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8/9/2019 CFX Laboratories 4 and 5 (2nd Version)- Simon Amboise and Vianney Kieken
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Laboratories 4 & 5Calculate the turbulent flow of nitrogen through thebend pipe with circular cross-section
Two ways in order to calculate the turbulent flow of
nitrogen : one in the bend pipe wit circular cross section,
the other with the half of the bend pipe with circular cross
section (we used the symmetry condition). Comparison
between these two methods.
AMBOISE Simon, KIEKEN Vianney
06/04/2008
Group: Monday 12-15
Simon Amboise
Vianney Kieken
Weronika Krakowiak
Jacek Kubas
Thomas Lehut
Adam Raczyoski
Mariusz Suwaa
Jarsaw TazbirBartsz Wgrwski
Krzysztof Wojciechowski
Pitr Wuka
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SommaireIntroduction : Pre-Processing..................................................................................................................3
Slve.....4
Post-Prcessing.......5
1) Plane of the pipe.................................................................................................................... 5
a) Static and total pressures ................................................................................................... 5
b) The Velocity ....................................................................................................................... 5
c) Static and Total temperature .............................................................................................. 6
2) Velocity vectors plots ............................................................................................................. 6
3) Streamlines (Total pressure) ................................................................................................... 8
4) Chart plot of static and total pressure along the axis of the pipe ............................................... 9
a) Static pressure along the axis of the pipe............................................................................. 9
b) Total pressure along the axis of the pipe ........................................................................... 10
5) Chart plot of velocity at the end of the straight part of the channel ................................ ........ 11
6) Quantitative analysis............................................................................................................ 12
a) Comparison between the velocities................................................................................... 12
b) Comparison with the pressure .......................................................................................... 12
c) Force values..................................................................................................................... 13
d) Static temperature ........................................................................................................... 13
e) Friction factors................................................................................................................. 14
1) Pressure drop in circular pipe ........................................................................................ 14
2) Blasius equation ........................................................................................................... 14
3) Conclusion: Friction factors ........................................................................................... 14
4) Results......................................................................................................................... 15
5) Hydrodynamic reaction calculation................................ ................................................ 16
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Introduction: Pre-processing
For the fourth and the fifth laboratories, we studied the same geometry (the dimensions are in mm):
There is a flow of nitrogen (N2) ideal gas. In the domain definition, we chose Total energy
heat transfer model and k- turbulence mdel.
We obtain these geometries; the geometry of the fifth laboratory is the half of the geometry of the
fourth laboratory (we will use the symmetry condition).
The boundary conditions at inlet:
total pressure: 1.2 bar,
total temperature: 20C,
turbulence intensity: 1%,
Laboratory 4 Laboratory 5
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The boundary conditions at outlet:
Mass flow rate corresponding to: Re = 50 000, 75 000, 100 000, 125 000, 150 000, 175 000.
Density of N2:
Known Data: Cp = 1040 J.K -1.kg-1 ; =n=1,4 ;
Mass flow rate:
Known Data: = 1.77 e-05 kg.m-1
.s-1
;
characteristic length L = Diameter = 20 mm
Re 50000 75000 100000 125000 150000 175000
Mass flow rate (kg.s-1) for lab.4 0,0139 0,0209 0,0278 0,0348 0,0417 0,0487
Mass flow rate (kg.s-1) for lab.5 0,0070 0,0104 0,0139 0,0174 0,0209 0,0243
The mass flow rate for the fifth laboratory is divided by two because we admit the symmetry.
We put the symmetry condition at the symmetry plane in the fifth laboratory.
Our convergence criteria is: MAX RES = 1. E-4
Solve
This picture is an example of the solve results.
There are 3 curves for the momentums, one
for the mass (and another for energy but it is
not shown here). These curves are nearly the
same for the fifth laboratory. For the both,
the RSM and MAX residuals can be considered
good (MAX residual is near 1. e-4). Theunbalanced mass, momentum and energy
flows are negligible (
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Post-Processing
1)Plane of the pipea) Static and total pressures
The graphics are the same for the two studies (lab 4 and lab 5).
Static Pressure (Re = 75000) Total Pressure (Re = 75000)
The total pressure picture shows us that the wall and the bend of the pipe increase the
energy wasting (drop of the total pressure, boundary layer, friction). The centrifugal strengths
generate a radial gradient of pressure: the static pressure picture shows us this statement. The static
pressure is higher in the outer radius of the bend than in the inner one.
b) The VelocityThe graphics are nearly the same for the two studies (lab 4 and lab 5).
The profile of the velocity is nearly the same as
the profile of total pressure. The velocity vectorcan be split in two components: one in the
stream wise direction and one in the cross
sectin (its called secndary flw). The flow
goes up to the outer radius of the bend.
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c) Static and Total temperatureThe graphics are the same for the two studies (lab 4 and lab 5).
