# Evaporation of a Liquid Droplet in a Hot Gas

(Difference between revisions)
 Revision as of 17:03, 1 June 2010 (view source) (Created page with 'Figure 9.3 describes the situation in which a liquid drop evaporates to a surrounding hot gas. Since the liquid droplet is surrounded by the hot gas, it is assumed that the liqui…')← Older edit Current revision as of 19:32, 3 June 2010 (view source) (One intermediate revision not shown) Line 1: Line 1: Figure 9.3 describes the situation in which a liquid drop evaporates to a surrounding hot gas. Since the liquid droplet is surrounded by the hot gas, it is assumed that the liquid droplet is spherical in shape and the evaporation is a spherically symmetric problem. Neglecting radiation effect, the energy equation for a liquid droplet is Figure 9.3 describes the situation in which a liquid drop evaporates to a surrounding hot gas. Since the liquid droplet is surrounded by the hot gas, it is assumed that the liquid droplet is spherical in shape and the evaporation is a spherically symmetric problem. Neglecting radiation effect, the energy equation for a liquid droplet is - + -
$\frac{\partial {{T}_{\ell }}}{\partial t}=\frac{{{\alpha }_{\ell }}}{{{r}^{2}}}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial {{T}_{\ell }}}{\partial r} \right)$ + {| class="wikitable" border="0" - (9.210)
+ |- + | width="100%" | +
$\frac{\partial {{T}_{\ell }}}{\partial t}=\frac{{{\alpha }_{\ell }}}{{{r}^{2}}}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial {{T}_{\ell }}}{\partial r} \right)$
+ |{{EquationRef|(1)}} + |} where the thermophysical properties of the liquid are assumed to be temperature-independent. The initial temperature of the droplet is where the thermophysical properties of the liquid are assumed to be temperature-independent. The initial temperature of the droplet is - + + {| class="wikitable" border="0" + |- + | width="100%" |
${{T}_{\ell }}(r,t)={{T}_{i}}\begin{matrix} [itex]{{T}_{\ell }}(r,t)={{T}_{i}}\begin{matrix} , & t=0 \\ , & t=0 \\ - \end{matrix}$ + \end{matrix}[/itex]
- (9.211)
+ |{{EquationRef|(2)}} + |} The symmetric condition at the center of the liquid droplet requires that The symmetric condition at the center of the liquid droplet requires that - + + {| class="wikitable" border="0" + |- + | width="100%" |
$\frac{\partial {{T}_{\ell }}}{\partial r}=0\begin{matrix} [itex]\frac{\partial {{T}_{\ell }}}{\partial r}=0\begin{matrix} , & r=0 \\ , & r=0 \\ - \end{matrix}$ + \end{matrix}[/itex]
- (9.212)
+ |{{EquationRef|(3)}} + |} Since the evaporation takes place on the surface of the liquid droplet, its surface temperature equals the saturation temperature corresponding to the partial pressure of the vapor at the interface, i.e., Since the evaporation takes place on the surface of the liquid droplet, its surface temperature equals the saturation temperature corresponding to the partial pressure of the vapor at the interface, i.e., - + + {| class="wikitable" border="0" + |- + | width="100%" |
$T(r,t)={{T}_{sat}}\begin{matrix} [itex]T(r,t)={{T}_{sat}}\begin{matrix} , & r={{r}_{I}} \\ , & r={{r}_{I}} \\ - \end{matrix}$ + \end{matrix}[/itex]
- (9.213)
+ |{{EquationRef|(4)}} + |} An energy balance at the surface of the liquid droplet accounts for the latent heat exchanged through conduction inside the drop and convection away from the droplet; that energy balance is An energy balance at the surface of the liquid droplet accounts for the latent heat exchanged through conduction inside the drop and convection away from the droplet; that energy balance is - + + {| class="wikitable" border="0" + |- + | width="100%" |
${{\rho }_{\ell }}{{h}_{\ell v}}\frac{d{{r}_{I}}}{dt}={{k}_{\ell }}\frac{\partial {{T}_{\ell }}}{\partial r}-h({{T}_{\infty }}-{{T}_{sat}})\begin{matrix} [itex]{{\rho }_{\ell }}{{h}_{\ell v}}\frac{d{{r}_{I}}}{dt}={{k}_{\ell }}\frac{\partial {{T}_{\ell }}}{\partial r}-h({{T}_{\infty }}-{{T}_{sat}})\begin{matrix} , & r={{r}_{I}} \\ , & r={{r}_{I}} \\ - \end{matrix}$ + \end{matrix}[/itex]
- (9.214)
+ |{{EquationRef|(5)}} + |} + + where h is the heat transfer coefficient between the gas and the liquid droplet. - where h is the heat transfer coefficient between the gas and the liquid droplet. If the liquid droplet is evaporating to the vapor-gas mixture, the mass balance at the interface can be written as If the liquid droplet is evaporating to the vapor-gas mixture, the mass balance at the interface can be written as - + -
${{\rho }_{\ell }}\frac{d{{r}_{I}}}{dt}=-{{\rho }_{g}}{{h}_{m}}\left( {{\omega }_{I}}-{{\omega }_{\infty }} \right)$ + {| class="wikitable" border="0" - (9.215)
+ |- + | width="100%" | +
${{\rho }_{\ell }}\frac{d{{r}_{I}}}{dt}=-{{\rho }_{g}}{{h}_{m}}\left( {{\omega }_{I}}-{{\omega }_{\infty }} \right)$
