Summary and tips Molecular Transport Phenomena for exam preparation

FORMULA SHEET MTP to sell 🤑 by Momi and Pablo-san✨

Important Things to remember

For a CISTR species: Volume not constant; no rxn
𝑑𝐶! 𝑉
= 𝐶!,!# 𝜑!# − 𝐶!,$%& 𝜑$%&
𝑑𝑡
𝑑𝐶! 𝑑𝑉
𝑉 + 𝐶! = 𝐶!,!# 𝜑!# − 𝐶!,$%& 𝜑$%&
𝑑𝑡 𝑑𝑡
Setting up Shell Balance:

1) Define Control Volume: 𝐴𝜕𝑥
'( '(
2) Conduction: -−𝜆 ') )
0 − -−𝜆 0
') )*+)
𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝐿𝑎𝑤
'
a. '&
:𝑝𝐶, 𝐴𝑇= = 𝑖𝑛 − 𝑜𝑢𝑡
3) Convection: [𝑣𝐴𝐶! ]) 𝑎𝑛𝑑 [𝑣𝐴𝐶! ])*+)
'-! '-!
4) Diffusion : -−𝐷𝐴 ') )
0 𝑎𝑛𝑑 -−𝐷𝐴 0
') )*+)
5) Set up the balance:
'(-! /')) '-! '-!
Example: '&
= -−𝐷𝐴 ')
0 − -−𝐷𝐴 0
') )*+)
+ [𝑣𝐴𝑐! ]) − [𝑣𝐴𝑐! ])*+)
)
6) Divide by 𝐴 and by 𝜕𝑥 :
𝜕(𝐶! ) 𝜕 𝜕𝐶! 𝜕𝐶! 𝜕
= FG−𝐷 H − G−𝐷 H I+ ([𝑣𝑐! ]) − [𝑣𝑐! ])*+) )
𝜕𝑡 𝜕𝑥 𝜕𝑥 ) 𝜕𝑥 )*+) 𝜕𝑥
7) Take limit 𝜕𝑥 → 0
𝜕(𝐶! ) 𝜕 𝜕𝐶! 𝜕𝑣𝑐!
= G𝐷 H−
𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑥
𝜕(𝐶! ) 𝜕 1 𝐶! 𝜕𝑐!
=𝐷 1
−𝑣
𝜕𝑡 𝜕𝑥 𝜕𝑥
8) Boundary conditions to consider: 𝐶! (𝑥 = 0, 𝑡) = 0 𝑎𝑛𝑑 𝐶! (𝑥 = 𝐿, 𝑡) = 𝐶!,2
+
Remember also: If Newtonian flow contributions: 𝜏)3 = −𝜂 +) 𝑣3 for instance O𝜏)3 P) 𝑑𝑦 −
O𝜏)3 P)*+) 𝑑𝑦

Volume of a Cylinder: 𝑉 = 𝜋𝑟 1 𝐿
4
Volume of a Sphere:𝑉 = 5 𝜋𝑟 5

,Chapter 3
o Explain how a gas is modeled based on the motion of individual gas molecules
o List and motivate the assumptions of the kinetic theory of gases.
o Derive macroscopic properties, such as temperature and pressure, from the kinetic theory of gases.
o Derive expressions for the mean free path length and the speed of gas molecules using a microscopic description.
o Use the Maxwellian distribution of speeds to calculate the mean speed, the most probable speed, and the root
mean square speed

6 #9 9 4
𝜌=7 =7 =7 :'
(Assuming 1 mol) and 𝑉6$; = 𝑉<&$6 𝑁/ with 𝑉<&$6 = 5
𝜋 𝑟5
"#$ "#$ %&#"


Continuum approximation (N=10^4 molecules per unit volume) ->𝜆6= ≪ 𝐿 (Mean free path
is much smaller than the land scale)

Kinetic gas theory -> 𝜆6= ≫ 𝑑 (Mean free path is much larger than the diameter of the
molecule)
> A:'
𝜆6= = 𝑁 =
√1@+ ( : B(

Derivation of velocity from the Gaussian distribution with a width sigma
"*(
6 DE F C) (
𝑓(𝑣) = X1@C 𝑒 (+) , 𝜎=X
)( 6


