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
𝑚"
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:
𝑚"
=
𝑚𝑥"
+
𝑘𝑔
"
𝑚𝑜
𝑠∗𝑚2
(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
𝑚"
• Rate of mass flux of “x” in the 𝑥 direction
•
𝑚𝑥"
𝑚"
𝑑𝑌𝑥
−𝜌𝒟𝑥𝑜
𝑑𝑥
= 𝑌𝑥 +
Advection
(Bulk Motion) Diffusion (due to concentration gradient)
• 𝑚" = 𝑚𝑥" + 𝑚𝑜"
• 𝒟𝑥𝑜 =Diffusion
coefficient
of x through o
3
2
𝑘𝑔 𝑚
𝑚
• Units 2
𝑚=
𝑚 𝑠 𝑘𝑔
𝑠
• Appendix D, pp. 707-9
• For gases, book shows that 𝜌𝒟𝑥𝑜 ~𝑇
1
2
𝑃0
o
oo
o
oo
o
o
o
Yo
x
Stefan Problem (no reaction)
x
L-
• One dimensional tube (Cartesian)
𝑌𝐴,∞
• Gas B is stationary: 𝑚𝐵" = 0
• Gas A moves upward 𝑚𝐴" > 0
YB
• Want to find this
YA
Y
𝑌𝐴,𝑖
B+A
A
•
𝑚𝐴"
=
• 𝑚𝐴" 1 −
•
𝑑𝑌𝐴
+
+ −𝜌𝒟𝐴𝐵
𝑑𝑥
𝑑𝑌𝐴
𝑌𝐴 = −𝜌𝒟𝐴𝐵
𝑑𝑥
"
𝐿
𝑌𝐴,∞ −𝑑𝑌𝐴
−𝑑𝑌𝐴 𝑚𝐴
=
;
𝑑𝑥 = 𝑌
0
1−𝑌𝐴 𝜌𝒟𝐴𝐵
𝐴,𝑖 1−𝑌𝐴
𝑌𝐴 𝑚𝐴"
"
𝑚𝐴
𝑑𝑥
𝜌𝒟𝐴𝐵
𝑚𝐵"
• 𝜌 = 𝑓𝑛 𝑥 but treat as constant
•
"
𝑚𝐴
𝐿
𝜌𝒟𝐴𝐵
= ln
1−𝑌𝐴,∞
1−𝑌𝐴,𝑖
Mass Flux of evaporating liquid A
8
6.908
•
𝑚𝐴"
=
1−𝑌𝐴,∞
𝜌𝒟𝐴𝐵
ln
𝐿
1−𝑌𝐴,𝑖
• For 𝑌𝐴,∞ = 0
•
"
𝑚𝐴
𝜌𝒟𝐴𝐵
𝐿
=
1
ln
1−𝑌𝐴,𝑖
(dimensionless)
6
𝑚𝐴"
m ( Y) 4
𝜌𝒟𝐴𝐵
𝐿
• 𝑚𝐴" increases slowly for small 𝑌𝐴,𝑖
• Then very rapidly for 𝑌𝐴,𝑖 > 0.95
• What is the shape of the 𝑌𝐴 versus x
profile?
2
0
0
0
0
0.2
0.4
0.6
𝑌𝐴,𝑖
Y
0.8
1
𝑌𝐴 𝑥 Profile Shape
1
•
"
𝑥
𝑚𝐴
𝑑𝑥
𝜌𝒟𝐴𝐵 0
•
"
𝑚𝐴
𝑥
𝜌𝒟𝐴𝐵
=
• but
0.99
"
𝑚𝐴
𝐿
𝜌𝒟𝐴𝐵
𝑥
𝐿
𝑌𝐴,𝑖 =0.9
• Ratio: =
YA ( x .05)
YA ( x .1)
𝑌
YA
𝐴 ( x𝑥.5)
•
0.6
𝑌𝐴,𝑖 =0.5
YA ( x .9)
YA ( x .99)
•
0.4
=
1−𝑌𝐴 (𝑥)
ln
1−𝑌𝐴,𝑖
𝑌𝐴,𝑖 =0.99
0.8
𝑌𝐴 (𝑥) −𝑑𝑌𝐴
