1. ME 475/675 Introduction to Combustion
Lecture 11
Mass Transfer, Stefan problem, Liquid/vapor interface boundary condition, Example problem 3.9

2. Announcements Midterm 1 HW 5 Due Friday, September 29]
Monday, October 2, 2017
8-10 AM, PE 102
Review Friday, September 29
HW 5 Due Friday, September 29]

3. Chapter 3 Introduction to Mass Transferx x x x x x x x x x x x x x x o o o x x x x x x x x o o o o x x x x x o o x x x x x o o o o x x x x o o o o x x o o o o o o o o o x o o o 1-
Mass
Fraction Y Yx Yo
𝑚 " x 0-
Consider two species, x and o
Concentration of “x” is larger on the left, of “o” is larger on the right
Species diffuse through (around) each other
They move from regions of high to low concentrations
Think of perfume in a room
Mass flux is driven by concentration difference (gradients)
Analogously, heat transfer is driven by temperature differences
There may also be bulk motion of the mixture (advection, wind)
Total mass flux rate: 𝑚 " = 𝑚 𝑥 " + 𝑚 𝑜 " 𝑘𝑔 𝑠∗ 𝑚 2 (sum of component mass fluxes)

4. Quantitative phenomena and expressions
𝑚 " Yx Yo
Mass
Fraction Y x
Rate of mass flux of “x” in the 𝑥 direction
𝑚 𝑥 " = 𝑌 𝑥 𝑚 " + −𝜌 𝒟 𝑥𝑜 𝑑 𝑌 𝑥 𝑑𝑥 ; 𝑚 " = 𝑚 𝑥 " + 𝑚 𝑜 "
Advection
(Bulk Motion) Diffusion (due to concentration gradient)
𝑚 " = 𝑚 𝑥 " + 𝑚 𝑜 "
𝒟 𝑥𝑜 = Diffusion coefficient of x through o
Units 𝑘𝑔 𝑚 2 𝑠 𝑚 3 𝑘𝑔 𝑚= 𝑚 2 𝑠
Appendix D, pp
For gases, book shows that 𝜌 𝒟 𝑥𝑜 ~ 𝑇 𝑃 0

5. Stefan Problem (no reaction)x
One dimensional tube (Cartesian)
Gas B is “stationary:” 𝑚 𝐵 " =0
Gas A moves upward 𝑚 𝐴 " >0
Constant, Want to find 𝑚 𝐴 "
𝑚 𝐴 " = 𝑌 𝐴 𝑚 𝐴 " + 𝑚 𝐵 " + −𝜌 𝒟 𝐴𝐵 𝑑 𝑌 𝐴 𝑑𝑥
𝑚 𝐴 " 1− 𝑌 𝐴 =−𝜌 𝒟 𝐴𝐵 𝑑 𝑌 𝐴 𝑑𝑥
𝑚 𝐴 " 𝜌 𝒟 𝐴𝐵 𝑑𝑥= −𝑑 𝑌 𝐴 1− 𝑌 𝐴 ; 0 𝐿 𝑚 𝐴 " 𝜌 𝒟 𝐴𝐵 𝑑𝑥 = 𝑌 𝐴,𝑖 𝑌 𝐴,∞ −𝑑 𝑌 𝐴 1− 𝑌 𝐴
𝒟 𝐴𝐵 𝜌=𝑓𝑛 𝑥 , but treat as constant
𝑚 𝐴 " 𝐿 𝜌 𝒟 𝐴𝐵 = ln 1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖
𝑌 𝐴,∞ L- YB YA
B+A Y
𝑌 𝐴,𝑖 A

6. Mass Flux of evaporating liquid A
𝑚 𝐴 " = 𝜌 𝒟 𝐴𝐵 𝐿 ln 1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖
Increases with 𝜌 𝒟 𝐴𝐵 and 𝑌 𝐴,𝑖
Decreases with 𝐿 and 𝑌 𝐴,∞
For 𝑌 𝐴,∞ =0 (strong wind at 𝑥=𝐿)
𝑚 𝐴 " 𝜌 𝒟 𝐴𝐵 𝐿 = ln 1 1− 𝑌 𝐴,𝑖 (dimensionless)
𝑚 𝐴 " increases slowly for small 𝑌 𝐴,𝑖
Then very rapidly for 𝑌 𝐴,𝑖 > 0.95
What is the shape of the 𝑌 𝐴 versus x profile?
𝑚 𝐴 " 𝜌 𝒟 𝐴𝐵 𝐿
𝑌 𝐴,𝑖

