ME 475/675 Introduction to Combustion Lecture 11

Announcements Midterm 1 HW 5 Due Friday, September 26, 2014 Review Friday, September 26 HW 5 Due Friday, September 26, 2014

Chapter 3 Introduction to Mass Transfer x 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 each other they move from regions of high to low concentrations Think of perfume in a room Mass flux is driven by concentration difference Analogously, heat transfer is driven by temperature differences There may also be bulk motion of the mixture (advection, like wind) Total rate of mass flux: 𝑚 " = 𝑚 𝑥 " + 𝑚 𝑜 " 𝑘𝑔 𝑠∗ 𝑚 2 (sum of component mass flux)

Chapter 3 Introduction to Mass Transfer x x x x x x x x x x x x x x x x o o x x x x o x x x o o o o x x x x x x x o o x x x x o o o o o x o o x o o x x x o o o o o o o o o x o o Yx Yo Mass Fraction Y 𝑚 " x Rate of mass flux of “x” in the 𝑥 direction 𝑚 𝑥 " = 𝑌 𝑥 𝑚 " + −𝜌 𝒟 𝑥𝑜 𝑑 𝑌 𝑥 𝑑𝑥 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. 707-9 For gases, book shows that 𝜌 𝒟 𝑥𝑜 ~ 𝑇 1 2 𝑃 0

Stefan Problem (no reaction) x One dimensional tube (Cartesian) Gas B is stationary: 𝑚 𝐵 " =0 Gas A moves upward 𝑚 𝐴 " >0 Want to find this 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 ; 𝑚 𝐴 " 𝜌 𝒟 𝐴𝐵 0 𝐿 𝑑𝑥 = 𝑌 𝐴,𝑖 𝑌 𝐴,∞ −𝑑 𝑌 𝐴 1− 𝑌 𝐴 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

Mass Flux of evaporating liquid A 𝑚 𝐴 " = 𝜌 𝒟 𝐴𝐵 𝐿 ln 1− 𝑌 𝐴,∞ 1− 𝑌 𝐴,𝑖 For 𝑌 𝐴,∞ =0 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

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 1 1− 𝑌 𝐴,𝑖 𝑥 𝐿 Large 𝑌 𝐴,𝑖 profiles exhibit a boundary layer near exit (large advection 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

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 = 1 1+ 1 𝜒 𝐴 −1 𝑀𝑊 𝐵 𝑀𝑊 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-Slapeyron Equation (page 18 eqn. 2.19) A+B Vapor 𝑌 𝐴,𝑖 Liquid A

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 𝜒 2 = 𝑃 2 𝑃 𝑇𝑜𝑡𝑎𝑙 = 𝑃 1 𝑃 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑝 ℎ 𝑓𝑔 𝑅 1 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 at P = 1 atm

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 25C. Estimate the evaporation rate of the n-hexane. (Hint: review the Clausius-Clapeyron relation a applied in Example 3.1)

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