𝑌𝐴,𝑖
1−𝑌𝑥
ln
= ln
1−𝑌𝐴,∞
1−𝑌𝐴,𝑖
1−𝑌𝐴 (𝑥)
1−𝑌𝐴,𝑖
1−𝑌𝐴,∞
ln
1−𝑌𝐴,∞
𝑥
ln
𝐿
1−𝑌𝐴,𝑖
𝑥
1−𝑌𝐴,∞ 𝐿
1−𝑌𝐴,𝑖
;
1−𝑌𝐴,𝑖
= ln
=
𝑥
1−𝑌𝐴,∞ 𝐿
1−𝑌𝐴,𝑖
𝑌𝐴,𝑖 =0.1
0
• 𝑌𝐴 𝑥 = 1 − 1 − 𝑌𝐴,𝑖
𝑌𝐴,𝑖 =0.05
0
0
0.2
𝑥
1−𝑌𝐴,∞ 𝐿
1−𝑌𝐴,𝑖
• For 𝑌𝐴,∞ = 0
0.2
0.4
𝑥
x
𝐿
0.6
0.8
1
1−𝑌𝐴 (𝑥)
1−𝑌𝐴,𝑖
1−𝑌𝐴 (𝑥)
1−𝑌𝐴,𝑖
• 𝑌𝐴 𝑥 = 1 − 1 − 𝑌𝐴,𝑖
0
= ln
1
1−𝑌𝐴,𝑖
𝑥
𝐿
• Large 𝑌𝐴,𝑖 profiles exhibit a boundary layer near
exit (large advection near interface)
Liquid-Vapor Interface Boundary Condition
•
A+B
Vapor
𝑌𝐴,𝑖
Liquid
A
•
•
𝑀𝐴
𝑁𝐴
𝑀𝑊𝐴
𝑀𝑊𝐴
At interface need 𝑌𝐴,𝑖 =
=
= 𝜒𝐴
𝑀𝑇𝑜𝑡𝑎𝑙
𝑁𝑇𝑜𝑡𝑎𝑙 𝑀𝑊𝑀𝑖𝑥
𝑀𝑊𝑀𝑖𝑥
𝑀𝑇𝑜𝑡𝑎𝑙
𝑁𝐴 𝑀𝑊𝐴 +𝑁𝐵 𝑀𝑊𝐵
𝑀𝑊𝑀𝑖𝑥 =
=
= 𝜒𝐴 𝑀𝑊𝐴 + 1 − 𝜒𝐴 𝑀𝑊𝐵
𝑁𝑇𝑜𝑡𝑎𝑙
𝑁𝑇𝑜𝑡𝑎𝑙
𝑀𝑊𝐴
1
So 𝑌𝐴,𝑖 = 𝜒𝐴
=
𝑀𝑊𝐵
1
𝜒𝐴 𝑀𝑊𝐴 + 1−𝜒𝐴 𝑀𝑊𝐵
1+
−1
• 𝜒𝐴 =
𝑃𝐴
𝑃𝑇𝑜𝑡𝑎𝑙
=
𝑃𝐴,𝑆𝑎𝑡
𝜒𝐴
𝑀𝑊𝐴
𝑃
• 𝑃𝐴,𝑆𝑎𝑡 = 𝑓𝑛(𝑇) 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
•
•
•
𝑃2 𝑑𝑃𝑆𝑎𝑡
𝑃1 𝑃𝑆𝑎𝑡
=
ℎ𝑓𝑔
𝑅
𝑇2 𝑑𝑇𝑆𝑎𝑡
2
𝑇1 𝑇𝑆𝑎𝑡
ℎ𝑓𝑔 1
1 𝑇2
1
=
−
=
−
𝑅
𝑇 𝑇1
𝑅 𝑇1
𝑇2
ℎ𝑓𝑔
𝑃2
𝑃1
1
𝜒2 =
=
𝑒𝑥𝑝
𝑃𝑇𝑜𝑡𝑎𝑙
𝑃𝑇𝑜𝑡𝑎𝑙
𝑅
𝑇1
𝑃2
ln
𝑃1
ℎ𝑓𝑔
−
1
𝑇2
• If given 𝑇1 , 𝑃1 , ℎ𝑓𝑔 𝑎𝑛𝑑 𝑇2 , we can use this to find 𝑃1
• Page 701, Table B: ℎ𝑓𝑔 , 𝑇𝐵𝑜𝑖𝑙 = 𝑇𝑆𝑎𝑡 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)
𝑌𝐴,∞
• Gas B is stationary 𝑚𝐵" = 0
YA
• but has a concentration gradient
YB
• Diffusion of B down = advection up
Y
YA,i
•
𝑚𝐵"
•
𝑌𝐵 𝑚𝐴"
=0=
=
𝑌𝐵 𝑚𝐴"
+
𝑚𝐵"
−
𝑑𝑌𝐵
𝜌𝒟𝐵𝐴
𝑑𝑥
𝑑𝑌𝐵
𝜌𝒟𝐵𝐴
𝑑𝑥
• 𝑚𝐴" = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
•
•
"
𝑚𝐴
𝑑𝑥
𝜌𝒟𝐵𝐴
"
𝑚𝐴
𝑥
𝜌𝒟𝐵𝐴
=
=
"
𝑥
𝑑𝑌𝐵 𝑚𝐴
;
𝑑𝑥
𝑌𝐵 𝜌𝒟𝐵𝐴 0
=
𝑌𝐵 (𝑥)
ln
; 𝑌𝐵 (𝑥)=𝑌𝐵,𝑖 𝑒
𝑌𝐵,𝑖
𝑌𝐵 (𝑥) 𝑑𝑌𝐵
𝑌𝐵,𝑖
𝑌𝐵
𝑚"𝐴
𝑥
𝜌𝒟𝐵𝐴