1. IX
Transport Phenomena

2. Transport Phenomena
Principle of Conservation of Energy
; Molecular Diffusion (mass)
; Heat Conduction (energy)
; Viscosity (momentum)
Molecular Diffusion
Fick’s Law
𝐽 = 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠
𝑚2 𝑠
𝑛 = 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠
𝑚3
𝐷 = 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑚2
𝑠−1
We define ‘positive’ to be on right hand
side. ∴ For the gas to diffuse to the right,
𝐽 𝑥 > 0,
𝜕𝑛
𝜕𝑥
< 0
Consider 𝑑𝑉 = 𝑆𝑑𝑥,
For 𝑁 = 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠,
∴
𝑁 𝑛𝑜. 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠
𝑆 𝑚2 𝑡 𝑠
= 𝐽
Suppose 𝐽 𝑥
′
> 𝐽 𝑥,
∴ 𝑑𝑁 = 𝐽 𝑥 − 𝐽 𝑥
′
𝑆𝑑𝑡
𝐽 𝑥 = −𝐷
𝜕𝑛
𝜕𝑥
𝐽 𝑥 𝐽 𝑥
′
𝑆
𝑑𝑥
Since 𝑑𝐽 𝑥 = 𝐽 𝑥
′
− 𝐽 𝑥,
𝑑𝑁 = 𝐽 𝑥 − 𝐽 𝑥
′
𝑆𝑑𝑡 = − 𝑑𝐽 𝑥 𝑆𝑑𝑡
𝑑𝑁
𝑑𝑡
= − 𝑑𝐽 𝑥 𝑆
𝑑𝑁
𝑑𝑡
= −
𝑑𝐽 𝑥
𝑑𝑥
𝑆𝑑𝑥 = −
𝑑𝐽 𝑥
𝑑𝑥
𝑑𝑉
Since
𝑁
𝑉
= 𝑛, 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦,
𝑑𝑛
𝑑𝑡
= −
𝑑𝐽 𝑥
𝑑𝑥
From Fick’s Law, 𝐽 𝑥 = −𝐷
𝜕𝑛
𝜕𝑥
𝜕𝑛
𝜕𝑡
= −
𝜕𝐽 𝑥
𝜕𝑥
= −
𝜕 −𝐷
𝜕𝑛
𝜕𝑥
𝜕𝑥
𝜕𝑛
𝜕𝑡
= 𝐷
𝜕2
𝑛
𝜕𝑥2
Diffusion Equation
In reality, the diffusion of gas is uneven.
However, we will learn only ‘stationary
diffusion’.
Stationary diffusion
𝜕𝑛
𝜕𝑡
= 0
Meaning that
𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 with respect to time.
𝜕𝑛
𝜕𝑡
= −
𝜕𝐽 𝑥
𝜕𝑥
𝜕𝑛
𝜕𝑡
= 𝐷
𝜕2 𝑛
𝜕𝑥2

3. Since
𝑑𝑛
𝑑𝑡
= −
𝑑𝐽 𝑥
𝑑𝑥
and
𝜕𝑛
𝜕𝑡
= 0,
𝑑𝐽 𝑥
𝑑𝑥
= 0
Meaning that
𝐽 𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥
or position on x − axis.
Consider Fick’s law 𝐽 𝑥 = −𝐷
𝜕𝑛
𝜕𝑥
,
For
𝑑𝐽 𝑥
𝑑𝑥
= 0,
𝐽 𝑥 = −𝐷
𝜕𝑛
𝜕𝑥
𝜕𝑛 = −
𝐽 𝑥
𝐷
𝜕𝑥
𝜕𝑛
𝑥
𝑥0
= −
𝐽 𝑥
𝐷
𝜕𝑥
𝑥
𝑥0
Since 𝐽 𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥,
𝑛 − 𝑛0 = −
𝐽 𝑥
𝐷
𝑥 − 𝑥0
Usually we take 𝑥0 = 0,
𝑛 = −
𝐽 𝑥
𝐷
𝑥 + 𝑛0
Stationary diffusion
𝜕𝑛
𝜕𝑡
= 0
𝑑𝐽 𝑥
𝑑𝑥
= 0
𝑛 = −
𝐽 𝑥
𝐷
𝑥 + 𝑛0
Heat Transfer
1). Heat Conduction
2). Heat Convection
3). Thermal Radiation
Heat Conduction
; occur when there is a difference in
temperature, from high to low
temperature.
Fourier’s Law
𝐽 𝐸 = 𝐸𝑛𝑒𝑟𝑦 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝐽
𝑚2 𝑠
𝐾 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦
𝐽
𝑚𝑠𝐾
𝐾 > 0
Consider 𝑚 = 𝜌𝑆𝑑𝑥,
Suppose 𝑑𝐽 𝐸 = 𝐽 𝐸
′
− 𝐽 𝐸,
𝑄 𝑟𝑒𝑚𝑎𝑖𝑛𝑒𝑑 = 𝑚𝐶 𝑑𝑇 = − 𝑑𝐽 𝐸 𝑆𝑑𝑡
𝜌𝑆𝑑𝑥 𝐶
𝑑𝑇
𝑑𝑡
= − 𝑑𝐽 𝐸 𝑆
𝜌𝐶
𝜕𝑇
𝜕𝑡
= −
𝜕𝐽 𝐸
𝜕𝑥
𝐽 𝐸 = −𝐾
𝜕𝑇
𝜕𝑥
𝐽 𝐸 𝐽 𝐸
′
𝑆
𝑑𝑥
𝜌𝐶
𝜕𝑇
𝜕𝑡
= −
𝜕𝐽 𝐸
𝜕𝑥

