# Surface Spray Cooling

When liquid drops are sprayed onto a hot surface, heat conduction causes the liquid to superheat and evaporate into a gas (see Figure 9.2). The cooling capacity of the process depends on the time required for an average-sized drop to cool. If the shape of the droplet can be assumed to be hemispherical during the evaporation process, the energy balance at the interface can be written as

${{\rho }_{\ell }}{{h}_{\ell v}}2\pi r_{I}^{2}\frac{d{{r}_{I}}}{dt}={{k}_{\ell }}\int_{0}^{\pi /2}{\frac{\partial {{T}_{\ell }}({{r}_{I}},\theta )}{\partial r}}2\pi r_{I}^{2}\cos \theta d\theta -h2\pi r_{I}^{2}\left( {{T}_{\infty }}-{{T}_{sat}} \right)$ (9.172)

where rI is the radius of the droplet. Equation (9.172) can be simplified as

${{\rho }_{\ell }}{{h}_{\ell v}}\frac{d{{r}_{I}}}{dt}={{k}_{\ell }}\int_{0}^{\pi /2}{\frac{\partial {{T}_{\ell }}({{r}_{I}},\theta )}{\partial r}}\cos \theta d\theta -h\left( {{T}_{\infty }}-{{T}_{sat}} \right)$ (9.173)

Analytical solution of eq. (9.173) is very difficult because the temperature distribution in the liquid droplet is two-dimensional in nature. A scale analysis similar to that in Lock (1994) is presented here to estimate the time required for droplet with an initial radius of Ri to evaporate completely.

The scale of the radius of the droplet, rI, is Ri. Thus the scale of the temperature gradient at the interface is

$\frac{\partial T({{r}_{I}},\theta )}{\partial r}\sim \frac{{{T}_{w}}-{{T}_{sat}}}{{{R}_{i}}}$ (9.174)

The scale analysis of eq. (9.173) yields

${{\rho }_{\ell }}{{h}_{\ell v}}\frac{{{R}_{i}}}{{{t}_{f}}}\sim {{k}_{\ell }}\frac{{{T}_{w}}-{{T}_{sat}}}{{{R}_{i}}}+h({{T}_{\infty }}-{{T}_{sat}})$ (9.175)

which can be rearranged as

$\frac{{{\rho }_{\ell }}{{h}_{\ell v}}}{{{k}_{\ell }}}\frac{R_{i}^{2}}{{{t}_{f}}}\sim ({{T}_{w}}-{{T}_{sat}})+\frac{h{{R}_{i}}}{{{k}_{\ell }}}({{T}_{\infty }}-{{T}_{sat}})$ (9.176)

Since the liquid droplet is very small, $h{{R}_{i}}/{{k}_{\ell }}$ is very small, so the second term on the right-hand side of eq. (9.175) is much smaller than the first. Equation (9.175) can be simplified by neglecting the second term on the right-hand side, i.e.,

${{\rho }_{\ell }}{{h}_{\ell v}}\frac{R_{i}^{2}}{{{k}_{\ell }}{{t}_{f}}}\sim ({{T}_{w}}-{{T}_{sat}})$ (9.177)

The physical significance of eq. (9.177) is that the latent heat of evaporation is balanced primarily by conduction in the liquid droplet, while the effect of convection on the surface of the droplet is negligible.

The time that it takes to completely evaporate the droplet with an initial radius of Ri is therefore estimated by

${{t}_{f}}\sim \frac{{{\rho }_{\ell }}{{h}_{\ell v}}R_{i}^{2}}{{{k}_{\ell }}\left( {{T}_{w}}-{{T}_{sat}} \right)}$ (9.178)

For the cases where the substrate heat flux beneath the droplet, q''w, is known, eq. (9.177) should be rewritten in terms of heat flux. The scale of the heat flux at the heating surface is

${{{q}''}_{w}}\sim \frac{{{k}_{\ell }}\left( {{T}_{w}}-{{T}_{sat}} \right)}{{{R}_{i}}}$ (9.179)

Combining eqs. (9.178) and (9.179) yields

${{t}_{f}}\sim \frac{{{\rho }_{\ell }}{{h}_{\ell v}}{{R}_{i}}}{{{{{q}''}}_{w}}}$ (9.180)

which indicates that a smaller drop will provide a larger cooling effect.

