Applied Thermodynamics and Heat Transfer

MCEN3000
Lecture Notes: Convection Heat Transfer
by
Prof. Tilak Chandratilleke
Department of Mechanical Engineering
Curtin University

1
 CONVECTION HEAT TRANSFER
Convection is the mechanism of heat removal from a heated surface due to displacement
of heated molecules of the medium.

Natural or Free Convection

• Fluid molecule displacement by buoyancy forces
due to temperature differential Natural or Free Convection
• Fluid circulation is called convection currents

Forced Convection
• Fluid molecule displacement is externally induced
by fan or blower
Forced Convection

Analytical prediction of convective heat transfer is prohibitively tedious because of the

inter-dependency of fluid temperature and velocity.
• Involves solution of Continuity, Navier-Stokes and Energy Equations (5 in total).
• No exact solution exists to these PDEs.
• Experimentally-based information with non-dimensional terms are used,
(heat transfer correlations).
• Convective process is simplified by defining “heat transfer coefficient”.

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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 CONVECTION HEAT TRANSFER
Convective Heat Transfer Coefficient
• Convective heat transfer coefficient
Units: W/m2.K
𝐴 • Obtained from empirical correlations
𝑄𝑜 = ℎ𝐴 𝑇𝑤 − 𝑇∞
𝑄𝑜 based on experimental data
Surface area
∙ 𝑇∞
𝑇𝑤 THERMAL RESISTANCE CONCEPT
1
𝑇𝑤 − 𝑇∞ = 𝑄
Solid Fluid ℎ𝐴 𝑜 𝑇𝑤 𝑄𝑜 𝑇∞ 𝑇𝑤 − 𝑇∞
𝑄𝑜 =
Thermal resistance, 𝑅 1
1 ℎ𝐴
𝑅=
ℎ𝐴
Analytical representation - Convective Boundary

𝐴 At convective boundary: 𝑄𝑤 = 𝑄𝑜
𝑄𝑤 𝑄𝑜 Conduction to wall Convection from wall

∙ 𝑇∞
𝑇𝑤 𝑑𝑇
𝑄𝑤 = −𝑘𝐴 = 𝑄𝑜 = ℎ𝐴 𝑇𝑤 − 𝑇∞
Solid 𝑥=𝐿 Fluid 𝑑𝑥 𝑥=𝐿
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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 CONVECTION HEAT TRANSFER
Convective Heat Transfer Coefficient
• Heat transfer correlations constructed of non-dimensional groups
using Buckingham Theorem (or similarity principles) Temperature-dependant properties
Evaluated at a Reference temperature
Forced Convection: Governing variables [ℎ, 𝑉, 𝐿, 𝜌, 𝜇, 𝑐𝑝 , 𝑘]

𝑓 ℎ, 𝑉, 𝐿, 𝜌, 𝜇, 𝑐𝑝 , 𝑘 = 0 ∅ 𝑁𝑢, 𝑅𝑒, 𝑃𝑟 = 0 𝑁𝑢 = 𝐶. 𝑅𝑒 𝑚 . 𝑃𝑟 𝑛

Nusselt ℎ𝐿 Reynolds 𝜌𝑉𝐿 Prandtl 𝑃𝑟 = 𝜇𝑐𝑝 𝑁𝑢 ℎ

𝑁𝑢 = 𝑅𝑒 =
Number 𝑘 Number 𝜇 Number 𝑘

Thermal coefficient of
Gravity 𝑇𝑤 − 𝑇∞
volumetric expansion
Temperature-dependant properties
Evaluated at a Reference temperature
Natural Convection: Governing variables [ℎ, 𝑔, 𝐿, 𝛽, ∆𝑇, 𝜌, 𝜇, 𝑐𝑝 , 𝑘]

