HEAT TRANSFER OPERATION
MIDTERM NOTE
I.
LAW OF HEAT TRANSFER:
1. Fundamental law:
Define: ALWAYS HAVE TO SATISFY, REGARDLESS OF SITUATION.
(Xài lúc nào cũng đúng)
a. Types of system consider:
_ Closed system:
+ Fix amount of matter (𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
+ No mass flows across the system boundaries (𝑚𝑎𝑠𝑠 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
+ Energy flow may occur across the boundaries (exchange heat, work to surrounding)
+ The boundaries may change with time (𝑉 𝑐ℎ𝑎𝑛𝑔𝑒)
_Open System - Control Volume
+ Volume of fixed size containing matter
+ Mass flows across the system boundaries (𝑚𝑎𝑠𝑠 𝑐ℎ𝑎𝑛𝑔𝑒)
+ Energy flow occur across the boundaries
+ The boundaries may change with time
b. Law of Conservation of Mass:
_For closed system : 𝑚𝑎𝑠𝑠 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (vô TH này thì mass trước = mass sau)
_ Fluid flow through a control volume:
𝑟𝑎𝑡𝑒 𝑚𝑎𝑠𝑠𝑖𝑛 − 𝑟𝑎𝑡𝑒 𝑚𝑎𝑠𝑠𝑜𝑢𝑡 − 𝑚𝑎𝑠𝑠 𝑎𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛𝑠𝑦𝑠𝑡𝑒𝑚 = 0
With:
𝑟𝑎𝑡𝑒 𝑚𝑎𝑠𝑠 =
𝑑𝑚
𝑑𝑉
𝑑𝑥
=𝜌
= 𝜌𝑑𝐴
= 𝜌𝐴𝑣 (1 − 𝐷𝑖𝑚𝑒𝑠𝑖𝑜𝑛, 𝑠𝑚𝑎𝑙𝑙 𝐴)
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑟𝑎𝑡𝑒 𝑚𝑎𝑠𝑠𝑖𝑛 = ∬
𝜌𝑣𝑖𝑛 𝑑𝐴𝑖𝑛 (𝑖𝑛𝑡𝑒𝑟𝑔𝑒𝑟𝑎𝑡𝑒𝑑 𝑎𝑏𝑜𝑣𝑒 𝑓𝑜𝑟 ℎ𝑜𝑙𝑑 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝑟𝑎𝑡𝑒 𝑚𝑎𝑠𝑠𝑖𝑛 = ∬𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝜌𝑣𝑜𝑢𝑡 𝑑𝐴𝑜𝑢𝑡
𝑚𝑎𝑠𝑠 𝑎𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛𝑠𝑦𝑠𝑡𝑒𝑚 =
𝜕
∭
𝜌𝑑𝑉
𝜕𝑡 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
Equation (1)
∬
𝜌𝑣𝑖𝑛 𝑑𝐴𝑖𝑛 − ∬
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝜌𝑣𝑜𝑢𝑡 𝑑𝐴𝑜𝑢𝑡 =
𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝜕
{∭
𝜌𝑑𝑉 }
𝜕𝑡
𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
(Description : Rate of mass tại 1 điểm là 𝜌𝐴𝑣 vói v theo phương x, do xét trong hệ vật
diện tich cố định, Rate of mass trên cả mặt của vật thì lấy nguyên hàm 2 phương y, z.
