Critical Droplet Radius for Dropwise Condensation

From Thermal-FluidsPedia

As mentioned before, upon formation in their nucleation sites, droplets grow only if they form with a radius that exceeds the equilibrium radius. The analysis that leads to the definition of the critical equilibrium radius is presented below. A good place to start this derivation lies in the Gibbs free energy minimum principle. The Gibbs free energy, G, arises from one of the Legendre transformations for closed systems states, which states that

$G = G (T, p) = E - T S + p V$	(1)

where T, p, V, E, S are the system temperature, pressure, volume, internal energy, and entropy, respectively. Equation (1) can be differentiated to obtain

$d G = d E - T d S - S d T + p d V + V d p$	(2)

Assuming the only work term is of the pdV type, and neglecting potential and kinetic terms, the fundamental thermodynamic relationship is

$dE\le TdS-pdV$	(3)

Figure 1 Contact angle and equilibrium of surface tensions of a liquid droplet embyro.

Substituting eq. (3) into eq. (2) yields

$dG\le -SdT+Vdp$	(4)

Assuming that the system pressure and temperature are held at fixed values, the above expression reduces to

$dG\le 0$	(5)

Along with the assumption that the system pressure and temperature are fixed, G becomes the availability of the system, Ψ:

$Ψ = G = E + p 0 V - T 0 S$	(6)

where p₀ and T₀ are the reservoir pressure and temperature, respectively. Availability is defined as the maximum amount of work one can get from a system as it comes into equilibrium with a large reference environment with pressure p₀ and temperature T₀. It follows from this analysis that for equilibrium to occur in the system, dΨ = 0. Furthermore, stable equilibrium corresponds to a minimum value of Ψ.

Carey (1992) considered a system with supersaturated vapor adjacent to a vertical wall that has an initial temperature and pressure, T_v and p_v. The initial availability Ψ_o of this system is given by

${{\Psi }_{0}}={{m}_{total}}{{g}_{v}}+{{({{A}_{sv}})}_{I}}{{\sigma }_{sv}}$	(7)

where m_total is the total mass of the system, and g_v is the specific Gibbs free energy of the vapor phase and is a function of T_v and P_v. The second term on the right-hand side represents the contribution of the work done by surface tension at solid-vapor interface on the availability, and (A_sv)_I is the initial surface area shared by the solid and vapor of the system.

Now consider that the wall temperature is lowered and droplets begin to form on the surface of the wall. The total availability of the system is the sum of the availabilities of the liquid ${{\Psi }_{\ell }}$ , the vapor Ψ_v, and the interface Ψ_I. The expressions for each are as follows:

${{\Psi }_{\ell }}={{m}_{\ell }}[{{g}_{\ell }}+({{p}_{v}}-{{p}_{\ell }}){{v}_{\ell }}]$	(8)

where the second term in the brackets on the right-hand side corrects for the difference between the vapor and liquid, and

${{\Psi }_{v}}=({{m}_{total}}-{{m}_{\ell }}){{g}_{v}}$	(9)

${{\Psi }_{I}}={{A}_{\ell v}}{{\sigma }_{\ell v}}+{{({{A}_{sv}})}_{f}}{{\sigma }_{sv}}+{{A}_{s\ell }}{{\sigma }_{s\ell }}$	(10)

where the three terms on the right-hand side represent the work done by surface tension at the liquid-vapor, solid-vapor, and solid-liquid interfaces. (A_sv)_f is the difference between the initial surface area shared by the solid-vapor interface and the surface area shared by the solid-liquid interface.