𝑓 ℎ, 𝑔, 𝐿, 𝛽, ∆𝑇, 𝜌, 𝜇, 𝑐𝑝 , 𝑘 = 0 ∅ 𝑁𝑢, 𝐺𝑟, 𝑃𝑟 = 0 𝑁𝑢 = 𝐶. 𝐺𝑟 𝑚 . 𝑃𝑟 𝑛

ℎ𝐿 𝜌2 𝑔𝛽∆𝑇𝐿3 Prandtl 𝜇𝑐𝑝

Nusselt
𝑁𝑢 =
Grashof 𝑃𝑟 = 𝑁𝑢 ℎ
Number 𝑘 Number 𝐺𝑟 = 𝜇2 Number 𝑘

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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 Forced Convection Heat Transfer Correlations

(a) Flow across a cylinder 𝑅𝑒 𝐶 𝑛

40 - 4000 0.683 0.466
𝑛 1
𝑉 𝐷 𝑁𝑢𝐷 = 𝐶. 𝑅𝑒𝐷 . 𝑃𝑟 3
4000 - 40000 0.193 0.618
40000 - 250000 0.0266 0.805

(b) Flow over a sphere

𝑉 𝐷 𝑁𝑢𝐷 = 0.97 + 0.68𝑅𝑒𝐷 0.5 𝑃𝑟 0.3 1 < 𝑅𝑒𝐷 < 2000

(c) Flow through a pipe (Dittus-Boelter Correlation) 𝑅𝑒𝐷 > 10000

0.8 0.7 < 𝑃𝑟 < 100

𝑉 𝐷 𝑁𝑢𝐷 = 0.023. 𝑅𝑒𝐷 . 𝑃𝑟 𝑛
𝑛 = 0.4 ℎ𝑒𝑎𝑡𝑖𝑛𝑔
𝑛 = 0.3 𝑐𝑜𝑜𝑙𝑖𝑛𝑔

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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 Natural Convection Heat Transfer Correlations

𝑅𝑎𝐿 = 𝐺𝑟𝐿 . 𝑃𝑟
(a) Horizontal plate (Rayleigh Number)

1
(hot face up) 𝑁𝑢𝐿 = 0.54 𝐺𝑟𝐿 . 𝑃𝑟 4 104 < 𝐺𝑟𝐿 . 𝑃𝑟 < 107
1
(hot face down) 𝑁𝑢𝐿 = 0.27 𝐺𝑟𝐿 . 𝑃𝑟 4 105 < 𝐺𝑟𝐿 . 𝑃𝑟 < 1010
𝑊
𝐴
Characteristic length, 𝐿 = , A = area and p = perimeter
𝑃
(b) Vertical plate

0.25
0.67 𝐺𝑟𝐿 . 𝑃𝑟
𝑁𝑢𝐿 = 0.68 + 4
𝑊
9 9 𝐺𝑟𝐿 . 𝑃𝑟 ≤ 109
0.492 16
1+ 𝑃𝑟

Characteristic length, 𝐿 = 𝑊

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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 Combined Conduction and Convection Thermal Network

𝑄o 𝑄o 𝑄o 𝑄o
𝑇1 𝑇2 𝑇3
𝑇𝑎 𝑅1 𝑇1 𝑅2 𝑇2 𝑅3 𝑇3 𝑅4 𝑇𝑏
𝑄o 𝑄o
𝑇𝑎 ∙ ∙ 𝑇𝑏 1 𝐿1 𝐿2 1
𝑅1 = 𝑅2 = 𝑅3 = 𝑅4 =
ℎ𝑎 ℎ𝑎 𝐴 𝑘1 𝐴 𝑘2 𝐴 ℎ𝑏 𝐴
ℎ𝑏
𝑘1 𝐿1 𝑘2 𝐿2
(Convective) (Conductive) (Conductive) (Convective)

