Chapter 1, Problem 1.68P

To determine

The maximum height at plane wall.

Answer to Problem 1.68P

The maximum height is $h=\sqrt{\frac{\Upsilon }{\rho g}}\cot \theta$ at plane wall.

Explanation of Solution

Given Information:

We have below mentioned data,

Slope as, $\frac{1}{R}=\frac{d^2\eta }{d{x}^2}$,

The height as, h

Surface tension as, $\Upsilon$

The pressure difference across the interface of water-air, $\Delta p=\rho g\eta$

The wetting contact angle as, $\theta \text{ at x=0}$ and

Horizontal surface as, $\eta =0\text{ as x}\to \infty$

We can understand that the given figure is representing a two-dimensional surface-tension with single curvature.

We can observe that the weight of film balances the surface-tension rise that results into,

$\begin{array}{l}\Delta p=\frac{\Upsilon }{R}\\ \rho g\eta =\Upsilon \frac{d^{}\eta }{d{x}^2}\\ \frac{d^{}\eta }{d{x}^2}-\frac{\rho g}{\Upsilon }\eta =0\end{array}$

Above equation has a well-known solution and known as second order differential equation. The solution is as follows,

$\eta ={\text{C}}_1\exp \left[\sqrt{\frac{\rho g}{\Upsilon }}x\right]\text{+}{\text{C}}_{\text{2}}\text{exp}\left[\text{-}\sqrt{\frac{\rho g}{\Upsilon }}\text{x}\right]\mathrm{..}(1)$

In above equation, the values of ${\text{C}}_1$ and ${\text{C}}_2$ are found constant from boundary equations.

We have horizontal surface as $\eta =0\text{ as x}\to \infty$, putting in equation (1),

$\begin{array}{l}\eta ={\text{C}}_1\exp \left[\sqrt{\frac{\rho g}{\Upsilon }}x\right]\text{+}{\text{C}}_{\text{2}}\text{exp}\left[\text{-}\sqrt{\frac{\rho g}{\Upsilon }}\text{x}\right]\\ 0={\text{C}}_1\exp \left[\sqrt{\frac{\rho g}{\Upsilon }}x\right]\text{+}{\text{C}}_{\text{2}}\text{exp}\left[\text{-}\sqrt{\frac{\rho g}{\Upsilon }}\text{x}\right]\\ \text{}{\text{for satisfying the eqaution, the value of C}}_1\text{must be zero}\text{.}\\ \therefore {\text{ C}}_1=0\text{}\text{ and}\\ \eta ={\text{C}}_{\text{2}}\text{exp}\left[\text{-}\sqrt{\frac{\rho g}{\Upsilon }}\text{x}\right]\end{array}$

But $\begin{array}{l}\eta =h\text{at x=0}\\ {\text{h =C}}_{\text{2}}\text{exp}\left[\text{-}\sqrt{\frac{\rho g}{\Upsilon }}\text{x}\right]\\ {\text{we have C}}_2=\text{ h}\\ \eta \text{ = h exp}\left[\text{-}\sqrt{\frac{\rho g}{\Upsilon }}\left(x\right)\right]\end{array}$

${\eta }_{x=0}=\text{h}\text{=}{\text{C}}_{\text{2}}\text{exp}\left(\text{0}\right),$ hence ${\text{C}}_{\text{2}}\text{=}\text{h}$

Meanwhile, the contact angle present in the above equation must be such that,

$\begin{array}{l}\frac{d\eta }{dx}|{}_{x=0}=-\cot \theta \\ -h\sqrt{\frac{\rho g}{\Upsilon }}\exp \left(-\sqrt{\frac{\rho g}{\Upsilon }}\left(x\right)\right)|{}_{x=0}=-\cot \theta \\ -h\sqrt{\frac{\rho g}{\Upsilon }}\exp \left(-\sqrt{\frac{\rho g}{\Upsilon }}\left(0\right)\right)|{}_{x=0}=-\cot \theta \\ -h\sqrt{\frac{\rho g}{\Upsilon }}=\cot \theta \\ h=\sqrt{\frac{\Upsilon }{\rho g}}\cot \theta \end{array}$

Conclusion:

The maximum height is $h=\sqrt{\frac{\Upsilon }{\rho g}}\cot \theta$ at plane wall.