Chapter 3 1

Chapter 3

One-Dimensional Steady-State Conduction

Chapter 3 2

One-Dimensional Steady-State Conduction

• Conduction problems may involve multiple directions and time-

dependent 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 3

The Plane Wall

Consider a simple case of one-dimensional conduction in a plane wall, separating two fluids of different temperature, without energy generation

• Temperature is a function of x

• Heat is transferred in the x-direction

Must consider– Convection from hot fluid to wall

– Conduction through wall

– Convection from wall to cold fluid

Begin by determining temperature distribution within the wall

qx

1,T

1,sT

2,sT

2,T

x

x=0 x=L

11, ,hT

22, ,hT

Hot fluid

Cold fluid

Chapter 3 4

Temperature Distribution

• Heat diffusion equation (eq. 2.4) in the x-direction for steady-state conditions, with no energy generation:

0

dx

dTk

dx

d

• Boundary Conditions: 2,1, )(,)0( ss TLTTT

• Temperature profile, assuming constant k:

1,1,2, )()( sss TL

xTTxT

Temperature varies linearly with x

qx is constant

(3.1)

Chapter 3 5

Thermal ResistanceBased on the previous solution, the conduction hear transfer rate can be calculated:

kAL

TTTT

L

kA

dx

dTkAq ss

ssx /2,1,

2,1,

Recall electric circuit theory - Ohm’s law for electrical resistance:

Similarly for heat convection, Newton’s law of cooling applies:

Resistance

e DifferencPotentialcurrent Electric

hA

TTTThAq S

Sx /1

)()(

And for radiation heat transfer:

Ah

TTTTAhq

r

surssursrrad /1

)()(

(3.2a)

(3.2b)

(3.2c)

Chapter 3 6

Thermal Resistance

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:

• We can use this electrical analogy to represent heat transfer problems using the concept of a thermal circuit (equivalent to an electrical circuit).

AhR

hAR

kA

LR

rradtconvtcondt

1,

1, ,,,

R

Tq overall

Resistance

Force DrivingOverall

Chapter 3 7

Thermal Resistance for Plane Wall

In terms of overall temperature difference:qx

1,T

1,sT

2,sT

2,T

xx=0 x=L

11, ,hT

22, ,hT

Hot fluid

Cold fluid

AhkA

L

AhR

R

TTq

tot

totx

21

2,1,

11

Ah

TT

kAL

TT

Ah

TTq ssssx

2

2,2,2,1,

1

1,1,

/1//1

Chapter 3 8

Composite Walls Express the following

geometry in terms of a an equivalent thermal circuit.

Chapter 3 9

Composite Walls What is the heat transfer rate for this system?

Alternatively

UAq

TRR

TUAq

ttot

x

1

where U is the overall heat transfer coefficient and T the overall temperature difference.

)]/1()/()/()/()/1[(

11

41 hkLkLkLhARU

CCBBAAtot

Chapter 3 10

Composite Walls

(a) Surfaces normal to the x-direction are isothermal

(b) Surfaces parallel to x-direction are adiabatic

For resistances in series: Rtot=R1+R2+…+Rn

For resistances in parallel:

Rtot=1/R1+1/R2+…+1/Rn

Chapter 3 11

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 12

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:

"",

x

BAct

q

TTR

See tables 3.1, 3.2 for typical values of Rt,c

Chapter 3 13

Alternative Conduction Analysis

• For steady-state conditions, no heat generation, one-dimensional heat transfer, qx is constant.

dx

dTxATkqx )()(

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:

T

T

x

xx

oo

dTTkxA

dxq )(

)(

Chapter 3 14

Example 3.3Consider a conical section fabricated from pyroceram. It is of circular cross section, with the diameter D=x, where =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

x1x2

x

Chapter 3 15

Radial Systems-Cylindrical Coordinates

Consider a hollow cylinder, whose inner and outer surfaces are exposed to fluids at different temperatures

Temperature distribution

Chapter 3 16

Temperature Distribution

• Heat diffusion equation (eq. 2.5) in the r-direction for steady-state conditions, with no energy generation:

01

dr

dTkr

dr

d

r

• Boundary Conditions: 2,21,1 )(,)( ss TrTTrT

• Temperature profile, assuming constant k:

2,221

2,1, ln)/ln(

)()( s

ss Tr

r

rr

TTrT

Logarithmic temperature distribution

(see previous slide)

• Fourier’s law: constdr

dTrLk

dr

dTkAqr )2(

Chapter 3 17

Thermal ResistanceBased on the previous solution, the conduction hear transfer rate can be calculated:

condt

ssssssx R

TT

Lkrr

TT

rr

TTLkq

,

2,1,

12

2,1,

12

2,1,

)2/()/ln()/ln(

2

In terms of equivalent thermal circuit:

)2(

1

2

)/ln(

)2(

1

22

12

11

2,1,

LrhkL

rr

LrhR

R

TTq

tot

totx

• Fourier’s law: constdr

dTrLk

dr

dTkAqr )2(

Chapter 3 18

Composite Walls

Express the following geometry in terms of a an equivalent thermal circuit.

Chapter 3 19

Composite Walls What is the heat transfer rate?

where U is the overall heat transfer coefficient. If A=A1=2r1L:

44

1

3

41

2

31

1

21

1

1lnlnln

11

hrr

rr

kr

rr

kr

rr

kr

h

U

CBA

alternatively we can use A2=2r2L, A3=2r3L etc. In all cases:

tRAUAUAUAU

144332211

Chapter 3 20

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 21

Spherical Coordinates

• 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?

• Fourier’s law:

dr

dTrk

dr

dTkAqr

)4( 2

Chapter 3 22

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 23

Equation 2.3-1 may be expressed in the integral form   = -

 For constant thermal conductivity, k

   Q =

 

=

Generally, this equation can be written in terms of 

Q =  where

  R =

2

1 24rr r

drQ

21 )(TT dTTk

21

21

/1/1

)(4

rr

TTk

2112

21( TT

rr

rr

condsphereR

TT

,

12

21

11

4

1

rrk

Chapter 3 24

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 25

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 26

Summary

• We obtained temperature distributions and thermal

resistances for problems involving steady-state, one-

dimensional conduction in orthogonal, cylindrical and

spherical coordinates, without energy generation

• Useful summary in Table 3.3