laplace equation surface tension derivationcoursera learner support team

The Laplace pressure reflects the energetic cost of creating additional interfacial area, as more surface area corresponds to additional unsatisfied bonds. Gravitation on the other hand stretches the drop from this spherical shape and the typical pear-like shape results. We write $-k^2$ for this constant, so that, \begin{equation} \frac{1}{Z} \frac{d^2 Z}{dz^2} = k^2, \tag{10.51} \end{equation}, \begin{equation} Z(z) = e^{\pm kz} \tag{10.52} \end{equation}, \begin{equation} Z(z) = \left\{ \begin{array}{l} \cosh(k z) \\ \sinh(k z) \end{array} \right. Appendix 2: Derivation of Young-Laplace and Kelvin Equations . Where the cross sectional area is large there is lower curvature resulting in lower pressure. Yet the property remains unchanged, that once the potential of the conductor is specified on each boundary, the solution to Laplace's equation formula between boundaries will be a unique solution. @9 3)*W=ee^)d=Q ]0wtJE%0# $7WY fwNMHcqY8AgEMesYe Fa'd>XvTUm&1zDsNx42BfR;5h>%Ht.0tCo"A,IAzBxb&Q:A>n has solutions that must blow up at $u=-1$ even when they are finite at $u = 1$, unless $\lambda$ is a nonnegative integer. And because the potential should not blow up at the centre of the sphere, we must also kill all $r^{-(\ell+1)}$ terms by setting $B^m_\ell = 0$. We also deduce that the relevant spherical harmonics will present themselves in the combinations identified previously; in particular, $Y^2_2$ and $Y^{-2}_2$ will come together to form $\frac{1}{2} \sin^2\theta \cos(2\phi)$. Beginning with the Laplacian in cylindrical coordinates, apply the operator to a potential function and set it equal to zero to get the Laplace equation. [15][9][16], A pendant drop is produced for an over pressure of p, A liquid bridge is produced for an over pressure of p. If the surface is to be in mechanical equilibrium, the two work terms as given must be equal, and on equating them and substituting in the expressions for dx and dy, the final result obtained is (1.20) Equation 11-7 is the fundamental equation of capillarity and is well known as Young-Laplace equation. The parameters $\alpha$ and $\beta$ are now determined in terms of the positive integers $n$ and $m$, and the factorized solutions become, \begin{equation} V_{n,m}(x,y,z) = \sin\Bigl( \frac{n\pi x}{a} \Bigr) \sin\Bigl( \frac{m\pi y}{a} \Bigr) \left\{ \begin{array}{l} e^{\sqrt{n^2+m^2}\, \pi z/a} \\ e^{-\sqrt{n^2+m^2}\, \pi z/a} \end{array} \right\} . Additional information, like the value of the total charge $q$, is therefore required for a unique solution. This exercise was actually carried out back in Sec.8.3, where we found that the boundary potential can be written as, \begin{equation} V(R,\theta,\phi) = V_0 \biggl[ \frac{1}{3} - \frac{1}{6} (3\cos^2\theta - 1) + \frac{1}{2} \sin^2\theta \cos(2\phi) \biggr]. Pin = Pressure Inside the Curved Surface ; Pout = Pressure Outside the Curved Surface ; = Surface Tension ; r = Radious of Curvature of the Curved Surface ; In order to minimize the surface area of a liquid, a formation of a curved suface occur. Pierre Simon Laplace followed this up in Mcanique Cleste[11] with the formal mathematical description given above, which reproduced in symbolic terms the relationship described earlier by Young. In general, the Laplace equation can be written as 2 f=0, where f is any scalar function with multiple variables. The excess of pressure is P i - P o. And once again the factorized solutions will gradually be refined by imposing the boundary conditions; because the box has six sides, there are six conditions to impose on the potential. Here, a typical boundary-value problem asks for $V$ between conductors, on which $V$ is necessarily constant. The Laplace equation formula was first found in electrostatics, where the electric potential V, is related to the electric field by the equation. Exercise 10.3: Prove that the expansion coefficients of the double sine Fourier series of Eq. As a second example of boundary-value problem, we wish to solve Laplace's equation $\nabla^2 V = 0$ inside a sphere of radius $R$, on which the potential is equal to $V(R,\theta,\phi) = V_0 \sin^2\theta \cos^2\phi$. Therefore the Laplace . Before the immersion the field was truly constant, but the arrival of the conductor distorts the electric field, because of the induced charge distribution on the surface; the final field is not quite uniform. (10.86) satisfies $\nabla^2 V = 0$ and becomes $V_0 \sin^2\theta \cos^2\phi$ when $r = R$. Incorporating this in Eq. Ans: The solution of the Laplace equation is the harmonic functions that are most widely used in many branches of engineering and Physics. for the unique solution to the boundary-value problem, expressed as an infinite sum of products of Bessel and exponential functions. The Laplace equation in Physics consists of two important properties. (10.29) in the case of the parallel plates. \tag{10.56} \end{equation}, This is a second-order differential equation for $S(s)$, and its form can be simplified by introducing the new variable $u := k s$. This gives rise to the same kind of contradiction that we encountered before in Sec.10.2, and to escape it we must declare that these functions are in fact constant. The Laplace equation is derived to make the calculations in Physics easier and it is named after the physicist Pierre-Simon Laplace. (10.63) gives, \begin{equation} V(s,z) = 2 V_0 \sum_{p=1}^\infty \frac{1}{\alpha_{0p} J_1(\alpha_{0p})} J_0(\alpha_{0p} s/R)\, e^{-\alpha_{0p} z/R} \tag{10.68} \end{equation}. The wall of the pipe at $s = R$ is maintained at $V = 0$, and the base of the pipe at $z = 0$ is maintained at $V = V_0$. For convenience we write $f(x) = \alpha^2 = \text{constant}$, or, \begin{equation} \frac{1}{X} \frac{d^2 