Department of Metallurgical Engineering

Heat Transfer Phenomena

Prof. Dr.: Kadhim F. Alsultani

CONDUCTION-CONVECTION SYSTEMS
The heat that is conducted through a body must frequently be removed (or delivered)
by some convection process. For example, the heat lost by conduction through a
furnace wall must be dissipated to the surroundings through convection. In heat-
exchanger 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 surroundings by
convection. Obviously, an analysis of combined conduction-convection systems is
very important from a practical standpoint.

Fin equation:
Consider the one-dimensional fin exposed to a surrounding fluid at a temperature T ∞
as shown in figure . The temperature of the base of the fin is T b. We approach the
problem by making an energy balance on an element of the fin of thickness dx as
shown in the figure. The defining equation for the convection heat-transfer coefficient
is recalled as q=hA (T w−T ∞ ) where the area in this equation is the surface area for
convection. Let the cross-sectional area of the fin be (Ac) and the perimeter be (P).
dT
q x =−kA c
Then the energy quantities are Energy in left face = dx ,
2
dT dT d T
q x +dx=−kA c ]x +dx=−kA c ( + 2 dx )
Energy out right face = dx dx dx

Energy lost by convection = hPdx (T−T ∞ )

Energy in left face= Energy out right face+ Energy lost by convection
d2 T 1
kA 2
dx=hp ( T− T ∞) dx ]÷
dx kA∗dx
d 2 T hp
= ( T −T ∞ )
dx 2 kA
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani

hp
let [θ=T −T ∞ ] and [ m2 = ]
kA
dT d (θ+T ∞ ) dθ
= =
dx dx dx
d2 T d2 θ
2
= 2
dx dx
d2 θ 2
2
=m θ (governor equation )
dx
d 2 2 2 2
=D D θ=m θ ( D −m )θ=0 D=±m
dx
θ=∑ CeDx
θ=C 1 e mx +C 2 e−mx [ general solution of governor equation ]
Can be applied to three types of fins, depending on physical situation
CASE 1 The fin is very long, and the temperature at the end of the fin is
essentially that of the surrounding fluid.
CASE 2 The end of the fin is insulated
CASE 3 The fin is of finite length and loses heat by convection from its end.
For case1 :- The boundary conditions are

at at [x = 0] θ(0 )=θ b=T b −T ∞

x =L=∞
Tfin tip=T∞ i.e (TL =T∞) then,
x=L
θ( L)=T L−T ∞=0 cond=conv
x=0 θ=θb
Appling this boundary conditions in general solution equation, yields
θ( 0)=C1 e-m ( 0)+C2 e m (0 )=C 1 +C 2 =θb

at x =L=∞ θ( L )=0=C 1 e m (∞)+ C2 e−m (∞)=C 1 ( 0 )+ C2 (∞ ) C2=0

θb =C 1 e−m( 0 )=C 1=θ b

θ( x )= θb e−mx
hp
θ( x )
=
T −T ∞ −mx
=e =e
−x
√ kA c

And the solution becomes θb T b −T ∞

For case2:- negligible heat loss from the fin tip (insulated fin tip ,qfin tip=0 )
the boundary conditions are
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani

at x = 0 θ(0 )=θ b 　
θ( 0)=C1 e-m ( 0)+ C2 e m (0 )=C 1 +C 2 =θb .. ..( a)

at x=L
∂θ
qfin(tip )=0 =0 |x= L
∂x
-mL m ( L)
0=−mC 1 e +mC2 e ]÷m.. . .. ..(b )

Solving equations (a and b) for C1 and C2 the solution is obtained as

θ( x ) e−mx e mx
= +
θb 1+e−2mL 1+e 2mL
θb θb
where C1= C2= then
1+ e −2mL
1+e 2mL
θ( x ) e−m ( x-L) e
m( x-L)
e
−m( x-L )
+e
m( x-L )
= −mL mL + −mL mL = −mL mL
θb e +e e +e e +e

