chapter 1 introduction 1.1 role of heat transfer...

1-1 Bachelor Degree in Mechanical Engineering

Heat Transfer (BDA 30603) N. Nordin, 2018

CHAPTER 1

INTRODUCTION

1.1 ROLE OF HEAT TRANSFER IN ENGINEERING

Heat transfer is one of the most important subjects for mechanical engineers,

as it occurs in a large number of engineering applications and by several

mechanisms.

Application of heat transfer :

a. Domestic application

Baking oven, bread toaster, boiling water.

b. Energy production

Steam power plant, gas turbine, and furnace.

c. Cooling of electronic equipment

Removal of heat that is generated due to flow as electric current

in electronic equipment.

d. Manufacturing / materials processing

Welding, brazing, soldering, foundry, machining, grinding.

e. Automobiles

Engines are cooled by radiator and passengers are cooled by

air condition.



ii. Some industrial applications of heat transfer :

a. Petroleum refining

Preheating of crude oil and fractional distillation

Evaporation of the various hydrocarbons

b. Sugar industry

Steam generation in boiler

Concentration of sugar cane juice (heating by steam)

c. Paper industry

Steam generation

Soaking of wood logs

Drying of paper on steam heated drums

d. Iron making

Smelting of core

Solidification of casting

e. Tire industry

Steam generation

Vulcanizing of rubber

Forming of tire in moulds



Example 1.1

The walls of furnace are covered with insulation. Heat is lost by convection from the

outer surface of insulation. The ambient is at 27 0C

The furnace is fired by natural gas having a calorific value of 30 MJ/m3 and a density

of 0.6 kg/m3. If gas costs RM1.00 /kg, calculate the cost of gas per kWh of heat loss

through the wall.

Solution:

1 kg natural gas ― RM 1

1 kWh ― RM ?

Assumptions:

1. Natural gas is assumed to be an ideal gas

2. There is no infiltration through the wall of furnace

Find heat released by burning 1kg of natural gas

Q = Cv = 30 x 103 kJ/m3

ρ 0.6 kg/m3

= 50000 kJ/kg

This means that 1kg of natural gas produces 50000 kJ of heat.

Conversion factors 1 kWh = 3600 kJ

Cv = 30 MJ/m3

ρ = 0.6 kg/m3

T∞ = 27 0C

Q



Heat transfer rate, by burning 1kg of natural gas

= Q = 50000 kJ

t = 3600 h

= 13.89 kWh

Amount of natural gas used to release 1kWh heat

1kg 13.89 kWh

? 1 kWh

m = 1 kWh x 1kg

13.89 kWh

= 0.072 kg

Costs of natural gas per kWh

1kg RM 1.00

0.072 kg ?

Costs of natural gas per kWh = 0.072 x RM 1.00

1kg

= RM 0.072

≈ 7 cents

Q

Q



1.2 RELATIONSHIP BETWEEN THERMODYNAMICS AND HEAT TRANSFER

1. Heat is the form of energy that can be transferred from one system to another

as a result of temperature difference.

2. Amount of heat, Q [J] can be determined by means of thermodynamic

analysis but no indication about how long the process will take.

Example 1.2

The initial conditions of gas are as follow:

Mass, m = 3.4 kg, volume, V = 0.92 m3 dan temperature, T = 17 0C. Gas is heated to

the temperarature of 147 0C. Calculate the amount of heat needed to heat up the gas

if the volume of the gas is constant.

Specific heat at constant volume, Cv = 0.72 kJ/kg.K

Solution:

Q = mCv [T2 – T1]

= 3.4 (0.72) [147 - 17]

= 318.24 kJ Only gives the idea of amount

heat being transferred.

318.24 kJ of heat needed to heat up the gas

Gas



Example 1.3

The initial conditions of gas are as follow:

Pressure, P = 293 kN/m², volume, V = 0.082 m³, temperature, T = 1800C. Gas is cooled

at constant pressure to the temperature of 24oC. By assuming R = 0.287 kJ/kg.K dan

Cp = 1.005 kJ/kg.K, calculate amount of heat needed.