4. Since 𝜌𝐶
𝜕𝑇
𝜕𝑡
= −
𝜕𝐽 𝐸
𝜕𝑥
,
And 𝐽 𝐸 = −𝐾
𝜕𝑇
𝜕𝑥
,
𝜌𝐶
𝜕𝑇
𝜕𝑡
= −
𝜕 −𝐾
𝜕𝑇
𝜕𝑥
𝜕𝑥
Stationary heat conduction
Meaning that
𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 with respect to time.
Meaning that
𝐽 𝐸 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥
or position on x − axis.
In the same way as stationary diffusion.
Stationary heat conduction
𝜕𝑇
𝜕𝑡
= 0
𝑑𝐽 𝐸
𝑑𝑥
= 0
𝑇 = −
𝐽 𝐸
𝐾
𝑥 + 𝑇0
𝜕𝐽 𝐸
𝜕𝑥
= 0
𝜌𝐶
𝜕𝑇
𝜕𝑡
= 𝐾
𝜕2
𝑇
𝜕𝑥2
𝜕𝑇
𝜕𝑡
= 0
𝑇 = −
𝐽 𝐸
𝐾
𝑥 + 𝑇0
Viscosity
; from 𝑝ℎ𝑖𝑔ℎ → 𝑝𝑙𝑜𝑤
𝐹𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 = −
𝑑𝑝
𝑑𝑡
Law of viscous flow
𝐽 𝑝 = 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑘𝑔
𝑚𝑠2
𝜂 = 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦
𝑁𝑠
𝑚2
Consider 𝑑𝑉 = 𝑆𝑑𝑥,
𝑑𝐽 𝑝 = 𝐽 𝑝
′
− 𝐽 𝑝
𝐽 𝑝 = −𝜂
𝜕𝑣 𝑦
𝜕𝑥
𝐽 𝑝 𝐽 𝑝
′
𝑆
𝑑𝑥

5. The change in momentum,
𝑁𝑚 𝑑𝑣 𝑦 = −𝑑𝐽 𝑝 𝑆𝑑𝑡
𝑁 = 𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑁𝑚
𝑑𝑣 𝑦
𝑑𝑡
= −
𝑑𝐽 𝑝
𝑑𝑥
𝑆𝑑𝑥
𝑁𝑚
𝑉
𝑑𝑣 𝑦
𝑑𝑡
= −
𝑑𝐽 𝑝
𝑑𝑥
𝜌
𝜕𝑣 𝑦
𝜕𝑡
= −
𝜕𝐽 𝑝
𝜕𝑥
If there is an external force, such as the
force along the banks of the river.
𝜏 = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠
Since 𝐽 𝑝 = −𝜂
𝜕𝑣 𝑦
𝜕𝑥
,
𝜌
𝜕𝑣 𝑦
𝜕𝑡
=
𝜕
𝜕𝑥
𝜏 − 𝐽 𝑝 =
𝜕
𝜕𝑥
𝜏 + 𝜂
𝜕𝑣 𝑦
𝜕𝑥
𝜕𝑣 𝑦
𝜕𝑡
=
𝜂
𝜌
𝜕2
𝑣 𝑦
𝜕𝑥2
+
1
𝜌
𝜕𝜏
𝜕𝑥
(Equation of motion of viscous flow)
𝜌
𝑑𝑣 𝑦
𝑑𝑡
= −
𝑑𝐽 𝑝
𝑑𝑥
𝜌
𝜕𝑣 𝑦
𝜕𝑡
=
𝜕
𝜕𝑥
𝜏 − 𝐽 𝑝
𝜕𝑣 𝑦
𝜕𝑡
=
𝜂
𝜌
𝜕2
𝑣 𝑦
𝜕𝑥2
+
1
𝜌
𝜕𝜏
𝜕𝑥
Stationary flow
Since
𝜕𝑣 𝑦
𝜕𝑡
= 0 𝑜𝑟 𝑎 𝑦 = 0,
From the 1st Law of Newton.
𝜏 = −𝜂
𝜕𝑣 𝑦
𝜕𝑥
Ex; A fluid flows inside a cylinder of
𝑟𝑎𝑑𝑖𝑢𝑠 = 𝑟, 𝑙𝑒𝑛𝑔𝑡ℎ = 𝑙,
𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑝,
𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 = 𝜂, 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝜌
Given that
𝜕𝑣 𝑦
𝜕𝑡
= 0,
𝜕𝑣 𝑦
𝜕𝑡
= 0
𝜕𝐽 𝑝
𝜕𝑥
= 0
𝜕2
𝑣 𝑦
𝜕𝑥2
= −
1
𝜂
𝜕𝜏
𝜕𝑥
𝐽 𝑝 = 𝜏
𝑣 𝑦
𝑎
𝑙

