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Heat midterm note

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HEAT TRANSFER OPERATION
MIDTERM NOTE
I.
LAW OF HEAT TRANSFER:
1. Fundamental law:
Define: ALWAYS HAVE TO SATISFY, REGARDLESS OF SITUATION.
(Xài lúc nào cΕ©ng Δ‘úng)
a. Types of system consider:
_ Closed system:
+ Fix amount of matter (π‘‘π‘œπ‘‘π‘Žπ‘™ π‘šπ‘œπ‘™π‘’π‘  = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘)
+ No mass flows across the system boundaries (π‘šπ‘Žπ‘ π‘  = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘)
+ Energy flow may occur across the boundaries (exchange heat, work to surrounding)
+ The boundaries may change with time (𝑉 π‘β„Žπ‘Žπ‘›π‘”π‘’)
_Open System - Control Volume
+ Volume of fixed size containing matter
+ Mass flows across the system boundaries (π‘šπ‘Žπ‘ π‘  π‘β„Žπ‘Žπ‘›π‘”π‘’)
+ Energy flow occur across the boundaries
+ The boundaries may change with time
b. Law of Conservation of Mass:
_For closed system : π‘šπ‘Žπ‘ π‘  = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ (vô TH này thì mass trΖ°α»›c = mass sau)
_ Fluid flow through a control volume:
π‘Ÿπ‘Žπ‘‘π‘’ π‘šπ‘Žπ‘ π‘ π‘–π‘› − π‘Ÿπ‘Žπ‘‘π‘’ π‘šπ‘Žπ‘ π‘ π‘œπ‘’π‘‘ − π‘šπ‘Žπ‘ π‘  π‘Žπ‘π‘’π‘šπ‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘›π‘ π‘¦π‘ π‘‘π‘’π‘š = 0
With:
π‘Ÿπ‘Žπ‘‘π‘’ π‘šπ‘Žπ‘ π‘  =
π‘‘π‘š
𝑑𝑉
𝑑π‘₯
=𝜌
= πœŒπ‘‘π΄
= πœŒπ΄π‘£ (1 − π·π‘–π‘šπ‘’π‘ π‘–π‘œπ‘›, π‘ π‘šπ‘Žπ‘™π‘™ 𝐴)
𝑑𝑑
𝑑𝑑
𝑑𝑑
π‘Ÿπ‘Žπ‘‘π‘’ π‘šπ‘Žπ‘ π‘ π‘–π‘› = ∬
πœŒπ‘£π‘–π‘› 𝑑𝐴𝑖𝑛 (π‘–π‘›π‘‘π‘’π‘Ÿπ‘”π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘ π‘Žπ‘π‘œπ‘£π‘’ π‘“π‘œπ‘Ÿ β„Žπ‘œπ‘™π‘‘ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’)
π‘π‘œπ‘›π‘‘π‘Ÿπ‘œπ‘™ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’
π‘Ÿπ‘Žπ‘‘π‘’ π‘šπ‘Žπ‘ π‘ π‘–π‘› = βˆ¬π‘π‘œπ‘›π‘‘π‘Ÿπ‘œπ‘™ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’ πœŒπ‘£π‘œπ‘’π‘‘ π‘‘π΄π‘œπ‘’π‘‘
π‘šπ‘Žπ‘ π‘  π‘Žπ‘π‘’π‘šπ‘’π‘™π‘Žπ‘‘π‘–π‘œπ‘›π‘ π‘¦π‘ π‘‘π‘’π‘š =
πœ•
∭
πœŒπ‘‘π‘‰
πœ•π‘‘ πΆπ‘œπ‘›π‘‘π‘Ÿπ‘œπ‘™ π‘£π‘œπ‘™π‘’π‘šπ‘’
Equation (1)
∬
πœŒπ‘£π‘–π‘› 𝑑𝐴𝑖𝑛 − ∬
π‘π‘œπ‘›π‘‘π‘Ÿπ‘œπ‘™ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’
πœŒπ‘£π‘œπ‘’π‘‘ π‘‘π΄π‘œπ‘’π‘‘ =
π‘π‘œπ‘›π‘‘π‘Ÿπ‘œπ‘™ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’
πœ•
{∭
πœŒπ‘‘π‘‰ }
πœ•π‘‘
πΆπ‘œπ‘›π‘‘π‘Ÿπ‘œπ‘™ π‘£π‘œπ‘™π‘’π‘šπ‘’
(Description : Rate of mass tαΊ‘i 1 Δ‘iểm là πœŒπ΄π‘£ vói v theo phΖ°Ζ‘ng x, do xét trong hệ vαΊ­t
diện tich cα»‘ Δ‘α»‹nh, Rate of mass trên cαΊ£ mαΊ·t của vαΊ­t thì lαΊ₯y nguyên hàm 2 phΖ°Ζ‘ng y, z.