${{({{A}_{sv}})}_{f}}={{({{A}_{sv}})}_{I}}-{{A}_{s\ell }}$	(11)

The areas are calculated directly from Fig. 1:

${{A}_{\ell v}}=2\pi {{r}^{2}}(1-\cos \theta )$	(12)

${{A}_{s\ell }}=\pi {{r}^{2}}(1-{{\cos }^{2}}\theta )$	(13)

The volume of the liquid droplet is

${{V}_{\ell }}=\frac{\pi {{r}^{3}}}{3}(2-3\cos \theta +{{\cos }^{3}}\theta )$	(14)

The change of the total availability of the system due to formation of the condensate embryo is then

$\begin{align} & \Delta \Psi =\left( {{\Psi }_{\ell }}+{{\Psi }_{v}}+{{\Psi }_{I}} \right)-{{\Psi }_{0}} \\ & \text{ }=\left[ {{m}_{\ell }}{{g}_{\ell }}+{{V}_{\ell }}({{p}_{v}}-{{p}_{\ell }}) \right.+({{m}_{total}}-{{m}_{\ell }}){{g}_{v}} \\ & \begin{matrix} {} & {} \\ \end{matrix}\left. +{{A}_{\ell v}}{{\sigma }_{\ell v}}+{{({{A}_{sv}})}_{f}}{{\sigma }_{sv}}+{{A}_{s\ell }}{{\sigma }_{s\ell }} \right]-\left[ {{m}_{total}}{{g}_{v}}-{{({{A}_{sv}})}_{I}}{{\sigma }_{sv}} \right] \\ & \text{ }={{m}_{\ell }}({{g}_{\ell }}-{{g}_{v}})+{{V}_{\ell }}({{p}_{v}}-{{p}_{\ell }}) \\ & \begin{matrix} {} & {} \\ \end{matrix}+{{A}_{\ell v}}{{\sigma }_{\ell v}}+[{{({{A}_{sv}})}_{f}}-{{({{A}_{sv}})}_{I}}]{{\sigma }_{sv}}+{{A}_{s\ell }}{{\sigma }_{s\ell }} \\ \end{align}$	(15)

Substituting eq. (11) into eq. (15), the change of availability becomes

$\Delta \Psi ={{m}_{\ell }}({{g}_{\ell }}-{{g}_{v}})+{{V}_{\ell }}({{p}_{v}}-{{p}_{\ell }})+{{A}_{\ell v}}{{\sigma }_{\ell v}}+{{A}_{s\ell }}({{\sigma }_{s\ell }}-{{\sigma }_{sv}})$	(16)

Substituting Young’s equation, eq. (1), into eq. (16), one obtains

$\Delta \Psi ={{m}_{\ell }}({{g}_{\ell }}-{{g}_{v}})+{{V}_{\ell }}({{p}_{v}}-{{p}_{\ell }})+({{A}_{\ell v}}-{{A}_{s\ell }}\cos \theta ){{\sigma }_{\ell v}}$	(17)

Substituting eqs. (12) and (13) into eq. (17), the following expression is obtained for a change in availability of the system:

$\begin{align} & \Delta \Psi =\Psi -{{\Psi }_{0}} \\ & ={{m}_{\ell }}({{g}_{\ell }}-{{g}_{v}})+{{V}_{\ell }}({{p}_{v}}-{{p}_{\ell }})+4\pi {{r}^{2}}{{\sigma }_{\ell v}}F \\ \end{align}$	(18)

where

$F=\frac{2-3\cos \theta +{{\cos }^{3}}\theta }{4}$	(19)

Considering that the pressure in the liquid droplet is related to the pressure in the vapor phase by

${{p}_{\ell }}-\text{ }{{p}_{v}}=2{{\sigma }_{\ell v}}/r$ and substituting eq. (14) into eq. (18), one obtains

$\Delta \Psi =\Psi -{{\Psi }_{0}}={{m}_{\ell }}({{g}_{\ell }}-{{g}_{v}})+\frac{4}{3}\pi {{r}^{2}}{{\sigma }_{\ell v}}F$	(20)

The mass of the embryo is

${{m}_{\ell }}=\frac{{{V}_{\ell }}}{{{v}_{\ell }}}=\frac{4}{3{{v}_{\ell }}}\pi {{r}^{3}}F$	(21)

Substituting eq. (21) into eq. (20), one obtains

$\Delta \Psi =\Psi -{{\Psi }_{0}}=\frac{4}{3{{v}_{\ell }}}\pi {{r}^{3}}F({{g}_{\ell }}-{{g}_{v}})+\frac{4}{3}\pi {{r}^{2}}{{\sigma }_{\ell v}}F$	(22)

