(Only Theory) Heat Mass Transfer

8/11/2019 (Only Theory) Heat Mass Transfer

1/35


2/35

Indias No.1

IES

Academy

25 - 1st Floor, Jia Sarai , Near IIT, New Delhi-110016 Ph: 011-26537570 , 9810958290 2

HEATM

ASSTRANSFER

MODES AND BASIC LAWS OF HEAT TRANSFER

The literature on heat transfer generally recognises

three distinct modes of heat transmission :

conduction , convection and radiation. These three

modes are similar in that a temperature differen-

tial must exist and the heat exchange is in be di-

rection of decreasing temperature. Each method

has, however, different physical picture and dif-

ferent controlling laws.

Conduction :Thermal conduction is a mechanism

of heat propagation from a region of higher tem-

perature to a region of low temperature with in a

medium (solid, liquid, or gaseours) or between dif-

ferent mediums in direct physical contact. Conduc-

tion does not involve any movement of macro-

scopic portions of matter relative to one another.The thermal energy may be transferred by means

of electrons which are free to move through the

lattice structure of the material. In addition, or al-

ternatively. It may be transferred as vibrational

energy in the lattice structure. Irrespective of the

exact mechanism, the observable effect of conduc-

tion is an equalization of temperature.

Consider the flow of heat along a metal rod,

one end of which is placed adjacent to a flame.

The elementary particles (molecules, atoms, elec-

trons) composing the rod, and which are in imme-diate vicinity of the flame, get heated. Because of

the resulting temperature growth, their kinetic en-

ergy increases and this puts them in a violent state

of agitation, and they start vibrating about their

mean positions. Consequently, these more active

particles collide with less active molecules lying

next to them. During collision, the less active par-

ticles also get excited, i.e.., thermal energy is im-

parted to them. The process is repeated for layer

after layer of molecules until the other end of therod is reached. Each layer of molecules is at a

slightly higher temperature than the preceding one,

i.e., a temperature gradient exists along the length

of the rod. The rate of heat flow between the two

ends depends upon the lengths of the rod, tempera-

ture difference the two ends, and the physical and

chemical composition of the bar material.

Fig 1

Fig : Conduction heat flow along a rod.

Since conduction is essentially due to random

molecular motion, the concept is termed as micro-

form of heat transfer and is usually referred to as

diffusion of energy. The rate equation for one-di-

mensional steady flow of heat by conduction is

prescribed by the Fourier Law :

......................1.2dt

Q kAdx

=

where, Q is the heat transfer rate. A is the area of

heat transfer surface. , dt is the temperature differ-

ence for a short perpendicular distance dx, and the

thermal conductivity k is a characteristic of the

surface material. Since the temperature gradient is

negative in the positive x-direction , the minus signin the equation gives positive heat flow.

If is the path length in the direction of heat flow

and

1 2( )t t

is the temperature difference, then

1 2( )kA t t

Q

=

............................1.3

The heat flux q is heat conducted per unit time per

unit area and is given by

1 2( )Q k t t

q A

= = ...........................1.4

Heat transfer in metal rods, in heat treatment of

steel forgings and through the walls of heat ex-

change equipment are some practical examples of

heat conduction.

Example:A 75 cm thick side will of an oven is

primarily made of insulation with a thermal con-

ductivity of 0.0345 kcal/m hr K (0.04 W/m K.).


3/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

Conditions on the inside of wall fix the tempera-

ture on that side at 420 K. The electric coil with in

the oven dissipate 31.4 kcal/hr (36.5 watts) of elec-

trical energy to made up for the heat loss through

the wall. Calculate the wall surface area, perpen-

dicular to heat flow, so that temperature on the other

side of the wall does not exceed 310 K.

Solution :Under the stipulations of one-dimen-

sional steady state heat conduction , the electrical

energy dissipation rate with in the oven must equal

the conduction heat conduction heat transfer rate

across the wall. That is :

1 2( )kA t t

Q

=

0.0345 (420 310)31.4 50.6

0.75

AA

= =

Hence , the required wall surface area,

231.4 0.6250.6

A m= =

Convection :Thermal convection is a process of

energy transport affected by the circulation or mix-

ing of a fluid medium (gas, liquid or a powdery

substance). Convection is possible only in a fluid

medium and is directly linked with the transport

of medium itself. Macroscopic particles of a fluid

moving in space cause the heat exchange, and thus

convection constitutes the microform of the heattransfer. The effectiveness of heat transfer by con-

vection depends largely upon the mixing motion

of the fluid.

With respect to origin, two types of con-

vection are distinguished; forced, and natural or

free convection.

In natural or free convention, the circula-

tion of the fluid medium is caused by buoyancy

effects, i.e., by the difference in the densities of

the cold and heated particles. Consider heat flow

from a hot plate to atmosphere. The stagnant layer

of air in the immediate vicinity of the plate gets

thermal energy by conduction. The energy thus

transferred serves to increase the temperature and

internal energy of the air particles. Because of tem-

perature rise these particles become less dense (and

therefore lighter) than the surrounding air. The

lighter air particles move upwards to a region of

low temperature where they mix with transfer a

part of their energy to the cold particles. Simulta-

neously the cold air particles descend downwards

to fill the space vacated by the hot air particles.

The circulation pattern, upward movement of the

warm air and the downward movement of the cool

air, is called the convection currents.