Solution:

PV = mRT

293 [0.082] = m (0.287) (180 + 273)

m = 0.185 kg

Q = mCp [T2 –T1]

= 0.185 [1.005] [24 - 180]

= -29kJ

However, in practice we are more concern about the rate of heat transfer [heat transfer

per unit time, J/s], than the amount of heat itself.

29 kJ out from the gas

Gas



1.2.1 First Law Thermodynamic

1. Relate to conservation of energy.

2. What’s stated in First Law Thermodynamic?

“The rate of energy transfer into a system be equal to the rate of the

energy of that system”

Or

“Jumlah tenaga yang dibekalkan kepada sistem oleh sekeliling adalah

sama dengan jumlah tenaga yang dilakukan oleh sistem ke atas

sekeliling.”

1.2.2 Second Law of Thermodynamic

1. Related to the direction of heat transfer.

2. Heat is being transferred in the direction of decreasing temperature.

3. The basic requirement for heat transfer is the presence of a temperature

difference.

There can be no heat transfer between two bodies that are at the same

temperature.

4. Temperature difference is the driving force for heat transfer.

∑Q = -Ve ∑W= +Ve

System

Boundary

Environment ∑Q = +Ve ∑W= -Ve



5. The rate of heat transfer depends on the magnitude of the temperature

gradient, [temperature difference per unit length]

Q

6. From 2nd Law of Thermodynamic,

“Heat can flow by itself only from a higher temperature to a lower

temperature”

7. Such flow of heat from higher to lower temperature is said to be

spontaneous.

8. The principle common to all of these is that a flow takes place from a higher

potential to a lower potential.

9. The potential may be temperature, level, pressure or voltage.

10. All these processes are said to be analogous - They all have the same

underlying principles.

11. A variety of processes undergo spontaneous change. Some examples:-

a) Radio-active decay

b) Flow of rivers downhill

c) Deflation of a balloon

d) Lightning strikes

12. Non-spontaneous process:-

a) The cup of coffee did not reach a high temperature spontaneously.

It has to be heated [ forced or driven]

b) The can of chilled coke warms and reaches equilibrium with the

surroundings spontaneously. But it has to be first chilled its

temperature down by consumption of electricity in a refrigerator.



c) An electric iron is unplugged will cool spontaneously but when

power is switched on, it will heat up. This heating is a driven

process.

13. Equilibrium – if two systems are at the same temperature, they are in

thermal equilibrium.

14. Steady state – the parameter such as temperature is not change with

time at a specified location.

15. Uniform state – the parameter such as temperature is not change with

position throughout a surface or region at a specified time.

Example 1.4

Sketch the temperature history of an electric iron that is switch on at zero time and

switched off after steady state has been reached.

Solution:

Pin

Electric iron

Assumptions:

1. Iron is modeled without automatic temperature control

2. Heat transfer coefficient, h is constant.

T∞

stored

released

Q

Q



Graph temperature versus time

1.3 HEAT AND OTHER FORMS OF ENERGY

1. Energy can exist in numerous forms such as thermal, mechanical, kinetic,

potential, electrical, magnetic, chemical and nuclear.

2. Microscopic energy – the forms of energy related to the molecular structure

of a system and the degree of the molecular activity. (i.e. kinetic and potential

energy)

3. The sum of all microscopic forms of energy [kinetic and potential energy] is

called the internal energy of a system, u [KJ/Kg]

i. The portion of internal energy of a system associated with the kinetic

energy is called as sensible energy or sensible heat.

“The average of velocity and the degree of activity of the molecules

proportional to the temperature”

Temperature ,velocity ,kinetic energy Internal Energy, U

ii. The internal energy is also associated with the intermolecular forces

between the molecules of the system.

