Chapter 3:
Steady Heat Conduction
Yoav Peles
Department of Mechanical, Aerospace and Nuclear Engineering
Rensselaer Polytechnic Institute
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Objectives
When you finish studying this chapter, you should be able to:
• Understand the concept of thermal resistance and its limitations,
and develop thermal resistance networks for practical heat
conduction problems,
• Solve steady conduction problems that involve multilayer
rectangular, cylindrical, or spherical geometries,
• Develop an intuitive understanding of thermal contact resistance,
and circumstances under which it may be significant,
• Identify applications in which insulation may actually increase
heat transfer,
• Analyze finned surfaces, and assess how efficiently and
effectively fins enhance heat transfer, and
• Solve multidimensional practical heat conduction problems
using conduction shape factors.
3.1 Steady Heat Conduction in Plane Walls
1) Considerable temperature
difference between the inner
and the outer surfaces of the
wall (significant temperature
gradient in the x direction).
2) The wall surface is nearly
isothermal.
Steady one-dimensional modeling approach is
justified.
• Assuming heat transfer is the only energy interaction
and there is no heat generation, the energy balance
can be expressed as
Zero for steady
operation
Rate of
heat transfer
into the wall
or
-
Rate of
heat transfer
out of the wall
=
Rate of change
of the energy
of the wall
=0
P
(3-1)
dE
Q in − Q out = wall = 0
dt
The rate of heat transfer through the
wall must be constant ( Q cond ,wall = constant ).
For the control volume (CV):
[入] - [出] + [生成量] = [增加量]
無
穩態
• Then Fourier’s law of heat conduction for the wall
can be expressed as
dT
Qcond , wall = −kA
dx
(W)
(3-2)
• Remembering that the rate of conduction heat transfer
and the wall area A are constant it follows
dT/dx=constant
the temperature through the wall varies linearly with x.
• Integrating the above equation and rearranging yields
T1 − T2
Qcond , wall = kA
L
(W)
(3-3)
Thermal Resistance ConceptConduction Resistance
• Equation 3–3 for heat conduction through a
plane wall can be rearranged as
T1 − T2
Qcond , wall =
Rwall
(W)
(3-4)
• Where Rwall is the conduction resistance
expressed as
Rwall
L
=
kA
( D C/W)
(3-5)
Analogy (類比) to Electrical Current Flow
• Eq. 3-5 is analogous to the relation for electric current
flow I, expressed as
V1 − V2
I=
Re
(3-6)
Heat Transfer
Electrical current flow
Rate of heat transfer
ÅÆ Electric current
Thermal resistance
ÅÆ Electrical resistance
Temperature difference ÅÆ Voltage difference
Thermal Resistance ConceptConvection Resistance
• Thermal resistance can also be applied to convection
processes.
• Newton’s law of cooling for convection heat transfer
rate ( Q conv = hAs (Ts − T∞ )) can be rearranged as
Ts − T∞
Qconv =
Rconv
(W) (3-7)
• Rconv is the convection resistance
Rconv
1
=
hAs
(D C/W) (3-8)
Thermal Resistance ConceptRadiation Resistance
• The rate of radiation heat transfer between a surface and
the surrounding
Ts − Tsurr
4
4
Qrad = εσ As Ts − Tsurr = hrad As (Ts − Tsurr ) =
(W)
Rrad
(
Rrad =
hrad
1
hrad As
)
(3-9)
(K/W)
(3-10)
Q rad
2
2
2
=
= εσ Ts + Tsurr (Ts + Tsurr ) (W/m ⋅ K)
As (Ts − Tsurr )
(
)
(3-11)
Thermal Resistance Concept
- Radiation and Convection Resistance
• A surface exposed to the surrounding might involves
convection and radiation simultaneously.
• The convection and radiation resistances are parallel to
each other.
• When Tsurr≈T∞, the radiation
effect can properly be
accounted for by replacing h
in the convection resistance
relation by
hcombined = hconv+hrad (W/m2K)
(3-12)
Thermal Resistance Network
• consider steady one-dimensional heat transfer through a
plane wall that is exposed to convection on both sides.
