2.1 INTRODUCTION The heat flux given in equation (1.4)

(1.4)

can be written in a more general (three dimensional) form as

(2.1)

or

(2.2)

dxdTk

AQq

.

zTk

yTj

xTikTkq

zyx kqjqiqq

2.1 INTRODUCTION…

Comparison of one dimensional and three dimensional heat flow

2.1 INTRODUCTION… Major objectives in a conduction analysis isto determine the temperature field,to determine the conduction heat flux,for a solid, to ascertain structural integrity

throughdetermination of thermal stresses, expansions, and deflections.

to optimize the thickness of an insulating material.

2.1 INTRODUCTION…

Fig. 2.1 Differential control volume, dx.dy.dz, for conduction analysis in Cartesian coordinates

Derivation of heat equation in rectangular coordinate

2.1 INTRODUCTION… The conservation of energy principle is used to

find the conduction (energy diffusion) equation.

Referring to Fig. 2.1,

dxdydztTρc

volumecontroltheinstoredenergyofrateE

dxdydzq

volumecontroltheingeneratedenergyofrateE

volumecontroltheleavingenergyofrateE

volumecontrolthetoenteringenergyofrateE

p

st

.

.

gen

.

out

.

in

.

2.1 INTRODUCTION… Applying the energy balance equation to the control

volume,(2.3)

(2.4)

But from Taylor series approximation, (2.5)

And from Fourier’s law (2.6)

stoutgenin EEEE....

dxdydztTcdxdyqdxdzqdydzq

dxdydzqdxdyqdxdzqdydzq

pdzzdyydxx

zyx

)(

)(.

zq

qqdyyq

qqdxxq

qq zzdzz

yydyy

xxdxx

,,

zTkq

yTkq

xTkq zyx

,,

2.1 INTRODUCTION… Substituting equations (2.5) and (2.6) in to

equation (2.4) and dividing by dxdydz gives(2.7)

Where = energy generated per unit volume ( ) ρ= density ( ) =specific heat capacity (J/kg.K)

Equation (2.7) is a general transient three dimensional diffusion equation in rectangular coordinate system with energy generation.

tTc

zTk

zyTk

yxTk

xq p

.

.q

pc

3/mkg

3/mW

2.1 INTRODUCTION… For materials with constant thermal conductivity

k, equation (2.7) can be written as

is called thermal diffusivity (m2/s ). It is clear from the above equation that the ability of a material to let heat pass through it increases with increasing thermal diffusivity. This can be due to a high thermal conductivity k or a low heat capacity of the material.

tT

tT

kc

zT

yT

xT

kq p

12

2

2

2

2

2.

pck

2.1 INTRODUCTION…

Fig. 2.2 Differential control volume, dr.rddz, for conduction analysis in cylindrical coordinates

Derivation of heat equation in cylindrical coordinate

2.1 INTRODUCTION… The heat flux in cylindrical coordinates is 

(2.8) Where

(2.9)Applying the principle of conservation of energy,

the energy diffusion equation can be obtained as

(2.10)

zTkT

rj

rTikTkq

1

zTkqT

rkq

rTkq zr

,,

tTc

zTk

zTk

rrTkr

rrq p

2

. 11

2.1 INTRODUCTION… For constant thermal conductivity,

(2.11)

Following similar procedure, the heat diffusion equation can be obtained in spherical coordinate system (Fig. 2.3).

For constant thermal conductivity(2.12)

tT

α1

tT

kρc

zT

φT

r1

rT

rT

r1

kq p

2

2

2

2

22

2.

tT

α1

tT

kρc

θT

r1

θT

tanθr1

φT

θsinr1

rT

rT

r2

kq p

2

2

222

2

222

2.

2.1 INTRODUCTION…

Fig. 2.3 Differential control volume, dr.rsinθdrdθ, for conduction analysis in spherical coordinates

Derivation of heat equation in spherical coordinate

2.1 INTRODUCTION… 2.1.1 Boundary and Initial Conditions

2.1 INTRODUCTION… Example 2.1Passage of an electric current through a long

conducting rod of radius ri and thermal conductivity kr results in a uniform volumetric heating at rate of . The conducting rod is wrapped in an electrically non-conducting cladding material of outer radius ro and thermal conductivity kc, and convection cooling is provided an adjoining flow. For steady state conditions, write appropriate forms of the heat equations for the rod and cladding. Express appropriate boundary conditions for the solution of these equations.

.q

2.1 INTRODUCTION…

Fig. Example 2.1

2.1 INTRODUCTION… Solution From Equation 2.10, the appropriate forms of the heat

equation are:In the Conducting Rod,

In the Cladding,

Appropriate boundary conditions are:symmetry

common temperature at interface

heat flux crossing interface area

convection heat transfer at outer surface

0.

drdTr

drd

kq

0

drdTr

drd

00

r

r

drdT

)()( icir rTrT

ri

cc

ri

rr dr

dTk

drdT

k

TrThdrdT

k ocro

cc )(

2.2 STEADY STATE CONDUCTION2.2.1 One Dimensional Steady State

ConductionIn "one-dimensional“ system:only one coordinate is needed to describe the

spatial variation of the dependent variables. temperature gradients exist along only a

single coordinate direction and heat transfer occurs exclusively in that direction.