Temperature, T

Driven Steady State Spontaneous

Heating Cooling Time, t

ON OFF



There will be a phase change if sufficient energy added to the system. The

internal energy associated with the phase of a system is called latent energy

and latent heat.

u Gas >Liquid>Solid

4. The internal energy, u represents the microscopic energy of a non-flowing fluid.

Kinetic

Energy = u

Stationary fluid Potential

Microscopic energy of a non-flowing fluid

5. The enthalpy, h represents the microscopic energy of a flowing fluid.

Flowing Fluid Energy = h = u + PV

Microscopic energy of a flowing fluid

Where,

PV = flow energy of the fluid [flow work]

* Energy needed to push a fluid and to maintain flow

P = pressure [N/m2] and V = specific volume [m3/kg]

Solid Liquid Gas



2. Specific heat represents the energy required to raise temperature of a unit

mass of a substance by one degree.

i. Specific heat at constant volume, Cv [kJ/kg.K]

ii. Specific heat at constant pressure, Cp [kJ/kg.K]

3. The specific heats, Cp and Cv are related to each other by :

Cp = Cv + R

1 kJ/kg °C = 1 kJ/kg.K [Identical]

since ΔT (°C) = ΔT (K)

4. The specific heat for an ideal gas depends only on temperature.

9. The differential changes in the internal energy, u and enthalpy, h of an ideal

gas can be expressed in terms of the specific heat as:

du = Cv dT and dh = Cp dT

constant volume constant pressure

Stationary fluid flowing fluid

Differential changes in the internal energy, u and enthalpy, h

10. The finite changes in the internal energy and enthalpy of an ideal gas during a

process can be expressed approximately by using specific heat values at the

average temperature.

Δu = Cv,avg ΔT and Δh = Cp,avg ΔT [kJ/kg]

1kJ

M= 1kg ∆T = 1°C

C = 1 kJ/kg.K



or

ΔU = m Cv,avg ΔT and ΔH = m Cp,avg ΔT [kJ]

Where,

m = mass of the system [kg]

11. Incompressible substance – A substance whose specific volume (or

density) does not change with temperature or pressure [i.e. liquid & solid].

12. The Cv and Cp values of incompressible substances are identical

Cp = Cv = Cavg

13. Thus, the change of incompressible of solids and fluids can be expressed as:

Δu = m Cavg ΔT

Where Cavg is evaluated at average temperature

14. The specific heat of several common gases, liquids and solids are given in

the steam table.

Q = The amount of heat transferred during the process [kJ]

= Heat transfer rate [kJ/s or kW]

Where = Q / Δt

q = Heat flux, the rate of heat transfer per unit area normal to the

direction of heat transfer [W/m2]

q = / A = 10W

3m

q = 10/15 W/m2

Example of heat flux

5m

Q

Q

Q

Q

qQQ



1.4 ENERGY BALANCE

Ein Eout

Heat Heat

Work Work

Mass Mass

– =

Ein – Eout = ΔEsystem [J]

In the rate form,

E in –

E out = dEsystem / dt [W]

In steady state condition, 0dt

dEsystem

E in –

E out = 0

E in =

E out

In heat transfer analysis, we are usually interested only in the forms of energy that can

be transferred as a result of temperature difference, that is, heat or thermal energy.

System

Net energy transfer by heat,

work, mass

Change in internal, potential, kinetic etc.

energies

Rate of net energy transfer by heat, work & mass

Rate of change in internal, kinetic, potential



The energy balance in that case, can be expressed as:

Qin – Qout + Egen = ΔEthermal,system

Where,

Egen = heat generation – nuclear, chemical, mechanical, electrical energies etc

1. 4.1 Energy Balance For Closed System (Fixed Mass)

1. The total energy, E for most systems encountered in practice consists of the

internal energy, u.

2. This is especially the case for stationary systems since they don’t involve any

changes in their velocity or elevation during a process.

Stationary closed system: Ein – Eout = Δu = m Cv ΔT

3. When the systems involve heat transfer only and no work interactions across

its boundary, the energy balance relation further reduces to:

Stationary closed system, no work: Q = m Cv ΔT [J]

Where,

Q = the net amount of heat transfer to or from the system [Qin – Qout]

Net heat transfer

Heat generation

Change in thermal energy of the system



1.4.2 Energy Balance for Steady Flow Systems

1. A large number of engineering devices involve mass flow in and out of system

water heater and car radiators.

