1. welcome
Towards a numerical model for the twin jet
interaction in air
K. Intzes,
Internship, Mechanical Engineering TU/e
Thermofluids Engineering group - EfAM

2. Introduction (1)
- Subject
● High volume manufacturing of liquid products
by Inkjet printing (ENTHALPY Project)
● Realized by a multinozzle printer head
● Liquid jets sprayed at close proximity
● Potential jet – jet interaction
● Study the factors causing jet – jet attraction

3. Introduction (2)
- Summary
● The physical problem
● Boundary layer model
● Numerical Navier Stokes solver
● Results
● Discussion

4. The physical problem (1)
Printing head - Multi nozzle configuration
Liquid columns flowing near each other in air.
Air boundary layers develop (due to skin drag)
at the liquid – air interface
Expansion, possible interaction - collision

5. The physical problem (2)
- Primary physical phenomena inside the
induced air flow field
● Momentum influx from both sides
● Air supply from the sides
● Air velocity change close to the jets in the upper
region pressure variations

7. The physical problem (3) – domain of
interest

8. The physical problem (3) – domain of
interest cont.

9. The physical problem (4)
- Assumptions and governing equations
● Neglect of jet instability, break up
● Approximate jets by moving walls
● Icompressible, laminar air (induced) flow
● Steady – State problem

10. The physical problem (4)
- Assumptions and governing equations cont.
● Incompressibility constraint :
● Incompressible Navier Stokes :
● Scale with jet gap d, jet velocity U and ρνU/d
(pressure scale). Dimensionless form :
∇⋅⃗u = 0
⃗u⋅∇ ⃗u = −ρ−1
∇ p+ν Δ⃗u
Re⃗u⋅∇ ⃗u = −∇ p+Δ⃗u , Re =
dU
ν

11. The physical problem (5)
- Boundary conditions
● Walls : upper wall
moving walls
⃗u = ⃗0
⃗u = U
● No inflow in x-direction● No inflow in x-direction
● Free lateral and bottom boundaries

13. Boundary layer model
- A preliminary investigation of the boundary layer
growth
● Analytical model based on a air flow profile of
predifined shape : exponential decay of the
velocity magnitude in the downstream direction
w.r.t. The distance from the interface
● Assumptions : steady state, uniaxial, laminar air
flow, block liquid velocity profile

15. Boundary layer model (2)
- Boundary conditions
● u = liquid (jet) velocity at the air – liquid
interface, u(z,0) = ul0
(z)
● u = 0 at the boundary layer (and beyond)
● At the interface
[∂2
u
∂ y
2
+
1
a(z)
∂u
∂ y ]∣y=0

16. Boundary layer model (3)
- equations
● Momentum balance
● Interface flux balance
● Liquid jet mass balance
2π ρa ∫
0
δ(z)
[α(z)+y]u
2
(z , y)dy+ρl π α
2
(z)ul (z) = ρl π α0
2
(z)ul0
d
dz
[π ρl α2
(z)ul
2
(z)] = 2π α(z)μa
∂u
∂ y∣y=0
π α2
(z)ul (z)=π α0
2
ul0

17. Boundary layer model (3)
- equations cont.
● Air velocity profile
● Closes system of equations and satisfies the
boundary conditions
● Is a generic solution to the Navier Stokes
equations
u(z , y) = ul (z)
(1−
1
β(z)
ln
(1+
y
α(z))) β(z) = ln
[1+
δ(z)
α(z)]

18. Boundary layer model (4)
● Single jet analysis
● To fit for the twin jet problem, two single jet
solutions are superimposed
● Estimation of the interaction distance
- Pitfalls of the superposition approach :
● Solution not linear
● Different boundary conditions active in the twin
jet problem
● Only one velocity component considered
● Pressure field obtained by Bernoulli

22. The numerical solver (1)
- A global approach
● Custom Finite Volume discretization numerical
algorithm
● Aim to resolve the induced air flow field in the
domain
Numerical implentation

