Unit 2 HT by Conduction

8/11/2019 Unit 2 HT by Conduction

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HEAT TRANSFERby

CONDUCTION

Prepared by,

Kumargaurao D PunaseAssistant Professor

Dept of Chemical Engg.COES, UPES

Dehradun


2/121

First mechanism - molecularinteraction (e.g. gas)

Greater motion of molecule

at higher energy level(temperature) imparts energyto adjacent molecules atlower energy levels

Second mechanism by free

electrons (e.g. solid)The mechanism of heat conduction

in different phases of substance

Fouriers law of Heat Conduction


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4/121

Steady state heat conduction

One directional heat flow

Isotropic and homogeneous material i.e. thermal conductivityhas a constant value in all the directions

Constant temperature gradient and a linear temperature profile

No internal heat generation

Assumptions of Fouriers law: Contd .


5/121

Summary: Fouriers Law

It is phenomenological, i.e. based on experimental evidence

Is a vector expression indicating that the heat flux is normal toan isotherm, in the direction of decreasing temperature

Applies to all states of matter

Defines the thermal conductivity, ie.

)/( xT

qk x

Contd .


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Chapter 2 6

Thermal Properties of Matter

Fouriers law for heat conduction:

The proportionality constant is a transport property , known asthermal conductivity k (units W/m.K)

the thermal conductivity of a material can be defined as therate of heat transfer through a unit thickness of the material

per unit area per unit temperature difference. The thermal

conductivity of a material is a measure of the ability of thematerial to conduct heat.

Usually assumed to be isotropic (independent of the direction

of transfer): k x=ky=kz=k

dxdT

k q x


7/121

7

Thermal Conductivity: Solids

Solid comprised of free electrons and atoms bound in lattice

Thermal energy transported through Migration of free electrons, k e Lattice vibrational waves, k l

l e k k k )(y,resistivitelectrical1

eek where

Contd .


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Thermal Conductivity: Gases

Intermolecular spacing is much larger Molecular motion is random Thermal energy transport less effective than in solids; thermal

conductivity is lower

Kinetic theory of gases:

cnk Where, n the number of particles per unit volume,the mean molecular speed and

the mean free path (average distance travelled before a collision) c

Contd .


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Thermal Conductivity: Liquids

Physical mechanisms controlling thermal conductivity not well

understood in the liquid state

Generally k decreases with increasing temperature (exceptions

glycerine and water)

k decreases with increasing molecular weight.

3/1

3/4

M Ac

k p

The parameter A depends on the temperature of the liquid;

The quantity (Ac p) is constant for all liquids

Contd .


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Thermal Conductivity: Insulators Low temperature insulation (cork, rock wool, glass wool and thermocole)

are used when the enclosures is at a temperature lower than the ambienttemperature and it is desired to prevent the enclosure from gaining heat.

High temperature insulation (asbestos, diatomaceous earth, magnesia

etc.) are used when the enclosures is at a temperature higher than theambient temperature and it is desired to prevent the enclosure fromloosing heat.

Super insulators include powders, fibres or multi-layer materials that

have been evacuated of all air

The low conductivity of insulating materials is due primarily to air that iscontained in the pores rather than the low conductivity of the solidsubstance.

Contd .


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Variation of Thermal Conductivity with temperatureand pressure

For most materials, the dependence of thermal conductivityon temperature is almost linear

T k k 10Where, k 0 is the thermal conductivity at 0 0C and

Is a constant value depending upon the material; may be positive ornegative depending on whether k increases or decreases withtemperature.

Thermal conductivity is very weakly dependent on pressurefor solids and liquids and essentially independent of pressurefor gases at standard atmospheric pressure

Contd .


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The variation of the thermal conductivity of the various solids,

liquids and gases with the temperature

Contd .

Adapted from Heat and MassTransfer A Practical Approach,Y.A. Cengel, Third Edition,McGraw Hill 2007.


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The range of thermal conductivity of various materials at room

temperature

Contd .

Adapted from Heat and MassTransfer A Practical Approach,Y.A. Cengel, Third Edition,McGraw Hill 2007.


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Thermal Diffusivity

The product C p , which is frequently encountered in heattransfer analysis, is called the heat capacity of a material.

Both the specific heat C p and the heat capacity C p representthe heat storage capability of a material.