TΖ°Ζ‘ng tα»± vα»›i mass accumulation nαΊΏu V Δ‘α»•i theo 3 phΖ°Ζ‘ng thì ta nguyên hàm theo 3
phΖ°Ζ‘ng Δ‘êr có Δ‘α»₯owc tα»•ng thay Δ‘α»•i của V)
Types of fluids
Steady state
(No accumulation)
Compressible fluid
Non-steady state
∬ πœŒπ‘£π‘–π‘› 𝑑𝐴𝑖𝑛 = ∬ πœŒπ‘£π‘œπ‘’π‘‘ π‘‘π΄π‘œπ‘’π‘‘
𝐢.𝑉
Incompressible fluid
(𝜌 = π‘π‘œπ‘›π‘ π‘‘)
𝐢.𝑉
∬ 𝑣𝑖𝑛 𝑑𝐴𝑖𝑛 = ∬ π‘£π‘œπ‘’π‘‘ π‘‘π΄π‘œπ‘’π‘‘
𝐢.𝑉
Equation (1)
Equation (1)
𝐢.𝑉
_In 1 dimensional : Remove term rate mass out of the integrate (Do A không Δ‘α»•i theo
phΖ°Ζ‘ng y,z; A tαΊ‘i Δ‘iểm vào và ra của mass là hαΊ±ng sα»‘)
Types of fluids
Steady state
(No accumulation)
πœŒπ‘–π‘› 𝑣𝑖𝑛 𝐴𝑖𝑛 = πœŒπ‘œπ‘’π‘‘ π‘£π‘œπ‘’π‘‘ π΄π‘œπ‘’π‘‘
Compressible fluid
𝑣𝑖𝑛 𝐴𝑖𝑛 = π‘£π‘œπ‘’π‘‘ π΄π‘œπ‘’π‘‘
Incompressible fluid
(𝜌 = π‘π‘œπ‘›π‘ π‘‘)
Non-steady state
πœŒπ‘–π‘› 𝑣𝑖𝑛 𝐴𝑖𝑛 − πœŒπ‘œπ‘’π‘‘ π‘£π‘œπ‘’π‘‘ π΄π‘œπ‘’π‘‘ = π‘Žπ‘π‘’π‘šπ‘’π‘™π‘Žπ‘‘π‘’
𝑣𝑖𝑛 𝐴𝑖𝑛 − π‘£π‘œπ‘’π‘‘ π΄π‘œπ‘’π‘‘ = π‘Žπ‘π‘’π‘šπ‘’π‘™π‘Žπ‘‘π‘’
c. Newton’s Second law of motion:
_For close system: ∑ 𝐹⃗ =
𝑑
𝑑𝑑
(π‘šπ‘£βƒ—) (π‘“π‘œπ‘™π‘œπ‘€ π‘‘β„Žπ‘’ π‘‘π‘–π‘Ÿπ‘’π‘π‘‘π‘–π‘œπ‘› π‘¦π‘œπ‘’ π‘β„Žπ‘œπ‘ π‘’)
_Fluid flow through a control volume:
π‘†π‘’π‘š π‘“π‘œπ‘Ÿπ‘π‘’ = π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ π‘šπ‘œπ‘šπ‘’π‘›π‘‘ π‘œπ‘’π‘‘ − π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ π‘šπ‘œπ‘šπ‘’π‘›π‘‘ 𝑖𝑛 + π΄π‘π‘’π‘šπ‘’π‘™π‘Žπ‘‘π‘’
With:
π‘Ÿπ‘Žπ‘‘π‘’ π‘šπ‘œπ‘šπ‘’π‘›π‘‘π‘’π‘š =
𝑑𝐹
π‘‘π‘š
𝑑π‘₯
= 𝑣⃗
= π‘£βƒ—πœŒπ‘‘π΄
= π‘£βƒ—πœŒπ΄π‘£π‘› (1 − π·π‘–π‘šπ‘’π‘ π‘–π‘œπ‘›, π‘ π‘šπ‘Žπ‘™π‘™ 𝐴)
𝑑𝑑
𝑑𝑑
𝑑𝑑
n
n
𝑣⃗: π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ π‘œπ‘“ π‘£π‘’π‘π‘‘π‘œπ‘Ÿ (π‘‘π‘–π‘Ÿπ‘’π‘π‘‘π‘–π‘œπ‘› 𝑒𝑠𝑒 π‘β„Žπ‘œπ‘ π‘’)