𝑇𝑎 − 𝑇𝑏
Overall resistance 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 𝑄𝑜 =
𝑅𝑒𝑞
(Not convective)
Overall Heat Transfer Coefficient 𝑄𝑜 = 𝑈𝐴 𝑇𝑎 − 𝑇𝑏 Structurally similar to 𝑄 = ℎ𝐴∆𝑇

𝑄o 𝑄o 𝑄o 𝑄o

ℎ𝑎 ℎ𝑏 𝑇𝑎 𝑅1 𝑅2 𝑅3 𝑇3 𝑅4 𝑇𝑏
𝑇1 𝑇2
𝑘1 𝑇𝑎 ∙ 𝑄𝑜
1 𝑟2 𝑟3
𝑟1 𝑇1 ∙ 𝑇𝑏 ln
𝑟1 ln 𝑟2 1
𝑇2
𝑅1 = 𝑅2 = 𝑅3 = 𝑅4 =
𝑘2 ℎ𝑎 2𝜋𝑟1 𝐿 2𝜋𝑘1 𝐿 2𝜋𝑘2 𝐿 ℎ𝑏 2𝜋𝑟3 𝐿
𝑟2
𝑇3
𝑟3 (Note circumferential area)

𝑇𝑎 − 𝑇𝑏 Overall
𝑈𝐴 = 𝑈1 𝐴1 = 𝑈2 𝐴2 = 𝑈3 𝐴3
𝑄𝑜 = 𝑄𝑜 = 𝑈𝐴 𝑇𝑎 − 𝑇𝑏 Heat Transfer
𝑅𝑒𝑞 2𝜋𝑟1 𝐿
Coefficient 2𝜋𝑟2 𝐿 2𝜋𝑟3 𝐿

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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 Optimum (Critical) Insulation Thickness : (Cylindrical systems)
𝑟1 , 𝑟2 Pipe radii (fixed) (𝑟3 − 𝑟2 ) Insulation thickness (variable)
𝑟3 Insulation radius (variable)
ℎ𝑎 ℎ𝑏
𝑄𝑜 𝑄o 𝑄o 𝑄o 𝑄o
𝑘𝑝 𝑇𝑎 ∙
𝑟1 𝑇1 ∙ 𝑇𝑏
𝑇2 𝑇𝑎 𝑅1 𝑇1 𝑅2 𝑇2 𝑅3 𝑇3 𝑅4 𝑇𝑏
𝑘𝑖 𝑟2 𝑟2 𝑟3
𝑇3 1 𝑟1 ln ln 𝑟2 1
𝑟3 𝑅1 = 𝑅2 = 𝑅3 = 𝑅4 =
(Constant) ℎ𝑎 2𝜋𝑟1 𝐿 (Constant) 2𝜋𝑘𝑝 𝐿 2𝜋𝑘𝑖 𝐿 ℎ𝑏 2𝜋𝑟3 𝐿
𝑟3
Maximum heat flow 𝑇𝑎 − 𝑇𝑏 ln 𝑟2 1
𝑄𝑜 = 𝑅3 + 𝑅4 = +
𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 2𝜋𝑘𝑖 𝐿 ℎ𝑏 2𝜋𝑟3 𝐿
𝑄𝑜
𝑅 𝑄𝑜 𝑟3
𝑅3 + 𝑅4 ln𝑟2 1
𝑅3 + 𝑅4 = + Taking, 𝑌 = (𝑅3 + 𝑅4 ) and
2𝜋𝑘𝑖 𝐿 ℎ 𝑟 2𝜋 𝑟3 𝐿 𝑟3
𝑏 2 𝑟2 𝑋=
𝑅3 𝑟2
1 𝑘2
𝑌= ln 𝑋 +
2𝜋𝑘𝑖 𝐿 𝑟2 ℎ𝑏 𝑋 (Point of inflection)
𝑅4
𝑑𝑌 1 1 𝑘2 𝑘𝑖
= − =0 𝑋=
𝑟3 𝑑𝑋 2𝜋𝑘𝑖 𝐿 𝑋 𝑟2 ℎ𝑏 𝑋 2 𝑟2 ℎ𝑏
1.0 𝑋∗ 𝑋=
𝑟2
No insulation Critical Insulation 𝑘𝑖 Critical insulation radius at which
∴ 𝑟3 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 = 𝑟𝑐 =
ℎ𝑏 heat flow through insulation is maximum
Note: Critical insulation radius is independent of pipe radii
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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
Optimum (Critical) Insulation Thickness : (Cylindrical systems)

