CHAPTER 2

CONDUCTION

2.1 PARTIAL DIFFERENTIAL EQUATION CONCEPT

2.1.1 Derivation of Partial Differential Equation

From energy balance, considered only in x-direction.

Heat
Heat transfer Heat transfer Energy change
generation
rate by - rate by + =
due to the rate of element
conduction at conduction at
volume of
‘x’ ‘x + Δx’
element

  E system
Q x - Q x+Δx + Egen = (2.1)
t

Where

Egen = ėgen (W/m³) x Volume (m³) = ėgen x AΔx [W] (2.2)

and

ΔEsystem = ΔUsystem = m C[Tt+Δ t – Tt].

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 = ρ V C[Tt+Δt – Tt]

= ρ AΔx C[Tt+Δt – Tt] (2.3)

Substitute Eq 2.2 and 2.3 in Eq 2.1.

 
Q X  Q X  X  e gan Ax  AX C
Tt t  Tt 
t

Dividing the equation by AΔx.

 
Q X  Q X  X T  Tt 
 e gen  C t  t
Ax t
Or

 


1  Q X  Q X  X   e gen  C Tt  t  Tt  (2.4)
A x  t
 

Taking the limit Δx 0 , Δt 0.

 
 
Q  Q  Q   T 
lim x0  X  X X     KA  (2.5)
 X
 X X  X 
 

and

lim x 0
Tt  t Tt  T
 (2.6)
t t

 Substitute Eq 2.5 and 2.6 in Eq 2.4.

1   T  T
  kA   e gen  C
A  X X  t

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A is constant, thus:

  T  T
X k X   e gen  C t

k is assumed to be constant as well, thus:-

  2T  T
k 2 
 e gen  C (2.7)
 X  t

Divide Eq 2.7 by k

  2T  e gen C T
 2 
 X  K K t

K
Where thermal diffusivity, 
C

Thus,

PDE for Plane wall is as following,

  2T  e gen 1 T
 2 
 X  K  t

Three (3) specified condition:-


(i) Steady state condition ; 0
t

  2T  e gen 1 T
 2   = Partial Differential Equation (PDE).
 X  K  t

 d 2T  e gen
 2 0 d = Ordinary Differential Equation (ODE)
 dX  K

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 
(ii) Transient condition (unsteady state);  0 , no heat generation
t

  2 T  1 T
 2
 X   t


(iii) Steady state, no heat generation;  0 , e gen  0
t

 d 2T 
 20
 dX 

2.1.2 Partial Differential Equation by Case

Assumptions.

1. One (1) dimensional heat conduction.

2. k is assumed to be constant.
No Case Partial Differential Equation

1. Plane Wall

  2T  egen 1 T
 X 2   K   t
 

Cylinder

1   T  egen 1 T
r  
r r  r  k  t
2.

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3. Sphere

r2 + =

2.2 THERMAL RESISTANCE CONCEPT

1. Analogy of ohm’s Law.

Electrical circuit Heat transfer circuit

I Q

Re Rt

V1 V2 T1 T2

V
I Electric Thermal
Re resistance resistance
V  V2
I 1
Re

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 2. If the circuit consists of more than one thermal resistance, [R1, R2, R3,
Rn], the total resistance needs to be found.

i. Series circuit

R1 R2 R3

Rt = R1 + R2 + R3

ii. Parallel Circuit

R1

R2

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 2.2.1 Thermal Resistance for Conduction - by Case.

Case Thermal Resistance , Rcond

1. Plane Wall Q = - k A dT
dx

Q ∫ dx = -k A ∫ dT

QL = -kA [T2 – T1]

Q = T1 – T2
L/kA

L
Thus ; Rcond,plane = (°C/W )
KA

2. Cylinder Q = - k A dT
dr

Q = - k [2πrL] dT
dr

Q ∫ 1 dr = -2 πkL ∫ dT
r

Q [ln r2/r1] = -2 πkL [ T2-T1]

Q = T1 – T2
[ ln r2/r1 ]
2 πkL

Thus;

ln( r 2 / r1)
Rcond,cylinder = (°C/W )
2kl

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 3. Sphere. Q = - k A dT
dr

Q = - k [4πr²] dT
dr

Q ∫ 1 dr = -4 πk ∫ dT
r²

Q [ -1 + 1] = -4 πk [T2 – T1]
r2 r1

Q [ -1 + 1 ] = 4 πk [T1 – T2]
r2 r1

Q [r2 – r1] = 4 πk [T1 – T2]

r1r2

Q = T1 – T2
[r2 – r1]
4 πk r1r2

Thus ;

(r 2  r1)
Rcond, sphere = (°C/W )
4kr1r 2

2.2.2 Thermal Resistance for Convection – by Case.

Q = h A [ Ts - T∞ ]

= [ Ts - T∞ ]
1 / hA R conv

Thus,

1
Rconv = (°C/W )
hA

Where :-
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 A cylinder = 2πrL = πDL

A sphere = 4πr² = πD

2.2.3 Thermal Resistance for Radiation – by Case.

Q = σ A [ Ts4 – Tsurr4 ] = hrad A [ Ts – Tsurr]

= Ts – Tsurr
1/ hrad A

Thus,

1
Rrad= (°C/W )
hradA

where

hrad = σ  [ Ts² + Tsurr²] [Ts + Tsurr]

(σ = 5.67 x 10-8 W/m2K4). Properties, emissivity

A cylinder = 2πrL = πDL

A sphere = 4πr² = πD²

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2.3 IMPORTANT TO BE REMEMBERED

 Partial Differential Equation Concept

  2T  egen 1 T
 2  *** ***** FOR PLANE WALL
 X  K  t

1   T  e 1 T
 r   gen  *** ***** FOR CYLINDER
r r  r  k  t

r2 + = *** ***** FOR SPHERE

 Thermal Resistance Concept

L
Rt cond , plane 
kA
r  r1
Rt cond , sphere  2
4kr1r2
r2
ln
r1
Rt cond , cylinder 
2kL
1
Rt conv 
hA
1
Rt rad 
hrad A

where hrad = σ[ Ts² + Tsurr²] [Ts + Tsurr]

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Example 2.1

Consider a large plane wall of thickness L = 0.2m, thermal conductivity, k =

1.2W/m , and surface area, A = 15m2. The two sides of the wall are maintained

at constant temperature of T1 = 120 , and T2 = 50 , respectively. Determine:

a) The variation of temperature within the wall and the value of temperature
at 0.1m
b) The rate of heat conduction through the wall under steady conditions, Q

Solution:

T1 = 120

T2 = 50

L = 0.2m

k = 1.2W/m

A = 15m2

T(x) = ?