7. 𝑌 𝐴 𝑥 Profile Shape 𝑚 𝐴 " 𝜌 𝒟 𝐴𝐵 0 𝑥 𝑑𝑥 = 𝑌 𝐴,𝑖 𝑌 𝐴 (𝑥) −𝑑 𝑌 𝐴 1− 𝑌 𝑥
𝑚 𝐴 " 𝜌 𝒟 𝐴𝐵 0 𝑥 𝑑𝑥 = 𝑌 𝐴,𝑖 𝑌 𝐴 (𝑥) −𝑑 𝑌 𝐴 1− 𝑌 𝑥
𝑚 𝐴 " 𝑥 𝜌 𝒟 𝐴𝐵 = ln 1− 𝑌 𝐴 (𝑥) 1− 𝑌 𝐴,𝑖
but 𝑚 𝐴 " 𝐿 𝜌 𝒟 𝐴𝐵 = ln 1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖
Ratio: 𝑥 𝐿 = ln 1− 𝑌 𝐴 (𝑥) 1− 𝑌 𝐴,𝑖 ln 1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖 ;
𝑥 𝐿 ln 1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖 = ln 1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖 𝑥 𝐿 = ln 1− 𝑌 𝐴 (𝑥) 1− 𝑌 𝐴,𝑖
1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖 𝑥 𝐿 = 1− 𝑌 𝐴 (𝑥) 1− 𝑌 𝐴,𝑖
𝑌 𝐴 𝑥 =1− 1− 𝑌 𝐴,𝑖 1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖 𝑥 𝐿
For 𝑌 𝐴,∞ =0
𝑌 𝐴 𝑥 =1− 1− 𝑌 𝐴,𝑖 − 𝑌 𝐴,𝑖 𝑥 𝐿
Large 𝑌 𝐴,𝑖 profiles exhibit a “boundary layer” near exit (large diffusion near interface)
𝑌 𝐴,𝑖 =0.99
𝑌 𝐴,𝑖 =0.9
𝑌 𝐴 𝑥
𝑌 𝐴,𝑖 =0.5
𝑌 𝐴,𝑖 =0.1
𝑌 𝐴,𝑖 =0.05
𝑥 𝐿

8. Liquid-Vapor Interface Boundary Condition
At interface need 𝑌 𝐴,𝑖 = 𝑀 𝐴 𝑀 𝑇𝑜𝑡𝑎𝑙 = 𝑁 𝐴 𝑁 𝑇𝑜𝑡𝑎𝑙 𝑀𝑊 𝐴 𝑀𝑊 𝑀𝑖𝑥 = 𝜒 𝐴 𝑀𝑊 𝐴 𝑀𝑊 𝑀𝑖𝑥
𝑀𝑊 𝑀𝑖𝑥 = 𝑀 𝑇𝑜𝑡𝑎𝑙 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝑁 𝐴 𝑀𝑊 𝐴 + 𝑁 𝐵 𝑀𝑊 𝐵 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝜒 𝐴 𝑀𝑊 𝐴 + 1− 𝜒 𝐴 𝑀𝑊 𝐵
So 𝑌 𝐴,𝑖 = 𝜒 𝐴 𝑀𝑊 𝐴 𝜒 𝐴 𝑀𝑊 𝐴 + 1− 𝜒 𝐴 𝑀𝑊 𝐵 = 𝜒 𝐴 −1 𝑀𝑊 𝐵 𝑀𝑊 𝐴 ; need 𝜒 𝐴
𝜒 𝐴 = 𝑃 𝐴 𝑃 𝑇𝑜𝑡𝑎𝑙 = 𝑃 𝐴,𝑆𝑎𝑡 𝑃
𝑃 𝐴,𝑆𝑎𝑡 =𝑓𝑛(𝑇) Saturation pressure at temperature T
For water, tables in thermodynamics textbook
Or use Clausius-Clapeyron Equation (page 18 eqn. 2.19)
A+B
Vapor
𝑌 𝐴,𝑖
Liquid A