X}{dx^2} = -\alpha^2. Remarkably, and this is not typical of such problems, the series of Eq. This led to what we now know as Young's law. 5 5. Exercise 10.1: Verify these results for the expansion coefficients $b_n$. Evaluating the potential of Eq. This requires finding the solution to the boundary-value problem specified by Laplace's equation $\nabla^2 V = 0$ together with the boundary conditions $V(x=0, y) = 0$, $V(x=L, y) = 0$, and $V(x,y=0) = V_0$. We shall need the curvilinear coordinates of Chapter 1, the special functions of Chapters 2, 3, 4, 5, and 6, and the expansion in orthogonal functions of Chapters 7, 8, and 9. In general, the potential V is independent of the variables x,y, and z and the differential equation must be integrated to explain the simultaneous dependence of the potential V on these three variables. [12][13] The part which deals with the action of a solid on a liquid and the mutual action of two liquids was not worked out thoroughly, but ultimately was completed by Carl Friedrich Gauss. According to the Young-Laplace equation, with a curved liquid . 1. The shape of liquid drop is governed by what is known as the Young-LaPlace equation. This is the case here also, as suggested by the fact that the coefficients decrease as $1/n$ with increasing $n$. This yields, \begin{equation} u^2 \frac{d^2 S}{du^2} + u \frac{dS}{du} + (u^2 - m^2) S = 0, \tag{10.57} \end{equation}, and comparison with Eq. The Laplace equation is derived to make the calculations in Physics easier and it is named after the physicist Pierre-Simon Laplace. (Boas Chapter 12, Section 7, Problem 1) Solve Laplace's equation inside a sphere of radius $R$ when the potential on the surface is given by $V(r=R,\theta) = 35\cos^4\theta$. second resultant force is due to surface tension on the wall. With $u = \cos\theta$, we are speaking of functions that become infinite at $\theta = \pi$, just like the $Y$ of Eq.(10.73). Potentials and conservative . brown mountain beach resort wedding. Writing the constant as $m^2$, we have that, \begin{equation} \frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2} = -m^2, \tag{10.46} \end{equation}, \begin{equation} \Phi(\phi) = e^{\pm im\phi} \tag{10.47} \end{equation}, \begin{equation} \Phi(\phi) = \left\{ \begin{array}{l} \cos(m\phi) \\ \sin(m\phi) \end{array} \right. Notably, these are used in electrostatics, gravitational, and fluid mechanics. If the surface tension is given by N, then the resultant force due to surface tension Ft is calculated as Ft = N2R. The $q/(4\pi \epsilon_0 r)$ term in the potential comes with a correction proportional to $r/R$, and this represents an irrelevant constant. Despite the passages of two centuries since the original derivation of the Laplace-Young equation, only two non-trivial exact solutions are known. This is the statement of the superposition principle, and it shall form an integral part of our strategy to find the unique solution to Laplace's equation with suitable boundary conditions. In the work of Laplace [1], he derived an equation to relate the pressure difference between interior and exterior of the liquid drop with the surface tension. (Boas Chapter 12, Section 2, Problem 11) Solve Laplace's equation for a potential $V(x,y)$ that satisfies the boundary conditions $V(x=0, y) = V_0$, $V(x=1,y) = 0$, $V(x, y=0) = V_0$, and $V(x, y=1) = 0$. A fourth boundary condition is implicit: the potential should vanish at $y = \infty$, so that $V(x, y=\infty) = 0$. Figure 3: Water rising inside a glass tube due to capillary action. The solution to this problem will be of the form of Eq. The form of the differential equation suggests the use of $r^\alpha$ as a trial solution, where $\alpha$ is a constant. Again we can go back and forth between the complex exponentials and the trigonometric functions, and the sign in front of $\beta^2$ can be altered by letting $\beta \to i\beta$. The classical Young-Laplace equation relates capillary pressure to surface tension and the principal radii of curvature of the interface between two immiscible fluids. In simpler terms, the equation helps describe the shape and behavior of . Suppose that $V_1$, $V_2$, $V_3$, and so on, are all solutions to Laplace's equation, so that $\nabla^2 V_j = 0$. The answer is no, because we are trespassing beyond the limits of the theorem. 2u=0,u is the velocity of the steady flow. This expansion is reminiscent of a sine Fourier series --- refer back to Sec.7.9 --- but the basis functions are functions of both $x$ and $y$. The second one is that $V = 0$ at $y=0$, and this eliminates $\cos(\beta y)$ from the factorized solutions. A widely used approach to calculate a minimum energy surface is by means of the Surface Evolver program.42 But several other approaches, both theoretical and numerical, have been used for studying Undergraduate Schedule of Dates which we shall insert within Laplace's equation. When these are nice planar surfaces, it is a good idea to adopt Cartesian coordinates, and to write. Any superposition of the form, \begin{equation} V =a_1 V_1 + a_2 V_2 + a_3 V_3 + \cdots, \tag{10.2} \end{equation}, where $a_j$ are constants, is also a solution, because, \begin{equation} \nabla^2 V = \nabla^2 \bigl( a_1 V_1 + a_2 V_2 + a_3 V_3 + \cdots \bigr) = a_1 \nabla^2 V_1 + a_2 \nabla^2 V_2 + a_3 \nabla^2 V_3 + \cdots = 0. which we claim is the unique solution to Laplace's equation that satisfies the stated boundary conditions. This mathematical operation is obtained in equation (2), the divergence of the gradient of a potential V is called the Laplacian equation. \tag{10.50} \end{equation}, Here we have that the function of $z$ on the left-hand side must be equal to the function of $s$ on the right-hand side. 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