Evaporation and Mass Transfer Fundamentals

Abstract

This chapter introduces fundamental mass transfer concepts through the modeling of the evaporation process that takes place at the free surface of a mug of coffee. An evaporative heat transfer coefficient, h_evap, is defined that can be included in any of the transient numerical simulations from earlier chapters (i.e., lumped 1-node, few node, multi-node). Evaporation is driven by concentration gradients, not a temperature difference, and defining an evaporation coefficient this way allows direct comparison of the relative effects of evaporation, convection, and radiation on the cooling of the free surface. As will become apparent, this approach makes sense only for cases when the liquid surface temperature differs significantly from ambient. The case where the coffee reaches thermal equilibrium with the environment and the slower evaporation continues is treated differently, and it will be seen that the coffee attains a temperature below ambient, a form of evaporative cooling.

Appendices

Workshop 13.1. Evaporation Experiments

During the first week of class, procure samples of clear tubing. Choose samples with a much larger length to diameter ratio than the value of 1.8 investigated in this chapter. Mount them vertically with an open end up top, and sealed at the bottom. Mark them with a scale. Find locations for them where they will not be disturbed, fill them to the top with water (eliminate all air bubbles trapped), and start recording L vs. t. Consider using different liquids, and note that if liquid mixtures are used with components of different volatility (i.e., isopropyl alcohol/water mixtures), that the composition will change with time. Modeling this effect requires some basic understanding of distillation-type processes. Compare results to theoretical predictions by the end of the semester.

Workshop 13.2. Surface Temperatures During Evaporation

Calculate the surface and coffee temperatures for an evaporating mug of coffee as a function of ambient temperature (assume zero relative humidity). Demonstrate that the effect becomes very large as the temperature approaches the boiling point of the liquid. Use the point source model as giving the largest effect modeled here. Alternatively, use published correlations for the mass transfer coefficients.

Workshop 13.3. A Cooling Mug of Coffee with Evaporation Effects

Include a lumped capacitance to the coffee node of Fig. 13.24 and include the effects of evaporation onto the basic cooling response of the coffee. Use a well-insulated mug and treat the thermal resistance between coffee and ambient (R_c,∞) as a constant. Use either an evaporative heat transfer coefficient (h_evap) or directly include the evaporation as a sink. Implement an implicit Euler method in which the surface temperature, evaporation rates, radiation, and natural convection are all calculated (iteratively) at each time step. Start with the coffee at its boiling point, and notice how much lower the surface temperature is than the average coffee temperature early on. Plot the dimensionless temperature vs. time on semi-logy axes to observe deviations from first-order behavior.

Appendix. Evaluating Dew Points and Mass Fractions

An important detail in evaluating evaporation rates is to relate the mass fractions in terms of input variables, which requires a little more background into modeling gas mixtures. The gas phase is modeled as being a mixture of two ideal gases, air and water vapor.

For humid air at temperature T, the partial pressure of water is given by:

$$ {p}_V={r}_h{p}_{sat}(T) $$

where r _h is the relative humidity.

The dew point temperature (T_DewPoint) of air with a relative humidity (r_H) at an ambient temperature (T_∞) is defined by the temperature at which the partial pressure of saturated vapor equals the partial pressure of vapor in the humid (but unsaturated) air. That is:

$$ {p}_{sat}\left({T}_{DewPoint}\right)={r}_h{p}_{sat}\left({T}_{\infty}\right) $$

The relationship between the saturation pressure and the temperature can be modeled using a curve fit of the form given by the Clausius–Clapeyron relation, namely:

$$ {p}_{sat}(T)=A{e}^{\frac{B}{T}} $$

where A and B are constants for a given species, and T is the absolute temperature. The values of A and B can be calculated by fitting two points taken from saturation tables in the vicinity of the problem in question. This curve for water was plotted in Fig. 13.2.

The partial pressure of water vapor is related to the concentration of water vapor. For n_v moles of water vapor at absolute temperature T occupying a volume V, the ideal gas law is:

$$ {p}_vV={n}_v\widehat{R}T $$

$$ \widehat{R}=8,314\ \frac{\mathrm{J}}{\mathrm{kmol}\;\mathrm{K}} $$

is the universal gas constant. Multiplying and dividing by the total number of moles and rearranging:

$$ {p}_v=\left(\frac{n_v}{n}\right)\left(\frac{n\widehat{R}T}{V}\right)={X}_vP $$

where P is the total pressure (the sum of all partial pressures). The mole fraction of vapor, X_v is the number of moles of vapor divided by the total number of moles in the gas mixture. The mass fraction and mole fractions are related to each other through the molecular weights of the mixture and vapor by the following word string:

$$ \left(\frac{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Vapor}}{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Mixture}}\right)=\left(\frac{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Vapor}}{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Mixture}}\right)\left(\frac{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Mixture}}{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Mixture}}\right)\left(\frac{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Vapor}}{\mathrm{Moles}\;\mathrm{of}\ \mathrm{Vapor}}\right) $$

$$ {Y}_v=\left({X}_v\right)\left(\frac{1}{{\mathrm{MW}}_{\mathrm{mix}}}\right)\left({\mathrm{MW}}_{\mathrm{vapor}}\right) $$

The molecular weight of the mixture is obtained from the following word string:

$$ \left(\frac{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Mixture}}{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Mixture}}\right)=\left(\frac{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Vapor}+\mathrm{Mass}\ \mathrm{of}\ \mathrm{Air}}{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Mixture}}\right) $$

$$ \left(\frac{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Mixture}}{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Mixture}}\right)=\left(\frac{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Vapor}}{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Mixture}}\right)\left(\frac{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Vapor}}{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Vapor}}\right)+\left(\frac{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Air}}{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Mixture}}\right)\left(\frac{\mathrm{Mass}\ \mathrm{of}\ \mathrm{Air}}{\mathrm{Moles}\ \mathrm{of}\ \mathrm{Air}}\right) $$

$$ M{W}_{mix}={X}_{vapor}M{W}_{vapor}+{X}_{air}M{W}_{air} $$

$$ {Y}_v=\left({X}_v\right)\left(\frac{{\mathrm{MW}}_{\mathrm{v}\mathrm{apor}}}{{\mathrm{X}}_{\mathrm{v}}{\mathrm{MW}}_{\mathrm{v}\mathrm{apor}}+\left(1-{X}_v\right)M{W}_{air}}\right) $$

$$ {Y}_v=\frac{1}{1+\left(\frac{1-{X}_v}{X_v}\right)\left(\frac{M{W}_{air}}{M{W}_{vapor}}\right)} $$