21.2.1 Normalized Capillary Heights

Using the capillary length, the capillary heights of tubes with different diameters can be calculated in Eq. 21.4. The results are depicted in Fig. 21.2. Fig. 21.2a shows the capillary height h_c normalized to $L_c^2$. The values shown are calculated for different wall contact angles Θ starting with Θ = 0° (perfectly wetting) as the topmost curve all the way down to Θ = 180° (perfectly non-wetting) in increments of 20°. For Θ = 90°, the line is horizontal.

The calculated values are valid for any fluid and depend only on the wall contact angle Θ of this fluid. Fig. 21.2b shows the capillary height h_c for water, again calculated for the different wetting cases.

21.2.2 Negative Capillary Pressure

One interesting aspect can be seen in Eq. 21.3 and Fig. 21.2: for contact angles of Θ > 90°, the capillary pressure and the capillary height are negative (see Fig. 21.1b). For these cases, the capillary will represent a pressure barrier, i.e., an overpressure is required to push the liquid into the capillary. This is the case, e.g., when pumping aqueous solutions through hydrophobic capillaries made from polytetrafluoroethylene, for example.

21.2.3 Meniscus Depth

We will discuss the fluid mechanics of the meniscus in section 21.3. However, for small capillary diameters, this meniscus can be adequately approximated by a circle (see Fig. 21.1). The depth of the meniscus h_r can be calculated by using the following geometrical relations

$\cos \text{Θ}r=\frac{d}{2}$

$\sin \text{Θ}r=r-h_r\to h_r=r\left(1-\sin \text{Θ}\right)$

Inserting Eq. 21.7 into Eq. 21.6 results in

$h_r=\frac{d}{2}\frac{1-\sin \text{Θ}}{\cos \text{Θ}}$

21.2.4 Capillary Number

The relevant dimensionless quantity to assess the influence of the capillary forces compared to the viscous forces of the fluid is the capillary number Ca (see section 9.9.10). In strongly porous materials, the capillary number is usually in the range of Ca < 1 × 10⁻⁵, in which case the viscous forces of the fluid can be ignored, and the flow is purely driven by capillary forces. Driving microfluidic flows with capillarity is one of the oldest and still most commonly used ways of controlling fluid motion.

21.3.1 Derivation Using the Surface Curvature

The first approach we will take is based on the curvature of the surface because of the Young-Laplace pressure drop across the curved interface. The shape of the meniscus is due to the balance of forces from gravitation and the Young-Laplace pressure drop at the curved surface. For a fluid segment dx, the force balance is

$\begin{array}{ll}{p}_{\text{gravity}}\hfill & =p_{\text{Young-Laplace}}\hfill \\ \frac{d\left(mg\right)}{dA}\hfill & =\frac{\gamma }{R\left(x\right)}\hfill \\ \frac{\rho gdV}{dA}\hfill & =\gamma \frac{d^{2}z\left(x\right)}{d{x}^2}\hfill \\ \rho gz\left(x\right)\hfill & =\gamma \frac{d^{2}z\left(x\right)}{d{x}^2}\hfill \\ \frac{\rho g}{\gamma }z\left(x\right)\hfill & =\frac{d^{2}z\left(x\right)}{d{x}^2}\hfill \end{array}$

where we used Eq. 3.103 to approximate the curvature. Noted that this equation is valid only if the changes in curvature $\frac{d^{2}z}{d{x}^2}=\frac{d}{dx}\left(\frac{dz}{dx}\right)$ are small because the function is a linearization of Eq. 3.102. As we will see, this approximation is incorrect, especially in close proximity to the wall where the meniscus is strongly curved. However, using the full equation of the surface curvature results in a differential equation that can only be solved numerically.

Eq. 21.9 is an ordinary second order differential equation with constant coefficients (see section 8.2.3), which is solved by a function of type

$z\left(x\right)=c_1{\text{e}}^{\sqrt{\frac{\rho g}{\gamma }}x}+c_2{\text{e}}^{-}{}^{\sqrt{\frac{\rho g}{\gamma }}x}$

The details of this solution can be found in section 8.2.3.6. We see that the surface smooths out the farther we move away from the vessel wall; therefore z (x) → 0 for x → ∞, in which case we find that c₁ = 0. The second integration constant is found for the boundary condition at x = 0 where $\frac{dz}{dx}=-\frac{1}{\tan \left({\text{Θ}}_0\right)}$ because $\frac{dx}{dz}=-\tan \left({\text{Θ}}_0\right)$ at x = 0. Therefore we find

