lecture 1 introduction 140115 [compatibility mode]

8/10/2019 Lecture 1 Introduction 140115 [Compatibility Mode]

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INTRODUCTION ANDBASIC CONCEPTS

CDB 2023: PROCESS HEAT TRANSFER

Jan Semester 2015

DR. YEONG YIN FONG

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Chapter 1: Introduction and Basic

Concepts Thermodynamics and Heat Transfer

Application of Heat Transfer in Process

Industries Heat Transfer Mechanisms

Units and Dimensions

Outline

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At the end of this session:

1) Understand how thermodynamics and heat transferare related to each other.

2) Understand the basic mechanisms of heat transfer,

which are conduction, convection, and radiation, andFourier's law of heat conduction, Newton's law ofcooling, and the StefanBoltzmann law of radiation.

3) Identify the mechanisms of heat transfer that occursimultaneously in practice.

Lesson Outcomes:

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Cengel, A. Y. and Ghajar, J. A., Heat and Mass

Transfer: Fundamentals and Applications, 5th Ed.

McGraw Hill 2015.

Holman, J. P. Heat Transfer, 10th Ed., McGraw Hill,

2009.

F. P. Incropera, D. P. Dewitt, T. L. Bergman, A. S.

Lavine. Fundamentals of Heat and Mass Transfer, 6th

Ed. Wiley, 2007.

Reference Books:

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Application of Hear Transfer in Process



Outline

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Heat: The form of energy that can be transferred fromone system to another as a result of temperaturedifference.

Thermodynamics is concerned with the amount of heattransfer as a system undergoes a process from one

equilibrium state to another.

Heat transfer deals with the determination of the rates

of such energy transfers as well as variation oftemperature.

Thermodynamics and Heat Transfer

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The transfer of energy as heat is always from the

higher-temperature medium to the lower-temperatureone (temperature difference).

Heat transfer stops when the two mediums reach thesame temperature.

The larger the temperature gradient/difference, thehigher the rate of heat transfer.

Heat can be transferred in three different modes:

i) conduction,

ii) convection, and

iii)radiation


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First law: The rate of energy transfer into a

system is equal to the rate of increase of theenergy of that system (also known as theconservation of energy principle: energy canneither be created nor destroyed; it can only

change forms).

Second law: The heat is transferred in thedirection of decreasing temperature.

Thermodynamic Laws

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The science of thermodynamics deals with theamount of heat transfer as a system undergoes aprocess from one equilibrium state to another, andmakes no reference to how long the process will take.

Where as in engineering, we are often interested inthe rate of heat transfer.

However, the law of thermodynamics lay theframework for the science of heat transfer.


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Outline

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Heat transfer is commonly encountered in engineeringsystems and other aspects of life.

The human body is constantly rejecting heat to itssurroundings.

The heating and air-conditioning system, refrigerator

or freezer, water heater, iron and even the computer,TV.

Heat transfer plays a major role in the design of many

devices ie., car radiators, solar collectors, variouscomponents of chemical plants.

Exchange of heat between two fluids is a widely usedunit operation in chemical process industries.

Applications of Heat Transfer

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Chapter 1: Introduction and BasicConcepts





Outline

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When two objects at different temperatures arebrought into contact, heat flows from the object at

the higher temperature to that at the lowertemperature.

Heat is thermal energy in transit due to a spatialtemperature difference, flowing from hightemperature to low temperature.

The mechanisms (modes) by which the heat mayflow are three: Conduction, Convection, andRadiation.

Heat Transfer Mechanism

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Conduction is the transferof energy from the more

energetic particles of asubstance to the adjacentless energetic ones as aresult of interactions

between the particles.

Conduction can take

place in solids, liquids, orgases.

Conduction

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Fouriers law of heat conduction: Heatflux is proportional to the temperaturegradient.

k is the thermal conductivity of thematerial, which measure of the abilityof a material to conduct heat.

Eq (1.21)(W)21.

x

TkA

x

TTkAQ

cond

=

=

In heat conduction analysis, Arepresents the area normal to the

direction of heat transfer.


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In metals, thermal conduction is by the motion of freeelectrons.

In poor conducting solids, thermal conduction is bythe momentum transfer between vibrating moleculesor atoms.

In liquid and gases, conduction occurs by randommotion of molecules, so called thermal collision anddiffusion.


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The mechanisms ofheat conduction indifferent phases of a

substance.


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The range of thermal conductivity of various materials at room

temperature


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The variation of the thermal conductivity of various solids, liquids, and

gases with temperature.


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The roof of an electrically heatedhome is 6m long, 8m wide, and0.25m thick, and is made of a flat

layer of concrete whose thermalconductivity is k = 0.8 W/m.K. Thetemperature of the inner and theouter surfaces of the roof one nightare measured to be 15C and 4 C,

respectively, for a period of 10hours. Determine:a) The rate of heat loss through theroof that night, and

b) The cost of that heat loss to thehome owner if the cost of theelectricity is $0.08/kWh.