A similar effect can also be demonstrated

by a hot-water heating system. Fig. where water

serves as the medium for carrying heat to all parts

of the building. Water is heated in the boiler in-

stalled at the base of the building. The hot water

becomes lighter, rises up in the left hand vertical

pipes and passes through the radiators fitted in dif-

ferent rooms of the building. The radiators get

heated and dissipate heat to the rooms. After los-

ing heat to the radiators, the water gets cooled and

returns back to the boiler through the pipe on theright. Convection currents are setup and the build-

ing

Fig 2

is kept warm continuously at a constant tempera-

ture. In this way, a constant circulation of water

through the pipes and through the radiator is main-

tained. Some other examples of free convectionare chilling effect of a cold wind on a warm body;

heat flow from a hot pavement to surrounding at-

mosphere and heating of air in a room by a stove;

cooling of billets in the atmosphere; heat exchange

on the outside of cold and warm pipes.

In forced convection, the flow of fluid is

caused by a pump, fan or by atmospheric winds.

These mechanical devices provide a definite cir-


4/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

cuit for the circulating currents and that speeds up

the heat transfer rate. Example of forced convec-

tion are; flow of water in condenser tubes, fluid

passing through the tubes of a heat exchanger; cool-

ing of internal combustion engine; air condition-

ing installation and nuclear reactors.

Regardless of the particular nature. The ap-

propriate rate equation for the convective heat trans-

fer between a surface and an adjacent fluid is pre-

scribed by Newtons Law of cooling :

1 2( )Q hA t t = ..........................1.5

where, Q is the convective heat flow rate, A is area

exposed to heat transfer, ts and tf are the surface

and fluid temperature respectively. The heat trans-

fer coefficient h depends upon the thermodynamic

and transport properties (e.g. density, viscosity,

specific heat and thermal conductivity of the fluid.)

the geometry of the surface the nature of fluid flow,

and the prevailing thermal conditions.

Convection mechanisms involving phase

changes leads to the important fields of boiling

(evaporation) and condensation.

Example :An oil cooler in a high performance

engine has an outside surface area 0.12 m2and a

surface temperature of 650C. The air rushes over

the surface of the cooler at a temperature of 300C

and gives rise to a surface coefficient of heat trans-

fer equal to 45.4 W/m2

K. Calculate the heat trans-fer rate from the cooler.

Solution :The conditions described imply a con-

vective process, that is , heat transfer from a solid

surface (the oil cooler) to an adjacent moving fluid

(the air passing over the cooler.) The rate of heat

transfer by convection from oil cooler to the air is

then

( )

45.4 0.12(65 30) 190.68

S FQ hA t t

W

=

= =

Radiation :Thermal radiation is the transmissionof heat in the form of radiant energy or wave mo-

tion from one body to another across an interven-

ing space. Unlike heat transfer by conduction and

convection, transport of thermal radiation does not

necessarily affect the material medium between the

heat source and the receiver. An intervening me-

dium is not even necessary and the radiation can

be affected through vacuum or a space devoid of

any matter. Radiation exchange, in fact, occurs most

effectively in vaccum. A material present between

the heat source and the receiver would either re-

duce or eliminate entirely the propagation of ra-

diation energy.

The mechanism of the heat flow by radiation con-

sists of three distinct phases :

(i) Conversion of thermal energy of the hot source

into electromagnetic waves :

All bodies above abosolute zero temperature are

capable of emitting radiant energy. Energy released

by a radiating surface is not continuous but is in

the form of successive and separate (discrete) pack-

ets or quanta of energy called photons. The pho-

tons are propagated through the space as rays; the

movement of swarm of photons is described as the

electromagnetic waves.

(ii) Passage of wave motion through-interveningspace :

The photons , as carries of energy, travel with un-

changed frequency in straight paths with speed

equal to that of light.

(iii) Transformation of waves into heat :When

the photons approach the cold receiving surface

there occurs reconversion of wave motion into ther-

mal energy which is partly absorbed, reflected or

transmitted through the receiving surface.

Thermal radiation is limited to range of wavelength

between 0.1 and 100 of the electromagnetic spec-

trum. Thermal radiations thus include the entire

visible and infrared, and a part of ultra violet spec-

trum. It is to be recognized that thermal radiation

is the transfer of energy by disorganized photon

propagation. In contrast, an organized photon en-

ergy such as radio transmission can be macroscopi-

cally identified and is not considered heat. Further,

emission of thermal radiations is associated with

thermally excited conditions which depend upon

the temperature and nature of the surface.The most vivid evidence of radiation heat

transfer is that represented by solar energy which

passes through inter stellar (conditions close to that

perfect vaccum) on its way to the earth space. So-

lar radiation plays an important part in the design

of heating and ventillating systems. Heat transfer

by radiation is encountered in boiler furnaces, bil-

let reheating furnaces and other types of heat ex-


5/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

change apparatus. The design and construction of

engines, gas turbines, nuclear reactors and solar

collectors is also significantly influenced by the

radiation heat transfer.

The basic rate equations for heat transfer

are based on Stefan-Boltzman Law :

4

b bE AT=

.....................1.6where, E

bis the energy radiated per unit time, T is

the absolute temperature of the surface, and b is

the Stefan-Boltzman constant.

8 2 2

8 2 4

5.67 10 /

4.86 10 /

b W m K

kcal m hr K

=

=

Equation 1.6 is essentially valid for an ideal radia-

tor or a black body - suffix b designates a black

surface. The radiant energy emitted by a real sur-

face is less than for an ideal emitter and is given byA

bE AT=

where is a radiative property of the surface andis called emissivity ; its value depends upon sur-

face characteristics and temperature. It indicates

how effectively the surface emits radiations com-

pared to an ideal or black body radiator. Normally

a body radiating heat is simultaneously receiving

heat from other bodies as radiation. Consider that

surface 1 at temperature T1is completely enclosed

by another black surface 2 at temperature T2. Thenet radiant heat transfer is

4 4

1 1 2( )bQ A T T =

....................1.8

Likewise, the net rate of heat transfer between the

real surface ( called gray surface) at temperature

T1to a surrouding black surface at temperature T

2

is

4 4

1 1 1 2( )bQ A T T = ...................1.9

The net exchange of heat between the two radiat-

ing surfaces is due to the fact that one at the higher

temperature radiates more and receives less energy

for its absorption. An isolated body which remains

at constant temperature emits just as much energy

by radiation as it receives.