If the embryo droplet has the exact equilibrium radius, re, in which the liquid droplet is in thermodynamic and mechanical equilibrium with the surrounding vapor,

${{g}_{\ell ,e}}={{g}_{v,e}}$	(23)

Therefore, the change of availability at equilibrium becomes

$\Delta {{\Psi }_{e}}=\frac{4}{3}\pi r_{e}^{2}{{\sigma }_{\ell v}}F$	(24)

The change of availability near the equilibrium radius can be obtained by expanding eq. (20) in the form of a Taylor series, i.e.,

$\Delta \Psi =\Delta {{\Psi }_{e}}+{{\left. \frac{\partial \Delta \Psi }{\partial r} \right\|}_{e}}(r-{{r}_{e}})+\frac{1}{2}{{\left. \frac{{{\partial }^{2}}\Delta \Psi }{\partial {{r}^{2}}} \right\|}_{e}}{{(r-{{r}_{e}})}^{2}}+\cdots$	(25)

The derivative of $ΔΨ$ with respect to r can be found from eq. (22), i.e.,

$\frac{\partial \Delta \Psi }{\partial r}=\frac{4}{3{{v}_{\ell }}}\pi {{r}^{3}}F\left( \frac{\partial {{g}_{\ell }}}{\partial r}-\frac{\partial {{g}_{v}}}{\partial r} \right)+\frac{4}{{{v}_{\ell }}}\pi {{r}^{2}}F({{g}_{\ell }}-{{g}_{v}})+\frac{8}{3}\pi r{{\sigma }_{\ell v}}F$	(26)

The second order derivative can be found by differentiating eq. (26), i.e.,

$\begin{align} & \frac{{{\partial }^{2}}\Delta \Psi }{\partial {{r}^{2}}}=\frac{4}{3{{v}_{\ell }}}\pi {{r}^{3}}F\left( \frac{{{\partial }^{2}}{{g}_{\ell }}}{\partial {{r}^{2}}}-\frac{{{\partial }^{2}}{{g}_{v}}}{\partial {{r}^{2}}} \right)+\frac{8}{{{v}_{\ell }}}\pi {{r}^{2}}F\left( \frac{\partial {{g}_{\ell }}}{\partial r}-\frac{\partial {{g}_{v}}}{\partial r} \right) \\ & \text{ }+\frac{8}{{{v}_{\ell }}}\pi rF({{g}_{\ell }}-{{g}_{v}})+\frac{8}{3}\pi {{\sigma }_{\ell v}}F \\ \end{align}$	(27)

Since the pressure and temperature of the vapor phase, $p v and T v$ , are fixed, the Gibbs free energy for the vapor phase remains the same near the equilibrium:

$g v = g v, e$	(28)

i.e.,

$\frac{\partial {{g}_{v}}}{\partial r}=0$	(29)

$\frac{{{\partial }^{2}}{{g}_{v}}}{\partial {{r}^{2}}}=0$	(30)

While the temperature of the liquid droplet is the same as the vapor temperature ( ${{T}_{\ell }}={{T}_{v}}$ ), the pressure in the liquid droplet is related to the pressure in the vapor phase by $\text{ }{{p}_{\ell }}={{p}_{v}}+2{{\sigma }_{\ell v}}/r$ . Therefore, the Gibbs free energy for the liquid phase is

${{g}_{\ell }}={{g}_{\ell ,e}}+d{{g}_{\ell }}$	(31)

where

$d{{g}_{\ell }}={{v}_{\ell }}d{{p}_{\ell }}=-\frac{2{{v}_{\ell }}{{\sigma }_{\ell v}}}{{{r}^{2}}}dr$	(32)

i.e.,

$\frac{\partial {{g}_{\ell }}}{\partial r}=-\frac{2{{v}_{\ell }}{{\sigma }_{\ell v}}}{{{r}^{2}}}$	(33)

and

$\frac{{{\partial }^{2}}{{g}_{\ell }}}{\partial {{r}^{2}}}=\frac{4{{v}_{\ell }}{{\sigma }_{\ell v}}}{{{r}^{3}}}$	(34)

