( )
*
+ ,
2-
. /
0
1 2 + u*B = u*
1κ
ln au*
ν
'
( )
*
+ , −
32κ
+ B'
( )
*
+ , .
The second part of this exercise is primarily algebraic. Start by rewriting the ratio Uav/u* in terms of
€
f , and rewriting the ratio u*a/ν in terms of Red and
€
f :
Uav
u*=
Uav
τ w ρ=
Uav12CfUav
2=
118 f
=8
f 1 2, and
€
au*ν
=2aUav
18 f
2ν=
132
f 1 2 Red .
Put these into the result of part a) and recall that ln(...) = ln(10)log10(...).
€
Uav
u*=1κln au*
ν
$
% &
'
( ) −
32κ
+ B →8
f 1 2=ln(10)κ
log10132
f 1 2 Red
$
% &
'
( ) −
32κ
+ B .
Rearrange to reach the desired form, and evaluate using the numbers provided above.
€
1f 1 2
=ln(10)8κ
log10 f 1 2 Red( ) +18−32κ
+ B − ln 32κ
$
% &
'
( ) =1.986log10 f 1 2 Red( ) −1.020.
This is the desired result.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.38. The cross-section averaged flow speed Uav in a wide channel of full height b may be written:
Uav ≡2b
U(y)dy0
b 2
∫ ,
where y is the vertical distance from the channel's lower wall. Turbulent channel flow has very little wake, and the viscous sublayer is very thin at high Reynolds number; therefore assume the log-law profile, U(y) = u* κ( ) ln yu* ν( )+B , holds throughout the channel to find
Uav ≅ u* 1 κ( ) ln bu* 2ν( )+B−1 κ#$ %& . Now use the definitions: Cf = τ w
12 ρUav
2 , Reb =Uavb ν , f = 4Cf = the Darcy friction factor, κ
= 0.41, B = 5.0, and switch to base-10 logarithms to reach: f −1 2 = 2.0 log10 Reb f1 2( )− 0.59 .
Solution 12.38. Start with given relationships and integrate:
Uav =2b
U(y)dy0
b 2
∫ ≅2b
u*
κln yu*
ν
#
$%
&
'(+u*B
)
*+
,
-.dy
0
b 2
∫
= 2νbκ
u*
νln yu*
ν
#
$%
&
'(dy
0
b 2
∫ +2u*Bb
dy0
b 2
∫ =2νbκ
ln β( )dβ0
bu* 2ν
∫ +2u*Bb
b2
= 2νbκ
β lnβ −β[ ]0
bu* 2ν+u*B =
2νbκ
bu*
2νln bu*
2ν#
$%
&
'(−
bu*
2ν#
$%
&
'(+u*B
= u*1κ
ln bu*
2ν#
$%
&
'(−
1κ+B
#
$%
&
'(.
The second part of this exercise is primarily algebraic. Start by rewriting the ratio Uav/u* in terms of
€
f , and rewriting the ratio u*b/2ν in terms of Reb and
€
f :
Uav
u*=
Uav
τ w ρ=
Uav12CfUav
2=
118 f
=8
f 1 2, and bu*
2ν=bUav
18 f
2ν=132
f 1 2 Reb .
Put these into the result of part a) and recall that ln(...) = ln(10)log10(...). Uav
u*=1κln bu*
2ν!
"#
$
%&−1κ+B →
8f 1 2
=ln(10)κ
log10132
f 1 2 Reb!
"#
$
%&−1κ+B .
Rearrange to reach the desired form, and evaluate using the numbers provided above. 1f 1 2
=ln(10)8κ
log10 f 1 2 Reb( )+ 18−1κ+B− ln 32
κ
"
#$
%
&'=1.986 log10 f 1 2 Red( )− 0.589 .
This is the desired result.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.39. For laminar flow, the hydraulic diameter concept is successful when the ratio f ⋅Uavdh ν( )duct f ⋅Uavd ν( )round pipe
is near unity. Show that this ratio is 1.5 when the duct is a
wide channel. Solution 12.39. For a channel with height b and width w, the hydraulic diameter from (12.103) is:
dh( )channel = 4bw
2b+ 2w= 2 b1+ b w
, and this simplifies to: (dh)channel ≈ 2b when w >> b.
The friction factor is appears in (12.102): Pu −Pd =12ρUav
2 ⋅Ldh⋅ f , so:
f ⋅Uavdhν
=2(Pu −Pd )dh
2
µUavL=2dh
2
µUav
−dpdx
#
$%
&
'( .