θ( x ) e m( L-x )+e -m (L-x )

=
Or θb e mL+e−mL a-Actual fin with convection at the tip

θ( x ) T ( x )−T ∞ cosh [m( L−x )]

θb
= =
T b−T ∞ coshmL
qfin Ac
qfin p
Case 3 the boundary conditions are
θ(0 )=θ b
at x = 0 　
qconduction(x=L)= qconvection(x=L)
Lc
|
∂θ ∂θ
−kAc ¿|x=L =hAc θ|x=L ¿−k |x=L=h(T L−T ∞ )|x=L ¿¿
∂x ∂x
|
T ( x )−T ∞ coshm( Lc−x )
=
T b −T ∞ coshmLc

All of the heat lost by the fin must be conducted into the base at x = 0.
Using the equations for the temperature distribution, we can compute the
dT
q=−kA ]
heat loss from dx x=0
An alternative method of integrating the convection heat loss could be
used:
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
L L
q=∫0 hP(T −T ∞ )dx=∫0 hPθ dx
In most cases, however, the first equation is
easier to apply.
−m( 0)
The heat flow for case 1 is
q=−kA (−mθb e )=√ hPkA c θ b
1 1
q=−kAθ b m( − )=√ hPkA c θ b tanhmL
For case 2, 1+ e−2mL 1+e +2mL

qAkPh
= √ Lmhnat
cθ b c For case 3

- :Fin efficiency

lautcA derrefsnart
taeh
η =f
taeHhcihw dluowderrefsnart
eb eritne
(fi nifaeraerew ta
esab pmet )
qfin
ηf =
qfinmax the fin efficiency for case1 infinite long fin
√ hPkAc (T b −T ∞ ) = 1 hp 1
η f ( verylongfin )=
hA fin (T b−T ∞ )
qfin
√=
L kAc mL
√ hPkAθ b tanhmL tanhmL
η f ( insulatedtipfin )= = =
q fin,max hPLθ b mL

where (z)is the depth of the fin and (t) is the thickness. Now, if the fin is
sufficiently deep, the term (2z) will be large compared with (2t ), and

2hz 2h
mL=
√ √
ktz
L=
kt
L

2h 3/2
Multiplying numerator and denominator by (L1/2 ) gives
(Lt) is the profile area of the fin, which we define as [Am = Lt] So that
mL=
√ kLt
L

2h 3/2
mL=
√
kAm
L
A corrected length Lc is then used in all the equations which apply for the case
of the fin with an insulated tip.
Ac z*t
Lc=L+ A c=z*t p=2( z+t ) if z >> t Lc=L+
p 2z
t
Lc=L+
2 ,The error which results from this approximation will be less than
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani
2
ht 1/2 1 πd /4
( ) ≤ Lc=L+ =L+d/4
8 percent when 2k 2 , πd
where Af in is the total surface area of the fin. Since[Afin =pL] for fins with
constant cross section.

Example (1) An aluminum fin [k = 200 W/m·℃] 3.0 mm thick and 7.5 cm long
protrudes from a wall. The base is maintained at 300℃, and the ambient
temperature is 50℃ with h = 10 W/m2·℃. Calculate the heat loss from the fin
per unit depth of material.
Solution:- We may use the approximate method of solution by extending the
fin a fictitious length t/2 and then computing the heat transfer from a fin with
insulated tip.

hP h(2z+2t )
Lc=L+t/2=7 . 5+0 . 15=7 . 65cm 　, m=
√ √
kA
=
ktz
=5 .774

q=( tanhmLc ) √ hPkA θb , A=(1)(3×10−3 )=3×10−3 m 2 [4 .65 in2 ]

And q = (5.774)(200)(3*10-3)(300-50)tanh[(5.774)(0.0765)] = 359 W/m
[373.5 Btu/h·ft]

