University of Hail
Faculty of Engineering
DEPARTMENT OF MECHANICAL ENGINEERING
ME 315 – Heat Transfer
Lecture notes
Chapter 3
One dimensional steady state heat conduction
and extended surfaces- Part 2
Prepared by : Dr. N. Ait Messaoudene
Based on:
“Introduction to Heat Transfer”
Incropera, DeWitt, Bergman, and Lavine,
5th Edition, John Willey and Sons, 2007.
2nd semester 2011-2012
Conduction with Thermal Energy Generation:
(One-Dimensional, Steady-State )
• Involves a local (volumetric) source of thermal energy due to conversion from another
form of energy in a conducting medium.
• The source may be uniformly distributed, as in the conversion from
electrical to thermal energy (Ohmic heating):
(3.38)
Other cases:
deceleration and absorption of neutrons in the fuel element of a nuclear reactor
exothermic chemical reactions occurring within a medium. (Endothermic reactions have
the inverse effect :a thermal energy sink)
conversion from electromagnetic to thermal energy may occur due to the absorption of
radiation within the medium. The process occurs, for example, when gamma rays are
absorbed in external nuclear reactor components (cladding, thermal shields, pressure
vessels, etc.) or when visible radiation is absorbed in a semitransparent medium.
Remember not to confuse energy generation with energy storage.
The Plane Wall
• Consider one-dimensional, steady-state conduction
in a plane wall of constant k, uniform generation,
and asymmetric surface conditions:
• Heat Equation:
(3.39)
• General Solution:
Apply the BC
The heat flux at any point in the wall may, of course, be determined by using Equation 3.46
with Fourier’s law.
Note that with generation the heat flux is no longer independent of x.
Symmetric Surface Conditions :
• Temperature Distribution:
(3.42)
or
(From
x=-L to +L )
with
The maximum temperature at the
midplane
One Surface Insulated:
at the plane of symmetry, (dT/dx)x=0 = 0
→ this plane may be represented by the adiabatic surface
→ Equation 3.42 also applies to plane walls that are
perfectly insulated on one side (x = 0) and maintained at a
fixed temperature Ts on the other side (x = L).
If the temperature of an adjoining fluid, T∞ , is known and
not Ts directly.
Overall energy balance on the wall →
(3.46)
Which can also be obtained by
applying a surface energy
balance at x=L
EXAMPLE 3.7
A plane wall is a composite of two materials, A and B. The wall of material A has uniform
heat generation q =1.5x106 W/m3, kA = 75W/m.K, and thickness LA = 50 mm. The wall material
B has no generation with kB = 150 W/m K and thickness LB = 20 mm. The inner surface of
material A is well insulated, while the outer surface of material B is cooled by a water stream
with T∞ =30 ⁰C and h = 1000 W/m2.K.
1. Sketch the temperature distribution that exists in the composite under steady-state
conditions.
2. Determine the temperature T0 of the insulated surface and the temperature T2 of the
cooled surface.
Assumptions:
1. Steady-state conditions.
2. One-dimensional conduction in x-direction.
3. Negligible contact resistance between walls.
4. Inner surface of A adiabatic.
5. Constant properties for materials A and B.
1. From the prescribed physical conditions, the temperature distribution in the composite is
known to have the following features, as shown:
(a) Parabolic in material A.
(b) Zero slope at insulated boundary.
(c) Linear in material B.
(d) Slope change kB/kA = 2 at interface.
(e) The temperature distribution in the water is characterized by large gradients near the
surface.
The temperature
distribution will have
the following aspect
T2 may be obtained by performing an energy balance on a CV about material B.
No generation; steady-state conditions ; consider a unit surface area:
( heat flux in at x = LA ) = (heat flux out due to convection at x = LA + LB )
(1)
q” may be determined by performing a second energy balance on a CV about material A.
Since the surface at x = 0 is adiabatic
(rate of energy generation) = ( heat flux in at x = LA )
For a unit surface area
(2)
Combining Equations 1 and 2
From Equation 3.48, T0 (at the insulated surface) is
(3)
where T1 may be obtained from the following thermal circuit:
(representing material B)
Where, for a unit surface area:
From equation (3)
Radial Systems
Heat generation may occur in a variety of radial geometries.
Cylindrical (Tube) Wall
Solid Cylinder (Circular Rod)
• Heat Equations:
Cylindrical
1 d  dT
 kr
r dr  dr
 