Evaluate this for laminar channel and round-pipe flows using the results from section 9.2:
f ⋅Uavdhν
"
#$
%
&'channel
=2(2b)2
µ −b2
12µdpdx
"
#$
%
&'
−dpdx
"
#$
%
&'=12 ⋅8 = 96 , and
f ⋅Uavdhν
"
#$
%
&'pipe
=2d 2
µ −d 2
32µdpdx
"
#$
%
&'
−dpdx
"
#$
%
&'= 32 ⋅2 = 64 .
Thus, the ratio is: f ⋅Uavdh ν( )channel f ⋅Uavd ν( )round pipe = 96/64 = 1.5.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.40. a) Rewrite the final friction factor equation in Exercise 12.38 in terms of the channel's hydraulic diameter instead of its height b. b) Using the friction factor-Reynolds number ratio given in Exercise 12.39, evaluate (12.107) for a wide channel. c) Are the results of parts a) and b) in good agreement? Solution 12.40. a) For a channel with height b and width w, the hydraulic diameter from (12.103) is:
dh( )channel = 4bw
2b+ 2w= 2 b1+ b w
, and this simplifies to: (dh)channel ≈ 2b when w >> b.
The final result of Exercise 12.38 is f −1 2 = 2.0 log10 Reb f1 2( )− 0.59 , which can be rewritten:
f −1 2 = 2.0 log1012Re2b f
1 2"
#$
%
&'− 0.59 = 2.0 log10 Redh f
1 2( )−1.19 .
b) For this problem, (12.107) implies
fchannel,turb−1 2 ≅ 2.0 log10
fpipe ⋅Redfchannel ⋅Redh
$
%&&
'
())laminar
⋅Redh ⋅ fchannel,turb−1 2
$
%&&
'
())− 0.80
= 2.0 log101
1.5⋅Redh ⋅ fchannel,turb
−1 2$
%&
'
()− 0.80
= 2.0 log10 Redh ⋅ fchannel,turb−1 2( )−1.15.
c) The only difference between the answers for parts a) and b) is 0.04 in the final subtracted constant. Thus, the agreement is good and this suggests that (12.107) provides a useful empirical correction for estimating friction factors in non-circular ducts.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.41. a) Simplify (12.114) when the roughness Reynolds number is large Reks >> 1 to show that CfR is independent of ν in the fully rough regime. b) Reconcile the finding of part a) with the results in Figure 12.25 which appear to show that CfR depends on Rex for all values of Reks. c) For this fully rough regime, compare CfR computed from (12.114) with the empirical formula provided in Schlichting (1979): CfR = 2.87+1.58 ⋅ log10 (x / ks )( )−2.5 . Solution 12.41. a) Eq. (12.114) is:
2CfR
≅1κln
HR 3.5( )Rex CfR 21+ 0.26Reks CfR 2
"
#$$
%
&''+B+
2Πκ
,
where Rex = Uex/ν and Reks = Ueks/ν. When Reks >> 1, the '1' in the denominator may be ignored, and this allows the argument of the natural log function to be simplified:
2CfR
≅1κln
HR 3.5( )Rex CfR 20.26Reks CfR 2
"
#$$
%
&''+B+
2Πκ
=1κln
HR 3.5( ) CfR 20.26
xks
"
#$$
%
&''+B+
2Πκ
,
where Rex/Reks = x/ks. b) Figure (12.25) provides curves of CfR at constant Reks vs. Rex. The ratio x/ks is not held constant on any of the plotted curves. Thus, the reader must examine points on different curves having the same value of x/ks. For example, consider x/ks = 102, where CfR = 1.16 for Rex/Reks = 105/103, 106/104, and 107/105. c) A plot of (12.114) (solid line) and Schlichting's formula (dashed line) vs. log10(x/ks) looks like:
Here, as in Example 12.10, HR = 1.3. The agreement between the two empirical formulae is generally good. At x/ks = 102, (12.114) is 3% higher than Schlichting's formula. At x/ks = 106, (12.114) is 6% lower Schlichting's formula.
0.001$
0.01$
0.1$
2$ 3$ 4$ 5$ 6$
CfR
log10(x/ks)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.42. Perhaps the simplest way to model turbulent flow is to develop an eddy viscosity from dimensional analysis and physical reasoning. Consider turbulent Couette flow with wall spacing h. Assume that eddies of size l produce velocity fluctuations of size
€
l ∂U ∂y( ) so that the turbulent shear stress correlation can modeled as:
€
−uv ∝ l2 ∂U ∂y( )2 . Unfortunately, l cannot be a constant because it must disappear near the walls. Thus, more educated guessing is needed, so for this problem assume ∂U/∂y will have some symmetry about the channel centerline (as shown) and try:
€
l = Cy for 0 ≤ y ≤ h/2 where C is a positive dimensionless constant and y is the vertical distance measured from the lower wall. With this turbulence model, the RANS equation for 0 ≤ y ≤ h/2 becomes:
€
U ∂U∂x
+ V ∂U∂y
= −1ρ
dp dx
+1ρ
∂τ xy
∂y where
€
τ xy = µ∂U∂y
+ ρC2y 2 ∂U∂y%
& '
(
) *
2
Determine an analytic form for U(y) after making appropriate simplifications of the RANS equation for fully developed flow assuming the pressure gradient is zero. Check to see that your final answer recovers the appropriate forms as y → 0 and C → 0. Use the fact that
€
U(y = h /2) =Uo 2 in your work if necessary.
