Laminar and Turbulent developing flow
with/without heat transfer over a flat plate
Introduction
The purpose of the project was to use the FLOLAB software to model the laminar and
turbulent flow over a flat plate. The results of the model under different flow conditions
were compared with the theoretical predictions. Detailed analysis of the data obtained
was done to characterize the flow parameters like Nusselt Number, Skin Friction
Coefficient, shear layer thickness, velocity and temperature. Effect of change in geometry
and initial flow conditions were also studied.
Theory
Boundary Layer Definition
1. Boundary layer thickness ( δ ): defined as the distance away from the surface where
the local velocity reaches to 99% of the free-stream velocity, that is u(y= δ )=0.99U∞.
Somewhat an easy to understand but arbitrary definition.
2. Displacement thickness ( δ * ): Since the viscous force slows down the boundary layer
flow, as a result, certain amount of the mass has been displaced (ejected) by the presence
of the boundary layer (to satisfy the mass conservation requirement). Imagine that if we
displace the uniform flow away from the solid surface by an amount, such that the flow
rate with the uniform velocity will be the same as the flow rate being displaced by the
presence of the boundary layer.
The essence of the “boundary layer approximation” is that the shear layer is thin, δ << x .
This is true if Re>>1.In the boundary layer u i.e. the velocity in the x direction scales
with L and v,the velocity in the y direction scales with δ .Therefore u>>v.Also the
∂u
∂u
∂v
∂v
following approximations like
<<
and
<<
are made to simplify the Navier
∂y
∂y
∂x
∂x
Stokes equations to the corresponding boundary layer forms.
∂u ∂u
+
= 0.....continuity
∂x ∂y
u
dU e
∂u
∂u
∂ 2u
+v
= Ue
+ ν 2 ......x − momentum
∂x
∂y
dx
∂y
Steady, incompressible, 2-D
laminar flow.
∂T
∂T
∂ 2T
∂u
+ v ) = k 2 + µ ( ) 2 .......energy
ρC p (u
∂x
∂y
∂y
∂y
The term U e
dU e
1 dp
=−
= 0 for flow over a flat plate as U e = constant.
dx
ρ dx
Blasius found a celebrated solution for Re>>1 with a similarity variable η = y
Ue
.
2νx
Using this we have
u = U e f ' (η )
v=
νU e
(ηf ' − f )
2x
Substituting these variables into the x-momentum equation of the boundary is going to
give us the following ODE as a function of η .
f ''' + ff '' = 0
Assuming no slip conditions we have u(x,0)=v(x,0)=0 and
The free stream merge condition u(x,∞)=Ue
These convert to
f ' (0) = f (0) = 0
f ' (∞ ) = 1
Solving the Blasius Equation we obtain the following results:Parameters
θ
Exact from Blasius
0.664
Re x
x
δ*
x
δ 99%
x
1.721
Re x
5
Re x
C f Re x
0.664
C D Re x
1.328
Definitions: For Laminar,
C f = 2τ w / ρU e
where τ w = µ
2
∂u
|w
∂y
L
1
C D ( L) = ∫ C f dx = 2C f ( L)
L0
Nu = q w x / k (Tw − Te ) = 0.332 Re x
For turbulence,
C f = 0.455 / ln 2 (0.06 Re x )
Nu = 0.0296 Re x
4/5
Pr 1 / 3 ...........0.6 < Pr < 60
Flow Conditions: LAMINAR FLOW: WATER: -
1/ 2
Pr 1 / 3
Re
Heat Transfer
Length of Plate
105
105
105
5*105
5*105
1500
1500
104
104
Tw=400 K,Tin=300 K
Tw=300 K,Tin=300 K
Tw=400 K,Tin=300 K
Tw=400 K,Tin=300 K
Tw=300 K,Tin=300 K
Tw=400 K,Tin=300 K
Tw=300 K,Tin=300 K
Tw=400 K,Tin=300 K
Tw=300 K,Tin=300 K
1
1
5
1
1
1
1
1
1
Re
Heat Transfer
Length of Plate
105
105
105
5*105
5*105
100
100
106
106
Tw=400 K,Tin=300 K
Tw=300 K,Tin=300 K
Tw=400 K,Tin=300 K
Tw=400 K,Tin=300 K
Tw=300 K,Tin=300 K
Tw=400 K,Tin=300 K
Tw=300 K,Tin=300 K
Tw=400 K,Tin=300 K
Tw=300 K,Tin=300 K
1
1
5
1
1
1
1
1
1
AIR : -
Re = 106 falls in the transition zone. To compare the laminar and the turbulent models
,we ran this case under both the conditions.