Example :A radiator in a domestic heating sys-

tem operates at a surface temperature of 600C. Cal-

culate the heat flux at the surface of the radiator if

it behaves as a black body.

Solution: The heat flux at the surfaces is the rate

at which radiant energy leaves the surfaces per unit

area.

4

8 2 25.67 10 (273 60) 697.2 /

b

Qq T

A

W m

= =

= + =

STEADY AND UNSTEADY HEAT TRANS-

FER

Any physical phenomenon generally involves a

changes of its physical properties. Likewise heat

exchange is also accompanied by space-time varia-

tion of temperature, and the analytical computa-

tions for the amount of heat exchange lie in deter-

mining a mathematical relation for the tempera-

ture field prescribed as

( , , , )t f x y z = .................1.10

Equation 1.10 refers to the entries set of tempera-

ture at all points of space studied at any instant of

time

.

Heat exchange between two systems may take place

under steady (stable) thermal conditions or under

unsteady (unstable) thermal condition, Steady state

implies that temperature at each point of the sys-

tem remains constant in the course of time, and it

is a function only of space co-ordinates.

( , , ); 0dt

t f x y zd

= =

....................1.11

Steady state results in a constant rate of heat ex-

change (heat influx equals heat effux), and there is

no change in the internal energy of the system dur-

ing such a process. Typical examples of steady state

heat transfer are : Cooling of an electric bulb by

surrounding atmosphere; heat flow from the prod-

ucts of combustion to water in the tubes of a boiler,

from the hot to cold fluid in a heat exchanger, and

from a refrigerated space to cooling surface of the

evaporator.Under unsteady thermal conditions, temperature of

the system changes continuous with time. Tempera-

ture is obviously a function of space and time co-

ordinates.

( , , , ); 0dt

i f x y zd

=

.........................1.12

Unsteady state results in heat transfer rate which


6/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

changes with time. Further, a change in tempera-

ture, indicates a change of internal energy of the

system. Energy storage is thus a part and parcel of

unsteady heat flow. Typical examples of unsteady

heat transfer are : Warm-up periods of furnaces,

boilers and turbines; cooling of castings in a

foundry; heat treatment and stress relieving of metal

castings.

A special kind of unsteady process is the

transient static wherein the system is subjected to

cyclic variations in the tempeature of its environ-

ment. The temperature at a particular point of the

system returns periodically to the same value; the

rate of heat flow and energy storage also undergo

periodic variations. Examples are : Heating or cool-

ing of the water of an I.C engine; heating or cool-

ing of the walls of a building during the 24 -hour

cycle of the day.Further , the heat transfer in a system may be in

one, two or more directions. In a one dimensional

heat flow, there is a single predominant direction

in which tempeature differential exists and obvi-

ously the heat flow takes place; heat flow in the

other two directions can be safety neglected. When

the temperature is a function of two co-ordinates ,

heat flow is two-dimensional . A three-dimensional

heat flow stipulates that temperature is a function

of three coordinates. A three-dimensional heat flow

stipulates that temperature is a function of three

co-ordinates, and consequently heat flow occurs

in all three directions.

Dimensionality of temperature field for

steady/unsteady heat flow can be mathematically

expressed as:

_____________________________________________

Type of heat flow Steady Unsteady

_______________________________________________

one-Dimensional t = f(x) t = f(x,

)

heat flowTwo-Deimensional t - f(x,y) t = f(x,y,

)

heat flow

Three-Dimensional t = f(x,y,z) t =

f(x,y,z,

)

het flow

________________________________________________

For simplicity, solutions to majority of heat trans-

fer problems are obtained by the one-dimensional

analysis.


7/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

FOURIER EQUATION

Conduction is primarily a molecular phenomenon

requiring temperature gradient as the driving force

. Experimental evidence does indicate that steady-

state one dimensional flow of heat by conduction

through a homogeneous material is given by the

Fourier Law.

...................2.1

dtQ kA

dx

Q dtq k

A dx

=

= =

The heat flux q (heat conducted per unit time per

unit area) flows along normal to area A in the di-

rection of decreasing temperature. The units oneach term are :

Q : rate of heat flow, kcal/hr or kJ/hr

A : area perpendicular to the direction of heat flow,

m2

dx : thickness of material along the path of heat

flow, m

dt : temperature difference between the two sur-

faces which heat is passing, degree Kelvin K or

degree centigrade C.

The ratio dt/dx represents the change in tempera-ture per unit thickness, i.e., the temperature gradi-

ent. The negative sign indicates that the heat flow

is in the direction of negative temperature gradi-

ent, and that serves to make the heat flow positive.

The proportionality factor k is called the heat con-

ductivity or thermal conductivity of the material.

Thermal conductivity of a material is one of its

transport properties. Others are the viscosity asso-

ciated with the transport of moment , and the dif-

fusion coefficient associated with the transport of

mass. Thermal conductivity provides an indicationof the rate at which heat energy is transferred

through a medium by the diffusion (conduction)

process. For a prescribed temperature gradient and

geometric parameters, the heat flow rate increase

with increasing thermal conductivity.

The Fourier law is essentially based on the follow-

ing assumptions :

FOURIER EQUATION AND THERMAL CONDUCTIVITY

Steady state conduction which implies that the

time rate of het flow between any two selected

points is constant with time. This also means that

the temperature of the fixed points within a heat

conducing body does not change with time :

( )t f

one-directional heat flow; only one space co-

ordinate is required to describe the temperature dis-

tribution within the heat conducting body; t = f(x).