T(x=0.1) = ?

=?

Assumptions:

1. 1 dimensional heat conduction

2. Steady state condition, ,

3. No heat generation, ,

4. k is assumed to be constant

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(1) Partial Differential Equation Concept

+ =

=0

= C1

T(x) = C1x + C2

Applying boundary conditions

At x = 0, T(x) = T1

T(x) = C1x + C2

T1 = C1(0) + C2

C2 = 120

At x =L, T(x) = T2

T(x) = C1x + C2

T2 = C1(L) + C2

50 = C1(0.2) + 120

C 1 = -350

Thus,

a) Temperature variation, T(x)

T(x) = C1x + C2

= -350x + 120

Temperature at x= 0.1m

T(x= 0.1) = -350 (0.1) +120

= 85

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 b) Rate of heat conduction,

Temperature gradient,

T(x) = -350x + 120

= -350

Thus, = -kA

= - 1.2 (15)(-350)

= 6300 W (

2. Thermal Resistance Concept

where

Rt = = = 0.0111

Thus,

=
To find:

= Temp distribution / temp gradient –Diff equation concept

Heat transfer rate – Diff equation & Thermal resistant

= 6300W concept

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Example 2.2

Consider a steam pipe of length L = 20m, inner radius r1 = 6cm, outer radius r2 =
8cm, and thermal conductivity k = 20W/m . The inner and outer surfaces of the

pipe are maintained at average temperature of T1 = 150 , and T2 = 60 ,

respectively. Obtain a general relation for the temperature distribution inside the
pipe under steady conditions, and determine the rate of heat loss from the steam
through the pipe.

Solution:

T1 = 150

T2 = 60

r1 = 6cm

r2 = 8cm

L = 20m

k = 20W/m

T(r) = ?

=?

Assumptions:

1. 1 dimensional heat conduction

2. Steady state condition, ,

3. No heat generation, ,
4. k is assumed to be constant

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1. Partial Differential Equation Concept

r + =

r =0

r =0

r = C1

T(r) = C1 ln r + C2

Applying boundary conditions

At r = r1, T(r) = T1

T(r) = C1 ln r + C2

T1 = C1 ln r + C2

150 = C1 ln [6x10-2] + C2

150 = -2.8134 C1+ C2 -------------------------------------------------------- (i)

At x =L, T(x) = T2

T(r) = C1 ln r + C2

T2 = C1 ln r2 + C2

60 = C1 ln [8x10-2] + C2

60 = -2.526 C1 + C2 ---------------------------------------------------------- (ii)

Eq (i) - (ii)

90 = -0.2874 C1

C1 = -313.152

and
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60 = -2.526 [ -313.152 ] + C2

C2 = -731

Thus,

a) Temperature distribution, T(r)

T(r) = C1 ln r + C2

= - 313.152 ln r + 731

b) Rate of heat loss,

Temperature gradient,

T(r) = - 313.152 ln r + 731

Thus,

= -kA

= -k [2

= - 20 [2 [- 313.152]

= 787 kW

2. Thermal Resistance Concept

Rt = ln (r2/r1)/2kL

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 =

= 1.1445 x 10-4

Thus,

= 786.4 kW

Example 2.3

Consider a spherical container of inner radius r1 = 8cm, outer radius r2 = 10cm,

and thermal conductivity k = 45W/m . The inner and outer surfaces of the

container are maintained at constant temperatures of T1 = 200 , and T2 = 80 ,

respectively, as a result of some chemical reactions occurring inside. Obtain a
general relation for the temperature distribution inside the shell under steady
conditions, and determine the rate of heat loss from the container.

Solution:

T1 = 200

T2 = 80

r1 = m

r2 =

k = 45W/m , T(r) = ?, =?

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 Assumptions:

1. 1 dimensional heat conduction

2. Steady state condition, ,

3. No heat generation, ,

4. k is assumed to be constant

1. Partial Differential Equation Concept

r2 + =

r2 =0

r2 =0

r2 = C1

T(r) = + C2

Applying boundary conditions

At r = r1, T(r) = T1

T(r) = + C2

T1 = + C2

200 = + C2

200 = -12.5 C1+ C2 --------------------------------------------------------- (i)

At x =L, T(x) = T2

T(r) = + C2

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T2 = + C2

80 = + C2

80 = -10 C1 + C2 --------------------------------------------------------- (ii)

Eq (i) - (ii)

120 = -2.5 C1

C1 = -48

and

80 = -10 [ -48 ] + C2

C2 = -400

Thus,

a) Temperature distribution, T(r)

T(r) = + C2

= + (-400)

= - 400

Temperature gradient,

T(r) = - 400

=-

Rate of heat loss,

= -kA

= -k [4

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 = - 45 [4 [-48] = 27.1 kW

2. Thermal Resistance Concept

Rt =

=4.4204 x 10-3

Thus,

= 27.1 kW

Example 2.4

Consider a 0.8m high and 1.5m wide glass window with the thickness 8mm and
thermal conductivity k = 0.78W/m Determine the steady rate of heat transfer

through this glass window and the temperature of its inner surface for a day
during which the room is maintained at 20 while the temperature of the outdoor

is -10 . Take the convection heat transfer coefficients on the inner and outer

surfaces on the window to be h1 = 10W/m2 and h2 = 40W/m2 , which includes

the effects of radiation.