Tương tự với mass accumulation nếu V đổi theo 3 phương thì ta nguyên hàm theo 3
phương đêr có đụowc tổng thay đổi của V)
Types of fluids
Steady state
(No accumulation)
Compressible fluid
Non-steady state
∬ 𝜌𝑣𝑖𝑛 𝑑𝐴𝑖𝑛 = ∬ 𝜌𝑣𝑜𝑢𝑡 𝑑𝐴𝑜𝑢𝑡
𝐶.𝑉
Incompressible fluid
(𝜌 = 𝑐𝑜𝑛𝑠𝑡)
𝐶.𝑉
∬ 𝑣𝑖𝑛 𝑑𝐴𝑖𝑛 = ∬ 𝑣𝑜𝑢𝑡 𝑑𝐴𝑜𝑢𝑡
𝐶.𝑉
Equation (1)
Equation (1)
𝐶.𝑉
_In 1 dimensional : Remove term rate mass out of the integrate (Do A không đổi theo
phương y,z; A tại điểm vào và ra của mass là hằng số)
Types of fluids
Steady state
(No accumulation)
𝜌𝑖𝑛 𝑣𝑖𝑛 𝐴𝑖𝑛 = 𝜌𝑜𝑢𝑡 𝑣𝑜𝑢𝑡 𝐴𝑜𝑢𝑡
Compressible fluid
𝑣𝑖𝑛 𝐴𝑖𝑛 = 𝑣𝑜𝑢𝑡 𝐴𝑜𝑢𝑡
Incompressible fluid
(𝜌 = 𝑐𝑜𝑛𝑠𝑡)
Non-steady state
𝜌𝑖𝑛 𝑣𝑖𝑛 𝐴𝑖𝑛 − 𝜌𝑜𝑢𝑡 𝑣𝑜𝑢𝑡 𝐴𝑜𝑢𝑡 = 𝑎𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒
𝑣𝑖𝑛 𝐴𝑖𝑛 − 𝑣𝑜𝑢𝑡 𝐴𝑜𝑢𝑡 = 𝑎𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒
c. Newton’s Second law of motion:
_For close system: ∑ 𝐹⃗ =
𝑑
𝑑𝑡
(𝑚𝑣⃗) (𝑓𝑜𝑙𝑜𝑤 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑦𝑜𝑢 𝑐ℎ𝑜𝑠𝑒)
_Fluid flow through a control volume:
𝑆𝑢𝑚 𝑓𝑜𝑟𝑐𝑒 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑢𝑡 − 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡 𝑖𝑛 + 𝐴𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒
With:
𝑟𝑎𝑡𝑒 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 =
𝑑𝐹
𝑑𝑚
𝑑𝑥
= 𝑣⃗
= 𝑣⃗𝜌𝑑𝐴
= 𝑣⃗𝜌𝐴𝑣𝑛 (1 − 𝐷𝑖𝑚𝑒𝑠𝑖𝑜𝑛, 𝑠𝑚𝑎𝑙𝑙 𝐴)
𝑑𝑡
𝑑𝑡
𝑑𝑡
n
n
𝑣⃗: 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑣𝑒𝑐𝑡𝑜𝑟 (𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑢𝑠𝑒 𝑐ℎ𝑜𝑠𝑒)
𝑛⃗⃗ ∶ 𝑉𝑒𝑐𝑡𝑜𝑟 𝑛𝑜𝑟𝑚𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
v in
𝑣𝑛 = |𝑣||𝑛|𝑐𝑜𝑠𝜃
_Fluid flow through in 1 direction:
∑ 𝐹𝑥 = − ∬ 𝑣𝑥 𝜌𝑣𝑛 𝑑𝐴𝑖𝑛 + ∬ 𝑣𝑥 𝜌𝑣𝑛 𝑑𝐴𝑜𝑢𝑡 +
𝐶𝑆
𝐶𝑆
𝜕
{∭ 𝜌𝑣𝑥 𝑑𝑉 }
𝜕𝑡
𝐶𝑉
(𝑣𝑛 là tích vector có phương của vector vận tốc, và vector vuông góc với mặt phẳng vật,
trong slide cth chung là 𝑣. 𝑛 thì tích có hướng này sẽ âm nếu v và n ngược chiều lúc đi
vào)
d. 1st Law of Thermodynamics:
_For closed system:
𝛿𝑄