𝑛⃗⃗ ∢ π‘‰π‘’π‘π‘‘π‘œπ‘Ÿ π‘›π‘œπ‘Ÿπ‘šπ‘Žπ‘™ π‘‘π‘œ π‘‘β„Žπ‘’ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’
v in
𝑣𝑛 = |𝑣||𝑛|π‘π‘œπ‘ πœƒ
_Fluid flow through in 1 direction:
∑ 𝐹π‘₯ = − ∬ 𝑣π‘₯ πœŒπ‘£π‘› 𝑑𝐴𝑖𝑛 + ∬ 𝑣π‘₯ πœŒπ‘£π‘› π‘‘π΄π‘œπ‘’π‘‘ +
𝐢𝑆
𝐢𝑆
πœ•
{∭ πœŒπ‘£π‘₯ 𝑑𝑉 }
πœ•π‘‘
𝐢𝑉
(𝑣𝑛 là tích vector có phΖ°Ζ‘ng của vector vαΊ­n tα»‘c, và vector vuông góc vα»›i mαΊ·t phαΊ³ng vαΊ­t,
trong slide cth chung là 𝑣. 𝑛 thì tích có hΖ°α»›ng này sαΊ½ âm nαΊΏu v và n ngược chiều lúc Δ‘i
vào)
d. 1st Law of Thermodynamics:
_For closed system:
𝛿𝑄
𝛿𝑑
−
π›Ώπ‘Š
𝛿𝑑
=
𝛿𝐸
𝛿𝑑
Usually no work involve: Heat change equal energy change:
𝛿𝑄 𝛿𝐸
=
𝛿𝑑
𝛿𝑑
2. Subsidiary law:
a. Fourier’s law of heat conduction:
π‘ž
πœ•π‘‡
= −𝐾
𝐴
πœ•π‘›
π‘ž
π‘Š
π‘ž: β„Žπ‘’π‘Žπ‘‘ π‘Ÿπ‘Žπ‘‘π‘’ (π‘Š π‘œπ‘Ÿ 𝐽/𝑠), : π»π‘’π‘Žπ‘‘ 𝑓𝑙𝑒π‘₯ (π‘Š/π‘š2 ), 𝐾: π‘‡β„Žπ‘’π‘Ÿπ‘šπ‘Žπ‘™ π‘π‘œπ‘›π‘‘π‘’π‘π‘‘π‘–π‘£π‘–π‘‘π‘¦ (
)
𝐴
π‘š. 𝐾
πœ•π‘‡
: π‘‡π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘›π‘‘ 𝑖𝑛 π‘π‘œπ‘Ÿπ‘‘π‘–π‘›π‘Žπ‘‘π‘’
πœ•π‘›
_If K constant: isotropic material
_One dimension, steady state situation:
+Infinite slab:
π‘ž
𝑇𝐻 − 𝑇𝐢
=𝐾
𝐴
𝑏
𝑇𝐻 /𝑇𝐢 ∢ π‘‘π‘’π‘šπ‘ π‘Žπ‘‘ β„Žπ‘œπ‘‘/π‘π‘œπ‘™π‘‘ 𝑠𝑖𝑑𝑒, 𝑏: π‘‘β„Žπ‘–π‘π‘˜ 𝑛𝑒𝑠𝑠 π‘œπ‘“ π‘‘β„Žπ‘’ π‘€π‘Žπ‘™π‘™/π‘ π‘™π‘Žπ‘ (π‘š)
v out
𝑇 π‘Žπ‘‘ π‘π‘’π‘Ÿπ‘‘π‘Žπ‘–π‘› π‘π‘œπ‘–π‘›π‘‘ π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ π‘₯ π‘“π‘œπ‘Ÿπ‘š β„Žπ‘œπ‘‘ π‘“π‘Žπ‘π‘’:
𝑇𝐻 − 𝑇
π‘₯
=
𝑇𝐻 − 𝑇𝐢 𝑏
+Infinite long hollow cylinder
π‘Ÿπ‘œ (π‘Ÿπ‘– ): π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘  π‘œπ‘’π‘‘π‘ π‘–π‘‘π‘’(𝑖𝑛𝑠𝑖𝑑𝑒), 𝐿: π‘™π‘’π‘›π‘”β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘¦π‘™π‘–π‘›π‘‘π‘’π‘Ÿ