9. Clausius-Clapeyron Equation (page 18)
Relates saturation pressure at a given temperature to the saturation conditions at another temperature and pressure
𝑃 1 𝑃 2 𝑑𝑃 𝑆𝑎𝑡 𝑃 𝑆𝑎𝑡 = ℎ 𝑓𝑔 𝑅 𝑇 1 𝑇 2 𝑑𝑇 𝑆𝑎𝑡 𝑇 𝑆𝑎𝑡 2 ; ℎ 𝑓𝑔 =ℎ𝑒𝑎𝑡 𝑜𝑓 𝑣𝑎𝑝𝑜𝑟𝑖𝑧𝑎𝑡𝑖𝑜𝑛 (page 701 for fuels)
ln 𝑃 2 𝑃 1 = ℎ 𝑓𝑔 𝑅 − 1 𝑇 𝑇 1 𝑇 2 = ℎ 𝑓𝑔 𝑅 1 𝑇 1 − 1 𝑇 2
Let 𝑇 1 = 𝑇 𝐵𝑜𝑖𝑙 and 𝑃 1 = 𝑃 𝐵𝑜𝑖𝑙 =1 𝑎𝑡𝑚 (tabulated for fuels on page 701)
Let 𝑃 2 = 𝑃 𝑆𝑎𝑡 that we are tying to find at temperature 𝑇 2 =T
ln 𝑃 𝑆𝑎𝑡 𝑃 𝐵𝑜𝑖𝑙 = ℎ 𝑓𝑔 𝑅 1 𝑇 𝐵𝑜𝑖𝑙 − 1 𝑇
𝜒 2 = 𝑃 𝑆𝑎𝑡 𝑃 𝑇𝑜𝑡𝑎𝑙 = 𝑃 𝐵𝑜𝑖𝑙 𝑃 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑝 ℎ 𝑓𝑔 𝑀𝑊 𝑅 𝑢 𝑇 𝐵𝑜𝑖𝑙 − 1 𝑇
If given 𝑇 𝐵𝑜𝑖𝑙 , 𝑃 𝐵𝑜𝑖𝑙 , ℎ 𝑓𝑔 𝑎𝑛𝑑 𝑇, we can use this to find 𝑃 𝑆𝑎𝑡
Page 701, Table B: ℎ 𝑓𝑔 , 𝑇 𝐵𝑜𝑖𝑙 at 𝑃 𝐵𝑜𝑖𝑙 =1 𝑎𝑡𝑚

10. Problem 3.9
Consider liquid n-hexane in a 50-mm-diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the liquid- air interface to the open end of the cylinder is 20 cm. Assume the diffusivity of n-hexane is 8.8x10-6 m2/s. The liquid n-hexane is at 25°C. Estimate the evaporation rate of the n-hexane. (Hint: review the Clausius-Clapeyron relation applied in Example 3.1)

12. Stefan Problem (no reaction)x
One dimensional tube (Cartesian)
Gas B is stationary 𝑚 𝐵 " =0
but has a concentration gradient
Diffusion of B down = advection up
𝑚 𝐵 " =0= 𝑌 𝐵 𝑚 𝐴 " + 𝑚 𝐵 " −𝜌 𝒟 𝐵𝐴 𝑑 𝑌 𝐵 𝑑𝑥
𝑌 𝐵 𝑚 𝐴 " =𝜌 𝒟 𝐵𝐴 𝑑 𝑌 𝐵 𝑑𝑥
𝑚 𝐴 " =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑚 𝐴 " 𝜌 𝒟 𝐵𝐴 𝑑𝑥= 𝑑 𝑌 𝐵 𝑌 𝐵 ; 𝑚 𝐴 " 𝜌 𝒟 𝐵𝐴 0 𝑥 𝑑𝑥 = 𝑌 𝐵,𝑖 𝑌 𝐵 (𝑥) 𝑑 𝑌 𝐵 𝑌 𝐵
𝑚 𝐴 " 𝑥 𝜌 𝒟 𝐵𝐴 = ln 𝑌 𝐵 (𝑥) 𝑌 𝐵,𝑖 ; 𝑌 𝐵 (𝑥)= 𝑌 𝐵,𝑖 𝑒 𝑚 𝐴 " 𝜌 𝒟 𝐵𝐴 𝑥
𝑌 𝐴,∞ L- YB YA Y
YA,i

13. Clausius-Clapeyron Equation (page 18)
Relates saturation pressure at a given temperature to the saturation conditions at another temperature and pressure
𝑃 1 𝑃 2 𝑑𝑃 𝑆𝑎𝑡 𝑃 𝑆𝑎𝑡 = ℎ 𝑓𝑔 𝑅 𝑇 1 𝑇 2 𝑑𝑇 𝑆𝑎𝑡 𝑇 𝑆𝑎𝑡 2
ln 𝑃 2 𝑃 1 = ℎ 𝑓𝑔 𝑅 − 1 𝑇 𝑇 1 𝑇 2 = ℎ 𝑓𝑔 𝑅 1 𝑇 1 − 1 𝑇 2 ; 𝑃 2 𝑃 1 =𝑒𝑥𝑝 ℎ 𝑓𝑔 𝑅 𝑇 1 − 1 𝑇 2
𝜒 2 = 𝑃 2 𝑃 𝑇𝑜𝑡𝑎𝑙 = 𝑃 1 𝑃 𝑇𝑜𝑡𝑎𝑙 𝑃 2 𝑃 1 = 𝑃 1 𝑃 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑝 ℎ 𝑓𝑔 𝑅 𝑇 1 − 1 𝑇 2
If given 𝑇 1 , 𝑃 1 , ℎ 𝑓𝑔 𝑎𝑛𝑑 𝑇 2 , we can use this to find 𝑃 1
Page 701, Table B: ℎ 𝑓𝑔 , 𝑇 𝐵𝑜𝑖𝑙 = 𝑇 𝑆𝑎𝑡 = 𝑇 1 at P = 1 atm = 𝑃 1