Estimating Evaporation Rate from Fundamental Principles Ethyl mercaptan (CH3CH2SH, CAS 75-08-1) is a liquid with an unpleasant skunk-like odor. The material is a strong oxidizer with a PEL of 0.5 PPM. A 55-gallon drum of the material is handled roughly and a small leak develops in a seam.

Question

Estimating Evaporation Rate from Fundamental Principles
Ethyl mercaptan ([latex]CH_3CH_2SH[/latex], CAS 75-08-1) is a liquid with an unpleasant skunk-like odor. The material is a strong oxidizer with a PEL of 0.5 PPM. A 55-gallon drum of the material is handled roughly and a small leak develops in a seam. A circular pool, approximately 10. m in diameter develops. The wind speed over the pool is estimated to be around 3.0 m/s.
To do: Estimate the evaporation rate from the spill ([latex]S_s[/latex], in kg/hr), using first principles rather than empirical equations such as Eq. (4-14).

[latex]S_s=0.3\left(\frac{U_a}{T_a} \right) ^{0.8}A^{0.9}M_jP_\mathbf{v}\left[\frac{(3.1+c_{molar,L}^{-0.33})^2}{T_a^{0.5}\left(\frac{1}{29}+\frac{1}{M_j} \right)^{0.5} } \right] ^{0.67}[/latex] (4-14)

Accepted Answer

It is assumed that the partial pressure of the contaminant at the liquid-gas interface is equal to the vapor pressure of the contaminant, [latex]P_{\mathbf{v},J}[/latex] the value of which can be found in Appendix A.8. It is also assumed that the characteristic length is the diameter of the pool, i.e. L = D. The temperature is 17.7 °C (290.85 K) and the total pressure is 1.00 atm (760. mm Hg).

ethyl mercaptan	air
[latex]P_{j,i} = p_{\mathbf{v},j}[/latex] = 400. mm Hg	ν = 15.89 × [latex]10^{-6}[/latex] m²/s
[latex]D_{j,a} = 0.9× 10^{-5}[/latex]m²/s	Pr = 0.707
[latex]M_j[/latex] = 62.1	[latex]U_{\infty}[/latex] = 3.0m/s
	[latex]P_{a,i} [/latex] = 760. - [latex]P_{j,i} [/latex] = 360. mm Hg

The presence of ethyl mercaptan in air just above the interface influences the Prandtl (Pr) and Schmidt (Sc) numbers. For a first approximation, this influence is neglected because the concentrations are not large. The following parameters are computed:

[latex]P_{am}=\frac{(760.-360.) mm Hg}{\ln \left(\frac{760.}{360.} ight) } =535.3 mm Hg[/latex] [latex]Re=\frac{U_\infty D}{ν} =\frac{3.0\frac{m}{s} (10.0 m)}{1.589 imes 10^{-5}\frac{m^2}{s} } =1.888 imes 10^6[/latex]

The criteria in Table 4.3 show that the flow is turbulent. Table 4.3 indicates that the Nusselt number can be calculated as

[latex]Nu=\left[0.037(Re)^{0.8}-871 ight] (Pr)^{0.33}=\left[0.037(1.888 imes 10^6)^{0.8}-871 ight] (0.707)^{0.33}=2685.[/latex]

Table 4.3 Vapor pressure, diffusion coefficient, and evaporation rate for several liquids; evaporation rate is also compared to that of carbon tetrachloride.

hydrocarbon	[latex]P_{\mathbf{v}}[/latex] (at [latex]T_{room}[/latex]) (mm Hg)	[latex]D_{ja} (10^{-5} m^2/S)[/latex]	evap. Rate (g/hr)	evap. rate relative to [latex]CCl_4[/latex]
acetone	180	0.83	7.95	1.11
carbon tetrachloride	91	0.62	7.16	1.00
benzene	75	0.77	3.90	0.54
toluene	20	0.71	1.20	0.17
acetic acid	11	1.06	0.75	0.10
chlorobenzene	12	0.62	0.23	0.03

