#4 Coffee Ring Effect

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Today we are going to talk about the first article which gave an explanation to the coffee ring effect, which is familiar to all of us. Have you ever spill your coffee on a napkin or a tissue paper? If you wait until the liquid evaporate, the granules suspended in the coffee would deposit in a ring pattern rather than deposit evenly. Why?

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When it talks to a mechanism which promotes the spreading of suspended particles, we could easily think of convection, gravitation, Marangoni effect caused by the gradient of surface tension, diffusion, electrostatics and so on. However, none of these could successfully explain the coffee ring effect. Deegan et al. thus come up with a mechanism utilizing the compensatory outward flow caused by the uneven evaporation of liquid.

Figure 1. Mechanism of outward compensatory flow during evaporation. This figure was originally the Figure 2(a) & (b) of the suggested reading.

As shown in Figure 1, solution consisting principally of water often have a small contact angle with most surface. Under this circumstances, there is less fluid near the peripheral of the liquid drop. If there were no compensatory outward flow from the central of the liquid drop, the diameter would decrease as time goes by. In contrary, if there is a compensatory outward flow to keep the forefront the same position, the flow could bring suspended particles outward and eventually cause a ring stain pattern.

Why shall we keep the forefront, or the contact line, fixed? This could be simply regarded as a geometric constraint, or it could be caused by the hydrophilic nature of the contact surface, which might be reluctant to lose the hydration energy.

In the suggested reading, the authors also proposed effective experiments to test their hypothesis. They formulated a simple model predicting the motion of suspended particles in liquid drop and observed it under microscope. They also checked their theory by forecasting the amount of water that would have evaporated within certain time.

Today there are some groups try to destroy the coffee ring effect to attain an ideal surface coating [1] while others try using it to concentrate the solute [2], just to name a few. There shall be even more applications to be discovered.

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Figure 2. Mechanism of compensatory outward flow with definition of terms.
This figure was originally the Figure 2(c) of the suggested reading.

Let’s see the cross-section of liquid drop in Figure 2 above. Assume that the height of liquid drop at r is h(r), mean outward flow velocity v(r), and evaporation rate per unit area J(r). Assume that at position r from the origin there is vapor concentration 𝜙(r) while there is vapor concentration 𝜙𝑠 on the surface of liquid and 𝜙∞ far away. According to Fick’s law:

$\vec{J}=-D\triangledown \phi$

$\frac{\partial \phi}{\partial t}=D\triangledown^2 \phi$

Under the steady state approximation, 𝜕𝜙/𝜕𝑡=0, so the concentration of water vapor satisfy Laplace equation. We should note that this is equivalent to solving the electrostatic problem having a charged conductor.

Figure 3. Equivalent electrostatics problem of original diffusion problem.

And now let’s solve the problem. Consider a cross-section. We first utilized the symmetrical properties of the problem to make it a full plane. We chose polar coordinate and defined the domain of our solution to be (𝜌cos𝜃,𝜌sin𝜃), 𝜌∈[0,∞], 𝜃∈[−𝛽/2,𝛽/2], in which 𝛽+2𝜃𝑐=2𝜋. The Laplace equation in polar coordinate reads

$\frac{1}{\rho}\frac{\partial}{\partial\rho}(\rho\frac{\partial\phi}{\partial\rho})+\frac{1}{\rho^2}\frac{\partial^2\phi}{\partial\theta^2}=0$

We solve this equation by separation of variables. We assume that 𝜙(𝜌,𝜃)=𝑅(𝜌)Θ(𝜃) and substitute it back to Laplace equation and we shall get

$\frac{\rho}{R}\frac{\textup{d}}{\textup{d}\rho}(\rho\frac{\textup{d}R}{\textup{d}\rho})+\frac{1}{\Theta}\frac{\textup{d}^2\Theta}{\textup{d}\theta^2}=0$

The 2 terms in the LHS of equation are the function of 𝜌 and 𝜃, respectively, but their sum is a constant. The only circumstance to make it possible is that both terms are in fact constant. That is

$\frac{\rho}{R}\frac{\textup{d}}{\textup{d}\rho}(\rho\frac{\textup{d}R}{\textup{d}\rho}) = \nu^2, \quad\frac{1}{\Theta}\frac{\textup{d}^2\Theta}{\textup{d}\theta^2}=-\nu^2$