Static Temperature (Re = 75000) Total Temperature (Re = 75000)
The highest static temperature is in the outer radius of the bend. The profiles of the
temperature are nearly the same as the profiles of the pressure. In the bend, where the pressure is
high, the temperature is high and vice versa.
2)Velocity vectors plotsThe graphics are nearly the same for the two studies (lab 4 and lab 5).
At the pipe inlet (Re = 75000) In front of bending (Re = 75000)
At the pipe inlet, the velocity is a constant. In front of bending, the velocity at the wall decreases but
the velocity in the inner radius of the bend increases.
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At downstream the bending, the tangential component of the velocity is very important on all the
cross section. During the straight pipe downstream the bending; the tangential component of thevelocity decreases and the maximum velocity is concentrating more and more in the outer radius of
the pipe.
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3)Streamlines (Total pressure)
On this picture we may bserve the mvement f the particles. At first, particles dnt rtate
(as it is shown by the ribbons). Their rotational movement starts in the bend part of the pipe, just
before the particles leave the straight part of the pipe. Some particles are not affected by the bend
(red streamlines). The others have their trajectory modified in the second straight part.
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4)Chart plot of static and total pressure along the axis of the pipea) Static pressure along the axis of the pipe
On this graph, for the first straight part of the pipe (0 to 0,05 m), we may observe the static
pressure drop, which is almost linear. There also appears the energy dissipation. For the curved part
of the pipe, we observe as well the pressure drop; however, the dissipation is greater than before. It
is caused by the appearance f the bending n the fluids path. For the second straight part of the
pipe (0,17 to 0,22 m), the static pressure drop as the same way as the first straight part of the pipe
(the drop is about 100 Pa during the 0,05 meters).
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b) Total pressure along the axis of the pipe
On this chart we may bserve the ttal pressures behavir. At the beginning, the line is
horizontal; there are no boundary layers, no secondary flows and no dissipation visible (along the
axis of the pipe). Due t the appearance f the bending n the fluids path, the dissipatin and thetotal pressure drop appear. Before the outlet, the total pressure increases lightly.
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5)Chart plot of velocity at the end of the straight part of thechannel
On this chart (fr Re= 75000), we may bserve the velcitys behavir. We have the lowest
velocity at the wall (22 m.s-1
for x = 0,09 m and 28 m.s-1
for x = 0,11m). Between x = 0,09 m and x =
0,095 m, the velocity increases. Then we may observe a plateau with the value is 46 m.s-1
until x =
0,102 m. Then the velocity increases again in the value 56 m.s-1
with another plateau. Due to the wall,
from x = 0,108 mm, we may observe a drop of the velocity. Due to the bend, the velocity at the
utlet isnt cnstant and its higher in the uter radius f the pipe.
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6)Quantitative analysisa)Comparison between the velocities
Fr each Reynldss number, the variatin between the inlets velcity and the utlets
velocity is lower than 5%. The velocity increases until the end of the bending part, after that, it
reduces lightly. We may bserve when the Reynldss number is twice higher; the velocity is lightly
higher than twice. For the two laboratories, the values are nearly the same.
b)Comparison with the pressureAt the inlet, we have the same boundary condition: Total pressure = 1,2 Bar.
Reynolds 50000 75000 100000 125000 150000 175000
Inlet 120001 120001 120002 120002 120003 120004
Bending1 119950 119889 119807 119700 119566 119396
Bending2 119871 119719 119514 119244 118906 118472
Outlet 119777 119519 119169 118708 118131 117382
Total pressure
The drop of the total pressure is higher for higher Reynldss number.
Reynolds 50000 75000 100000 125000 150000 175000Part 1 62 138 244 387 574 825
Part 2 93 204 358 567 839 1213
Part 3 82 176 306 484 720 1054
Drop of the static pressure for each part
The drop of the pressure (so the looses) are the most important in the bend. Due to the
length of the second straight pipe, the drop of the static pressure is higher in the second straight pipe
than in the first straight pipe.
0
20
40
60
80
100
120
140
1 2 3 4
Velocitym/s
Re = 50000
Re = 75000
Re = 100000
Re=125 000
Re = 150000
Re = 175000
First s traight part bending part Second straight part
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c)Force values
Fr each labratries and cmpnents, the Reynldss number hasnt significant effect n
the frces values. The Z component of the forces is negligible. The Y component is the opposite of
the X component. Due to the symmetry conditions, the X and Y components for the laboratory 5 are
the half of the X and Y components for the laboratory 4. On this graph, we already multiply by two
the lab 5s values. We can see that graphs match.
d)Static temperature
The static temperature decreases during the first straight part of the pipe and the bending
part of the pipe. Then, in the second straight part of the bend, the static temperature lightly
increases. For a higher Reynoldss number, the drp f the static pressure is higher.