But C p expresses it per unit mass whereas C p expresses it perunit volume, as can be noticed from their units J/kgC andJ/m 3C, respectively.

Thermal diffusivity is the ratio of the thermal conductivity to

the heat capacity:

pck

storedHeatconductedHeat

(m 2/s)

Contd .


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Note that the thermal conductivity k represents how well amaterial conducts heat, and the heat capacity C p represents

how much energy a material stores per unit volume.

Therefore, the thermal diffusivity of a material can be viewed asthe ratio of the heat conducted through the material to the heatstored per unit volume.

A material that has a high thermal conductivity or a low heatcapacity will obviously have a large thermal diffusivity. Thelarger the thermal diffusivity, the faster the propagation of heat

into the medium.

A small value of thermal diffusivity means that heat is mostlyabsorbed by the material and a small amount of heat will beconducted further.

Contd .


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Introduction to development of 3D equation

d


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Contd .

Adapted from Heat and Mass Transfer A Practical Approach, Y.A. Cengel, Third Edition,McGraw Hill 2007.

C d


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Contd .

C d


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Contd .

C td


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Contd .



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The Conduction Rate Equation

In reality we must account for heat transfer in three dimensions Temperature is a scalar field T(x,y,z) Heat flux is a vector quantity. In Cartesian coordinates:

z y x qqq k jiq

for isotropic medium z T k q

yT k q

xT k q z y x ,,

T k z

T

y

T

x

T

k

k jiqWhere three dimensional del operator in Cartesian coordinates:

z y x k ji

C td


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Energy Conservation Equation st

st g out in E dt

dE E E E

dz z dy ydx xout qqq E z y xin qqq E

Contd .

Contd


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E=mcpT=( V)cpT= (dxdydz)c pT

dy

dx

qx qx+dx

x

y

q KA T

x K dydz

T

x

q q dq q q

xdx

x x x

x dx x x x x

( )

dx

dy

dz

Contd .

Contd


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where from Fouriers law

z

T dxdyk

z

T kAq

yT

dxdz k yT

kAq

x

T dydz k

x

T kAq

z z

y y

x x

)(

)(

)(

dz z

qqq

dy y

qqq

dx

x

qqq

z

z dz z

y ydy y

x xdx x

Energy in

Energy out

Contd .

Contd


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Thermal energy generation due to an energy source : Manifestation of energy conversion process (between thermal

energy and chemical/electrical/nuclear energy)

Positive (source) if thermal energy is generatedNegative (sink) if thermal energy is consumed

)(dV dz dydxqq E g

Energy storage term Represents the rate of change of thermal energy stored in the

matter in the absence of phase change.

)( dz dydxt T

c E p st

t T c p /is the time rate of change of the sensible (thermal) energy ofthe medium per unit volume (W/m 3)

q is the rate at which energy is generated per unit volume of themedium (W/m 3)

Contd .

Contd


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dxdydz t T

C dxdydz qqqqqqq pdz z dy ydx x z y x

dxdydz t

T C dxdydz qdz

z

qdy

y

qdx

x

q p

z y x

z

T kdxdyq

yT kdxdz q

xT

kdydz q

z

y

x

st st

g out in E dt dE

E E E Contd .

Contd


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dxdydz t

T C dxdydz qdxdydz

z

T k

z dxdydz

y

T k

ydxdydz

x

T k

x P

t T C q

z T k

z yT k

y xT k

x P

Heat (Diffusion) Equation: at any point in the medium the rate of

energy transfer by conduction in a unit volume plus thevolumetric rate of thermal energy must equal to the rate ofchange of thermal energy stored within the volume.

Contd .

Contd


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dx x x qqdx

x

T k

x

t T

k q

z T

yT

xT

1

2

2

2

2

2

2

Net conduction heat flux into the controlled volume

If the thermal conductivity is constant.

Where = k /( Cp) is the thermal diffusivity

t T

C q z T

k z y

T k

y xT

k x P

Contd .

Contd


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0

q z T

k z y

T k

y xT

k x

0)( dxdT

k dxd

Under steady-state condition, there can be no change in theamount of energy storage.

If the heat transfer is one-dimensional and there is noenergy generation, the above equation reduces to

Under steady-state, one-dimensional conditions with noenergy generation, the heat flux is a constant in the

direction of transfer.

