Temporary Divergence: Diffusivity: That Smell-Methyl Mercaptan (Methanethiol), Odorless Natural Gas, Odorless Propane, and Even Flatulence

Temporary Divergence: Diffusivity: That Smell-Methyl Mercaptan (Methanethiol), Odorless Natural Gas, Odorless Propane, and Even Flatulence

Lynyrd Skynyrd. That Smell. youtube.com[online]. 2013. Available from: http://youtu.be/ZDB-yswOrzc

Diffusivity of Methyl Mercaptan Versus Methane and Propane

Methyl mercaptan[3-6], “methanethiol”, is the byproduct of many natural processes. Flatulence is one example[7]. Because of its odor threshold, 1 ppb has been reported[4], methanethiol is also added to odorless natural gas, methane, and odorless propane for detection purposes. Apparently, it is used as a communication warning system in mining operations as well[4].

In this blog post, I will be comparing the diffusivity of the polar chemical methanethiol to the non-polar chemicals methane and propane in air. I have heard reports that the diffusivity of methanethiol is significantly greater than methane and propane. See bottom of post for diffusivities.Since reference[1] has tabular values for methane and propane in appendix B, I will use the tabular values and the Chapman-Enskog equation to calculate diffusivity values for methane and propane. For methanethiol, I will use Fuller, et al equation and tabular values for the atoms making up methanethiol, $CH_3-SH$

Chapman-Enskog Equation. From reference[1], the average absolute error of this “theoretical equation” is 7.9%

$D_{AB} = \frac{3}{16} \frac{(\frac{4 \pi \kappa T}{M_{AB}})^{1/2}}{n \pi \sigma_{AB}^2 \Omega_D} f_D$

If $f_D$ is chosen as unity and “n” expressed by ideal-gas law

$D_{AB} = \frac{0.00266 T^{3/2}}{P M_{AB}^{1/2} \sigma_{AB}^2 \Omega_D}$

For Non-polar gases: Methane and Propane

Methane $CH_4$ in air at 25$\textdegree$C and 1 atmosphere (atm)

$M_{AB} = 2[\frac{1}{M_A} + \frac{1}{M_B}]^{-1}$

$M_A = M_{CH_4} = 1(MW_C) + 4(MW_H) = 1(12.01) + 4(1.008) = 16.0 \frac{g}{mol}$

$M_B = M_{air} = 29.0 \frac{g}{mole}$

$M_{AB} = 2[\frac{1}{M_A} + \frac{1}{M_B}]^{-1} = 2[\frac{1}{16.0} + \frac{1}{29.0}]^{-1} = 20.6$

Need $\Omega_D$

Neufield, et al.: $\Omega_D = \frac{A}{(T^*)^B} + \frac{C}{\exp{((D)(T^*))}} + \frac{E}{\exp{((F)(T^*))}} + \frac{G}{\exp{((H)(T^*))}}$

$T^* = \frac{\kappa T}{\epsilon_{AB}}$

From appendix B[1]

Methane: $\sigma = 3.758 \AA; \frac{\epsilon_A}{\kappa} = 148.6 K$

Air: $\sigma = 3.711 \AA; \frac{\epsilon_B}{\kappa} = 78.6 K$

$\sigma_{AB} = \frac{\sigma_A + \sigma_B}{2} = \frac{3.758 + 3.711}{2} = 3.735 \AA$

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{(148.6)(78.6)} = 108.1 K$

$\frac{\epsilon_{AB}}{\kappa T} = \frac{108.1 K}{298 K} = 0.363$

$T^* = \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.363} = 2.76$

$\Omega_D = \frac{A}{(T^*)^B} + \frac{C}{\exp{((D)(T^*))}} + \frac{E}{\exp{((F)(T^*))}} + \frac{G}{\exp{((H)(T^*))}} =$

$\Omega_D =\frac{1.06036}{(2.76)^{0.15610}} + \frac{0.19300}{\exp{((0.47635)(2.76))}} + \frac{1.03587}{\exp{((1.52996)(2.76))}} + \frac{1.76474}{\exp{((3.89411)(2.76))}} =$

$\Omega_D = 0.972$

Diffusivity of Methane in Air:

$D_{AB} = \frac{0.00266 T^{3/2}}{P M_{AB}^{1/2} \sigma_{AB}^2 \Omega_D} = \frac{0.00266 (298)^{3/2}}{1(20.6)^{1/2} (3.735)^2 (0.972)} = \frac{13.68}{61.54} = 0.222 \frac{cm^2}{s}$

Propane $CH_3CH_2CH_3$ in air at 25$\textdegree$C and 1 atmosphere (atm)

$M_{AB} = 2[\frac{1}{M_A} + \frac{1}{M_B}]^{-1}$

$M_A = M_{CH_3CH_2CH_3} = 3(MW_C) + 8(MW_H) = 3(12.01) + 8(1.008) = 44.1 \frac{g}{mol}$

$M_B = M_{air} = 29 \frac{g}{mol}$

$M_{AB} = 2[\frac{1}{44.1} + \frac{1}{29}]^{-1} = 35.0$

Need $\Omega_D$

Neufield, et al.: $\Omega_D = \frac{A}{(T^*)^B} + \frac{C}{\exp{((D)(T^*))}} + \frac{E}{\exp{((F)(T^*))}} + \frac{G}{\exp{((H)(T^*))}}$

$T^* = \frac{\kappa T}{\epsilon_{AB}}$

From appendix B[1]

Propane: $\sigma_A = 5.118 \AA; \frac{\epsilon_A}{\kappa} = 237.1 K$

Air: $\sigma_B = 3.711 \AA; \frac{\epsilon_B}{\kappa} = 78.6 K$

$\sigma_{AB} = \frac{\sigma_A + \sigma_B}{2} = \frac{5.118 + 3.711}{2} = 4.42 \AA$