Solution 12.42. In Couette flow V will be zero because the fluid is confined and ∂/∂x = 0 because the flow is homogeneous in this direction. Thus, with zero pressure gradient, the RANS BL
equation becomes very simple:
€
0 =1ρ
∂τ xy∂y
. Integrate this equation from lower wall to find:
€
τ xy (y) − τw = 0 where τw is the wall shear stress. Now substitute in the model equation for the shear stress to find a quadratic equation for the velocity gradient:
€
µ∂U∂y
+ ρC2y 2 ∂U∂y$
% &
'
( )
2
− τw = 0, which implies
€
∂U∂y
=−µ ± µ2 + 4ρτwC
2y 2
2ρC2y 2.
Here, a positive velocity gradient is expected for the lower half of the channel so the “+” sign should be chosen. A bit of work or use of a table of integrals produces:
€
U(y) =1C
τwρ
1− 1+ 4ρτwC2y 2 µ2
2C ρτw y µ( )+ ln 2C ρτw
yµ
%
& ' (
) * + 1+
4ρτwC2y 2
µ2
+ , -
. -
/ 0 -
1 -
2
3 4 4
5
6 7 7
Now check to see that this model outcome has the appropriate behavior. As y → 0, the velocity should be linear in y. To find this behavior in the equation above, expand the square roots using the Taylor series
€
1+ ε =1+ 12ε + ... for ε << 1.
€
U(y) =1C
τwρ1− (1+ 2ρτwC
2y 2 µ2 + ...)2C ρτw y µ( )
+ ln 2C ρτwyµ
%
& ' (
) * +1+
2ρτwC2y 2
µ2 + ...+ , -
. / 0
1
2 3
4
5 6 . ($)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Simplify the first term, expand the natural logarithm using
€
ln(1+ ε) = ε + ... for ε << 1, and ignore terms involving y2 or higher powers:
€
U(y) ≅ 1C
τwρ
−C ρτwyµ
&
' ( )
* + + 2C ρτw
yµ
&
' ( )
* +
,
- .
/
0 1 =
τwµy ,
and this is the expected form. As C → 0, the same result is recovered. In addition, the fact:
€
U(y = h /2) =Uo 2, can be used to link τw and Uo by evaluating ($) at y = h/2. And finally, when
€
4ρτwC2y 2
µ2 >>1 a log-layer is produced by this turbulence model.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.43. Incompressible, constant-density-and-viscosity, fully-developed, pressure-gradient-driven, turbulent channel flow is often used to test turbulence models for wall-bounded flows. Thus, for this flow, investigate the following simplified mixing-length model for the Reynolds shear stress:
€
− # u # v = βy τw ρ ∂U ∂y( ) for 0 ≤ y ≤ h/2 where y is measured from the lower wall of the channel, β is a positive dimensionless constant, τw = wall shear stress (a constant), and ρ = fluid density. a) Use this turbulence model, the fully-developed flow assumption
€
U = U(y) ˆ e x , the assumption
of a constant downstream pressure gradient, and the x-direction RANS mom. equ.,
€
U ∂U∂x
+ V ∂U∂y
= −1ρ∂P∂x
+ ν∂ 2U∂x 2
+∂ 2U∂y 2
&
' (
)
* + −
∂∂x
, u 2( ) − ∂∂y
, u , v ( )
to find: U(y) = u*β
1+ 2νβu*h
!
"#
$
%&ln 1+
βu*yν
!
"#
$
%&−2yh
(
)*
+
,- for 0 ≤ y ≤ h/2 where u* = τ w ρ .
b) Does this velocity profile have the proper gradient at y = 0 and y = h/2? c) Show that this velocity profile returns to a parabolic flow profile as
€
β → 0. d) How should the constant β be determined? Solution 12.43. a) When the flow is fully developed (U = U(y)ex) the only non-zero field gradient in the flow direction is ∂P/∂x, so the Reynolds-averaged x-direction momentum equation simplifies to:
0 = − 1ρ∂P∂x
+ν∂ 2U∂y2
−∂∂y
"u "v( ) ≅ − 1ρdPdx
+∂∂y
ν∂U∂y
+βuτ y∂U∂y
$
%&
'
()
where the second approximate equality follows from the given turbulence model with
€
uτ = τw ρ . Here all the vertical gradients are switched to total derivatives because y is the only independent variable. Integrate the last form of the equation once in the y-direction:
1ρdPdx
y+ const =ν ∂U∂y
+βuτ y∂U∂y
.