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Solution:

A=

L= m

k = 0.78 W/mk

T 1= 20 , h1 = 10W/m2

T 2= -10 , h2 = 40W/m2

T1 = ?

=?

Assumptions:

1. 1 dimensional heat conduction

2. Steady state condition, ,

3. No heat generation, ,
4. k is assumed to be constant

2. Partial Differential Equation Concept

+ = =0

=0

= C1

T(x) = C1x + C2, = C1

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Applying boundary conditions

At x = 0, T(x) = T1

T(x) = C1x + C2

T1 = C1(0) + C2

T1 = C2

and

h1[T 1- T1] = -k

10[20- T1] = - 0.78 C1

200 - 10T1 = - 0.78 C1

- 0.78 C1 + 10T1 =200

- 0.78 C1 + 10C2 =200 ------------------------------------------------------ (i)

At x =L, T(x) = T2

T(x) = C1x + C2

T2 = C1( ) + C2

T2 = C1 + C2

and

h2[T2- T 2] = -k

40[T2 + 10] = - 0.78 C1

40T2 + 400 = - 0.78 C1

40 [ C1 + C2 ] + 400 = - 0.78 C1

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0.32 C1 + 40C2 + 400 = - 0.78 C1

-1.1 C1 – 40 C2 = 400 ------------------------------------------------------ (ii)

From (ii)

C1 =

Substitute C1 in (i)

-0.78 + 10 C2 = 200

0.7091 [400 +40 C2] + 10 C2 = 200

283.64 + 28.364 C2 + 10 C2 = 200

C2 = -2.2

and

C1 =

= -283.64

Thus,

Temperature distribution, T(x)

T(x) = C1x + C2

= -283.64x + (-2.2)

= -283.64x – 2.2

Temperature gradient,

= - 283.64

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Rate of heat loss,

= -kA

= - 0.78 [1.2 [-283.64]

= 265 W

Inner temperature, [T(x) = 0]

T(x) = C1x + C2

T(x=0) = -283.64 (0) + (-2.2)

= -2.2

3. Thermal resistance concept

R1 = = = 0.0833

R2 = = =

R3 = = = 0.02083

Rt = R1 + R2 + R3

= 0.0833 + + 0.02083

= 0.1128

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Thus,

= 266 W

266 =

T2 = - 2.2

Example 2.5

3 layers of materials are used to build the store as shown in

figure. The fireclay bricks with thickness, t = 3m and thermal
conductivity, kb = 1.34 W/moC are covered with inside and
outside of the store with plaster, t = 0.3m and kp = 1.40 W/moC.
The temperature and convection heat transfer coefficient inside
the store are T ,1 = 28˚C and h1 = 10 W/m2˚C whereas outside T  2 = 32˚C and
h2 =40 W/m2˚C. The store shape is semi-sphere with r1=5m. Determine:-

a) heat transfer rate

b) inner and outer surface temperature; Ti , To

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Solution

t b = 3m

k b= 1.34 W/m˚C

t p = 0.3m

k p= 1.40 W/m˚C

T ,1 = 28˚C

T , 2 = 32˚C

h1 = 10 W/m2˚C

h2 =40 W/m2˚C

r1=5m

4r 2
A= = 2πr2 (semi sphere)
2

Assumptions:

1. 1 dimensional heat conduction

2. Steady state condition, ,

3. No heat generation, ,

4. k is assumed to be constant

5. Radiation is negligible.

Where,

1 1 1
Ri  = = = 6.365 x 10-4˚C/ W
hi A hi (2ri ) 10(2 )52
2

r2  r1 1 r r
  2 1
4K r1 r2 2 2Kr1 r2

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 r2  r1 5 .3  5
R1  = = 1.287 x 10-3˚C/W
2 K p r1r2 2 1.4 (5.3)(5)

r3  r2 8.3  5.3
R2   = 8.1x 10-3˚C/W
2K b r2 r3 21.34 8.35.3

r4  r3 8.6  8.3
R3   = 4.78x 10-4˚C/W
2 K p r4 r3 2 1.40 8.6 8.3

1 1 1
Ro  = = = 5.38 x 10-5˚C/ W
h2 A h2 (2r4 ) 40(2 )8.6
2 2

Thus,

Rt = Ri + R1+ R2+ R3+ R0

= 6.365 x 10-4 + 1.287 x 10-3 + 8.1x 10-3 +4.78x 10-4 + 5.38 x 10-5

= 0.010555 ˚C/ W

T ,1  T , 2 28  32
=   379 W (outside to inside)
Rt 0.01055

T ,1  Ti To  T , 2
= , Ti  28.2 ˚C and = , T0  31.98 ˚C
Rt R0

Example 2.6

A 2 kW resistance heater wire whose thermal conductivity is k = 15 W/m˚C has a

diameter of D= 4mm and a length of L= 0.5m, and is used to boil water. If the
outer surface temperature of the resistance wire is Ts = 105˚C, determine :

a) Temperature distribution, T(r)

b) Maximum temperature, Tmax

c) Sketch temperature profile.