𝛿𝑡
−
𝛿𝑊
𝛿𝑡
=
𝛿𝐸
𝛿𝑡
Usually no work involve: Heat change equal energy change:
𝛿𝑄 𝛿𝐸
=
𝛿𝑡
𝛿𝑡
2. Subsidiary law:
a. Fourier’s law of heat conduction:
𝑞
𝜕𝑇
= −𝐾
𝐴
𝜕𝑛
𝑞
𝑊
𝑞: ℎ𝑒𝑎𝑡 𝑟𝑎𝑡𝑒 (𝑊 𝑜𝑟 𝐽/𝑠), : 𝐻𝑒𝑎𝑡 𝑓𝑙𝑢𝑥 (𝑊/𝑚2 ), 𝐾: 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑣𝑖𝑡𝑦 (
)
𝐴
𝑚. 𝐾
𝜕𝑇
: 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑖𝑛 𝑐𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒
𝜕𝑛
_If K constant: isotropic material
_One dimension, steady state situation:
+Infinite slab:
𝑞
𝑇𝐻 − 𝑇𝐶
=𝐾
𝐴
𝑏
𝑇𝐻 /𝑇𝐶 ∶ 𝑡𝑒𝑚𝑝 𝑎𝑡 ℎ𝑜𝑡/𝑐𝑜𝑙𝑑 𝑠𝑖𝑑𝑒, 𝑏: 𝑡ℎ𝑖𝑐𝑘 𝑛𝑒𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙/𝑠𝑙𝑎𝑏 (𝑚)
v out
𝑇 𝑎𝑡 𝑐𝑒𝑟𝑡𝑎𝑖𝑛 𝑝𝑜𝑖𝑛𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑥 𝑓𝑜𝑟𝑚 ℎ𝑜𝑡 𝑓𝑎𝑐𝑒:
𝑇𝐻 − 𝑇
𝑥
=
𝑇𝐻 − 𝑇𝐶 𝑏
+Infinite long hollow cylinder
𝑟𝑜 (𝑟𝑖 ): 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑢𝑡𝑠𝑖𝑑𝑒(𝑖𝑛𝑠𝑖𝑑𝑒), 𝐿: 𝑙𝑒𝑛𝑔ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟
𝑇𝑜 (𝑇𝑖 ): 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑢𝑡𝑠𝑖𝑑𝑒(𝑖𝑛𝑠𝑖𝑑𝑒)
𝑟𝑖
𝑇𝑖
𝑞
𝑞
𝜕𝑇
𝑞
𝜕𝑟
𝑞
𝑟𝑖
=
= −𝐾
↔
∫
= −𝐾 ∫ 𝜕𝑇 ↔
𝑙𝑛 ( ) = −𝐾(𝑇𝑖 − 𝑇𝑜 )
𝐴 2𝜋𝑟𝐿
𝜕𝑟
2𝜋𝐿
𝑟
2𝜋𝐿
𝑟𝑜
𝑟𝑜
𝑇𝑜
↔𝑞=−
2𝜋𝐾𝐿(𝑇𝑖 − 𝑇𝑜 )
𝑟
𝑙𝑛 (𝑟𝑖 )
𝑜
𝑇 𝑎𝑡 𝑐𝑒𝑟𝑡𝑎𝑖𝑛 𝑟 ∶
𝑇 − 𝑇𝑖
𝑙𝑛(𝑟/𝑟𝑖 )
=
𝑇𝑖 − 𝑇𝑜 𝑙𝑛(𝑟𝑜 /𝑟𝑖 )
_At not steady state:
Coordinate
Cartesian (slab)
Cylindrical
Sphere
x
𝑥=𝑥
𝜕𝑇 𝜕𝑇
↔
=
𝜕𝑛 𝜕𝑥
𝑥 = 𝑟 𝑐𝑜𝑠𝜃 = 𝑟
↔ 𝜕𝑥 = 𝜕𝑟
𝜕𝑇 𝜕𝑇
↔
=
𝜕𝑥 𝜕𝑟
(radial heat transfer)
y
𝑦=𝑦
𝜕𝑇 𝜕𝑇
↔
=
𝜕𝑛 𝜕𝑦
z
𝑧=𝑧
𝜕𝑇 𝜕𝑇
↔
=
𝜕𝑛 𝜕𝑧
𝑦 = 𝑟 𝑠𝑖𝑛𝜃 = 𝑟𝜃
↔ 𝜕𝑦 = 𝑟𝜕𝜃
𝜕𝑇 1 𝜕𝑇
↔
=
𝜕𝑦 𝑟 𝜕𝜃
𝑧=𝑧
𝜕𝑇 𝜕𝑇
↔
=
𝜕𝑛 𝜕𝑧
𝐴𝑡 𝑠𝑚𝑎𝑙𝑙 𝜃: 𝑠𝑖𝑛𝜃 ≈ 𝜃, 𝑐𝑜𝑠𝜃 ≈ 1
𝑦 = 𝑟 𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛𝜑 = 𝑟𝜃𝜑
𝑥 = 𝑟 𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠𝜑 = 𝑟𝜃
↔ 𝜕𝑦 = 𝑟𝜃 𝜕𝜑
↔ 𝜕𝑥 = 𝑟𝜕𝜃
𝜕𝑇
1 𝜕𝑇
𝜕𝑇 1 𝜕𝑇
↔
=
↔
=
𝜕𝑦 𝑟𝜃 𝜕𝜑
𝜕𝑥 𝑟 𝜕𝜃
b. Newton’s law of cooling
:
𝑞
= ℎ(𝑇𝑊 − 𝑇𝑓 )
𝐴
𝑧 = 𝑟 𝑐𝑜𝑠𝜃 = 𝑟
𝜕𝑇 𝜕𝑇
↔
=
𝜕𝑧 𝜕𝑟
𝑇𝑊 : 𝐻𝑒𝑎𝑡 𝑜𝑓 𝑜𝑏𝑗𝑒𝑐𝑡, 𝑇𝑓 : ℎ𝑒𝑎𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑, ℎ: 𝐻𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 (