+ |{{EquationRef|(6)}} + |} where ${{h}_{m}}$ is the mass transfer coefficient,  ${{\rho }_{g}}$ is the density of the mixture, and $\omega$ is the mass fraction of the evaporating component in the mixture. where ${{h}_{m}}$ is the mass transfer coefficient,  ${{\rho }_{g}}$ is the density of the mixture, and $\omega$ is the mass fraction of the evaporating component in the mixture. - Analytical solution of the problem defined by eq. (9.210) – (9.215) will yield a very complex expression of the final results. Since the liquid droplet is usually very small, one can assume that the heat convected to the droplet surface is balanced by the latent heat of evaporation, so heat conduction to the liquid droplet can be neglected. This assumption is valid because the bulk temperature of the droplet will quickly approach the saturation temperature [[#References|(Lock, 1994)]], i.e., + Analytical solution of the problem defined by eq. (1) – (6) will yield a very complex expression of the final results. Since the liquid droplet is usually very small, one can assume that the heat convected to the droplet surface is balanced by the latent heat of evaporation, so heat conduction to the liquid droplet can be neglected. This assumption is valid because the bulk temperature of the droplet will quickly approach the saturation temperature [[#References|(Lock, 1994)]], i.e., - + + {| class="wikitable" border="0" + |- + | width="100%" |
$\frac{d{{r}_{I}}}{dt}=-\frac{h({{T}_{\infty }}-{{T}_{sat}})}{{{\rho }_{\ell }}{{h}_{\ell v}}}\begin{matrix} [itex]\frac{d{{r}_{I}}}{dt}=-\frac{h({{T}_{\infty }}-{{T}_{sat}})}{{{\rho }_{\ell }}{{h}_{\ell v}}}\begin{matrix} , & r={{r}_{I}} \\ , & r={{r}_{I}} \\ - \end{matrix}$ + \end{matrix}[/itex]
- (9.216)
+ |{{EquationRef|(7)}} + |} which is subject to the following initial condition: which is subject to the following initial condition: - + + {| class="wikitable" border="0" + |- + | width="100%" |
${{r}_{I}}={{R}_{i}}\begin{matrix} [itex]{{r}_{I}}={{R}_{i}}\begin{matrix} , & t=0 \\ , & t=0 \\ - \end{matrix}$ + \end{matrix}[/itex]
- (9.217)
+ |{{EquationRef|(8)}} + |} - Integrating eq. (9.216) and considering its initial condition, eq. (9.217), the transient radius of the liquid droplet is obtained. + Integrating eq. (7) and considering its initial condition, eq. (8), the transient radius of the liquid droplet is obtained. - + -
${{r}_{I}}={{R}_{i}}-\frac{h({{T}_{\infty }}-{{T}_{sat}})t}{{{\rho }_{\ell }}{{h}_{\ell v}}}$ + - (9.218)
+ - In arriving at eq. (9.218), the heat transfer coefficient was assumed to be constant throughout the entire evaporation process. The time required to evaporate the droplet completely, tf, can be obtained by setting ${{r}_{I}}$ equal to zero in eq. (9.218), i.e., + {| class="wikitable" border="0" - + |- -
${{t}_{f}}=\frac{{{\rho }_{\ell }}{{h}_{\ell v}}{{R}_{i}}}{h({{T}_{\infty }}-{{T}_{sat}})}$ + | width="100%" | - (9.219)
+
${{r}_{I}}={{R}_{i}}-\frac{h({{T}_{\infty }}-{{T}_{sat}})t}{{{\rho }_{\ell }}{{h}_{\ell v}}} + |{{EquationRef|(9)}} + |} - The time required to evaporate the droplet completely can also be estimated from the mass balance equation (9.215). By following a procedure similar to that which obtained eq. (9.219), one obtains + In arriving at eq. (9), the heat transfer coefficient was assumed to be constant throughout the entire evaporation process. The time required to evaporate the droplet completely, tf, can be obtained by setting [itex]{{r}_{I}}$ equal to zero in eq. (9), i.e., - + -
${{t}_{f}}=\frac{{{\rho }_{\ell }}{{R}_{i}}}{{{\rho }_{g}}{{h}_{m}}\left( {{\omega }_{I}}-{{\omega }_{\infty }} \right)}$ + - (9.220)
+ - Combining eqs. (9.219) – (9.220), one obtains + - + {| class="wikitable" border="0" -
${{\omega }_{I}}-{{\omega }_{\infty }}=\frac{h}{{{h}_{m}}}\frac{\text{J}{{\text{a}}_{g}}}{{{\rho }_{g}}{{c}_{{{p}_{g}}}}}$ + |- - (9.221)
+ | width="100%" | +
${{t}_{f}}=\frac{{{\rho }_{\ell }}{{h}_{\ell v}}{{R}_{i}}}{h({{T}_{\infty }}-{{T}_{sat}})}$
+ |{{EquationRef|(10)}} + |} + + The time required to evaporate the droplet completely can also be estimated from the mass balance equation (6). By following a procedure similar to that which obtained eq. (10), one obtains + + {| class="wikitable" border="0" + |- + | width="100%" | +
${{t}_{f}}=\frac{{{\rho }_{\ell }}{{R}_{i}}}{{{\rho }_{g}}{{h}_{m}}\left( {{\omega }_{I}}-{{\omega }_{\infty }} \right)}$
+ |{{EquationRef|(11)}} + |} + + Combining eqs. (10) – (11), one obtains + + {| class="wikitable" border="0" + |- + | width="100%" | +
${{\omega }_{I}}-{{\omega }_{\infty }}=\frac{h}{{{h}_{m}}}\frac{\text{J}{{\text{a}}_{g}}}{{{\rho }_{g}}{{c}_{{{p}_{g}}}}}$
+ |{{EquationRef|(12)}} + |} where ''Jag'' is the Jakob number, defined as where ''Jag'' is the Jakob number, defined as - + -
$\text{J}{{\text{a}}_{g}}=\frac{{{c}_{pg}}({{T}_{\infty }}-{{T}_{sat}})}{{{h}_{\ell v}}}$ + {| class="wikitable" border="0" - (9.222)
+ |- + | width="100%" | +
$\text{J}{{\text{a}}_{g}}=\frac{{{c}_{pg}}({{T}_{\infty }}-{{T}_{sat}})}{{{h}_{\ell v}}}$
+ |{{EquationRef|(13)}} + |} The above analysis is based on the assumption that conduction to the liquid droplet is negligible. Evaporation of a liquid droplet with consideration of the effect of conduction heat loss will be investigated in the following example 9.8. The above analysis is based on the assumption that conduction to the liquid droplet is negligible. Evaporation of a liquid droplet with consideration of the effect of conduction heat loss will be investigated in the following example 9.8. Line 84: Line 138: ==References== ==References== + + Elperin, T., Fominykh, A., and Krasovitov, B., 2005, “Modeling of Simultaneous Gas Absorption and Evaporation of Large Droplet,” Proceedings of 2005 International Mechanical Engineering Congress and Exposition, Orlando, FL, Nov. 5-11, 2005. + + Lock, G.S.H., 1994, Latent Heat Transfer, Oxford Science Publications, Oxford University, Oxford, UK.