Case (a): 𝑟𝑐 > 𝑟2

• For this case, the heat flow through the insulation is more
𝑄𝑜 than from the bare pipe until insulation radius r = r*.
• The insulation is effective only when r > r*. This is useful
for applications when it is necessary to promote heat flow
while having an insulation of radius r1 < r < r*.

e.g. Insulation on a wire carrying an electrical current

𝑟2 𝑟𝑐 𝑟∗ 𝑟
Outside pipe wall

Case (b): 𝑟𝑐 < 𝑟2

𝑄𝑜 Outside pipe wall

• For this case, heat flow is immediately reduced for any

insulation added to the outside pipe surface.
Useful for insulating pipes carrying steam or coolant

𝑟𝑐 𝑟2 𝑟
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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 EXTENDED SURFACES OR FINS
There are many engineering applications requiring heat dissipation from a solid body to
fluid (liquid or gas) medium by convection. Some examples are:

Electronic cooling

I.C. engine cooling Pipe heating or cooling

Heat sink

A common element in all these applications

Fins or extended surfaces

𝑄𝑜 = ℎ𝐴 𝑇𝑤 − 𝑇∞

Provide Higher
increased area heat dissipation

Standard fin configurations

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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 FIN ANALYSIS – Uniform fin (Steady One-dimensional)
𝐴 𝑄𝑠 𝑄𝑜 = 𝐹𝑖𝑛 𝑏𝑎𝑠𝑒 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤
𝑄𝑠 = 𝐶𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑓𝑟𝑜𝑚 𝑓𝑖𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
𝑄𝑜
𝑄𝑡 𝑄𝑡 = 𝐶𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑓𝑟𝑜𝑚 𝑓𝑖𝑛 𝑡𝑖𝑝

𝐿 𝑇𝑜 = 𝐹𝑖𝑛 𝑏𝑎𝑠𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒

𝑇𝑜
𝑇∞ = 𝐴𝑚𝑏𝑖𝑒𝑛𝑡 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒

𝑇𝑡 𝑇𝑡 = 𝐹𝑖𝑛 𝑡𝑖𝑝 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒

𝑇∞ 𝐴 = 𝐹𝑖𝑛 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎
𝐿 = 𝐹𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ
𝑥
𝑝 = 𝐹𝑖𝑛 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 (involved in heat transfer)
Basic Definitions
𝒂 𝑭𝒊𝒏 𝑬𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒄𝒚
𝐴𝑐𝑡𝑢𝑎𝑙 𝑓𝑖𝑛 𝑏𝑎𝑠𝑒 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑓𝑎𝑙𝑙𝑖𝑛𝑔 𝑓𝑖𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑄 𝑎𝑐𝑡𝑢𝑎𝑙
𝐹𝑖𝑛 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝜂 =
𝐹𝑖𝑛 𝑏𝑎𝑠𝑒 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑖𝑛 𝑖𝑑𝑒𝑎𝑙 𝑓𝑖𝑛 (𝑢𝑛𝑖𝑓𝑜𝑟𝑚 𝑓𝑖𝑛 𝑡𝑒𝑚𝑒𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒) 𝑄 𝑚𝑎𝑥

𝒃 𝑭𝒊𝒏 𝑬𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆𝒏𝒆𝒔𝒔 𝑄 𝑚𝑎𝑥 = ℎ(𝑝𝐿 + 𝐴)(𝑇𝑜 − 𝑇∞ )