𝑊
)
𝑚2 𝐾
c. Stefan-Boltzmann Law:
𝑞
= 𝜎𝑇 4
𝐴
𝑊
𝐵𝑡𝑢
𝜎 = 5.67 × 10−8 ( 2 4 ) = 1.7134 (
) : 𝑆𝑡𝑒𝑓𝑎𝑛 − 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑚 𝐾
ℎ. 𝑓𝑡 2 ℉4
d. Thermal resistance: (nhiệt trở, na ná điện trở)
Infinite slab
Thermal resistance
(𝑍𝑡ℎ )
Infinite hollow
Cylinder
𝑙𝑛(𝑟𝑜 /𝑟𝑖 )
2𝜋𝑅𝐾𝐿
𝑏
𝐾𝐴
Surface of liquid
1
ℎ𝐴
If the wall have only 1 way to transfer heat (many layers wall): similar to resistance put
in series:
𝑍𝑡ℎ,𝑡𝑜𝑡𝑎𝑙 = ∑ 𝑍𝑡ℎ
If the wall have more than 1 way to transfer heat similar to resistance put in parallel:
(Ex: The wall have 1 nail stick through its, heat can transfer in two paths: Through nail or
through wall)
1
𝑍𝑡ℎ,𝑡𝑜𝑡𝑎𝑙
=∑
1
𝑍𝑡ℎ,𝑝𝑎𝑡ℎ 𝑖
(1 𝑝𝑎𝑡ℎ 𝑚𝑎𝑦 ℎ𝑎𝑣𝑒 𝑚𝑎𝑛𝑦 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑠𝑒𝑟𝑖𝑒𝑠)
We can calculate the resistance of the path to find temperature of cold side knowing hot
side and heat rate:
𝑞=
∆𝑇
𝑍𝑡ℎ,𝑡𝑜𝑡𝑎𝑙
Ex: Calculate the temp at intersection of 2, 3 layers in 3 layers wall
𝑍𝑡ℎ,𝑡𝑜𝑡𝑎𝑙 = 𝑍1 + 𝑍2
Z1
Z2
Z3
Or if do not have A we can call calculate: Overall heat transfer coefficient
𝑈 = 𝐴𝑍𝑡ℎ,𝑡𝑜𝑡𝑎𝑙
II.
The Differential Equation of Heat Conduction:
There are several step to derive equation of heat conduction as problem give:
Step 1: Chose the shape:
_Slab with 3D: (x,y,z)
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕𝑇
(𝐾. ) +
(𝐾. ) + (𝐾. ) + 𝑞̅ = 𝜌𝐶𝑝
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕𝑡
With: 𝐶𝑝 (𝐽/𝐾𝑔. 𝐾): 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦
𝑊
𝑞̅ ( 3 ) : ℎ𝑒𝑎𝑡 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝑚
_ 3D cylindrical coordinate system: (𝑟, 𝜃, 𝑧)
1𝜕
𝜕𝑇
1 𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕𝑇
(𝐾. 𝑟 ) + 2
(𝐾. ) + (𝐾. ) + 𝑞̅ = 𝜌𝐶𝑝
𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝜕𝜃
𝜕𝜃
𝜕𝑧
𝜕𝑧
𝜕𝑡
_3D spherical coordinate system (𝑟, 𝜃, 𝜑)
1𝜕
𝜕(𝑟𝑇)
1
𝜕
𝜕𝑇
1
𝜕
𝜕𝑇
𝜕𝑇
(𝐾.