Equation (2.7) for one dimensional steady state heat conduction with no energy generation in rectangular coordinate system reduces to(2.14) 0

dxdTk

dxd

2.2 STEADY STATE CONDUCTION…Integration of equation (2.14) gives us

constant slope (2.15) and

linear variation (2.16)The constants C1 and C2 are obtained from boundary

conditions.Similarly for cylindrical coordinate system (equation

(2.11)) reduces to(2.17)

Integration of equation (2.17) gives (2.18)

1CkC

dxdT

21)( CxCxT

01

drdTkr

drd

r

21 )ln()( CrCrT

2.2 STEADY STATE CONDUCTION…2.2.2 Thermal Resistance Concept

Consider equation (2.16) applied to the slab shown in Fig.2.4. Applying boundary conditions

The heat transfer rate in the slab becomes

  

(2.19)Where R is called thermal resistance.

LTTC

dxdTandTC

LTTC 12

11212

1 ,

kALR

RTTQ

RTT

kALTT

LTTkA

LTTkA

dxdTkAQ

,21.

21212112.

Fig. 2.4 Thermal Resistance Concept

L

2.2 STEADY STATE CONDUCTION…Defining resistance as the ratio of a driving potential to the

corresponding transfer rate, the conduction thermal resistance can be written as,(2.20)

For electrical conduction in the same system, Ohm’s law provides an electrical resistance of the form(2.21)

We can also write a thermal resistance for convection heat transfer from Newton’s Law of cooling

And the convection thermal resistance is 

(2.22)

kAL

Q

TTRcond

.21

AL

IEERelect

21

RTT

hA

TTQorTThAQ ss

convsconv

)(1

)()(

..

hAQ

TTR

conv

sconv

1.

2.2 STEADY STATE CONDUCTION…The thermal resistance concept is more appropriate for

heat transfer problems with composite materials in series or parallel as shown in Fig. 2.5.

Fig. 2.5 Composite wall in series and corresponding thermal circuit

2.2 STEADY STATE CONDUCTION…The heat transfer rate for this system may be expressed as

(2.23) 

(2.24)

 Alternatively,  

(2.25)

In terms of an overall heat transfer coefficient,(2.26)

Where ΔT is the overall temperature difference.

RT

RTT

Q

4,1,.

AhAkL

AkL

AkL

Ah

TTQ

C

C

B

B

A

A

41

4,1,.

11

Ah

TT

AkLTT

AkL

TT

AkL

TT

Ah

TTQ s

C

C

s

B

B

A

A

ss

4

4,4,4,33221,

1

1,1,.

11

TUAQ .

2.2 STEADY STATE CONDUCTION…From equations (2.23) and (2.26) we see that UA = 1 /ΣR. Hence, for the

composite wall of Fig. 2.5,

(2.27)

In general, we may write

(2.28)

The thermal resistance concept can be applied to radial systems too (Fig.2.6). Assuming heat transfer only in the radial direction, equation (2.18) can be

used to obtain an expression for the thermal resistance.

Introducing the boundary conditions T(r1)=Ts,1 and T(r2)=Ts,2 , the constants C1 and C2 can be obtained and equation (2.18) becomes

(2.29)

41

1111

hkL

kL

kL

hRA

U

C

C

B

B

A

A

.

1

Q

TAU

RRtot

21 )ln()( CrCrT

2,2

2

1

2,1, lnln

)( sss T

rr

rr

TTrT

2.2 STEADY STATE CONDUCTION…The conduction heat transfer rate in the cylinder will

be obtained from Fourier’s law, 

(2.30) Equation (2.30) shows that conduction thermal

resistance in cylindrical coordinate system is given by (2.31)

1

2

2,1,

2

1

2,1,.

ln2

ln22

rrTT

Lk

rrr

TTrLk

drdTrLk

drdTkAQ ssss

RT

LkrrTT

Q ss

2

ln1

2

2,1,.

Lkrr

R2

ln1

2

2.2 STEADY STATE CONDUCTION…For the hollow cylinder shown in Fig. 2.6, the total thermal

resistance can be written as(2.32)

  outConvCondinConv

tot

RRRLrhLk

rr

LrhR

,,

22

1

2

11 21

2

ln

21

Fig. 2.6 Thermal resistance of a hollow cylinder

2.2 STEADY STATE CONDUCTION…Example 2.2 Uninsulated thin-walled pipe of 100mm diameter

is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions the pipe wall achieves a temperature of -150C and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is 30C and a convection coefficient of 2000W/m2K is maintained at the inner surface of the ice (k1.94W/mK), which is at 00C, what is the thickness of the ice layer?

2.2 STEADY STATE CONDUCTION…

Fig. Example 2.2

2.2 STEADY STATE CONDUCTION…SolutionPerforming an energy balance for a control

surface about the ice/water interface, it follows that, for a unit length of pipe,

Dividing both sides of the equation by r2,

The equation is satisfied by r2/r1=1.114, in which case r1=0.05m/1.114=0.045m, and the ice layer thickness is

krroTsiTsTTrh

qq

isii

condconv

2/)/ln(,,))(2(

12,.1

097.03

1505.02000

94.1,,)/()/ln(

,.212

12

xTT

oTsiTsrhk

rrrr

isii

mmmrr 5005.012

2.2 STEADY STATE CONDUCTION…2.2.3 Thermal Contact Resistance

Surfaces of solids are practically rough with numerous picks and valleys. When two or more such surfaces are pressed together, the picks form good contact(conductor) and the valleys form voids filled with air(insulator), Fig. 2.7(b).