𝐴𝑐𝑡𝑢𝑎𝑙 𝑓𝑖𝑛 𝑏𝑎𝑠𝑒 ℎ𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑓𝑎𝑙𝑙𝑖𝑛𝑔 𝑓𝑖𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑄 𝑎𝑐𝑡𝑢𝑎𝑙

𝐹𝑖𝑛 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒𝑛𝑒𝑠𝑠 𝜖 =
𝐻𝑒𝑎𝑡 𝑓𝑙𝑜𝑤 𝑡ℎ𝑜𝑢𝑔ℎ 𝑓𝑖𝑛 𝑏𝑎𝑠𝑒 𝑎𝑟𝑒𝑎 (𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑖𝑛 𝑎𝑡𝑡𝑎𝑐ℎ𝑒𝑑) 𝑄 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑖𝑛

𝑄 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑖𝑛 = ℎ𝐴(𝑇𝑜 − 𝑇∞ ) 11

MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 FIN ANALYSIS – Uniform fin (Steady One-dimensional)
𝐴 𝑑𝑄𝑠 (𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 𝑙𝑜𝑠𝑠) Heat balance to element:

𝑄𝑜 𝑄𝑥 𝑄𝑥 + 𝑑𝑄𝑥 𝑄𝑥 = 𝑄𝑥 + 𝑑𝑄𝑥 + 𝑑𝑄𝑠

(incremental)
𝑑𝑄 𝑠 = ℎ(𝑝𝑑𝑥)(𝑇 − 𝑇∞ ) Convective heat loss
𝑥=0 𝑥=𝐿
𝑇𝑜
𝑑𝑇 Fourier heat conduction
𝑇𝑡 𝑄𝑥 = −𝑘𝐴
𝑑𝑥
𝑇∞
𝑑𝑄𝑥 𝑑2 𝑇
𝑥 𝑥 + 𝑑𝑥 𝑥 = −𝑘𝐴
𝑑𝑥 𝑑𝑥 2 Fin equation

𝑑𝑄𝑥 𝑑2 𝑇 𝑑2 𝑇 ℎ𝑝
𝑄𝑥 = 𝑄𝑥 + 𝑑𝑥 + 𝑑𝑄𝑠 −𝑘𝐴 𝑑𝑥 + ℎ(𝑝𝑑𝑥)(𝑇 − 𝑇∞ ) = 0 − (𝑇 − 𝑇∞ ) = 0
𝑑𝑥 𝑑𝑥 2 𝑑𝑥 2 𝑘𝐴
Let:
ℎ𝑝2 𝑑2 𝑇 ℎ𝑝 𝑑2 𝜃
𝜃 = (𝑇 − 𝑇∞ ) and 𝑚 = ∴ − (𝑇 − 𝑇∞ ) = 0 − 𝑚2 θ = 0
𝑘𝐴 𝑑𝑥 2 𝑘𝐴 𝑑𝑥 2
Excess fin temperature Trial solution: 𝜃 = 𝐶 𝑒 𝑚𝑥 + 𝐷 𝑒 −𝑚𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝐶 𝑎𝑛𝑑 𝐷 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠
above ambient