Forces
-50
-40
-30
-20
-10
0
10
20
30
40
50
50000 70000 90000 110000 130000 150000 170000
Reynolds
Forces
(N)
X component, lab 4
X com onent lab 5
Z com onent lab 4
Y component, lab 4
Y component, lab 5
Static temperature
75000
100000
125000
175000
175000
50000 Re5000050000
75000 Re
100000 Re
125000 Re
125000
150000 Re150000
175000 Re
292,5
292,6
292,7
292,8
292,9
293
293,1
293,2
1 2 3 4Position along the pipe
Statictemperature(K)
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e)Friction factors1) Pressure drop in circular pipe
Linear losses:2
..
.
2Lv
Dp
With: p: pressure lss *Pa+
: density [kg/m]
: friction factor [Pa/m]
v: velocity [m/s]
L: length
D: hydraulic diameter
Local losses:
2
.2
vp
With: : local friction factor
2) Blasius equation25,0
Re316,0
This empiric equation will validate our answer, or not.
3) Conclusion: Friction factors
Lv
Dp
..
.2
2
2
.
.2
v
p
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4) Resultsa) Friction factors in both of straight part of the pipe
We may conclude that there exist a significant dependence between the pressure loss
(friction factor) and the Reynolds number.
For the first straight part of the pipe and for the second straight part of the pipe, the friction
factor visibly decreases as the Reynolds number value increases.
The friction factor of the first straight part of the pipe is higher than the friction factor of the
second straight part of the pipe.
We can introduce the equations for these two friction factors on this data range:
- For the first straight part of the pipe: = 2E-19 Re + 1E-13 Re - 6E-08 Re + 0,0371.
- For the first straight part of the pipe: = 7E-19 Re + 1E-14 Re - 5E-08 Re + 0,024.
We may observe the results are practically the same for both laboratories. Moreover, the Blasius
equation graph is very close to the results. This verification allows us to think that results are good.
Linear losses
0,01
0,015
0,02
0,025
0,03
0,035
0,04
50000 70000 90000 110000 130000 150000 170000 190000
Reynolds
Part 1, lab 5
Part 2, lab 4
Part 2, lab 5Blasius
Part 1, lab 4
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b) Local friction factor for the bending part of the pipe
We may conclude that there exist a significant dependence between the friction factor and
the Reynolds number.
For the bending part of the pipe, the friction factor visibly decreases as the Reynolds number
value increases. This friction factor is higher than the friction factors of both straight part of the pipe.
So there are more loses in the bending part of the pipe than in the straight parts of the pipe.
We may observe the results are practically the same for both laboratories.
5) Hydrodynamic reaction calculationIn order to calculate the hydrodynamic reaction, we have this formula:
Rg - hydrodynamic reaction in N
m - mass flow rate in kg/s
V1 velocity on the inlet in m/s
V2 velocity on the outlet in m/s
n1 vector normal to the surface of the inlet
n2 vector normal to the surface of the outlet
p1 pressure on the inlet in Pa
p2 pressure on the outlet in Pa
A1 cross-section area of the inlet in m2A2 cross-section area of the outlet in m2
Local losses
lab 4
lab 5
1,10E-01
1,12E-01
1,14E-01
1,16E-01
1,18E-01
1,20E-01
1,22E-01
1,24E-01
1,26E-01
1,28E-01
50000 70000 90000 110000 130000 150000 170000 190000
Reynolds
LossCoefficient
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ANSYS CFX results:
group 1 2 3 4 5 6
Reynolds 50000 75000 100000 125000 150000 175000
X 37,8837 38,1723 37,3511 37,6049 37,3511 40,562
Y -37,8147 -38,0233 -37,1243 -37,7618 -37,1243 -39,8252
Z 0,000000 -0,000008 -0,003011 -0,002942 -0,003011 0,000054
Resultant (N) 53,5268742 53,8785285 52,6623046 53,2924202 52,6623046 56,8447218
Calculation results:
group 1 2 3 4 5 6
Reynolds 50000 75000 100000 125000 150000 175000
Mass flow
rate (kg.s-1)
0,0139 0,0209 0,0278 0,0348 0,0417 0,0487
Velocity Inlet 32,3319 48,9851 65,8733 83,7188 102,3 122,584
Outlet 32,8797 49,9411 66,4922 86,0098 105,708 127,741
Static pressur Inlet 119283 118361 117057 115291 113056 110191
Outlet 119046 117843 116149 113853 110923 107099
area m Rx 37,92327 38,20799 38,60582 39,13315 39,78350 40,58736
0,000314159 Ry -37,85643 -38,06524 -38,33777 -38,76112 -39,25551 -39,86713
Resultante 53,58436 53,93341 54,40766 55,08019 55,89027 56,89220
Comparison between these two ways:
The results are nearly the same for the two ways. For Re=50000, 75000 and 175000, the
variation is less than 1%. For Re = 100000, 125000 and 150000, the variation is between 3% and 7%.
So we may consider that the formula calculation results check the ANSYS CFX results.
-40
-30
-20
-10
0
10
20
30
40
50
60
50000 70000 90000 110000 130000 150000 170000
X CFX
X calculation
Y CFX
Y calculation
Resultant CFX
Resultant calculation