(a) Ideal (b) Actual

Fig. 2.7 Thermal Contact Resistance

2.2 STEADY STATE CONDUCTION…Thus, an interface offers some resistance to heat transfer,

and this resistance per unit interface area is called the thermal contact resistance, Rc, given as(2.33)

 It can also be expressed in the form of Newton’s law of

cooling as (2.34)Where hc= thermal contact conductance A=apparent interface area

∆Tinterface=effective temperature difference at interfaceThe thermal contact resistance is given by

(2.35)

gapcontact QQQ...

erfacec TAhQ int

.

AQ

Th

R erface

cc

/

1.int

2.2 STEADY STATE CONDUCTION…Table 2.2 Thermal contact conductance of some metal surfaces in air

2.2 STEADY STATE CONDUCTION…2.2.4 Critical Thickness of Insulation

When a plane surface is covered with insulation, the rate of heat transfer always decreases.

However, the addition of insulation to a cylindrical or spherical surface increases the conduction resistance but reduces the convection resistance because of the increased surface area.

The critical thickness of insulation corresponds to the condition when the sum of conduction and convection resistances is a minimum. The rate of heat transfer from the insulated pipe to the surrounding air can be expressed as (Fig. 2.8)

 (2.36)

LhrLkrr

TTQ

2

1

2

1.

21

2

ln

2.2 STEADY STATE CONDUCTION…The variation of with the outer radius of the

insulation r2 is plotted in Fig. 2.9. The value of r2 at which reaches a maximum is determined from the requirement that . Solving this for r2 yields the critical radius of insulation for a cylinder to be

(2.37)Similarly the critical radius of insulation for a

sphere is given by

(2.38)

.Q

.Q

0/ 2

.drQd

hkrcr

hkrcr

2

2.2 STEADY STATE CONDUCTION…From Fig. 2.8 it can be seen that insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r2<rcr .

Fig. 2.8 Insulated pipe exposed to convection from external Fig. 2.9 Variation of heat transfer

rate with insulation thickness

2.2 STEADY STATE CONDUCTION…2.2.5 Optimum Thickness of Insulation

Insulation does not eliminate heat transfer but it merely reduces it.

The thicker the insulation, the lower the rate of heat transfer but the higher the cost of insulation.

Therefore, there should be an optimum thickness of insulation corresponding to a minimum combined cost of insulation and heat lost (Fig. 2.10).

The total cost, which is the sum of insulation cost and lost heat cost, decreases first, reaches a minimum, and then increases.

The thickness corresponding to the minimum total cost is the optimum thickness of insulation, and this is the recommended thickness of insulation to be installed.

2.2 STEADY STATE CONDUCTION…

Fig. 2.10 Optimum Insulation Thickness

2.3 EXTENDED SURFACES The term extended surface is used to describe a

system in which the area of a surface is increased by the attachment of fins.

A fin accommodates energy transfer by conduction within its boundaries, while its exposed surfaces transfer energy to the surroundings by convection or radiation or both.

Fins are commonly used to augment heat transfer from electronic components, automobile radiators, engine and compressor cylinders, control devices, and a host of other applications.

2.3 EXTENDED SURFACES …

Fig. 2.11 Use of fin to enhance heat transfer from a plane wall

2.3 EXTENDED SURFACES …To determine the heat transfer rate associated with a fin, we

must first obtain the temperature distribution along the fin. The following assumptions in determining the temperature

distribution:one-dimensional conduction in the x direction,Steady-state conditions,Constant thermal conductivity,Negligible radiation from the surface,heat generation effects are absent, and Convection heat transfer coefficient h is uniform over the

surface.

2.3 EXTENDED SURFACES …

Fig. 2.12 Fin element used for analysis

2.3 EXTENDED SURFACES …Applying the conservation of energy requirement to the

differential element of Fig.2.12, we obtain (2.39)But from Fourier’ Law,

(2.40)The conduction heat transfer at x+dx can be expressed as

(2.41)

Inserting equation (2.40) in to equation (2.41) (2.42)

The convection heat transfer will be expressed as(2.43)

convdxxx QdQQ...

dxdTkAQ cx

.

dxdxQd

QQ xxdxx

...

dxdx

dxdTkAd

dxdTkAQ

c

cdxx

)(.

))((.

TTdAhQd sconv

2.3 EXTENDED SURFACES …Then equation (2.36) becomes,

Or (2.44)

For fins with constant cross-section Ac, the element surface area dAs=Pdx (where P is fin perimeter) and equation (2.44) becomes,

(2.45)

))(()(

TTdAhdxdx

dxdTkAd

dxdTkA

dxdTkA s

c

cc

0))((

TTdxdA

kh

dxdTA

dxd s

c

0)(2

2

TTkAhP

dxTd

c

2.3 EXTENDED SURFACES …Let (x)=(T(x)-T∞)then since T∞ is constant. Equation (2.45)

becomes,

Or (2.46)

Where . The general solution of differential equation (2.46) is

(2.47)

 The constants C1 and C2 are obtained from boundary conditions.