Evaluating this equation at y = 0 determines the constant:
€
const = ν dU dy( )y= 0 = τw ρ .
Rearrange the equation to find: . The parametric format of this equation
can be simplified by using a CV to get a simple relationship between τw and dP/dx. Conserve horizontal (x) momentum in a stationary rectangular CV that encloses all the fluid in the channel between x and x + Δx. Here the horizontal velocity profile is steady and unchanged between x and x + Δx, so the unsteady and flux terms are zero. Thus COMOx simplifies to:
€
0 = P(x)h − P(x + Δx)h − 2τwΔx , which is a balance of pressure forces on the vertical CV sides and skin-friction forces on the horizontal CV sides. When this CV equation is rearranged and the limit as
€
Δx→ 0 is taken, it becomes
€
dP dx = −2τw h . Thus, the differential equation for U(y) can be rewritten:
dUdy
=τ w 1− 2y h( )ρ ν +βuτ y( )
=τ w 1− 2y h( )
ρ ν + 12 βuτh 2y h( )( )
=τ w −2y h+1( )
12 ρβuτh 2y h+ 2ν βuτh( )
€
dUdy
=y dP dx( ) + τwρ ν + βuτ y( )
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Setting
€
η = 2y h , and integrating produces:
€
U(y) =τwρβuτ
−η +1η + 2ν βuτh∫ dη =
τwρβuτ
−1+1+ 2ν βuτhη + 2ν βuτh
)
* +
,
- . ∫ dη
€
=uτβ
−2yh
+ 1+2νβuτh
&
' (
)
* + ln
2yh
+2νβuτh
&
' (
)
* +
&
' (
)
* + + const
The constant can be evaluated by requiring U(0) = 0.
€
0 =uτβ
−0 + 1+2νβuτh
&
' (
)
* + ln
2νβuτh
&
' (
)
* +
&
' (
)
* + + const or
€
const = −uτβ1+
2νβuτh
&
' (
)
* + ln
2νβuτh
&
' (
)
* +
Thus,
€
U(y) =uτβ
−2yh
+ 1+2νβuτh
&
' (
)
* + ln
2yh
+2νβuτh
&
' (
)
* + − ln
2νβuτh&
' (
)
* +
,
- .
/
0 1
&
' ( (
)
* + + , or
€
=uτβ
−2yh
+ 1+2νβuτh
&
' (
)
* + ln 1+
βuτ yν
&
' (
)
* +
&
' (
)
* +
Here it should be noted that as the Reynolds number of the flow increases the importance of the turbulence model increases. This fact can be ascertained by noting how U(y) depends on
€
uτh ν .
b) Yes; from above,
€
dUdy
=τw 1− 2y h( )ρ ν + βuτ y( )
equals
€
τwµ
at y = 0, and equals 0 at
€
y =h2
.
c) Use a two term expansion of the natural log function for small β in the answer for part a) and complete the algebraic multiplications of terms
€
limβ →0
U(y) = limβ →0
uτβ
−2yh
+ 1+2νβuτh
'
( )
*
+ , ln 1+
βuτ yν
'
( )
*
+ ,
'
( )
*
+ ,
€
= limβ →0
uτβ
−2yh
+ 1+2νβuτh
'
( )
*
+ , βuτ yν
−12βuτ yν
'
( )
*
+ , 2
+ ...'
( ) )
*
+ , ,
'
( ) )
*
+ , ,
€
= limβ →0
uτβ
−2yh
+βuτ yν
−12βuτ yν
'
( )
*
+ , 2
+2νβuτh
βuτ yν
−2νβuτh
12βuτ yν
'
( )
*
+ , 2
+ ...'
( ) )
*
+ , ,
Cancel like terms and simplify:
€
= limβ →0
uτ2hν
yh1− y
h−β ...( )
'
( )
*
+ , =
uτ2hν
yh1− y
h'
( )
*
+ , = −
h2
2µdPdx
yh1− y
h'
( )
*
+ ,
The final equality is the parabolic flow velocity profile for pressure gradient driven flow between parallel plates; it follows from the prior expression via the substitution:
€
dP dx = −2τw h = −2ρ uτ2 h .