Results Laminar Case
Velocity Analysis:
Fluid: Water
V in let velo c ity= 0.0 8954 m /sec ,R e= 1000 00
0 .0 25
X velo c ity at inle t
X velo c ity at X= 0 .2 m
X velo c ity at X= 0 .4 m
X velo c ity at X= 0 .6 m
0.02
X velo c ity at X= 0 .8 m
X velo c ity at X= 0 .9 m
X velo c ity at X= o utlet
0 .0 15
Y
S h ear s tre s s a nd h enc e C f
de c rea s es as X
inc reas e s.T he value o f
D elta inc reas es .
0.01
0 .0 05
0
0
0.0 5
0.1
X v elo city
0.15
R e = 1 5 0 0 ,V in le t= 0 .0 0 1 3 4 5 m /s
0 .6
X v e lo c ity a t in le t
X v e lo c ity a t X = 0 .2
0 .5
X v e lo c ity a t X = 0 .4
X v e lo c ity a t X = 0 .6
X v e lo c ity a t X = 0 .8
0 .4
X v e lo c ity a t X = 0 .9
Y
X v e lo c ity a t o u tle t
0 .3
0 .2
0 .1
0
0
0 .0 0 0 5
0 .0 0 1
0 .0 0 1 5
0 .0 0 2
X v e lo c ity
Observations:
- Shear stress and hence Cf decreases as x increases, this is because as Reynolds #
increases with x, Cf goes down.
- δ becomes thicker as x increases.
- As x increases we observe that the velocity exceeds the inlet velocity. It shows a
problem in Flowlab’s approach in solving the velocity profile equations.
- In the case of the low Re # (1500), δ is much thinner than a higher Re #
Fluid: Air
Re=10^5 velocity=1.34m/s
0.02
0.018
inlet
x=0.2
x=0.4
x=0.6
x=0.9
x=0.8
outlet
0.016
y direction
0.014
0.012
0.01
0.008
0.006
0.004
0.002
0
-0.4
0.1
0.6
1.1
1.6
velocity
Re=10^2 velocity=0.00134m/s
0.5
0.45
inlet
x=0.2
x=0.4
x=0.6
x=0.8
x=0.9
outlet
0.4
y direction
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.0005
0.001
0.0015
0.002
velocity
Observations:
- Same as in the case of water
- For Re=100,i.e under creep flow the velocity profiles does not obey the Boundary
layer theory any more.
Analysis of the Coefficient of Skin Friction:
Fluid: Air/Water
skin fric tio n coefficient W a te r
0.1
0.09
C f R e=100000
C f R e=1500
0.08
C f R e=500000
C f R e=10000
0.07
Cf
0.06
0.05
0.04
0.03
0.02
0.01
0
0
0.2
0.4
0.6
X
0.8
1
1.2
Skin friction coefficient Water
0.7
0.6
0.5
Re=10^2
Re=10^5
Re=5*10^5
Re=10^6
Cf
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
1.2
x direction
Observations
- With the increase of x the value of Cf decreases as Rex increases. Cf ∝ 1/√Rex
- As the value of inlet velocity is decreased so that the Re decreases the skin
friction coefficient goes up.
- For low Re i.e at 1500 or 100 the Cf does not follow the relation C f Re x =0.664
as predicted by the Boundary layer solution by Blasius.
Cf W ater at Re=10^5
0.03
Cf Air at Re=10^5
0.03
0.025
0.025
Re=10^5
0.02
Cf Re=100000
Cf
Cf
0.02
0.015
0.015
0.01
0.01
0.005
0
0.005
0
0.2
0.4
0.6
x
0
0
0.2
0.4
0.6
X
0.8
1
1.2
0.8
1
1.2
Observation
- For two different fluids water and air with varying viscosity and density if the
Re is maintained constant at a value say 10^5 then Cf gives the same profile at
different x as Cf depends on Re only. This shows that Re drives the problem in
incompressible fluid flow.