02

2

ckAhP

dxd

022

2

mdxd

ckAhPm

mxmx eCeCx 21)(

2.3 EXTENDED SURFACES …The boundary conditions used occur at fin

base and tip.A. Boundary condition at fin base is specified

temperature condition. This temperature is usually assumed to be known.

(2.48)From equation (2.47),

(2.49)

TTx bb )0(

21 CCb

2.3 EXTENDED SURFACES …B. The boundary condition at fin tip has three optionsI. Infinitely long fin(L→∞,Ttip=T∞)

(x=L)=Ttip-T∞=0=C1emL+C2e-mL

But as L→∞, e-mL=0 and C1=0 and C2=b (from equation (2.49))So, for an infinitely long fin equation (2.47) becomes

or (2.50)

xkAhP

bmx

bceex

)(

xkAhP

b

cex

)(

2.3 EXTENDED SURFACES …The heat removed by the fin at base is

(2.51)

00

.

x

cx

cbase dxdkA

dxdTkAQ

cbcbcbase kA

hPkAmkAQ .

cbbase hPkAQ .

2.3 EXTENDED SURFACES …II. Negligible heat loss from fin tip

(2.52)Combining equations (2.49) and (2.52), and

solving for the constants C1 and C2, (2.53)

The heat loss from fin base is(2.54)

0Lxdx

d

021

mLmL

Lx

eCeCmdxd

)cosh(

)(cosh)(mL

xLmx

b

)tanh(.

mLhPkAQ cbbase

2.3 EXTENDED SURFACES …III. Convection from fin tip

(2.55)Or

(2.56)Solving for the constants C1 and C2

(2.57)The corresponding heat loss from fin base will

be,(2.58)

LxcLxc dxdTkATThAQ

)(.

LxcLxc dxdkAhA

)sinh(/)cosh(

)(sinh/)(cosh)(mLmkhmL

xLmmkhxLmx

b

)sinh()/()cosh()cosh()/()sinh(.

mLmkhmLmLmKhmLhPkAQ cbbase

2.3 EXTENDED SURFACES …Example 2.3 A brass rod 100mm long and 5mm in diameter

extends horizontally from a casting at . The rod is in an air environment with and . What is the temperature of the rod 25, 50 and 100mm from the casting? Take thermal conductivity of brass to be k=133W/m.K.

2.3 EXTENDED SURFACES …

2.3 EXTENDED SURFACES …SolutionBased on the assumption of convection heat

loss from fin tip, the temperature distribution, from equation (2.57), has the form

The temperatures at the prescribed location are tabulated belowx(m) T(0C) 0.025 156.5 0.05 128.9 0.10 107.0

)sinh(/)cosh(

)(sinh/)(cosh)(mLmkhmL

xLmmkhxLmx b

2.3 EXTENDED SURFACES …2.3.1 Fin Effectiveness

The performance of fins is judged on the basis of the enhancement in the heat transfer relative to the no-fin case. The performance of fins expressed in terms of the fin effectiveness fin is defined as (Fig. 2.13)(2.59)

In any rational design the value of fin should be as large as possible, and in general, the use of fins may rarely be justified unless fin 2.

)T(ThAQ

Q

Q

AareaofsurfacethefromratetransferHeatAareabaseoffinthe

fromratetransferHeat

εbb

fin

.

finno

.fin

.

b

bfin

2.3 EXTENDED SURFACES …

Fig. 2.13 Fin Effectiveness

2.3 EXTENDED SURFACES …2.3.2 Fin Efficiency

Another measure of fin thermal performance is provided by the fin efficiency, fin.

The maximum driving potential for convection is the temperature difference between the base (x = 0) and the fluid, b=Tb-T∞.

Hence the maximum rate at which a fin could dissipate energy is the rate that would exist if the entire fin surface were at the base temperature.

However, since any fin is characterized by a finite conduction resistance, a temperature gradient must exist along the fin and the above condition is an idealization.

A logical definition of fin efficiency is therefore

(2.60)

Where Afin is the surface area of the fin. )T(ThAQ

Q

Q

bfin

fin

.

.fin

.

fin

max

2.3 EXTENDED SURFACES …2.3.3 Proper Length of Fin

The temperature of a fin drops along the fin exponentially and reaches the environment temperature at some length.

The part of the fin beyond this length does not contribute to the heat transfer.

Designing such an extra long fin results in material waste, excessive weight and increased size and cost.

2.3 EXTENDED SURFACES …

Fig. 2.14 Proper length of fin

0 2 4 60

0.5

1

tanh mL( )

mL

Fig. 2.15 Variation of heat transfer from a fin relative to that from relatively long fin

2.3 EXTENDED SURFACES …To get the sense of the proper length of a fin, we

compare the heat transfer from a fin of finite length to the heat transfer from an infinitely long fin with the same conditions. (2.61)

This ratio becomes unity for mL2.5 as can be seen from Fig. 2.15. Therefore, gives proper length of a fin and the designer should make proper compromise between heat transfer performance and fin size.

)tanh()(

)tanh()(mL

TThpkA

mLTThpkA

Q

QRatioTransferHeat

bc

bc

finlong

.fin

.

mL 5.2

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION

Fig. 2.16 Conduction in plane wall with uniform energy generation

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…Consider a one dimensional heat flow through plane

wall of Fig. 2.16.For constant thermal conductivity k, equation (2.7)

reduces to (2.62)

The general solution is(2.63)

Where C1 and C2 are the constants of integration obtained from boundary conditions.

