1. Saffman-Taylor Instability of Hele-Shaw Cell
Chen Shu, Xizhi Cao, Enkhsanaa Sommers

2. Introduction
Our group focuses on theoretically analyzing two-flow Saffman Taylor instability in a Hele-Shaw cell with numerical methods.
A Hele-Shaw cell consists with two parallel plates are spaced infinitesimally thin. Less viscous flow injected from above or below into the more vicious fluid which is sitting between plates. The boundary condition of this flow is defined by surface tension and curvature.
In our case we will apply perturbation on Hele- Shaw cell and calculate expected number of fingers.
We compare numerical result with the experimental result
Fluid #1
Fluid #2

3. Derive Darcy’s law Navier-Stokes equation: 𝑢∗𝛻 𝑢=− 1 𝜌 𝛻P+ν∗ 𝛻 2 𝑢
𝜕𝑢 𝜕𝑡 + +𝑔
(1)
𝜕𝑢 𝜕𝑡 =0 (Since steady state uniform flow)
𝑔 =0 gravity term is ignored since we are dealing with 2-D system

4. Derive Darcy’s law 𝑢∗𝛻 𝑢=− 1 𝜌 𝛻P+ν∗ 𝛻 2 𝑢 (2)
The equation reduced to:
𝑢∗𝛻 𝑢=− 1 𝜌 𝛻P+ν∗ 𝛻 2 𝑢 (2)
Evaluate inertial term(IT) and viscous term(VT)
𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑡𝑒𝑟𝑚 : 𝑢∗𝛻 𝑢=𝑢 𝜕𝑃 𝜕𝑥 +𝑣 𝜕𝑄 𝜕𝑦 +𝑤 𝜕𝑅 𝜕𝑧 ~ 𝑂 𝑈 2 𝐿
𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑡𝑒𝑟𝑚 : v∗ 𝛻 2 𝑢=𝑣∗ 𝜕 2 𝑢 𝜕 𝑥 2 , 𝜕 2 𝑢 𝜕 𝑦 2 , 𝜕 2 𝑢 𝜕 𝑧 𝜕 2 𝑢 𝜕 𝑥 2 ~𝑂 𝑈 𝐿 𝜕 2 𝑢 𝜕 𝑦 2 ~𝑂 𝑈 𝐿 𝜕 2 𝑢 𝜕 𝑧 2 ~𝑂( 𝑈 ℎ 2
h<<L we can ignore x and y axes
Reynold Number= 𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑡𝑒𝑟𝑚 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑡𝑒𝑟𝑚 ≪1, 𝑡ℎ𝑒𝑛 𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑡𝑒𝑟𝑚 𝑖𝑠 𝑖𝑔𝑛𝑜𝑟𝑒𝑑.

5. Derive Darcy’s law
Since, 𝑈 2 𝐿 𝑣∗ 𝑈 ℎ 2 = 𝑈𝐿 𝑣 ∗ ( ℎ 𝐿 ) 2 << 1 , so 𝑢∗𝛻 𝑢~0
𝑢∗𝛻 𝑢=− 1 𝜌 𝛻P+ν∗ 𝛻 2 𝑢
0= 1 𝜌 𝛻P+ν∗ 𝜕 2 𝑢 𝜕 𝑧 (3)
𝛻P=𝜇∗ 𝜕 2 𝑢 𝜕 𝑧 𝜕𝑃 𝜕𝑥 =𝜇∗ 𝜕 2 𝑢 𝜕 𝑧 2 , 𝜕𝑃 𝜕𝑦 =𝜇∗ 𝜕 2 𝑣 𝜕 𝑧 2 , 𝜕𝑃 𝜕𝑧 =𝜇∗ 𝜕 2 𝑤 𝜕 𝑧 2
Pressure change in z direction is small enough to neglect

6. Derive Darcy’s law 𝑢=− 1 2𝜇 ∗𝛻P∗ 𝑧 2 + 𝑐 1 𝑧+ 𝑐 0 (4)
Boundary condition: u 0 = u h =0
𝑢=− ℎ 2 12𝜇 𝛻P (5)

7. Laplace-Young BC: General form: ∆P=𝛾𝜅 In our case:
Pinner
Pouter R2 R1
General form:
∆P=𝛾𝜅
In our case:
R2 is big enough so that Pouter can be ignored
𝑷 𝒊𝒏𝒏𝒆𝒓 =𝜸𝜿 𝑷 𝒐𝒖𝒕𝒆𝒓 =𝟎
𝛾 - surface tension
𝜅 - curvature

8. Saffman-Taylor Instability
Saffmen Taylor instability occurs when a less viscous fluid is injected into a more viscous fluid.
 It can also occur driven by gravity (without injection) if the interface is horizontal separating two fluids of different densities with the heavier fluid on the top.
In the rectangular configuration the system evolves until a single finger (the Saffman– Taylor finger) forms.
In the radial configuration the pattern grows forming fingers by successive tip-splitting.

