Intro to Conduction

7/28/2019 Intro to Conduction

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Chapter 1

Conduction

(Material presented in this chapter are based on those in Chapters 2 & 3,Fundamentals of Heat and Mass Transfer, Fifth Edition by Incropera andDeWitt)

1.1 Introduction to conduction

Transport of energy in a medium due to temperature gradient is called con-duction. The medium can in general be solid, liquid or gas. The mechanismthat governs the conduction process is the random movement of atoms andmolecules.

1.2 Fouriers law

Conduction is governed by Fouriers law. Fouriers law provides a relationshipbetween the heat transfer flux and the temperature gradient, that is themanner by which the temperature varies in the medium. Fouriers law isdeveloped based on the observed phenomena and not from first principles.

Consider the system shown in Fig. (1.1) where the temperature at one

end of the control volume (say A) is T1 and at the other end (say B) isT2 with T1 T2. If the heat transfer rate is qx and the cross-sectional areaof heat transfer is A, intuition suggests that the heat transfer rate has tobe proportional to the cross-sectional area and the temperature differenceacross the control volume and inversely proportional to the length of the

9


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10 CHAPTER 1. CONDUCTION

Figure 1.1: 1-D heat conduction cartoon

control volume. Hence,

qx AT

x(1.1)

As x 0 and if the proportionality constant is k, which is the thermal

conductivity in W/(m.K), then,

qx = kAT

x(1.2)

Note that the minus sign in Eq. (1.2) signifies the heat flow along the negativetemperature gradient (see Fig. (1.2).

1.2.1 Heat flux

Often heat flux is required for estimating the extent of heat transfer. Heatflux is defined as the heat transfer rate per cross-sectional area of heat trans-

fer, that is, the area in the direction normal to the heat transfer. (In general,heat flow is typically normal to the constant temperature surface.) Heat fluxis given by

qx =qxA

= kT

x(1.3)


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1.2. FOURIERS LAW 11

Figure 1.2: Direction of heat transfer.

Heat flux need not necessarily be isotropic. Therefore, for heat transferthrough a 3-D cartesian co-ordinate system,

q = kT = k iT

x+ j

T

x+ k

T

z = iq

x + jq

y + kq

x (1.4)

or, in general, for any arbitrary geometry, if n is the outward normal, then

qn = kT

n(1.5)

We have so far assumed that the thermal conductivity, k is independent ofposition, that, the medium is isotropic. This is not always the case. So, ifthe system is anisotropic, then kx, kx, kz exists.

1.2.2 Thermal conductivity

Thermal conductivity of solids is larger than that of liquids and that of liquidsis larger than that of gases, that is,

ksolid > kliquid > kgas (1.6)


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Solid state

The thermal conduction in solids consists of electronic component and latticecomponent. Accordingly, the thermal conductivity can be written as

ksolid = ke + kl (1.7)

where ke is the electronic component which is inversely proportional to theelectrical resistivity and kl is the lattice component.

In case of pure metals, ke kl and hence the lattice component is notimportant. However, in case of non-metals, kl can be important and dependson the lattice arrangement.

Insulating systems

Insulating systems prevent/minimize transfer of heat through is and thereforerequire materials of low thermal conductivity. Several multi-phase systemssuch as those with solid and air media can be used for insulation purposes.For example, foams, flakes etc. The effective thermal conductivity, kef f ofan insulating systems is proportional to the thermal conductivity of solid k,radiative properties of the second phase, volume fraction.

Fluid state

Molecules more random than in solids. Thermal energy transport lesser thanin solids. Kinetic theory of gases to explain thermal conductivity dependenceon temperature, pressure and chemical species.

kgases nc (1.8)

where n is the number of particles/unit volume, c is the mean molecularspeed, and is the mean free path.

As the temperature increases, the kg increases, whereas, as molecularweight increases, the kg decreases. As n P,

1

P. Therefore kg is

independent of pressure.

Non-metallic liquids

k decreases with increase in T.


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1.3. HEAT DIFFUSION EQUATION 13

1.2.3 Thermal diffusivity

Thermal diffusivity, denoted usually by is the ability of a material to con-duct thermal energy relative to its ability to store the energy, that is,

=k

Cp(1.9)

1.3 Heat diffusion equation

1.3.1 Cartesian coordinates

Consider the differential element shown in Fig. (1.3) in cartesian co-ordinates.