Speed is the magnitude of the velocity (3 components)

𝑐 1 = 𝑣)1 + 𝑣31 + 𝑣G1

Considering a spherical shell volume with a fraction of molecules with speed c between c
and c+dc
H/ *+H/ H. *+H. H- *+H- I*+I
Z Z Z 𝑓:𝑣) , 𝑣3, 𝑣G =𝑑𝑣) 𝑑𝑣3 𝑑𝑣G = Z 𝑓:𝑣) , 𝑣3, 𝑣G =4𝜋𝑐 1 𝑑𝑐
H/ H. H- I

5 6H-( 6H.( 6H/( 5 6MH-( *H.( *H/( N
𝑚 1 DE F KE1C) ( L DE F 𝑚 1
KE
1C) (
L
1C) ( 1C) (
𝑓:𝑣) , 𝑣3, 𝑣G = = F I 𝑒 𝑒 𝑒 =F I 𝑒
2𝜋𝑘J 𝑇 2𝜋𝑘J 𝑇
By combining the three formulas: Maxwell-Boltzmann speed distribution function
0 "1(
6 ( 1 DE F
𝑓(𝑐) = 4𝜋 _1@C ` 𝑐 𝑒 (+) ,
)(

, Velocity and speed are normalized functions

O ( O (
𝑚 DE
6H
F 𝑚 DE
6H
F 𝑚 2𝜋𝑘J 𝑇
Z 𝑓(𝑣)𝑑𝑣 = a 𝑒 1C ) ( =a Z 𝑒 1C) ( 𝑑𝑣 = a a =1
EO 2𝜋𝑘J 𝑇 2𝜋𝑘J 𝑇 EO 2𝜋𝑘J 𝑇 𝑚

O
( 𝜋
Z 𝑒 E<) 𝑑𝑥 = X
EO 𝑎

O O 5 O ( 5 (
𝑚 1 DE
6I
F 𝑚 1 DE
6I
F
Z 𝑓(𝑐)𝑑𝑐 = Z 4𝜋 F I 𝑐 1 𝑒 1C) ( 𝑑𝑐 = 4𝜋 F I Z 𝑐 1 𝑒 1C) ( 𝑑𝑐
EO EO 2𝜋𝑘J 𝑇 2𝜋𝑘J 𝑇 EO
5 5
𝑚 1 √𝜋 2𝑘J 𝑇 1
= 4𝜋 F I F I =1
2𝜋𝑘J 𝑇 4 𝑚
O
( √𝜋
Z 𝑥 1 𝑒 E<) 𝑑𝑥 = 5
EO 4 𝑎1
Derivation of speeds

Mean speed
O O ( #!
𝑐 = ∫EO𝑐 𝑓(𝑐)𝑑𝑐 with ∫EO 𝑥 1#*> 𝑒 E<) 𝑑𝑥 = 1<234

8𝑘J 𝑇
𝑐=a
𝜋𝑚

Root mean squared speed

O O ( (1#E>)‼ @
𝑐Q6R = X∫EO 𝑐 1 𝑓(𝑐)𝑑𝑐 with ∫EO 𝑥 1# 𝑒 E<) 𝑑𝑥 = 1234 < 2
X< and it is worth noting that
(1#)!
(2𝑛 − 𝑎)‼ =
12 #!


3𝑘J 𝑇
𝑐Q6R = a
𝑚

Most probable speed
5 (
𝑑𝑓(𝑐) 𝑚 1 𝑑 6I
1 DE1C) (F
= 4𝜋 F I h𝑐 𝑒 i
𝑑𝑐 2𝜋𝑘J 𝑇 𝑑𝑐
( ( ( (
𝑑 DE
6I
F DE
6I
F 𝑚 DE1C6I
F DE
6I
F 𝑚𝑐 1
h𝑐 1 𝑒 1C) ( i = 2𝑐 𝑒 1C) ( − 𝑐 5 𝑒 ) ( = 𝑐𝑒 1C) ( h2 − i
𝑑𝑐 𝑘J 𝑇 𝑘J 𝑇