π‘‡π‘œ (𝑇𝑖 ): π‘‘π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ π‘œπ‘’π‘‘π‘ π‘–π‘‘π‘’(𝑖𝑛𝑠𝑖𝑑𝑒)
π‘Ÿπ‘–
𝑇𝑖
π‘ž
π‘ž
πœ•π‘‡
π‘ž
πœ•π‘Ÿ
π‘ž
π‘Ÿπ‘–
=
= −𝐾
↔
∫
= −𝐾 ∫ πœ•π‘‡ ↔
𝑙𝑛 ( ) = −𝐾(𝑇𝑖 − π‘‡π‘œ )
𝐴 2πœ‹π‘ŸπΏ
πœ•π‘Ÿ
2πœ‹πΏ
π‘Ÿ
2πœ‹πΏ
π‘Ÿπ‘œ
π‘Ÿπ‘œ
π‘‡π‘œ
↔π‘ž=−
2πœ‹πΎπΏ(𝑇𝑖 − π‘‡π‘œ )
π‘Ÿ
𝑙𝑛 (π‘Ÿπ‘– )
π‘œ
𝑇 π‘Žπ‘‘ π‘π‘’π‘Ÿπ‘‘π‘Žπ‘–π‘› π‘Ÿ ∢
𝑇 − 𝑇𝑖
𝑙𝑛(π‘Ÿ/π‘Ÿπ‘– )
=
𝑇𝑖 − π‘‡π‘œ 𝑙𝑛(π‘Ÿπ‘œ /π‘Ÿπ‘– )
_At not steady state:
Coordinate
Cartesian (slab)
Cylindrical
Sphere
x
π‘₯=π‘₯
πœ•π‘‡ πœ•π‘‡
↔
=
πœ•π‘› πœ•π‘₯
π‘₯ = π‘Ÿ π‘π‘œπ‘ πœƒ = π‘Ÿ
↔ πœ•π‘₯ = πœ•π‘Ÿ
πœ•π‘‡ πœ•π‘‡
↔
=
πœ•π‘₯ πœ•π‘Ÿ
(radial heat transfer)
y
𝑦=𝑦
πœ•π‘‡ πœ•π‘‡
↔
=
πœ•π‘› πœ•π‘¦
z
𝑧=𝑧
πœ•π‘‡ πœ•π‘‡
↔
=
πœ•π‘› πœ•π‘§
𝑦 = π‘Ÿ π‘ π‘–π‘›πœƒ = π‘Ÿπœƒ
↔ πœ•π‘¦ = π‘Ÿπœ•πœƒ
πœ•π‘‡ 1 πœ•π‘‡
↔
=
πœ•π‘¦ π‘Ÿ πœ•πœƒ
𝑧=𝑧
πœ•π‘‡ πœ•π‘‡
↔
=
πœ•π‘› πœ•π‘§
𝐴𝑑 π‘ π‘šπ‘Žπ‘™π‘™ πœƒ: π‘ π‘–π‘›πœƒ ≈ πœƒ, π‘π‘œπ‘ πœƒ ≈ 1
𝑦 = π‘Ÿ 𝑠𝑖𝑛 πœƒ π‘ π‘–π‘›πœ‘ = π‘Ÿπœƒπœ‘
π‘₯ = π‘Ÿ 𝑠𝑖𝑛 πœƒ π‘π‘œπ‘ πœ‘ = π‘Ÿπœƒ
↔ πœ•π‘¦ = π‘Ÿπœƒ πœ•πœ‘
↔ πœ•π‘₯ = π‘Ÿπœ•πœƒ
πœ•π‘‡
1 πœ•π‘‡
πœ•π‘‡ 1 πœ•π‘‡
↔
=
↔
=
πœ•π‘¦ π‘Ÿπœƒ πœ•πœ‘
πœ•π‘₯ π‘Ÿ πœ•πœƒ
b. Newton’s law of cooling
:
π‘ž
= β„Ž(π‘‡π‘Š − 𝑇𝑓 )
𝐴
𝑧 = π‘Ÿ π‘π‘œπ‘ πœƒ = π‘Ÿ
πœ•π‘‡ πœ•π‘‡
↔
=
πœ•π‘§ πœ•π‘Ÿ
π‘‡π‘Š : π»π‘’π‘Žπ‘‘ π‘œπ‘“ π‘œπ‘π‘—π‘’π‘π‘‘, 𝑇𝑓 : β„Žπ‘’π‘Žπ‘‘ π‘œπ‘“ 𝑓𝑙𝑒𝑖𝑑, β„Ž: π»π‘’π‘Žπ‘‘ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿ π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ (
π‘Š
)
π‘š2 𝐾
c. Stefan-Boltzmann Law:
π‘ž
= πœŽπ‘‡ 4
𝐴
π‘Š
𝐡𝑑𝑒
𝜎 = 5.67 × 10−8 ( 2 4 ) = 1.7134 (
) : π‘†π‘‘π‘’π‘“π‘Žπ‘› − π΅π‘œπ‘™π‘‘π‘§π‘šπ‘Žπ‘› π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
π‘š 𝐾
β„Ž. 𝑓𝑑 2 ℉4