Contd .

Contd


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Boundary and Initial Conditions Heat equation is a differential equation:

Second order in spatial coordinates: Need 2 boundary conditions First order in time: Need 1 initial condition

Boundary Conditions

1) B.C. of first kind (Dirichlet condition):

x

T(x,t)

Ts

Constant Surface Temperature At x=0, T(x,t)=T(0,t)=T s

x=0

Contd .

Contd


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Boundary and Initial Conditions2) B.C. of second kind (Neumann condition): Constant heat flux at the

surface

1. Finite heat flux

xT

xT

k q x

s slopeconst0

"

0slope00

"

xT

xT

k q x

s

xT(x,t)

x

T(x,t)

qx = q s

2. Adiabatic surface qx=0

Contd .

Contd .


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Boundary and Initial Conditions3) B.C. of third kind: When convective heat transfer occurs

at the surface

)],0([0

t T T h x

T k

x

T(x,t)

T(0,t)

x

hT ,

= q conduction q convection

Contd .

Contd .


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Contd .



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Cylindrical coordinates

Contd .


35/121

Cylindrical coordinates

z T

k z qT

r k

qr T

k r q ";";"

)1

(" Z T

k T

r j

r T

ik T k q

When the del operator is expressed in cylindrical coordinates,the general form of the heat flux vector , and hence the FouriersLaw, is

Contd .


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Heat Conduction Equation

In cylindrical coordinates:

t T

cq z T

k z

T k

r r T

kr r r p

211


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Spherical coordinates

Contd .


38/121

Spherical coordinates

T

r k

qT

r k

qr T

k r q ";sin";"

)sin11

("

T r

k T

r j

r T

ik T k q

When the del operator is expressed in spherical coordinates,the general form of the heat flux vector , and hence the FouriersLaw, is

Contd .


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Heat Conduction Equation

In spherical coordinates:

t T

cqT

k r

T k

r r T

kr r r

p

sinsin

1

sin

11222

22


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Energy balance for the wall

rate ofheat transferinto the wall

rate ofheat transferout of the wall

rate of changeof the energyof the wall

- =

dt dE

QQ wall out in

0dt

dE wall

consQ wall cond ,

steady operation; since there is no change in the

temperature of the wall with time at any point

The rate of heat transfer through the wall is constant

If there is no heat generation


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FOURIERS LAW OF HEAT CONDUCTION

wall cond Q , dx

dT kAQ wall cond , (W)

and A constant, then

dxdT constant also

Temperature through the wall varies linearly withx. Temperature distribution in the wall understeady conditions is a straight line.

2

1,0

T

T T wall cond

L

x kAdT dxQ

LT T

kAQ wall cond 21

,



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THERMAL RESISTANCE

wall wall cond R

T T Q 21,

kA L

Rwall

(W)

(0C / W)

Depends on the geometry andthe thermal properties of themedium

e RV V I 21 A L R ee

e R21 V V e

Electrical resistance

Voltage differenceacross the resistance

Electricalconductivity

Contd .


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NEWTONS LAW OF COOLING FORCONVECTION HEAT TRANSFER RATE

)( T T hAQ S S conv

conv

S conv R

T T Q

S

conv

hA

R1

conv R

h

Convection resistance of surface

(W)

(0C / W)

Convection heat transfercoefficient


Contd .


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RADIATION

rad

surr S surr S S rad rad R

T T T T AhQ

)(

S rad rad Ah

R1

surr s surr s surr S S

rad rad T T T T T T A

Qh 22)(

rad convcombined hhh

)( 44 surr S S rad T T AQ

Contd .


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The thermal resistance network for heat transfer through a plane wallsubjected to convection on both sides and the electrical analogy

THERMAL RESISTANCE NETWORK



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ONE DIMENSIONAL STEADY HEATFLOW

Rate of

heat convection

from the wall

Rate of

heat convection

into the wall

Rate of

heat conduction

through the wall

= =

)()( 22221

111 T T Ah LT T

kAT T AhQ

Ah

T T

kA L

T T

Ah

T T Q

2

2221

1

11

/1//1

2,

2221

1,

11

convwall conv RT T

RT T

RT T

Q

adding the numerators and denominators

total RT T

Q 21


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Total Thermal Resistance

total RT T

Q 21

Ah Ak L

Ak L

Ah R

R R R R R

total

convwall wall convtotal

22

2

1

1

1

2,2,1,1,

11



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Thermal Contact Resistance

gapcontact QQQ

erfacec T AhQ int

erfacec T

AQh

int

/(W/m 2 0C)

(m 2 0C/ W) AQ

T

h R erface

cc

/

1 int

hC: thermal contact conductance


Contd .