$\frac{\epsilon_{AB}}{\kappa} = \sqrt{(\frac{\epsilon_A}{\kappa})(\frac{\epsilon_B}{\kappa})} = \sqrt{(237.1)(78.6)} = 136.5 K$

$\frac{\epsilon_{AB}}{\kappa T} = \frac{136.5}{298} = 0.458$

$T^* = \frac{\kappa T}{\epsilon_{AB}} = \frac{1}{0.458} = 2.18$

$\Omega_D = \frac{A}{(T^*)^B} + \frac{C}{\exp{((D)(T^*))}} + \frac{E}{\exp{((F)(T^*))}} + \frac{G}{\exp{((H)(T^*))}} =$

$\Omega_D = \frac{1.06036}{(2.18)^{0.15610}} + \frac{0.19300}{\exp{((0.47635)(2.18))}} + \frac{1.03587}{\exp{((1.52996)(2.18))}} + \frac{1.76474}{\exp{((3.89411)(2.18))}} =$

$\Omega_D = 1.05$

Diffusivity of Propane in Air:

$D_{AB} = \frac{0.00266T^{3/2}}{PM_{AB}^{1/2} \sigma_{AB}^2 \Omega_D} = \frac{0.00266(298)^{3/2}}{1 (35.0)^{1/2} (4.42)^2 (1.05)} = \frac{13.68}{121.36} =$

$D_{AB} = 0.113 \frac{cm^2}{s}$

For polar molecule $CH_3-SH$, will use Fuller, et al. equation. From reference[1], the absolute relative error of this equation is 5.4%. Authors report an average absolute error of about 4% when using

Fuller, et al.

$D_{AB} = \frac{0.00143 T^{1.75}}{PM_{AB}^{1/2}[(\sum \nu)_A^{1/3} + (\sum \nu)_B^{1/3}]^2}$

T = 25$\textdegree$C = 298 K; P = 1 atm

$M_{AB} = 2[\frac{1}{M_A} + \frac{1}{M_B}]^{-1}$

$M_A = M_{CH_3SH} = 1(C) + 4(H) + 1(S) = 1(12.01) + 4(1.008) + 1(32.07) = 48.11 \frac{g}{mol}$

$M_B = M_{air} = 29 \frac{g}{mol}$

$M_{AB} = 2[\frac{1}{48.11} + \frac{1}{29}]^{-1} = 18.1$

Summation of “Atomic and Structural Diffusion Volume Increments from table 11-1[1]

$(\sum \nu)_A = (\sum \nu)_{CH_3SH} = 1 (C) + 4(H) + 1(S) = 1(15.9) + 4(2.31) + 1(22.9) = 40.04$

$(\sum \nu)_B = (\sum \nu)_{air} = 19.7$

$D_{AB} = \frac{0.00143T^{1.75}}{PM_{AB}^{1/2}[(\sum \nu)_A^{1/3} + (\sum \nu)_B^{1/3}]^2} = \frac{0.00143(298)^{1.75}}{1(18.1)^{1/2}[(40.0)_A^{1/3} + (19.7)_B^{1/3}]^2} =$

Diffusivity of methanethiol in air

$D_{AB} = \frac{30.56}{159.38} = 0.192 \frac{cm^2}{s}$

Diffusivities in air in decreasing order

Methane: $D_{AB} = 0.222 \frac{cm^2}{s}$

Methanethiol: $D_{AB} = 0.192 \frac{cm^2}{s}$

Propane: $D_{AB} = 0.113 \frac{cm^2}{s}$

At a detection threshold of 1 part per billion (ppb) and the above diffusivities, one might detect methanethiol prior to experiencing propane. In truth, there is an equation that takes “mixture” into account but I do not know the percent mixture of each component[2].

Equation for mixture

$D_{1-mixture} = \frac{1}{\frac{z_2}{D_{1-2}} + \frac{z_3}{D_{1-3}} + .... + \frac{z_n}{D_{1-n}}}$

$z_n$ is the mole fraction of component “n” in the gas mixture evaluated on a component-1-free basis

$z_2 = \frac{y_2}{y_2 + y_3 + ... + y_n}$

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• Wait until 2nd attempt

References:

[1] Poling, Bruce E.; Prausnitz, John M.; O’Connell, John P. (2001) The Properties of Gases and Liquids, Fifth Edition. New York: Mcgraw-Hill.

[2] Welty, James R.; Wicks, Charles E.; Wilson, Robert E. (1984) Fundamentals of Momentum, Heat, and Mass Transfer, third edition. New York: John Wiley & Sons.

[3] ScienceBlogs. Molecule of the day. Methanethiol (They put that in, you know), March 2009. scienceblogs.com[online]. 2013. Available from: http://scienceblogs.com/moleculeoftheday/2009/03/18/methanethiol-they-put-that-in/

[4] Wikipedia. Methanethiol. Also known as methyl mercaptan. en.wikipedia.org[online]. 2013. Available from: http://en.wikipedia.org/wiki/Methanethiol

[5] NCBI.PubChem Substance. Methanethiol-Substance Summary (SID 3699). Also known as Methylmercaptan (CAS: 74-93-1). pubchem.ncbi.nlm.nih.gov[online]. 2013. Available from: http://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?sid=3699

[6] US National Institute of Standards and Technology. NIST. Methanethiol. webbook.nist.gov[online]. 2013. Available from: http://webbook.nist.gov/cgi/cbook.cgi?ID=74-93-1&Units=SI

[7] Wikipedia. Flatulence. en.wikipedia.org[online]. 2013. Available from: http://en.wikipedia.org/wiki/Flatulence