2. These sorts of devices are modeled as control volume and analyzed under

steady operating conditions, ΔEcv = 0.

3. The flow of a fluid through a pipe or duct can often be approximated to be one

dimensional.

4. All properties are assumed to be uniform at any cross-section normal to the

flow direction, and properties are assumed to have bulk average values over

the entire cross-section.

5. The mass flow rates of a fluid flowing in a pipe or duct can be expressed as:

m = ρ v Ac [kg/s]

where,

ρ = fluid density [kg/m3]

v = average velocity of fluid [m/s]

Ac = cross-sectional area of the pipe or duct [m2]

6. The volume flow rate of a fluid flowing in a pipe or duct can be expressed as:

v = v Ac =

m / ρ [m3/s]

7. For a steady flow system with one inlet and one outlet, the rate of mass flow

into the control volume must be equal to the rate of mass flow out of it.

m in =

m out =

m

8. The energy balance for steady flow system is:

Q =

m Δh =

m Cp ΔT [kJ/s]



Example 1.5

A 10cm diameter copper ball is to be heated from 100°C to an average temperature

of 150°C in 30 min. Taking the average density and specific heat of copper in this

temperature range to be ρ=8950 kg/m³ and Cp=0.395 KJ/kg°C, respectively,

determine :

a) The total amount of heat transfer to the copper ball, Q

b) The average rate of heat transfer to the ball,

c) The average heat flux.

Solution:

Copper ball

Assumption:

1. Heat is being transferred uniformly to the ball.

2. Constant properties can be used for the copper at average temperature.

Properties:

ρ=8950 kg/m³ and Cp= Cavg = 0.395 KJ/kg°C are taken at average temperature.

a) Q = ?

Q = Δu =m Cavg [T₂ –T₁]

m = ρV= 8950[π(0.1)³/6]

Q

D =10cm

T1 = 100°C

T2 = 150°C

t = 30 min =1800 sec

V = πD³/6 and A = πD²

Sphere



Thus,

Q = 4.687(0.395)(150-100)

= 92.6 kJ

It means that 92.6 kJ of heat needs to be transferred to the copper ball to heat the

ball from 100°C to 150°C.

b)

avg = Q/Δt

= 92.6/1800

= 0.0514 KJ/s

= 51.4 W

c)

q

q avg = avg/A

= avg/πD²

= 51.4/π (0.1)²

= 1636 W/m²

Q

Q

Q

Q



Example 1.6

1.2 kg of liquid water initially at 15°C is to be heated to 95°C in a teapot equipment

with a 1200W electric heating element inside. The teapot is 0.5 kg and has an average

specific heat of 0.7kJ/kg.K. Taking the specific heat of water to be 4.18 kJ/kg.K and

disregarding any heat loss from the teapot, determine how long will take for the water

to be heated.

Solution:

Assumptions:

1. Heat loss from teapot is negligible.

2. Steady state condition - constant properties.

Properties:

Cwater = 4.18 kJ/kg.k , Cteapot = 0.7 kJ/kg.K

From energy balance :

Ein – Eout = ΔEsystem ------------------------------------------( 1 )

No mass and work interactions, only heat transfer involves in the system, thus:

mwater = 1.3 kg mteapot = 0.5 kg

T1 = 15 °C Δt = ? T2 = 95°C

P = 1200W

Teapot



Qin – Qout = ΔEsystem stationary fluid system

Q = ΔEsystem

Q = Δuwater + Δuteapot --------------------------------------( 2 )

Q = [ mCΔT ]water + [ mCΔT ]teapot

= [1.2 x 4.18 x (95-15)] + [0.5 x 0.7 x (95-15)]

= 401.28 + 28

= 429.3 kJ

It means that, 429.3 kJ of heat needed to raise the temperature of water from 15°C to

95°C.

