Workshop 8: Evaporation and stagnant-gas diffusion

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Attempt these problems before opening the worked solutions.

1 Methanol emission from an open tank

Planar diffusion through stagnant air

A tank (diameter = 1m, height = 3 m) containing liquid methanol (MW = 32 g/mol) has its top open to the atmosphere. The level of methanol in the tank is maintained at 0.5 m. The gas space inside the tank is stagnant and the methanol vapours are immediately dispersed once they exit the tank. Engineers on‐site are concerned that this open tank may be emitting a considerable amount of methanol vapour.

a. What is the emission rate of methanol vapour from the tank (in kg methanol/day$\mathrm{kg\ methanol/day}$) when the temperature of the tank is 30 ∘C$30\,{}^\circ\mathrm{C}$? State all assumptions and boundary conditions.

b. If the temperature of the tank is raised to 40 ∘C$40\,{}^\circ\mathrm{C}$, what is the new emission rate?

Data: Vapour pressure of MeOH = 163 mmHg (30 ∘C$30\,{}^\circ\mathrm{C}$); 265 mmHg (40 ∘C$40\,{}^\circ\mathrm{C}$)

𝐷𝐴𝐵$D_{AB}$ (methanol/air) @ 298 K = 0.162 × 10−4$10^{-4}$ m2/s$\mathrm{m^2/s}$.

Worked solution

Figure 1: The open tank has the same one-dimensional gas-space model as this evaporation tube. The diffusion distance is the tank height minus the maintained liquid depth.

Data

Quantity	Symbol	Value
Tank diameter	𝑑$d$	1.0 m$1.0\ \mathrm m$
Gas depth above maintained liquid level	𝐿$L$	3.0 −0.5 =2.50 m$3.0-0.5=2.50\ \mathrm m$
Atmospheric pressure	𝑃$P$	101325 Pa =760 mmHg$101325\ \mathrm{Pa}=760\ \mathrm{mmHg}$
Methanol molar mass	𝑀𝐴$M_A$	0.032 kg mol−1$0.032\ \mathrm{kg\,mol^{-1}}$
Reference diffusivity	𝐷𝐴𝐵(298)$D_{AB}(298)$	1.62 ×10−5 m2 s−1$1.62\times10^{-5}\ \mathrm{m^{2}\,s^{-1}}$
Saturation pressure at 30 and 40 degrees Celsius	𝑝𝑠𝑎𝑡$p_{sat}$	163,265 mmHg$163,265\ \mathrm{mmHg}$

We need the total daily emission from the open tank at each temperature.

Assumptions

• Steady one-dimensional diffusion through the gas space.
• Uniform temperature in each calculation and ideal-gas behaviour.
• Constant pressure, maintained liquid level and saturated vapour at the surface.
• Zero net air flux through the liquid; dry gas at the tank mouth.
• Natural convection, wind penetration and transport through the walls are neglected.

Calculations

Let A denote methanol and B denote air. Coordinate 𝑠$s$ points upwards from the liquid. Zero net air flux permits Stefan flow of the mixture, so use

𝑁𝐴=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑠,𝑐=𝑃𝑅𝑇$$N_A=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{ds},\qquad c=\frac{P}{RT}$$(1)

Here 𝑅 =8.314 Pa m3 mol−1 K−1$R=8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}$. The boundary values are 𝑦𝐴(0) =𝑝𝑠𝑎𝑡/𝑃$y_A(0)=p_{sat}/P$ and 𝑦𝐴(𝐿) =0$y_A(L)=0$. At the same pressure, estimate the change in diffusivity from

𝐷𝐴𝐵(𝑇)=𝐷𝐴𝐵(298)(𝑇298)3/2$$D_{AB}(T)=D_{AB}(298)\left(\frac{T}{298}\right)^{3/2}$$

Integrate Eq. 1 across the gas space and multiply the flux by the open cross-sectional area, molar mass and seconds per day.

The maintained liquid level fixes the gas diffusion distance throughout each calculation.

Since 𝑁𝐴$N_A$ is constant over the planar gas space, integration gives

𝑁𝐴=𝑐𝐷𝐴𝐵𝐿ln⁡1−𝑦𝐴(𝐿)1−𝑦𝐴(0).$$N_A=\frac{cD_{AB}}{L}\ln\frac{1-y_A(L)}{1-y_A(0)}.$$(2)

At 30 degrees, 𝑇 =303.15 K$T=303.15\,\mathrm K$,

𝑦𝐴(0)=163760=0.214474,$$y_A(0)=\frac{163}{760}=0.214474,$$

𝑐 =40.2021 mol m−3$c=40.2021\,\mathrm{mol\,m^{-3}}$ and 𝐷𝐴𝐵 =1.66218 ×10−5 m2 s−1$D_{AB}=1.66218\times10^{-5}\,\mathrm{m^2\,s^{-1}}$.

Using Eq. 2 gives 𝑁𝐴 =6.45246 ×10−5 mol m−2 s−1$N_A=6.45246\times10^{-5}\,\mathrm{mol\,m^{-2}\,s^{-1}}$.