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Solution

P = 2000W

K = 15 W/m˚C

r = 0.002m

L = 0.5m

Ts = 105˚C

Assumptions:

1. 1 dimensional heat conduction

2. Steady state condition,

3. With heat generation

(1) Partial differential equation concept

1   T  egen 1 T
r  
r r  r  k  t

1 d  dT  e
 r    gen
r dr  dr  k

d  dT  egen r
r  
dr  dr  k

dT egen r 2
r =  +C1........................................................ (i)
dr 2k

Applying boundary condition

dT
at r = 0, =0
dr

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Thus substitute in (i)
0 = 0 + C1
C1 = 0

Therefore,

dT egen r 2
r = 
dr 2k
dT egen r
=
dr 2k
egen r 2
T(r) =   C2 ................................................................ (ii)
4k

Applying boundary condition

at r = 0.002m, T(r)= 105

Thus substitute in (ii)

egen 0.002 
2

105 =   C2
415

where,
P
egen =
V
2000
=
 (0.002 ) 2 (0.5)
= 0.318 x 109 W/m3

Thus

0.318 x 109 0.002 

2
C2 = 105 +
154
= 105 + 21.2
= 126

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a) Temperature distribution, T(r)

egen r 2
T(r) =   C2
4k

0.318 x 10 9 2
=  r  C2
415 

0.318 x 109 2
=  r  126
415 
= -5.3 x 106 r2 + 126

b) Maximum temperature, Tmax

Tmax is at r = 0m

Tmax = 126˚C

c) Sketch temperature profile.

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Example 2.7

Consider the base plate of a 1200 W household iron that has a thickness of
L = 0.5cm, base area of A = 300cm2, and thermal conductivity of k = 15 W/m˚C.
The inner surface of the base plate is subjected to uniform heat flux generated by
the resistance heaters inside, and the outer surface loses heat to the
surroundings at T  = 20˚C by convection. Taking the convection heat transfer
coefficient to be h = 80 W/m2˚C and disregarding heat loss by radiation, obtain an
expression for the variation of temperature in the base plate, and evaluate the
temperatures at the inner and the outer surfaces.

Solution

= 1200W

L = 0.5 x 10-2

A = 300cm2

T = 20 ˚C

h = 80 W/m2˚C

T(x) = ?

T1 = ?

T 2= ?

Assumptions:

1. 1 dimensional heat conduction

2. Steady state condition, ,

3. No heat generation, ,
4. k is assumed to be constant
5. Radiation is negligible
6. Heat is uniformly being transferred.

2-31 Bachelor Degree in Mechanical Engineering

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(1) Partial Differential Equation Concept

+ = =0

=0

= C1

T(x) = C1x + C2, = C1

Applying boundary conditions

At x = 0, T(x) = T1

T(x) = C1x + C2

T1 = C1(0) + C2

T1 = C2

and,

d T 0 
= k
dx

where,
Pe 1200
q   40000 W/m2
A 300  10 4

dT 0 
 C1
dx
Thus,

40000 = -15 C1
40000
C1 =
 15
= -2666.67

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At x=L , T(x)=T2

T(x) = C1 X + C2

T2 = C1 (0.5X10-2)+C2

T2= 0.5X10-2(-2666.67)+C2

T2= -13.33 + C2

And

qcond = qconv

-k

-k C1 = h

-15 [-2666.67]=80[(-13.33+C2)-20]

40000 = 80 [-33.33+C2]

C2 = 533.33

Thus,

Temperature distribution, T (X)

T(x) = C1x + C2

= -2666.67x + 533.33

Inner temperature , T(x=0)=T1

T1 = 533.33 oC

Outer temperature , T(x=L)=T2

T2 = -2.666.67 (0.5X10-2) + 533.33

=520 oC

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Counter check

Heat transfer rate, Q

Q = -kA

Where,

= -2666.6

Thus,

Q = -15 (300X10-4)(-2666.67)

=1200W

(2) Thermal Resistance Concept

T∞, h

Q= Pe = 1200 W

Q1 = Q2

Where

R1 = = = 0.0111 ̊ c /W

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R2 = = = 0.417 ̊ c /W

Thus,

Q=

1200 =

T1= 533.7 oC

And

Q = (T2-T)/R2

1200 = (T2-20)/0.417

T2 = 520 oC

Example 2.8

A 3m internal diameter spherical tank made of 2 cm thick stainless steel (k=15

W/m oC) is used to store iced water at T∞1 = 0 oC. The tank is located in a room
whose temperature is T∞2 = 22 oC. The walls of the room are also at 22 ̊C. The
outer surface of the tank is black and heat transfer between the outer surface of
the tank and the surroundings is by natural convection and radiation. The
convection heat transfer coefficient at the inner and the outer surfaces of the tank
are h1= 80 W/m2 ̊C and h2= 10 W/m2 C,
̊
respectively. Determine the rate of heat
transfer to the iced water in the tank.

Solution:

T∞,2, h2 K = 15 W/m

r1 = d1 / 2 = 3/2 =1.5m

T = 2 cm

2-35 Bachelor Degree in Mechanical Engineering

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 r2 = d2 / 2 = 3.04/2 =1.52m

T∞,1 = 0

T∞,2 = 22

ε=1

h1=80 W/m2 ̊C

h2 = 10W/m2 ̊C

=?

Assumptions

1. 1 dimensional heat conduction Q=?

2. Steady state condition,

3. No heat generation
4. K is assumed to be constant
5. Heat is uniformly transferred

Thermal resistance concept is practical to use instead of different equation

concept since the problem involves multi boundary conditions.

Ri = = = = 4.420X10-4 o
C/W

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R1 = = = 4.653X10-5 oC /W

Rrad =

Where

hrad = ε [Ts2 + Tsurr2 ] [Ts + Tsurr]

= (5.67X10-8)(1)[T22 + 2952] [T2 + 295]

Assume T2 = 10 oC = 283K

hrad = 5.67x10-8 [2832+2952] [283 + 295]

o
=5.477W/m C

Thus,

Rrad = = = 6.288 X10-3 oC /W

R0 = = = = 3.444X10-3 o
C /W

Parallel Circuit

= +

= +

Re = 2.225 X 10-3 o
C/W

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Thus,

Rt = Ri + R1 + Re

=4.42X10-4 + 4.653X10-4 + 2.225X10-3

=2.714X10-3 oC/W

Thus,

Heat Transfer rate, Q

Q=

= -8107W = 8107 W (outside to inside)

