Two basic topics of heat transfer have been covered up by me based on the famous books of :- 1) John H. Lienhard (Professor Emeritus, University of Houston) 2) J.P. Holman (Professor, Southern Methodist University) 3) Prabal Talukdar (Associate Professor, IIT, India)

1. SMYTH_WORKS FINS
INS EQUATION, UNSTEADY C
UMPED HEAT CAPACITY SYSTEM
LUMPED
Das Buch ist schlecht und
CONDITION &
YSTEM
die Lehrer ist wirklich Streber
9/1/2014
Streber, auch!

2. 2
 General Equations for a One-Dimensional Fin:
We take a consideration that,
this is a steady-state heat transfer
process. Heat flows through an
elemental cross-section.
Where,
δx = Length of the cross-section
δAS = Surface area
AC = Cross-sectional area
h = Heat transfer coefficient
Tf = Temperature of the fluid.
Convection occurs at the surface and hereby, writing down the heat-balance equation in words:
Heat Flow (into element) = Heat flow (out of element) + Heat transfer (into surroundings)
Or, QX = QX+δX + h. δAS .( T-Tf ) --------------------------------------------------------------(i)
From Fourier’s Law:
QX = -kAC
--------------------------------------------------------------------------------------------- (ii)
From Taylor’s Series, using equation (ii) we get:
QX+δX = QX +
Smyth_Works
(-kAC
) δx ------------------------------------------------------------------ (iii)
So, combining equation (i) (iii), this becomes:
(kAC
) δx - h. δAS .( T-Tf ) = 0 -----------------------------------------------------------(iv)
The left-sided term is identical to the result for a plane wall. The difference here is that the area is not
constant with ‘x’.

3. 3
So, using the product-rule to multiply out the left-sided term, gives us:
kAC .
+ k.
.
- ℎ.
.(T-Tf ) = 0
→
+
Smyth_Works
.
.
-

4. .
.
.(T-Tf ) = 0 ---------------------------------------------(v)
From figure,
AS = P.x [here, x = length of the whole fin; P = perimeter of the fin]
→
= P -----------------------------------------------------------------------------------------------------(vi)
Putting the value of equation (vi) into (v), we get:
+
.
.
-

5. .
. .(T-Tf ) = 0
→
-

6. .
. .(T-Tf ) = 0 -------[Let, the fins has uniform cross-section; so,
≈ 0]------(vii)
→
- m2Ɵ = 0 ; Let, m=

7. .
.
and, Ɵ=( T-Tf ).
It is called the general equation for one-dimensional fins.
 Solution of
- m2Ɵ = 0 :
General solution of the above equation is:
Ɵ = C1e – mx + C2e mx --------------------- [C1 C2 = Constants; depend on the boundary condition]
→ƟO = (C1+C2) -------------------------------[But, C2 = 0 and thus C1 = ƟO ]
→Ɵ = ƟO .e – mx
→ Ɵ
Ɵ = e – mx ------------------------------------------------------------------------------------------------(viii)
→
= e – mx
Therefore, Ɵ
Ɵ
=
= e – mx ; it is the solution for one-boundary condition.

8. 4
Smyth_Works
Solutions for above equation in a Tabular form
Criteria Boundary
Condition
Solution Heat Transfer
Case
001
(i) Fin is very long (x= ∞)
(ii) Tend of fin=Tfluid
(surrounding fluid)
at, x= 0 → Ɵ= Ɵ0
at, x= ∞ → Ɵ= 0
Ɵ
Ɵ
=
= e – mx q = √ℎPA .Ɵ
Case
002
(i) Fin is of finite length
(ii) Loses heat by
convection from its end.
N/A
(Holman-p43)
Case
003
Tend of fin = insulated at, x= 0 → Ɵ= Ɵ0
at, x= L → Ɵ
= 0
Ɵ
Ɵ
= !# [%(')]
!#(%') q = √ℎPA .Ɵ.tanh (mL)
 Fin efficiency =
*+,-./ 01., 2.345121
61./ 01., 2.345121
(01., 789+8 7:-/ ;1 ,2.345121)
= ŋf
If the entire area is at base temperature,
then, ŋf =
√=* .Ɵ?.,.38 (@A)
BCƟ?
 Fins : Was ist das?
In the study of heat transfer,
“A fin is a surface that extends from an object to increase the rate of heat transfer to or from
the environment by increasing convection.”
By-----
(1) increasing the temperature difference between the object and the environment,
(2) increasing the convection heat transfer coefficient,
(3) increasing the surface area of the object
……the heat transfer can be increased.
But,
Sometimes it is not economical or feasible to change the first two options. Adding a fin to an
object, however, increases the surface area and can sometimes be an economical solution to heat
transfer problems.