q 0

Spherical Wall (Shell)
Solid Sphere
Spherical
1 d  2 dT  
 kr
q 0
r 2 dr 
dr 
For a long solid cylinder with constant k, the heat equation becomes:
(3.49)
Separating variables and assuming uniform generation, it may be integrated to obtain
once
(3.51)
twice
Prescribed surface
temperature
B.C: symmetry at the center
Evaluating T0 at the centerline,
we can obtain a ND form
•
•
Overall energy balance  Eout  Eg  0
Same result is obtained with a surface energy balance Ein  Eout  0 qcond  ro   qconv
•
•
Where qcond is evaluated using Fourier’s Law at the surface and qconv using Newton’s Law of
cooling
• Solution for Uniform Generation in a Solid Sphere of Constant k
with Convection Cooling:
Temperature Distribution
Surface Temperature
•
dT
q r3
kr

 C1
dr
3
Overall energy balance:
2
•
•
•
•
 Ts  T 
 Eout  Eg  0
q r 2 C1
T 
  C2
6k
r
q ro
3h
Or from a surface energy balance:
dT
|r  0  0  C1  0
dr
Ein  Eout  0  qcond  ro   qconv
•
•
q ro 2
T  ro   Ts  C2  Ts 
6k
•
q ro 2  r 2 
T r  
1    Ts
6k  ro 2 
•
qcond   k
dT
dr

r  r0
qconv  h(Ts  T )
qr0
3
•
qr
 Ts  T  o
3h
A summary of temperature distributions is provided in Appendix C for
plane, cylindrical and spherical walls, as well as for solid cylinders and spheres.
Note how boundary conditions are specified and how they are used to obtain surface
temperatures.
You are required to become familiar with the contents of the appendix.
EXAMPLE 3.8
Consider a long solid tube, insulated at the outer radius r2 and cooled at the inner radius r1,
with uniform heat generation q (W/m3) within the solid.
1. Obtain the general solution for the temperature distribution in the tube.
2. In a practical application a limit would be placed on the maximum temperature that is
permissible at the insulated surface (r = r2). Specifying this limit as Ts,2, identify appropriate BC
that could be used to determine the arbitrary constants appearing in the general solution and
determine these constants.
3. Determine the heat removal rate per unit length of tube.
4. If the coolant is available at a temperature T∞, obtain an expression for the convection
coefficient that would have to be maintained at the inner surface to allow for operation
at prescribed values of Ts,2 and q .
Assumptions:
1. Steady-state conditions.
2. One-dimensional radial
conduction.
3. Constant properties.
4. Uniform volumetric heat
generation.
5. Outer surface adiabatic.
1. To determine T(r), the appropriate form of the heat equation, Equation 2.24, must be
solved. For the prescribed conditions, this expression reduces to Equation 3.49, and
the general solution is given by Equation 3.51. Hence, this solution applies in a cylindrical
shell, as well as in a solid cylinder
2. Two BC are needed to evaluate C1 and C2, and in this problem it is appropriate to specify both
conditions at r2: prescribed temperature limit and adiabatic outer surface
(1)
(2)
Using Equations 3.51 , 1 and 2 it follows that
(3)
(4)
Hence, from Equation 4 and 3 :
(5)
(6)
Substituting Equations 5 and 6 into the general solution, Equation 3.51, it follows that
(7)
3. The heat removal rate may be determined by obtaining the conduction rate at r1 or by
evaluating the total generation rate for the tube. From Fourier’s law.
Hence, substituting from Equation 7 and evaluating the result at r1,
(8)
4. Applying the energy conservation requirement, Equation 1.13, to the inner surface, it
follows that
or
Hence
(10)
where Ts,1 may be obtained by evaluating Equation 7 at r = r1.
Problem 3.91 Thermal conditions in a gas-cooled nuclear reactor
with a tubular thorium fuel rod and a concentric
graphite sheath: (a) Assessment of thermal integrity
.
for a generation rate of q  108 W/m.3 (b) Evaluation of
temperature distributions in the thorium
and graphite
•
for generation rates in the range 108  q  5x10
. 8
Schematic:
Assumptions: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Constant
properties, (4) Negligible contact resistance, (5) Negligible radiation, (6) Adiabatic surface at r1.
Properties:
Table A.1, Thorium: Tmp  2000K; Table A.2, Graphite: Tmp  2300K.
Analysis: (a) The outer surface temperature of the fuel, T2 , may be determined from the rate equation
q 
 
