# Evaporation in an Inclined Microchannel

The analysis in the preceding subsection dealt with evaporation of a liquid to pure vapor when the evaporation was driven by heat transfer. Chakraborty and Som (2005) analytically investigated heat transfer in an evaporating thin liquid film moving slowly along the walls of an inclined microchannel, and this work will be introduced in this subsection.

Figure 9.28 Evaporation in an inclined microchannel (Chakraborty and Som, 2005; Reprinted with permission from Elsevier).

Figure 9.28 shows the lower half of the microchannel, in which quiescent air with a temperature ${{T}_{\infty }}$ and mass fraction ${{\omega }_{\infty }}$ is in the core of the channel while liquid film forms on the microchannel wall. It is assumed that the transport phenomena are two-dimensional and that the diffusion in the axial direction is negligible. The temperature distribution in the liquid film is assumed to be linear along the y-direction. The momentum equation that governs the liquid flow is

${{\mu }_{\ell }}\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}=\frac{\partial {{p}_{\ell }}}{\partial x}-{{\rho }_{\ell }}g\sin \theta =-F(x)$ (9.290)

and is subject to the following boundary conditions

u = 0 at y = 0 (9.291)
$\frac{\partial u}{\partial y}=0\text{ at }y=\delta$ (9.292)

Integrating eq. (9.290) twice and considering eqs. (9.291) and (9.292) yields

$u=\frac{1}{2{{\mu }_{\ell }}}F(x)(2\delta y-{{y}^{2}})$ (9.293)

The energy equation in the vapor phase is

${{V}_{\delta }}\frac{\partial {{T}_{v}}}{\partial y}={{\alpha }_{v}}\frac{{{\partial }^{2}}{{T}_{v}}}{\partial {{y}^{2}}}$ (9.294)

where Vs is the Stefan velocity due to evaporation at the liquid surface. Equation (9.294) is subject to the following boundary conditions

Tv = Tδ at y = δ (9.295)
${{T}_{v}}={{T}_{\infty }}\text{ at }y=h$ (9.296)

The solution of eq. (9.294) with boundary conditions eqs. (9.295) and (9.296) is

$\frac{{{T}_{v}}-{{T}_{\delta }}}{{{T}_{\delta }}-{{T}_{\infty }}}=\frac{\exp ({{V}_{\delta }}y/{{\alpha }_{v}})-\exp ({{V}_{\delta }}\delta /{{\alpha }_{v}})}{\exp ({{V}_{\delta }}\delta /{{\alpha }_{v}})-\exp ({{V}_{\delta }}h/{{\alpha }_{v}})}$ (9.297)

Similarly, the mass fraction of the vapor is

$\frac{\omega -{{\omega }_{\delta }}}{{{\omega }_{\delta }}-{{\omega }_{\infty }}}=\frac{\exp ({{V}_{\delta }}y/{{D}_{v}})-\exp ({{V}_{\delta }}\delta /{{D}_{v}})}{\exp ({{V}_{\delta }}\delta /{{D}_{v}})-\exp ({{V}_{\delta }}h/{{D}_{v}})}$ (9.298)

where Dv is the mass diffusivity of the vapor in the air.

The mass balance at the interface is

$\frac{d\Gamma }{dx}=-{{\rho }_{v}}{{V}_{\delta }}$ (9.299)

where $\Gamma =\int_{0}^{\delta }{{{\rho }_{\ell }}udy}$ is the mass flow rate per unit width. Considering the velocity profile in eq. (9.293), the mass balance becomes

${{\delta }^{3}}{F}'(x)+3{{\delta }^{2}}F(x)\frac{d\delta }{dx}=-3{{\nu }_{\ell }}{{\rho }_{v}}{{V}_{\delta }}$ (9.300)

Assuming the interface is impermeable to nonevaporating species, the mass balance at the interface can be written as