For Validation,

Q=

-8107 =

T2 = 3.96 oC (not satisfy the previous assumption which is T2 = 10 oC )

Assume T2 = 4 oC = 277K

hrad = 5.67x10-8 [2772+2952] [277 + 295]

o
=5.311W/m C

Thus,

Rrad =

= 6.484x10-3 o
C/W

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 = +

= +

Re = 2.249 X 10-3 o
C/W

Rt = Ri + R1 + Re

= 4.42X10-4 + 4.653X10-5 + 2.249X10-3

= 2.738X10-3 o
C/W

Thus,

Heat Transfer rate, Q

Q=

= -8035W = 8035 W (outside to inside)

For Validation,

Q=

-8107 =

T2 = 3.93 = 4 oC (satisfy the previous assumption)

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Example 2.9

A 3-m high and 5-m wide wall consists of long 16 cm x 22 cm cross section
horizontal bricks (k = 0.72 W/m oC) separated by 3 cm thick plaster layers (k =
0.22 W/m oC). There are also 2 cm thick plaster layers on each side of the brick
and a 3 cm thick rigid foam (k= 0.026 W/m oC) on the inner side of the wall. The
indoor and the outdoor temperatures are 20 oC and -10 oC, respectively and the
convection heat transfer coefficient on the inner and the outer sides are h1=10
W/m2 o
C and h2 = 25W/m2 o
C , respectively. Assuming one-dimensional heat
transfer and disregarding radiation, determine the rate of heat transfer through
the wall.
Plaster
Solution

A =3 X 5 = 15m2

Kp = 0.22W/m

1.5cm
Kf = 0.026W/m

Foam
22cm T∞,1 = 20 ,h1=10
Brick
W/m2 ̊C q

T∞,2 = -10 , h2 =25

25W/m2 ̊C 1.5cm

=?

3cm 2cm 16cm 2cm

Assumptions:

1. 1 dimensional heat conduction

2. Steady state condition

3. K is assumed to be constant
4. No heat generation
5. Radiation is negligible

Where,

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 o
Ri = = = 0.4 C/W

o
R1 = = = 4.6 C/W

o
R2 = R6 = = = 0.364 C/W

R3 = R5 = = = 48.485 o C/W

R4 = = = 1.01 o C/W

o
Ro = = = 0.16 C/W

Parallel circuit

R3

R4

R5

= + +

= + +

Re = 0.9696 oC/W

Thus,

Ri R1 R2 Re R6 Ro
6

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Rt = Ri + R1 + R2 + Re + R6 + Ro

= 0.4 + 4.6 + 0.364 +0.9696 + 0.364 +0.16

= 6.85 o C/W

Thus,

Q=

= 4.38 W (per 0.25 m2)

For whole area of the wall, A= 3X 15 = 15m2

4.38W  0.25m2

Q ? = 15m2

Thus,

Q=

= 263W

2.3 CRITICAL RADIUS OF INSULATION

1. Adding insulation will increase thermal resistance.

2. The thicker the insulation, the lower the heat transfer.
From Ohm’s Law analogy for heat transfer ;

T
=
Rt

Where

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 L
Rt cond , plane 
KA
r  r1
Rt cond , sphere  2
4 Kr1r2
r2
ln
r1
Rt cond , cylinder 
2 KL
1
Rt conv 
hA
1
Rt rad 
hrad A

3. However, the scenario stated above is only true when the plane wall case
is considered, since the heat transfer area, A is always constant.

L1
R1 
K1 A
L2
Rins 
K2 A
1
R0 
hA

4. Adding insulation to a cylindrical or a spherical shape is a different matter.

5. As a result of adding insulation to a cylindrical or a spherical ;-
 The thermal resistance due to conduction increase.
 The thermal resistance due to convection decrease as the outer
surface increase.

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 r2
ln
r1
R1 
2  K1 L
r3
ln
r2
Rins  [Rconv decrease as A increase]
2 K 2 L
1
R0 

h 2 r3 l 
Rt = R1 + R ins + R0

Thus,

T1  T
=
Rt

max

bare

rcr

6. Therefore, the heat transfer from pipe may increase or decrease, depending
on which effect dominates.

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7. Critical radius of insulation depends on the thermal conductivity and external
convection heat transfer.

8. From the above graph:-

 r2 < rcr = the heat transfer from the cylinder or sphere increase
 r2 = rcr = the heat transfer rate is maximum
 r2 > rcr = the heat transfer from the cylinder or sphere decrease

Example 2.10

Calculate the critical radius of insulation for asbestos [k = 0.17W/m˚C]

surrounding a pipe and exposed to room air at 20˚C with h = 3.0W/m 2˚C.
Calculate the heat loss from a 200C, 5.0cm diameter pipe when covered with the
critical radius of insulation and without insulation.

Solution

k = 0.17W/m˚C

h = 3.0W/m2˚C

T  = 20˚C

T1 = 200˚C

r1 = 0.025m

ins and without insulation = ?

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 max

0.17
=
3

= 0.0567 m rcr r2

Heat loss when pipe is covered with r2 = rcr of insulation.

T1  T
=
Rt

where,

Rt = R ins + Rconv

rcr
ln
r1 1 1
Rins  Rt conv  =
2K L hA h(2rcr L)

ln 0.0567 1
= 0.025 =
2 (0.17) L h(2 (0.0567 ) L)

0.7666 0.9355
= ˚C/W = ˚C/W
L L

Thus

Rt = R ins + Rconv

0.7666 0.9355
= +
L L

1.702
= ˚C/W
L

T1  T
=
Rt

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Q 200  20
=
L 1.702

= 105.7 W/m

Heat loss without insulation

T1  T
=
Rt

105.7 W/m
1 1 1
Rt conv  = = max
hA h(2r1L) 3(2 (0.025 ) L )

2.122 bare
=
L

Thus, r1 rcr

200  20
Qwihoutinsulation =
2.122
L

Qwihoutinsulation
 84.8 W/m r2 > rcr
L

In order to decrease the heat transfer rate from the pipe, outside rtadius,r2 must
exceed rcr.