where Rtot
T2  T

Rtot
1n  r3 / r2 
2 k g

1
 0.0185 m  K/W
2 r3h
The heat rate may be determined by applying an energy balance to a control surface about the fuel
•
•
element,
Eout  E g
or, per unit length,
•
•
E out  E g
Since the interior surface of the element is essentially adiabatic, it follows that
q  q   r22  r12   17,907 W/m
•
Hence,
  T  17,907 W/m  0.0185 m K/W   600K  931K
T2  qRtot
With zero heat flux at the inner surface of the fuel element, Eq. C.14 yields
q r22  r12  q r12  r2
T1  T2 
1

1n 
4kt  r22  2kt  r1


•
•

  931K  25K  18K  938K

<
Since T1 and T2 are well below the melting points of thorium and graphite, the prescribed
operating condition is acceptable.
(b) The solution for the temperature distribution in a cylindrical wall with generation is
•
q r22  r 2 
Tt  r   T2 
1  
4kt  r22 
• 2

2

q
r
r
1n  r2 / r 
2
1

1  2   T2  T1  

 4kt  r 
 1n  r2 / r1 

2



(C.2)
Boundary conditions at r1 and r2 are used to determine T1 and T2 .
r  r1 :
r  r2 :
 q• r 2  r 2 

2
1
k
1  2   T2  T1  
•
 4kt  r2 

qr1


q1  0 

2
r11n  r2 / r1 
 qr• 2  r 2 

2
1
k
1  2   T2  T1  
•
 4kt  r2 

q r2


U 2 T2  T  

2
r21n  r2 / r1 
 
U 2   A2 Rtot
1
 
  2 r2 Rtot
1
(C.14)
(C.17)
(3.32)
The following results are obtained for temperature distributions in the graphite.
2500
Temperature, T(K)
2100
1700
1300
900
500
0.008
0.009
0.01
0.011
Radial location in f uel, r(m)
qdot = 5E8
qdot = 3E8
qdot = 1E8
•
8
3
Operation at q  5x10 W/m is clearly unacceptable since the melting point of
thorium would be exceeded. To prevent softening of the material, which would occur
•
below the melting point, the reactor should not be operated much above q  3x108 W/m.3
The small radial temperature gradients are attributable to the large value of kt .
Using the value of T2 from the foregoing solution and computing T3 from the surface condition,
q 
2 k g T2  T3 
(3.27)
1n  r3 / r2 
the temperature distribution in the graphite is
Tg  r  
r
T2  T3
1n    T3
1n  r2 / r3   r3 
(3.26)
Temperature, T(K)
2500
2100
1700
1300
900
500
0.011
0.012
0.013
0.014
Radial location in graphite, r(m)
qdot = 5E8
qdot = 3E8
qdot = 1E8
•
Operation at q  5x108 W/m3 is problematic for the graphite. Larger temperature gradients
are due to the small value of k g .
Heat Transfer from Extended Surfaces
An extended surface (also know as a combined conduction-convection system or a fin) is a
solid within which heat transfer by conduction is assumed to be one dimensional, while heat is
also transferred by convection (and/or radiation) from the surface in a direction transverse to
that of conduction.
the most frequent application is one in
which an extended surface is used
specifically to enhance heat transfer
between a solid and an adjoining fluid. Such
an extended surface is termed a fin.
The thermal conductivity of the fin material
can have a strong effect on the temperature
distribution along the fin and therefore
influences the degree to which the heat
transfer rate is enhanced. Ideally, the fin
material should have a large thermal
conductivity to minimize temperature
variations from its base to its tip. In the limit
of infinite thermal conductivity, the entire
fin would be at the temperature of the base
surface, thereby providing the maximum
possible heat transfer enhancement.
Some typical fin configurations:
Straight fins of (a) uniform and (b) non-uniform cross sections; (c) annular fin, and
(d) pin fin of non-uniform cross section.
General Conduction Analysis
Assuming one-dimensional (even though conduction within the fin is actually 2D), steady-state
conduction in an extended surface of constant conductivity k and uniform cross-sectional area
Ac , with negligible radiation and no generation .
Applying the conservation of energy
requirement, Equation 1.11c, to the
differential
element
with
and
also
or
3.61
This result provides a general form of the energy equation for an extended surface (sometimes
called the fin equation). Its solution for appropriate BC’s provides the temperature distribution,
which may be used to calculate the conduction rate at any x.
Fins of Uniform Cross-Sectional Area
Ac is a constant and As = Px, where As is
the surface area measured from the base
to x and P is the fin perimeter.