0.

2

2

kq

dxTd

212

.

2)( CxCx

kqxT

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…For the prescribed boundary conditions

T(-L)=Ts,1 and T(L)=Ts,2

Solving for constantsand

The temperature distribution becomes(2.64)

Consider the long solid cylinder of Fig.2.17

LTT

C ss

21,2,

1

2

.

1,2,2 22

LkqTT

C ss

221

2)( 1,2,1,2,

2

22.

ssss TTLxTT

Lx

kLqxT

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…For constant thermal conductivity k, equation

(2.10) reduces to(2.65)

And the temperature distribution will be(2.66)

The constants of integration C1 and C2, are obtained by applying the boundary conditions

and

01.

drdTr

drd

rkq

212

.

)ln(4

)( CrCrkqrT

so TrT )( 00

rdr

dT

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…

Fig. 2.17 Conduction in a solid cylinder with uniform energy generation

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…The constants will be

andAnd the temperature distribution becomes

(2.67)

01 C 2

.

2 4 os rkqTC

so

o Trr

krq

rT

2

22.

14

)(

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…Example 2.4A nuclear fuel element of thickness 2L is covered with a

steel cladding of thickness b. Heat generated within the nuclear fuel at a rate removed by a fluid at T∞, which adjoins one surface and is characterized by a convection coefficient h. The other surface is well insulated, and the fuel and steel have thermal conductivities of kf and ks, respectively. a. Obtain an equation for the temperature distribution T(x) in

the nuclear fuel. Express your results in terms of , kf, L,W ks, h and T∞.

b. Sketch the temperature distribution T(x) for the entire system.

.q

.q

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…

Fig. Example 2.4

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…Solution(a) The heat equation for the fuel is,

And the corresponding temperature distribution is

The insulated wall at x = -(L+b) dictates that the heat flux at x = - L is zero (for an energy balance applied to a control volume about the wall, Ein=Eout=0).

)(0.

2

2

LxLkq

dxTd

f

212

.

2)( CxCx

kqxT

f

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…Hence

(a)

The value of Ts,1 may be determined from the energy conservation requirement that or on a unit area basis

ff kLqCorCL

kq

dxdT

.

11

.

0)(

2

.

2

.

2)( Cx

kLqx

kqxT

ff

convcondg QQE...

TThTTbk

Lq ssss

2,2,1,

.)2(

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…Hence,

(b)Combining equations (a) and (b),

(c)This gives

The temperature distribution for (-L≤x≤L) is

ThLq

kLbqTs

s)2()2(

..

1,

2

2...

1, 23)2()2()( C

kLqT

hLq

kLbqTLT

fss

T

kL

hkbLqC

fs 2322.

2

T

kL

hkbLqx

kLqx

kqxT

fsff 2322

2)(

..

2

.

2.4 CONDUCTION WITH THERMAL ENERGY GENERATION…(b) The temperature distribution is shown in

the following three regions

tconsdxdTbLxL

xwithdxdTLxL

TTdxdTLxbL

tan:

:

,0:)( max

2.5 TRANSIENT CONDUCTIONPractical problems in heat transfer, in general,

involve the variation of temperature with Position (x,y,z) and Time (t).

Analysis of transient heat conduction is more complicated than that of steady state conduction and making simplifying assumptions is more appropriate.Lumped capacitance method andOne dimensional assumptions.

2.5 TRANSIENT CONDUCTION…2.5.1 Lumped Capacitance SystemThe temperature variations within some bodies remain

essentially uniform at a given time during a heat transfer process (Fig. 2.18).

Such bodies are said to act like a ‘lump’. Consider a hot metal forging that is initially at a

uniform temperature Ti and is quenched by immersing it in a liquid of lower temperature T∞<Ti (Fig.2.18).

The essence of the lumped capacitance method is the assumption that the temperature of the solid is spatially uniform at any instant during the transient process (T=f(t)).

2.5 TRANSIENT CONDUCTION…

Fig. 2.18. Cooling of a hot metal forging

2.5 TRANSIENT CONDUCTION…The variation of temperature of the hot metal

with time can be obtained by applying energy balance within a time interval of dt.

(2.68)(2.69)

Rearranging equation (2.69) and noting that dT=d(T-T∞), since T∞ is constant, we obtain(2.70)

dtinbodytheofcontentenergytheindecreasedtduringbodythefromtransferHeat

Outconv EQ..

dtdTmCTThA p )(

dtmChA

TTdT

p

2.5 TRANSIENT CONDUCTION…Integration of equation (2.70) gives us

Or(2.71)

The temperature of the metal gradually decreases and eventually equals the ambient air temperature.

t

p

T

T

dtmChA

TTdT

i 0

tmChA

TTTtT

pi

)(ln

tmChA

i

peTTTtT

)(

2.5 TRANSIENT CONDUCTION…The rate of convection heat transfer is

(2.72)Total amount of heat transfer from the hot

body to the surrounding, time 0-t, is(2.73)

The maximum amount of heat transfer is (2.74)

])([.