The surfaces in the y-and z-direction are perfectly

insulated.

bounding surfaces are isothermal in character,

i.e, constant and uniform temperature are main-tained at the two faces.

isotropic and homogenous material , i.e., ther-

mal conductivity has a constant value in all the di-

rections.

constant temperature gradient and a linear tem-

perature profile.

no internal heat generation.

Some essential features of the Fourier relation are

enumerated below :

Fourier law is valid for all matter regardless of

its state; solid, liquid or gas

Fourier law is a vector expression indicating

that heat flow rate is normal to an isotherm and is

in the direction of decreasing tempeature.

Fourier law cannot be derived from first prin-ciple ; it is a generalization based on experimental

evidence.

Fourier law helps to define the transport prop-

erty k,i.e., the thermal conductivity of the heat con-

ducting medium.

Assuming dx = 1m; A = 1m2and dt = 10; , we ob-

tain


8/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

Q = k

Hence thermal conductivity may be defined as the

amount of heat conducted per unit time across unit

area and through unit thickness, when a tempera-

ture difference of unit degree is maintained across

the bounding surfaces.

The units of thermal conductivity are worked out

form the Fourier law written in the form.

Q dxk

A dt=

Thus

2

2

1[ ] / deg

deg

1 m[ ] / deg

deg

kcal mk kcal m hr

hr m

kJk kJ m hr

hr m

= =

= =

The unit kJ/m- hr-deg could also be specified as J/m-s-deg or W/m-deg and this is actually done while

quoting the numerical values of thermal conduc-

tivity. Following conversion factors help to con-

vert the thermal conductivity from MKS system

of units into SI units.

1.0 / deg 4.186 / deg

1.163 / deg

kcal m hr kJ m hr

W m

=

=

THERMAL RESISTANCE

Observations indicate that in systems involving

flow of fluid, heat and electricity, the flow quan-

tity is directly proportional to the driving and in-

versely proportional to the flow resistance. In a

hydraulic circuit, the pressure along the path is the

driving potential and roughness of the pipe is re-

sistance of the flow material . Likewise tempera-

ture difference constitutes the driving force for heat

conduction through a medium.

From Ohms law

potential (dV)( )resistance (Re)

voltagecurrent ielectrical

=

and from Fourier;s law

temperature potential (dt)flow rate (Q)=

resistance (dx/kA)heat

thermal

Obviously there is a one-one correspondence be-

tween the flow of electric current and heat i.e.,

- electrical current (amperes) is analogous to ther-

mal heat flow rate ( kJ/hr.)

- electric voltage (volts) corresponds to thermal

temperature difference (degree Kelvin).

- electric resistance (ohms) is analogous to quan-

tity dx/kA. This quantity is called thermal resis-

tance.

Thermal resistance, Rt = (dx/kA), is expressed in

the unit hr-deg/kcal ir s-deg/J or deg/W. The recip-

rocal of thermal resistance is called thermal con-

ductance and it represents the amount of heat con-

ducted through a solid wall of area A and thick-

ness dx when a temperature difference of unit de-

gree is maintained across the bouding surfaces.

Fig 3

Fig: Concept of thermal resistance

Sometimes the heat conducting capacity of a given

physical system is expressed in terms of unit ther-

mal resistance rtand unit thermal conductance c

t

1c =

rt

dx kr and

k dx= =

1 2

:.

( )

dtQ kA

dx

t tkA

dx

=

=

1 2

1 2

( )

( ).....................2.1

t

A t t

r

cA t t a

=

=

GENERAL HEAT CONDUCTION EQUA-TION

The objective of conduction analysis is two fold :

(i) to determine the temperature distribution , i.e.,

variation of temperature with time and position,

and

(ii) to make computations for heat transfer etc. :

Fourier law of heat conduction is essentially valid

for heat flow under uni-directional and steady state


9/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

conditions. However in many practical cases the

temperature may be function of space co-ordinate

as well as time. Recourse is then made to three-

dimensional heat flow equations which consider

both non-uniformly of temperature and any irregu-

larity in the boundary of the surface. To accom-

plish this task an elemental volume is taken , the

relevant energy transfer processes are identified and

the appropriate rate equations are introduced. So-

lution of the resulting differential equations yield

the temperature distribution. Fourier rate equation

is then invoked to workout the heat transfer rate

through the conducting medium.

Cartesian Co-ordinates:Consider the flow of heat

through an infinitesimal volume element oriented

into a three-dimensional co-ordinate system fig.

The side dx, dy and dz have been taken parallel to

the x, y and z axis respectively.Let t represent the temperature at the left face of

the differential control volume. Since area of this

face can be made arbitrarily small, the temperature

t may be assumed uniform over the entire surface.

The temperature changes along the x-direction and

the rate of change is given by /t x . Then changeFig 4

Fig: Conduction analysis in Cartesian co-

ordinates

of temperature through distance dx will be

( )/ .t x dx

This temperature change has been

graphically illustrated in Fig. Therefore the tem-

perature on the right face, which lies at a distance

dx from the left face will be ( )/ .t t x dx + . For

non-strophic materials there will also be a change

in thermal conductivity as heat flows through the

control volume.

Fig 5

Fig : Change in temrpertaure as a function of

distance.

The general conduction equation can be set up by

applying Fourier equation in each cartesian direc-

tion, and then applying the energy conservation

requirement. If kx represents the thermal conduc-

tivity at the left face, then quantity of heat flowing

into the control volume through this face during

time interval

d

is given by :

Heat influx

( ) ......................2.6x x

tQ k dy dz d

x

=

During the same time interval the heat flow out of

the right face of the control volume will be,

Heat efflux

( ) ..................2.7x dx x x

Q Q Q dxd

+

= +

Equation 2.7 simply states that the x-component

of heat transfer rate at (x + dx) is equal to value ofthis component at x plus the amount by which it

changes with respect to x times dx.