9. Exact solution to circular BC
Pinner
Pouter R2 R1
𝛻 2 P= r 2 𝜕 2 𝑃 𝜕 𝑟 2 +𝑟 𝜕𝑃 𝜕𝑟 + 𝜕 2 𝑃 𝜕 𝜃 2 =0
𝑃=𝑐∗𝑙𝑛𝑟+𝑑
At t=0 , 𝑷 𝑹𝟏 =𝜸𝜿 𝑷 𝑹𝟐 =𝟎, 𝜅 0 = 1 𝑅 1
𝑃 0 = 𝛾 𝑅 1 1− 𝑙𝑛 𝑟 𝑅 1 𝑙𝑛 𝑅 2 𝑅 1
𝑢 0𝑟 =− ℎ 2 12𝜇 𝛻P= ℎ 2 12𝜇 𝛾 𝑅 1 𝑟 1 𝑙𝑛 𝑅 2 𝑅 1

10. Linear stability analysis:
Perturbation:
𝑃= 𝑃 0 +𝜀 𝑃 1 (𝑟,𝜃) Pressure (R1)
𝑢= 𝑢 0 𝑟 +𝜀 𝑢 1 (𝑟,𝜃) Velocity
𝜅= 𝜅 0 +𝜀 𝜅 Curvature
𝑅 1 (𝜃,𝑡)= 𝑅 1 (𝑡)(1+𝜀𝜂 𝜃,𝑡 )
𝜂 𝜃,𝑡 =𝑁(𝑡)cos(𝑚𝜃)

11. Linear stability analysis:
𝐵𝑦 𝐷𝑒𝑓𝑖𝑛𝑎𝑡𝑖𝑜𝑛: 𝜅= 𝑟 𝑟 ′ 2 −𝑟 𝑟 ′′ ( 𝑟 2 + 𝑟′ 2 ) R 1 (t)(1+εη θ,t
𝜅 = 1 𝑅 1 (1−𝜀𝜂−𝜀 𝜂 𝜃𝜃 ) => (𝜅= 𝜅 0 +𝜀 𝜅 1 )
O(1): 𝜅 0 = 1 𝑅 1
O(𝜀): 𝜅 1 =− 𝜂+ 𝜂 𝜃𝜃 𝑅 1

12. Linear stability analysis:
𝑃= 𝑃 0 +𝜀 𝑃 1 (𝑟,𝜃)
(P 0 +ε P 1 ) | R 1 (1+εη) = −γκ | R 1 1+εη =−γ 1 R 1 −ε 𝜂+ 𝜂 𝜃𝜃 𝑅 1
P 1 fits in Laplace equation so that P 1 = P m cos 𝑚𝜃 𝑟 −𝑚 .
ln 1+ε𝜂 ≈ε𝜂, 𝜂 𝜃𝜃 =− 𝑚 2 𝜂
P 1 𝑟,𝜃 = γ R 1 ( 1 𝑙𝑛 𝑅 2 𝑅 1 +(1− 𝑚 2 )) 𝑟 −𝑚 𝜂
u r 𝑟,𝜃 | R 1 (1+εη) = ℎ 2 12𝜇 { 𝛾 𝑅 𝑙𝑛 𝑅 2 𝑅 1 +𝜀𝑚 γ 𝑅 𝑙𝑛 𝑅 2 𝑅 − 𝑚 2 } 𝜂