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Thermal contact resistance is inverse of thermalcontact conduction,

Depends on

Surface roughness, Material properties,

Temperature and pressure at interface, Type of fluid trapped at interface

Contd .


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Effect of metalliccoatings on thermalcontact conductance

For soft metals withsmoot surfaces at highpressures

Thermal contactconductance

Thermal contactresistance



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Overall heat transfer coefficient

q1 = convection

q2 = conduction

q3 = convection

q = q 1 = q2 = q3

The overall heat transfer bycombined conduction andconvection is frequently expressed interms of an overall heat transfercoefficient U

overall T UAq

?U

Contd


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)( T T hAq wconv

hAT T

q wconv /1

hA Rconv /1

Convection Boundary condition

Conduction

)( 12 T T xkA

qcond

Ak x

Rn

ncond

kA xT T

qcond /21

Contd

Contd


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)()()( 222111 B A T T AhT T xkA

T T Ahq

AhkA x Ah

T T q B A

21 /1//1

R Ahk xhU

11/1//1

1

21

Overall heat transfer coefficient

Contd

21

111hk

xhU


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Thermal resistance networkthrough a two-layer plane

T UAQ

total RUA

1



55/121

THERMAL RESISTANCE NETWORKS

)11

)((21

212

21

1

2121 R R

T T R

T T R

T T QQQ

total RT T

Q 21

21

111 R R Rtotal

21

21

R R R R Rtotal

Resistances are parallel



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COMBINED SERIES-PARALLEL ARRANGEMENT

total RT T Q 1

convconvtotal R R R R R R

R R R R 321

21312

11

11 Ak

L R

22

22 Ak

L R

33

33 Ak

L R 3

1hA Rconv



57/121

Problem:

Two slabs, each 100mm thick and made of materials with thermalconductivities of 16 and 200 W/m 0C, are placed in contact whichis not perfect. Due to roughness of surfaces, only 40% of area is incontact and air fills 0.02 mm thick gap in the remaining area.If the extreme surfaces of the arrangement are at temperaturesof 250 0C and 30 0C, determine the heat flow through thecomposite system, the contact resistance and temperature dropin contact.

Take thermal conductivity of air as 0.032 W/m 0C and assume thathalf of the contact is due to either metal.

1 D Steady Heat Transfer in the Plane Wall


58/121

For an infinite flat plate (y, z

infinite),find the steady statetemperature profile with thetemperature at x=0 remainsat T 1 and the temperature atx=L remains at T 2

X=0 X=L

T1

T2

1-D, Steady Heat Transfer in the Plane Wall

The Governing Differential Equation

t T

C q z T

k z y

T k

y xT

k x P

=0 =0 =0 =0

T is not function of y and z

H t E tiContd .


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022

xT ---

Heat Equation :

0

d dT

k dx dx

022

dxT d k

Contd .


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Integrate eq.

Integrate again,

T=C1.x+C2

Apply the boundary conditions:

T1=C2 --- T2=C1.L+C2 ---

1C dxdT

B. C. T=T1 at x=0B. C. T=T2 at x=L

Contd .


61/121

Putting, into

L x

T T T T

L x

T T T T

T x L

T T T

LT T C T LC T

12

1

121

112

121112

).(

.

.


62/121

HEAT CONDUCTION IN CYLINDERS AND SPHERES

Steady-state heat conduction

Heat is lost from a hot-water

pipe to the air outside in theradial direction.

