This document presents the derivation of the one-dimensional steady-state fin equation for a long fin. It defines key terms like fin, temperature at the base (Tb) and surrounding fluid (T∞). The energy balance equation is shown for a differential volume element of the fin. Solving the resulting ordinary differential equation gives the temperature distribution equation as T(x) - T∞ = (Tb - T∞)e-mx, where m is a parameter involving heat transfer coefficient and fin properties. The heat flow equation is also derived. Two example problems are included to demonstrate calculating heat loss from long fins using the derived equations.

1.  Submitted by,
Registration no-
 2013334038
 2013334039
 2013334040
 2013334041
 Submitted to,
Mr. Chowdhuury Md
Lutfur Rahman
Assistant
Professor,
Department of
Industrial and
Production
Engineering,
PRSENTATION ON ONE
DIMENSIONAL STEADY STATE FIN
EQUATION (LONG FIN)

2. ONE DIMENSIONAL
STEADY STATE
FIN EQUATION FOR LONG FIN

3. FINS
When heat transfer by convection between a surface and the fluid surrounding it can be
increased by attaching to the surface thin strips of metal called fins.
Figure: Fin
In heat change applications a finned tube
arrangement might be used to remove heat from a
hot liquid. The heat transfer from the liquid to the
finned tube is by convection. The heat is conducted
through the material and finally dissipated to the
surrounding by convection.

4. ONE DIMENSIONAL STEADY STATE FIN EQUATION
(LONG FIN)
Figure: Nomenclature for the derivation of
one dimensional fin equation.
Consider one dimensional fin exposed to a
surrounding fluid at temperature T∞. The
temperature at the base (x=0) of fin 𝑻˳ .
To develop a one dimensional steady state
equation for fin of uniform cross- section.
Let us consider a differential volume
element of thickness dx. The surface area
for convection, of which is pdx∕hpdx(T-
T∞).

5. ONE DIMENSIONAL STEADY STATE FIN EQUATION
(LONG FIN)
Figure: Nomenclature for the derivation of
one dimensional fin equation.
The steady state energy balance equation for this
volume element is:
qx = qx+dx + dqconvec
qx = −𝑲𝑨
𝒅𝑻
𝒅𝒙
qx+dx = −𝑲𝑨
𝒅𝑻
𝒅𝒙
+
𝒅
𝒅𝒙
𝒅𝑻
𝒅𝒙
𝒅𝒙
qconvec = 𝒉𝒑𝒅𝒙 𝑻 𝒙 − 𝑻∞
𝒅 𝟐 𝑻
𝒅𝒙 𝟐 −
𝒉𝒑
𝒌𝑨
𝑻 𝒙 − 𝑻∞ = 𝟎; When, 𝑻 𝒙 − 𝑻∞ = 𝜭 𝒙
Energy in left face = Energy out in right face + Energy lost by convection

6. ONE DIMENSIONAL STEADY STATE FIN EQUATION
(LONG FIN)
Figure: Nomenclature for the derivation of
one dimensional fin equation.
And m=√
𝒉𝒑
𝑲𝑨
𝒅 𝟐 𝜽 𝒙
𝒅𝒙 𝟐
− 𝒎 𝟐 𝜽 𝒙 = 𝟎
Which is one dimension linear homogeneous equation.
AT x=0, 𝜽 = 𝜽˚ = 𝑻 − 𝑻∞
At 𝒙 = ∞ , 𝜽 = 𝟎 = 𝑻∞ − 𝑻∞
For homogeneous equation the solution is:
𝜽(𝒙) = 𝑪1e-mx +C2 emx = T(x) –T(∞)
𝜽 𝒙 = 𝑻 𝑿 − 𝑻 ∞

7. ONE DIMENSIONAL STEADY STATE FIN EQUATION
(LONG FIN)
Figure: Nomenclature for the derivation of
one dimensional fin equation.
At x = 0, 𝜽 𝒙 = 𝜽 = C1 At 𝒙 = ∞, 𝜽 𝒙 = 𝟎, C2=0
θ(x)= θ˳e-mx
𝜽
(
𝒙
)
𝜽˳
=e-mx
𝑻 𝑿 −𝑻∞
𝑻−𝑻∞
=e-mx
This is the temperature distribution equation for long fin.
Heat Flow
Q = -KA
𝒅𝜽
𝒅𝒙
𝒂𝒕 𝒙 = 𝟎
= KAm 𝜽
= 𝜽˳ 𝒉𝒑𝑲𝑨
This is the heat flow equation for long fin.

8. Problem
 A steel rod diameter D=2cm, length L=25 Cm and thermal conductivity K=50w∕m˳c is exposed to ambient air at T=20°c
with a heat transfer coefficient h=64w∕m2ᵒ-c. If one end of the rod is maintained at temperature of 120ᵒ.calculate heat
loss from the rod.
m2 =
𝒉𝒑
𝒌𝑨
=
𝒉𝑫𝝅
𝝅 𝟒 𝑫 𝟐 𝒌
=
𝟒𝒉
𝒌𝑫
=
𝟒×𝟔𝟒
𝟓𝟎×𝟎.𝟎𝟐
m=16 mL=16× 𝟎. 𝟐𝟓 = 𝟒
q=𝜽√𝒉𝒑𝑲𝑨=𝜽√{𝝅𝑫
𝝅𝑫 𝟐
𝟒
𝑲𝒉}
=(120-20)
𝝅
𝟐
√{(𝟎. 𝟎𝟐) 𝟑 × 𝟓𝟎 × 𝟔𝟒}
=25.1 w (Answer).
Solution

9. Exercise
 Consider a long, slender copper rod of diameter D=1 cm and thermal conductivity K=380W/(mᵒc), with one end
thermally attached to a wall at 200ᵒc.Heat is dissipated from the rod by convection with a heat transfer coefficient
h∞=15W/𝒎 𝟐
°𝒄 . Determine the heat transfer rate from the rod Into the surrounding air at T∞=30ᵒ.
m2 =
𝒉𝒑
𝒌𝑨
=
𝒉𝑫𝝅
𝝅 𝟒 𝑫 𝟐 𝒌
=
𝟒𝒉
𝒌𝑫
=
𝟒×𝟏𝟓
𝟑𝟖𝟎×𝟎.𝟎𝟏
m = 3.98 ML=3.98× 𝟏 = 𝟑. 𝟗𝟖
q=𝜽 𝒉𝒑𝑲𝑨=𝜽√{𝝅𝑫
𝝅𝑫 𝟐
𝟒
𝑲𝒉
=(200-30)
𝝅
𝟐
√{(𝟎. 𝟎𝟏) 𝟑
× 𝟑𝟖𝟎 × 𝟏𝟓}
= 20.16W (Answer).
Solution