Accumulation of heat in the elemental volume due

to heat flow in the x-direction is given by the dif-

ference between heat influx and heat effiux. Thus

the heat accumulation due to heat flow in x-direc-

tion is


10/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

( )

( )

x x x x

x

dQ Q Q Q dxd

Q dxx

= +

=

2 (

....................2.8x

tk dydz d dx

x x

tk dx dy dz d

x x

=

Likewise the heat accumulation in the control vol-

ume due to heat flow along the y-and z-directions

will be:

dy dz d ............2.9y yt

dQ k dxy y

=

dy dz d ............2.10x zt

dQ k dxz z

=

Sume of heat accumulations as prescribed by equa-

tions 2.8 , 2.9 and 2.10 gives the total heat stored

in the elemental volume due to heat flow along all

the coordinate axes.

Total or net accumulation of heat is equal to

dy dz dx y zt t t

k k k dx

x x y y z z

+ +

......... . 2.11

There may be heat sources inside the control vol-

ume due to nuclear fission, flow of electric current

in the coils of electric motors and generators, and

ohmic heating of the material. If qgis the heat gen-

erated per unit volume and per unit time, then the

total heat generated in the control volume equal to

dy dz d .........................2.12g

q dx

The total heat accumulated in the lattice due to heat

flow along all the co-ordinate axes Eq. 2.11 andthe heat generated within the lattice (Eq. 2.12) to-

gether serve to increase the thermal energy of the

lattice. This increase in thermal energy is reflected

by the time rate of change in the heat capcity of the

control volume and is given by :

t( dy dz) c ..............2.13dx d

where is the density and c is the specific heat of

the material. Thus from energy balance consider-

ations :

dy dz d

dy dz d dy dz c ..........2.14

x y z

g

t t tk k k dx

x x y y z z

tq dx dx d

+ + +

=

Dividing both sides by dx dy dz d

.

x y z g c

t t t t k k k q

x x y y z z

+ + + =

............2.14

or , using the vector operator ,

.( ) ...................2.14g

tk t q c a

+ =

Equation 2.14 represents a volumetric heat balance

which must be satisfied at each point for self-gen-

erating, unsteady state three-dimensional heat flow

through a non-dimensional heat flow through non-

isotropic material. This expression known as the

general heat conduction equation, equation, estab-

lishes in differential form the relationship between

the time and space variation of temperature at any

point of the solid through which conduction takes

place. It should be noted that the heat generation

term qgmay be a function of position or time, or

both.

Homogenous and isotropic material :A homoge-

neous material implies that the properties, i.e., den-

sity , specific heat and thermal conductivity of the

material are same everywhere in the material sys-

tem. Isotropic means that these properties are not

directional characteristic of the material i.e, they

are independent of the orientation of the surface.

Therefore for an isotropic and homogenours mate-

rial, thermal conductivity is same at every pointand in all directions. In that case k

x= k

y= k

z= k

and the diffusion equation 2.14 takes the form.

2 2 2

2 2 2

1........2.15

gqt t t c t t

x y z k k

+ + + = =

The quantity /k pc== is called the thermal dif-

fusively, and it represent a physical property of the

material of which the solid is composed. Thermal


11/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

diffusivity is an important characteristic quantity

for unsteady conduction situations. By using the

Laplacian operator

2

, the above equation 2.15

may be written as :

2 1 .................2.15g

t

q ta

k

+

Equation 2.15 governs the temperature distribution

under unsteady heat flow through a homogeneous

and isotropic material.

Different cases of particular interest are :

(i) In many situations there is no dependence of

temperature on time. Conduction then occurs in

the steady state, and the heat flow equation reduces

to :

2 2 2

2 2 20

gqt t t

x y z k

+ + + =

or2 0

gqt

k + = (Poissons equation ) ......2.16

In the absence of internal heat generation or re-

lease of energy with in the body, equation 2.16 fur-

ther reduces to :2 2 2

2 2 2

2

0

0 (Laplace equation)........2.17

t t t

x y z

or t

+ + =

=

(ii) Unsteady state heat flow with no internal heat

generation gives

2 2 2

2 2 2

2

1

1 (Fourier equation)........2.18

t t t t

x y z

tor t

+ + =

=

(iii) For one-dimensional and steady state heat flow

with no internal het generation, the general con-

duction equation takes the form :2

20; 0.............2.19

d dt d t k

dx dx dx

= =

Solution of these equations for any specific bound-

ary conditions will yield the temperature distribu-

tion in the conducting material.

Thermal diffusivity.The following reflections can

be made with regard to this physical property of

the conducting material :

(i) Thermal diffusivity of a material is the ratio

of its thermal conductivity k to the thermal storage

capacity pc. The heat storage capacity essentially

represents thermal capacitance or thermal inertia

of the material, i.e., its sluggishness to conduct heat.

A high value of thermal diffusivity could result ei-

ther from a high value of thermal conductivity or

form low value of thermal conductivity or form

low value of thermal capacity. Liquids have a low

thermal conductivity, high thermal inertia and

hence a small thermal diffusivity. Metals posses

high thermal conductivity, low thermal inertia and

hence a large thermal diffusivity.

(ii) Thermal diffusivity indicates the rate at which

heat is distributed in a material , and this rate de-

pends not only on the conductivity but also on the

rate at which heat energy can be stored. An insightinto equation 2.15 would reveal that thermal

diffusivity, higher would be the rate of change of

temperature at any point of the material.