## Current revision as of 19:32, 3 June 2010

Figure 9.3 describes the situation in which a liquid drop evaporates to a surrounding hot gas. Since the liquid droplet is surrounded by the hot gas, it is assumed that the liquid droplet is spherical in shape and the evaporation is a spherically symmetric problem. Neglecting radiation effect, the energy equation for a liquid droplet is

 $\frac{\partial {{T}_{\ell }}}{\partial t}=\frac{{{\alpha }_{\ell }}}{{{r}^{2}}}\frac{\partial }{\partial r}\left( {{r}^{2}}\frac{\partial {{T}_{\ell }}}{\partial r} \right)$ (1)

where the thermophysical properties of the liquid are assumed to be temperature-independent. The initial temperature of the droplet is

 ${{T}_{\ell }}(r,t)={{T}_{i}}\begin{matrix} , & t=0 \\ \end{matrix}$ (2)

The symmetric condition at the center of the liquid droplet requires that

 $\frac{\partial {{T}_{\ell }}}{\partial r}=0\begin{matrix} , & r=0 \\ \end{matrix}$ (3)

Since the evaporation takes place on the surface of the liquid droplet, its surface temperature equals the saturation temperature corresponding to the partial pressure of the vapor at the interface, i.e.,

 $T(r,t)={{T}_{sat}}\begin{matrix} , & r={{r}_{I}} \\ \end{matrix}$ (4)

An energy balance at the surface of the liquid droplet accounts for the latent heat exchanged through conduction inside the drop and convection away from the droplet; that energy balance is

 ${{\rho }_{\ell }}{{h}_{\ell v}}\frac{d{{r}_{I}}}{dt}={{k}_{\ell }}\frac{\partial {{T}_{\ell }}}{\partial r}-h({{T}_{\infty }}-{{T}_{sat}})\begin{matrix} , & r={{r}_{I}} \\ \end{matrix}$ (5)

where h is the heat transfer coefficient between the gas and the liquid droplet.