Appropriate in and out heat transfer rates are depicted in the figure (Fig.(1.3)).

Figure 1.3: Differential element in cartesian coordinates for heat balance.

The energy balance for the element is

qx + qy + qz qx+dx qy+dy qz+dz + qdxdydz = CpT

tdxdydz (1.10)

where, q is the rate at which heat is generated per unit volume, is densityof the medium and Cp is the specific heat of the material.

After algebra and substituting Fouriers law in all three dimensions, Eq.(1.10) will take the form

x

kx

T

x

+

y

ky

T

y

+

z

kz

T

z

+ q = Cp

T

t(1.11)


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If we assume isotropic properties for the medium, then Eq. (1.11) reduces

tok

2T

x2+

2T

y2+

2T

z2

+ q = Cp

T

t(1.12)

that is,

2T +q

k=

1

T

t(1.13)

If the heat transfer is only in one direction, that is, if it is a 1-D problemthen Eq. (1.12) reduces to

2T

x2+

q

k=

1

T

t(1.14)

1.3.2 Cylindrical coordinates

The heat flux in cylindrical coordinates is

qr = kT = k

i

T

r+ j

1

r

T

+ k

T

z

(1.15)

The heat balance cylindrical coordinates similar to that in cartesian co-ordinates (Eq. (1.12) is

1r

r

kr T

r

+ 1

r2

k T

+

z

k T

z

+ q = Cp T

t(1.16)

1.3.3 Spherical coordinates

The heat flux in spherical coordinates is

qr = kT = k

i

T

r+ j

1

r

T

+ k

1

r sin

T

z

(1.17)

The heat balance spherical coordinates similar to that in cartesian coor-

dinates (Eq. (1.12) is1

r2

r

kr2

T

r

+

1

r2 sin2

k

T

+

1

r2 sin

k sin

T

+q = Cp

T

t(1.18)


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1.4. 1-D STEADY STATE CONDUCTION 15

1.3.4 Boundary conditions

Three different types of boundary conditions may be imposed on the heatbalance.

Dirichlet boundary condition: T(0, t) = TS where is TS is the constantsurface temperature.

Constant or finite flux boundary condition (Neumann boundary con-dition):

Finite heat flux

kT

x

|x=0 = q

s

Adiabatic or insulated surface

T

x|x=0 = 0

Convection surface conditions (Mixed or robin):

kT

x|x=0 = h[T T(0, t)]

, that is, flux at the boundary is equal to the flux at heat exchanged/entering

at the boundary.

1.4 1-D Steady state conduction

1.4.1 Plane wall

Consider the plane wall shown in Fig. (1.4). Assume steady state conditionsand the temperature at x = 0 to be Ts,1 and x = L to be Ts,2 and thecorresponding bulk temperature to be T,1 in the hot fluid and T,2 in thecold fluid. Assume the corresponding convection heat transfer coefficients to

be h1 and h2.Assuming no heat generation, that is q = 0, the heat balance will

readd

dx

k

dT

dx

= 0 (1.19)


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Figure 1.4: Heat transfer through a plane wall.

and the associated boundary conditions are

T(0) = Ts,1 & T(L) = Ts,2 (1.20)

After solving Eqs (1.19 - 1.20), the temperature along the plane wall isgiven by

T = (Ts,2 Ts,1)x

L+ Ts,1 (1.21)

and the heat transfer rate is given by

qx = kAdT

dx=

KA

L(Ts,1 Ts,2) (1.22)

Fig. (1.4) shows the temperature profile in the wall (obtained using Eq.(1.21)).

1.5 Thermal resistance

Thermal resistance is the resistance offered by a system to heat flow. Ageneral framework for detecting the thermal resistance for a given system is


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1.5. THERMAL RESISTANCE 17

developed here. Note that thermal resistance concept developed here can be

used only when the heat transfer rate is constant.A thermal resistance concept is useful for solving complex problems andthe associated thermal resistance networks ease the design calculations.

1.5.1 Thermal resistance offered by the wall

The resistance offered by the wall for heat flow is call the thermal resistance.It is defined as the temperature difference divided by the heat transfer rate.In mathematical form,

Rt,cond =

Ts,1 Ts,2qx =

L

kA (1.23)

1.5.2 Comparison with electrical resistance

Heat conduction has an analogy with the electrical conduction, which canbe realized by looking into the underlying resistances. Electrical conductionresistance is given by

Re =Es,1 Es,2

I=

L

A(1.24)

where, I is the current, E the voltage and the electrical resistivity.

Electrical resistance in Eq. (1.24) is the of same form as the thermalconduction resistance in Eq. (1.23).

1.5.3 Thermal resistance offered on either ends

At either boundaries of the plane wall, the wall is exposed to hot or coldfluid. The heat transfer (due to convection) from and to the hot or cold fluidto and from the plane wall is governed by Newtons law of cooling:

Rt,conv =Ts T

q=

1

hA(1.25)

where h is the convection heat transfer coefficient. h can also be an effectiveconstant for both convection and radiation. Similarly, separate h may existfor convection and radiation.


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As there is no heat generation, the heat transfer rate throughout

the system will remain constant.Therefore,

qx =T,1 Ts,1

1/(h1A)=

Ts,1 Ts,2L/(kA)

=Ts,2 T,2

1/(h2A)(1.26)

The overall temperature gradient in the system (starting from the hotfluid plane wall cold fluid) is T,1 T,2. Hence,

qx =T,1 T,2

Rtot(1.27)

where the total resistance, Rtot is given by

Rtot =1

h1A+

L

kA+

1

h2A(1.28)

Thermal resistance concept developed here can be used only when theheat transfer rate is constant.

1.5.4 Thermal resistance network for plane wall

A thermal resistance network for the plane wall (Fig. (1.4)) can be con-structed as shown in Fig. (1.5) using the expressions derived in Eqs (1.23),(1.25). The resistances due to cooling on either sides of the wall and due to

Figure 1.5: Thermal resistance network for a plane wall.

thermal conduction in the solid wall can be constructed in series. The totalresistance will be sum of the individual resistances.

1.6 Composite walls

Consider the composite wall shown in Fig. (1.6a) consisting of three walls,viz. A, B, and C. The temperature profile, which is linear with respect to x


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1.6. COMPOSITE WALLS 19

Figure 1.6: (A) Schematic of a composite plane wall. (B) Thermal resistancenetwork for the composite wall.

in each of the walls and relevant heat transfer coefficients are presented in thefigure. A resistance network can be constructed to mimic the heat transferthrough the composite wall as shown in Fig. (1.6b). The heat transfer rateis given by

qx =T,1 T,4

Rtot(1.29)

where the total resistance, Rtot is given by

Rtot =1

h1A+

LAkAA

+LB

kBA+

LCkCA

+1

h4A(1.30)

and A is the cross-section area.Note that all the resistances are in series. In principle, resistances can

also be in parallel depending upon the configuration of the parallel wall.(Note that the heat transfer rate can also be expressed in terms of the

resistance in each element similar to that presented in Eq. (1.26) for a planewall.)

1.6.1 Overall heat transfer coefficient

Similar to the heat transfer coefficient in Newtons law of cooling, a overallheat transfer coefficient U for conduction can be defined. The heat transfer


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rate in terms of the U is

qx = U AT (1.31)where,

U =1

RtotA=

11

h1+ LA

kA+ LB

kB+ LC

kC+ 1

h4

(1.32)

In general,

Rtot = Rt =T

q=

1

U A(1.33)

1.6.2 Contact resistance

So far, we assumed that the interface between the surfaces offer negligible

resistance. This assumption may not be valid under all situations, especiallyduring heat transfer via non-planar surfaces. In order to account for theresistance offered by the interface between walls A and B (Fig. (1.7)), contactresistance can be defined as

Rt,contact =TA TB

qcontact(1.34)

Figure 1.7: Contact between two non-planar surfaces.

Contact resistances are mainly due to surface roughness and there existsno good theory to predict these.


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1.7. ALTERNATIVE CONDUCTION ANALYSIS 21

1.7 Alternative conduction analysis

In all the earlier discussions, we considered situations where the heat transferrate and heat flux were constant along the direction in which the energy istransferred. This is due to the fact that the cross-sectional area remainedconstant throughout the system, the thermal conductivity was assumed con-stant and there was no heat generation in the medium. In this section, wewill consider the cases in which the heat transfer rate is constant and theflux is not as surface area varies and the thermal conductivity varies withtemperature and with no heat generation. The heat transfer rate is nowgiven by

qx = kAdT

dx

= k(T)A(x)dT

dx

(1.35)

If we assume no source and sink and steady-state, then the heat transferrate is constant whereas the heat flux is not constant. Heat transfer rate canbe expressed as

qx

xx0

dx

A(x)=

TT0

k(T)dT (1.36)

1.7.1 Trapezium

Consider the case of a trapezium (see Fig. (1.8)). The cross-section areaA is not constant, that is, A

xand k(T). The geometry can be much more

complicated than a trapezium.The heat transfer rate is now given by

qx = kAdT

dx(1.37)

where area is A = D2/4. If we assume D = ax, then

qx = ka2x2

4

dT

dx(1.38)

If we assume constant thermal conductivity, then Eq. (1.38) can be inte-grated over the whole domain to obtain

xx1

4qxa2x2

dx =

TT1

kdT (1.39)


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Figure 1.8: Variable cross-sectional area.

4qxa2

1

x+

1

x1

= k(T T1) T(x) = T1

4qxa2k

1

x1

1

x

(1.40)

At one of the boundaries, x = x2, T = T2. Therefore,

qx =a2k(T1 T2)

4

1

x1 1

x2

(1.41)Substituting for the flux qx (Eq. (1.41)) into Eq. (1.40) leads to

T(x) = T1 + (T1 T2)

1

x 1

x11

x1 1

x2

(1.42)

1.7.2 Radial systems - Cylinder

Consider the case of heat conduction in the radial direction in the wall of anannulus (see Fig. (1.9a)). Note that the area of heat transfer changes alongthe radial direction. However the heat transfer rate remains constant alongthe radius whereas the heat flux does not.


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Figure 1.9: Heat conduction through a cylindrical geometry.

The heat transfer rate of this system is given by

qr = kAdA

dr= k(2rL)

dT

dr(1.43)

The mathematical model that represents the heat conduction in this sys-tem is

1

r

d

dr

2rLdT

dr

= 0 1

r

d

dr

rdT

dr

= 0 (1.44)

The corresponding boundary conditions are

T(r = r1) = Ts,1; T(r = r2) = Ts,2 (1.45)

Solution for the Eqs (1.44) and (1.45) is

T(r) =Ts,1 Ts,2

ln r1r2

ln

r

r2

+ Ts,2 (1.46)

The temperature profile inside the annulus is presented in Fig. (1.9b).Compare this with the profile obtained in a planar wall. Note that the tem-perature profile in cartesian coordinates is linear whereas that in cylindricalcoordinates is logarithmic.


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Thermal resistance

Heat transfer rate is given by

qr =2Lk(Ts,1 Ts,2)

ln

r2r1

(1.47)and the thermal resistance is given by

Rt,cond =ln

r2r1

2Lk

(1.48)

Thermal resistance network

Heat transfer through the annulus can be represented in a thermal resistancenetwork. The resistance due to conduction in the annulus is given by Eq.(1.48) and the resistance due to convection of the fluid flowing inside andoutside the annulus is given by

Rt,conv,1 = Rt,conv,2 =1

h22r2L(1.49)

Using these resistances, the overall resistance network can be constructed asshown in Fig. (1.9c).

The heat transfer rate in terms of the total resistance offered by thesystem is given by

qr =T,1 T,2

1

h12r1L+ 1

h22r2L+

lnr2

r1

2Lk

(1.50)

1.7.3 Composite cylindrical wall

Consider the composite wall shown in Fig. (1.10a). The resistance networkfor this composite wall is in Fig. (1.10b).

If U is overall heat transfer coefficient for the composite wall, U is givenby

U =1

1

h1+ r1

kAln

r2r1

+ r2

kBln

r3r2

+ r3

kCln

r4r3

+ 1

h4

r1r4

(1.51)


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Figure 1.10: (a) Schematic of a composite cylindrical wall. (b) Resistancenetwork for the system in (a).

1.7.4 Spherical wall

Heat transfer rate in spherical annulus is

qr = k(4r2)

dT

dr(1.52)

Integrating Eq. (1.52), we obtain

qr =4k(Ts,1 Ts,2)

1

r1 1

r2

(1.53)

and the thermal conduction resistance is given by

Rt,cond =1

4k

1

r1

1

r2

(1.54)


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1.8 Conduction with thermal energy genera-

tionAll the cases we considered so far, we assumed no heat source/sink. En-ergy generation may be due to absorption of neutrons during a certain nu-clear/exothermic reaction occurring within the medium. Another sourcecould be due to electrical heating of the system in which case thermal en-ergy generation will be due to conversion of electrical to thermal energy.Generation of heat can strongly affect the conduction of heat in the solid.

In case of electrical energy source, the generation term will be governedby

Eg = I2R (1.55)

and the generation rate is given by

q =EgV

=I2Re

V(1.56)

where, I is the current generated, Re the electrical resistance and V thevolume of the system.

1.8.1 Plane wall

Consider a plane wall (Fig. 1.11) of width 2L in which q amount of energy isgenerated uniformly. Assume the temperature of the fluid flowing on eitherside is T,1 and T,2 and the corresponding heat transfer coefficients to beh1 and h2.

The heat balance for this system is

d2T

dx2+

q

k= 0 (1.57)

subject to the boundary conditions

T(L) = Ts,1; T(L) = Ts,2 (1.58)

Solving Eqs (1.57)-(1.58) gives the temperature profile:

T(x) =qL2

2k

1

x2

L2

+

Ts,2 Ts,12

x

L+

Ts,1 + Ts,22

(1.59)


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1.8. CONDUCTION WITH THERMAL ENERGY GENERATION 27

Figure 1.11: Plane wall with heat generation.

If Ts,1 = Ts,2 = Ts then Eq. (1.59) reduces to

T(x) =qL2

2k

1

x2

L2

+ Ts (1.60)

In many systems, it is of practical importance to estimate the maximal tem-perature inside the wall. The maximal temperature inside the wall will beat x = 0 for this problem and is given by

T(0) = T0 = qL

2

2k + Ts (1.61)

that is, dTdx|x=0 = 0. Note that this is also the temperature at the symmetry

point or at the adiabatic surface. (The maximal temperature occurring atx = 0 is due to the similar boundary condition on either boundaries, thatis at x = L. Such a symmetry will be broken when different boundarycondition combinations are used.)

Using Eq. (1.61), Eq. (1.60) can be rewritten as

T(x) T0Ts T0

= x

L2

(1.62)

Relationship between Ts and T

The temperature of the fluid Tis an observable, whereas it is difficult tomeasure Ts. Therefore, for design purposes, it is useful to relate the two


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temperatures. At x = L,

k dTdx|x=L = h(Ts T) (1.63)

k

qL

k

= h(Ts T) Ts = T +

qL

k(1.64)

1.8.2 Radial systems

Consider a cylinder (Fig. 1.12) in which heat is being generated at a rateq and the heat from the cylinder is being lost to the surrounding cold fluidwhich is at a constant temperature T.

Figure 1.12: Heat conduction through a cylinder with heat generation.

The energy balance for this system leads to the following model equation

1

r

rr T

r + q

k

= 0 (1.65)

which is subject to the boundary conditions

dT

dr|r=0 = 0; T(r0) = Ts (1.66)


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1.8. CONDUCTION WITH THERMAL ENERGY GENERATION 29

The dependence of the temperature on the radial position can be obtained

by solving the model equations (Eqs 1.65 - 1.66) and is given by

T(r) TsT0 Ts

= 1

r

r0

2(1.67)

As Ts is not a convenient observable, it is useful to relate this temperaturewith the fluid temperature. Using the overall balance

q(r20L) = h(2r0L)(Ts T) (1.68)

the relationship between Ts and T is

Ts = T + qr0

2h(1.69)


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Intro to Conduction