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Young-Laplace equation

  1. Apr 9, 2014 #1
    Hello there,

    Can anyone explain me how to get to the Young-Laplace equation for surfaces with two radii of curvature from the Young-Laplace equation for axisymmetric surfaces?
    Take for instance the case in which a droplet is trapped between two plates, where we assume that the height between the two plates is much smaller than the radius of the pancake shaped droplet.

    Thanks,
    Rudy
     
  2. jcsd
  3. Apr 9, 2014 #2
    Hi RudyL90. Welcome to Physics Forums!!!!

    I'm a little confused. The situation you are describing is axisymmetric. Are you asking how to derive the Young-Laplace equation for this situation, or are you asking how to get the principle radii of curvature for this situation?

    Chet
     
  4. Apr 9, 2014 #3
    Hi Chet,

    Thanks for your fast reply. How do I get for instance for the situation that I described the radii of curvature from the Young-Laplace equation? How do I recognize which part of the equation belongs to R1 and which does to R2?

    Rudy
     
  5. Apr 9, 2014 #4
    You can't get the individual radii of curvature from the Young-Laplace equation. All you can do is get the sum of their reciprocals (and to do this, you need to know the pressure difference across the interface in advance). The radii of curvature are obtained from the differential geometry of the surface. These radii are then used to calculate the local pressure difference across the interface using the Young_Laplace equation. I don't know whether this answers your question or not.

    Chet
     
  6. Apr 9, 2014 #5

    Andy Resnick

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    Heh... that is a tricky one. It arises from the jump condition of normal stress across an interface, just as Young's equation for the three-phase line is obtained. Sources for this typically have a title similar to "generalized theory of capillarity", Neumann and Slattery are two good authors to check.

    I've also seen a derivation based on hydrostatics.

    In any case, under the appropriate simplifications, the pressure jump is proportional to the mean curvature, so boundary conditions are required to be more specific: pinning the contact line to a circle of known radius, for example. Gravity acts in one direction but not others, so the two principal radii can be different and also vary with height.
     
  7. Apr 9, 2014 #6
    If the height between the surfaces is small, you can neglect gravity. Let's use cylindrical polar coordinates, with z = 0 at the plane half-way between the plates, and with the axis at the centerline of the squished drop. A position vector from the origin to any point on the free surface is given by

    [tex]\vec{r}=r\vec{i}_r+z\vec{i}_z[/tex]

    We are going to represent the radial and axial locations on the free surface parameterically as a function of the arc length parameter s measured along the arc from the mid plane. Thus, r = r(s) and z = z(s), with r(0) = rmax, and z(0) = 0. In addition, [tex](ds)^2=(dr)^2+(dz)^2[/tex]
    This equation can be satisfied automatically if we make use of the contour angle ø of the arc, defined such that
    [tex]dr=-sin(\phi(s))ds[/tex]
    [tex]dz=cos(\phi(s))ds[/tex]
    with the initial condition [tex]\phi(0)=0[/tex]
    By introducing the angle ø(s) into the analysis, we have reduced the number of dependent variables from two to one. Once the function ø(s) is specified, the entire shape of the free surface is established (by integrating the above differential equations for r and z, subject to the initial conditions on r and z at s = 0).

    A differential position vector within the surface is given by:

    [tex]d\vec{r}=\vec{i}_rdr+\vec{i}_zdz+\vec{i}_θrdθ=[-sin\phi\vec{i}_r+
    cos\phi\vec{i}_z]ds+rdθ\vec{i}_θ[/tex]
    Let's do a force balance on a small "rectangular" window of the free surface between s and s+Δs, and θ and θ+Δθ. The sides of the rectangle are of length Δs and rΔθ. The membrane force acting on the side at s+Δs is given by [itex]σ[-sin\phi\vec{i}_r+
    cos\phi\vec{i}_z]_{s+Δs}rΔθ[/itex], where σ is the surface tension. The membrane force acting on the side at s+Δs is given by [itex]-σ[-sin\phi\vec{i}_r+
    cos\phi\vec{i}_z]_srΔθ[/itex]. The membrane force on the side at θ+Δθ is given by [itex]σ[\vec{i}_θ]_{θ+Δθ}Δs[/itex]. The membrane force on the side at θ is given by [itex]-σ[\vec{i}_θ]_{θ}Δs[/itex]. The area of the window is rΔsΔθ. The unit normal vector to the window is given by [itex]cos\phi\vec{i}_r+sin\phi\vec{i}_z[/itex]. So, the differential force balance on the window is given by:

    [tex](p_{in}-p_{out})(cos\phi\vec{i}_r+sin\phi\vec{i}_z)rΔsΔθ+σ[-sin\phi\vec{i}_r+
    cos\phi\vec{i}_z]_{s+Δs}rΔθ-σ[-sin\phi\vec{i}_r+
    cos\phi\vec{i}_z]_srΔθ+σ[\vec{i}_θ]_{θ+Δθ}Δs-σ[\vec{i}_θ]_{θ}Δs=0[/tex]
    If we divide this equation by rΔsΔθ and take the limit as Δs and Δθ approach zero, we obtain:
    [tex](p_{in}-p_{out})(cos\phi\vec{i}_r+sin\phi\vec{i}_z)+σ\frac{∂[-sin\phi\vec{i}_r+
    cos\phi\vec{i}_z]}{∂s}+\frac{σ}{r}\frac{∂\vec{i}_θ}{∂θ}=0[/tex]
    If we evaluate the derivatives in the above equation, we obtain:
    [tex](p_{in}-p_{out})(cos\phi\vec{i}_r+sin\phi\vec{i}_z)-σ(cos\phi\vec{i}_r+sin\phi\vec{i}_z)\frac{d\phi}{ds}-\frac{σ}{r}\vec{i}_r=0[/tex]
    If we take the dot product of this equation with the unit normal to the free surface, we obtain:
    [tex](p_{in}-p_{out})=σ\left(\frac{d\phi}{ds}+\frac{1}{r}cos\phi \right)[/tex]
    From this equation, we can see that the principal radii of curvature are [itex]\frac{1}{d\phi/ds}[/itex] and [itex]\frac{r}{cos\phi}[/itex]

    Hope this helps.

    Chet
     
    Last edited: Apr 9, 2014
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