Съдържание

APhO 2009, Теория — Задача 3. The Leidenfrost Phenomenon

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Всички материали от APhO 2009

Theoretical competition, 26 April 2009

Внимание: Бележка към темата

The complete archive solution originals for all three questions are now included. Their printed competition headings match the question packet; organizer-host provenance was not independently established.

Obtained source packet

All question and solution originals print “Theoretical competition”, 26 April 2009, and the APhO2009 Bangkok, Thailand logo. The three separate archive keys match the three complete questions and all 21 native numbered requests. The complete archive solutions are reproduced, including three methods for 1.7 and two methods for 2.6. No separate theory answer sheet, marking-scheme file, common instruction sheet or constants companion was obtained from the complete theory folder; no explicit total points or time limit is printed. Native per-part marks are retained. Archive appearance alone is not proof of organizer-host provenance.

Условие

The purpose is to estimate the lifetime of a (hemispherical) drop of a liquid sitting on top of a very thin layer of vapour which is thermally insulating the drop from the very hot plate below.

Perspective sketch of a hemispherical drop on a flat plate; labels: drop, very hot surface, thin layer of vapour (a thin layer between the drop and the plate). — Figure 1

It will be assumed here that the flow of vapour underneath the drop is streamline and behaves as a Newtonian fluid of viscosity coefficient $\eta$ and of thermal conductivity $\mathcal{K}$ . The specific latent heat of vaporization of the liquid is $\ell$ . And for a Newtonian fluid we have the shear stress $\dfrac{F}{A} = \eta \times$ the rate of shear $\dfrac{dv}{dz}$ where $v$ is the flow velocity and $z$ is the perpendicular distance to the direction of flow, and the direction of $F$ is tangential to the surface area $A$ .

Cross-section of a hemispherical drop of radius R at distance b above the hot surface, with vertical axis Z through centre O, gravity arrow g, a dotted mid-plane of the vapour layer, the velocity v at height z above the mid-plane, and radial distances r and δr marked along the hot surface. — Figure 2

$v$ is the velocity of vapour in the radial direction at the height $z$ above the mid-plane. The pressure $P$ inside the vapour must be higher towards the centre O. This will result in the out-flowing of vapour and in force that holds the drop against the pull of gravity. The thickness of vapour layer under thermal and mechanical equilibria is $b$ .

For a Newtonian flow of vapour we can approximate that

\frac{d}{dz}v = \frac{z}{\eta}\frac{d}{dr}P

3.1) Show that

v(z) = \frac{z^2}{2\eta}\frac{d}{dr}P + C

where $C$ is an arbitrary constant of integration. [0,5 т.]

3.2) Refer to figure 2, find the value of $C$ in terms of $\eta, \dfrac{d}{dr}P,$ and $b$ using the boundary condition $v = 0$ for $z = \pm\dfrac{b}{2}$ . [0,5 т.]

3.3) Calculate the volume rate of flow of vapour through the cylindrical surface defined by $r$ . (Hint: the cylinder is of radius $r$ and of height $b$ underneath the drop). [1 т.]

3.4) By assuming that the rate of production of vapour of density $\rho_{\mathrm{V}}$ is due to heat flow from the hot surface to the drop, find the expression for the pressure $P(r)$ . Use $P_{\mathrm{a}}$ to represent the atmospheric pressure, and use $\Delta T$ for the temperature difference between those of the hot surface and of the drop. Assume that the system has reached the steady state. [2 т.]

3.5) Calculate the value of $b$ by equating the weight of the drop to the net force due to pressure difference between the bottom and the top of the drop. The density of the drop is $\rho_0$ . [2 т.]

3.6) Now, what is the total rate of vaporization? [2 т.]

3.7) Assume that the drop maintains a hemispherical shape, what is the life-time of the drop? [2 т.]

Решение

The Leidenfrost Phenomenon — archive solution original

3.1) As given in the problem

\frac{dv}{dz}=\left(\frac1\eta\frac{dP}{dr}\right)\cdot z.\qquad\text{(i)}

Integrating (i) with respect to $z$ , we get

v(z)=\left(\frac1{2\eta}\frac{dP}{dr}\right)\cdot z^2+C.\qquad\text{(ii)}\quad\text{(0.5 point)}

3.2)

v\left(\frac b2\right)=0=\left(\frac1{2\eta}\frac{dP}{dr}\right)\cdot\left(\frac b2\right)^2+C.\qquad\text{(iii)}

Therefore

C=-\frac{b^2}{8\eta}\frac{dP}{dr}.\quad\text{(0.5 point)}

Note that $C$ is not a real constant; its value depends on $\frac{dP}{dr}$ which is a function of $r$ .

3.3) Let $Q$ be the volume rate of flow of the vapour through the cylindrical surface of $2\pi rb$ .

\delta Q=v(z)\cdot2\pi r\delta z\quad\text{(0.3 point)}

where from (ii) and (iii):

v(z)=\left(\frac1{2\eta}\frac{dP}{dr}\right)\cdot\left[z^2-\frac{b^2}{4}\right].\qquad\text{(iv)}

Therefore

Q=2\int_{z=0}^{b/2}v(z)\cdot2\pi r\,dz=\left(\frac{2\pi r}{\eta}\frac{dP}{dr}\right)\int_{z=0}^{b/2}\left[z^2-\frac{b^2}{4}\right]dz,

Q=-\frac{\pi rb^3}{6\eta}\frac{dP}{dr}.\qquad\text{(v)}\quad\text{(0.7 point)}

3.4) The total rate of heat flow from the area $\pi r^2$ of the hot surface to the drop is $\frac{\pi r^2\mathcal K\Delta T}{b}$ . We assume that this heat goes into vaporizing the drop.

