- #1

member 428835

Hi PF!

Suppose a 2D rectangular channel with wall-normal perpendicular to the downward gravity force ##F## rests is static equilibrium. What shape would the liquid take?

Looking at the equations of motion, I think the following would have to be satisfied: $$\frac{d P}{dy} = -\rho F\\

\Delta P = \sigma k_1$$

1) Eulers equation for 1D (since gravity not in horizontal ##x##-direction)

2) Laplace pressure jump where ##\Delta P## is pressure jump from inside to outside liquid interface and 2D implies curvature ##k_2 = 0##, thus not listed.

Integrating Eulers eq yields ##P = -\rho F y + c##.

If we call the meniscus ##\Gamma(x)##, then it's curvature must be ##k_1(x) = \Gamma''(x) / (1+\Gamma(x)^2)^{3/2}##. Evaluating the pressure along the meniscus and applying the Laplace pressure jump implies $$\sigma \frac{\Gamma ''(x)}{(1+\Gamma'(x)^2 )^{3/2}} = \rho F \Gamma(x)$$ subject to $$\Gamma'(0) = -\cot \theta\\ \Gamma'(1) = 0$$ where I assume the channel has length 2. But how is this equation solved; the IC's are both at the first derivative.

For help purposes, seems like a good question for @Chestermiller and @boneh3ad though anyone's thoughts are welcomed.

Suppose a 2D rectangular channel with wall-normal perpendicular to the downward gravity force ##F## rests is static equilibrium. What shape would the liquid take?

Looking at the equations of motion, I think the following would have to be satisfied: $$\frac{d P}{dy} = -\rho F\\

\Delta P = \sigma k_1$$

1) Eulers equation for 1D (since gravity not in horizontal ##x##-direction)

2) Laplace pressure jump where ##\Delta P## is pressure jump from inside to outside liquid interface and 2D implies curvature ##k_2 = 0##, thus not listed.

Integrating Eulers eq yields ##P = -\rho F y + c##.

If we call the meniscus ##\Gamma(x)##, then it's curvature must be ##k_1(x) = \Gamma''(x) / (1+\Gamma(x)^2)^{3/2}##. Evaluating the pressure along the meniscus and applying the Laplace pressure jump implies $$\sigma \frac{\Gamma ''(x)}{(1+\Gamma'(x)^2 )^{3/2}} = \rho F \Gamma(x)$$ subject to $$\Gamma'(0) = -\cot \theta\\ \Gamma'(1) = 0$$ where I assume the channel has length 2. But how is this equation solved; the IC's are both at the first derivative.

For help purposes, seems like a good question for @Chestermiller and @boneh3ad though anyone's thoughts are welcomed.

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