The time needed to supply 429.3 kJ of heat is

P = 1.2 kW = 1.2 kJ/s

1.2 kJ 1 sec

429.3 kJ ?

Δt = ( 429.3 kJ x 1 sec )/ (1.2 kJ )

= 358 sec ≈ 6 min



Example 1.7

A 5m long section of an air heating system of a house passes through an unheated

space in basement. The cross-section of the rectangular duct of the heating system is

20cm × 25cm. Hot air enters the duct at 100kPa and 60°C at an average velocity of

5m/s. The temperature of the air in the duct drops to 54°C as a result of heat loss to

the cool space in the basement. Determine the rate of heat loss from the air in the duct

to the basement under steady conditions. Also determine the cost of heat loss per hour

if the house is heated by a natural gas furnace that has an efficiency of 80%, and the

cost of the natural gas in that area is $1.60/therm. (1therm = 105500kJ)

Solution:

T2 = 54˚C

P = 100kPa

T1 = 60˚C εfurnace = 0.80

V = 5m/s

Rectangular duct

Assumptions:

1. Steady operating condition,

m in =

m out =

m

2. Constant properties can be used and taken at average temperature.

3. Natural gas is treated to be an ideal gas. PV = mRT.



Properties:

Average temperature = = 57°C

From table A-15

Cp @ T = 57°C = 1.007kJ/kg.K

From Energy Balance,

Ein – Eout = ∆Esystem ………….(1)

No mass and work interactions, only heat transfer involves in the system, thus:-

Qin – Qout = ∆Esystem--------→ steady flow system

Net heat transfer rate, in =

m Cp∆T…………..(2)

Where;

m = ρAV……………..(3)

As natural gas,it is assumed to be an ideal gas;

PV = mRT

P =

P =ρRT

ρ =

=

= 1.046kg/m3

Substitute ρ into (3)

m = ρAv

= 1.046 (0.20×0.25) × 5

= 0.2616kg/s



Thus,

=

m Cp∆T

= 0.2616 (1.007)(54-60)

= -1.58kJ/s @kW

This means in 1sec, 1.58kJ amount of heat loss through the ducting.

1sec -----------→ 1.58kJ

3600sec-------→ =?

=

= 5688kWh ≈ 5688kJ/h

1kWh = 3600kJ ------------→ conversion factor

Heat loss of using furnace with 80% efficiency

=

= 7110kWh

Thus, the cost of heat loss by using 80% efficiency of furnace;

Cost of natural gas = $1.60/therm

The cost of heat loss = Rate of heat loss × unit cost of energy input

= 7110kWh × ×

= $0.1078 ≈ 10.8 cents per hour



Example 1.8

Consider a house that has a floor space of 200 m2 and an average height of 3 m at

1500 m evaluation where the standard atmospheric pressure is 84.6 kPa. Initially the

house is at a uniform temperature at 10oC .Now ,the electric heather is turned on, and

the heather runs until the temperature of air in the house to average value of 200C

Determine the amount of energy transferred to the air, assuming:-

(a) The house is air-tight and thus no air escape during the heating process---Fixed

mass system (Cv)

(b) Some air escape through the cracks as the heated air in the house expands at

constant pressure –---steady flow system (Cp)

Also determine the cost of this heat for each case if the cost of electricity in that area

is $0.075/kWh.

Solution:

Afloor =200m2

Patm= 84.6kPa H

H = 3m

T1=100C

T2=200C

House

Assumption:

1. Steady operating condition-Constant properties could be used and taken at

average temperature.

2. Air is assumed to be an ideal gas

3. No heat generation, work and mass transfer.

Properties:

Tavg = = 15oC

From Table A-15

Cp @ T =150C =1.007kJ/kg.K, R=0.287kJ/kg.K



As air is an ideal gas ,

Cp = Cv + R

Cv = Cp – R

=1.007-0.287

=0.72kJ/kg.K

(a) Close system-Fixed mass

From Energy Balance .