Rate of
heat convection
into the wall
Q = h1 A (T∞ ,1 − T1 )
T1 − T2
= kA
L
= h2 A (T2 − T∞ ,2 )
(3-13)
=
Rate of
heat conduction
through the wall
=
Rate of
heat convection
from the wall
Rearranging and adding
⎧T∞ ,1 − T1 = Q ⋅ Rconv ,1
⎪
+ ⎨T1 − T2 = Q ⋅ Rwall
⎪
⋅R
−
=
T
T
Q
2
∞
,2
conv ,2
⎩
T∞ ,1 − T∞ ,2 = Q ( Rconv ,1 + Rwall + Rconv ,2 ) = Q ⋅ Rtotal
where
T∞,1 − T∞,2
Q=
Rtotal
Rtotal = Rconv ,1 + Rwall + Rconv ,2
(W)
(3-15)
1
L
1 D
=
+
+
( C/W)
h1 A kA h2 A
(3-16)
• It is sometimes convenient to express heat transfer
through a medium in an analogous manner to
Newton’s law of cooling as
Q = UA∆T
(W)
(3-18)
• where U is the overall heat transfer coefficient.
• Note that
1
UA =
Rtotal
D
( C/K)
(3-19)
Multilayer Plane Walls
• The rate of steady heat transfer through a two-layer
composite wall can be expressed through Eq. 3-15 where
the total thermal resistance is
Rtotal = Rconv ,1 + Rwall ,1 + Rwall ,2 + Rconv ,2
1
L1
L2
1
=
+
+
+
h1 A k1 A k 2 A h2 A
(3-22)
3.2 Thermal Contact Resistance
• In reality surfaces have some roughness.
• When two surfaces are pressed against each other, the
peaks form good material contact but the valleys form
voids filled with air.
• As a result, an interface contains
numerous air gaps of varying sizes
that act as insulation because of the
low thermal conductivity of air.
• Thus, an interface offers some
resistance to heat transfer, which
is termed the thermal contact
resistance, Rc.
• The value of thermal contact resistance
depends on the
–
–
–
–
surface roughness,
material properties,
temperature and pressure at the interface,
type of fluid trapped at the interface.
• Thermal contact resistance is observed to
decrease with decreasing surface roughness
and increasing interface pressure.
• The thermal contact resistance can be
minimized by applying a thermally conducting
liquid called a thermal grease.
3.3 Generalized Thermal Resistance Network
• The thermal resistance concept can be used to solve
steady heat transfer problems that involve parallel
layers or combined series-parallel arrangements.
• The total heat transfer of two parallel layers
⎛ 1
T1 − T2 T1 − T2
1 ⎞
+
= (T1 − T2 ) ⎜ + ⎟
Q = Q1 + Q2 =
R1
R2
⎝ R1 R2 ⎠
1
(3-29)
Rtotal
⎛ 1
R1 R2
1
1 ⎞
= ⎜ + ⎟ → Rtotal =
Rtotal ⎝ R1 R2 ⎠
R1 + R2
(3-31)
Combined Series-Parallel Arrangement
The total rate of heat transfer
Through the composite system
T1 − T∞
Q=
Rtotal
(3-32)
where
Rtotal = R12 + R3 + Rconv
R1R2
=
+ R3 + Rconv (3-33)
R1 + R2
L3
L1
L2
1
R1 =
; R2 =
; R3 =
; Rconv =
k1 A1
k2 A2
k3 A3
hA3
(3-34)
3.4 Heat Conduction in Cylinders
Consider the long cylindrical layer
Assumptions:
– the two surfaces of the cylindrical
layer are maintained at constant
temperatures T1 and T2,
– no heat generation,
– constant thermal conductivity,
– one-dimensional heat conduction.
Fourier’s law of heat conduction
dT
Qcond ,cyl = − kA
dr
(W)
(3-35)
dT
(3-35)
Qcond ,cyl = − kA
(W)
dr
Separating the variables and integrating from r=r1,
where T(r1)=T1, to r=r2, where T(r2)=T2
r2
∫
Q cond ,cyl
dr = −
T2
∫
kdT
(3-36)
A
T =T1
Substituting A =2πrL and performing the integrations
give
T1 − T2
(3-37)
Qcond ,cyl = 2π Lk
ln ( r2 / r1 )
r = r1
Since the heat transfer rate is constant
T1 − T2
Qcond ,cyl =
Rcyl
(3-38)
Thermal Resistance with Convection
Steady one-dimensional heat transfer through a
cylindrical or spherical layer that is exposed to
convection on both sides
T∞ ,1 − T∞ ,2
Q=
Rtotal
(3-32)
where
Rtotal = Rconv ,1 + Rcyl + Rconv ,2 =
ln ( r2 / r1 )
1
1
=
+
+
( 2π r1L ) h1 2π Lk ( 2π r2 L ) h2
(3-43)
Multilayered Cylinders
• Steady heat transfer
through multilayered
cylindrical or spherical
shells can be handled
just like multilayered plane.