)+ 2
(𝐾. 𝑠𝑖𝑛𝜑 ) + 2 2
(𝐾. ) + 𝑞̅ = 𝜌𝐶𝑝
𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝑠𝑖𝑛𝜑 𝜕𝜑
𝜕𝜑
𝑟 𝑠𝑖𝑛 𝜑 𝜕𝜃
𝜕𝜃
𝜕𝑡
Step 2: Chose the condition: (more than 1 usually applied)
Condition
K Constant
Absence of
heat generation
Meaning
Ex: in 3D slab
𝜕
𝜕𝑇
𝜕
𝜕𝑇
𝜕
𝜕𝑇
(𝐾. ) +
(𝐾. ) + (𝐾. )
𝜕𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑦
𝜕𝑧
𝜕𝑧
𝜕 𝜕𝑇
𝜕 𝜕𝑇
𝜕 𝜕𝑇
= 𝐾{ ( )+
( ) + ( )}
𝜕𝑥 𝜕𝑥
𝜕𝑦 𝜕𝑦
𝜕𝑧 𝜕𝑧
2
= 𝐾∇ 𝑇
𝑞̅ = 0
Equation change
𝑞̅ 1 𝜕𝑇
∇2 𝑇 + =
𝐾 𝛼 𝜕𝑡
With
𝐾
𝛼=
; 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑡𝑦
𝜌𝐶𝑝
(𝑚2 /𝑠)
Steady state
In 1 direction
𝜕𝑇
=0
𝜕𝑡
Ex: 3D cylindrical: Assume temperature
change only on r coordinate (radial heat
transfer)
𝜕𝑇 𝜕𝑇
=
=0
𝜕𝜃 𝜕𝑧
With (𝑢𝑣)′ = 𝑢′𝑣 + 𝑢𝑣′
In (*)
𝜕
: 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑎𝑙 𝑤𝑖𝑡ℎ 𝑟,
𝜕𝑟
Step 3: find 𝑻 𝒂𝒏𝒅
Into equation in I.2.b
(𝐾∇2 𝑇 = 0)
1𝜕
𝜕𝑇
𝑞̅ 1 𝜕𝑇
(𝑟 ) + =
𝑟 𝜕𝑟 𝜕𝑟
𝐾 𝛼 𝜕𝑡
1 𝜕𝑇
𝜕 2𝑇
↔ ( + 𝑟 2 ) + ⋯ (∗)
𝑟 𝜕𝑟
𝜕𝑟
𝜕𝑇
𝑎𝑛𝑑 𝑟 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑟
𝜕𝑟
𝝏𝑻
𝝏𝒙
𝒕𝒐 𝒂𝒑𝒑𝒍𝒊𝒆𝒅 𝒊𝒏 𝒃𝒐𝒖𝒅𝒂𝒓𝒚 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 ∶
Ex: for 1D slab (condition 4), steady state (condition 2) and K constant (condition 1)
General equation become:
𝜕 2 𝑇 𝑞̅
𝜕 2𝑇
𝑞̅
+
=
0
↔
=
−
𝜕𝑥 2 𝐾
𝜕𝑥 2
𝐾
→
𝜕𝑇
𝑞̅
= − 𝑥 + 𝐶1
𝜕𝑥
𝐾
→ 𝑇(𝑥) = −
𝑞̅ 𝑥 2
+ 𝐶1 𝑥 + 𝐶2
𝐾 2
Ex: for 1D hollow cylinder (condition 4), steady state (condition 2) and K constant
(condition 1)
1𝜕
𝜕𝑇
𝑞̅
1 𝜕𝑇
𝜕 2𝑇
𝑞̅
(𝐾. 𝑟 ) + = 0 ↔ ( + 𝑟 2 ) + = 0
𝑟 𝜕𝑟
𝜕𝑟
𝐾
𝑟 𝜕𝑟
𝜕𝑟
𝐾
1 𝜕𝑇 𝜕 2 𝑇
𝑞̅
↔
+ 2 =−
𝑟 𝜕𝑟 𝜕𝑟
𝐾
↔
↔𝑟
𝜕𝑇
𝜕 2𝑇
𝑞̅
+ 𝑟 2 = − 𝑟 (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑟 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒)
𝜕𝑟
𝜕𝑟
𝐾
𝜕𝑇
𝑞̅ 𝑟 2
𝜕𝑇
𝑞̅ 𝑟
𝑞̅ 𝑟 2
=−
(𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 2 𝑠𝑖𝑑𝑒) ↔