23. The numerical solver (2)
- Computational grid
● Standard staggered Finite Volume grid
● Velocity nodes control volumes shifted by half
grid size w.r.t. the scalar contol volumes

25. The numerical solver (3)
- Discretization
● Continuity : around scalar control volumes
x-direction velocity nodes : u
y-direction velocity nodes : v
z-direction velocity nodes : w
ui+1, J , K −ui , J , K
dx
+
vI , j+1, K −vI , j , K
dy
+
wI , J ,k−wI , J ,k+1
dz
= 0

26. The numerical solver (3)
- Discretization cont.
● Momentum equations : around staggered
control volumes in each respective direction
Central Difference scheme. For x-direction :
0 = −
pI , J ,K − pI−1, J , K
dx
+
ui−1, J , K−2ui , J , K+ui+1, J , K
dx2
+
ui , J −1, K−2ui , J , K+ui , J +1, K
dy2
+
ui , J ,K−1−2ui , J , K+ui , J , K +1
dz2
−
Re
dx (1
2
(ui+1, J , K+ui , J ,K )
1
2
(ui+1, J , K +ui , J , K)−
1
2
(ui , J , K+ui−1, J , K )
1
2
(ui , J , K +ui−1, J , K ))−
Re
dy (1
2
(vI −1, j+1, K+vI , j+1, K )
1
2
(ui , J +1, K+ui , J , K )−
1
2
(vI −1, j , K +vI , j , K)
1
2
(ui , J , K +ui , J−1, K ))−
Re
dz (1
2
(wI−1, J ,k+wI , J ,k )
1
2
(ui , J , K+ui , J , K −1)−
1
2
(wI −1, J ,k+1+wI , J , k+1)
1
2
(ui , J , K +ui , J , K +1))

28. The numerical solver (4)
- Boundary conditions
● Walls (moving walls and upper printer head) : no
slip
● z-y planes :
no flow in x-direction u = 0
zero tangential shear :
∂v
∂ x
=
∂ w
∂ x
= 0

29. The numerical solver (4)
- Boundary conditions cont.
● Lateral boundaries (x-y planes) :
zero normal velocity gradient and tangential
shear :
● Bottom boundary (x-z plane) : similarly
● Introduced by virtual points and discretized by
centrall diference
∂u
∂ z
=
∂v
∂ z
=
∂ w
∂ z
= 0
∂u
∂ y
=
∂ v
∂ y
=
∂ w
∂ y
= 0

30. The numerical solver (5)
- Iterative linear solver
● System of nonlinear discretized equations :
L : Discretization of Stokes equation, x :
degrees of freedom (velocity and pressure
nodes), C(x) : discretization of transport term,
b : boundary conditions
L x+C (x)−b = 0

31. The numerical solver (5)
- Iterative linear solver cont.
● Linearized system : Ax = b
● ILU(0) (Incomplete LU factorization with no fill-in)
left preconditioned restarted GMRES
● Fast converging, numerically robust, flexible

32. The numerical solver (6)
- Discrete Penalty Function Method
● Big zero block in the discretization matrix
● ILU(0) not applicable
● Modify continuity discretization as such :
● ε small (but not too small !) = 10-8
ui+1, J , K−ui , J , K
dx
+
vI , j+1, K−vI , j , K
dy
+
wI , J ,k−wI , J , k+1
dz
+εpI , J , K = 0

34. The numerical solver (7)
- Picard linearization iteration
● Non linearity requires convergence of the
solution inside iteration
● Linearized transport term : at iteration level k+1
● Iteration initiated with the solutionof the Stokes
eqation
● Stop when and
⃗u⋅∇ ⃗u = ⃗uk
⋅∇ ⃗uk+1
∣∣xk+1
−xk
∣∣inf < tol
∣∣L xk+1
+C(xk+1
)−b∣∣inf < tol