d) The constant β would need to be determined by fitting a sample profile to an experimental result.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.44. The model equations for the two-equation "k-ε" turbulence model, (12.124) and (12.126), include 5 empirical constants. One of these, Cε2, can be estimated independently of the others by fitting a solution of the model equations to experimental results for the decay of the turbulent kinetic energy, e , downstream of a random grid placed at the inlet of a wind-tunnel test section. The development of this estimate is further simplified by use of a coordinate system that translates with the average flow velocity in the wind-tunnel. In these translating coordinates Ui = 0, and e and ε are both functions of time t alone. a) Simplify (12.124) and (12.126) for random grid turbulence when Ui = 0. b) Assume e (t) follows a power-law solution, e = eot
−n , where eo and n are positive constants, and determine a formula for the model constant Cε2 in terms of n. c) The experimental value of n is approximately 1.3, so the part b) formula then predicts Cε2 = 1.77, which is below the standard value (Cε2 = 1.92 from Launder & Sharma, 1974). Provide at least two reasons that justify this discrepancy.
Solution 12.44. a) When Ui = 0 and e & ε are both functions of time t alone, all the terms with spatial derivatives (∂/∂xj) drop out. Thus, the simplified model equations are:
dedt
= −ε and dεdt
= −Cε2ε( )2
e,
where d/dt has replaced ∂/∂t because t is the only independent variable.
b) Use the first equation to eliminate ε from the second to find: − d2edt2
= −Cε21e−dedt
"
#$
%
&'2
,
Insert the suggested solution, e = eot−n , simplify & divide out common factors, and solve for Cε2.
−eo(−n)(−n−1)t−n−2 = −Cε2
1eot
−n (+n)2eo
2t−2n−2 → (n)(n+1) =Cε2 (n)2 , or Cε2 =1+
1n
.
c) Two reasons that justify the difference are as follows. (i) The actual flow includes spatial variations in e and ε so the time-only simplified model equations do not precisely apply to grid turbulence. (ii) The five model constants have been tuned to produce acceptable outcomes in many different flows and this requires trade-offs to be made between accuracy and breadth of applicability. For example, while Cε2 = 1.77 might be appropriate for decaying turbulent flows (like grid turbulence), fitting the "k-ε" turbulence model to other flows involving mean-flow shear, bounding surfaces, and turbulence production may require a larger value for this model constant.
Ut!
random grid!
U!
grid turbulence!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.45. Derive (12.127) from (12.35) with constant density using the definition equalities in (12.128), (12.129), and (12.131). Solution 12.45. Start from (12.35) with constant density (α = 0): ∂uiuj∂t
+Uk
∂uiuj∂xk
+∂uiujuk∂xk
= −uiuk∂Uj
∂xk−ujuk
∂Ui
∂xk−1ρui∂ p∂x j
+uj∂ p∂xi
"
#$$
%
&''− 2ν
∂ui∂xk
∂uj∂xk
+ν∂ 2
∂xk2 uiuj
Rearrange the terms so that the first four are the same as in (12.127): ∂uiuj∂t
+Uk
∂uiuj∂xk
= −uiuk∂Uj
∂xk−ujuk
∂Ui
∂xk− 2ν ∂ui
∂xk
∂uj∂xk
−1ρui∂ p∂x j
+uj∂ p∂xi
"
#$$
%
&''+ν
∂ 2
∂xk2 uiuj −
∂uiujuk∂xk
The fifth term is εij as defined by (12.128). The remaining terms on the right may be rewritten to extract a ∂/∂xk differentiation from everything but the pressure-rate-of-strain tensor.
−1ρui∂ p∂x j
+uj∂ p∂xi
"
#$$
%
&''+ν
∂ 2
∂xk2 uiuj −
∂uiujuk∂xk
= − 1ρ
∂∂x j
ui p− p∂ui∂x j
+∂∂xi
uj p− p∂uj∂xi
"
#$$
%
&''+
∂∂xk
ν∂∂xk
uiuj −uiujuk"
#$
%
&'
= − 1ρ
∂∂xk
ui pδ jk − p∂ui∂x j
+∂∂xk
uj pδik − p∂uj∂xi
"
#$$
%
&''+
∂∂xk
ν∂∂xk
uiuj −uiujuk"
#$
%
&'
= ∂∂xk
ν∂∂xk
uiuj −uiujuk −ui pρδ jk −
uj pρδik
"
#$$
%
&''+
1ρ
p ∂ui∂x j
+ p∂uj∂xi
"
#$$
%
&''
=Mij + Nij
The first equality above follows from the product rule of differentiation and the fact that spatial differentiation and averaging are independent operations that can be performed in either order. Thus, the rearranged version of (12.35) for constant density is:
∂uiuj∂t
+Uk
∂uiuj∂xk
= −uiuk∂Uj
∂xk−ujuk
∂Ui
∂xk−εij +Mij + Nij ,
where (12.129) has been used for Mij and (12.131) has been used for Nij. This final equation is is (12.127)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.46. Derive (12.132) by taking the divergence of the constant-density Navier-Stokes momentum equation, computing its average, using the continuity equation, and then subtracting the averaged equation from the instantaneous equation. Solution 12.46. Start with the constant density NS-momentum equation with the nonlinear term written in flux form [see (4.23)]:
∂ !ui∂t
+∂∂x j
!ui !uj( ) = − 1ρ∂ !p∂xi
+ν∂ 2 !ui∂x j∂x j
,
where as in (12.24) a tilde indicates the full value of a dependent field variable. Take the divergence of this equation (i.e. apply ∂/∂xi to both sides of the equation) to reach:
∂∂t∂ !ui∂xi
+∂∂xi
∂∂x j
!ui !uj( ) = − 1ρ
∂ 2 !p∂xi∂xi
+ν∂ 2
∂x j∂x j
∂ !ui∂xi
The first and final terms are zero in an incompressible flow, leaving ∂ 2 !p∂xi∂xi
= −ρ∂∂xi
∂∂x j
!ui !uj( ) .