While eqs. (9.178) and (9.180) provide the order of magnitude of the time required to completely evaporate the droplet, quantitative estimation of the evaporation time is often also desirable. A simple approximate analysis will be done below by estimating the conduction in the liquid droplet by the following correlation:

${{q}_{d}}={{k}_{\ell }}\bar{A}\frac{{{T}_{w}}-{{T}_{sat}}}{{\bar{\delta }}}$ (9.181)

where $\bar{A}$ and $\bar{\delta }$ are the average cross-sectional area of heat conduction and the average path length of the conduction, respectively. For a hemispherical droplet, the contact area between the droplet and the heated wall is ${{A}_{w}}=\pi r_{I}^{2}$ and the interfacial area of the droplet is ${{A}_{I}}=2\pi r_{I}^{2}$. Thus, we can take the average conduction area as

$\bar{A}=\frac{1}{2}({{A}_{w}}+{{A}_{I}})=\frac{3}{2}\pi r_{I}^{2}$ (9.182)

The average path length for conduction is

$\bar{\delta }=\frac{V}{{\bar{A}}}=\frac{(2/3)\pi r_{I}^{3}}{(3/2)\pi r_{I}^{2}}=\frac{4}{9}{{r}_{I}}$ (9.183)

Therefore, the conduction in the liquid droplet becomes

${{q}_{d}}=\frac{27}{8}\pi {{k}_{\ell }}{{r}_{I}}({{T}_{w}}-{{T}_{sat}})$ (9.184)

Replacing the first term on the right-hand side of eq. (9.173) with eq. (9.184), and dropping the second term on the right-hand side of eq. (9.173) (see the above scale analysis), the energy balance for the hemispherical droplet becomes

${{\rho }_{\ell }}{{h}_{\ell v}}2\pi r_{I}^{2}\frac{d{{r}_{I}}}{dt}=\frac{27}{8}\pi {{k}_{\ell }}{{r}_{I}}({{T}_{w}}-{{T}_{sat}})$ (9.185)

which can be rearranged as

${{r}_{I}}\frac{d{{r}_{I}}}{dt}=\frac{27}{16}\frac{{{k}_{\ell }}({{T}_{w}}-{{T}_{sat}})}{{{\rho }_{\ell }}{{h}_{\ell v}}}$ (9.186)

which is subject to the following initial conditions:

${{r}_{I}}={{R}_{i}}\begin{matrix} , & t=0 \\ \end{matrix}$ (9.187)

Integrating eq. (9.186) and considering its initial condition, eq. (9.187), one obtains the transient radius of the liquid droplet:

${{r}_{I}}=\sqrt{R_{i}^{2}-\frac{27}{8}\frac{{{k}_{\ell }}({{T}_{w}}-{{T}_{sat}})t}{{{\rho }_{\ell }}{{h}_{\ell v}}}}$ (9.188)

The time required for the droplet to evaporate completely can be obtained by letting rI in eq. (9.188) equal zero, i.e.,

${{t}_{f}}=\frac{8{{\rho }_{\ell }}{{h}_{\ell v}}R_{i}^{2}}{27{{k}_{\ell }}({{T}_{w}}-{{T}_{sat}})}$ (9.189)

which agrees with the results obtained by scale analysis, eq. (9.178).

The above analysis is otherwise simplistic because it assumes that the shape of the droplet is hemispherical. In fact, the shape of the droplet depends on the velocity with which it impacts the wall as well as the wettability of the liquid droplet on the wall. If the contact area between the droplet and the wall is larger (due to higher impacting velocity or good wettability), the time for conduction through the liquid is shorter and the life of the drop will be shorter as well.