Comparison of Flowlab Cf values with Theoretical (water)
R e = 1 5 0 0 ,V in le t= 0 .0 0 1 3 4 5 m /s
Wall skin friction coefficient
W a ll s kin fric tio n ,R e= 10 00 00 ,V in le t= 0 .0 89 54 m /
0 .0 09
0 .1 8
0 .0 08
0 .1 6
0 .0 07
w all s k in fric tio n
0 .0 06
b la us ius
p red ic tio n
0 .0 05
0 .1 4
s k in f r ic t i o n
c o e f f i c ie n t
0 .1 2
P r e d ic t e d s k i n
f r i c t io n
0 .1
0 .0 04
0 .0 8
0 .0 03
0 .0 6
0 .0 02
0 .0 4
0 .0 01
0 .0 2
0
0
0
0 .2
0 .4
0.6
X
0.8
1
0
0 .5
1
X
Observations
- Cf matches with the theoretical value quite closely for Re=10^5.But at lower
values of Re say 1500 the boundary layer theory has got large discrepancies.
So the theoretical values do not match the flolab generated results
1 .5
Comparison of Flolab Cf values with Theoretical values(Air)
Cf Re=10^5
0.009
0.008
0.007
Cf
0.006
0.005
flowlab
theoretical
0.004
0.003
0.002
0.001
0
0
0.2
0.4
0.6
0.8
1
1.2
x direction
Cf Re=10^2
0.7
0.6
0.5
flowlab
0.3
theoretical
Cf
0.4
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
1.2
x direction
Observations:
- At low Reynolds number the boundary layer theory breaks down,so the value of Cf does
not match with that predicted by B.L. The deviations are appreciable.
Nusselt Number with/without heat transfer (water)
N u,vario us R eyn old s n um ber,w ith heat tran sfer
500
450
400
350
R e=100000
R e=10000
300
R e=500000
Nu
R e=1500
250
200
150
100
50
0
0
0.2
0.4
0.6
0.8
1
1.2
X
Observations
1/ 2
- Nu = q w x / k (Tw − Te ) = 0.332 Re x Pr 1 / 3 for laminar case. Here for a constant Pr as Re
increases with x,the nusselt number increases as well as √Rex
Nusselt num ber w ithout heat transfer
350
300
250
Re=100000
Nusselt
Re=1500
200
150
100
50
0
0
0.2
0.4
0.6
0.8
1
1 .2
X
Observations
- With Twall=Te the profile doesn’t obey the Polhaussen Pr1/3 law.
Comparison of Nu with theoretical values(water)
W a ll n u s s e lt n u m b e r ,R e = 1 0 0 0 0 0 ,T w = 3 0 0 K , T i n le t = 4 0 0 K
250
Nusselt number
200
150
100
W a ll n u s s e lt n u m b e r
N u s s e lt n u m b e r p r e d ic te d b y
p o lh a u s s e n
50
0
0
0 .2
0 .4
0 .6
0 .8
1
X (m )
Observations
- The value from Flolab closely matches the theoretical value as calculated by
1/ 2
Polhaussen. Nu = q w x / k (Tw − Te ) = 0.332 Re x Pr 1 / 3
Nusselt no Re=100000 without heat transfer
350
The mismatch is due to the
fact that at Tw=Te,the
theoretical expression for
nusselt number
0.332*(Rex)^0.5*(Pr)^0.3333
is not valid
300
Nusselt Number
250
200
plate-wall
predicted
150
100
50
0
0
0.2
0.4
0.6
Y
0.8
1
1.2
Nusselt Number for Air
Nu number with heat transfer
350
300
Nu number
250
Re=10^2
Re=10^5
Re=5*10^5
Re=10^6
200
150
100
50
0
0
0.2
0.4
0.6
0.8
1
1.2
x direction
Nu number no heat transfer
500
400
Nu
300
Re=10^2
Re=10^5
Re=5*10^5
Re=10^6
200
100
0
0
0.2
0.4
0.6
-100
x direction
0.8
1
1.2
Nu number at Re=10^5 with heat transfer for
air
250
120
200
100
150
80
R e=100000
Nu
Nu
N u at R e=10^5 w ith h eat transfer W ater
100
60
Re=10^5
flowlab
40
50
20
0
0
0
0.5
1
X
1 .5
0
0.2
0.4
0.6
0.8
x direction
Observations
- Nu varies as Pr1/3.Pr of air is approx=0.72 and Pr for Water=6.So the
Nuwater/Nuair=2.02.From the chart it is obvious that The flolab results do agree
with these for a constant Re=10^5
1
1.2
Temperature Profiles for Water
T e m p e r a t u r e p r o f ile s R e = 1 0 ^ 5 w it h h e a t t r a n s f e r
0 .0 1 5
0 .0 1 4
0 .0 1 3
0 .0 1 2
0 .0 1 1
0 .0 1
T e m p e r a t u r e a t d if f e r e n t X lo c a t io n s
