### Chapter 3

```Chapter 3
One-Dimensional Steady-State Conduction
Chapter 3
1
One-Dimensional Steady-State Conduction
• Conduction problems may involve multiple directions and timedependent conditions
• Inherently complex – Difficult to determine temperature distributions
• One-dimensional steady-state models can represent accurately
numerous engineering systems
• In this chapter we will
 Learn how to obtain temperature profiles for common geometries
with and without heat generation.
 Introduce the concept of thermal resistance and thermal circuits
Chapter 3
2
The Plane Wall
Consider a simple case of onedimensional conduction in a plane
wall, separating two fluids of different
temperature, without energy
generation
Cold fluid
T ,1
T,2 , h2
Ts ,1
• Temperature is a function of x
Ts , 2
• Heat is transferred in the x-direction
Must consider
– Convection from hot fluid to wall
T,1,h1
qx
T , 2
– Conduction through wall
– Convection from wall to cold fluid
 Begin by determining temperature
distribution within the wall
Chapter 3
Hot fluid
x=0
x=L
x
3
Temperature Distribution
• Heat diffusion equation (eq. 2.4) in the x-direction for steady-state
conditions, with no energy generation:
d  dT 
k
0
dx  dx 
• Boundary Conditions:
 qx is constant
T (0)  Ts,1, T ( L)  Ts,2
• Temperature profile, assuming constant k:
x
T ( x)  (Ts,2  Ts,1 )  Ts,1 (3.1)
L
 Temperature varies linearly with x
Chapter 3
4
Thermal Resistance
Based on the previous solution, the conduction hear transfer rate can
be calculated:

Ts,1  Ts,2 
dT kA
Ts,1  Ts,2  
qx  kA

dx L
L / kA
(3.2a)
Similarly for heat convection, Newton’s law of cooling applies:
(TS  T )
qx  hA(TS  T ) 
1 / hA
(3.2b)
And for radiation heat transfer:
qrad
(Ts  Tsur )
 hr A(Ts  Tsur ) 
1 / hr A
(3.2c)
 Recall electric circuit theory - Ohm’s law for electrical resistance:
Electriccurrent 
Chapter 3
PotentialDifference
Resistance
5
Thermal Resistance
• We can use this electrical analogy to represent heat transfer problems
using the concept of a thermal circuit (equivalent to an electrical circuit).
OverallDrivingForce Toverall
q

Resistance
R

 Compare with equations 3.2a-3.2c
 The temperature difference is the “potential” or driving force for the
heat flow and the combinations of thermal conductivity, convection
coefficient, thickness and area of material act as a resistance to this
flow:
Rt ,cond 
Chapter 3
L
1
1
, Rt ,conv 
, Rt ,rad 
kA
hA
hr A
6
Thermal Resistance for Plane Wall
Cold fluid
T ,1
T,2 , h2
Ts ,1
qx 
Ts , 2
T,1,h1
Hot fluid
qx
x=0
x
T , 2
x=L
T,1  Ts ,1
1 / h1 A
L / kA

Ts ,2  T,2
1 / h2 A
In terms of overall
temperature difference:
qx 
Rtot
Chapter 3

Ts ,1  Ts ,2
T,1  T,2
Rtot
1
L
1



h1 A kA h2 A
7
Composite Walls
 Express the following
geometry in terms of a
an equivalent thermal
circuit.
Chapter 3
8
Composite Walls
 What is the heat transfer rate for this system?
Alternatively
qx  UAT
Rtot 

T
1
Rt 

q
UA
where U is the overall heat transfer coefficient and T the overall
temperature difference.
1
1
U

Rtot A [(1 / h1 )  ( LA / k A )  ( LB / kB )  ( LC / kC )  (1 / h4 )]
Chapter 3
9
Composite Walls
(a) Surfaces normal to the xdirection are isothermal
 For resistances in series:
Rtot=R1+R2+…+Rn
 For resistances in parallel:
Rtot=1/R1+1/R2+…+1/Rn
Chapter 3
(b) Surfaces parallel to xdirection are adiabatic
10
Example (Problem 3.15 textbook)
Consider a composite wall that includes an 8-mm thick hardwood
siding (A), 40-mm by 130-mm hardwood studs (B) on 0.65-m centers
with glass fiber insulation (D) (paper faced, 28 kg/m3) and a 12-mm
layer of gypsum (vermiculite) wall board (C).
 What is the thermal resistance associated with a wall that is 2.5 m
high by 6.5 m wide (having 10 studs, each 2.5 m high?)
(Note: Consider the direction of heat transfer to be downwards, along
the x-direction)
Chapter 3
11
Contact Resistance
The temperature drop
across the interface
between materials may be
appreciable, due to surface
roughness effects, leading
to air pockets. We can
define thermal contact
resistance:
Rt",c
T A  TB

q "x
See tables 3.1, 3.2 for
typical values of Rt,c
Chapter 3
12
Alternative Conduction Analysis
When area varies in the x direction and k is a function of temperature,
Fourier’s law can be written in its most general form:
dT
qx  k (T ) A( x)
dx
• For steady-state conditions, no heat generation, one-dimensional heat
transfer, qx is constant.
 qx
Chapter 3

x
xo
dx

A( x)
T

k (T )dT
To
13
Example 3.3
Consider a conical section fabricated from pyroceram. It is of circular
cross section, with the diameter D=ax, where a=0.25. The small end
is at x1=50 mm and the large end at x2=250 mm. The end
temperatures are T1=400 K and T2=600 K, while the lateral surface is
well insulated.