Substituting eqs. (29), (30), (32), and (33) into eqs. (26) and (27), one obtains

$\frac{\partial \Delta \Psi }{\partial r}=\frac{4}{{{v}_{\ell }}}\pi {{r}^{2}}F({{g}_{\ell }}-{{g}_{v}})$	(35)

$\frac{{{\partial }^{2}}\Delta \Psi }{\partial {{r}^{2}}}=-8\pi F{{\sigma }_{\ell v}}+\frac{8}{{{v}_{\ell }}}\pi rF({{g}_{\ell }}-{{g}_{v}})$	(36)

At equilibrium, eqs. (35) and (36) become

${{\left. \frac{\partial \Delta \Psi }{\partial r} \right\|}_{e}}=0$	(37)

${{\left. \frac{{{\partial }^{2}}\Delta \Psi }{\partial {{r}^{2}}} \right\|}_{e}}=-8\pi F{{\sigma }_{\ell v}}$	(38)

Substituting eqs. (37) and (38) into eq. (25), the change of availability near the equilibrium radius becomes

$\Delta \Psi =\frac{4}{3}\pi r_{e}^{2}{{\sigma }_{\ell v}}F-4\pi {{\sigma }_{\ell v}}F{{(r-{{r}_{e}})}^{2}}+\cdots$	(39)

Figure 2 Variation of the system availability with droplet radius.

It can be seen from the above equation that ΔΨ is at its maximum at r = r_e, and, therefore, it is shown once again that at the equilibrium radius the droplet is in unstable equilibrium (see Fig. 2). However, to maintain equilibrium dΨ = dG < 0, i.e., the system tends naturally to achieve the lowest Gibbs free energy value. Therefore, as the droplet increases in radius, the availability of the system decreases and the droplet is in equilibrium. Also, it should be pointed out that if a droplet forms with a radius smaller than equilibrium it spontaneously destroys itself. This can also be seen by the minimum Gibbs free energy principle. If the liquid droplet forms with a radius below the equilibrium radius and tries to grow, ΔΨ > 0; therefore, the system is increasing in availability, which is highly unstable. For the availability to decrease and approach equilibrium, the liquid droplet would need to continuously decrease in size until it disappears. This phenomenon can also be seen in Fig. 2. The unstable equilibrium radius is at the top of the curve. If a droplet with such a radius loses one molecule, it will need to continuously decrease in size to meet the ΔΨ< 0 criterion. If the droplet gains one molecule, it will need to continuously increase in size to meet the ΔΨ< 0 criterion.

An expression to determine the minimum equilibrium radius of a newly formed droplet that will not spontaneously disappear was developed in thermodynamics at the interfaces; the result is expressed by

${{r}_{\min }}=\frac{2\sigma {{v}_{\ell }}}{{{R}_{g}}T\ln [{{p}_{v}}/{{p}_{sat}}(T)]}$	(40)

Considering the Clapeyron equation (2.168) with $T v = T 0$ , $T = T w,$ and ${{T}_{v}}{{T}_{w}}\approx T_{w}^{2}$ , the expression for minimum equilibrium radius size becomes

${{r}_{\min }}=\frac{{{D}_{\min }}}{2}=\frac{2{{v}_{\ell }}\sigma {{T}_{v}}}{{{h}_{\ell v}}({{T}_{v}}-{{T}_{w}})}$	(41)

where ${{h}_{\ell v}}$ is the latent heat of energy for the vapor-to-liquid conversion, and $σ$ is the surface tension of the condensing fluid.

References

Faghri, A., and Zhang, Y., 2006, Transport Phenomena in Multiphase Systems, Elsevier, Burlington, MA

Faghri, A., Zhang, Y., and Howell, J. R., 2010, Advanced Heat and Mass Transfer, Global Digital Press, Columbia, MO.

Retrieved from "http://www.thermalfluidscentral.org/encyclopedia/index.php/Critical_Droplet_Radius_for_Dropwise_Condensation"

Critical Droplet Radius for Dropwise Condensation

From Thermal-FluidsPedia

References

Views

Personal tools

Navigation

Search