TtThAQ

)(

])([tT

Tipp

i

TtTmCdTmCQ

][][max TTmCTTmCQ ipip

2.5 TRANSIENT CONDUCTION…Validity of lumped capacitance methodThe lumped capacitance analysis gives us a simple

and convenient way of analyzing transient heat transfer problems.

But this method is ideal since it assumes uniform temperature distribution within a body at an instant and it is better to know when to use it.

Before establishing a criterion for the validity of the method, it is worthy to define the terms known as characteristic length, Lc, and Biot number, Bi.

2.5 TRANSIENT CONDUCTION…(2.75)

(2.76)The Biot number can also be expressed as

Or

AVLc

khL

B ci

BodythewithinTransferHeatConductionBodytheofSurfacetheatTransferHeatConvection

TLk

ThB

c

i

BodytheofSurfacetheatcesisConvectionBodythewithincesisConduction

hkLB c

i tanRetanRe

/1/

2.5 TRANSIENT CONDUCTION…The lumped capacitance analysis assumes

uniform temperature distribution within the body. This is true when the conduction resistance

within the body is zero, which is in turn true only when the Biot number is zero.

Therefore, the lumped capacitance analysis is exact only when Bi=0 and approximate when Bi>0.

From experience the lumped capacitance method is, in general, acceptable if Bi≤0.1.

2.5 TRANSIENT CONDUCTION…2.5.2 Transient heat conduction in large plane

walls, long cylinders and spheresA more realistic assumption than lumped

capacitance method is the case where temperature varies with time and position in one dimension, T(x, t) or T(r, t), which is applicable for large plane walls, long cylinders and spheres (Fig. 2.19).

If heat transfer takes place between the environment (h and Ti>T∞) and the large plane wall (initially at Ti ) of Fig. 2.19(a) by convection, the temperature at the surface of the wall drops.

2.5 TRANSIENT CONDUCTION…

(a) Large plane wall (b) Long cylinder (c) SphereFig. 2.19 Bodies where one dimensional temperature variation can be assumed

2.5 TRANSIENT CONDUCTION…

Fig. 2.20 Transient temperature variation in a large plane wall

2.5 TRANSIENT CONDUCTION…For the large plane wall the diffusion

equation (2.7) becomes,

(2.77)The solution of equation (2.77) results in

infinite series which are inconvenient and time consuming to evaluate.

Therefore, the solutions are presented in tabular or graphical form.

tTc

xTk

x p

tT

xT

1

2

2

2.5 TRANSIENT CONDUCTION…Before presenting the solutions graphically,

some parameters need to be nondimensionalzed to reduce number of parameters.

timeessDimensionlnumberFourier

t)coefficientransferheatless(DimensionnumberBiot

centerfromdistanceessDimensionl

etemperaturessDimensionl),(

),(

2LtkhLBi

LxX

TTTtxT

txi

2.5 TRANSIENT CONDUCTION…For Fourier number , the infinite series

solutions of equation (2.77) can be approximated by taking only the first terms of the series. These solutions are given below,(2.78)

(2.79)

(2.80)

2.0,cos),(),( 11

21

LxeA

TTTtxTtx

iwall

2.0,),(),( 11

21

oo

icyl r

rJeATTTtrTtr

2.0,sin

),(),(1

1

1

21

o

o

isph

rrrr

eATTTtrTtr

2.5 TRANSIENT CONDUCTION…The constants A1 and 1 are obtained from table

2.3. The function J0 is the zeroth-order Bessel function

of the first kind whose value can be determined from Table 2.4.

The temperature of the body changes from Ti to T∞ at the end of the transient heat conduction.

The maximum amount of heat transfer during this process can be obtained by(2.81)

)()(max ipip TTVCTTmCQ

2.5 TRANSIENT CONDUCTION…The fraction of heat transfer within time t is

obtained by the following equations for the large wall, long cylinder and sphere.

(2.82)

(2.83)

(2.84)

1

11

max

)sin(1

21

eA

QQ

wall

1

111

max

)(212

1

JeA

QQ

cylinder

31

1111

max

)cos()sin(31

21

eAQQ

wall

2.5 TRANSIENT CONDUCTION…Example 2.5A load of peas at a temperature of 250C is to be

cooled down in a room at constant air temperature of 10C.

(a) How long the peas will require to cool down to 20C when the surface heat transfer coefficient of the pea is 5.81W/m2K?

(b) What is the temperature of the peas after a lapse of 10 minutes from the start of cooling?

(c) What air temperature must be used if the peas were to be cooled down to 50C in 30 minutes? The peas are supposed to have an average diameter of 8 mm, their density is 750kg/m3 and specific heat 3.35 kJ/kgK.

2.5 TRANSIENT CONDUCTION…SolutionThe problem can be solved by making use of

the lumped capacitance method, neglecting any variation of temperature within the peas due to its small diameter. From equation (2.68)

(a) Solving for time t,

tmChA

i

peTTTtT

)(

min6.30183512512ln

008.0**81.5

3350*6008.0**750

)(ln 2

3

sTTTtT

hAmC

ti

p

2.5 TRANSIENT CONDUCTION…(b) From equation (2.65)

(c)

CTeTeTTTtT t

mChA

i

p 0

600*3350*

6008.0*

*750

008.0**81.5

48.9600353.0125

1)600()(3

2

CTeTTe

TTTtT t

mChA

i

p 0

1800*3350*

6008.0*

*750

008.0**81.5

08.4044.0255)(

3

2

2.6 NUMERICAL METHODSMost of the practical problems encountered

in engineering involve:Complicated geometries,Complex boundary conditions, andVariable properties.