d. Thermal resistance: (nhiệt trở, na ná Δ‘iện trở)
Infinite slab
Thermal resistance
(π‘π‘‘β„Ž )
Infinite hollow
Cylinder
𝑙𝑛(π‘Ÿπ‘œ /π‘Ÿπ‘– )
2πœ‹π‘…πΎπΏ
𝑏
𝐾𝐴
Surface of liquid
1
β„Žπ΄
If the wall have only 1 way to transfer heat (many layers wall): similar to resistance put
in series:
π‘π‘‘β„Ž,π‘‘π‘œπ‘‘π‘Žπ‘™ = ∑ π‘π‘‘β„Ž
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
π‘π‘‘β„Ž,π‘‘π‘œπ‘‘π‘Žπ‘™
=∑
1
π‘π‘‘β„Ž,π‘π‘Žπ‘‘β„Ž 𝑖
(1 π‘π‘Žπ‘‘β„Ž π‘šπ‘Žπ‘¦ β„Žπ‘Žπ‘£π‘’ π‘šπ‘Žπ‘›π‘¦ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ 𝑖𝑛 π‘ π‘’π‘Ÿπ‘–π‘’π‘ )
We can calculate the resistance of the path to find temperature of cold side knowing hot
side and heat rate:
π‘ž=
βˆ†π‘‡
π‘π‘‘β„Ž,π‘‘π‘œπ‘‘π‘Žπ‘™
Ex: Calculate the temp at intersection of 2, 3 layers in 3 layers wall
π‘π‘‘β„Ž,π‘‘π‘œπ‘‘π‘Žπ‘™ = 𝑍1 + 𝑍2
Z1
Z2
Z3
Or if do not have A we can call calculate: Overall heat transfer coefficient
π‘ˆ = π΄π‘π‘‘β„Ž,π‘‘π‘œπ‘‘π‘Žπ‘™
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)
πœ•
πœ•π‘‡
πœ•
πœ•π‘‡
πœ•
πœ•π‘‡
πœ•π‘‡
(𝐾. ) +
(𝐾. ) + (𝐾. ) + π‘žΜ… = πœŒπΆπ‘
πœ•π‘₯
πœ•π‘₯
πœ•π‘¦
πœ•π‘¦
πœ•π‘§
πœ•π‘§
πœ•π‘‘
With: 𝐢𝑝 (𝐽/𝐾𝑔. 𝐾): 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 β„Žπ‘’π‘Žπ‘‘ π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦
π‘Š
π‘žΜ… ( 3 ) : β„Žπ‘’π‘Žπ‘‘ π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›
π‘š
_ 3D cylindrical coordinate system: (π‘Ÿ, πœƒ, 𝑧)
1πœ•
πœ•π‘‡
1 πœ•
πœ•π‘‡
πœ•
πœ•π‘‡
πœ•π‘‡
(𝐾. π‘Ÿ ) + 2
(𝐾. ) + (𝐾. ) + π‘žΜ… = πœŒπΆπ‘
π‘Ÿ πœ•π‘Ÿ
πœ•π‘Ÿ
π‘Ÿ πœ•πœƒ
πœ•πœƒ
πœ•π‘§
πœ•π‘§
πœ•π‘‘
_3D spherical coordinate system (π‘Ÿ, πœƒ, πœ‘)
1πœ•
πœ•(π‘Ÿπ‘‡)
1
πœ•
πœ•π‘‡
1
πœ•
πœ•π‘‡
πœ•π‘‡
(𝐾.