The emitting area is

𝐴=𝜋(1.0)24=0.785398 m2,$$A=\frac{\pi(1.0)^2}{4}=0.785398\,\mathrm{m^2},$$

hence

˙𝑚𝐴=𝑀𝐴𝐴𝑁𝐴=1.62168×10−6 kg s−1,$$\dot m_A=M_A A N_A =1.62168\times10^{-6}\,\mathrm{kg\,s^{-1}},$$𝑚𝐴,𝑑𝑎𝑦=86400˙𝑚𝐴=0.1401 kg day−1.$$m_{A,day}=86400\dot m_A =0.1401\,\mathrm{kg\,day^{-1}}.$$

At 40 degrees, 𝑇 =313.15 K$T=313.15\,\mathrm K$, 𝑦𝐴(0) =0.348684$y_A(0)=0.348684$, 𝑐 =38.9183 mol m−3$c=38.9183\,\mathrm{mol\,m^{-3}}$ and 𝐷𝐴𝐵 =1.74510 ×10−5 m2 s−1$D_{AB}=1.74510\times10^{-5}\,\mathrm{m^2\,s^{-1}}$.

The same integration yields

𝑁𝐴=1.16479×10−4 mol m−2 s−1𝑚𝐴,𝑑𝑎𝑦=0.2529 kg day−1$$N_A=1.16479\times10^{-4}\,\mathrm{mol\,m^{-2}\,s^{-1}} \qquad m_{A,day}=0.2529\,\mathrm{kg\,day^{-1}}$$

The predicted emission increases by a factor of 1.805.

Increased saturation vapour pressure is the main cause, with a smaller diffusivity increase and a slight gas-concentration decrease.

These rates require the assumed stagnant gas space; ventilation or buoyant circulation would change the model.

Answer

The methanol emission is 0.140 kg day−1$0.140\ \mathrm{kg\,day^{-1}}$ at 30 ∘C$30\,{}^\circ\mathrm C$ and 0.253 kg day−1$0.253\ \mathrm{kg\,day^{-1}}$ at 40 ∘C$40\,{}^\circ\mathrm C$, under the stagnant-gas model.

2 Evaporation of a suspended water droplet

Spherical diffusion and a shrinking liquid balance

Consider the evaporation of water droplets (diameter ~ 0.1 mm) in still air at 50 ∘C$50\,{}^\circ\mathrm{C}$, 1 atm pressure, with 20% relative humidity. As the droplets are very small, it can be considered that their terminal velocity is negligible, i.e. they are suspended in air. The droplet temperature can be assumed to be at 50 ∘C$50\,{}^\circ\mathrm{C}$ and its vapour pressure is 12 kPa. If the diffusion coefficient of water vapour in air is 0.288 cm2/s$\mathrm{cm^2/s}$,

a. Calculate the initial rate of evaporation of a single droplet, assuming steady state condition.

b. Determine the time required for the water droplet to evaporate completely.

Worked solution

Figure 2: The droplet evaporates into air with a fixed far-field humidity. Its surface concentration is saturated, while its radius decreases as liquid is lost. The dashed outer boundary is taken to infinity in this problem.

Data

Property	Value
Total pressure 𝑃$P$	101325 Pa$101325\,\mathrm{Pa}$
Temperature 𝑇$T$	323.15 K$323.15\,\mathrm K$
Water saturation vapour pressure 𝑝𝑠𝑎𝑡$p_{sat}$	12000 Pa$12000\,\mathrm{Pa}$
Water-air diffusivity 𝐷𝐴𝐵$D_{AB}$	0.288 cm2 s−1$0.288\,\mathrm{cm^2\,s^{-1}}$
Liquid water density 𝜌𝐿$\rho_L$	1000 kg m−3$1000\,\mathrm{kg\,m^{-3}}$
Water molar mass 𝑀𝐴$M_A$	0.018 kg mol−1$0.018\,\mathrm{kg\,mol^{-1}}$

Assumptions

Assume spherical symmetry, negligible droplet velocity, constant properties and no air dissolving in the droplet. The gas profile adjusts much faster than the droplet shrinks, so treat it as steady at each instantaneous radius 𝑎(𝑡)$a(t)$. An external heat supply maintains the liquid and gas at 50 degrees; ignore evaporative cooling. Far-field humidity stays fixed in an effectively infinite gas domain. At 𝑟 =𝑎$r=a$, 𝑦𝑠 =𝑝𝑠𝑎𝑡/𝑃$y_s=p_{sat}/P$. At 𝑟 →∞$r\to\infty$, 𝑦∞ =0.20𝑝𝑠𝑎𝑡/𝑃$y_\infty=0.20p_{sat}/P$. Relative humidity is a ratio of water partial pressure to saturation pressure, not a gas mole fraction.