$\frac{-{{D}_{v}}{{\left. \frac{\partial \omega }{\partial y} \right|}_{y=\delta }}}{1-{{\omega }_{\delta }}}=-{{V}_{\delta }}$ (9.301)

Substituting eq. (9.298) into eq. (9.301) yields

${{V}_{\delta }}=\frac{{{D}_{v}}}{h-\delta }\ln \left( \frac{1-{{\omega }_{\infty }}}{1-{{\omega }_{\delta }}} \right)$ (9.302)

The energy balance at the interface is

$-{{k}_{v}}{{\left. \frac{\partial {{T}_{v}}}{\partial y} \right|}_{y=\delta }}+{{k}_{\ell }}{{\left. \frac{\partial {{T}_{\ell }}}{\partial y} \right|}_{y=\delta }}=-{{\rho }_{v}}{{h}_{\ell v}}{{V}_{\delta }}$ (9.303)

Substituting eq. (9.297) into eq. (9.303) and considering the linear temperature profile in the liquid film, the interfacial temperature becomes

${{T}_{\delta }}=\frac{-{{\rho }_{v}}{{h}_{\ell v}}{{V}_{\delta }}+{{k}_{\ell }}{{T}_{w}}/\delta -{{T}_{\infty }}f}{{{k}_{\ell }}/\delta -f}$ (9.304)

where

$f={{k}_{v}}\frac{{{V}_{\delta }}}{{{\alpha }_{v}}}\frac{\exp ({{V}_{\delta }}\delta /{{\alpha }_{v}})}{\exp ({{V}_{\delta }}\delta /{{\alpha }_{v}})-\exp ({{V}_{\delta }}h/{{\alpha }_{v}})}$ (9.305)

The pressure in the liquid film is

${{p}_{\ell }}={{p}_{\infty }}-\frac{\sigma }{R(x)}+\frac{A}{{{\delta }^{3}}}$ (9.306)
Figure 9.29 Local Nusselt number (Chakraborty and Som, 2005; Reprinted with permission from Elsevier).

where ${{p}_{\infty }}$ is the vapor pressure, R(x) is the radius of curvature at x, and the last term on the right-hand side accounts for the effect of disjoining pressure.

Figure 9.29 Local Nusselt number (Chakraborty and Som, 2005; Reprinted with permission from Elsevier). Therefore, the function F(x) in eq. (9.290) becomes

$F(x)=-\frac{\sigma }{{{R}^{2}}}\frac{dR}{dx}+\frac{3A}{{{\delta }^{4}}}\frac{d\delta }{dx}+{{\rho }_{\ell }}g\sin \theta$ (9.307)

Substituting eqs. (9.302) and (9.304) into eq. (9.300) and considering the variation of ${{\omega }_{\delta }},\text{ }{{\text{T}}_{\delta }},\text{ and }{{h}_{\ell v}}$, a fourth order ordinary differential equation of δ can be obtained. The boundary conditions of the ordinary differential equation include δ = δ0 at x = 0 and $\delta =d\delta /dx={{d}^{2}}\delta /d{{x}^{2}}=0\text{ at }x\to \infty$. The liquid film thickness can be obtained by a fourth order Runge-Kutta method. The local Nusselt number is then obtained by

$N{{u}_{x}}=\frac{{{h}_{x}}x}{{{k}_{\ell }}}=\frac{x}{\delta }$ (9.308)

Figure 9.29 shows the variation of the local Nusselt number along the axial direction for different δ0 / h. It can be seen that the Nusselt number increases with increasing x and eventually reaches a very high value. This represents the location where dryout occurs and the liquid film becomes very small. The Nusselt number increases with decreasing initial film thickness δ0 / h, but its effect becomes insignificant for δ0 / h < 0.01.

## References

Chakraborty, S., and Som, S.K., 2005, “Heat Transfer in an Evaporating Thin Liquid Film Moving Slowly Along the Walls of an Inclined Microchannel,” International Journal of Heat Mass Transfer, Vol. 45, pp. 2801-2805.