=−
→ 𝑇(𝑟) = −
+ 𝐶1
𝜕𝑟
𝐾 2
𝜕𝑟
𝐾2
𝐾 4
Step 4: Find boundary condition for solving C1 and C2 (normally maximum 2 is
require):
Boundary condition
Temperature of surface is maintain at 𝑇𝑠𝑢𝑟
(𝑥 = 𝑏, 𝑟 = 𝑟𝑜 )
Knowing heat flux (q/A) at surface
(𝑥 = 𝑏, 𝑟 = 𝑟𝑜 )
Special case for heat Insulation surface
flux
Thermal symmetry
(Heat flow in/out from
2 side to center)
Convection Boundary Condition
(Have fluid flow to cool/heat surface with
temperature 𝑇𝑓 and h)
𝑇(𝑏) = 𝑇𝑠𝑢𝑟
(𝑇𝑠𝑢𝑟 is a number, position of surface
depend on how you put the origin and
coordinate)
𝜕𝑇
𝑞/𝐴 = ( )
𝜕𝑥 𝑥=𝑏
𝜕𝑇
( )
=0
𝜕𝑥 𝑥=𝑏
𝜕𝑇
( )
=0
𝜕𝑥 𝑥=𝑏/2
(b is the thickness of wall, if chose x =0 at
center change the x=b/2 by x=0)
𝜕𝑇
−𝐾 ( )
= −ℎ(𝑇𝑓 − 𝑇𝑥=𝑏 )
𝜕𝑥 𝑥=𝑏
Step 5: Put C1 and C2 to equation.
(TO SEE HOLD SOLVING CAN SEE TAKE HOME PRACTICE)
III.
Unsteady State Heat Conduction:
1. Some dimension for use:
𝐵𝑖𝑜𝑡 𝑛𝑢𝑚𝑏𝑒𝑟: 𝐵𝑖 =
ℎ𝐿
(𝑅𝑎𝑡𝑖𝑜 𝑜𝑓 𝑍𝑡ℎ 𝑜𝑓 𝑠𝑜𝑙𝑖𝑑 𝑎𝑛𝑑 𝑍𝑡ℎ 𝑜𝑓 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝐾
𝐹𝑜𝑢𝑟𝑖𝑒𝑟 𝑁𝑢𝑚𝑏𝑒𝑟: 𝑋 =
𝛼=
𝛼𝑡
𝐾𝑡
=
(𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑙𝑒𝑠𝑠 𝑜𝑓 𝑡𝑖𝑚𝑒)
𝐿2 𝜌𝐶𝑝 𝐿2
𝐾
; 𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑡𝑦 (𝑚2 /𝑠)
𝜌𝐶𝑝
1
𝐵𝑖𝑜𝑡
𝑥
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛: 𝑛 =
𝐿
𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒: 𝑚 =
Shape
Infinite slab
Infinite slab 1 side insulate
L (Characteristic dimension)
Thickness (b) divide by 2 (b/2)
Treat as infinite slab with thickness: 2b
(double thickness of original slab) cal equal
b
Radius (r)
𝑉
𝐿=
𝐴𝑠𝑢𝑟
Infinite cylinder/sphere
Other shape
2. Way to solve:
Step 1: Define the shape: infinite slab, infinite cylinder, sphere, or cubic,…
If not find K, you can look for the material and find in Appendix H.