Let say,

r2 = 0.5m ; r2 > rcr

Rt = R ins + Rconv

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 r2
ln
r1 1
= +
2 K L h(2r2 L)

ln 0.5 1
= 0.025 +
2 (0.17) L 3(2 (0.5)( L ))

2.804 0.106
= +
L L

2.910
= ˚C/W
L

Thus,

Qins 200  20

L 2.91

= 61.85 W/m

The thickness of insulation is tins = r2 - r1 = 0.5-0.025 = 0.475m= 47.5cm, which is

too thick. Other insulation materials with low thermal conductivity should be
found, otherwise the cost will be high.

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2.4 FINS

2.4.1 Fins Applications

Cooled Motorcycle Engine with Fins

Bristle-Fin Surface on Condenser Tubes

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 Pentium 3 processor and attached heat sink

Pentium 4 with fin and fan draws air through it.

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2.4.2 Fins Analysis

1. The practical way to increase heat transfer rate is by increasing the

surface area.
2. This can be done by attaching to the surface, extended surface called
fins.
3. Assumptions made when analyzing fins problem are:
a) Steady state condition
b) No heat generation in the fin
c) Thermal conductivity, k is constant
d) Convection heat transfer coefficient, h is uniform and constant.

From energy balance,

= (2.8)

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Where:

conv = ( 2.9 )

Where:

P = perimeter

Substituting (2.9 ) in (2.8)

= +

+ = 0 ________________________( 3)

Taking the limit as ∆x 0, gives

+ =0

From Fourier Law

= -k dT/dx

Substituting in ( 1 )

dT/dx] + hP =0

dT/dx] - hP =0

d2T/dx2 - =0

- =0 ( 2.10)

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Where

At the fin base, =

The general solution for equation (2.10) is

Governing equation for fin

- =0 (2.11)

Where

The general solution for equation (2.11 ) is

= (2.12)

In order to find the value of and , two boundary conditions are at least

needed.

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 Base
Tip

0
X

Base

 Specified temperature

Tip

 Specified temperature
 Negligible heat loss (adiabatic)
 Convection
 Convection and radiation

Boundary condition at the fin base

At x = 0

= = ( 2.13 )

i) Infinitely Long Fin

For a sufficiency long fin of uniform cross section (Ac=constant), the temperature
of the fin tip approaches the environment temperature, T∞.

At x=L, L→∞

Where,

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 (2.14)

From (2.12)

, where L ∞

This condition is only satisfied by function , but not by since it tends to

be infinity as x get larger.

T T

Tb Tb

T∞ T∞
x --
x
L L

Thus,

(x)= (2.15)

Where from (2.13) at

Thus the temperature distribution for infinitely long fin is,

x=be-mx (2.16)

Where,

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Thus,

(2.17)

The heat transfer rate from the entire fin can be determined from Fourier’s Law,

(2.18)

From (2.17)

(2.19)

Subtitute (2.19) into (2.18)

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 (2.20)

ii) Negligible Heat loss from the fin tip (Adiabatic fin tip, )

(2.21)

The application of the boundary condition (at the fin base and fin tip) on the
general solution yields, after some manipulations, this relation for the temperature
distribution:

(2.22)

The heat transfer rate from the fin

(2.23)

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iii ) Convection or combined convection and radiation from fin tip.

L
COMPLICATED
0 TO BE SOLVED

Actual fin Convection

MUCH EASIER TO
0
BE SOLVED

Equivalent fin

Insulated

Corrected fin length,

(2.24)

Fins subjected to convection at their tips can be treated as fins with insulated tips
by replacing the actual fin length by the corrected length in equation (2.22) and
(2.23).

And,

Where:

t = thickness,

d = diameter

Multiplying equation (2.24) by the perimeter gives,

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Afin

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2.4.3 Fin Efficiency

The fin problems are commonly analyzed under steady state condition, which
there is no temperature difference along the fin. This is an ideal process:

Ideal Process

Heat is transferred from the fin to the surrounding by convection:

! Remember the side wall transfer heat to the fin by conduction product. However,
in practice, the temperature of the fins drops along the fin.

Actual Process

This causes the heat transfer rate decrease as the temperature

difference reduce.

The efficiency of the fin,

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Thus,

Where,

Total surface area of the fin

Efficiency of the fin

Heat transfer coefficient (convection)

Base temperature

Surrounding temperature

The efficiency of very long fins,

, where

The efficiency of fins with adiabatic tips,

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The efficiency of fins subjected to convection

Where,

The efficiency of other geometries can be obtained from the manual or text book.

The mathematical functions I and K that appear in some of these relations are the
modified Bessel functions, and their values are given in Table 3.4 (page 166)
(Cengel Y.A., 2006).

** Refer to HEAT AND MASS TRANSFER A Practical approach, YUNUS A.

CENGEL**

2.4.4 Fin Effectiveness

The fin effectiveness can be defined as follow,

Fin effectiveness,

Where,

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 The addition of the fins to the surface does not affect heat

transfer at all.

The fins act as insulation, slowing down the heat transfer

from the surface.

The fins enhance heat transfer from the surface.

The fin efficiency and fin effectiveness ( respectively are related to the

performance of the fin, but they are different quantities.

and can be related by:

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2.4.5 Proper Length of a Fin

The determination of the appropriate length of the fin is the important step in
designing a fin. The temperature drops along the fin exponentially and reaches
the environment temperature exponentially. The part of the fin beyond this length
does not contribute to heat transfer since it is at the temperature of the
environment. Designing such an ‘extra long fin’ is out of the question since it
results in material waste, excessive weight and increased size and thus
increased cost with no benefit in return.