Equalisation of temperature would then proceed

at a higher rate in materials having large thermal

diffusivity.

(iii) Temperature distribution indicates state is be-

ing governed both by thermal conductivity as well

as by thermal storage capacity. In contrast, during

steady state heat conduction (Eq. 2.16) thermal

conductivity is the only property of the medium

which influences the temperatures distribution.

(iv) The relative magnitude of thermal diffusively

is a measure of the rapidity with which the mate-

rial responds to sudden temperature changes in the

surrounding. Metals and gases have relatively large

value of

and their response to temperature

changes is quite rapid. The non-metallic solids and

liquids respond slowly to temperature changes be-

cause of their relatively small value of thermal

diffusivity.


12/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

COMPOSITE SLAB AND EQUIVALENT RESISTANCE

A composite plane wall is shown in figure (a) be-

low.

Fig 1

In the Figure there are three layers of different

materials with different thermal conductivity and

widths. The temperature at various interfaces are

noted in the figure. In a steady flow, we assume

that the heat condition through all the materials is

same. This can be used to write the following equa-

tion

2 1

3 2

4 3

( )

( )

( )

A

A

B

B

C

C

T Tq k A

x

T Tk A

x

T Tk A

x

=

=

=

From the above equations we can obtain

4 1( )

CA B

A B C

T Tq

xx x

k A k A k A

=

+ +

also( )

A B C

Tq

R R R

=

+ +

Again electrical analogy circuit can be obtained as

shown in figure below:

Fig 2

In general , the heat flow and electrical current flow

can be represented as

thermal potential differenceflow=

resistanceHeat

thermal

Combined heat transfer process

In most of the engineering applications, heat is

transferred in successive steps by similar or differ-ent mechanisms. In a particular location, the trans-

fer may occur by one mechanism only or by more

than one mechanism working in parallel. For in-

stance in the case of the heating of water in a tube

laid in a combustion chamber of a boiler, see fig-

ure. the water will receive heat directly from the

products of combustion which may contain hot

gases which emit and absorb radiation. The heat

will flow by combination of different modes

through successive steps as indicated below.

Fig 3

Step I:Transfer of heat from hot gas to the outer

surfaces of area A of the tube by convection and

radiation working in parallel.


13/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

( )

1 2 1 1 2

1 2

1 2

1

( ) ( )

( )( )

c r

c

c r

q q q

h A t t h A t t

A h h t t

t t

R

= +

= +

= +

=

where1

R is the resistance to heat flow through step

(1) =

( )1

c rh h A

+

and t1and t

2are temperature of

hot gas and that of outside surface of the tube re-

spectively.

Step II:Conduction of heat through the metallic

wall of the tube.

( )2 3kA

q t tl

= ( )2 3K t t= ( )2 32

t t

R

=

where R2is the resistance to heat flow in step II

and t3is the inner wall temperature.

Step III.Finally , the heat will pass to the water

from the inner wall by convection.

( )

( )

3 4

3 4

3

' 'c

q q h A t t

t t

R

= =

=

where R3is the resistances to heat flow through

step III and t4is the water temperature within the

tube.


14/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

Heat Transfer with Fins

This is a case of heat in combined modes of con-

duction and convection.

Consider the one dimensional fin exposed to a sur-

rounding fluid at a temperature ofmT . The tem-

perature of the base of the fin is To. See figure

Fig 4

The problem of heat transfer can be approached by

making energy balance on an element of the fin of

length dx at a distance x as shown in figure.

Thus

IN OUT lost

left face right face by convection

Energy Energy Energy = +

Convective heat transfer is calculated as

( )conq hA T T =

Let the cross-sectional area of the fin be A and pe-

rimeter be P. Then the energy quantities are

Energy IN left face x fdT

q kdx

=

Energy OUT right face xx dx

dTq dx kA

dx +

+ =

2

2

dT d T kA dx

dx dx

= +

Energy lost by convection

( )conq hPdx T T =

It should be noted that the differential area for con-

vection is the product of the perimeter of the fin

and the differential length dx. When we combine

the quantities, the energy balance yields.

( )2

2

10

d hPT T

dx kA = =

Let

T T =

then the above equation becomes,

2

20

d hP

dx kA

=

To solve the above equation, we need two differ-

ent boundary conditions. One common condition

is

x = 0o o

T T at = = =

(i.e) at the base, tempera-

ture is

oT

or

o =

and we can assume three differ-

ent physical situations.

Case (i)

The fin is assumed to be very long so that the tem-

perature at the tip of the fin is essentially that of

the surrounding fluid. The boundary conditions for

the case is( ) 0

( ) 0

oa at x

b at x L or

= =

= =

Let

2

'

hPm

kA=

the above differential equation yields

a general solution as

mx

1 2

mxC e C e = +

where C1and C

2are constants and they are calcu-

lated by applying the above conditions

o

T-T( )

T

mx

o

mx

e

or e

T

=

=

and the rate of heat through the fin is given by

0

fin

x

dTq kA

dx =

=

{ }0

( ) mx

o xkA m T T e

=

=

= kAm ( )oT T

HEAT DISSIPATION FROM EXTENDED SURFACES


15/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

( )oPhkA T T =

Case (ii)

For this case, the fin is of finite length and loses

heat by convection from its end to the ambient air,

equal to the heat conducted by the fin - and the

solution for the heat flow equation will be (same

as before)

1 2

mx mxC e C e = +

with the new boundary conditions

0oT T at x = =

and

LT T at x L = =

( )Lx L

dTk h T T

dx

=

=

[ Heat conducted at x= L

is the same convected by the

end to the ambient air.]