0 .0 0 9
T e m p e ra tu re a t X = 0 .2 m
T e m p e ra tu re a t X = 0 .4 m
0 .0 0 8
Y
T e m p e ra tu re a t X = 0 .6 m
T e m p e ra tu re a t X = 0 .8 m
0 .0 0 7
T e m p e ra tu re a t X = 0 .9 m
T e m p e ra tu re a t X = 1 m
0 .0 0 6
0 .0 0 5
0 .0 0 4
0 .0 0 3
0 .0 0 2
0 .0 0 1
0
290
300
310
320
330
340
350
360
370
380
390
te m p e ra tu re
Observations:
- The profiles closely matches the profile as found by polhaussen
- The temperature is almost constant when Tw=Tin
400
410
Re=10^5 without heat transfer
300.001
300
T
299.999
inlet
x=0.2
299.998
x=0.4
299.997
x=0.6
x=0.8
299.996
x=0.9
outlet
299.995
299.994
299.993
0
0.1
0.2
0.3
0.4
0.5
0.6
y direction
T profile at Re=1500 w ith heat transfer
410
390
370
T at X=0
T at X=0.2
350
T at X=0.4
T
T at X=0.6
T at X=0.8
330
T at X=0.9
T at X=1
310
290
270
250
0
0.01
0.02
0.03
0.04
0.05
Y
0.06
0.07
0.08
0.09
0.1
Effect of length change of the plate (water)
W a l l f r ic t io n , r e = 1 0 0 0 0 0
0 .0 0 9
0 .0 0 8
0 .0 0 7
0 .0 0 6
W a ll f r ic t io n a t L = 5 m
w a ll fr ic tio n a t L = 1 m
Cf
0 .0 0 5
0 .0 0 4
0 .0 0 3
0 .0 0 2
0 .0 0 1
0
0
0 .2
0 .4
0 .6
0 .8
1
1 .2
X
N u s s e lt n u m b e r a t R e = 1 0 ^ 5
250
200
n u s s e lt n u m b e r fo r L = 5 m
Nusselt No
N u s s e lt w ith L = 1 m
150
100
50
0
0
1
2
3
X /L
4
5
Observations
- if the length of the plate and inlet velocity is changed such that Re is constant
then Nu and Cf does not change at all as evident from the above graphs
Temperature Profiles for Air
Temperature Re=10^2
410
390
inlet
x=0.2
x=0.4
x=0.6
x=0.8
x=0.9
outlet
Temperature
370
350
330
310
290
270
250
0
0.1
0.2
0.3
0.4
0.5
y direction
Re=10^5 Temperature
300.001
300
inlet
Temperature
299.999
x=0.2
299.998
x=0.4
299.997
x=0.6
299.996
x=0.8
x=0.9
299.995
outlet
299.994
299.993
0
0.1
0.2
0.3
0.4
0.5
0.6
y direction
Observations:
- The profiles closely matches the profile as found by polhaussen
- The temperature is almost constant when Tw=Tin
Re=106 is a typical case of transition, although the laminar model shows large
inaccuracies for Re=5*105.So we have used both the laminar as well as the turbulent
model to run a case where the fluid is air for Re=105.
Cf Re=10^6
0.007
0.006
Cf
0.005
0.004
laminar flow
0.003
turbulent flow
0.002
0.001
0
0
0.5
1
x direction
Nu number Re=10^6
1800
1600
1400
Nu
1200
1000
lamina flowr
turbulent flow
800
600
400
200
0
0
0.2
0.4
0.6
0.8
X direction
1
1.2
Observations:
- Turbulent model predicts a much higher Nu and Cf than the laminar counterpart. We
believe that at 106 the turbulent model must be the better one to solve the flow problem.
Sample velocity contour as generated by FLOLAB at different X locations
Contour plot from flolab
Vector Plot from flolab
Streamlines as plotted by flolab
Turbulent Theory : Approach suggested by Kestin and Persen is to assume that the streamwise
velocity u in the boundary layer is correlated by inner variables
u ( x, y )
yv * ( x)
= u + ≈ f ( y + ), y + =
*
ν
v ( x)
These velocity assumptions when substituted directly into the streamwise
boundary layer momentum equation with dU e / dx = 0
∂u
∂u 1 ∂τ
u
+v
=
∂x
∂y ρ ∂y
The turbulent boundary layer begins at x=0 and the integral evaluated
algebraically
λ
Re x = ∫ Gdλ
0
G in its practical range 20<λ<40 is G(λ)=8.0e0.48λ
0.455
ln (0.06 Re x )
So from Prandtl’s seventh power law yields
0.058
is not accurate. The inner variable approach can be extended readily
Cf ≈
1/ 5
Re x
to variable pressure gradients.
