An Analytical Approach to Model the Effect of Evaporation on Oscillation Amplitude of Liquid Drops in Gaseous Environment

Abstract

The combined effect of evaporation and oscillation of liquid drops in gaseous stagnant environment is analytically modelled. Mechanical energy and mass balances are used to derive the time evolution of drop size and amplitude of oscillation. Two approaches, based on different assumptions about the kinetic energy distribution inside the drop, are used to evaluate the energy loss due to evaporation. Conditions for oscillation damping by evaporation are derived. Application of the model to the case of water, acetone and n-dodecane drops evaporating in hot air shows a non neglectful decrease of drop lifetime, with respect to non-oscillating drops.

Appendix

The surface of a spheroid can be calculated as $A = 4\uppi\,R_{d}^{2} \beta$, where [13]

$$ \beta = \frac{A}{{4\pi R_{d}^{2} }} = \frac{1}{{2\varepsilon^{2/3} }}\left\{ {\begin{array}{*{20}l} {1 + \frac{{\varepsilon^{2} }}{{\sqrt {1 - \varepsilon^{2} } }}\frac{1}{2}\log \left( {\frac{{1 + \sqrt {1 - \varepsilon^{2} } }}{{1 - \sqrt {1 - \varepsilon^{2} } }}} \right)} \hfill & {{\text{oblate (}}\varepsilon { < 1)}} \hfill \\ {1 + \frac{{\varepsilon^{2} }}{{\sqrt {\varepsilon^{2} - 1} }}\arctan \left( {\sqrt {\varepsilon^{2} - 1} } \right)} \hfill & {{\text{prolate (}}\varepsilon { > 1)}} \hfill \\ \end{array} } \right.. $$

(A1)

When evaluating the time-integrals over a cycle, it is convenient to observe first that, given any function of β(t), the integral can be broken into the sum of two integrals, over the two half-periods where the drop is in a prolate and oblate shape, respectively. Each of these integrals can in turn be split into the integral from the spherical shape (β = 1) to the maximum deformation (β = β_max) and back. Changing the variable of integration from t to β, the integrals over time can be transformed into integral over β, from 1 to β_max, for the two cases (oblate and prolate):

$$ \int\limits_{0}^{{t_{0} }} {f\left( \beta \right)dt } = 2\int\limits_{\begin{subarray}{l} \quad 1 \\ prolate \end{subarray} }^{{\beta _{{\max }} }} {\frac{{f\left( \beta \right)}}{{\left| {\dot{\beta }} \right|}}d\beta } + 2\int\limits_{\begin{subarray}{l} \quad 1 \\ oblate \end{subarray} }^{{\beta _{{\max }} }} {\frac{{f\left( \beta \right)}}{{\left| {\dot{\beta }} \right|}}d\beta } $$

(A2)

with $\dot{\beta } = \frac{d\beta }{dt}$. If the integral contains explicitly ε, equation (A1) can be used to substitute ε by ε(β) and, if $\dot{\varepsilon }$ appears explicitly, then the variable transformation

$$ \dot{\varepsilon } = \frac{d\varepsilon }{dt} = \frac{d\varepsilon }{d\beta }\frac{d\beta }{dt} = \frac{{\dot{\beta }}}{{\beta^{{\prime }} }} $$

(A3)

where $\beta^{{\prime }} = \frac{d\beta }{d\varepsilon }$, allows to eliminate it in favour of β and β′. From Eqs. (A1) and (10), the derivatives $\dot{\beta }$ and β′ can be written as function of β:

$$ \dot{\beta } = \frac{d\beta }{dt} = \omega_{2} f\left( {\beta ,\beta_{{\max} } } \right);\quad \begin{array}{*{20}l} {\beta ' = \pm \frac{{3 + \varepsilon^{2/3} \beta \left( {\varepsilon^{2} - 4} \right)}}{{3\varepsilon^{5/3} \left( {1 - \varepsilon^{2} } \right)}}} \hfill \\ { - {\text{oblate, + prolate}}} \hfill \\ \end{array} . $$

(A4)

To calculate the integral $\bar{\varGamma } = \frac{1}{{t_{0} }}\int\limits_{0}^{{t_{0} }} {{\varGamma }(t)dt}$ , first apply the transformation (A2) and then from Eqs. (A4) and (4) one obtains:

$$ \bar{\varGamma } = \frac{1}{\pi }\left[ {\int\limits_{\begin{subarray}{l} \quad 1 \\ prolate \end{subarray} }^{{\beta_{{\max} } }} {\frac{{{\varGamma }\left( \beta \right)}}{{\left| {f\left( {\beta ,\beta_{{\max} } } \right)} \right|}}d\beta } + \int\limits_{\begin{subarray}{l} \quad 1 \\ oblate \end{subarray} }^{{\beta_{{\max} } }} {\frac{{\Gamma \left( \beta \right)}}{{\left| {f\left( {\beta ,\beta_{{\max} } } \right)} \right|}}d\beta } } \right] $$

(A5)

where the symbol Γ(β) is used for $\Gamma \left[ {\varepsilon \left( \beta \right)} \right]$, showing that $\bar{\varGamma}$ is only a function of β_max and it is independent of the oscillation frequency. To calculate the integral $\bar{e}_{K} = \frac{1}{{t_{0} }}\int\limits_{0}^{{t_{0} }} {e_{K} (t)dt}$, where e_K(t) is given by Eq. (16), the integrand can be transformed, using Eq. (A3), (A4) and (A1), to a function of β; then, applying again transformation (A2) and using Eq. (4):

$$ \bar{e}_{K} = \frac{{R_{d}^{2} \omega_{2}^{2} }}{36\pi }\left[ {\int\limits_{\begin{subarray}{l} \quad 1 \\ prolate \end{subarray} }^{{\beta_{{\max} } }} {\frac{I\left( \beta \right)}{{\beta \varepsilon^{10/3} \left( \beta \right)}}\frac{{f\left( {\beta ,\beta_{{\max} } } \right)}}{{\beta^{{{\prime }2}} }}d\beta } + \int\limits_{\begin{subarray}{l} \quad 1 \\ oblate \end{subarray} }^{{\beta_{{\max} } }} {\frac{I\left( \beta \right)}{{\beta \varepsilon^{10/3} \left( \beta \right)}}\frac{{f\left( {\beta ,\beta_{{\max} } } \right)}}{{\beta^{{{\prime }2}} }}d\beta } } \right] $$

(A6)

where I(ε) was transformed in a function of β using again equation (A1). Again the term in square brackets is a function of β_max only and

$$ \bar{e}_{K} = R_{d}^{2} \omega_{2}^{2} F_{s} \left( {\beta_{{\max} } } \right). $$

(A7)