9. 5
 Unsteady State Condition :
When,
a solid body suddenly subjected to a change in environment,
sometimes elapse belong an equilibrium temperature.
Condition will prevail; this is called transient problem.
Mathematically,
Ɵ
ƟD
= E
DE
Smyth_Works
= F
G
K . L [MN
Σ
I
JI
O ]PQ .sinIG
R'
where, n = 1, 3, 5, …………up to (2n+1)th term.
 Equation for Lumped Heat Capacity System (LHCS) :
Lumped heat capacity system assumes that, resistance of heat conduction is so small compared to the resistance of
heat convection. Mathematically,
Rcond. Rconv.
i.e. Internal resistance of any body is negligible in comparison with the external resistance and there will be a
major temperature gradient along the surface.
Let,
A=Surface area
h=heat transfer coefficient
T∞=Ambient temperature
Bi 0.1 and T = T(t) [function of time]
Applying conservation of energy,
during time interval dt :
Ein + Eg – Eout = ΔE -----------------------(i)
Where,
Ein = Energy added,
Eg = Energy generated,
Eout = Energy removed,
ΔE = Energy change.

10. 6
Assuming that, the body in the above figure is losing heat and no heat generation is occurred.
So, Eg = Ein = 0 and therefore equation (i) becomes,
Smyth_Works
- Eout = ΔE ----------------------------------------------------- (ii)
Neglecting radiation and assuming that heat is removed by convection,
Eout = hA ( T - T∞ ) ------------------------------------------------ (iii)
For incompressible materials,
ΔE = ρCV
S -------------------------------------------------------------------- (iv)
where,
ρ = density
C = specific heat
V = volume
Evaluating values from equation (iii) (iv) into equation (ii),
- hA ( T - T∞ ) = ρCV
S
→ - hA ( T - T∞ ) = ρCV ( T )
S [ replacing , T = T - T∞ ]
→ ( T )
(T) = -

11. .S
ρCV ----------------------------------------------------------- (v)
This is the lumped-capacity equation for all bodies exchanging heat by convection and also valid
for Bi 0.1 . When, Bi 0.1 then, the initial condition : T − TJ = T−TJ
If,
Boundary condition : when, t = 0 then, T = T0
when, t 0 then, T = T∞
and, T - T∞ = T0 - T∞
Then, integrating equation (v) we get:
ln ( ZZT
ZZT
) = - #[.
]^_
→ ( ZZT
ZZT
) = e ab.c
def --------------------------------------------------------------- (vi)
Introducing new dimensionless temperature, Ɵ = ( ZZT
ZZT
)
And time constant, τ = ]^_

12. Rewritten equation (vi) will be like this, Ɵ = Lg
h .

13. 7
 Applicability of LHCS :
Biot number, Bi =
Smyth_Works
(O
i)
(O
j)
= klmnopqrlm sturuqvmpt
klmwtpqrlm sturuqvmpt =

14. (x
y)
[Bi 0.1]
where, z
= Characteristic length = L.
 Transient Heat Flow in a Semi-Infinite Solid :
Let us consider,
there is a semi-infinite solid shown in figure beside,
maintained at some initial temperature = Ti ,
suddenly lowered surface temperature = T0 .
So, the differential equation for temperature distribution
T(x,τ) is:
{Z
{ =
P.{Z
{|
----------------------------------------------- (i)
The boundary conditions are: T(x,0) = Ti
T(0,τ) = T0 [for τ 0]
Then the solution of equation (i) will be:
(,|) Z
D
= erf
R√PQ
Where, the Gauss error function is defined as:
R√PQ = R
√G ∫/R√PQ Lŋ
erf
.dŋ
Here, ŋ = Dummy variable (i.e. ŋ = x, y, z, etc)

15. 8
Smyth_Works
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