Equation 3.61 reduces to:
To simplify the form of this equation, we transform the dependent variable by defining an
excess temperature as
where T∞ is constant:
where
The general solution is of the form
Boundary conditions:
The second condition, specified at the fin tip (x = L), may correspond to one of four different
physical situations summarized in Table 3.4.
Case A: Convection heat transfer from the fin
tip (with specified h)
Applying an energy balance to a control surface
about the tip, we obtain
or
Using the general solution:
and
Solving for the constants, it may be shown, after some manipulation, that
Amount of heat transferred from the entire fin:
This trend is a consequence of the reduction in the
conduction heat transfer qx(x) with increasing x due to
continuous convection losses from the fin surface.
Or: qx = qx = heat transferred by convection from the whole fin surface Af (including the tip)
Which will lead to the same result
Summary of the results for the 4 different conditions at the fin tip
(3.70)
(3.72)
(3.75)
(3.76)
(3.77)
(3.78)
(3.79)
(3.80)
EXAMPLE 3.9
A very long rod 5 mm in diameter has one end maintained at 100⁰C. The surface of the rod is
exposed to ambient air at 25 ⁰ C with a convection heat transfer coefficient of 100 W/m2.K.
1. Determine the temperature distributions along rods constructed from pure copper,
2024 aluminum alloy, and type AISI 316 stainless steel. What are the corresponding
heat losses from the rods?
2. Estimate how long the rods must be for the assumption of infinite length to yield an
accurate estimate of the heat loss.
Assumptions:
1. Steady-state conditions.
2. One-dimensional conduction along the rod.
3. Constant properties.
4. Negligible radiation exchange with
surroundings.
5. Uniform heat transfer coefficient.
6. Infinitely long rod.
1. Subject to the assumption of an infinitely long fin, the temperature distributions are
determined from Equation 3.79, which may be expressed as
where m = (hP/kAc) 1/2 =(4h/kD)1/2. Substituting for h and D, as well as for the thermal
conductivities of copper, the aluminum alloy, and the stainless steel, respectively, the
values of m are 14.2, 21.2, and 75.6 m-1.
The temperature distributions may then be computed and plotted as follows:
There is little additional heat transfer
associated with extending the length
of the rod much beyond 50, 200, and
300 mm, respectively, for the
stainless steel, the aluminum alloy,
and the copper.
From Equation 3.80, the heat loss is
For copper, aluminum alloy and stainless steel, respectively, the heat rates are
qf = 8.3 W; 5.6 W and 1.6 W.
Fin Performance Parameters
fins are used to increase the heat transfer by increasing the effective surface area.
However, the fin itself represents a conduction resistance from the original surface.
no assurance that the heat transfer rate will be increased through the use of fins.
An assessment of this matter may be made by evaluating the fin effectiveness εf .It is defined as
the ratio of the fin heat transfer rate to the heat transfer rate that would exist without the fin.
Fin Effectiveness:  f 
qf
hAc , bb
where Ac,b is the fin cross-sectional area at the base
(3.81)
Fins are justified if εf > 2 in general. For the case of an infinite fin,
Another parameter is the Fin Resistance:
Also, the Fin Efficiency:  f 
qf
q f ,max

qf
hAf b
For a straight fin of uniform cross section
and an adiabatic tip,
(3.83)
>2 if
(kP/hAc) > 4.
;using
(3.86)
Consider a triangular fin:
1/ 2
2
Af  2w  L2   t / 2  


Ap   t / 2  L
1 I1  2mL 
f 
mL I 0  2mL 
Expressions for the fin efficiency are provided in Table 3.5 for common geometries.
Fin Arrays: Overall Surface Efficiency
An overall surface efficiency is
defined
Representative
arrays of
(a) rectangular
(b) annular fins.
where qt is the total heat rate
from the surface area At
associated with both the fins
and the exposed portion of
the base (prime surface).
If there are N fins in the array, each of surface area Af, and
the area of the prime surface is designated as Ab, then:
At  NAf  Ab
Number of fins
(3.99)
Area of exposed base (prime surface)
– Total heat rate:
qt  N f hAf b  hAbb  o hAtb 
b
(3.100)
Rt , o
– Overall surface efficiency and resistance:
o  1 
Rt , o 
NAf
b
qt
At

