3. Introduction
Whenever the available surface is found inadequate to transfer the required
quantity of heat with available temp. drop & convective heat transfer
coefficient heat transfer coefficient , extended surface/fin are use.
In many Engineering Application , large quantities of heat have to be
dissipated from small areas.
The fins increase the effective area of the surface there by increasing the
heat transfer by convection.
3
5. Common application of finned surface
Cooling of electronics components
Condensers & economisers of thermal power plant
Dry type cooling tower
Air cooled cylinders of Compressor , I.C. Engines
Electric motors and Transformers .
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7. Heat flow through ‘Rectangular fins’
𝑑2ߐ
𝑑𝑥2 - 𝑚2𝛳 = 0
Where , m =
𝑝 ∗ ℎ
𝐴𝑐𝑠 ∗ 𝑡
Above equation is general form of the heat dissipation via fin to surroundings.
θ ( x ) ≡ T ( x ) − T∞
The general solution is of the form
θ ( x ) = C 1 e mx + C 2 e − mx
T (0 ) = T∞
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8. 8
Heat conduction in to the element at plane x,
𝑄𝑥 = -k 𝐴𝑐𝑠 (
𝑑𝑡
𝑑𝑥
)𝑥
Heat conducted out of element at plane (x+dx)
𝑄𝑥+𝑑𝑥 = -k 𝐴𝑐𝑠 (
𝑑𝑡
𝑑𝑥
)𝑥+𝑑𝑥
Heat Convection from fin to environment ,
𝑄𝑐𝑜𝑛𝑣 = h A (t – ta )
= h ( p dx)(t-ta)
Applying energy balance
𝑄𝑥 = 𝑄𝑥+𝑑𝑥 + 𝑄𝑐𝑜𝑛𝑣
-k 𝐴𝑐𝑠 (
𝑑𝑡
𝑑𝑥
)𝑥 = -k 𝐴𝑐𝑠 (
𝑑𝑡
𝑑𝑥
)𝑥+𝑑𝑥 + h( p dx)(t-ta)
Expended (
𝑑𝑡
𝑑𝑥
)𝑥+𝑑𝑥 according to taylor series
Neglecting higher order term ,
k 𝐴𝑐𝑠 (
𝑑2𝑡
𝑑𝑥2)𝑥 dx = h (p dx)(t-ta)
𝑑2𝑡
𝑑𝑥2 -
ℎ 𝑝
𝑘 𝐴𝑐𝑠
(t-ta) = 0
9. 9
According to above Equation ;
𝑑2ߐ
𝑑𝑥2 - 𝑚2𝛳 = 0
Where , m =
𝑝 ∗ ℎ
𝐴𝑐𝑠 ∗ 𝑡
Solution of above liner & homogenious second order differential eqn.
θ ( x ) = C 1 e mx + C 2 e − mx
t - ta = C 1 e mx + C 2 e − mx
1
st
boundry condition x=0 , t=t0
t0 - ta = C 1 e m(0) + C 2 e − m(0)
11. Heat dissipation from infinitely long fin (
𝓁→ 𝛼)
Heat dissipation can be determined in either of the two ways :
(1) By considering the heat which is transmitted by convection from the surface of fin to the
surrounding ,or
(2) By considering the heat flow across the base by conduction.
Considering the second method.
Qfin = k * Acs * m (to – ta)
Put ; m =
𝑝 ∗ ℎ
𝐴𝑐𝑠 ∗ 𝑡
Qfin = k × Acs (to – ta)
𝑝 ∗ ℎ
𝐴𝑐𝑠 ∗ 𝑡
Qfin = 𝑝 × ℎ × Acs × 𝑘 × (to – ta)
11
12. Heat dissipation from a fin insulated at
the tip
X=0,ߐ=ߐ0 &
X=L,𝑑𝑡
𝑑𝑥=0
θ ( x ) = C 1 e mx + C 2 e − mx
Applying 1st Boundry condition
ߐ0 = c1 + c2 &
𝑑𝑡
𝑑𝑥 = m C 1 e mx +(-m) C 2 e − mx
Applying 2nd Boundry condition
C 1 e mx - C 2 e − mx =0
Now solving for C1 & C2;
C1 = 𝜃0
e − mx
e mx + e − mx
C2 =
e − mx
e mx + e − mx
𝜃
0 12
13. Putting the value of C1 & C2 in equation ;
ߐ =
e − mx
e mx + e − mx
e mx 𝜃0 +
e mx
e mx + e − mx
e − mx 𝜃0
𝜃
𝜃0
=
cos ℎ { 𝑚 𝑙−𝑥 }
cos ℎ (𝑚𝑙)
Rate of heat dissipation
𝑄𝑓𝑖𝑛 = -k 𝐴𝑐𝑠
𝑑𝑡
𝑑𝑥
𝑑𝑡
𝑑𝑥
= -m (to-ta) tanh (ml)
Substitute value of m ;
Qfin = 𝑝 × ℎ × Acs × 𝑘 × (to – ta) × tanh(m l )
13
14. Efficiency and effectiveness of fin
The fins are used to increase the heat transfer rate by increasing the heat
transfer area.
The following factors need consideration for the optimum design of fins;
Total cost
Manufacturing difficulties
Complexity associated with fins
Weight (in case of aeroplanes & automobile)
Pressure drop caused
Space consideration
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15. Fin effectiveness (∈fin)
The fin effectiveness is defined as the ratio of fin heat transfer rate to that of
which would occur from the surface on which the fin was attached.
∈fin = ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑤𝑖𝑡ℎ 𝑓𝑖𝑛
ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑖𝑛
=
𝑄 𝑤𝑖𝑡ℎ 𝑓𝑖𝑛
𝑄 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑓𝑖𝑛
consider fin with infinite length,
∈fin =
𝑝𝑘
𝐴𝑐𝑠 ℎ
15
16. Efficiency of fin (𝜂fin)
The efficiency of a fin is defined as the ratio of the actual heat transferred by
the fin to the maximum heat transferable by fin, if entire fin area is at base
temperature.
𝜂fin = Q𝑓𝑖𝑛
Q𝑚𝑎𝑥
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17. Efficiency of fin (𝜂fin)
a) For a infinitely long fins ,
𝜂fin =
1
𝑚𝑙
b) For a fin with finite length and top of the fin is insulated ,
𝜂fin =
tanh(𝑚𝑙)
(𝑚𝑙)
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