Heat transfer from a longpipe is one dimensional


A LONG CYLINDERICAL PIPE


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STEADY STATE OPERATION

dr dT

kAQ cy l cond ,

Fouriers law of conduction

cyl cond Q , constant

2

1

2

1

, T

T T

r

r r

cyl cond

kdT dr A

Q

rL A 2

)/ln(2 1221, r r

T T Lk Q cyl cond

cyl cyl cond

R

T T Q 21,

Lk

r r Rcy l

2

)/ln( 12

There are flow of gases on bothComposite cylindrical wall


64/121

sides of the wall, which remainat temp T i & To. Looking for(1) temperature profile(2) heat flux requited to

maintain steady state

r1 r2

r3

Ti

T1 T2

T3 To

p y

))(2(

)()/ln(

2

)()/ln(

2))(2(

33

3223

2

2112

1

11

oo

ii

T T Lr h

T T r r Lk

T T r r

Lk T T Lr hQ

Contd .


65/121

65

oi

oi

Lhr Lk r r

Lk r r

Lhr

T T Q

32

23

1

12

1 21

2)/ln(

2)/ln(

21

ResistanceTotalDifferenceeTemperatur OverallQ

I fi it C li dFor steady radial flow of heat through the


66/121

Infinite Cylindery g

wall of a hollow cylinder. The inner surfaceis maintained at temperature T 1, whileouter surface is kept at temperature T 2.

Find the temperature profile across thewall and the heat flux required to maintainsteady state.T1

T2

r1

r2

T is not function of and z

t T

cq z T

k z

T k

r r T

kr r r p

2

11

Governing equation in cylindrical coordinate=0 =0 =0 =0

The GDE isContd .


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The GDE is

0)(1

r

T r

r r k

21

1

1

ln

0)(

cr cT

r

c

r

T

cr

T r

r

T r

r

---

Contd .


68/121

68

Apply boundary conditions

2212

2111

ln

ln

cr cT

cr cT

B.C. r=r 1, T=T1 r=r 2, T=T2

Contd .


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)/ln(

)/ln()ln(ln

21

211

21121121

r r T T c

r r cr r cT T

222122122

2221212

2221

212

lnlnlnlnln)()ln(ln

ln.)/ln(

r T r T r T r T ccr T T r r T

cr r r

T T T

Contd .


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70

)ln)(ln()ln)(ln(

)ln)(ln()ln(ln)ln(ln

lnlnlnlnlnlnln

221212

22121221

2221221221

21

r r T T r r T T

r r T T r r T r r T

r T r T r T r T r r r

T T T

)/ln()/ln(

21

2

21

2

r r r r

T T T T

The variation of temperature within the cylinder is determined to be

STEADY STATE HEAT


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Ti

To ri

ro

dr dT

kAQ

24 r A

dr

dT r k Q 24

oi

o

i

T

T

r

r dT

r

dr

k

Q24

io

o

i

i

o

o

i

T T

r

r

T

T

r

r T

r k Q

dT r dr

k Q 1

44 2

STEADY STATE HEATCONDUCTION FOR

SPHERES

TContd .


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RT

Q

oi

oi

r r

T T k Q

11)(4

k r r R oi

4

11

k r r

r r R

oi

io

sph 4OR

Contd .


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k r r r r

R sph21

12

4

sph sphcond R

T T Q 21,

including convection

22

221

12

41

4 hr k r r r r

Rtotal

total RT T

Q 1

Temperature distribution inside the spherical shell


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p p

t T cqT k

r T k

r r T kr

r r p

sinsin1

sin11 222

22

=0 =0 =0 =0

01 2

2

r

T kr

r r

02

r T

r r

21

r C

dr dT

21)( C

r

C r T

1)(C

Contd .


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22

1222

21

1111

)(

)(

C r C

T T r T

C r

T T r T

12

11222

21

12

211

r r T r T r

C

T T

r r

r r C

12

112221

12

21)(r r

T r T r T T

r r r r r

r T

Applying BoundaryConditions

The constantsobtained are as:

The variation of temperature within the spherical shell isdetermined to be


76/121

CRITICAL RADIUS OFINSULATION

)2(1

2)/ln(

2

12

11

Lr h Lk r r

T T R RT T

Qconvins

CYLINDER

However, as r 2 A 0 convection heat transfer resistance of surface

=>Q

As r2 thickness of wall heat resistance in the wall => Q

The heat transfer from the pipe may increase or decrease,Contd .


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0/ 2 dr Qd

The heat transfer from the pipe may increase or decrease,depending on which effect dominates.