Since such problems cannot be solved analytically, the need for numerical solution methods, especially in cases of multidimensional problems, is inevitable.

2.6 NUMERICAL METHODS…2.6.1 Finite Difference Equation

There are several types of numerical methods. Some are:Finite Difference Method,Finite Element Method,Boundary Element Method, andControl Volume Method.

Because of its ease of application, the finite-difference method is well suited for an introductory treatment of numerical techniques.

2.6 NUMERICAL METHODS…A numerical solution enables determination of the

temperature at only discrete points. The first step in any numerical analysis must,

therefore, be to select these points. This is done by subdividing the medium of

interest into a number of small regions and assigning to each a reference point that is at its center.

The reference point is frequently termed as nodal point (or simply a nod), and the aggregate of points is termed a nodal network, grid, or mesh.

2.6 NUMERICAL METHODS…The nodal points are designated by a numbering

scheme that, for a two-dimensional system, may take the form shown in Fig.2.21.

The x and y locations are designated by the m and n indices, respectively.

The temperature of node (m, n) is assumed to be the average of the surrounding shaded area.

The accuracy of a numerical analysis is increased by increasing the number of nodes (fine nodes).

But the increased number of nodes requires more computing time and capacity.

2.6 NUMERICAL METHODS…

Fig. 2.21 Nodal network of two-dimensional conduction

2.6 NUMERICAL METHODS…The finite-difference equation for a node can be

obtained by applying conservation of energy to a control volume about the nodal region.

Since the actual direction of heat flow (into or out of the node) is often unknown, it is convenient to formulate the energy balance by assuming that all the heat flow is into the node.

For steady-state conditions with no generation, the appropriate form of equation (2.3) is(2.85) 0

. inE

2.6 NUMERICAL METHODS…There are different finite difference equations for

interior and boundary nodes. 1. For interior node (m, n) of Fig. 2.22, the finite

difference equation can be obtained, assuming unit depth, as

for x=y, the above equation simplifies to (2.86)

0),(),1(

.

),(),1(

.

),()1,(

.

),()1,(

. nmnmnmnmnmnmnmnm QQQQ

01.1.1.1. ),(),1(),(),1(),()1,(),()1,(

xTT

ykxTT

ykyTT

xkyTT

xk nmnmnmnmnmnmnmnm

04 ),(),1(),1()1,()1,( nmnmnmnmnm TTTTT

2.6 NUMERICAL METHODS…

Fig. 2.22 Interior node for finite difference equation formulation

2.6 NUMERICAL METHODS…2. Internal corner node with convection (Fig.

2.23)

for x=y, the above equation simplifies to

(2.87)

0.

),(),1(

.

),(),1(

.

),()1,(

.

),()1,(

. convnmnmnmnmnmnmnmnm QQQQQ

01.22

1.

1.2

1.2

1.

),(),(),1(

),(),1(),()1,(),()1,(

nmnmnm

nmnmnmnmnmnm

TTyxhxTT

yk

xTTyk

yTTxk

yTT

xk

0232)()(2 ),(),1()1,(),1()1,(

TkxhT

kxhTTTT nmnmnmnmnm

2.6 NUMERICAL METHODS…

Fig. 2.23 Internal corner node with convection

2.6 NUMERICAL METHODS…3. Plane surface node with convection (Fig. 2.24)

for x=y, the above equation simplifies to

(2.88)

0.

),(),1(

.

),()1,(

.

),()1,(

. convnmnmnmnmnmnm QQQQ

01.1.1.2

1.2 ),(

),(),1(),()1,(),()1,(

nm

nmnmnmnmnmnm TTyhxTT

ykyTTxk

yTTxk

0222)2 ),()1,(),1()1,(

TkxhT

kxhTTT nmnmnmnm

2.6 NUMERICAL METHODS…

Fig. 2.24 Plane surface node with convection

2.6 NUMERICAL METHODS…4. External corner node with convection (Fig.

2.25)

for x=y, the above equation simplifies to(2.89)

0.

),(),1(

.

),()1,(

. convnmnmnmnm QQQ

01.22

1.2

1.2 ),(

),(),1(),()1,(

nmnmnmnmnm TTxyh

xTTyk

yTTxk

0212) ),()1,(),1(

TkxhT

kxhTT nmnmnm

2.6 NUMERICAL METHODS…

Fig. 2.25 External corner node with convection

2.6 NUMERICAL METHODS…5. Plane surface node with heat flux (Fig. 2.26)

for x=y, the above equation simplifies to(2.90)

0.

),(),1(

.

),()1,(

.

),()1,(

. fluxnmnmnmnmnmnm QQQQ

0''1.1.1.2

1.2

),(),1(),()1,(),()1,(

qyhxTT

ykyTTxk

yTTxk nmnmnmnmnmnm

0''242 ),()1,(),1()1,(

kxqTTTT nmnmnmnm

2.6 NUMERICAL METHODS…

Fig. 2.26 Plane surface node with heat flux

2.6 NUMERICAL METHODS…2.6.2 Solution of the finite difference

equationsThe equations obtained for each type of node

reduce the heat transfer problem to solving of system of linear equations, which can be written in matrix notation as,(2.91)

Where [A] is coefficient matrix, {T} is vector of nodal temperatures and {C} is vector of constants obtained from boundary conditions.