35. The numerical solver (8)
- Convergence stability and speed
● Under relaxation :
0 < a ≤ 1
● Continuation : Reynolds number increazed
incremantaly
Re Re / n, n positive integer
xk+1
= α xk+1
+(1−α) xk

36. The numerical solver (9)
- Algorithm overview
● 1. Data input
● 2. Grid
● 3. Discretization of The Stokes equation
● 4. Solution of the linear system
● 5. Discretization of transport terms (linearized)
● 6. Iterate full Navier Stokes
● 7. output

37. Numerical nesults (1)
- Results for the Stokes problem
● Air at temperature : 20 o
C
● Moving wall velocity : -12 m/s
● Jet gap : 300μm
● Reynolds number : 79.42

38. Velocity magnitude contour in the x – y mid-plane of the domain.
Incoming jet velocity = - 12 m/s, jet gap = 300 μm

39. Velocity magnitude contour in the x – z top-plane of the domain.
Incoming jet velocity = - 12 m/s, jet gap = 300 μm

40. Fig. 17 : Pressure contour in the x – y mid-plane of the domain.
Incoming jet velocity = - 12 m/s, jet gap = 300 μm

41. Pressure contour in the x – z top-plane of the domain. Incoming jet
velocity = - 12 m/s, jet gap = 300 μm

42. Numerical nesults (2)
- Results for the Stokes problem
● Air at temperature : 20 o
C
● Moving wall velocity : -12 m/s
● Jet gap : 500μm
● Reynolds number : 79.42

43. Velocity magnitude contour in the x – y mid-plane of the domain.
Incoming jet velocity = - 12 m/s, jet gap = 500 μm

44. Pressure contour in the x – y mid-plane of the domain. Incoming jet
velocity = - 12 m/s, jet gap = 500 μm

45. Numerical nesults (3)
- Results for the Stokes problem
● Air at temperature : 20 o
C
● Moving wall velocity : -4 m/s
● Jet gap : 300μm
● Reynolds number : 26.47

46. Velocity magnitude contour in the x – y mid-plane of the domain.
Incoming jet velocity = - 4 m/s, jet gap = 300 μm

47. Pressure contour in the x – y mid-plane of the domain. Incoming jet
velocity = - 4 m/s, jet gap = 300 μm

48. Numerical nesults (4)
- Results for the Navier Stokes problem
● Air at temperature : 20 o
C
● Moving wall velocity : -0.5 m/s
● Jet gap : 300μm
● Reynolds number : 3.31

49. Velocity magnitude contour in the x – y mid-plane of the domain.
Incoming jet velocity = - 0.5 m/s, jet gap = 300 μm

50. Pressure contour in the x – y mid-plane of the domain. Incoming jet
velocity = - 0.5 m/s, jet gap = 300 μm

51. Numerical nesults (5)
- Results for the Navier Stokes problem
● Air at temperature : 20 o
C
● Moving wall velocity : -2 m/s
● Jet gap : 300μm
● Reynolds number : 13.27

52. Velocity magnitude contour in the x – y mid-plane of the domain.
Incoming jet velocity = - 2 m/s, jet gap = 300 μm

53. Pressure contour in the x – y mid-plane of the domain. Incoming jet
velocity = - 2 m/s, jet gap = 300 μm

54. Numerical nesults (2) - discussion
● Velocity boundary conditions succesful
● Development of boundary layer expands far
downstream
● No stagnate flow region

55. Numerical nesults (3) - discussion
● Pressure field unrealistic
● Numerical results sound from a numerical point
of view only. Comparison of cases reveals trend
● Pressure virtual grid points
● No distinct driving force

56. Closing remarks
● Sudden velocity change near jet walls in the
upper region gives rise to pressure gradients
● Reconsider boundary conditions
● Challenging physical problem equals
challenging numerical problem
Well thought out boundary conditions !