$\begin{array}{ll}\frac{dx\left(x=0\right)}{dx}=-c_2\sqrt{\frac{\rho g}{\gamma }}\hfill & =-\frac{1}{\tan \left({\text{Θ}}_0\right)}\hfill \\ c_2\hfill & =\frac{1}{\tan \left({\text{Θ}}_0\right)\sqrt{\frac{\rho g}{\gamma }}}\hfill \end{array}$

which leads to the final solution

$z\left(x\right)=\frac{1}{\tan \left({\text{Θ}}_0\right)\sqrt{\frac{\rho g}{\gamma }}}{\text{e}}^{-\sqrt{\frac{\rho g}{\gamma }}x}$

$z\left(x\right)=\frac{L_e}{\tan \left({\text{Θ}}_0\right)}{\text{e}}^{-\frac{x}{L_c}}$

The meniscus therefore decays exponentially. In Eq. 21.12 we used the capillary length (see Eq. 21.5), which is characteristic length scale for the decay.

As already stated, this result is a crude approximation because the meniscus near the wall is strongly curved, which is in contradiction to our initial assumption of a not-strongly curved meniscus. This becomes especially obvious for liquid that wets the vessel wall well. For ${\text{Θ}}_0\to \frac{\pi }{2}$ (perfect wetting) Eq. 21.12 will converge to infinity, which is (obviously) incorrect.

21.3.2 Derivation Without Linearization

In order to find the correct solution, we need to apply the surface curvature without linearizing it. As already stated, this results in a nonlinear ordinary differential equation that is extremely difficult to solve analytically.

However, there are a couple of ways of solving this problem. Looking at Fig. 21.3, we first note the following relations

$dx=\cos \phi ds$

$dz=-\sin \phi ds$

$ds=d\phi R\to \frac{1}{R}=\frac{d\phi }{ds}$

Using Eq. 21.15, we can rewrite Eq. 21.9 to

$\begin{array}{ll}\frac{\rho g}{\gamma }z\hfill & =\frac{d\phi }{ds}\hfill \\ \frac{1}{L_c^2}z\hfill & =\frac{d\phi }{ds}\hfill \end{array}$

this time applying the nonlinearized form $\frac{d\phi }{ds}$ to express the surface curvature. We now take the derivative with respect to s and apply Eq. 21.14

$\begin{array}{ll}\frac{1}{L_c^2}\frac{dz}{ds}z\hfill & =\frac{d^2\phi }{d{s}^2}\hfill \\ \frac{1}{L_c^2}\sin \phi \hfill & =\frac{d^2\phi }{d{s}^2}\hfill \end{array}$

We now apply a small trick in order to convert the double integral $\frac{d^2\phi }{d{s}^2}$ to a single integral (see section 3.3.2) by exploiting the fact that

$\frac{d^2\phi }{d{s}^2}\frac{d\phi }{ds}=\frac{d}{ds}\left(\frac{1}{2}\frac{d{\phi }^2}{ds}\right)$

which we can employ after multiplying Eq. 21.16 by $\frac{d\phi }{ds}$

$\begin{array}{l}\frac{1}{L_c^2}\sin \phi \frac{d\phi }{ds}=\frac{d^2\phi }{d{s}^2}\frac{d\phi }{ds}\hfill \\ \frac{1}{L_c^2}\sin \phi \frac{d\phi }{ds}=\frac{d}{ds}\left(\frac{1}{2}\frac{d{\phi }^2}{ds}\right)\hfill \\ \frac{1}{L_c^2}\sin \phi d\phi =d\left(\frac{1}{2}\frac{d{\phi }^2}{ds}\right)\hfill \end{array}$

Eq. 21.17 can be integrated to result in

$-\frac{1}{L_c^2}\cos \phi +c_1=\frac{1}{2}\frac{d{\phi }^2}{ds}$

The integration constant is determined by the fact that for ϕ → 0: $\frac{d\phi }{ds}=0$, in which case c₁ = L_c, which results in

$\begin{array}{ll}\frac{2}{L_c^2}\left(1-\cos \phi \right)\hfill & =\frac{d{\phi }^2}{ds}\hfill \\ \pm \frac{\sqrt{2}}{L_c}\sqrt{2\frac{1-\cos \phi }{2}}\hfill & =\frac{d\phi }{ds}\hfill \end{array}$

Here we can exploit the fact that $\pm \sqrt{\frac{1-\cos \phi }{2}}=\text{sin}\frac{\phi }{2}$ (see Eq. 3.35), which results in

$\frac{2}{L_c}\sin \frac{\phi }{2}=\frac{d\phi }{ds}$

Eq. 21.19z contains the implicit solution to the contour of the meniscus. It is expressed as a function of the angle φ and the path variable s. As we will see, it is analytically possible but very tedious to derive the function z (x) from Eq. 21.19. It is significantly easier to derive two functions from it, i.e., x (φ) and z (φ). Using these two equations, it is possible to calculate the contour implicitly by finding x and y as functions of the angle φ. The value range for φ is given by the initial contact angle ${\phi }_0=\frac{\pi }{2}-{\text{Θ}}_0$ and φ → 0 for x → ∞.