13. Linear stability analysis:
Using the algebraic manipulation:
𝑑 𝑑𝑡 ( 1 2 𝑅 1 𝑅 1 )= 𝑅 1 ∗ 𝑑 𝑅 1 𝑑𝑡
𝑅 1 𝑅 1 ′ 1+2𝜀𝜂 + 𝑅 1 2 𝜀 𝜂 ′ = ℎ 2 12𝜇 𝛾 𝑅 1 { 1 𝑙𝑛 𝑅 2 𝑅 1 +𝜀𝑚 1 𝑙𝑛 𝑅 2 𝑅 − 𝑚 2 } 𝜂
O (1): ℎ 2 12𝜇 𝛾 𝑅 𝑙𝑛 𝑅 2 𝑅 1 = 𝑅 1 𝑅 1 ′
O (𝜀): ℎ 2 12𝜇 𝑚 γ R 𝑙𝑛 𝑅 2 𝑅 − 𝑚 2 𝜂=2 𝑅 1 𝑅 1 ′ 𝜂+ 𝑅 1 2 𝜂′
𝑅 1 ∗ u r 𝑟,𝜃 | R 1 (1+εη)
Left hand Side

14. Linear stability analysis:
𝜂 ′ = 𝑅 1 ′ 𝑅 1 (𝑚+ ℎ 2 12𝜇 γ 𝑅 𝑅 1 ′ 𝑚 1− 𝑚 2 −2)𝜂
𝜎= 𝑚+ ℎ 2 12𝜇 γ 𝑅 𝑅 1 ′ 𝑚 1− 𝑚 2 −2
𝜂= 𝜂 0 exp(𝜎𝑡)
exp⁡(𝜎𝑡) is the growth rate of 𝜂.
If 𝜎>0 the perturbation is unstable (expand), otherwise stable (shrink).
In order to find the best perturbation, we need to find the largest 𝜎

15. Linear stability analysis:
Let 𝑎= ℎ 2 γ 12𝜇 𝑅 𝑅 1 ′
𝜎=(𝑚+𝑎𝑚 1− 𝑚 2 −2)
𝑚 𝑚𝑎𝑥 = ( 12𝜇 ℎ 𝑅 𝑅 1 ′ γ +1)
𝜎 vs m

16. Relate Flow and 𝑹 𝟏 ′ 𝑄= 𝜋 𝑅 1 2 ℎ ′ 𝑅 1 ′ = 𝑄 2𝜋 𝑅 1 ℎ
𝑄= 𝜋 𝑅 1 2 ℎ ′
𝑅 1 ′ = 𝑄 2𝜋 𝑅 1 ℎ
𝑚 𝑚𝑎𝑥 = ( 12𝜇 ℎ 2 𝑄 𝑅 1 2𝜋ℎγ +1)
Paterson:
𝑚= ( 𝑅𝑄 2𝜋 𝑀 2 γ +1 )

17. How do we find number of fingers from the experiment result

18. Comparison between theoretical and experimental result
Width of Spacers (mm)
Mass (g)
Volume Inside Syringe (mL)
Number of Fingers (Measure)
Number of Fingers (Theoretical)
Flow Rate
Relative Error
0.32
500
1.2 29 40
1.26E-07
0.275
1000 32 51
2.00E-07
1500 37 79
4.89E-07
2000 88
2.58E-06
0.52
200 18 19
1.20E-07 22 21
1.47E-07 24 27
2.43E-07 42
5.89E-07
0.83 11 8
1.04E-07
0.375 12
1.30E-07
0.5 14
2.46E-07

19. Result Analysis Spacing 0.32mm has a large error.
Small spacing creates a lot of fingers and many of them are small which give us hard time to count
Spacing 0.52mm has the closest simulation result.
0.52mm minimize the errors come from experiment equipment
Spacing 0.83mm has a large error.
the theoretical value for number of fingers are small so a little error will create a large relative error, also the process is quick so it’s hard to tell the exact time
Error grows when mass gets bigger.
Flow rate grows in a really rapid way while the equation only handle it as 𝑄

20. Comparison between theoretical and experimental result

21. Possible causes of error
Time error
Counting error
Error due to the Experiment equipment
Our model is not accurate enough

22. References: References:
[1] Acheson, D. J. (1990), Elementary Fluid Dynamics, Oxford Applied Mathematics and Computing Science Series, Oxford University Press,
[3] Kondic, Lou (2014), Linear Stability Analysis of two phase Hele­-Shaw Flow, Unpublished.
[4] Paterson, Lincoln (1981) Radial Fingering in a Hele­Shaw Cell, Department of Engineering Physics, Research School of Physical Sciences, The Australian National University, <
[4]
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