1   f 
1
o hAt
(3.102)
(3.103)
Equivalent Thermal Circuit :
EXAMPLE 3.10
The engine cylinder of a motorcycle is constructed of 2024-T6 aluminum alloy and is of height
H= 0.15 m and outside diameter D = 50 mm. Under typical operating conditions the outer
surface of the cylinder is at a temperature of 500 K and is exposed to ambient air at 300 K, with
a convection coefficient of 50 W/m2. K. Annular fins are integrally cast with the cylinder to
increase heat transfer to the surroundings. Consider five such fins, of thickness t=6 mm, length
L=20 mm, and equally spaced. What is the increase in heat transfer due to use of the fins?
Assumptions:
1. Steady-state conditions.
2. One-dimensional radial conduction
in fins.
3. Constant properties.
4. Negligible radiation exchange with
surroundings.
5. Uniform convection coefficient over
outer surface (with or without fins).
Problem 3.116: Assessment of cooling scheme for gas turbine blade.
Determination of whether blade temperatures are less
than the maximum allowable value (1050 °C) for
prescribed operating conditions and evaluation of blade
cooling rate.
Schematic:
Assumptions: (1) One-dimensional, steady-state conduction in blade, (2) Constant k, (3)
Adiabatic blade tip, (4) Negligible radiation.
Analysis: Conditions in the blade are determined by Case B of Table 3.4.
(a) With the maximum temperature existing at x=L, Eq. 3.75 yields
T  L   T
Tb  T
m   hP/kAc 
1/ 2

1
cosh mL

 250W/m2  K  0.11m/20W/m  K  6  104 m2
mL = 47.87 m-1  0.05 m = 2.39

1/ 2
= 47.87 m-1
From Table B.1, coshmL=5.51 . Hence,
T  L   1200 C  (300  1200) C/5.51  1037 C
and, subject to the assumption of an adiabatic tip, the operating conditions are acceptable.
(b) With M   hPkAc 
1/ 2
b

2
 250W/m  K  0.11m  20W/m  K  6  10
Eq. 3.76 and Table B.1 yield
qf  M tanh mL  517W  0.983  508W
Hence,
qb  qf  508W
4
m
2
  900 C  517W ,
1/ 2
Problem 3.132: (a)Determination of maximum allowable power qc for a 20mm
x 20mm electronic chip whose temperature is not to exceed
85⁰C when the chip is attached to an air-cooled heat sink with
N=11 fins of prescribed dimensions.
Tc = 85oC
W = 20 mm
R”t,c= 2x10-6 m2-K/W
k = 180 W/m-K
L b= 3 mm
Lf = 15 mm
Air
Too = 20oC S
h = 100 W/m2-K
t
Rt,b
Tc
qc
Rt,c
 = 1.8 mm
Assumptions: (1) Steady-state, (2) One-dimensional heat transfer, (3) Isothermal chip, (4)
Negligible heat transfer from top surface of chip, (5) Negligible temperature rise for air flow,
(6) Uniform convection coefficient associated with air flow through channels and over outer
surface of heat sink, (7) Negligible radiation.
Too
Rt,o
Analysis: (a) From the thermal circuit,
T  T
Tc  T
qc  c

R tot
R t,c  R t,b  R t,o
2
6 2
2
R t,c  R t,c / W  2  10 m  K / W /  0.02m   0.005 K / W
   0.003m /180
R t,b  L b / k W
2
W/mK
 0.02m 2
 0.042 K / W
From Eqs. (3.103), (3.102), and (3.99)
R t,o 
1
,
o h A t
o  1 
N Af
1  f  ,
At
A t  N Af  A b
-4
2
Af = 2WLf = 2  0.02m  0.015m = 6  10 m
2
2
-3
-4
2
Ab = W – N(tW) = (0.02m) – 11(0.182  10 m  0.02m) = 3.6  10 m
-3 2
At = 6.96  10 m
1/2
With mLf = (2h/kt)
2
1.17, tanh mLf = 0.824 and Eq. (3.87) yields
f 
tanh mLf 0.824

 0.704
mLf
1.17
o = 0.719,
Rt,o = 2.00 K/W, and
qc 
-3
1/2
Lf = (200 W/m K/180 W/mK  0.182  10 m)
85  20  C
 0.005  0.042  2.00  K / W
 31.8 W
(0.015m) =