ME 475/675 Introduction to

CombustionLecture 12

Droplet evaporation, Example 3.2

Announcements

• HW 4 due Now• Midterm 1• September 28, 2015, 8-10 AM, PE 104• Review Friday, September 25

• HW 5 Due Friday, September 25, 2015

Droplet Spherical Evaporation• Liquid A is evaporating: find and • Gas B is “stagnant” ;

•

𝑌 𝐴 , ∞

𝑌 𝐴 , ∞

𝑌 𝐴 , 𝑠

0 10 20 30 40 500

1

2

3

43.932

0

m1 By( )

500 By

YA1 r YAs( ) 11 YAs( )

1 YAs( )1

1

r

Transfer Number, (based on mass fraction )

• ; • • is driving potential for mass transfer • If then and

• If and do not change with time• Then decreases as decreases

�̇�𝐴

4𝜋 𝑟𝑠 𝜌𝒟𝐴𝐵

𝐵𝑌

Droplet Diameter versus time • Mass Conservation:

• , • Evaporation Const.

• Constant slope for versus • Confirmed by experiment • Droplet life

Example 3.2 (page 99) Turn in next time for EC• In mass-diffusion-controlled evaporation of a fuel droplet, the droplet surface

temperature is an important parameter. Estimate the droplet lifetime of a 100-mm-diameter n-dodecane droplet evaporating in dry nitrogen at 1 atm if the droplet temperature is 10 K below the dodecane boiling point. • Repeat the calculation for a temperature 20 K below the boiling point, and

compare the results. • For simplicity, assume that, in both cases, the mean gas density is that of nitrogen

at a mean temperature of 800 K. Use this same temperature to estimate the fuel vapor diffusivity. The density of liquid dodecane is 749 kg/m3.

• Find:___• Given:___

Liquid-Vapor Interface Boundary Condition

• At interface need

• For fuels (A): Page 701, Table B: , at A+B

Vapor

𝑌 𝐴 , 𝑖

LiquidA

If

• Increases rapidly as • Same shape as Stefan

Problem (Cartesian, last lecture)

0 0.2 0.4 0.6 0.80

2

4

6

86.908

0

m Y( )

10 Y

YA x Yi( ) 1 1 Yi( )1

1 Yi

x

𝑌 𝐴 , 𝑠

�̇�𝐴

4𝜋 𝑟𝑠 𝜌𝒟𝐴𝐵

Stefan Problem Mass Flux of evaporating liquid A

• For • (dimensionless)• increases slowly for small • Then very rapidly for > 0.95

0 0.2 0.4 0.6 0.80

2

4

6

86.908

0

m Y( )

10 Y𝑌 𝐴 , 𝑖

�̇�𝐴}} over {{ { } rsub { }} over { }𝜌 𝒟 𝐴𝐵 𝐿 ¿¿

versus r profiles (for )• : • But: • Ratio:

•

• For

2 4 6 8 100

0.2

0.4

0.6

0.8

1

4.652 10 3

YA1 r .05( )

YA1 r .1( )

YA1 r .5( )

YA1 r .9( )

YA1 r .95( )

111 r𝑟𝑟𝑠

𝑌 𝐴 , 𝑠=0.95

𝑌 𝐴 , 𝑠=0.5

Stefan Problem• but

• Ratio: ;

• For

• Large profiles exhibit a boundary layer near exit (large advection near interface)

0 0.2 0.4 0.6 0.80

0.2

0.4

0.6

0.8

10.99

0

YA x .05( )

YA x .1( )

YA x .5( )

YA x .9( )

YA x .99( )

10 x𝑥𝐿

=0.99

=0.9

=0.5

=0.1

=0.05