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Cryogenic Flow in
Corrugated Pipes
2nd CASA Day: April 23, 2009
Patricio Rosen
Outline
• Motivation
• Pipe Flow Preliminaries
• Fluid Flow Equations
• CFD Navier Stokes
• CFD k-e Turbulence Model
• Conclusions
• Further work
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PAGE 1
Outline
• Motivation
• Pipe Flow Preliminaries
• Fluid Flow Equations
• CFD Navier Stokes
• CFD k-e Turbulence Model
• Conclusions
• Further work
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PAGE 2
Motivation
• Corrugated Pipes/Hoses
• Portable
• Flexible
• Several Application Areas
• LNG Transport
• Development of DTSE (Dual
Tank Stirling Engine)
• Goal
• Describe and Predict Flow
Behavior in Corrugated
Hoses in an Efficient Way
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PAGE 3
Outline
• Motivation
• Pipe Flow Preliminaries
• Fluid Flow Equations
• CFD Navier Stokes
• CFD k-e Turbulence Model
• Conclusions
• Further work
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PAGE 4
Preliminaries
• Darcy Weisbach Equation
L ½v2
z
P1 ¡ P2 = : ¢ PL = f
4R
P1
vz
R
L
• Straight Pipe
• Poiseuille Flow
µ
¶
2
r
vz = 2vz 1 ¡
R2
64
f =
Re
Re :=
4R½vz
¹
• Non-straight Pipes?
• Roughness
• Actual Shape
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PAGE 5
P2
Moody Diagram
• Experimental Results
• Moody Diagram
• Fully Turbulent
• Colebrook Equation
 e 
 D
1
2.51
 2  log10    
f
 3.7 Re  f








e
D
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PAGE 6
LNG Composite Hose
• Moody Prediction
• Taking corrugation as
roughnes
f= 0.045
• Measurements
• Water
• LNG
f= 0.058
f= 0.13
•Moody Diagram is a poor
indicator for the friction factor
•Find a better Alternative (CFD)
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PAGE 7
Outline
• Motivation
• Pipe Flow Preliminaries
• Fluid Flow Equations
• CFD Navier Stokes
• CFD k-e Turbulence Model
• Conclusions
• Further work
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PAGE 8
Fluid Flow Equations
Navier-Stokes
r ¢v = 0
h
1 @P
vr i
v ¢r vr = ¡
+ º r 2 vr ¡
½ @r
r2
1 @P
vz = ¡
+ º r 2 vz
½ @z
Periodicity
v (R(z);
z) = 0
v (r; 0) = v (r; L )
P (r; 0) = P (r; L ) + kL
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• Assumptions
• Incompressible Flow
• Steady Flow
• Gravity negligible
• Isothermal Flow
• One-phase
• Cylindrical and
Periodic Hose
• Fixed Wall
• No Swirl
• Cylindrical Symmetry
Set up saves lots of computation time
PAGE 9
Analytic Expression for DFF
• From continuity
v ¢r vz = r ¢(vz v ) ¡ vz r ¢v = r ¢(vz v )
• Rewrite z-momentum 1
r ¢(v vz ) = ¡ r ¢(P ez ) + º r ¢(r vz ):
½
• Using Divergence Theorem
µZ
¶
Z
1
@vz
¢ PL := P i n ¡ P ou t =
P n z dS ¡ ¹
dS :
j¡ i n j
@
n
¡
¡
Pressure “Friction“
• For Poiseuille8¹Flow
vL
¢ PL =
:
R2
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f =
PAGE 10
Skin “Friction“
64¹
64
=
:
D ½v
Re
Outline
• Motivation
• Pipe Flow Preliminaries
• Fluid Flow Equations
• CFD Navier Stokes
• CFD k-e Turbulence Model
• Conclusions
• Further work
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PAGE 11
CFD Navier Stokes
Setting
P = : P~ + f z z
8
>
<
>
:
@P
@r
=
@P
@z
=
@P~
@r
@P~
@z
fz
fz
fz
fz
+ fz
Discretization
Velocity
Pressure
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PAGE 12
Navier Stokes
Re=2.13
Re=213
Re=676
Re=2713
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PAGE 13
Validation Several Corrugations (Re=213.5)
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PAGE 14
Forces at Wall
Re=213
Re=2713
Pressure and Viscous Forces
scale with Re in Laminar Regime
and Skin Friction Dominates
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PAGE 15
Friction Factor
Same Friction Factor as
for Straight Pipes in
Laminar Regime
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PAGE 16
Outline
• Motivation
• Pipe Flow Preliminaries
• Fluid Flow Equations
• CFD Navier Stokes
• CFD k-e Turbulence Model
• Conclusions
• Further work
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PAGE 17
CFD k-e Turbulence Model
Re=177
Re=843
Re=1.737e4
Re=3860
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PAGE 18
Several Periods
Re=177
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Re=847
PAGE 19
Re=3860
Re=177
Re=3860
At High Reynolds Numbers the Pressure
Forces become Dominant
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PAGE 20
Friction Factor
One Period
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Several Periods
PAGE 21
Outline
• Motivation
• Pipe Flow Preliminaries
• Fluid Flow Equations
• CFD Navier Stokes
• CFD k-e Turbulence Model
• Conclusions
• Further work
\Mathematics Department
PAGE 22
Conclusions
•Correct Prediction of
the Friction Factor
(One phase, adiabatic flow)
Problem Solved?
•Sensibility of Results
(needs validation)
•Cryogenic Liquids not
yet manageable
•Expensive computation
time for dynamic flow
computations
4 hours Computation
(NS Example)
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PAGE 23
Outline
• Motivation
• Pipe Flow Preliminaries
• Fluid Flow Equations
• CFD Navier Stokes
• CFD k-e Turbulence Model
• Conclusions
• Further work
\Mathematics Department
PAGE 24
Towards a 1D Model
vz (r; z) = vz (z) + v^z (r; z)
Z
¡ in
1
vz (r; z)dA
¡ (z)
¡ (Z
z)
1
P(z) =
P dA
j¡ (z)j
vz :=
¡ out
¡ ( z)
d
(R 2 vz ) = 0
dz
Z
d
1 d
d R ( z)
2
(R 2 v2 ) = ¡
(R 2 P ) ¡ 2
v^2 (r; z)dr + P (R(z); z)R`(z)R(z)+
z
z
dz
½dz
dz 0
½
·
2º
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¸
@vz
@v
(R(z); z) ¡ R(z) z (R(z); z) R(z)
@r
@z
PAGE 25
Thanks for your attention!
RANS
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PAGE 27
k-e Model
Summary
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PAGE 28
FEM for Navier Stokes
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PAGE 29
Navier Stokes Weak Form
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PAGE 30
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