T 1 ,T

2 , k, L, h, r

i are constant terms, therefore:

The variation of Q with the outerradius of the insulation r 2 is plotted:

The value of r 2 at which Q reachesa maximum is determined from therequirement that

Adapted from Heat and Mass Transfer A PracticalApproach, Y.A. Cengel, Third Edition, McGraw Hill2007.

Contd .


78/121

222

2

21

2

2

21

2

2

11.

1ln

1

)(2

1ln

1)(2

0

r hkr

r hr r

k

T T L

dr hr r

r k

T T Ld

dr dQ

o

o

oi

oi

o

c

o

o

h

k r

r hkr

r hkr

,2

222

222

11

011

Critical radius of insulation for cylinderContd .


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hk

r cylinder cr ,Thermal conductivity

External convection heattransfer coefficient

y

Critical radius of insulation for shpere

hk

r spherecr 2

,

CHOSING INSULATION THICKNESSContd .


80/121

cr

cr

cr

r r

r r

r r

2

2

2

Q max

Before insulation check for critical radius

CHOSING INSULATION THICKNESS

Contd .


81/121

Contd .


82/121

Optimum thickness of insulation

Conduction with Generation


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Conduction with GenerationThermal energy may be generated or consumed due to

conversion from some other energy form.If thermal energy is generated in the material at the expense ofsome other energy form, we have a source: is +ve Deceleration and absorption of neutrons in a nuclear reactor

Exothermic reactions Conversion of electrical to thermal energy:

V R I

V

E q

e g 2 where I is the current, R e the electrical

resistance, V the volume of the medium

If thermal energy is consumed we have a sink : is -ve

Endothermic reactions

q

q

The Plane Wall

0 L


84/121

Consider one-dimensional,steady-state conduction in aplane wall of constant k, withuniform generation, andasymmetric surface conditions:

Heat conduction equation

02

2

k

q

dx

T d

The Governing Differential Equation

t

T C q

z

T k

z y

T k

y x

T k

x P

=0 =0 =0

2 qTd Contd .


85/121

02 k q

dxT d

222

2

dxk qT d

k q

dxT d

1c xdxk

qdT

Boundary Conditions:

2,1, )(,)0( s s T LT T T

General Solution:

212

2C xC x

k

qT

Temperature ProfileContd .


86/121

p

1,1,2,2 )(2)( s s s T L x

T T x Lxk

q xT

Profile is parabolic.Heat flux not independent of x

0 L

1,2 sT C Lk

q

L

T T C s s

2

)( 1,2,1

1,1,2, )(2)( s s s T x

L

T T x L

k q

xT

OR

Symmetrical DistributionContd .


87/121

When both surfaces are maintained at a commontemperature, T s,1= Ts,2 = Ts 0 L

1,1,2, )(2)( s s s T x

L

T T x L

k q

xT

sT x x Lk q

xT 2

)(

Maximum temperature within a symmetric systemContd .


88/121

0 L

sT x x L

k

q xT

2)(

0dxdT

The maximum temperature at anyposition x can be obtained as:

02

2 x L

k

q

dx

dT

02k q or02 x L

2

L x

At x=L/2, T = Tmax

Contd .


89/121

, max

sT x x Lk q xT 2)(

sT L L

Lk

qT

222(max)

Putting x=L/2,

sT Lk q

T 28

(max)

Heat transfer then occurs towards both surfaces and for

Contd .


90/121

Heat transfer then occurs towards both surfaces, and foreach surface it is given by:

L x xdxdT kAQ

or0

q

ALQ

x Lk q

kAQ L x x

2

22 or0

Heat transfer for both surfaces,

q ALQ

=(volume of conducting medium)(heat generating capacity)

Heat conducted to the wall surface is finally dissipated to the

Contd .