1. Derive an expression for the temperature distribution T(x) in symbolic
form, assuming one-dimensional conditions. Sketch the temperature
distribution
2. Calculate the heat rate, qx, through the cone.
T2
T1
x2
Chapter 3
x1
x
14
Radial Systems-Cylindrical Coordinates
Consider a hollow cylinder, whose inner and outer surfaces are
exposed to fluids at different temperatures
Temperature distribution
Chapter 3
15
Temperature Distribution
• Heat diffusion equation (eq. 2.5) in the r-direction for steady-state
conditions, with no energy generation:
1 d  dT 
 kr
0
r dr  dr 
• Fourier’s law:
qr  kA
• Boundary Conditions:
dT
dT
 k (2rL)
 const
dr
dr
T ( r1 )  Ts,1, T ( r2 )  Ts,2
• Temperature profile, assuming constant k:
(Ts ,1  Ts ,2 )
 r 
T (r) 
ln   Ts ,2  Logarithmic temperature distribution
(see previous slide)
ln(r1 / r2 )  r 2 
Chapter 3
16
Thermal Resistance
Based on the previous solution, the conduction hear transfer rate can
be calculated:
• Fourier’s law:
qx 
dT
dT
qr  kA
 k (2rL)
 const
dr
dr
2Lk Ts ,1  Ts ,2 
ln(r2 / r1 )

Ts,1  Ts,2 
ln(r2 / r1 ) /(2Lk )

Ts ,1  Ts ,2 

Rt ,cond
 In terms of equivalent thermal circuit:
qx 
Rtot
Chapter 3
T,1  T,2
Rtot
1
ln(r2 / r1 )
1



h1 (2r1L)
2kL
h2 (2r2 L)
17
Composite Walls
 Express the following
geometry in terms of a
an equivalent thermal
circuit.
Chapter 3
18
Composite Walls
 What is the heat transfer rate?
where U is the overall heat transfer coefficient. If A=A1=2r1L:
1
U
1 r1
r
r
r
r
r
r 1

ln 2  1 ln 3  1 ln 4  1
h1 k A r1 k B r2 kC r3 r4 h4
alternatively we can use A2=2r2L, A3=2r3L etc. In all cases:
1
U1 A1  U 2 A2  U 3 A3  U 4 A4 
Rt

Chapter 3
19
Example (Problem 3.37 textbook)
A thin electrical heater is wrapped around the outer surface of a long
cylindrical tube whose inner surface is maintained at a temperature of
5°C. The tube wall has inner and outer radii of 25 and 75 mm
respectively, and a thermal conductivity of 10 W/m.K. The thermal
contact resistance between the heater and the outer surface of the
tube (per unit length of the tube) is R’t,c=0.01 m.K/W. The outer
surface of the heater is exposed to a fluid of temperature –10°C and a
convection coefficient of h=100 W/m2 .K.
 Determine the heater power per unit length of tube required to
maintain the heater at To=25°C.
Chapter 3
20
Spherical Coordinates
• Fourier’s law:
dT
qr  kA
dr
2 dT
 k (4r )
dr
• Starting from Fourier’s law, acknowledging that qr is constant,
independent of r, and assuming that k is constant, derive the equation
describing the conduction heat transfer rate. What is the thermal
resistance?
Chapter 3
21
For steady-state, one dimensional conditions with no heat generation;
The appropriate form of Fourier’s equation is
Q = -k A dT/dr
= -k(4πr2) dT/dr
Note that the cross sectional area normal to the heat flow is
A= 4πr2
(instead of dx) where r is the radius of the sphere
Chapter 3
22
Equation 2.3-1 may be expressed in the integral form
T2
Q r 2 dr
= - T1 k (T )dT
r1
4
r2
For constant thermal conductivity, k
4k (T 1  T 2)
Q=
1 / r1  1 / r 2
=
r1r 2
(T 1  T 2
r 2  r1
Generally, this equation can be written in terms of
Q = T 2  T1
Rsphere, cond
where
R =
Chapter 3
1 1 1
  
4k  r1 r 2 
23
Example:
Consider a hollow steel sphere of inside radius r1 = 10 cm and outside
radius, r2 = 20 cm. The thermal conductivity of the steel is k = 10 W/moC.
The inside surface is maintained at a uniform temperature of T1 = 230 oC
and the outside surface dissipates heat by convection with a heat transfer
coefficient h = 20 W/m2oC into an ambient at T = 30oC. Determine the
thickness of asbestos insulation (k=0.5 W/mK) required to reduce the heat
loss by 50%.
Chapter 3
24
Example (Problem 3.69 textbook)
One modality for destroying malignant tissue involves imbedding a
small spherical heat source of radius ro within the tissue and
maintaining local temperatures above a critical value Tc for an
extended period. Tissue that is well removed from the source may be
assumed to remain at normal body temperature (Tb=37°C).
 Obtain a general expression for the radial temperature distribution in
the tissue under steady-state conditions as a function of the heat rate
q.
 If ro=0.5 mm, what heat rate must be supplied to maintain a tissue
temperature of T>Tc=42°C in the domain 0.5<r<5 mm? The tissue
thermal conductivity is approximately 0.5 W/m.K.
Chapter 3
25
Summary
• We obtained temperature distributions and thermal
resistances for problems involving steady-state, onedimensional conduction in orthogonal, cylindrical and
spherical coordinates, without energy generation
• Useful summary in Table 3.3
Chapter 3
26
```