Ein-Eout = Esystem

Qin-Qout = Usystem

Q = mCv T……………..(1)

Where

PV = mRT

m =

m =

m = 625 kg

Q = mCv T

= 625(0.72)[20-10]

= 4500kJ

This means 4500kJ of heat needed to rise the temperature of air in the house from

100C to 200C

1kWh 3600 kJ

? 4500 kJ

So,



=

= 1.25 kWh

The cost of energy,

1 kWh $0.075

1.25 kWh $ ?

Cost of energy =

= $ 0.094

(b) Steady flow system

From energy balance

Ein-Eout = Esystem

Qin-Qout = Hsystem

Q = Hsystem =mCp T……………..(2)

Q = mCp T

= 625(1.007)[20-10]

= 6294kJ

This means 6294kJ of heat needed to rise the temperature of air in the house from

100C to 200C

1kWh 3600 kJ

? 6294 kJ

So,

=

= 1.75 kWh



The cost of energy

1 kWh $0.075

1.75 kWh $ ?

Cost of energy =

= $ 0.131

1.5 HEAT TRANSFER MECHANISMS

1. Heat is the forms of energy that can be transferred from one system to another as

a result of temperature different

2. Heat is transferred from the higher temperature medium to the lower temperature,

and heat transfer stops when the two medium reach the same temperature

[Equilibrium].

3. Heat can be transferred in three different modes:-

a) Conduction

b) Convection

c) Radiation

1.5.1 Conduction

1. Conduction is the transfer of energy from the more energetic particles of a

substance to the adjacent less energetic ones as a result of interaction between

the particles.

2. Conduction can take place in solid, liquid and gases

Solid Liquid Gas

Mechanisms of Conduction



Solid= Lattice vibration, Flow of free electrons

Liquid & Gas = Molecular collisions, Molecular diffusion

3. The rate of conduction depends on :-

a) Geometry of the medium ,A

b) The thickness of the medium, t

c) The temperature difference, T

d) Material of the medium, k

t

T1

Where

α

=k A

4. The rate of heat conduction through a plane layer is proportional to the

temperature difference across the layer and the heat transfer area, but is

inversely proportional to the thickness of the layer.

5. Fourier’s Law is used to analyze heat transfer problem which is due to

conduction. It is stated by Fourier’s Law;

[W]

T2 A



Where;

k = the constant of proportionality, thermal conductivity of the

material [W/mk]

A = the area normal to the direction of heat transfer [m²]

= Temperature difference [K]

= Thickness [m]

6. In the limiting case of 0, the equation reduces to:

[W]

Where;

= Temperature gradient

-ve sign = show that heat is transferred in the direction of decreasing

Temperature [2ndLaw of Thermodynamics]

7. The purpose of conduction analysis is to find the temperature

distribution/gradient in the medium.

e.g. :

Temperature distribution,

Temperature gradient,

8. The major purpose of obtaining temperature distribution/gradient is to obtain

the more practical result i.e, the heat transfer rate.

1.5.1.1 Conduction Properties

Thermal conductivity, k [W/mk]

Thermal conductivity, k is a measure of a material’s ability to conduct heat.

*Specific heat is a measure of material’s ability to store heat.



e.g: Find thermal conductivity, k and specific heat, for water and iron at

room temperature.

= 4.18 kJ/kg , = 0.45 kJ/kg ,

= 0.607 W/m = 80.2 W/m

Water can store heat greater than iron, but it is a poor conductor relative to

iron.

Thermal conductivity can be defined as the rate of heat transfer through a unit

thickness through a unit thickness of the material per unit area per unit

temperature different.

k good heat conductor

k poor heat conductor (insulator)

Thermal conductivities of materials vary with temperature.

The temperature depends of thermal conductivity cause considerable

complexity in conduction analysis.

In analysis, thermal conductivity k is always assumed to be constant and taken

at the average temperature.

Thermal Diffusivity, [m²/s]

Thermal Diffusivity, represent how fast heat diffuses through a material.

[m²/s]

k , - The larger the thermal diffusivity, the faster the propagation of

heat into a medium.

k , - A small value of the thermal diffusivity means that heat is

mostly absorbed by the material and a small amount of heat is conducted

further.