Insert the Reynolds decomposition, !ui =Ui +ui and !p = P + p where a capital letter indicates and average field variable, and a lower case letter indicates a turbulent fluctuation, to find:
∂ 2 (P + p)∂xi∂xi
= −ρ∂∂xi
∂∂x j
(Ui +ui )(Uj +uj )( ) , or
∂ 2P∂xi∂xi
+∂ 2p∂xi∂xi
= −ρ∂∂xi
∂∂x j
UiU j +Uiuj +uiU j +uiuj( ) . ($)
Average this equation to reach: ∂ 2P∂xi∂xi
= −ρ∂∂xi
∂∂x j
UiU j −uiuj( ) .
Subtract this from ($), ∂ 2p∂xi∂xi
= −ρ∂∂xi
∂∂x j
Uiuj +uiU j +uiuj −uiuj( ) .
Performing the xj-differentiation on the first two term on the right side: ∂ 2p∂xi∂xi
= −ρ∂∂xi
∂Ui
∂x juj +Ui
∂uj∂x j
+∂ui∂x j
U j +ui∂Uj
∂x j
"
#$$
%
&''− ρ
∂∂xi
∂∂x j
uiuj −uiuj( ) .
The second and fourth terms inside the large parentheses are zero for incompressible flow. Performing the xi-differentiation on remaining terms in the large parentheses leads to:
∂ 2p∂xi∂xi
= −ρ∂Ui
∂x j
∂uj∂xi
− ρ∂ui∂x j
∂Uj
∂xi− ρ
∂∂xi
∂∂x j
uiuj −uiuj( ) .
Since both indices are summed over, i and j can be exchanged in the second term on the right thereby making it identical to the first term on the right. Making this change leads to the final form:
∂ 2p∂xi∂xi
= −2ρ ∂Ui
∂x j
∂uj∂xi
− ρ∂∂xi
∂∂x j
uiuj −uiuj( ) ,
and this is (12.132).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.47. Using the Green's function given in Exercise 12.14 and the properties of homogeneous turbulence, formally solve (12.132) and then use (12.131) to reach (12.133). Solution 12.47. From Exercise 12.14, the Green's function solution of the Poisson equation ∂2p ∂xi
2 = f (x j ) , is:
p(x j ) = −1
4π1
(x j − yj )2
all y∫ f (yj )d
3y = − 14π
1x− yall y
∫ f (y)d3y ,
where the second form merely involves changes in notation. For the present purposes, the inhomogeneous term in (12.132) is:
f = −2ρ ∂Ui
∂x j
∂uj∂xi
− ρ∂∂xi
∂∂x j
uiuj −uiuj( ) ,
thus the formal solution of (12.132) is:
p(x) = − 14π
1x− yall y
∫ −2ρ ∂Ui
∂yj
∂uj∂yi
− ρ∂∂yi
∂∂yj
uiuj −uiuj( )#
$%%
&
'((d
3y .
Using this formal solution, the pressure-rate-of-strain tensor Nij may be written:
Nij ≡pρ
∂ui∂x j
+∂uj∂xi
#
$%%
&
'((=
14π
∂ui∂x j
1x− yall y
∫ 2∂Uk
∂yl
∂ul∂yk
+∂∂yk
∂∂yl
ukul −ukul( )#
$%
&
'(d3y
14π
∂uj∂xi
1x− yall y
∫ 2∂Uk
∂yl
∂ul∂yk
+∂∂yk
∂∂yl
ukul −ukul( )#
$%
&
'(d3y.