𝑑𝜃 𝑑2𝜃 𝑑2𝜃
= 𝑚𝐶 𝑒 𝑚𝑥 + (−𝑚)𝐷 𝑒 −𝑚𝑥 = 𝑚2 𝐶 𝑒 𝑚𝑥 + 𝑚2 𝐷 𝑒 −𝑚𝑥 = 𝑚2 𝐶 𝑒 𝑚𝑥 + 𝐷 𝑒 −𝑚𝑥
𝑑𝑥 𝑑𝑥 2 𝑑𝑥 2
𝑑2𝜃
− 𝑚2 θ = 0 ( ∴ Solution)
𝑑𝑥 2 12
MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 FIN ANALYSIS – Uniform fin (Steady One-dimensional)
Solving fin equation (depends on fin tip condition):
𝐴
CASE (a): Thermally-long fin (physically may not be long)
𝑄𝑜
𝑄𝑡 𝑇𝑡 → 𝑇∞ 𝑇ℎ𝑢𝑠, 𝑄𝑡 → 0 (𝑛𝑎𝑡𝑢𝑟𝑎𝑙𝑙𝑦)
𝑥=0 𝑥=𝐿
𝑇𝑜 𝑥=0 𝑇 = 𝑇𝑜 𝑜𝑟 𝜃 = 𝜃𝑜
Boundary conditions: 𝑥 → ∞ 𝑇 → 𝑇 𝑜𝑟 𝜃 = 0
∞
𝑇𝑡
𝑇∞ 𝜃 = 𝐶 𝑒 𝑚𝑥 + 𝐷 𝑒 −𝑚𝑥 𝐶 = 0 𝐷 = 𝜃𝑜
𝑒 𝑚𝑥 → ∞ with 𝑥 → ∞ ∴ 𝐶 = 0
𝑥
Fin (thermally-long) temperature profile
𝑑𝑇
∴ 𝜃 = 𝜃𝑜 𝑒 −𝑚𝑥 (𝑇 − 𝑇∞ ) = (𝑇𝑜 − 𝑇∞ ) 𝑒 −𝑚𝑥 𝑄𝑜 = −𝑘𝐴 𝑥=0
𝑑𝑥 𝑥=0
𝑄𝑜 = −𝑘𝐴(−𝑚)(𝑇𝑜 − 𝑇∞ ) 𝑒 −𝑚𝑥2 ℎ𝑝
𝑚 =
THERMAL RESISTANCE CONCEPT (Thermally-long fin) 1 𝑘𝐴
𝑄𝑜 = ℎ𝑝𝑘𝐴 2 (𝑇𝑜 − 𝑇∞ )
𝑇𝑜 𝑄𝑜 𝑇∞ 1
−
𝑇𝑜 − 𝑇∞ (𝑇𝑜 − 𝑇∞ )= 𝑄𝑜 ℎ𝑝𝑘𝐴 2
𝑄𝑜 =
𝑅𝑓
1 Ohms Law
−
𝑅𝑓 = ℎ𝑝𝑘𝐴 2
Electrical circuits
𝑉=𝐼𝑅

13
MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 FIN ANALYSIS – Uniform fin (Steady One-dimensional)
Solving fin equation (depends on fin tip condition):
𝐴
CASE (b): Fin with insulated tip (occurs due to geometry)
𝑄𝑜
𝑄𝑡 𝑄𝑡 = 0 (𝑎𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐)
𝑥=0 𝑥=𝐿 Boundary conditions: 𝑥=0 𝑇 = 𝑇𝑜 𝑜𝑟 𝜃 = 𝜃𝑜
𝑇𝑜
𝑑𝑇
𝑥 = 𝐿 𝑄𝑡 = −𝑘𝐴 =0
𝑑𝑥 𝑥=𝐿
𝑇𝑡
𝑇∞ 𝜃 = 𝐶 𝑒 𝑚𝑥 + 𝐷 𝑒 −𝑚𝑥

𝑒 −𝑚𝐿 𝑒 𝑚𝐿
𝑥 𝐶 = 𝑚𝐿 𝐷 = 𝑚𝐿
Fin (insulated tip) temperature profile 𝑒 + 𝑒 −𝑚𝐿 𝑒 + 𝑒 −𝑚𝐿
𝑒 𝑚(𝐿−𝑥) + 𝑒 −𝑚(𝐿−𝑥) 𝑑𝑇
(𝑇 − 𝑇∞ ) = (𝑇𝑜 − 𝑇∞ ) 𝑄𝑜 = −𝑘𝐴
𝑒 𝑚𝐿 + 𝑒 −𝑚𝐿 𝑑𝑥 𝑥=0