)+ 2
(𝐾. π‘ π‘–π‘›πœ‘ ) + 2 2
(𝐾. ) + π‘žΜ… = πœŒπΆπ‘
π‘Ÿ πœ•π‘Ÿ
πœ•π‘Ÿ
π‘Ÿ π‘ π‘–π‘›πœ‘ πœ•πœ‘
πœ•πœ‘
π‘Ÿ 𝑠𝑖𝑛 πœ‘ πœ•πœƒ
πœ•πœƒ
πœ•π‘‘
Step 2: Chose the condition: (more than 1 usually applied)
Condition
K Constant
Absence of
heat generation
Meaning
Ex: in 3D slab
πœ•
πœ•π‘‡
πœ•
πœ•π‘‡
πœ•
πœ•π‘‡
(𝐾. ) +
(𝐾. ) + (𝐾. )
πœ•π‘₯
πœ•π‘₯
πœ•π‘¦
πœ•π‘¦
πœ•π‘§
πœ•π‘§
πœ• πœ•π‘‡
πœ• πœ•π‘‡
πœ• πœ•π‘‡
= 𝐾{ ( )+
( ) + ( )}
πœ•π‘₯ πœ•π‘₯
πœ•π‘¦ πœ•π‘¦
πœ•π‘§ πœ•π‘§
2
= 𝐾∇ 𝑇
π‘žΜ… = 0
Equation change
π‘žΜ… 1 πœ•π‘‡
∇2 𝑇 + =
𝐾 𝛼 πœ•π‘‘
With
𝐾
𝛼=
; π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘Žπ‘™ 𝑑𝑖𝑓𝑓𝑒𝑠𝑖𝑑𝑦
πœŒπΆπ‘
(π‘š2 /𝑠)
Steady state
In 1 direction
πœ•π‘‡
=0
πœ•π‘‘
Ex: 3D cylindrical: Assume temperature
change only on r coordinate (radial heat
transfer)
πœ•π‘‡ πœ•π‘‡
=
=0
πœ•πœƒ πœ•π‘§
With (𝑒𝑣)′ = 𝑒′𝑣 + 𝑒𝑣′
In (*)
πœ•
: 𝑖𝑠 π‘‘π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘‘π‘Žπ‘™ π‘€π‘–π‘‘β„Ž π‘Ÿ,
πœ•π‘Ÿ
Step 3: find 𝑻 𝒂𝒏𝒅
Into equation in I.2.b
(𝐾∇2 𝑇 = 0)
1πœ•
πœ•π‘‡
π‘žΜ… 1 πœ•π‘‡
(π‘Ÿ ) + =
π‘Ÿ πœ•π‘Ÿ πœ•π‘Ÿ
𝐾 𝛼 πœ•π‘‘
1 πœ•π‘‡
πœ• 2𝑇
↔ ( + π‘Ÿ 2 ) + β‹― (∗)
π‘Ÿ πœ•π‘Ÿ
πœ•π‘Ÿ
πœ•π‘‡
π‘Žπ‘›π‘‘ π‘Ÿ 𝑖𝑠 π‘“π‘’π‘›π‘π‘‘π‘–π‘œπ‘› π‘œπ‘“ π‘Ÿ
πœ•π‘Ÿ
𝝏𝑻
𝝏𝒙
𝒕𝒐 π’‚π’‘π’‘π’π’Šπ’†π’… π’Šπ’ π’ƒπ’π’–π’…π’‚π’“π’š π’„π’π’π’…π’Šπ’•π’Šπ’π’ ∢
Ex: for 1D slab (condition 4), steady state (condition 2) and K constant (condition 1)
General equation become:
πœ• 2 𝑇 π‘žΜ…
πœ• 2𝑇
π‘žΜ…
+
=
0
↔
=
−
πœ•π‘₯ 2 𝐾
πœ•π‘₯ 2
𝐾
→
πœ•π‘‡
π‘žΜ…
= − π‘₯ + 𝐢1
πœ•π‘₯
𝐾
→ 𝑇(π‘₯) = −
π‘žΜ… π‘₯ 2
+ 𝐢1 π‘₯ + 𝐢2
𝐾 2
Ex: for 1D hollow cylinder (condition 4), steady state (condition 2) and K constant
(condition 1)
1πœ•
πœ•π‘‡
π‘žΜ…
1 πœ•π‘‡
πœ• 2𝑇
π‘žΜ…
(𝐾. π‘Ÿ ) + = 0 ↔ ( + π‘Ÿ 2 ) + = 0
π‘Ÿ πœ•π‘Ÿ
πœ•π‘Ÿ
𝐾
π‘Ÿ πœ•π‘Ÿ
πœ•π‘Ÿ
𝐾
1 πœ•π‘‡ πœ• 2 𝑇
π‘žΜ…
↔
+ 2 =−
π‘Ÿ πœ•π‘Ÿ πœ•π‘Ÿ
𝐾
↔
↔π‘Ÿ
πœ•π‘‡
πœ• 2𝑇
π‘žΜ…
+ π‘Ÿ 2 = − π‘Ÿ (π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘¦ π‘Ÿ π‘“π‘œπ‘Ÿ π‘’π‘Žπ‘β„Ž 𝑠𝑖𝑑𝑒)
πœ•π‘Ÿ
πœ•π‘Ÿ
𝐾
πœ•π‘‡
π‘žΜ… π‘Ÿ 2
πœ•π‘‡
π‘žΜ… π‘Ÿ
π‘žΜ… π‘Ÿ 2
=−