Combine terms with like integrands:
Nij ≡pρ
∂ui∂x j
+∂uj∂xi
#
$%%
&
'((=
12π
∂Uk
∂ylall y∫ ∂ui
∂x j+∂uj∂xi
#
$%%
&
'((∂ul∂yk
#
$%
&
'(d3yx− y
+ 14π
∂ui∂x j
+∂uj∂xi
#
$%%
&
'((∂ 2ukul∂yk∂yl
#
$%
&
'(
all y∫ d3y
x− y
− 14π
∂ui∂x j
+∂uj∂xi
#
$%%
&
'((∂ 2ukul∂yk∂yl
#
$%
&
'(
all y∫ d3y
x− y.
The final term is zero because the average of the fluctuating strain rate (the first factor inside the last integral) is zero. This leaves:
Nij =1
2π∂Uk
∂ylall y∫ ∂ui
∂x j+∂uj∂xi
#
$%%
&
'((∂ul∂yk
#
$%
&
'(d3yx− y
+1
4π∂ui∂x j
+∂uj∂xi
#
$%%
&
'((∂ 2ukul∂yk∂yl
#
$%
&
'(
all y∫ d3y
x− y. (&)
For homogeneous turbulence, ∂Uk/∂yl will be independent of position. Thus, it can be taken outside the integrand in the first term and this simplification leads to (12.133):
Nij =1
2π∂Uk
∂xl
∂ui∂x j
+∂uj∂xi
"
#$$
%
&''∂ul∂yk
"
#$
%
&'
all y∫ d3y
x− y+
14π
∂ui∂x j
+∂uj∂xi
"
#$$
%
&''∂ 2ukul∂yk∂yl
"
#$
%
&'
all y∫ d3y
x− y. (12.133)
When the turbulence is inhomogeneous, then (&) is correct and (12.133) represents a local approximation.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.48. Turbulence largely governs the mixing and transport of water vapor (and other gases) in the atmosphere. Such processes can sometimes be assessed by considering the conservation law (12.34) for a passive scalar. a) Appropriately simplify (12.34) for turbulence at high Reynolds number that is characterized by: an outer length scale of L, a large-eddy turnover time of T, and a mass-fraction magnitude of Yo. In addition, assume that the molecular diffusivity κm is at most as large as
€
ν = µ ρ = the fluid’s kinematic viscosity. b) Now consider a simple model of how a dry turbulent wind collects moisture as it blows over a nominally flat water surface (x1 > 0) from a dry surface (x1 < 0). Assume the mean velocity is steady and has a single component with a linear gradient,
€
U j = (Sx2,0,0), and use a simple gradient diffusion model:
€
−u j # Y = ΔUL 0,∂Y ∂x2,0( ) , where ΔU and L are (constant) velocity and length scales that characterize the turbulent diffusion in this case. This turbulence model allows the turbulent mean flow to be treated like a laminar flow with a large diffusivity = ΔUL (a turbulent diffusivity). For the simple boundary conditions:
€
Y (x j ) = 0 for x1 < 0,
€
Y (x j ) =1 at x2 = 0 for x1 > 0, and
€
Y (x j )→ 0 as
€
x2 →∞, show that
€
Y (x1,x2,x3) = exp − 19ζ
3( )ξ
∞
∫ dζ exp − 19ζ
3( )0
∞
∫ dζ where
€
ξ = x2S
ΔULx1
$
% &
'
( )
1 3
for x1,x2 > 0.
Solution 12.48. a) Start with (12.34):
€
∂Y ∂t
+ U j∂Y ∂x j
=∂∂x j
κm∂Y ∂x j
− u j % Y &
' ( (
)
* + + , but not all the terms
are needed at high Reynolds number. Using the length, time, and mass-fraction scales, set
€
Y * = Y Yo ,
€
U j* =U jT L ,
€
t* = t T ,
€
x j* = x j L , and
€
u j " Y *
= u j " Y LYo /T( ), and insert these into (12.34) to find:
€
Yo
T∂Y *
∂t*+
LT
Yo
LU j* ∂Y *
∂x j* =κm
Yo
L2∂ 2Y *
∂x j*∂x j
* −1L
LT
Yo∂∂x j
* u j % Y * .
Divide by Yo and multiply by T to find that the terms on the left side are of order unity along with the second term on the right.