𝑒 𝑚𝐿 − 𝑒 −𝑚𝐿 1
THERMAL RESISTANCE CONCEPT (Insulated fin tip) 𝑄𝑜 = 𝑚𝐿 ℎ𝑝𝑘𝐴 2 (𝑇𝑜 − 𝑇∞ )
𝑒 + 𝑒 −𝑚𝐿
𝑇𝑜 𝑄𝑜 𝑇∞ 𝑒 𝑚𝐿 + 𝑒 −𝑚𝐿 1
𝑇𝑜 − 𝑇∞ (𝑇𝑜 − 𝑇∞ ) = 𝑄𝑜 ℎ𝑝𝑘𝐴 −
2
𝑄𝑜 = 𝑒 𝑚𝐿 − 𝑒 −𝑚𝐿
𝑅𝑓
𝑒 𝑚𝐿 + 𝑒 −𝑚𝐿 −
1 Ohms Law
𝑅𝑓 = 𝑚𝐿 ℎ𝑝𝑘𝐴 2 Electrical circuits
𝑒 − 𝑒 −𝑚𝐿 𝑉=𝐼𝑅
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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 FIN ANALYSIS – Uniform fin (Steady One-dimensional)
Solving fin equation (depends on fin tip condition):
𝐴
CASE (c): Fin with convective tip (naturally occurs)
𝑄𝑜
𝑄𝑡 𝑄𝑡 = ℎ𝐴 𝑇𝑡 − 𝑇∞
𝑥=0 𝑥=𝐿 Boundary conditions: 𝑥=0
𝑇 = 𝑇𝑜 𝑜𝑟 𝜃 = 𝜃𝑜
𝑇𝑜 𝑑𝑇
𝑥 = 𝐿 𝑄𝑡 = −𝑘𝐴 = ℎ𝐴 𝑇𝑡 − 𝑇∞
𝑑𝑥 𝑥=𝐿
𝑇𝑡 𝜃 = 𝐶 𝑒 𝑚𝑥 + 𝐷 𝑒 −𝑚𝑥
𝑇∞
ℎ ℎ
𝑒 −𝑚𝐿
1− 1+ 𝑒 𝑚𝐿
𝑚𝑘 𝑚𝑘
𝑥 𝐶=
ℎ ℎ
𝐷=
ℎ ℎ
Fin (insulated tip) temperature profile 1+ 𝑒 𝑚𝐿 + 1 − 𝑒 −𝑚𝐿 1+ 𝑒 𝑚𝐿 + 1 − 𝑒 −𝑚𝐿
𝑚𝑘 𝑚𝑘 𝑚𝑘 𝑚𝑘
ℎ ℎ
1+ 𝑒 𝑚(𝐿−𝑥) + 1 − 𝑒 −𝑚(𝐿−𝑥) 𝑑𝑇
𝑚𝑘 𝑚𝑘
(𝑇 − 𝑇∞ ) = (𝑇𝑜 − 𝑇∞ ) 𝑄𝑜 = −𝑘𝐴
ℎ ℎ 𝑑𝑥
1+ 𝑒 𝑚𝐿 + 1 − 𝑒 −𝑚𝐿 𝑥=0
𝑚𝑘 𝑚𝑘
ℎ ℎ
1+ 𝑒 𝑚𝐿 − 1 − 𝑒 −𝑚𝐿 1
THERMAL RESISTANCE CONCEPT (Convective fin tip) 𝑚𝑘 𝑚𝑘
𝑄𝑜 = ℎ𝑝𝑘𝐴 2 (𝑇𝑜 − 𝑇∞ )
ℎ ℎ
1+ 𝑒 𝑚𝐿 + 1 − 𝑒 −𝑚𝐿
𝑚𝑘 𝑚𝑘
𝑇𝑜 𝑄𝑜 𝑇∞ ℎ ℎ
𝑇𝑜 − 𝑇∞ 1+
𝑚𝑘
𝑒 𝑚𝐿 + 1 −
𝑚𝑘
𝑒 −𝑚𝐿 1
𝑄𝑜 = (𝑇𝑜 − 𝑇∞ ) = 𝑄𝑜
ℎ ℎ
ℎ𝑝𝑘𝐴 −
2
𝑅𝑓 1+ 𝑒 𝑚𝐿 − 1 − 𝑒 −𝑚𝐿
𝑚𝑘 𝑚𝑘
ℎ ℎ
1+ 𝑒 𝑚𝐿 + 1 − 𝑒 −𝑚𝐿 1
𝑚𝑘 𝑚𝑘 −
𝑅𝑓 = ℎ𝑝𝑘𝐴 2
Ohms Law
ℎ ℎ
1+ 𝑒 𝑚𝐿 − 1 − 𝑒 −𝑚𝐿
𝑚𝑘 𝑚𝑘 𝑉=𝐼𝑅 Electrical circuits
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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University
 Typical Fin Situations/Applications
𝑄2
𝑏
𝑇∞ 𝑎 𝑅𝑜 𝑄f 𝑃o = 𝑄 1 + 𝑄 2
𝑃𝑜
𝑄1 𝑄1 = 𝑄f + 𝑄f
𝑇𝑜 𝑃𝑜 𝑅𝑓
𝑝 = 2𝑎 + 𝑏 𝑅𝑓
𝑇𝑜 𝑅𝑗 𝑇𝑗 𝑇∞