6. Since
𝜕𝑣 𝑦
𝜕𝑡
= 0, 𝑎 𝑦 = 0
From the 1st Law of Newton 𝐹 = 0.
∴ 𝐽 𝑝 = 𝜏
𝜏 = −𝜂
𝜕𝑣 𝑦
𝜕𝑟
Since 𝑝 =
𝐹
𝐴
𝐹 = 𝑝𝐴 = 𝑝 𝜋𝑟2
𝜏 =
𝐹
𝐴
=
𝑝 𝜋𝑟2
2𝜋𝑟𝑙
=
𝑝𝑟
2𝑙
Since 𝜏 = −𝜂
𝜕𝑣 𝑦
𝜕𝑟
,
𝜕𝑣 𝑦 = −
𝜏
𝜂
𝑑𝑟 = −
𝑝
2𝜂𝑙
𝑟𝑑𝑟
𝜕𝑣 𝑦 = −
𝑝
2𝜂𝑙
𝑟𝑑𝑟
𝜕𝑣 𝑦
𝑣 𝑦
0
= −
𝑝
2𝜂𝑙
𝑟𝑑𝑟
𝑟
𝑎
𝑣 𝑦 =
𝑝
4𝜂𝑙
𝑎2
− 𝑟2
𝑣 𝑦
𝑎
𝑙
𝑟
Mass flow rate 𝑄
𝑑𝑄
𝑘𝑔
𝑠
= 𝜌 ∙ 𝑣 𝑦 𝑑𝑠
𝑘𝑔
𝑚3
𝑚
𝑠
𝑚
∴ 𝑑𝑄 = 𝜌 ∙ 𝑣 𝑦 𝑑𝑠
Since 𝑣 𝑦 =
𝑝
4𝜂𝑙
𝑎2
− 𝑟2
,
𝑑𝑄 = 𝜌 ∙
𝑝
4𝜂𝑙
𝑎2
− 𝑟2
𝑑𝑠
𝑑𝑄
𝑄
0
= 𝜌 ∙
𝑝
4𝜂𝑙
𝑎2
− 𝑟2
𝑑𝑠
𝑎
0
𝑄 =
𝜌𝜋𝑝𝑎4
8𝜂𝑙
Poiselle’s law
Mean free path, 𝒍
𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
𝑛 =
𝑁
𝑉
=
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
𝑄 =
𝜌𝜋𝑝𝑎4
8𝜂𝑙
𝑙 =
1
4 2𝜋𝑟2 𝑛

7. Molecular diffusion
𝑣 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑙 = 𝑚𝑒𝑎𝑛 𝑓𝑟𝑒𝑒 𝑝𝑎𝑡ℎ
𝑟 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
𝑛 =
𝑁
𝑉
=
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
Since 𝑣 =
8
𝜋
𝑘𝑇
𝑚
,
𝑣 ∝ 𝑇
And since 𝑙 =
1
4 2𝜋𝑟2 𝑛
,
𝑙 ∝
1
𝑛
Since 𝑃𝑉 = 𝑁𝑘𝑇
𝑛 =
𝑁
𝑉
=
𝑃
𝑘𝑇
From 𝑙 ∝
1
𝑛
and 𝑛 =
𝑃
𝑘𝑇
,
𝑙 ∝ 𝑇
𝐽 𝑥 = −𝐷
𝜕𝑛
𝜕𝑥
𝐷 =
1
3
𝑣 𝑙
𝑣 =
8
𝜋
𝑘𝑇
𝑚
𝑙 =
1
4 2𝜋𝑟2 𝑛
Since 𝑣 ∝ 𝑇 and 𝑙 ∝ 𝑇
When 𝑃 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Heat conduction
𝑣 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑙 = 𝑚𝑒𝑎𝑛 𝑓𝑟𝑒𝑒 𝑝𝑎𝑡ℎ
𝑛 =
𝑁
𝑉
=
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
𝑘 = 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛′
𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Viscosity
𝑛 =
𝑁
𝑉
=
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
𝑚 =
𝑀
𝑁
=
𝑚𝑎𝑠𝑠 𝑜𝑓 1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
𝐷 ∝ 𝑇3/2
𝐽 𝐸 = −𝐾
𝜕𝑇
𝜕𝑥
𝐾 =
1
2
𝑣 𝑛𝑘𝑙
𝐽 𝑝 = −𝜂
𝜕𝑣 𝑦
𝜕𝑥
𝜂 =
1
3
𝑣 𝑛𝑚𝑙 =
1
3
𝑣 𝜌𝑙