Step 2: Define the shape in component infinite slab and infinite cylinder:
Shape
Finite cylinder
Box (cubic)
Component
Infinite of cylinder and infinite slab
3 infinite slabs different (same) thickness
Step 3: Calculate Biot Number: (if more than 1 component Cal for each component)
_If 𝐵𝑖 < 0.1: Neglect internal temperature gradient
ℎ𝐴𝑡
𝑇 − 𝑇𝑓
𝜃
−
=
= 𝑒 𝜌𝐶𝑝 𝑉 = 𝑒 −𝐵𝑖×𝑋
𝜃𝑜 𝑇𝑜 − 𝑇𝑓
_ If 𝐵𝑖 > 0.1:
𝜃
Step 4: Define 𝜃 𝒐𝒇 that shape:
𝑜
3
𝜃
𝜃
𝐸𝑥: ( )
= [( )
]
𝜃𝑜 𝑐𝑢𝑏𝑖𝑐
𝜃𝑜 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑒 𝑠𝑙𝑎𝑏
𝜃
Step 5: Calculate Biot, Fourier, m, n for each component and find for 𝜃 in Appendix F:
𝑜
If there are no line near with you’re calculate m:
Ex: m = 0.6, but in chart only have 0.5 and 0.75:
𝜃
𝜃
𝑜
𝑜
You can find 𝜃 for 0.5 and 0.75, assume m and 𝜃 are linear:
𝜃
= 𝑎𝑚 + 𝑏
𝜃𝑜
And use 2 point at 0.5 and 0.75 you are found to calculate a and b
𝜃
Step 6: Found the 𝜃 of the shape and calculate T
𝑜

There are some problem like give you T and make you find t:
𝜃
Step 1: You will calculate 𝜃 of the shape and write it component:
𝑜
Ex:
𝜃
𝜃
𝜃
( )
=( )
( )
𝜃𝑜 𝑓𝑖 𝑐𝑦𝑙
𝜃𝑜 𝑖𝑛𝑓𝑖 𝑐𝑦𝑙 𝜃𝑜 𝑖𝑛𝑓𝑖 𝑠𝑙𝑎𝑏
Step 2: Then you calculate the Fourier number of component through time:
Ex:𝑋𝑖𝑛𝑓𝑖 𝑠𝑙𝑎𝑏 = 𝑎 × 𝑡,
𝑋𝑖𝑛𝑓𝑖 𝑐𝑦𝑙 = 𝑏 × 𝑡
Step 3: Then you will guess :)))))) any relation you want: (CHOSE ONLY 1 COMPONENT)
Ex
𝜃
𝜃
𝜃
𝜃
( )
= √( )
𝑜𝑟 ( )
= 1/2 ( )
𝜃𝑜 𝑖𝑛𝑓𝑖 𝑐𝑦𝑙
𝜃𝑜 𝑓𝑖 𝑐𝑦𝑙
𝜃𝑜 𝑖𝑛𝑓𝑖 𝑐𝑦𝑙
𝜃𝑜 𝑓𝑖 𝑐𝑦𝑙
(𝑌𝑜𝑢 𝑐𝑎𝑛 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 𝑐ℎ𝑜𝑠𝑒 𝑏𝑎𝑦 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑒𝑎𝑐ℎ 𝐿 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑡)
𝜃
Step 4: Then put it in(𝜃 )
𝑜
of the shape to calculate each of component, put in appendix
𝑖𝑛𝑓𝑖 𝑐𝑦𝑙
of find time by each component.
𝜃
𝜃
𝑜
𝑜
Step 5Take that time to find other component𝜃 . Then cal (𝜃 )
𝑠ℎ𝑎𝑝𝑒
𝜃
Step 6: If 𝜃 of the shape not fit, adjust t and try again
𝑜
3. Fin:
a. Some equation:
We have the fin with:
𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝐿 , 𝐴𝑡𝑖𝑝 𝑖𝑠 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑖𝑝 𝑎𝑛𝑑 𝑃 𝑖𝑠 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑖𝑝
𝑀 = √ℎ𝑃𝐾𝐴𝑡𝑖𝑝 ,
𝑓𝑖𝑛 𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑛𝑒𝑠𝑠: 𝜀𝑓 =
𝑚=√
ℎ𝑃
𝐾𝐴𝑡𝑖𝑝
𝑞𝑓
ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑤𝑖𝑡ℎ 𝑓𝑖𝑛
=
𝑞
ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑡ℎ𝑒 𝑓𝑖𝑛
𝐾𝑃
𝜀𝑓 = √
𝑡𝑎𝑛ℎ(𝑚𝐿) (𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐 𝑓𝑖𝑛)
ℎ𝐴𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝑓𝑖𝑛 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦: 𝑛𝑓 =
𝑞𝑟𝑒𝑎𝑙
𝑞𝑟𝑒𝑎𝑙
=
𝑞𝑖𝑑𝑒𝑎𝑙 (ℎ𝑒𝑎𝑡 𝑎𝑡 𝑡𝑖𝑝 𝑒𝑞𝑢𝑎𝑙 𝑎𝑡 𝑏𝑎𝑠𝑒) ℎ𝐴𝑠𝑢𝑟 (𝑇 − 𝑇𝐿 )
b. Calculation:
_For only pin fin:
_For circular and other strainght fin:
(𝑟𝐿 − 𝑟𝑜 )