Constants C1and C

2are calculated with these con-

ditions and, substituting the constants, the tempera-

ture distribution along the length of the fin is given

by

( )

( )

/ 2( / )

21

mL x Lm L x L

mLo

T T e e

T T e

whereh mk

h mk

+=

The above equation can also expressed in terms of

hyperbolic function as:

hcosh m(L-x)+ sinh ( )

mkh

cosh mL+ sinhmk

o

m L xT T

T TmL

=

The rate of heat transfer is given by,

0

2

21.1

tanh

( )

1 tanh mL

fin

x

mL

mL

o

dTq kA

dx

ePhAke

hmL

mkPhAk T T

h

mk

=

=

+=

+

= +

Case (iii)

In this case, the end of the fin is assumed to be

insulated (i.e) 0dT

dx= at x = L the boundary condi-

tions are

( ) 0oi at x = =

and

( ) od

ii at x Ldx

= =

thus

( )1 2

1 20

o

mL mL

C C

m C e C e

= +

= +

Solving these equations, C1and C

2and be calcu-

lated - the solution for temperature distribution is

2 2

cosh ( )

1 1 cosh

mx mx

mL mL

o

e e m L x

e e mL

= + =

+ +

the heat transfer in this case is given by

( )0 tanhfinq h p kA mL=

In many applications straight fins are used all along

the length of the wall, instead of a number of spines.

Such fins are generally fins are generally of rect-

angular profile and are called straight rectangular

fins as shown in figure.

Fig 5

The width of the fin in Z, thickness of fin is t, then

A = Zt2( )

2

P Z t

Z if Z t

= +

= >>

then

( )2

rect

hP hm

kA kt = =

For a circular fin


16/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

( )2

2

2

cir

h rm

k r

h

kr

=

=............................12

Fin effectiveness

This is also called fin efficiency. The objective offin is only to increase the area of the heat transfer

surface. The temperature at the base of the fin will

be maximum and will be decreasing gradually to-

wards the tip. The fin would have dissipated maxi-

mum amount of heat if its temperature all along

the length was the same as that at the base. Since

this cannot be achieved, the fin efficiency cannot

be at maximum efficiency.

In order to study the thermal performance of a fin,

we need a definition as fin effectiveness or fin ef-

ficiency.

Fin efficiency = fin

transferred across fin surface

heat transfer across equal maintained

at a constant temperature equal to that of base

Heat

Theoretical=

Design of in system

Before the actual design of in system, it is impor-

tant to recognise the various conditions for which

the finned surface has advantage over the unfinned

surface. The probable conditions are

1. space consideration, limiting certain dimensions

of fins,

2. weight considerations and

3. cost considerations.

The quantity of heat dissipated by a fin to its sur-

roundings goes on increasing as the length, but

there is a limit on the length of the fin, imposed by

the manufacturing difficulties, stability and so on.

The mathematical limit for the design is 0.dqdL

= It

may also be be reminded that for short fins, heat

flow becomes tow-dimensional and all our assump-

tions and expressions given earlier become use-

less.

The consideration of the weight of the fin is par-

ticularly important in the case of automobile and

aircraft practice. In all such cases the requirements

is the maximum heat exchange with minimum

weight of the heat exchange system. Triangular or

parabolic shaped fins may be used depending upon

the specific conditions imposed. Any design of a

fin system, should always be aimed at maximum

fin efficiency and minimum material requirement.

It is generally a compromise of several aspects in

the design.


17/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

Heat exchangers are generally classified accord-

ing to the direction of flow of the hot and cold flu-

ids with respect to each other, or according to the

temperature distribution of the two fluids along the

exchanger length. Thus, we may have the follow-

ing types of heat exchangers:

(a) Parallel-flow exchanger

(b) Counter-flow exchanger

(c) Cross-flow exchanger

(d) Condenser or evaporator.

(a) Parallel-flow Heat Exchanger

In this type of heat exchanger the hot and cold flu-

ids flow in the same directions, hence the name

parallel-flow. Such an exchanger is shown sche-

matically in figure (a) and (b). In a parallel flowexchanger, the temperature difference between the

hot and cold fluids keeps as decreasing from inlet

to exit. Many devices, such as water heaters and

oil coolers, etc. belong to this class.

Fig 6

(b) Counter-Flow Heat Exchanger

In this case, as schematically shown in figure (a)

and (b), the two fluids flow through the exchange

in opposite directions hence the name, counter-

flow. A common type of counter flow exchanger is

shown in figure (d).The temperature distributionin a counter-flow exchanger is shown in figure (c).

Fig 7

(c) Cross-Flow Exchanger

This exchanger is shown in figure. Here the two

fluids flow at right angles to each other. Two dif-

ferent arrangements of this exchanger are com-

monly used. In one case, each of the fluids is un-

mixed as it flows through the exchanger. As a re-

sult, the temperature of the fluids leaving the ex-

changer are not uniform. An automobile radiator

is an example of this type of exchanger.

In the other case, one fluid is perfectly mixed while

the other is unmixed as it flows through the ex-

changer. This type is shown in figure The air flow-

ing over the tube bank is mixed while the hot gases

flowing inside the separate tubes are not mixed.

Fig 8

(d) Condenser of Evaporator

In a condenser, fluid remains at constant tempera-

ture throughout the exchanger while the tempera-

ture of the colder fluid gradually increases from

inlet to exit, Similarly, in an evaporator, the boil-

ing fluid (cold fluid) remains at constant tempera-

ture while the hot fluid temperature gradually de-

Heat Exchangers


18/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

creases. The temperature distribution in these ex-

changes are shown in figure (a) and (b). Since the

temperature of one of the fluid remains constants,

it is immaterial whether the two fluids flow in the

same direction or in opposite directions.