Circumferential fins of rectangular profile

L= r2 − r1 Lc = L +t/2 r2c= r1 + Lc Am= tLc

Example (2) Aluminum fins 1.5 cm wide and 1.0 mm thick are placed on a 2.5-
cm-diameter tube to dissipate the heat. The tube surface temperature is 170℃,
and the ambient-fluid temperature is 25℃.Calculate the heat loss per fin for h =
130 W/m2·℃ .assume k=200W/m.oC for aluminum.
Solution. For this example we can compute the heat transfer by using the fin-
efficiency curves . The parameters needed are
Lc = L + t / 2 = 1.5 + 0.05 = 1.55 cm , r1 = 2.5/2 = 1.25 ,r2c = r1 + Lc = 1.25 +
1.55=2.80 cm
r2c/ r1 = 2.80 / 1.25 = 2.24, Am = t(r2c - r1 ) = (0.001)(2.8 – 1.25)(10-2) = 1.55*10-5
m2
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani

h 1/2 130
L3/2
c (
kA m
From figηf = 82 percent.
√
) =(0 .0155 )3/2
(200)(1.55×10−5 )
=0 .396

qf=hAf(Tb-T∞) ,Af=surface area Af=2π(rc22-r1)

q max =2 π ( r 22c−r 21 ) h( T 0 −T ∞ )=2 π ( 2. 82 −1 .25 2 )( 10−4 )( 130)( 170−25)=74 .35W [ 253. 7Btu/h ]
q act =(0 . 82)(74. 35)=60 . 97W [208 Btu/h ]
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani

Tb Tb

FIN EFFECTIVENESS :- Ab
qfin
qfin Ab
The performance of the fins is judged on the basic of the enhancement in heat

q withfin η fin A fin hθb

ε fin= =
transfer relative to the no fin case q withoutfin hA b θb For the insulated-tip
fin. Where Afin=pL is the total surface area of the fin and Ab=A is the base area.
ɛfin=1 indicated that the addition of fins to the surface does not affect heat transfer.
ɛfin<1 indicated that the fin actually acts as insulation slowing down the heat transfer
from the surface. This situation can occur when fins made of low thermal conductivity
materials are used .
ɛfin>1 indicated that the fins are enhancing heat transfer from the surface.

THERMAL CONTRCT RESISTANCE

A B
T 1 −T 2 A T 2 A −T 2 B T 2 B−T 3 q q
q=k A A = =k B A
Δx A 1/hC A ΔxB
XA XA
T 1−T 3 (a)

q= T
or　　 Δx A /k A A+1/hc A +Δx B /k B A
T1
T2A
1/hc A is called the thermal contact
T2B
resistance and hc is called the contact
T3
coefficient. (b)
1 2 3 X

Example :Two 3.0cm diameter 304

stainless-steel bars, 10 cm long, have ground surfaces and are exposed to air
with a surface roughness of about 1 μm. If the surfaces are pressed together
with a pressure of 50 atm and the two-bar combination is exposed to an overall
temperature difference of 100℃, calculate the axial heat flow and temperature
drop across the contact surface.
Department of Metallurgical Engineering
Heat Transfer Phenomena
Prof. Dr.: Kadhim F. Alsultani

Solution:- The overall heat flow is subject to three thermal resistances, one
conduction resistance for each bar, and the contact resistance. For the bars,

Δx (0. 1)( 4 )
Rth = = =8 . 679
kA (16 .3 )π (3×10−2 )2 ℃/W
−4
1 (5 .28×10 )( 4 )
Rc = = =0 .747
From Table the contact resistance is hc A π (3×10−2 )2 ℃/W
The total thermal resistance is thereforeΣRth = (2)(8.679) + 0.747 = 18.105
ΔT 100
q= = =5 . 52W [18 . 83 Btu/h ]
∑ R 18 . 105
and the overall heat flow is th
The temperature drop across the contact is found by taking the ratio of the
contact resistance to the total thermal resistance:
Rc (0 . 747)(100 )
ΔT c = ΔT = =4 . 13
∑ R th 18 .105
℃　　[39.43℉]
In this problem the contact resistance represents about 4 percent of the total
resistance.