(π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘™ 2 𝑠𝑖𝑑𝑒) ↔
=−
→ 𝑇(π‘Ÿ) = −
+ 𝐢1
πœ•π‘Ÿ
𝐾 2
πœ•π‘Ÿ
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 π‘‡π‘ π‘’π‘Ÿ
(π‘₯ = 𝑏, π‘Ÿ = π‘Ÿπ‘œ )
Knowing heat flux (q/A) at surface
(π‘₯ = 𝑏, π‘Ÿ = π‘Ÿπ‘œ )
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) = π‘‡π‘ π‘’π‘Ÿ
(π‘‡π‘ π‘’π‘Ÿ is a number, position of surface
depend on how you put the origin and
coordinate)
πœ•π‘‡
π‘ž/𝐴 = ( )
πœ•π‘₯ π‘₯=𝑏
πœ•π‘‡
( )
=0
πœ•π‘₯ π‘₯=𝑏
πœ•π‘‡
( )
=0
πœ•π‘₯ π‘₯=𝑏/2
(b is the thickness of wall, if chose x =0 at
center change the x=b/2 by x=0)
πœ•π‘‡
−𝐾 ( )
= −β„Ž(𝑇𝑓 − 𝑇π‘₯=𝑏 )
πœ•π‘₯ π‘₯=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:
π΅π‘–π‘œπ‘‘ π‘›π‘’π‘šπ‘π‘’π‘Ÿ: 𝐡𝑖 =
β„ŽπΏ
(π‘…π‘Žπ‘‘π‘–π‘œ π‘œπ‘“ π‘π‘‘β„Ž π‘œπ‘“ π‘ π‘œπ‘™π‘–π‘‘ π‘Žπ‘›π‘‘ π‘π‘‘β„Ž π‘œπ‘“ π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’)
𝐾
πΉπ‘œπ‘’π‘Ÿπ‘–π‘’π‘Ÿ π‘π‘’π‘šπ‘π‘’π‘Ÿ: 𝑋 =
𝛼=
𝛼𝑑
𝐾𝑑
=
(π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘  π‘œπ‘“ π‘‘π‘–π‘šπ‘’)
𝐿2 πœŒπΆπ‘ 𝐿2
𝐾
; π‘‘β„Žπ‘’π‘Ÿπ‘šπ‘Žπ‘™ 𝑑𝑖𝑓𝑓𝑒𝑠𝑖𝑑𝑦 (π‘š2 /𝑠)
πœŒπΆπ‘
1
π΅π‘–π‘œπ‘‘
π‘₯
π‘…π‘’π‘™π‘Žπ‘‘π‘–π‘£π‘’ π‘π‘œπ‘ π‘–π‘‘π‘–π‘œπ‘›: 𝑛 =
𝐿
π‘…π‘’π‘™π‘Žπ‘‘π‘–π‘£π‘’ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’: π‘š =
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=
π΄π‘ π‘’π‘Ÿ
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 𝐡𝑖 < 0.1: Neglect internal temperature gradient
β„Žπ΄π‘‘
𝑇 − 𝑇𝑓
πœƒ
−
=
= 𝑒 πœŒπΆπ‘ 𝑉 = 𝑒 −𝐡𝑖×𝑋
πœƒπ‘œ π‘‡π‘œ − 𝑇𝑓
_ If 𝐡𝑖 > 0.1:
πœƒ
Step 4: Define πœƒ 𝒐𝒇 that shape:
π‘œ
3
πœƒ
πœƒ
𝐸π‘₯: ( )
= [( )
]
πœƒπ‘œ 𝑐𝑒𝑏𝑖𝑐
πœƒπ‘œ 𝑖𝑛𝑓𝑖𝑛𝑖𝑑𝑒 π‘ π‘™π‘Žπ‘
πœƒ
Step 5: Calculate Biot, Fourier, m, n for each component and find for πœƒ in Appendix F:
π‘œ
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:
πœƒ
πœƒ
π‘œ
π‘œ
You can find πœƒ for 0.5 and 0.75, assume m and πœƒ are linear:
πœƒ
= π‘Žπ‘š + 𝑏
πœƒπ‘œ
And use 2 point at 0.5 and 0.75 you are found to calculate a and b
πœƒ
Step 6: Found the πœƒ of the shape and calculate T
π‘œ
ο‚·