• The steady heat transfer rate through a three-layered
composite cylinder of length L with convection on both
sides is expressed by Eq. 3-32 where:
Rtotal = Rconv ,1 + Rcyl ,1 + Rcyl ,3 + Rcyl ,3 + Rconv ,2 =
ln ( r2 / r1 ) ln ( r3 / r2 ) ln ( r4 / r3 )
1
1
=
+
+
+
+
2π Lk 2
2π Lk3
( 2π r1L ) h1 2π Lk1
( 2π r2 L ) h2
(3-46)
3.5 Critical Radius of Insulation (略)
• Adding more insulation to a wall or to the attic
always decreases heat transfer.
• Adding insulation to a cylindrical pipe or a spherical
shell, however, is a different matter.
• Adding insulation increases the conduction resistance
of the insulation layer but decreases the convection
resistance of the surface because of the increase in the
outer surface area for convection.
• The heat transfer from the pipe may increase or
decrease, depending on which effect dominates.
• A cylindrical pipe of outer radius r1
whose outer surface temperature T1 is
maintained constant.
• The pipe is covered with an insulator
(k and r2).
• Convection heat transfer at T∞ and h.
• The rate of heat transfer from the insulated pipe to the
surrounding air can be expressed as
T1 − T∞
T1 − T∞
Q=
=
(3-37)
Rins + Rconv ln ( r2 / r1 )
1
+
2π Lk
h ( 2π r2 L )
• The variation of the heat transfer rate with the outer
radius of the insulation r2 is shown
in the figure.
• The value of r2 at which Q
reaches a maximum is
determined by
dQ
=0
dr2
• Performing the differentiation
and solving for r2 yields
rcr ,cylinder
k
=
h
(m) (3-50)
• Thus, insulating the pipe may actually increase the
rate of heat transfer instead of decreasing it.
3.6 Heat Transfer from Finned Surfaces
• Newton’s law of cooling
Q conv = hAs (Ts − T∞ )
• Two ways to increase the rate of heat transfer:
– increasing the heat transfer coefficient,
– increase the surface area Æfins
• Fins are the topic of this section.
Fin Equation
Under steady conditions, the energy balance on this
volume element can be expressed as
Rate of heat
Rate of heat
conduction into = conduction from the
the element at x
element at x+∆x
+
Rate of heat
convection from
the element
Q cond , x = Q cond , x +∆x + Q conv
where
Q conv = h ( p∆x )(T − T∞ )
Substituting and dividing by ∆x, we obtain
Q cond , x +∆x − Q cond , x
∆x
+ hp (T − T∞ ) = 0
(3-52)
Taking the limit as ∆x → 0 gives
dQ cond
+ hp (T − T∞ ) = 0
dx
(3-53)
From Fourier’s law of heat conduction we have
dT
Qcond = − kAc
dx
(3-54)
Substitution of Eq. 3-54 into Eq. 3–53 gives
d ⎛
dT ⎞
⎜ kAc
⎟ − hp (T − T∞ ) = 0
dx ⎝
dx ⎠
(3-55)
For constant cross section and constant thermal conductivity
Where
d 2θ
2
−m θ =0
2
dx
θ = T − T∞
(3-56)
hp
; m=
kAc
• Equation 3–56 is a linear, homogeneous, second-order differential
equation with constant coefficients.
• The general solution of Eq. 3–56 is
θ ( x) = C1e mx + C2 e − mx
(3-58)
• C1 and C2 are constants whose values are to be determined from the
boundary conditions at the base and at the tip of the fin.
