1. Tony Bougebrayel, PE, PhD Engineering Analyst Swagelok Co.
CFD Prediction of Liquid Flow through a 12-Position Modular Sampling System
Tony Bougebrayel, PE, PhD
Engineering Analyst
Swagelok Co.

2. AGENDA
How is the driving pressure consumed?
Why do liquids require more driving pressure?
Predicting driving pressure for a conventional system
What is CFD?
CFD application to a 12-position modular system
Results: CFD vs. Actual
Conclusion

3. How is the driving pressure consumed?
Momentum Loss:
Pipe size reduction
Control Components (valves, filters, check valves, meters, gages…)
Entry and exit effects (velocity profile)
Contraction/Expansion
Directional Changes (elbows, Ts..)
Potential Energy: Height
Viscous Losses: Boundary Layer formation
Turbulent Energy
Modular systems experience Momentum, Viscous, and Turbulent losses

4. ¤ ¤ Driving Liquids Flow in a straight pipe
Darcy’s equation: P = x f x  x L x Q2 / d5
fα Re,ε(Re =  U d/)
↑ Re ↓ f↑ P↑
 ↑ Re ↑ f↓ P↑
10x increase in  yields 71% increase in P
10x increase in  yields in 580% increase in P
Density is dominant in straight pipes
f values taken for smooth pipes flowing at 104 and 105 Re

5. For Non-Uniform Geometry
Driving Liquids
For Non-Uniform Geometry
Navier-Stokes Equations (Incompressible, Laminar, in 3D Cartesian Coordinates)
Piezometric pressure gradient
Viscous terms
Momentum terms
Local
acceleration
Both Density and Viscosity affect 2nd order terms

6. Conventional System: Predicting Driving Pressure
Bernoulli’s Equation (mechanical energy along a streamline)
z p1/1 + v12/2g = z p2/2 + v22/2g + hL
Potential Pressure Kinetic Total Energy Energy Energy Head Loss
Where, hL = K v2 / 2g
Ki = f L / D (Ki: Flow Resistance)
Ktotal = Ki
L / D: Equivalent pipe length for non-pipes i.e. valves, fittings
Ki = f L / D (This is Darcy’s equation for pipes)
Fitting
L/D
Globe Valve
340
Lift Check Valve
600
Ball Valve 6
Tee- Branch flow 60
Elbow- 90
Bend r/D = 20 50
Flow resistance approach in systems design

7. Conventional System: Predicting Driving Pressure
Q, ml/min
300 OD
1/4"
Pipe friction, f
0.0308
Wall Thickness
0.065
Non-Pipe friction, f_T
0.0379
Component
Quantity Ki
f*Ki K
90-Elbow 20 60
2.27
45.4
Check Valves 1
600
22.71
22.7
Globe Valves 8
500
18.93
151.4
90-Bends, r/d=8 24
.91
18.2
Flow-thru-branch 6
13.6
Pipe, inch
120
30.8
K_total
282.1
h, in
265.8
P, psi
9.6
K values are empirical
Courtesy of Exxon Mobil

8. Pressure Required for a MPC
Empirical Approach (Cv or K):
Cv = 29.9 d2 / k1/2
(1/Cv-total)2 = Σ (1/Cv-i)2
Testing
CFD
Flow
Cv-5
Cv-4
Cv-3
Cv-total < Cv-i
Cv-2
Cv-1

9. What is CFD?
A numerical approach to solving the Governing flow equations over any Geometry and Flow conditions
CFD is used to solve the general form of the flow equations

10. CFD – The Governing Equations
F= d(MU)/dt = u(u/x)dxdy + v(u/y)dxdy
pdy
[p+(p/x)dx]dy
C.V.
(u/y)dx|y
(u/y)dx|y+dy w
Differential Control Volume dx dy 1 x y
External Forces
[u+ (u/x)dx]2dy
 [u+(u/y)dy][v+(v/y)dy]dx
u2dy
C.V.
uvdx
Change in Momentum
The flow equations are based on the conservation laws

11. CFD – The Governing Equations
Navier-Stokes Equations for an Incompressible, Laminar flow
Note: These equations can actually be reduced into Bernoulli’s
Local
acceleration
Inertia terms
Viscous terms
Piezometric pressure gradient
Continuity equation
The N.S. eqs. are highly elliptical and impossible to solve manually

12.     CFD – How does it Work? 
Solve: y + y = 0 (1st order PDE)
for 0 x  1
From Taylor’s: 
-yi + (1+ x )yi+1 = 0 (3)
Plug into (1): XN X1
Discrete Domain
(Eq. 2): Discretized, Algebraic Equation
For a structured grid:
x = Xi+1 - Xi
Apply equation (3) to the 1-D grid at nodes 1,2,3:  y1 y2   y3  y4
-y1 + (1+ x )y2 = 0 (i=1) (4)
-y2 + (1+ x )y3 = 0 (i=2) (5)
-y3 + (1+ x )y4 = 0 (i=3) (6)
Equations 4, 5, & 6 are 3 equations with 4 unknowns
The B.C. y1=1 completes the system of equations y y x 1 j i x
Convert the PDE into an Algebraic equation

13. Accuracy is grid dependent
What is CFD?
Next, we write the system of equations in a matrix form: [A]{y}={0}
y1 = (BC)
y2 = (4)
y3 = (5)
y4 = (6)
-1 (1+ x )
(1+ x )
(1+ x )
To solve, is to find [A]-1
Much CFD work revolves around optimizing the inversion process
(strongly tridiagonal matrix – Iterative approach)
Accuracy is grid dependent

14. CFD – Application to Current System
Check Valve
Switching Valve
Pressure
Pressure
Toggle Shut-off
Pneumatic Switching Valve
Pneumatic Shut-off
ManualShut-off
PneumaticShut-off
Toggle Shut-off
Toggle Shut-off
Flow
Flow

15. CFD – Application to Current System
Build the Geometry

16. CFD – Application to Current System
Extract the Fluid volume

17. CFD – Application to Current System
Create the Mesh: 3.2 million cells

18. CFD – Application to Current System
Set Boundary Conditions
Solve

19. Conventional Calculated
Results
Pressure required to drive 300 cc/min through the 12-position system, psi
MPC
Tested
CFD Predictions
Conventional Calculated
Water
15.6
16.9
9.6
Diesel
17.9
15.1
10.8
Gasoline
12.5
11.7
7.1
Pressure required to drive liquid samples through modular systems are in line with available pressure

20. Results: CFD vs. Actual
CFD predictions are very accurate when fluid characteristics are known

21. Results: Density vs. ViscositySG
, cP
65 F 1
Diesel Fuel 100 F
.85
1.69
Unleaded Gasoline
.73
.47
SGdiesel < SGwater but Qwater > Qdiesel
Viscosity effects are more prominent than density effects in modular systems
Testing conducted by Colorado Engineering Experiment Station Inc.

22. Results: Density vs. ViscositySG
 = /, cSt
65 F 1
Diesel Fuel 100 F
.85 2
Unleaded Gasoline
.73
.64
ΔPfluid/ΔPwater ≈ (fluid/water)0.5
The Kinematic viscosity compares relatively well to pressure

23. Conclusion
Reasonable pressure required to drive typical liquid samples through NeSSITM systems
CFD can be employed to accurately predict flow under different conditions
The Kinematic viscosity of the liquid sample is a good indicator of its pressure requirement

24. Questions?