Example 1.5: The Cost of Heat Lossthrough a Roof

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Given: k = 0.8 W/m.K, A = 6 m x 8 m = 48 m2, T1= 15oC and T2 = 4

oC,

Determine:

a) The rate of heat transfer

The steady rate of heat transfer through the roofis:

Solution

25

Wm

Cm

Km

W

x

TTkAQ

cond1690

25.0

)415(48

.8.0

0221

.

=

=

=

Eq (1.21)(W)21

.

x

TTkAQ

cond

=


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b) The cost of that heat loss

For 10 hours period, the amount of heat lost and itcost:

Cost = amount of energy x unit cost of energy= 16.9 kWh x $ 0.08/kWh

= $1.35

16.9kWh10h1.69kW.

=== tQQ cond


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The convective flux is proportional to the difference

between the surface temperature and the fluid

temperature, referred as Newtons law of cooling.

where Q = heat flow rate, As= surface area of heat

transfer, h = heat transfer coefficient (W/m2.K), Ts= surface

temperature, = temperature of the fluid

)W()(.

= TThAQ ssconv Eq (1.24)

T

Heat transferfrom a hotsurface to air byconvection.


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Forced convection: If the fluid is forced to flow overthe surface by external means such as a fan, pump,or the wind.

Natural (or free) convection: If the fluid motion is

caused by buoyancy forces that are induced bydensity differences due to the variation of temperaturein the fluid.

Natural and Forced Convection

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The cooling of a boiled egg by forcedand natural convection.


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Given: wire = 2m long, 0.3cm diameter, voltage drop= 60V, 1.5 A. room temperature = 15 oC, surface

temperature of wire =152 oC.

Determine h.

When steady operating conditions are reached, therate of heat loss from the wire equals the rate ofheat generation in the wire as a result of resistance

heating. That is

Solution

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W90A1.5V60VI..

==== generatedEQ


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The surface area of the wire is:

Newtons law at cooling for convection heat

transfer is express as

Rearrange Eq 1.24, the convection heat transfercoefficient is determined to be:

2m0.01885m)(0.003m)(2DL === sA

)W()(.

= TThAQ ssconv Eq (1.24)

Km

W9.34)15152)(m(0.01885

W90

)(22

.

=

=

=

CTTA

Qho

ss

conv


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Radiation is the energy

emitted by matter in the

form of electromagneticwaves (or photons) as a

result of the changes in the

electronic configurations of

the atoms or molecules.

Does not require the

presence of an intervening

medium. Example: energy of the sun

reaches the earth.

Radiation

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Th i di ti fl itt d b b d t


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The maximum radiation flux emitted by a body at

temperature T is given by Stefan-Boltzmann law

where Ts is absolute temperature in kelvins, As is the

surface area is the Stefan-Boltzmann constant [ =

5.6697 x 10-8 W/(m2 . K4)].

The idealised surface that emits radiation at this maximum

rate is called blackbody.

(W)4

max,

.

ssemit TAQ = Eq (1.25)


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The radiation emitted by

all real surfaces is less

than the radiation emitted

by a blackbody at the

same temperature and is

expressed as:

where , emissivity lies

between 0 and 1

(W)4.

ssemit TAQ =

Eq (1.26)


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When a surface of emissivity and surface area As ata temperature Ts is completely enclosed by a much

larger surface at temperature Tsurr, the net rate of

radiation heat transfer between these two surfaces is:

(W))T-( surr44

.

ssrad TAQ = Eq (1.28)


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Given: Tsurr,winter = 10oC, Tsurr,summer = 25

oC, Ts = 30oC, As

= 1.4 m2

= 0.95 (Table 1-6), = 5.6697 x 10-8 W/m2 . K4

The net rates of radiation heat transfer form the body tothe surrounding walls, ceiling, and floor in winter andsummer are: (from equation 1.28)

Solution

40W152

K])27310(273)[(30

)m4.1)(KW/m107(0.95)(5.6

)T-(

444

2428-

surr44

winterrad,

.

=

++

=

= ss TAQ


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W40.9

K])27325(273)[(30

)m4.1)(KW/m107(0.95)(5.6

)T-(

444

2428-

surr44

summerrad,

.

=

++

=

= ss TAQ

Simultaneous Heat Transfer


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Heat transfer is only by conduction in opaque (dense)solids, but by conduction and radiation in semitransparentsolids a solid may involve conduction and radiation but

not convection.

A solid may involve heat transfer by convection and/orradiation on its surfaces exposed to fluid or other surfaces.

For examples: the outer surface of a cold piece of rock willwarm up in a warmer environment as a result of

i) heat gain by convection (from the air)

ii) radiation (from the sun or the warmer surroundingsurface)

But the inner part of the rock will warm up as this heat

transferred to the inner region of the rock by conduction.

Simultaneous Heat TransferMechanism

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Heat transfer is by conduction


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Heat transfer is by conductionand possibly by radiation in a

still fluid (no bulk fluid motion)and by convection and radiationin a flowing fluid.