We could then solve the problem:

$\textup{if}\quad \nu=0, \quad R=a_0+b_0\rho, \quad \Theta = A_0+B_0\theta\\ \forall \nu\neq 0, \quad R=a_{\nu}\rho^\nu+b_{\nu}\rho^{-\nu}, \quad \Theta=A_{\nu}\cos(\nu\theta)+B_{\nu}\sin(\nu\theta)$

There are some limitation on the coefficients. To make the domain of θ unlimited to 0~2π, 𝐵0 shall be 0. If we require the solution to include the origin, then all 𝑏𝜈 equal 0. We could then express our solution as

$\phi(\rho, \theta)=a_0+\sum_{\nu}a_{\nu}\rho^\nu\cos(\nu\theta+\alpha_\nu)$

Considering our boundary conditions, we’ll have

$\forall \rho\geq 0, \quad \theta=-\frac{\beta}{2}\lor \frac{\beta}{2}, \quad \phi=\phi_s\\ \forall \rho\geq 0, \quad \theta\in [-\frac{\beta}{2},\frac{\beta}{2}], \quad \phi(\rho, \theta)=\phi(\rho, -\theta)$

We could therefore express our solution to be

$\phi(\rho, \theta)=a_0+\sum_{\nu}a_{\nu}\rho^\nu\cos(\nu\theta)$

And

$\nu\frac{\beta}{2}=\frac{2m+1}{2}\pi, \quad m=0,1,2,...$

Our solution could now be expressed as

$\phi(\rho,\theta)=\phi_s+\sum_{m=0}^{\infty}a_m\rho^{\frac{(2m+1)\pi}{\beta}}\cos(\frac{(2m+1)\pi}{\beta}\theta)$

If we only consider the place near the origin, or 𝜌 →0, then we neglect the terms other than m=0:

$\phi \approx \phi_s+a_0\rho^{\frac{\pi}{\beta}}\cos(\frac{\pi}{\beta}\theta)$

The magnitude of vapor flow would then be

$|\vec{J}|=D\frac{\pi}{\beta}a_0\rho^{\frac{\pi}{\beta}-1}\propto \rho^{-\frac{\pi-2\theta_c}{2\pi-2\theta_c}} \propto (R-r)^{-\lambda}, \quad \lambda=\frac{\pi-2\theta_c}{2\pi-2\theta_c}$

To maintain the contact line of liquid drop, the magnitude of compensatory outward flow should be comparable to the evaporation rate, or

$\overline{v_r}\sim |\vec{J}|\propto (R-r)^{-\lambda}$

Now let’s consider how long would it take for water move from rt to the boundary R:

$t = \int_{r_t}^{R}\frac{dr}{\overline{v_r}}\sim(R-r_t)^{1+\lambda}$

We then consider the amount of water lies farther than r from the central of liquid drop. Let’s call it M(r, t). Because the cross-section of liquid droplet is approximately a triangle, we then assume

$M(r,0)\propto (R-r)^2$

Besides, the water at R at time t came from water farther than rt. That means

$M(R,t)=M(r_t,0)\propto t^{\frac{2}{1+\lambda}}$

Given the contact angle 𝜃𝑐=.25 rad, the authors predicted the mass evaporated at time t follows a power law

$(M-M_0)\propto t^{\frac{2}{1+\lambda}}=t^{1.37}$

Which perfectly matched the observed power.

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Suggested Reading:

R. D. Deegan, O. Bakajin, T. F. Dupont, G. Huber, S. R. Nagel, T. A. Witten. Capillary flow as the cause of ring stains from dried liquid drops. Nature 389, 827-829 (1997)

Reference:

[1] Yunker, P. J., Still, T., Lohr, M. A. & Yodh, A. G. (2011). Suppression of the coffee-ring effect by shape-dependent capillary interactions. Nature 476: 308-311.

[2] Wong, T. S., Chen, T. H., Shen, X., & Ho C. M. (2011). Nanochromatography driven by the coffee ring effect. Anal. Chem. 83(6): 1871-1873.

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