Boundary Conditions
Several boundary conditions are typically employed at the
fin tip :
•
At the fin base
–
Specified temperature boundary condition, expressed
as: θ(0)= θb= Tb-T∞
(a) Infinitely Long Fin (Tfin tip = T)
• For a sufficiently long fin,
the temperature at the fin
tip approaches the
ambient temperature
Boundary condition:
θ(L→∞) = T(L) - T∞ = 0
• When x→∞ , so does emx→∞
C1=0
•
@ x = 0: emx = 1
C2= θb
• The temperature distribution:
T ( x) − T∞
−x
− mx
=e =e
Tb − T∞
hp / kAc
(3-60)
• heat transfer from the entire fin
dT
Q = − kAc
dx
= hpkAc (Tb − T∞ )
x =0
(3-61)
(b) Adiabatic Tip
• Boundary condition at fin tip:
dθ
dx
=0
x=L
(3-63)
• After some manipulations, the temperature
distribution:
T ( x) − T∞ cosh m ( L − x )
=
Tb − T∞
cosh mL
(3-64)
• heat transfer from the entire fin
dT
Q = − kAc
dx
= hpkAc (Tb − T∞ ) tanh mL
x =0
(3-65)
(c & d) Convection (or Combined Convection
and Radiation) from Fin Tip
• A practical way of accounting for the heat loss from
the fin tip is to replace the fin length L in the relation
for the insulated tip case by a corrected length
defined as
Lc=L+Ac/p (3-66)
• For rectangular and cylindrical
fins Lc is
• Lc,rectangular=L+t/2
• Lc,cylindrical =L+D/4
Fin Efficiency
• To maximize the heat transfer from a fin the
temperature of the fin should be uniform (maximized)
at the base value of Tb
• In reality, the temperature drops along the fin, and thus
the heat transfer from the fin is less
• To account for the effect we define
a fin efficiency
Q fin
Actual heat transfer rate from the fin
η fin = =
Q
Ideal heat transfer rate from the fin
fin ,max
or
if the entire fin were at base temperature
Q fin = η finQ fin ,max = η fin hAfin (Tb − T∞ )
(3-69)
Fin Efficiency
• For constant cross section of very long fins:
ηlong , fin
Q fin
=
Q
=
fin ,max
hpkAc (Tb − T∞ )
hAfin (Tb − T∞ )
1 kAc
1 (3-70)
=
=
L hp mL
• For constant cross section with adiabatic tip:
ηadiabatic , fin
Q fin
=
Q
fin ,max
=
hpkAc (Tb − T∞ ) tanh aL
hAfin (Tb − T∞ )
tanh mL
=
mL
(3-71)
Fin Effectiveness
• The performance of the fins is judged on the basis of the
enhancement in heat transfer relative to the no-fin case.
• The performance of fins is expressed
in terms of the fin effectiveness εfin
defined as
Heat transfer rate
ε fin
Q fin
Q fin
= =
=
Qno fin hAb (Tb − T∞ )
(3-72)
from the fin of base
area Ab
Heat transfer rate
from the surface
of area Ab
Remarks regarding fin effectiveness
• The thermal conductivity k of the fin material should
be as high as possible. It is no coincidence that fins
are made from metals.
• The ratio of the perimeter to the cross-sectional area
of the fin p/Ac should be as high as possible.
• The use of fins is most effective in applications
involving a low convection heat transfer coefficient.
The use of fins is more easily justified when the
medium is a gas instead of a liquid and the heat
transfer is by natural convection instead of by forced
convection.
Overall Effectiveness
• An overall effectiveness for a
finned surface is defined as the
ratio of the total heat transfer
from the finned surface to the
heat transfer from the same
surface if there were no fins.
Q fin
ε fin ,overall = Q
no fin
=
h ( Aunfin + η fin Afin )
hAno fin
(3-76)
Proper Length of a Fin
• An important step in the design of a fin is the
determination of the appropriate length of the fin once
the fin material and the fin cross section are specified.
• The temperature drops along
the fin exponentially and
asymptotically approaches the
ambient temperature at some
length.
3.7 Heat Transfer in Common Configurations
• Many problems encountered in practice are two- or
three-dimensional and involve rather complicated
geometries for which no simple solutions are
available.
• An important class of heat transfer problems for
which simple solutions are obtained encompasses
those involving two surfaces maintained at constant
temperatures T1 and T2.
• The steady rate of heat transfer between these two
surfaces is expressed as
Q=Sk(T1=T2)
(3-79)
• S is the conduction shape factor, which has the
dimension of length.
Table 3-7