91/121

a s T T hAq AL2

Heat conducted to the wall surface is finally dissipated to thesurrounding atmosphere at temperature T a

Lh

qT T a s 2

OR

Substituting this value of wall temperature, the temperature

distribution in term of surrounding atmospheric temperature T a

x x Lk

q L

hq

T xT a 22)(

Contd


92/121

Note that at the plane of symmetry:

0q" 00

0

x xdx

dT

Equivalent to adiabatic surface

Cylinder with uniform heat generation


93/121

Solid Cylinder Temperature distribution within a solid cylinder

224)( r Rk qT r T s

R

Ts

h,Ta

Temperature distribution is parabolic andthe maximum temperature T max occurring at the centre (r=0) of the rod isgiven by

2max 4

Rk

qT T s

The temperature distribution in term of surrounding atmospherictemperature T a

2242

r Rk

q R

hq

T T a

Hollow Cylinder with temperatures specified at the inside andt id f


94/121

outside surfaces

Temperature distribution within a thickness of hollow cylinder

2

1

121

2221

2211

ln

ln

44)(

r r

r r

r r k

qT T r r

k q

T r T

Sphere with uniform heat generation


95/121

Temperature distribution within a solid cylinder

226)( r Rk qT r T s R

Ts

h,Ta

Temperature distribution is parabolic and themaximum temperature T max occurring at thecentre (r=0) of the rod is given by

2max 6

Rk

qT T s

The temperature distribution in term of surrounding atmospherictemperature T a

2263

r Rk

q R

hq

T T a

Heat Transfer from Extended Surfaces


96/121

An extended surface (also know as a combined conduction-convectionsystem or a fin) is a solid within which heat transfer by conduction isassumed to be one dimensional , while heat is also transferred by convection (and/or radiation ) from the surface in a direction transverse to that ofconduction.

Anatomy of A STRIP FIN

io n

Contd .


97/121

thickness

x

x

Direction of Heat Conduction

D i r e c t i o n o

f H e a t C o n v e c t i o

Basic Geometric Features of Fins

Contd .


98/121

profilePROFILE AREA

cross-section

CROSS-SECTION AREA

Basic Geometric Features of Fins

GARDNER-MURRAY ANALYSIS : ASSUMPTIONSContd .


99/121

Steady state one dimensional conduction Model.No Heat sources or sinks within the fin .Thermal conductivity is constant and uniform in all directions.Heat transfer coefficient is constant and uniform over fin faces.Surrounding temperature is constant and uniform.Base temperature is constant and uniform over fin base.Fin width much smaller than fin height or length.No bond resistance between fin base and prime surface.

Heat flow off fin proportional to temperature excess.

Why Fins are needed?

Contd .


100/121

Temperature gradient dT/dx ,

Surface temperature, T, Are expressed such that T is a function of x only.

Newtons law of cooling

Two ways to increase the rate of heat transfer: increasing the heat transfer coefficient , increase the surface area fins

conv s sQ hA T T

Extended surfaces may exist in many situations but are commonly used as

Contd .


101/121

fins to enhance heat transfer by increasing the surface area available forconvection (and/or radiation).

Some typical fin configurations:

Straight fins of (a) uniform and (b) non-uniform cross sections; (c) annularfin, and (d) pin fin of non-uniform cross section.


102/121

Innovative Fin Designs

Steady One-dimensional Conduction through Fins


103/121

Steady One dimensional Conduction through Fins

Adapted from Heat and Mass Transfer A Practical Approach, Y.A. Cengel, Third Edition,

McGraw Hill 2007.

Contd .


104/121

Energy Balance on Volume Element (fin)Contd .


105/121

Energy Balance on Volume Element (fin)

rate of heatconduction intothe element at x

rate of heatconduction fromthe element at

x+x

rate of heatconvection fromthe element

+=

))(( T T x P hQ S conv

conv x xcond xcond QQQ ,,

0)(,, T T xhP QQ xcond x xcond

Expressing the surface area, A s, in terms of width, x, times the perimeter P

QQ

Contd .


106/121

0 x

0)( T T hP dxdQ cond

0)(,, T T hP x

QQ xcond x xcond

Taking limit x tends to zero and using the definition of derivative:

dxdT

kAQ ccond

0)(

TThPdT

kAd

Contd .


107/121

0)(

T T hP dx

kAdx c

At constant A C and kSolution is;

C kAhp

m 2 T T

0)(22

T T kAhP

dxT d

c

Putting following constants as:

2d

Contd .


108/121

022

mdxd

mxmx eC eC x 21)(

Equation (A) is a linear, homogeneous, second-order differentialequation with constant coefficients.

The general solution of Eq. (A) is

(A)

Boundary Conditions

Contd .