∴ Cf for flat plate ≅
2
Local Skin friction on a smooth flat plate for a turbulent flow showing several
theories
The transition occurs from laminar to turbulent Re=5e5.
TURBULENT FLOW CONDITIONS: WATER
Re
Heat Transfer
2.23*109
2.23*109
5*106
5.586*108
107
107
107
1500
Tw=400 K,Tin=1000 K
Tw=400 K,Tin=400 K
Tw=300 K,Tin=300 K
Tw=300 K,Tin=400 K
Tw=300 K,Tin=1000 K
Tw=300 K,Tin=400 K
Tw=300 K,Tin=400 K
Tw=300 K,Tin=300 K
Re
Heat Transfer
107
5*106
107
106
5*106
107
107
107
Tw=300 K,Tin=400 K
Tw=300 K,Tin=400 K
Tw=300 K,Tin=400 K
Tw=300 K,Tin=300 K
Tw=300 K,Tin=300 K
Tw=300 K,Tin=300 K
Tw=300 K,Tin=1000 K
Tw=300 K,Tin=400 K
Length of
Plate
1
1
1
1
1
1
5
1
AIR
Length of
Plate
1
1
1
1
1
1
1
5
Results Turbulent Case
Velocity Analysis:
Fluid: Water
Re=5*10^6,Vinlet=4.477m/s,T=300-300)
0.1
0.09
y
0.08
inlet
0.07
0.2L
0.06
0.4L
0.05
0.6L
0.04
0.8l
0.03
0.9l
outlet
0.02
0.01
0
3.5
3.7
3.9
4.1
4.3
4.5
4.7
velocity
Re=2.23*10^9,Vinlet=2000m/s,T=1000-400)
0.1
0.09
0.08
inlet
y
0.07
0.2l
0.06
0.4l
0.05
0.6l
0.04
0.8l
0.9l
0.03
outlet
0.02
0.01
0
1900
2000
velocity
Observations:
- Shear stress and hence Cf decreases as x increases, this is because as Reynolds #
increases with x, Cf goes down.
- δ becomes thicker as x increases.
- As x increases we observe that the velocity exceeds the inlet velocity. It shows a
problem in Flowlab’s approach in solving the velocity profile equations.
- The difference with the laminar case is that the shear layer in turbulence is much
thinner than its laminar counterpart.
Air:
Re= 10^7, V inlet= 134.2 m/s, T= 300-400
0.1
0.09
0.08
Inlet
0.07
x=0.2
Y
0.06
x=0.4
0.05
x=0.6
0.04
x=0.8
x=0.9
0.03
Outlet
0.02
0.01
0
90
95
100
105
110
115
120
125
130
135
140
Velocity
Re= 5*10^6, Vinlet=67.14286, T=300-400
0.04
0.035
0.03
Inlet
x=0.2
Y
0.025
x=0.4
0.02
x=0.6
x=0.8
0.015
x=0.9
Outlet
0.01
0.005
0
30
35
40
45
50
Velocity
55
60
65
70
Analysis of the Coefficient of Skin Friction:
Fluid: Air/Water
Cf vs Reynolds
0.005
0.004
5*10^6
0.003
Cf
10^7
5.5*e^8
0.002
2.23*e^9
0.001
0
0
0.2
0.4
0.6
0.8
1
1.2
x
Cf Vs Reynolds
0.007
0.006
0.005
Re 5*10^6
Re 10^6
Re 10^7
Cf
0.004
0.003
0.002
0.001
0
0
0.2
0.4
0.6
X
0.8
1
1.2
Observations
- With the increase of x the value of Cf decreases as Rex increases.
0.455
- Cf ∝ 2
ln (0.06 Re x )
- As the value of inlet velocity is decreased so that the Re decreases the skin
friction coefficient goes up.
Re=10^7,Vinlet=8.954m/s water
0.0045
0.004
0.0035
Cf
0.003
0.0025
flowlab
theoretical
0.002
0.0015
0.001
0.0005
0
0
0.2
0.4
0.6
0.8
1
1.2
x
Cf
Re= 10^7, Vinlet=134.28 m/s air
0.0045
0.004
0.0035
0.003
0.0025
0.002
0.0015
0.001
0.0005
0
Flowlab
Prandtl
0
0.2
0.4
0.6
X
0.8
1
1.2
Observations:
As the Re is the same the values of Cf remains unaltered immaterial of the fluid.
