Conduction

1) Steady state conduction- One dimension

2) Steady state conduction- Multiple dimension

3) Unsteady state conduction

Ch 2

Steady State Conduction-

One Dimension (1D)

Point of view Application of Fourier’s law of heat conduction to

calculation of heat flow in simple 1D system

(1) Plane Wall

(2) Cylinders

(3) Spherical

1D The temp. in the body is a function only of radial distance and independent of azimuth angle/

axial distance 方位

Radial Systems

徑向一維

(1) The Plane Wall

Integrated Fourier’s law

# If k varies with temp. according linear relation

, the heat flow become;

(2-1)

(2-2)

歐姆定律

# If 3 material (multilayer wall) involved, heat flow become;

Note: The heat flow must be SAME through all section

# The heat flow rate also can be represented as resistance network;

(Different conceptual view point for Fourier’s law)

Electrical analog circuit

(2-3)

Consider Heat transfer rate as flow and combination of as a resistance to this flow. The temp. is the potential function of the heat flow. So that, the Fourier equation may be written as:

3 wall side by side act as 3 thermal resistance in series

Electrical analog circuit: used to solve more complex problem (series and parallel thermal resistance)

Thermal resistance (°C/W)

(2-4)

(2-5)

Electrical analog circuit

Parallel

Series

Insulation & R value • The performance of insulation R value, define as

• Guide to choose insulating material in terms of their application and allowable temperature range

(2) Cylinders

From Fourier’s Law

Integrate this with the

boundary conditions

The heat flow rate

Sec.2-4 p. 29

(2-7)

(2-8)

(2) Cylinders

Thermal resistant for cylinder is

Multiple cylindrical sections

(2) Cylinders

Thermal-resistance concept for multiple-layer cylindrical walls

= Thermal-resistance concept for plane wall

So that,

the heat flow rate

(2-9)

(3) Spheres

The heat flow rate

Prove this equation!!

(2-10)

Example 1:

(Multilayer plane wall conduction)

An exterior wall of a house may be approximated by a

100 mm layer of common brick (k= 0.7 W/m.°C)

followed by a 40mm layer of gypsum plaster (k=0.48

W/m.°C). What thickness of loosely packed rock wool

insulation (k=0.065 W/m.°C) should be added to reduce

the heat loss (or gain) through the wall by 80%?

Answer: ∆xinsulation= 0.0584 m

Example 2: (Multilayer Cylindrical System)

A thick-walled tube of stainless steel [18% Cr, 8% Ni, k=19

W/m.°C ] with 2 cm inner diameter (ID) and 4 cm outer

diameter (OD) is covered with a 3 cm layer of asbestos

insulation [k=0.2 W/m.°C ]. If the inside wall temperature of

the pipe is maintained at 600 °C, calculate the heat loss per

meter of length. Also calculate the tube-insulation interface

temperature.

Answer: Tinsulation= 596.05 °C

Convection Boundary Conditions

Newton rate equation

/Newton’s Law of cooling

So that, an electric-resistance analogy for convection

process become:

(2-11)

The Overall Heat Transfer Coefficient, U

Consider:

Plane wall expose to a hot fluid A on 1 side and

a cooler fluid B on the other side

So that, the heat flow is express by

The overall hate transfer rate become;

Overall temp. difference

The sum of the thermal resistances

Sec 2-5

(2-12)

The Overall Heat Transfer Coefficient, U

1/ h A represent the convection resistance;

∆x/ k A represent the conduction resistance

The overall heat transfer (conduction + convection) can be expressed in term

of an overall heat transfer coefficient, U defined by relation:

The Overall Heat Transfer Coefficient

U also related to the R-value:

Where,

A: Area for the heat flow

(2-13)

單位: w/m2oC

The performance of insulation

Consider:

Hollow cylinder exposed to a convection

environment on its inner and outer surfaces with TA

and TB the two fluid temp. The area for convection is

not same for both liquids (depend on the inside tube

diameter and wall thickness

The Overall Heat Transfer Coefficient, U

The overall hate transfer rate become;

Overall temp. difference

The sum of the thermal resistances

(2-14)

The Overall Heat Transfer Coefficient, U

Ai & Ao: Inside & outside

surface areas of the inner tube

The Overall Heat Transfer Coefficient (Hollow cylinder)

based on:

1) Inside area of the tube Ai

2) Outside area of the tube, Ao

(2-15)

(2-16)

The Overall Heat Transfer Coefficient, U

The general notion (plane wall or cylinder coordinate system)

is that;

Rth Thermal resistance

Info P35 Table 2-2

Some typical value of U for heat exchanger are given in table.

Some value of U for common types of building construction

system also given in table and employed for calculation involving

the heating and cooling buildings.

Example 3

A house wall may be approximated as two 1.2 cm

layers of fiber insulating board, an 8.0 cm layer of

loosely packed asbestos, and a 10 cm layer of common

brick. Assuming convection heat transfer coefficient of

12 W/m2. °C on both sides of the wall, calculate the

overall heat transfer coefficient for this arrangement.

Answer: U= 0.6221 W/m2. °C

Example 4

A wall is constructed of a section of stainless steel

[k=16 W/m. °C] 4.0 mm thick with identical layers of

plastic on both sides of the steel. The overall heat

transfer coefficient, considering convection on both

sides of the plastic, is 120 W/m2.°C. If the overall temp.

different across the arrangement is 60 °C, calculated

the temperature difference across the stainless steel.

Answer: Tss= 18 °C

p.38 exap2-4 exap.2-5

Critical Thickness of Insulation

Consider:

A layer of insulation which might be installed around

a circular pipe. The inner temp. of the insulation is

fixed at Ti and the outer surface is exposed to a

convection environment at T∞.

The heat transfer in the thermal

network term

Sec.2--6

絕熱臨界厚度

Critical Thickness of Insulation

The result is (The critical-radius-of

insulation concept)

Manipulated this equation to determine the outer radius of insulation, ro, which

will maximize the heat transfer. The maximization condition is :

1) If ro < critical radius value, means:

The critical-radius-of insulation

concept

Critical Thickness of Insulation

Concept

The heat transfer will be increased by adding more insulation thickness

2) If ro > critical radius value, means:

The heat transfer will be decrease by adding more insulation thickness

絕熱臨界半徑

Example 5

A 1.0 mm diameter wire is maintained at a temp. of

400 °C and exposed to a convection environment at

40 °C with h= 120 W/m2. °C. Calculated the thermal

conductivity that will just cause an insulation

thickness of 0.2 mm to produce a “critical radius”.

Answer: k= 0.084 W/m.°C

Assignment 2

Tutorial 1 Book: J.P. Holman

1)2-1

2)2.2

3)2-5

4)2-6

5)2-7

6)2-8

7)2-9

8)2-17 and 2-18

9)2-26

10)2-62

Book: J.P. Holman

1) 2-4

2) 2-13

3) 2-19

4) 2-20

5) 2-21

6) 2-22

7) 2-23

8) 2-27

9) 2-31

10) 2-60

Heat Source Systems

Situation:

The system generated heat internally. Confine

our discussion to 1D system (the temp. is a

function of only 1 space coordinate) which is:

1)Plane Wall

2)Cylinder

3)Hollow Cylinder

1) Plane wall with heat sources Consider:

The plane wall with uniformly distributed heat sources.

The thickness of the wall in the x direction is 2 L

Assumed:

1)The heat flow as 1D

2)The heat generated per unit volume is

3)Thermal conductivity, k does not vary with temp.

From steady-state 1D heat flow with heat sources

For the boundary conditions

(The temp. on either side of the wall)

Integrated * equation with the boundary condition. So, the general solution become;

*

1) Plane wall with heat sources

General solution

C1 = 0 Because the temp. must be the SAME on each side of the wall.

C2 = T0 The temp. T0 at the midplane (x=0)

The solution obtain the temp. distribution

OR

“Parabolic

distribution”

OR

1) Plane wall with heat sources

Midplane temp. T0 ?? (Obtained through an energy balance)

At steady state conditions:

Differenced to get the temp. gradient at

the wall, dT/dx

The total heat generated = The heat lost at the faces

From parabolic

distribution :

So that

Simplify steady state condition become

1) Plane wall with heat sources

Midplane temp. T0

The same result for midplane temp. could be obtain by substituting

into this equation (The temp. distribution):

2) Cylinder with heat sources Consider:

Cylinder radius, R with uniformly distributed heat sources and constant thermal

conductivity.

Assumed:

The cylinder is sufficiently long so that the temp. as function of radius only.

The appropriate differential equation obtained by neglecting the axial, azimuth

and time dependent term

From steady-state 1D heat flow in

cylinder coordinates with heat sources

For the 1st boundary conditions

At steady state conditions:

The total heat generated = The heat lost at the surface

2) Cylinder with heat sources 2nd boundary conditions

(The temp. function must

be continuous at the

centre of the cylinder), so

could specify that

From steady-state 1D heat flow

in cylinder coordinates with

heat sources

Rewrite

Note that :

Substitute: Integrated

Integration yield and

2) Cylinder with heat sources

From 2nd boundary condition

Thus,

*

*

From 1st boundary condition

**

**

Thus,

The final solution for the temp. distribution is then

(substitute C1 & C2)

Or

Dimensionless form

2) Cylinder with heat sources

The final solution for the temp. distribution is then

(substitute C1 & C2)

Where T0 is the temperature at r = 0. So, the temp. distribution become:

3) Hollow cylinder with heat sources

For hollow cylinder with uniformly distributed heat sources, the boundary condition

would be

The general solution same with cylinder

Used boundary condition to get C1 and C2

Where the constant, C1 given by

Basic steps involved in the solution of heat transfer problems

Thermal Contact Resistance Situation:

2 solid bars brought into contact, with the sides of the bars insulated so that

heat flow only in the axial direction. The material may have different thermal

conductivities, but if the side are insulated, the heat flux must be the same

through both materials under steady-state conditions.

The temp. drop at plane 2 (the contact

plane between 2 material) because of

“Thermal Contact Resistance”

An energy balance on the 2 materials:

Thermal Contact Resistance

Where,

Thermal Contact Resistance

Contact Coefficient

This factor extremely important in a

no. of application because of the

many heat transfer situations involve

mechanical joining of 2 materials.

Thermal Contact Resistance Physical mechanism

Examining a joint in more detail

No real surface is perfectly smooth and actual surface roughness is believed to play a

central role in determining the contact resistance.

There are 2 principle contributions to the heat transfer at the joint:

1) The solid-to-solid conduction at the spots of contact

2) The conduction through entrapped gases in the void space created by the contact

Concept:

2nd factor is believed to represent the major resistance to

heat flow because thermal conductivity of the gas is quite

small compare of solid

Thermal Contact Resistance

So, designating the contact area, AC and the void area, AV, the Heat Flow

across the joint may write as:

Contact Coefficient

: Thickness of the void space

: Thermal conductivity of the fluid which fills the void space

: Total cross-sectional area of the bars

Assignment 3

Tutorial 2

Book: J.P. Holman

1)2-30

2)2-55

3)2-117

Book: J.P. Holman

1)2-44

2)2-45

3)2-46