CTA

2.6 NUMERICAL METHODS…

nnnn

n

n

aaa

aaa

aaa

A

.............

....

....

21

22221

11211

nT

TT

T

.

.

.2

1

nC

CC

C

.

.

.2

1

2.6 NUMERICAL METHODS…Equation (2.91) can be solved using either

the matrix inversion or the iterative methods. In the matrix inversion method, the nodal temperatures will be obtained from

(2.92)A good example of iterative methods of

solving linear system of equations is the Gauss-Seidel Iteration method.

CAT 1

2.6 NUMERICAL METHODS…Consider the following system of equations

for explanation of the solution procedure.

a) Solve each equation for one of the variables (one with larger coefficient) in terms of other variables,

132452

103

321

321

321

xxxxxx

xxx

)()2/()13()(5/)24()(310

213

312

321

cxxxbxxxaxxx

2.6 NUMERICAL METHODS…b) Make initial guess for each unknown,Let x2=0 and x3=0c) Using equations from step 1, find new values for each

unknown,Using the initial guess and equation (a) of step 1,

x1=10+3(0)-0=10Using the updated value x1=-5 and equation (b) of step

1, x2=(4-2(10)-0)/5=-3.2From equation (c), x3=(-13+10-(-3.2))/(-2)=-0.1d) Repeat step 3 until a desired convergence criterion is

satisfied.

2.6 NUMERICAL METHODS…Example 2.6 Consider the square channel shown in the sketch operating

under steady-state conditions. The inner surface of the channel is at a uniform temperature of 600K, while the outer surface is exposed to convection with a fluid at 300K and a convection coefficient of 50W/mK.

a) Beginning with properly defined control volume, derive the finite-difference equations for interior and boundary nodes. Due to symmetric nature of the problem, take one eighth of the geometry with Δx=Δy=0.01m. Calculate the temperatures for all nodes.

b) Calculate the heat loss per unit length from the channel.

2.6 NUMERICAL METHODS…

Fig. Example 2.6

2.6 NUMERICAL METHODS…Solution(a) Define control volumes about the nodes

taking advantage of symmetry where appropriate and performing energy balances, , with Δx=Δy=0.01m. The one eighth geometry of the channel is meshed in to nine nodes as shown below.

0..

outin EE

2.6 NUMERICAL METHODS…

2.6 NUMERICAL METHODS…Node 1:

)1(1505.2

2

0

0222

0

152

152

11512

11512

.

15

.

12

.

TTT

TkxhT

kxhTT

TTkxhTTTT

TTxhxTTxk

xTTxk

QQQ Conv

2.6 NUMERICAL METHODS…Node 2:

)2(30052

022

0

2361

2232621

.

23

.

26

.

21

.

TTTTgivesequationthissolving

TTxhxTTxk

xTT

xkxTTxk

QQQQ Conv

2.6 NUMERICAL METHODS…Node 3: Similar to Node 2,

Node 4:)3(30052 3472 TTTT

)4(1505.1

022

0

43

443

.

43

.

TTgivesequationthissolving

TTxhxTTxk

QQ Conv

2.6 NUMERICAL METHODS…Node 5:

)5(60042

022

0

561

585651

58

.

56

.

51

.

TTTgivesequationthissolving

xTTxk

xTTxk

xTTxk

QQQ

2.6 NUMERICAL METHODS…Node 6: Interior node

)6(600404

int83.2

6752

69752

TTTTTTTTT

nodeeriorforequationFrom

2.6 NUMERICAL METHODS…Node 7:

)7(02

0

0

763

7673

76

.

73

.

TTTgivesequationthissolvingxTT

xkxTT

xk

QQ

2.6 NUMERICAL METHODS…Equations (1) through (7) can be solved

simultaneously using either matrix inversion method or Gauss-Seidel iteration method. But since the number of equations is few, the matrix inversion method can be used.

(8)

026004

600421505.1

3005230052

1505.2

763

6752

561

43

3472

2361

152

TTTTTTT

TTTTT

TTTTTTTT

TTT

2.6 NUMERICAL METHODS…Equation (8) can be written in matrix form as

[A]{T}={C} where

2,1,0,0,1,0,01,4,1,0,0,1,00,2,4,0,0,0,1

0,0,0,5.1,1,0,02,0,0,1,5,1,00,2,0,0,1,5,1

0,0,1,0,0,1,5.2

A

7

6

5

4

3

2

1

TTTTTTT

T

0600600150300300150

C

,

,

2.6 NUMERICAL METHODS…The temperatures can be obtained using matrix

conversion as, {T}=[A]-1{C}The solution will be,

112.443092.492571.503755.362133.394684.421102.430

7

6

5

4

3

2

1

TTTTTTT

T

2.6 NUMERICAL METHODS…b) Referring to Fig. Example 2.6 Solution (a), the

heat loss from the channel is eight times the heat loss from the one eighth portion shown.

m3746.946W/

))300755.362(5.0)300133.394()300684.421()300102.430(5.0(03.0508))(5.0)()()(5.0(8 4321

.

xxTTTTTTTThAQloss