There are some problem like give you T and make you find t:
πœƒ
Step 1: You will calculate πœƒ of the shape and write it component:
π‘œ
Ex:
πœƒ
πœƒ
πœƒ
( )
=( )
( )
πœƒπ‘œ 𝑓𝑖 𝑐𝑦𝑙
πœƒπ‘œ 𝑖𝑛𝑓𝑖 𝑐𝑦𝑙 πœƒπ‘œ 𝑖𝑛𝑓𝑖 π‘ π‘™π‘Žπ‘
Step 2: Then you calculate the Fourier number of component through time:
Ex:𝑋𝑖𝑛𝑓𝑖 π‘ π‘™π‘Žπ‘ = π‘Ž × π‘‘,
𝑋𝑖𝑛𝑓𝑖 𝑐𝑦𝑙 = 𝑏 × π‘‘
Step 3: Then you will guess :)))))) any relation you want: (CHOSE ONLY 1 COMPONENT)
Ex
πœƒ
πœƒ
πœƒ
πœƒ
( )
= √( )
π‘œπ‘Ÿ ( )
= 1/2 ( )
πœƒπ‘œ 𝑖𝑛𝑓𝑖 𝑐𝑦𝑙
πœƒπ‘œ 𝑓𝑖 𝑐𝑦𝑙
πœƒπ‘œ 𝑖𝑛𝑓𝑖 𝑐𝑦𝑙
πœƒπ‘œ 𝑓𝑖 𝑐𝑦𝑙
(π‘Œπ‘œπ‘’ π‘π‘Žπ‘› β„Žπ‘Žπ‘£π‘’ π‘‘β„Žπ‘’ π‘Ÿπ‘–π‘”β„Žπ‘‘ π‘β„Žπ‘œπ‘ π‘’ π‘π‘Žπ‘¦ π‘π‘œπ‘›π‘ π‘–π‘‘π‘’π‘Ÿ π‘‘β„Žπ‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘œ π‘œπ‘“ π‘’π‘Žπ‘β„Ž 𝐿 π‘œπ‘“ π‘’π‘Žπ‘β„Ž π‘π‘œπ‘šπ‘π‘œπ‘›π‘’π‘‘)
πœƒ
Step 4: Then put it in(πœƒ )
π‘œ
of the shape to calculate each of component, put in appendix
𝑖𝑛𝑓𝑖 𝑐𝑦𝑙
of find time by each component.
πœƒ
πœƒ
π‘œ
π‘œ
Step 5Take that time to find other componentπœƒ . Then cal (πœƒ )
π‘ β„Žπ‘Žπ‘π‘’
πœƒ
Step 6: If πœƒ of the shape not fit, adjust t and try again
π‘œ
3. Fin:
a. Some equation:
We have the fin with:
𝑒π‘₯𝑑𝑒𝑛𝑑𝑒𝑑 𝐿 , 𝐴𝑑𝑖𝑝 𝑖𝑠 π‘Žπ‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‘β„Žπ‘’ 𝑑𝑖𝑝 π‘Žπ‘›π‘‘ 𝑃 𝑖𝑠 π‘π‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ π‘œπ‘“ π‘‘β„Žπ‘’ 𝑑𝑖𝑝
𝑀 = √β„Žπ‘ƒπΎπ΄π‘‘π‘–π‘ ,
𝑓𝑖𝑛 𝑒𝑓𝑓𝑒𝑐𝑑𝑖𝑣𝑒𝑛𝑒𝑠𝑠: πœ€π‘“ =
π‘š=√
β„Žπ‘ƒ
𝐾𝐴𝑑𝑖𝑝
π‘žπ‘“
β„Žπ‘’π‘Žπ‘‘ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿ π‘€π‘–π‘‘β„Ž 𝑓𝑖𝑛
=
π‘ž
β„Žπ‘’π‘Žπ‘‘ π‘‘π‘Ÿπ‘Žπ‘›π‘ π‘“π‘’π‘Ÿ π‘€π‘–π‘‘β„Žπ‘œπ‘’π‘‘ π‘‘β„Žπ‘’ 𝑓𝑖𝑛
𝐾𝑃
πœ€π‘“ = √
π‘‘π‘Žπ‘›β„Ž(π‘šπΏ) (π‘Žπ‘‘π‘–π‘Žπ‘π‘Žπ‘‘π‘–π‘ 𝑓𝑖𝑛)
β„Žπ΄π‘ π‘’π‘Ÿπ‘“π‘Žπ‘π‘’
𝑓𝑖𝑛 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦: 𝑛𝑓 =
π‘žπ‘Ÿπ‘’π‘Žπ‘™
π‘žπ‘Ÿπ‘’π‘Žπ‘™
=
π‘žπ‘–π‘‘π‘’π‘Žπ‘™ (β„Žπ‘’π‘Žπ‘‘ π‘Žπ‘‘ 𝑑𝑖𝑝 π‘’π‘žπ‘’π‘Žπ‘™ π‘Žπ‘‘ π‘π‘Žπ‘ π‘’) β„Žπ΄π‘ π‘’π‘Ÿ (𝑇 − 𝑇𝐿 )
b. Calculation:
_For only pin fin:
_For circular and other strainght fin:
(π‘ŸπΏ − π‘Ÿπ‘œ )
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