Deriving x (φ). We will start by deriving x (φ). For this we replace ds in Eq. 21.19 using Eq. 21.13, resulting in

$\frac{2}{L_c}dx=\frac{\cos \phi }{\sin \frac{\phi }{2}}d\phi$

which can be integrated to result in

$\frac{2}{L_c}x+c_2=4\cos \frac{\phi }{2}+2\ln \left(\frac{1}{\sin \frac{\phi }{2}}-\frac{1}{\tan \frac{\phi }{2}}\right)$

The integral is not straightforward, and it is best to locate it in a suitable integral table or by using, e.g., Maple. The integration constant c₂ can be found for x = 0: φ = φ₀, in which case we find

$c_2=4\cos \frac{{\phi }_0}{2}+2\ln \left(\frac{1}{\sin \frac{{\phi }_0}{2}}-\frac{1}{\tan \frac{{\phi }_0}{2}}\right)$

This allows us to rewrite Eq. 21.19 as

$\begin{array}{l}\frac{2}{L_c}x=4\cos \frac{\phi }{2}+2\ln \left(\frac{1}{\sin \frac{\phi }{2}}-\frac{1}{\tan \frac{\phi }{2}}\right)-\left(4\cos \frac{{\phi }_0}{2}+2\ln \left(\frac{1}{\sin \frac{{\phi }_0}{2}}-\frac{1}{\tan \frac{{\phi }_0}{2}}\right)\right)\hfill \\ \frac{1}{L_c}x=2\cos \frac{\phi }{2}+\ln \left(\frac{1}{\sin \frac{\phi }{2}}-\frac{1}{\tan \frac{\phi }{2}}\right)-\left(2\cos \frac{{\phi }_0}{2}+\ln \left(\frac{1}{\sin \frac{{\phi }_0}{2}}-\frac{1}{\tan \frac{{\phi }_0}{2}}\right)\right)\hfill \\ x=L_c\left(2\cos \frac{\phi }{2}+\ln \left(\frac{1}{\sin \frac{\phi }{2}}-\frac{1}{\tan \frac{\phi }{2}}\right)-\left(2\cos \frac{{\phi }_0}{2}+\ln \left(\frac{1}{\sin \frac{{\phi }_0}{2}}-\frac{1}{\tan \frac{{\phi }_0}{2}}\right)\right)\right)\hfill \end{array}$

Eq. 21.21 allows us to find the values for x for a given angle φ.

Deriving z (φ). We now turn to finding the function z (φ), which can be derived by replacing ds in Eq. 21.19 using Eq. 21.14, resulting in

$\frac{2}{L_c}dz=\frac{\sin \phi }{\sin \frac{\phi }{2}}d\phi$

which can be integrated as

$\begin{array}{l}\frac{2}{L_c}z=4\sin \frac{\phi }{2}\hfill \\ x=2\sqrt{L_c}\sin \frac{\phi }{2}\hfill \end{array}$

Again, the integral $\frac{\text{sin}\phi }{\text{sin}\frac{\phi }{2}}d\phi$ is best located in an integration table or by using an algebra tool. Eq. 21.22 is the sought equation. We now have two equations to calculate the values x (φ) and z (φ) and thereby implicitly find the meniscus contour.

21.3.3 Meniscus Contours

One important aspect that we can already see when looking at Eq. 21.21 and Eq. 21.22 is that they are both scaled to the capillary length L_c. Therefore the derived equations are correct for all liquids, irrespective of their physical properties because for each liquid, the capillary length is a constant. Therefore it is not necessary to calculate the meniscus shape for different liquids. If the wetting contact angle Θ₀ at the wall is known, the profile can be derived from Eq. 21.21 and Eq. 21.22.

Fig. 21.4 shows the calculated profiles for the meniscus at different wetting angles at the wall, starting from Θ₀ = 0° (perfectly wetting) as the topmost profile all the way to Θ₀ = 180° (perfectly non-wetting) as the bottommost profile in increments of 20°. Please note that Eq. 21.21 is discontinuous for Θ₀ = 90°, which is why this profile is not shown. However, for Θ₀ = 90°, there is no meniscus formation, and the surface remains flat. Fig. 21.4a shows the calculated meniscus shapes for water at STP conditions. As stated, the profiles will look identical for all liquids as long as Θ₀ is known. Fig. 21.4b shows this general meniscus shape normalized to the capillary length, the latter of course being different for different liquids.