Fig 9

Temperature distribution in Condensing ofEvaporating Heat Exchanges


19/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

An important parameter in the design of heat ex-

changers is the overall heat transfer coefficient U,

between the two fluids. A general expression for U

can be easily obtained as follows: Consider a

double pipe heat exchanger in which one fluidflows through the inner pipe and the other fluid

through the annular space between the two pipes.

Let L = Length of heat exchanger, m

ri= Inside radius of inner pipe, m

or

= Outside radius of inner pipe, m

iA

= Inside surface area of inner pipe

2(2 ),ir L m

oA = Outside surface area of inner pipe

2(2 ),or L m

ih = Film coefficient of heat transfer at inside sur-

face of inner pipe, W/m2

k.

0h

= Film coefficient of heat transfer at outside

surface of inner pipe,

2/W m k

.

wk

= Thermal conductivity of inner pipe wall, W/

mK

it

= Temperature of fluid flowing through the in-

ner pipe, 0C. ot

= Temperature of fluid flowing through the an-

nular space between the two pipes

o C

.

iR

= Thermal resistance of fluid film at the inside

surface of inner pipe, K/W.

oR

= Thermal resistance of fluid film at the out-

side surface of inner pipe , K/W

wR

= Thermal resistance of inner pipe wall, K/W.

The rate of heat transfer between the two fluids is

given by

i ot tqR

=

where

i o wR R R R= + +

since

1i

i i

RA h

=

( / )

2

n o iw

w

l r rR

Lk=

1o

o o

RA h

=

Hence1 1

2

i o

o

i i

i i w o o

t tqr

nr A

A h Lk A h

=

+ +

.....................3

If Uiand U

odenotes respectively the overall heat

transfer coefficient basedon unit area ofthe inside

and outside surfaces of the inner pipe, then

( ) ( )i i i o o o i oq A U t t A U t t = =

From Equations (1) and (2)

1

ln

12

i

oi

i

ii w

o o

Ur

Ar

Ah LkA h

=

= +

and 1

1

1

( / )1 1

2

oo o n o

i w o

UA A l r r

A h Lk h

=+ + ................4

Since 2i iA r L= and

2o o

A r L=

Equations (3) and(r) can also be written as :

1............5

1 1i

i o in

i w i o o

Ur r r

lh k r r h

=

+ + +

and

1............6

1 1i

o o o

i i w i o

Ur r r

lnr h k r h

=

+ +

In most heat exchangers some scale formation will

take place on one or on both sides of the heat trans-

fer surface after the heat exchanger has been in use

for some time. This introduces two additional re-

sistance in the heat flow path. Thus the total ther-

mal resistance becomes.

i si w so oR R R R R R= + + + +

OVERALL HEAT TRANSFER COEFFICIENT


20/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

where

siR

= Thermal resistance due to scale for-

mation on inside surface of inner pipe,

2 /M K Watt

.

soR

= Thermal resistance due to scale formation

on outside surface of inner pipe

2 /m K watt

.

Since it is difficult to ascertain accurtain accurately

the thickness and thermal conductivity of the scaleformed, the effect of the scale deposit on heat flow

is generally taken into account by specifying an

equivalent scale heat transfer coefficient hs.

The reciprocal of the scale heat transfer coefficient

is called the fouling factor. If hsiand n

sodenote the

heat transfer coefficients for the scale formed on

the inside and outside surface, respectively, of the

inner pipe, then,

1....................7

1...........8

si

i si

so

o so

RA h

RA h

=

=

and

( )

1 1 1 1

2

i o

on

i

i i i si w o so o o

t tq

rl

r

Ah Ah Lk A h A h

=

+ + + +

( )

11 1 1

2

i o

oi n

i i i

i si w o so o o

t tq

rAl

r A A

h h Lk A h A h

=

+ + + +

or

1..............11

1 1 1 1ln

i

i i i i

i si o o o so o o

U

r r r r h h r r r h r h

=

+ + +

1

ln1 1 1 1

2

o

o

o

io o

i i i si w so o

Ur

ArA A

A h A h Lk h h

=

+ + + +

or

1

1 1 1 1o

o o o o

i i i si w i so o

Ur r r r

inr h r h k r h h

=

+ + + +

The fouling factors

1

sh

for some repensentative

applications are listed in table 1 given below:

Table 1

Fouling Factors

________________________________________

Fluid Fouling Factor

21 /s

m K Wh

_______________________________________

Distilled water 0.000086

Sea water 0.000172Well water 0.000344

Treated boiler feed water 0.000172

Liquid gasoline 0.000086

Refrigerant liquids, brine 0.000172

or oil-bearing steam

Refrigerant vapours 0.000344

Fuel oil and crude oil 0.00086

Diesel exhaust gas 0.000172

_________________________________________


21/35

Indias No.1

IES

Academy


HEATM

ASSTRANSFER

(a) Forced convection over flat plates and through

tubesCase (1): Pipe and tube flow. (Dittus-Boelter rela-

tion)

Nu = 0.023 Re0.8Prn

where n = 0.4 for heating

= 0.3 for cooling

Case (2) : Flow across cylinders

0.52 0.3(0.35 0.56Re ) Pr Nu +

in the range 1 510 Re 10 <

(iii)

0.8274.82 0.0185Nu Pe= +

for the range3 5 2 4

3.6 10 Re 9.05 10 10 10and Pe < < < 109

0.330.1( Pr)Nu Gr=

Case (3): Horizontal cylinders2 0.250.53( Pr )Nu Gr=

Case (4): Inclined surfaces at

0.250.56( Pr cos )Nu Gr =

where 0 5 1188 10 Prcos 10and Gr <

Date post:	02-Jun-2018
Category:	Documents
Upload:	phaneufdl
View:	233 times
Download:	0 times

(Only Theory) Heat Mass Transfer

Documents