The Schmidt number is

[latex]Sc=\frac{\mu }{D_{ja} ho } =\frac{V}{D_{ja}} =\frac{1.589 imes 10^{-5}\frac{m^2}{s} }{9.0 imes 10^{-6}\frac{m^2}{s} } =1.766[/latex]

The mass transfer coefficient (kc) can be computed from Eq. (4-80), where [latex]b_1[/latex] = 0.33,

[latex]K_G=Nu\frac{D_{ja}}{L} \left(\frac{Sc}{Pr} ight) ^{b_1}\frac{P}{P_{am}} \frac{1}{R_uT}[/latex] (4-80)

[latex]K_G=Nu\frac{D_{ja}}{D} \left(\frac{Sc}{Pr} ight) ^{0.33}\frac{P}{P_{am}} \frac{1}{R_uT}[/latex]

[latex]=2685\frac{9.0 imes 10^{-6}\frac{m^2}{s} }{10.0 m} \left(\frac{1.766}{0.707} ight) ^{0.33}\frac{760. mm Hg}{535.3 mm Hg} \frac{1}{8.314\frac{KJ}{K\cdot mol\cdot K}(290.85 K) } \left(\frac{KJ}{m^3Kpa} ight) [/latex]

which yields

[latex]K_G=1.92 imes 10^{-6}\frac{Kmol}{m^2s\cdot Kpa}[/latex]

The molar flux [latex](N_j)[/latex] can be computed from Eq. (4-67), assuming that the ethyl mercaptan partial pressure in the far field is zero [latex](P_{j,\infty =0})[/latex],

[latex]N_j=K_G(P_{j,1}-P_{j,\infty })[/latex] (4-67)

[latex]N_j=K_G(P_{j,1}-P_{j,\infty })=1.92 imes 10^{-6}\frac{Kmol}{m^2s\cdot Kpa} (400. mm Hg-0)\left(\frac{101.3 Kpa}{760. mm Hg} ight) =1.02 imes 10^{-4}\frac{Kmol}{m^2s}[/latex]

Finally, the total evaporation rate from the pool is obtained from Eq. (4-46),

[latex]\dot{m} _{evap,j}=N_jM_jA[/latex] (4-46)

[latex]\dot{m} _{evap,j}=N_jM_jA=1.02 imes 10^{-4}\frac{Kmol}{m^2s} 62.1\frac{Kg}{Kmol} \frac{\pi }{4} (10.0 m)^2\left(\frac{3600 s}{hr} ight) =1.79 imes 10^3\frac{Kg}{hr}[/latex]

or approximately 1800 kg/hr.

Discussion: If the empirical equation, Eq. (4-14), were used instead to estimate the evaporation rate, a value of 1670 kg/hr would have been obtained. The evaporation rate computed from fundamental principles is about 5% larger than the value computed from the empirical equation.
This agreement is excellent, considering the numerous approximations and assumptions! An explanation for the discrepancy may lie in the fact that the equation used to compute the Nusselt number assumes that the velocity and concentration boundary layers begin at the leading edge of the pool. In the actual case, air passing over the ground has a fully established boundary layer at the leading edge of the pool, and the equation for the Nusselt number may not apply to the actual case.

Question 4.12: Estimating Evaporation Rate from Fundamental Principles Ethy...

Related Answered Questions

Verified Answer:

Estimation of the Diffusion Coefficient of Toluene in Air Toluene (C6H5OH, CAS 108-88-3) is a common solvent with a sweet, pungent odor and a PEL equal to 200 PPM. From handbooks and Appendix A.10 the following data can be found: ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Evaporation of Volatile Compounds from a Stagnant Waste Lagoon A waste lagoon is 25. m × 40. m × 3.5 m deep. It contains 100. mg/L of benzene (Mb = 78.0) and 100. mg/L of chloroform (Mc = 119.0) in water. The air and liquid temperatures are 25.0 °C and the wind speed is 1.70 m/s at z = 10. m. ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer: