CHAPTER 2

Conduction

1) Steady state conduction2) Unsteady state conduction

(1) Conduction

• Temperature different: T1 and T2 [K]

• Thermal conductivity: k [W m-1 K-1]

• Heat transfer area: A [m2]

• Thickness of the material: x [m]

• Heat transfer rate by conduction: qc [??]

• Heat flux: q/A [??]

Fourier’s Law

(The rate of heat flow by conduction is proportional to the area measured normal to thedirection of heat flow, and to temperature gradient in that direction.)

Heat transfer by conduction from a high temperature region to the low

temperature region

The driving force of this heat transfer is the temperature gradient

Fourier’s Law

Thermal conductivity, k

Experiment measurement made to

determine the value of k (different

material, different k)

The k value is a physical property

of each solid, liquid and gas

material

k is strongly temp-dependent

Unit W m-1 K-1

The numerical value of k indicates

how fast heat will flow in a

given material.

If molecules move fasters (gas),

the faster they will transport

energy.

So, value of k depends on the

molecules structure (gas, liquid or

solid) K of various materials at 0 °C

Thermal conductivity, k

Why we should know k of each

material???

(2) ConvectionNewton’s law of cooling

Temperature different between wall and fluid: Tw and Tf [K]

Convective heat transfer coefficient: hcv [W m-2 K-1]

Heat transfer area: A [m2]

Heat transfer rate by convection: qcv [??]

Heat flux: q/A [??]

Also referred to as the “Newton rate equation” or “Newton’s law of

cooling”

(3) RadiationSteafan-Boltzmann law

This is most commonly used law of thermal radiation. It states that the

thermal radiation heat flux emitted from a black surface is proportional

to the fourth power of the absolute temperature of the surface.

Q/A =σT4

where σ is the constant of proportionality and is called the Steafan-

Boltzmann constant. In SI units, it has a value of 5.67 x 10-8 W/(m2K4).

Plancks law, Wien’s law and Kirchoffs law are also used.

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 functiononly of radial distance andindependent of azimuth angle/axial distance

Radial Systems

(1) The Plane Wall

Integrated Fourier’s law

If k varies with temp.

according linear relation, the heat flow become;

# 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

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)

Electrical analog circuit

Parallel

Series

Insulation & R value

The performance of insulation R value, define as

The units for R is °C m2 /W

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

(2) Cylinders (Radial System)

From Fourier’s Law

For a cylinder with length very large compared to diameter, it may be assumed that the heat flows only in a radial direction

Thermal resistant for cylinder is

Boundary conditions

The heat flow rate

Electrical analog circuit

Multiple cylindrical sections

Thermal-resistance concept for

multiple-layer cylindrical walls =

Thermal-resistance concept for

plane wall

So that, the heat flow rate

Electrical analog circuit

(3) Spheres

The heat flow rate

Prove this equation!!

Convection Boundary Conditions

Newton rate equation /Newton’s Law of cooling

So that, an electric-resistance analogy for convection process become:

Example 1: (Multilayer plane wall conduction)

An exterior wall of a house may be approximated

by a 4-in layer of common brick (k= 0.7 W/m.°C)

followed by a 1.5-in 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. °C

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

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

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

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

The Overall Heat Transfer Coefficient, U

The general notion (plane wall or cylinder coordinate system) is that;

Rth Thermal resistance

Info

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

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

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

Calculate the critical radius of insulation for

asbestos [k =0.17 W/m◦C] surrounding a pipe

and exposed to room air at 20◦C with h=3.0 W/m2

◦C. Calculate the heat loss from a 200◦C, 5.0-cm-

diameter pipe when covered with the critical

radius of insulation and without insulation

Answer: q/L with insulation =105.7W/m

q/L without insulation =84.8W/m

Example 5

Book: J.P. Holman

1) 2-8

2) 2-9

3) 2-17

4) 2-18

5) 2-26

6) 2-62

Problems