109/121

Several boundary conditions are typically employed: At the fin base

Specified temperature boundary condition, expressedas: (0)= b= T b-T

At the fin tip1. Specified temperature

2. Infinitely Long Fin3. Adiabatic tip

4. Convection (andcombined convection).

Contd .


110/121

Base ( x = 0) condition 0 b bT T

Tip ( x = L) conditions A. :Conve ti |c on / x Lkd dx h L

B. :A / |diabati 0c x Ld dx

Fixed temper C. :ature L L D. (I >2.65): 0nfinite fin mL L Fin Heat Rate:

0| f f c x A sd

q kA h x dAdx


111/121

Solutions of Differential EquationContd .


112/121

)( T T hPkAq bc f

Fin Efficiency

To maximize the heat transfer from a fin the temperature

Fin Performance


113/121

To maximize the heat transfer from a fin the temperatureof the fin should be uniform (maximized) at the base value

of T b In reality, the temperature drops along the fin , and thus

the heat transfer from the fin is less To account for the effect we define

a fin efficiency

or,max

fin fin

fin

Q

Q Actual heat transfer rate from the fin

Ideal heat transfer rate from the finif the entire fin were at base temperature

,max ( ) fin fin fin fin fin bQ Q hA T T

For constant cross section of very long fins:

Contd .


114/121

For constant cross section of very long fins:

For constant cross section with adiabatic tip:

, ,max

1 1 fin c b clong fin

fin fin b

Q hpkA T T kAQ hA T T L hp mL

, ,max

tanh

tanh

fin c badiabatic fin

fin fin b

Q hpkA T T mL

Q hA T T

mLmL

Afin = P*L

Fin Effectiveness


115/121

The performance of the fins is judged on the basis of the

enhancement in heat transfer relative to the no-fin case. The performance of fins is expressed

in terms of the fin effectiveness fin

defined as

fin fin

finno fin b b

Q Q

Q hA T T

Heat transfer ratefrom the surface

of area A b

Heat transfer ratefrom the fin of base

area A b

with , and / f ch k A P

Fin Arrays


116/121

Representative arrays of(a) rectangular and(b) annular fins.

Total surface area:t f b A NA A

Number of fins Area of exposed base ( prime surface)

Define terms: A b: base area exposed to coolantAf : surface area of a single finAt: total area including base area and total finned surface,

N: total number of fins

( ) ( )t b f b b f f bq q Nq hA T T N hA T T

Overall fin efficiency for an array of fins: Contd .


117/121

Overall surface efficiency and resistance :

,1b

t o

t o t

Rq hA

1 1 f o f t

NA

A

bt

t t o hA

qq

q

max

OR

[( ) ]( ) [ (1 )]( )

[1 (1 )]( ) ( )

Define overall fin efficiency: 1 (1 )

t f f f b t f f b

f t f b O t bt

f O f

t

h A NA N A T T h A NA T T

NAhA T T hA T T A

NA

A

Total heat rate:

,

bt f f b b b o t b

t o

q N hA hA hA R

Contd .


118/121

,,

,

,

1( ) where

Compare to heat transfer without fins

1( ) ( )( )

where is the base area (unexposed) for the finTo enhance heat transfer

Th

bt t O b t O

t O t O

b b b f b

b f

t O

T T q hA T T R

R hA

q hA T T h A NA T T hA

A A A

Oat is, to increase the effective area . t A

=Ab+NAb,f

Thermal Resistance ConceptContd .


119/121

T1 T

Tb T2

T1 T Tb T2

L1 t

R1=L1/(k 1A)

Rb=t/(k bA)

)/(1, Ot Ot hA R

1 1

1 ,b t O

T T T T q R R R R

A=Ab+NAb,f

Topics to be referred by the students from Textbooks


120/121

Topics to be referred by the students from Textbooks

Shape factor

Effect of variable conductivity

Logarithmic mean radius


121/121

References:

Heat Transfer A Practical Approach, Y.A. Cengel, Second Edition, Tata McGraw Hill2003.

Heat and Mass Transfer A Practical Approach, Y.A. Cengel, Third Edition, TataMcGraw Hill 2007.

Heat Transfer , P. K. Nag, First Edition, Tata McGraw Hill 2002.

Heat and Mass Transfer , D.S. Kumar, Sixth Edition, S.K. Kataria & Sons 2004.

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Unit 2 HT by Conduction

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