€
∂Y *
∂t*+ U j
* ∂Y *
∂x j* =κm
TL2
∂ 2Y *
∂x j*∂x j
* −∂∂x j
* u j % Y *
Given that κm is at most as large as ν, the coefficient of the first term on the right side is at most as large as 1/Re, where Re is the Reynolds number =
€
L2 νT ; thus, this term can be dropped compared to the three that are of order unity. b) For steady mean flow, the approximate Reynolds-averaged scalar transport equation becomes:
€
U j∂Y ∂x j
= −∂∂x j
u j $ Y ( ).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Insert the given mean velocity U1 = Sx2, and turbulent diffusion model
€
−u2 # Y = ΔUL ∂Y ∂x2( ) :
€
Sx2∂Y ∂x1
= ΔUL ∂2Y ∂x2
2 . (@)
Introduce the similarity variable
€
ξ = x2S
ΔULx1
$
% &
'
( )
1 3
, and note that:
€
∂Y ∂x1
=dY dξ
∂ξ∂x1
= −ξ3x1
dY dξ
,
€
∂Y ∂x2
=dY dξ
∂ξ∂x2
=S
ΔULx1
%
& '
(
) *
1 3dY dξ
, and
€
∂ 2Y ∂x2
2 =S
ΔULx1
$
% &
'
( )
2 3d2Y dξ 2
.
Thus, the approximate scalar transport equation becomes:
€
−Sx23x1
ξdY dξ
= ΔUL SΔULx1
%
& '
(
) *
2 3d2Y dξ 2
, which simplifies to
€
−13ξ 2
dY dξ
=d2Y dξ 2
after some manipulation of the factors. Integrate and exponentiate this last equation to find:
€
dY dξ
= Aexp − 19ξ 3
$
% &
'
( ) , and integrate again to reach
€
Y (ξ) = A exp − 19ζ
3( )dζ0
ξ∫ + B ,
where ζ is just an integration variable. The boundary conditions are: (i)
€
Y (x j ) = 0 for x1 < 0, (ii)
€
Y (x j ) =1 at x2 = 0 for x1 > 0, and (iii)
€
Y (x j )→ 0 as
€
x2 →∞. Condition (i) can be set immediately since
€
Y = 0 is a solution of the field equation (@). Condition (ii) implies:
€
Y (0) = A exp − 19ζ
3( )dζ0
0∫ + B =1, or 0+B = 1 for x1 > 0.
Condition (iii) implies:
€
Y (∞) = A exp − 19ζ
3( )dζ0
∞∫ + B = 0 , or 0+B = 1 for x1 > 0.
Therefore, B = 1 and
€
A = − exp − 19ζ
3( )dζ0
∞∫( )
−1, so
€
Y (ξ) =1−exp − 1
9ζ3( )dζ0
ξ∫exp − 1
9ζ3( )dζ0
∞∫=
exp − 19ζ
3( )dζ0
∞∫ − exp − 1
9ζ3( )dζ0
ξ∫
exp − 19ζ
3( )dζ0
∞∫=
exp − 19ζ
3( )dζξ
∞∫exp − 1
9ζ3( )dζ0
∞∫.
Thus, the final form is:
€
Y (x1,x2,x3) =0 for x1 < 0,x2 > 0
exp − 19ζ
3( )ξ
∞
∫ dζ exp − 19ζ
3( )0
∞
∫ dζ for x1,x2 > 0
'
( )
* )
+
, )
- ) where
€
ξ = x2S
ΔULx1
$
% &
'
( )
1 3
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.49. Estimate the Monin–Obukhov length in the atmospheric boundary layer if the surface stress is 0.1 N/m2 and the upward heat flux is 200 W/m2. Solution 12.49. First compute the friction velocity:
u* =τ wρ=
0.1Pa1.23kgm−3 = 0.289ms
−1 .
The heat flux determines
€
w " T
€
w " T =QρCp
=200Wm−2
1.2kgm−3(1004m2s−2K−1)= 0.166ms−1K .
Thus, the the Monin–Obukhov length is:
€
LM =u*3
καgw $ T =
(0.289ms−1)3
(0.41)(3.4 ×10−3K−1)(9.81ms−2)(0.166ms−1K)=10.6m .
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 12.50. Consider one-dimensional turbulent diffusion of particles issuing from a point source. Assume a Gaussian Lagrangian correlation function of particle velocity
€
r(τ) = exp −τ 2 tc2{ },
where tc is a constant. By integrating the correlation function from τ = 0 to ∞, find the integral time scale Λt in terms of tc. Using the Taylor theory, estimate the eddy diffusivity at large times t/Λt >> 1, given that the rms fluctuating velocity is 1 m/s and tc = 1 s. Solution 12.50. Start from the definition of the integral time scale:
€
Λ t = r(τ)dτ0
∞
∫ = exp − τ2
tc2
'
( )
*
+ , dτ
0
∞
∫ = exp − τ2
tc2
'
( )
*
+ , dτ
0
∞
∫ = tc exp −β 2( )dβ0
∞
∫ = tcπ2
= 0.886tc .
From (12.129),
€
DT ≅ u2Λ t = (1ms−1)2 ⋅ 0.886(1s) = 0.886m2s−1 .
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