1.5.2 Convection

Convection is the mode of energy transfer between a solid surface and the adjacent

liquid or gas that is in motion, and it involves the combined effects of conduction and

liquid motion.

Mechanism of convection

There are two types of convection:

i. Forced convection

The fluid is forced to flow over the surface by external means such

as at pump, blower, and etc.

ii. Natural convection

The fluids motion is caused by buoyancy forces that are induced by

density different due to the variation of temperature in the fluid.

Heat transfer process that involves change of phase of a fluid are also considered to

be convection because of the fluid motion induced during the process example Boiling

and condensation process.

Convection heat transfer can be expressed by Newton’s Law of cooling as;

[W]



Where

= Convection heat transfer coefficient [W/m² ]

Surface area though which convection heat transfer takes place [m²]

= Surface temperature [

= Fluid temperature [

Convection heat transfer coefficient is influenced by;

iii. Surface geometry

iv. The nature of the fluid

v. Properties of the fluid

vi. Bulk fluid velocity

is not a property of the fluid.[ i.e we cannot get the value of from the manual]. It

needs to be determined first before convection heat transfer problem could be

solved.

Typical values of convection heat transfer coefficient,

Type of convection [W/m² ]

Free convection of gases 2-25

Free convection of liquids 10-1,000

Forced convection of gases 25-250

Forced convection of liquids 50-20,000

Boiling & Condensation 2500-100,000

1.5.3 Radiation

Radiation is the energy emitted by matter in the form of electromagnetic waves (or

photons) as a result of the changes in the electronic configurations of the atoms or

molecules.



The transfer of heat by radiation does not require the presence of an intervening

medium- e.g. sun emissions.

Thermal radiation - The form of radiation emitted by body because of their

temperature.

Stefan- Boltzman Law is used to solve thermal radiation problem.

[W]

where,

= Stefan- Boltzman constant, 5.67 x 108 W/m

42 K

= Emissivity of the surface, 0 1 , for Blackbody, 1

1.5.3.1 Radiation Properties

Emissivity, , 0 1

This property provides a measure of how efficiently a surface emits energy

relative to a blackbody.

for blackbody is 1 – Blackbody is a perfect emitter.

This property depends strongly on the surface material and finish; and represent

active value are provided in the manual.

Absorbtivity, , 0 1

The fraction of the radiation energy incident on a surface that is absorbed by

the surface.

Blackbody is a perfect absorber ( )1 absorbs the entire radiation incident

on it.

Thermal radiation incident upon a body (or medium) may be partially reflected,

partially absorbed and the remainder transmitted.



P + 1

Mechanism of radiation

When a surface of emissivity, and surface area, A s at a thermodynamic

temperature, T s is completely enclosed by a much larger (or black) surface at

thermodynamic temperature, T surr separately by a gas (such as air) that does not

intervene with radiation. The net rate of radiation heat transfer between these two

surfaces is given by

[W]

Radiation heat transfer between a surface and the surfaces surrounding it



1.6 COMBINATION OF HEAT TRANSFER MODES

Even there are three modes of heat transfer, i.e. conduction, convection and radiation;

a medium may involve only two of them simultaneously.



Combination of heat transfer

Medium Combination of heat transfer

1. Opaque solids

2. Semitransparent solids

3. Solid exposed to a fluid or other

surfaces

4. Still fluid (no bulk fluid motion)

5. Flowing fluid

conduction

conduction and radiation

convection and/or radiation

conduction and radiation

convection and radiation

1.7 IMPORTANT TO BE REMEMBERED



Example 1.9

The roof of an electrically heated home is 6m long, 8m wide and 0.25m thick, and is

made of a flat layer of concrete whose thermal conductivity is CmWk ./8.0 (Figure

1-9). The temperatures of the inner and the outer surfaces of the roof one night are

measured to be 15 C and 4 C , respectively, for a period of 10 hours. Determine:-

(a) The rate of heat loss through the roof that night and

(b) The cost of that heat loss to the home owner if the cost of electricity is $0.08/kWh.