High

Low

high low

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 2mm diameter

L=20mm

1mm

Plate fin
Outer
radius, r2

L=20mm
(r2-r1)

r2
=20mm

Fins Geometries

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Example 2.11

A cup holds coffee at 80 oC. A spoon, 150mm long, is half-immersed in the coffee.
The other half of the spoon is exposed to air at 30oC. The cross-sectional area and
periphery of the spoon are 7mm2 and 95mm respectively and maybe taken to be
constant. The heat transfer coefficients on the coffee side and air side are 500W/m 2K
and 15 W/m2K respectively.

Which material should be used for a teaspoon, silver or stainless steel?

(ksilver=427.5W/mK, kstainless=14.6W/mK)

Solution

Coffee side Air side

Tip

0.075m 0.075m

0.15m

Ac=7x10-6m2
P=9.5x10-3m
Ksilver=427.5W/mK
Kstainless=14.6W/mK

Assumption
1. Spoon is assumed to be a convecting tip fin case.
2. K and h are assumed to be constant

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Silver, =427.5W/mK

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At coffee side

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At air side

Base
Coffee side Air side

Tcoffee=79.62oC

k silver=427.5W/mK
k stainless=14.6W/mK

Stainless steel, k=14.6W/mK

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2-70 Bachelor Degree of Mechanical Engineering
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At coffee side

At air side

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Table 2.1: Temperature with Silver Spoon and Stainless Steel Spoon
Temperature(⁰ C) Silver spoon Stainless steel spoon

Ttip coff 79.62oC 80oC

Ttip base 76.15oC 72.66oC

T`tip air 70.6oC 35.05oC

It is proven that stainless steel spoon is more suitable to be used as it is cold to be

hold compare to silver spoon. This is apparently because silver spoon conducts more
heat than stainless steel spoon. (Kss=14.6W/mK and Ksilver=426.7W/mK).

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Example 2.12

In a long cylindrical fin of 10mm diameter, the temperature is measured at three

points along its length, each 10cm apart. The measured temperatures are 80, 70 at x1
and x2. The ambient is at 30oc and heat transfer coefficient to the ambient is
20W/m2K. Calculate the temperature at x3 and the thermal conductivity of the fin
material.

Tb T∞=30oc

80oc 70oc T3

0.1m 0.1m

X1 X2 X3

Solution

D = 10×10-3m

∆x = 10×10-2m

At x = x1, T1 = 80oc

x = x2, T2 = 70oc

T∞ = 30oc

h = 20W/m2k

T3 = ? , k = ?

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Assumption

- Fin is assumed to be an infinite long fin.

For very long fin case

Assuming x1 is the base of the fin, x = 0.1m

ln 0.8 = -0.1m

m = 2.231

k = 1606W/mk

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T3 = 62oc

Example 2.13

Consider the fin of Example 2.12. Instead of an infinite fin, take it as 50mm long with
Tbase of 80oc and a convecting tip. Calculate the fin heat transfer for

a) Fin with convecting tip, and

b) An adiabatic fin with length correction
c) What is the % error in the heat flow

Solution

T∞=30oc

h=20W/m2k
Tb=80oc
k=1606W/mk
L=0.05m
D=0.01m

Assumption

- Convection heat transfer fin problem

a) Fin with convecting tip

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Where,

M=

=(

= 14.08

= 2.232

L = 0.05m

mL = 2.232 × 0.05

= 0.1116

Thus,

= 1.703W

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 b) An adiabatic fin with length correction

Lc = L + D/4

= 0.05 + 0.01/4

= 0.0525

Where

Thus,

= 1.64W

= 3.5%

As the error is small, we can just simplify the convecting fin tip problem by
considering adiabatic tip case with length correction.

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Example 2.14

A room heater consists of fin tubes inside which saturate steam condenses at 100 oc.
The tubes are made of copper, 30mm outer diameter. Annular fins of aluminium,
1.5mm thick, are fitted tightly on the tube. The fin pitch is 5mm. Air at 30 oc flowing
past the fins gives rise to a heat transfer coefficient of 120W/m 2K. If the latent heat of
steam is 2500kJ/kg, find the amount of steam condensed per hour over one meter
length of pipe.

Solution

r1

r2
L

H t

Tb = 100oc

r1 = 0.015m, r2 = 0.0275m

L = 0.0125m

t = 1.5×10-3m

H = 5×10-3m

h = 120W/m2k

T∞ = 30oc

Latent heat = 2500kJ/kg

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Assumption

- Convection heat transfer fins problem

As we know,

Where,

Heat transfer rate from the fin,

Lc = L + t/2

= 0.0125 + 0.0015/2

= 0.01325m

r2c = r2 + t/2

= 0.0275 + 0.0015/2

= 0.02825m

Ap = Lct

= 0.01325(0.0015)

= 1.9875×10-5m2

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 = 0.243

Thus, from figure 2.3

At ℰ = 0.243, r2c/r1 = 1.88, ηfin = 0.95

Afin = Lateral Surface + Tip Surface

= 2π(r2c2 – r12) + 2πr2ct ------------ t is relatively small

= 2π(0.028252 – 0.0152)

= 3.6×10-3m2

Therefore

= 0.95(120)(3.6×10-3)(100-30)

= 28.728W

Heat transfer rate from the unfinned surface,

Aunfin = 2πr1H = πD1H

= 2π(0.015)(0.005)

= 4.713×10-4m2

Thus,
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 = 3.959W

Total heat transfer rate of the fins for 1m tube length

= (28.728 + 3.959)153

= 5001W

Amount of steam condensed is

= 2.0×10-3kg/s

= 7.2kg/h

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Example 2.15

An Aluminium fin [k = 200W/moC] 3.0 mm thick and 7.5 cm long protrudes from a
wall, as in figure below. The base is maintained at 300oC, and the ambient
temperature is 50oC with h = 10W/m² oC. Calculate the heat loss from the fin per unit
depth of material.