If the liquid droplet is evaporating to the vapor-gas mixture, the mass balance at the interface can be written as

 ${{\rho }_{\ell }}\frac{d{{r}_{I}}}{dt}=-{{\rho }_{g}}{{h}_{m}}\left( {{\omega }_{I}}-{{\omega }_{\infty }} \right)$ (6)

where hm is the mass transfer coefficient, ρg is the density of the mixture, and ω is the mass fraction of the evaporating component in the mixture.

Analytical solution of the problem defined by eq. (1) – (6) will yield a very complex expression of the final results. Since the liquid droplet is usually very small, one can assume that the heat convected to the droplet surface is balanced by the latent heat of evaporation, so heat conduction to the liquid droplet can be neglected. This assumption is valid because the bulk temperature of the droplet will quickly approach the saturation temperature (Lock, 1994), i.e.,

 $\frac{d{{r}_{I}}}{dt}=-\frac{h({{T}_{\infty }}-{{T}_{sat}})}{{{\rho }_{\ell }}{{h}_{\ell v}}}\begin{matrix} , & r={{r}_{I}} \\ \end{matrix}$ (7)

which is subject to the following initial condition:

 ${{r}_{I}}={{R}_{i}}\begin{matrix} , & t=0 \\ \end{matrix}$ (8)

Integrating eq. (7) and considering its initial condition, eq. (8), the transient radius of the liquid droplet is obtained.

 ${{r}_{I}}={{R}_{i}}-\frac{h({{T}_{\infty }}-{{T}_{sat}})t}{{{\rho }_{\ell }}{{h}_{\ell v}}}$ (9)

In arriving at eq. (9), the heat transfer coefficient was assumed to be constant throughout the entire evaporation process. The time required to evaporate the droplet completely, tf, can be obtained by setting rI equal to zero in eq. (9), i.e.,

 ${{t}_{f}}=\frac{{{\rho }_{\ell }}{{h}_{\ell v}}{{R}_{i}}}{h({{T}_{\infty }}-{{T}_{sat}})}$ (10)

The time required to evaporate the droplet completely can also be estimated from the mass balance equation (6). By following a procedure similar to that which obtained eq. (10), one obtains

 ${{t}_{f}}=\frac{{{\rho }_{\ell }}{{R}_{i}}}{{{\rho }_{g}}{{h}_{m}}\left( {{\omega }_{I}}-{{\omega }_{\infty }} \right)}$ (11)

Combining eqs. (10) – (11), one obtains

 ${{\omega }_{I}}-{{\omega }_{\infty }}=\frac{h}{{{h}_{m}}}\frac{\text{J}{{\text{a}}_{g}}}{{{\rho }_{g}}{{c}_{{{p}_{g}}}}}$ (12)

where Jag is the Jakob number, defined as

 $\text{J}{{\text{a}}_{g}}=\frac{{{c}_{pg}}({{T}_{\infty }}-{{T}_{sat}})}{{{h}_{\ell v}}}$ (13)

The above analysis is based on the assumption that conduction to the liquid droplet is negligible. Evaporation of a liquid droplet with consideration of the effect of conduction heat loss will be investigated in the following example 9.8.

Elperin et al. (2005) numerically investigated simultaneous heat and mass transfer during evaporation and condensation on the surface of a stagnant droplet in the presence of admixtures that contain noncondensable solvable gas. The equations for Stefan velocity and the transient radius of the droplet were obtained by using conservation of species mass at droplet surface with absorption and evaporation at the surface accounted for. Their results showed that droplet evaporation rate, droplet temperature, interfacial absorbate concentration, and rate of mass transfer are highly interdependent.

## References

Elperin, T., Fominykh, A., and Krasovitov, B., 2005, “Modeling of Simultaneous Gas Absorption and Evaporation of Large Droplet,” Proceedings of 2005 International Mechanical Engineering Congress and Exposition, Orlando, FL, Nov. 5-11, 2005.

Lock, G.S.H., 1994, Latent Heat Transfer, Oxford Science Publications, Oxford University, Oxford, UK.