Calculations

Derive 𝑟2𝑁𝐴 =constant$r^2N_A=\text{constant}$ from a spherical gas shell balance. Integrate Eq. 1 radially using 𝑁𝐵 =0$N_B=0$. Then use the shrinking liquid balance

−𝜌𝐿4𝜋𝑎2𝑑𝑎𝑑𝑡=𝑀𝐴˙𝑛𝐴,$$-\rho_L4\pi a^2\frac{da}{dt}=M_A\dot n_A,$$

where ˙𝑛𝐴 =4𝜋𝑟2𝑁𝐴$\dot n_A=4\pi r^2N_A$ is the molar evaporation rate. Use the instantaneous rate as the radius decreases; do not divide the initial mass by the initial rate to obtain the lifetime.

The spherical area increases outward while the total molar evaporation rate stays constant.

A steady shell balance gives

𝑑(𝑟2𝑁𝐴)𝑑𝑟=0.$$\frac{d(r^2N_A)}{dr}=0.$$

Write 𝑟2𝑁𝐴 =𝐶$r^2N_A=C$ and integrate the flux law between 𝑎$a$ and infinity:

𝐶𝑎=𝑐𝐷𝐴𝐵Λ,Λ=ln⁡1−𝑦∞1−𝑦𝑠,˙𝑛𝐴=4𝜋𝑎𝑐𝐷𝐴𝐵Λ.$$\begin{aligned} \frac{C}{a}&=cD_{AB}\Lambda,\\ \Lambda&=\ln\frac{1-y_\infty}{1-y_s},\\ \dot n_A&=4\pi a cD_{AB}\Lambda. \end{aligned}$$(3)

Convert 𝐷𝐴𝐵 =2.88 ×10−5 m2 s−1$D_{AB}=2.88\times10^{-5}\,\mathrm{m^2\,s^{-1}}$ and

𝑎0=𝑑02=5.00×10−5 m.$$a_0=\frac{d_0}{2}=5.00\times10^{-5}\,\mathrm m.$$

The boundary fractions are 𝑦𝑠 =0.118431$y_s=0.118431$ and 𝑦∞ =0.0236862$y_\infty=0.0236862$, so Λ =0.102081$\Lambda=0.102081$ and 𝑐 =37.7140 mol m−3$c=37.7140\,\mathrm{mol\,m^{-3}}$.

Using the initial radius in Eq. 3 gives

˙𝑛𝐴(0)=4𝜋(5.00×10−5)(37.7140)(2.88×10−5)(0.102081)$$\dot n_A(0)=4\pi(5.00\times10^{-5})(37.7140)(2.88\times10^{-5})(0.102081)$$

Evaluating the initial rate gives

˙𝑛𝐴(0)=6.96655×10−8 mol s−1,$$\dot n_A(0)=6.96655\times10^{-8}\,\mathrm{mol\,s^{-1}},$$˙𝑚𝐴(0)=𝑀𝐴˙𝑛𝐴(0)=1.2540×10−9 kg s−1.$$\dot m_A(0)=M_A\dot n_A(0) =1.2540\times10^{-9}\,\mathrm{kg\,s^{-1}}.$$

Substitute the instantaneous rate into the liquid balance, cancel 4𝜋𝑎$4\pi a$ and integrate:

−𝜌𝐿𝑎𝑑𝑎𝑑𝑡=𝑀𝐴𝑐𝐷𝐴𝐵Λ,𝑎2(𝑡)=𝑎20−2𝑀𝐴𝑐𝐷𝐴𝐵Λ𝜌𝐿𝑡,𝑡𝑓=𝜌𝐿𝑎202𝑀𝐴𝑐𝐷𝐴𝐵Λ=0.6263 s.$$\begin{aligned} -\rho_L a\frac{da}{dt}&=M_AcD_{AB}\Lambda,\\ a^2(t)&=a_0^2-\frac{2M_AcD_{AB}\Lambda}{\rho_L}t,\\ t_f&=\frac{\rho_La_0^2}{2M_AcD_{AB}\Lambda}\\ &=0.6263\,\mathrm s. \end{aligned}$$

The initial liquid mass is 5.2360 ×10−10 kg$5.2360\times10^{-10}\,\mathrm{kg}$.

Dividing it by the initial mass rate gives 0.4175 s and underestimates the lifetime because the rate decreases with radius.

The gas diffusion time

𝑎20𝐷𝐴𝐵=8.68×10−5 s$$\frac{a_0^2}{D_{AB}}=8.68\times10^{-5}\,\mathrm s$$

is much shorter than 𝑡𝑓$t_f$, which supports the quasi-steady approximation.

Complete evaporation is the 𝑎 →0$a\to0$ limit of this continuum model.

Cooling and molecular-scale effects are outside the assumptions.

Answer

The initial mass evaporation rate is 1.25 ×10−9 kg s−1$1.25\times10^{-9}\ \mathrm{kg\,s^{-1}}$ per droplet. The shrinking-sphere model gives a lifetime of 0.626 s$0.626\ \mathrm s$, with the droplet temperature maintained at 50 ∘C$50\,{}^\circ\mathrm C$.