Hence

\rho Q\ell=\frac{\pi r^2\mathcal K\Delta T}{b},

and using (v) we get

\frac{dP}{dr}=-\left(\frac{6\eta\mathcal K\Delta T}{\rho_{\mathrm V}\ell b^4}\right)\cdot r.\qquad\text{(vi)}\quad\text{(0.4 point)}

This gives

P(r)=-\left(\frac{3\eta\mathcal K\Delta T}{\rho_{\mathrm V}\ell b^4}\right)\cdot r^2+B.\quad\text{(0.4 point)}

where $B$ is an arbitrary constant whose value can be found by applying the boundary condition $P(R)=P_{\mathrm a}$ , the atmospheric pressure. Hence

B=P_{\mathrm a}+\left(\frac{3\eta\mathcal K\Delta T}{\rho_{\mathrm V}\ell b^4}\right)\cdot R^2,\qquad\text{(vii)}\quad\text{(0.4 point)}

and

P(r)=P_{\mathrm a}+\left(\frac{3\eta\mathcal K\Delta T}{\rho_{\mathrm V}\ell b^4}\right)\cdot(R^2-r^2).\qquad\text{(viii)}\quad\text{(0.8 point)}

3.5) The net force due to pressure is in the upward direction and of magnitude

f=\int_{r=0}^R[P(r)-P_{\mathrm a}]2\pi r\,dr=\frac{3\pi\eta\mathcal K\Delta TR^4}{2\rho_{\mathrm V}\ell b^4}.\qquad\text{(ix)}\quad\text{(1.0 point)}

The weight of the drop is $\frac23\pi R^3\rho_0g$ , where $\rho_0$ is the density of liquid. Therefore

\frac23\pi R^3\rho_0g=\frac{3\pi\eta\mathcal K\Delta TR^4}{2\rho_{\mathrm V}\ell b^4},

b=\left(\frac{9\eta\mathcal KR\Delta T}{4\rho_0\rho_{\mathrm V}\ell g}\right)^{1/4}.\qquad\text{(x)}

Note that

\frac{3\eta\mathcal K\Delta T}{\rho_{\mathrm V}\ell b^4}=\frac43\frac{\rho_0g}{R}.\qquad\text{(xi)}\quad\text{(1.0 point)}

3.6) Use equations (xi) and (viii) to obtain

P(r)=P_{\mathrm a}+\left(\frac43\frac{\rho_0g}{R}\right)\cdot(R^2-r^2),\qquad\text{(xii)}

\frac{dP(r)}{dr}=-\left(\frac83\frac{\rho_0g}{R}\right)\cdot r.\qquad\text{(xiii)}\quad\text{(0.8 point)}

Then use (v) to calculate the total mass-rate of vaporization $Q\rho_{\mathrm V}$ at $r=R$ :

Q\rho_{\mathrm V}=\left(\frac{2\pi b^3R}{12\eta}\right)\left(\frac83\frac{\rho_0g}{R}\right)R\rho_{\mathrm V}=\left(\frac{4\pi\rho_{\mathrm V}\rho_0gR}{9\eta}\right)b^3

=\left(\frac{4\pi\rho_{\mathrm V}\rho_0gR}{9\eta}\right)\left(\frac{9\eta\mathcal KR\Delta T}{4\rho_0\rho_{\mathrm V}\ell g}\right)^{3/4}

=\left(\frac{4\pi^4\mathcal K^3\rho_{\mathrm V}\rho_0g(\Delta T)^3}{9\eta\ell^3}\right)^{1/4}\cdot R^{7/4}=\beta R^{7/4}.\qquad\text{(xiv)}\quad\text{(1.2 points)}

3.7) The life-time $(\tau)$ of the drop is to be found from

\frac{d}{dt}\left(\frac23\pi R^3\rho_0\right)=-Q\rho_{\mathrm V}=-\beta R^{7/4},

R^{1/4}\frac{dR}{dt}=-\frac\beta{2\pi\rho_0},

\int_R^0R^{1/4}\,dR=-\int_0^\tau\frac\beta{2\pi\rho_0}\,dt,\quad\text{(1.0 point)}

\tau=\frac{8\pi\rho_0}{5\beta}R^{5/4}=\frac85\left(\frac{9\eta\rho_0^3\ell^3}{4\mathcal K^3\rho_{\mathrm V}g(\Delta T)^3}\right)^{1/4}\cdot R^{5/4}.\quad\text{(1.0 point)}

Printed source notation

The dynamo key says that $\gamma$ must be positive for current growth but prints $\mu_0Na^2\omega/(2\ell)-R\ge0$ and labels the equality value $\omega_{\min}$ . The original inequality and value are preserved; at equality the printed exponential has zero exponent. In the Leidenfrost key the first heat-balance equation prints bare $\rho$ , whereas subsequent equations use $\rho_{\mathrm V}$ ; both native forms are retained. The rolling-cylinder key changes the directions of $\psi$ and $\phi$ for 1.7 as explicitly stated, then returns to equation (i) in 1.8.

Оригинал в Архива: th_problems.pdf · решение от архива: 3_sol.pdf · допълнителни данни: 2_sol.pdf · допълнителни данни: 3_sol.pdf