Solution:

?

Assumptions:

1. Steady state conditions for whole night – constant properties could be used.

2. Heat is uniformly being transferred over the entire surface of the roof.

Properties:

CmWkroof

./8.0

(a)

6m

8m

0.25m

CT 151

CT 42

hrt

CT

CT

mt

mxAroof

10

4

15

25.0

4886

2

1

2

Q

Cost = ?

Q

t

TTkAQ 21



(b) Cost of heat transfer

Heat loss within 10 hours

= 1.69(10)

= 16.9 kWh

Cost of heat transfer = 16.96kWh ($0.08/kWh)

= $1.35

25.0

415)48(8.0

kW69.1

6.1689

Q



Example 1.10

The wall of an industrial furnace is constructed from 0.15m thick fireclay brick having

a thermal conductivity of 1.7W/m.K. measurements made during steady-state

operation reveal temperatures of 1400 and 1500k at the inner and outer surfaces,

respectively. What is the rate of heat loss through a wall that is 0.5m by 1.2m on a

side?

Solution:

Assumptions:

1. Steady state conditions.

2. Constant thermal conductivity.

3. One-dimensional conduction through the wall.

4. Heat is uniformly being transferred.

Properties:

k=1.7W/m.K

Analysis:

x

TkAQ

cond

Q

1.2m 0.5m

0.15

m

Brick

wall

0.15m

?

Q

KT 14001

KT 11502



x

TTkA 12

x

TTkA 21

15.0

115014002.15.07.1 x

W1700

Example 1.11

A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15 C , as shown

in below. Heat is generated in the wire as a result of resistance heating, and the surface

temperature of the wire is measured to be 152 C in steady operation. Also, the voltage

drop and electric current through the wire are measured to be 60V and 1.5A,

respectively. Disregarding any heat transfer by radiation, determine the convection

heat transfer coefficient for heat transfer between the outer surface of the wire and the

air in the room.

Solution:

Assumptions:

1. Steady operating conditions.

2. Radiation heat transfer is negligible.

Analysis:

radQ WxIVE generated 905.160

22 01885.0)2)(103.0( mxdLAs

AI

vV

md

mL

5.1

60

3.0

2

60

V

convQ

1.5A

CTs

152CT 15



CmWh

h

TThAQ ssconv

./9.34

)15152)(01885.0(90

)(

2

Example 1.12

It is a common experience to feel “chilly” in winter and “warm” in summer in our homes

even when the thermostat setting is kept the same. This is due to the so called

“radiation effect” resulting from radiation heat exchanger between our bodies and the

surrounding surfaces of the walls and the ceiling.

Consider a person standing in a room maintained at 22 C at all times. The inner

surfaces of the walls, floors, and the ceiling of the house are observed to be at an

average temperature of 1 C in winter and 25 C in summer. Determine the rate of

radiation heat transfer between this person and the surrounding surfaces if the

exposed surface area and the average outer surface temperature of the person are

1.4 C and 30 C , respectively

Solution:

CT erwsurr

10int,

CT summersurr

25,

radQ ?

The rates of radiation heat transfer between a person and the surrounding surfaces at

specified temperature are to be determined in summer and winter.

32C

radQ

surrT

CT

mA

s

s

30

4.1 2



Assumptions:

1. Steady operating conditions.

2. Heat transfer by convection is not considered.

3. The surrounding surfaces are at a uniform temperature.

Properties:

The emissivity of a person is = 0.95 (Table 1-6, Cengel, Y.A., 2006).

Analysis:

)( 44

int, surrsserwrad TTAQ

= (0.95)(5.67 810x )(1.4)[ (30+273)4

- (10+273)4

]

= 152W

)( 44

, surrsssummerrad TTAQ

= (0.95)(5.67810x )(1.4)[ (30+273)

4- (25+273)

4]

= 40.9W

chapter 1 introduction 1.1 role of heat transfer...

Documents