Report

ME 475/675 Introduction to Combustion Lecture 11 Announcements • Midterm 1 • September 29, 2014 • Review Friday, September 26 • HW 5 Due Friday, September 26, 2014 Chapter 3 Introduction to Mass Transfer xx x xx x x xx x x Mass 1Fraction Y 0- x x x x x x x x x x x x x oo o o x x x xx x xx o x x x x xx o o o o o x o o o o o oo x o o o " Yx o oo o oo o o o Yo • 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: " = " + " ∗2 (sum of component mass flux) x Chapter 3 Introduction to Mass Transfer xx x xx x x xx x x Mass Fraction Y x x x x x x x x x x x x x oo o o x x x xx x xx o x x x x xx o o o o o x o o o o o oo x o o o Yx " • Rate of mass flux of “x” in the direction • " " − = + Advection (Bulk Motion) Diffusion (due to concentration gradient) • " = " + " • =Diffusion coefficient of x through o 3 2 • Units 2 = • Appendix D, pp. 707-9 • For gases, book shows that ~ 1 2 0 o oo o oo o o o Yo x Stefan Problem (no reaction) x L- • One dimensional tube (Cartesian) ,∞ • Gas B is stationary: " = 0 • Gas A moves upward " > 0 YB • Want to find this YA Y , B+A A • " = • " 1 − • + + − = − " ,∞ − − = ; = 0 1− , 1− " " " • = but treat as constant • " = ln 1−,∞ 1−, Mass Flux of evaporating liquid A 8 6.908 • " = 1−,∞ ln 1−, • For ,∞ = 0 • " = 1 ln 1−, (dimensionless) 6 " m ( Y) 4 • " increases slowly for small , • Then very rapidly for , > 0.95 • What is the shape of the versus x profile? 2 0 0 0 0 0.2 0.4 0.6 , Y 0.8 1 Profile Shape 1 • " 0 • " = • but 0.99 " , =0.9 • Ratio: = YA ( x .05) YA ( x .1) YA ( x.5) • 0.6 , =0.5 YA ( x .9) YA ( x .99) • 0.4 = 1− () ln 1−, , =0.99 0.8 () − , 1− ln = ln 1−,∞ 1−, 1− () 1−, 1−,∞ ln 1−,∞ ln 1−, 1−,∞ 1−, ; 1−, = ln = 1−,∞ 1−, , =0.1 0 • = 1 − 1 − , , =0.05 0 0 0.2 1−,∞ 1−, • For ,∞ = 0 0.2 0.4 x 0.6 0.8 1 1− () 1−, 1− () 1−, • = 1 − 1 − , 0 = ln 1 1−, • Large , profiles exhibit a boundary layer near exit (large advection near interface) Liquid-Vapor Interface Boundary Condition • A+B Vapor , Liquid A • • At interface need , = = = + = = = + 1 − 1 So , = = 1 + 1− 1+ −1 • = = , • , = () Saturation pressure at temperature T • For water, tables in thermodynamics textbook • Or use Clausius-Slapeyron Equation (page 18 eqn. 2.19) Clausius-Clapeyron Equation (page 18) • Relates saturation pressure at a given temperature to the saturation conditions at another temperature and pressure • • • 2 1 = ℎ 2 2 1 ℎ 1 1 2 1 = − = − 1 1 2 ℎ 2 1 1 2 = = 1 2 ln 1 ℎ − 1 2 • If given 1 , 1 , ℎ 2 , we can use this to find 1 • Page 701, Table B: ℎ , = 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 liquidair 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 L- • One dimensional tube (Cartesian) ,∞ • Gas B is stationary " = 0 YA • but has a concentration gradient YB • Diffusion of B down = advection up Y YA,i • " • " =0= = " + " − • " = • • " " = = " ; 0 = () ln ; ()=, , () , "