Assumptions:

1. Steady operating condition.

2. Thermal conductivity is constant.
3. Convection heat transfer coefficient is uniform over the entire surface of the
fins.
4. Radiation is negligible.

Figure Example 1

Solution:

1. Equation Method:

The approximate method is used by extending the fin a fictitious length t/2 and the
computing the heat transfer from a fin with insulated tip.

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Where,

Thus,

2. Graphical method:

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From straight rectangular fin graph in the manual,

Thus,

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3. Table method:

From straight rectangular fin table in the manual,

Where,

Thus,

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Example 2.16

Aluminium fins 1.5cm wide and 1mm thick are placed on a 2.5cm diameter tube to
dissipate the heat. The tube surface is 170oC, and the ambient-fluid temperature is
25⁰ C. Calculate the heat loss per fin for h = 130 W/m² oC. Assume k = 200 W/moC
for aluminium.

Assumptions:

1. Steady operating condition.

2. Thermal conductivity is constant.
3. h is uniform over the entire surface.
4. Radiation is negligible.

Given:

Find:

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Solution:

Where,

From figure 3.43

Where,

Thus,

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Multi Number of Fins

Solution (apply formulae):

Thus, overall effectiveness,

Where,

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 (one fin)

Example 2.17

Steam in a heating system flows through tubes whose outer diameter is D1 = 3cm and
whose are maintained at a temperature of 120oC. Circular Aluminium alloy fins
of outer diameter D2 = 6cm and constant thickness t = 2mm are

attached to the tube, as shown in figure below. The space between the fins is 3mm,
and thus there are 200fins per meter length of the tube. Heat is transferred to the
surrounding air at , with a combined heat transfer coefficient of

. Determine the increase in heat transfer from the tube per meter of
its length as a result of adding fins.

Given:

Assumptions:

1. Steady operating condition.

2. k is constant.
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 3. h is uniform over the entire surface of the fin.
4. Radiation is negligible.
Solution:
Heat transfer when there are no fins,

Heat transfer when fins are added,

Graphical Method (from figure 3)

Thus,

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Thus, effectiveness,

From,

(t is too small)

Thus,

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Heat transfers from the unfinned portion,

Thus,

(1m = 200fins)

Therefore,

The increase in heat transfer as a result of the addition of the fin:

Thus, overall effectiveness,

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2.5 TRANSIENT HEAT CONDUCTION

1) In the previous chapter, we considered heat conduction under steady

conduction where the temperature of a body at any points does not change
with time.
2) In this chapter the variation of temperature with time as well as position is
considered transient heat conduction.
3) Lumped system analysis- the temperature of a body varies with time but
remains uniform ( i.e does not change with position) at any time given.

30˚C 30˚C

30˚C 30˚
30˚C 30˚C C

4) In lumped system the analysis, the temperature can be taken to be

function of time only, T(t).

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 From Energy balance,

Heat transfer into the The increase in the an

body during dt = energy of the body during
dt

h As [T∞ - T(t)] dt = mCpdT

hAs[T∞ - T(t)] dt = ρV Cpd[T -T∞]

(2.26)

Where,

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Equation 2.26 enables us to determine the temperature T(t) of a body at time t, or
alternatively the time t required for the temperature to reach a specified value T(t).

As the T(t) obtained,

The rate of convection heat transfer between the body and its environment

Q = h* As* [T∞ - T(t)]

The total amount of the heat transfer between the body and the surrounding
during dt (i.e from t=0 to t)

Q = m* Cp* [T(t) - T(i)]

The maximum heat transfer between the body and surrounding is

Qmax = m* Cp* [T∞ - T(i)]

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2.6 Criteria for Lumped System Analysis

Lumped system analysis is only applicable if ;

Where,

The biot number can be viewed as the ratio of the convection at the surface to
conduction within the body

Where,

h = convection heat transfer coefficient

k = thermal conductivity

Lc = characteristic length

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Example 2.18

The temperature of a gas stream is to be measured by a thermocouple whose

junction can be approximately as a 1 mm diameter sphere. The properties of the
junction are k = 35W/m˚c, ρ = 8500 kg/m3 and Cp = 320 J/kg˚c and the convection
heat transfer coefficient between the junction and the gas is h = 210W/m 2˚c.
Determine how long it will take for the thermocouple to read 99% of the initial
temperature difference.

Gas Junction

T∞, h

D = 1mm
T(t)
Solution

D= 0.001m

k = 35W/m˚c

ρ = 8500 kg/m3

Cp = 320 J/kg˚c

h = 210W/m2˚c

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Assumptions

Temperature vary with time

Temperature is uniform over the entire surface of the sphere.
Thermal conductivity, K and heat transfer coefficient are constant.
Radiation is neglected.

Thus,

As Biot number, Bi is less than 0.1, lumped system analysis is applicable.

In order to read 99% of the initial temperature differences, Ti - T∞ between the

junction and the gas,

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Where

Where

Thus, 10s at least needed for the temperature of the thermocouple junction approach
99 percent of the initial junction – gas temperature difference.

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Example 2.19

A steel ball (C= 0.46 kJ/kg.˚c, k=35W/m˚c) 5.0 cm in diameter and initially at a
uniform temperature of 450˚c is suddenly placed a controlled environment in which
the temperature is maintained at 100˚c. The convection heat transfer coefficient is
10W/m2˚c. Calculate the time required for the ball to attain a temperature 150˚c.
(ρsteel = 7800kg/m3)

Ti , T(t)

T∞ , h

Solution

C= 0.46 kJ/kg.˚c

k=35W/m˚c

d= 0.05m

Ti= 450˚c

T∞ = 100˚c

h= 10W/m2˚c

T(t)= 150˚c

t=?

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Assumptions

1) Temperatures varies with time

2) Temperature is uniform over the entire surface of the steel ball
3) k and h are constant
4) Radiation is negligible

Biot number, Bi

As Bi ≤ 0.1, lumped system analysis is applicable.

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Where,

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