Comparison of Flowlab Cf values with Theoretical (water)
Re=2.2*10^9,Vinlet=2000m/s,water
0.002
Cf
0.0015
flowlab
0.001
theoretical
0.0005
0
0
0.2
0.4
0.6
0.8
1
1.2
x
Cf
Re= 10^7, Vinlet=134.28 m/s, air
0.0045
0.004
0.0035
0.003
0.0025
0.002
0.0015
0.001
0.0005
0
Flowlab
Prandtl
0
0.2
0.4
0.6
0.8
1
1.2
X
Observations
- Cf matches with the theoretical value quite closely for all Re values.
Nusselt Number with/without heat transfer (water)
Nu
Nu vs Re(with heat transfer)
6000000
5500000
5000000
4500000
4000000
3500000
3000000
2500000
2000000
1500000
1000000
500000
0
10^7
5.586e^8
2.233e^9
0
0.2
0.4
0.6
0.8
1
1.2
x
Nu vs Re(without heat transfer)
5000000
4500000
4000000
3500000
Nu
3000000
10^7
5.583e^8
2.23e9
2500000
2000000
1500000
1000000
500000
0
0
0.2
0.4
0.6
x
0.8
1
1.2
Observations:
4/5
Nu = 0.0296 Re x Pr 1 / 3 ...........0.6 < Pr < 60 for turbulent case. Here for a constant Pr as
Re increases with x, the nusselt number increases as well as Re4/5.Without heat transfer ,
the empirical law of Nu is not appropriate.
Re=2.2*10^9,Vinlet=2000m/s,water without heat transfer
5000000
4500000
4000000
3500000
Nu
3000000
flowlab
2500000
theoretical
2000000
1500000
1000000
500000
0
0
0.2
0.4
0.6
0.8
1
1.2
x
Re=10^7,Vinlet=8.954m/s,water with heat transfer
45000
40000
35000
Nu
30000
25000
Flowlab
20000
Theory
15000
10000
5000
0
0
0.2
0.4
0.6
x
0.8
1
1.2
Observations:
4/5
Nu = 0.0296 Re x Pr 1 / 3 ...........0.6 < Pr < 60 for turbulent case. For without heat transfer
case the deviation from the empirical law is much more substantial than the case with
heat transfer.
Fluid Air:
Nu Vs Re Temp with heat transfer
14000
12000
Nu
10000
Re 10^6
Re 5*10^6
Re 10^7
8000
6000
4000
2000
0
0
0.2
0.4
0.6
0.8
1
1.2
X
Re= 10^7, Vinlet= 134.28m/s, with heat tranfer
14000
12000
Nu
10000
8000
Flowlab
Theory
6000
4000
2000
0
0
0.2
0.4
0.6
X
0.8
1
1.2
Temperature Profiles for Water
Re=2.233*10^9,Vinlet=2000m/s,T=400-1000
0.1
0.09
0.08
inlet
y
0.07
0.2l
0.06
0.4l
0.05
0.6l
0.04
0.8l
0.9l
0.03
outlet
0.02
0.01
0
395
397
399
401
403
405
407
409
Temp
Re=10^7, Vinlet=8.954m/s,with heat transfer
0.1
0.09
0.08
inlet
0.2l
0.4l
0.6l
0.8l
0.9l
outlet
0.07
y
0.06
0.05
0.04
0.03
0.02
0.01
0
560
570
580
590
600
610
temp
620
630
640
650
Observations: -
-
The thickness of the thermal boundary layer in the case of turbulence reduces
as compared to the laminar case.
Fluid Air
Re= 5*10^6, Vinlet= 67.14, T= 300-400
0.1
0.09
0.08
inlet
0.07
x=0.2*l
Y
0.06
x=0.4*l
0.05
x=0.6*l
0.04
x=0.8*l
x=0.9*l
0.03
outlet
0.02
0.01
0
290
300
310
320
330
340
350
Temp
Observations: -
-
The thickness of the thermal boundary layer in the case of turbulence reduces
as compared to the laminar case.
Conclusions: -
-
The Flowlab shows fairly good agreement in all respects as compared to the
theoretical values obtained by the boundary layer theory.
However it is not possible to get δ* , δ ,θ directly from the flowlab results.
The overshoot in the velocity profiles above the inlet velocity also couldn’t
be explained .
Y
Full solution of the Navier Stokes
equation.
Boundary layer solution.
x
The Flowlab solves the full Navier Stokes Equation