When deal with the heattransfer through a fluid, wehave either conduction orconvection, but not both.

Heat transfer through a vacuumis by radiation only sinceconduction or convectionrequires the presence ofmaterial medium.

Although there are three mechanisms ofheat transfer, a medium may involve

only two of them simultaneously.

Example 1-10: Heat Loss From a


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Consider a person standingin a breezy room at 20 oC.

Determine the total rate ofheat transfer from thisperson if the exposedsurface area and the

average outer surfacetemperature of the personare 1.6 m2 and 29 oC,

respectively, and theconvection heat transfercoefficient is 6 W/m2.K

Example 1 10: Heat Loss From aPerson

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Given: Room temperature (surrounding) = 20oC, surface temperature of the person = 29 oC,

surface area = 1.6 m2 , convection heat transfercoefficient = 6 W/m2.K

Determine: The total rate of heat transfer from aperson by both convection, Qconv, (Eq 1.24) andradiation, Qrad, (Eq 1.28) to the surrounding air

and surfaces at specified temperatures.

Solution

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The rate of heat transfer through convection (Eq 1 24):


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The rate of heat transfer through convection (Eq 1.24):

The rate of heat transfer through radiation (Eq 1.28):

The total heat transfer from the body

= 86.4 +81.7 = 168 W

W86.4

C20)-)(29mK)(1.6W/m(6

)(o22

.

=

=

= TThAQ ssconv

W81.7

K])27320(273)[(29

)m6.1)(KW/m107(0.95)(5.6

)T-(

444

2428-

surr44

.

=

++

=

= ssrad TAQ

Example 1-11: Heat Transfer between


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Consider steady heat transferbetween two large parallel plates atconstant temperature of T1 = 300 Kand T

2

= 200 K that are L =1 cmapart. Assuming the surfaces to beblack (emissivity =1), determine therate of heat transfer between theplates per unit surface area

assuming the gap between theplates is

(a) filled with atmospheric air

(b) evacuated

(c) filled with urethane insulation(d) filled with superinsulation that

has an apparent thermalconductivity of 0.00002 W/m.K

Example 1 11: Heat Transfer betweenTwo Isothermal Plates

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Table A-15,pg 924


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Table A 15,pg 924

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-23oC

(b) When the air space between the plates is evacuated


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(b) When the air space between the plates is evacuated,there will be no conduction or convection, and the onlyheat between the plates will be radiation, therefore,

QTotal= Qrad = 369 W

(c) The urethane blocks direct radiation heat transferbetween the plates. Thermal conductivity for urethane is0.026 W/m.K (Table A-6). The rate of heat transfer:

QTotal= Qcond = 260 W

W260m01.0

K)200300(1m

Km

W026.0

221.

=

=

=

x

TTkAQ

cond


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(d) Given, apparent thermal conductivity of superinsulation,k 0 00002 W/ K


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k= 0.00002 W/m.K.

Note that the layer of the superinsulation prevent any directradiation heat transfer between the plates

W2.0m01.0

K)200300(1m

Km

W00002.0

2

21

..

=

=

==

x

TTkAQQ

condTotal

Outline


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Chapter 1: Introduction and BasicConcepts


Application of Heat Transfer in ProcessIndustries

Heat Transfer Mechanisms


Outline

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Units are scales used to quantify the dimensions in a


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standard way.

Systems of units: CGS, SI, and American EngineeringSystem.

Base units are the units for the base dimensions.

Multiple units multiples or fractions of base units. E.g.,minutes, hours, milliseconds, all of which are defined in

terms of the base unit of time, second.

Derived units units for the derived dimensions.a) By multiplying and dividing base or multiple units (e.g.,

cm2

, ft/min, kg.m/s2

). Derived units of this type arereferred to as compound units.b) As defined equivalents of compound units

(1 lbf = 32.174 lbm.ft/s2).

SI and CGS Systems of Units


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SI and CGS are metric systems of units.

The base units in SI (International System of Units) are:

meter (m) for length, kilogram (kg) for mass, second (s) fortime, Kelvin (K) for temperature, and mole (mol) for the

amount of substance.

CGS system : almost identical to SI, the difference being

that gram (g) and centimeters (cm) are used instead of

kilograms & meters as the base units of mass and length.

SI has gained widespread use in international scientific

and engineering community.

SI and CGS Systems of Units

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American/British Engineering System


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Base units: foot (ft) for length, pound-mass (lbm) formass, second (s) for time.

Still widely used in the United States.

The derived unit of force is pound-force (lbf), which isdefined as 1 lbf = 32.174 lbm ft/s2.

Derived Units of Force in Various Systems

American/British Engineering System

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Units in Heat Transfer


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Q (heat flow) J/s or W

q (heat flux) J/(s-m2) or W/m2

k (thermal conductivity) W/m oC or W/m K

h (heat transfer coefficient) W/m2 oC or W/m2 K

(Stefan-Boltzmann constant) W/(m2 . K4)

Units in Heat Transfer

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End of Chapter

lecture 1 introduction 140115 [compatibility mode]

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