𝑇𝑗 𝑄f
𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑖𝑐 𝑐𝑜𝑜𝑙𝑖𝑛𝑔 𝐿𝑜𝑛𝑔 𝑓𝑖𝑛

𝐿𝑜𝑛𝑔 𝑜𝑟 𝑄f
𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑣𝑒 𝑓𝑖𝑛
𝑄o 1
𝑇∞ 𝑇𝑜 𝑝=2m 𝑅
6 𝑓 1
(two convective 𝑅
6 𝑢𝑓
surface lengths of 1 m each) 𝑇𝑜 𝑅1 𝑇1 𝑅2 𝑇2 𝑇∞

𝑄 uf
𝐹𝑖𝑛𝑛𝑒𝑑 𝑝𝑖𝑝𝑒 (𝐻𝑜𝑡 𝑜𝑟 𝑐𝑜𝑙𝑑 𝑓𝑙𝑢𝑖𝑑)
𝐻𝑜𝑡 𝑠𝑝𝑜𝑡𝑠 𝐹𝑖𝑛 𝑤𝑖𝑡ℎ 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑒𝑑 𝑡𝑖𝑝
(𝑑𝑢𝑒 𝑡𝑜 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦)
𝐹𝑖𝑛 𝑤𝑖𝑡ℎ 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑒𝑑 𝑡𝑖𝑝 𝐹𝑖𝑛 𝑤𝑖𝑡ℎ 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑒𝑑 𝑡𝑖𝑝
(𝑑𝑢𝑒 𝑡𝑜 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦) (𝑑𝑢𝑒 𝑡𝑜 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑦)

𝑇𝑜 𝑇𝑜

𝑀𝑒𝑡𝑎𝑙 𝑟𝑜𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑤𝑎𝑙𝑙𝑠 𝑆𝑜𝑙𝑎𝑟 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑜𝑟 − 𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑟 𝑝𝑙𝑎𝑡𝑒

(fuse wire) 𝑆𝑞𝑢𝑎𝑟𝑒 𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑎𝑙 𝑟𝑒𝑠𝑖𝑠𝑡𝑜𝑟
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MCEN300 Lecture Notes by Prof. Tilak Chandratilleke, Department of Mechanical Engineering, Curtin University