ADVANCED HEAT TRANSFER

MCEG 5223

UNIT 4. THEORY AND FUNDAMINTALS OF THERMAL CONDUCTIVITY

Objectives: The objectives of this unit are (1) to familiarize the student with ranges of

measured values of thermal conductivities of solids, liquids and gases; (2) to

examine the fundamental transport mechanisms giving rise to thermal

conductivity; and (3) to briefly discuss theoretical methods of determining

thermal conductivities.

The student should develop a "feel" for magnitudes of thermal

conductivities for various types of materials, and he/she should understand the

contributing mechanisms in a given type of material and their effects on apparent

thermal conductivity.

THERMAL CONDUCTIVITY -- EXPERIMENTAL VALUES

Values of thermal conductivity for some selected solids, liquids, and gases are given in

Table xxx, page xx of the text. Notice in particular that the values for metals are for pure metals

--- design applications usual1y involve alloys.

Note the range of values (W/m-K):

Solids (metallic): 34.7 - 4l8.0

Solids (nonmetallic): l.05 - 41.6

Liquids: 0.0728 - 8.2l

Gases: 0.0066 - 0. I 75

Roughly speaking, there is an order-of-magnitude between the range for each successively

increasing group beginning with the lowest, gases.

THE NATURE OF THERMAL CONDUCTIVITY

The fundamental transport mechanisms in each of these are briefly discussed in the

following outline:

A. Solids

In general, thermal energy may be transported in solids by means of lattice waves

(phonons) and free (valence) electron drift. A phonon is a quantum of energy in the form of a

thermoelastic wave of a given (fixed) frequency -- this is analogous to a photon in

electromagnetic radiation theory. Transport by either lattice waves or by free electrons may be

predominant, or they may both be of the same order of magnitude as we will see in the special

cases to follow.

A.1. Nonmetallic Solids

In nonmetallic solids, there are no free electrons, and the total thermal energy transport is

by lattice waves (i.e., vibration of the crystalline lattice). Kinetic theory for these solids results in

A major problem is that the mean free path length is difficult to obtain (or predict). We will next;

consider two subcases of nonmetallic solids.

A.1.a. Ideal Crystals - For nonmetallic ideal crystals, there is no phonon interference (wave

interference), and approaches the dimension of the specimen, d. Then

and the thermal conductivity is a function of the specimen size! Even in ideal crystals, however,

there are at least two phonon-phonon types of collisions -- one of these, the Umklapp process, is

an anharmonic interaction resulting in a limiting of the mean free path .

A.1.b. Actual Crystals (nonmetallic) - For actual crystals, in addition to the UmKlapp process,

there are:

Elastic scattering

Inelastic scattering

Impurity scattering

Mosaic structure scattering

They all arise because of the small structural anomalies in the crystal lattice. All of these

interfere with wave vibrations, i.e., they "scatter" phonons and thereby decrease k.

Some important non-metallic crystals are

glassy or amorphous materials -- in these, atoms or molecules are distributed in a

manner lacking symmetry or periodicity.

refractory materials -- these are mixtures of crystalline and amorphous substances.

In these, k depends upon the thermal conductivities and void fractions of the

components of the mixture

Other -- building and insulating materials, plastics, rubber, leather, fibrous,

powders, and granular materials. These are mixtures of solid amorphous materials

or amorphous materials with fluids. In these, a measured thermal conductivity is

the Apparent Thermal Conductivity. This measured value may depend upon

convection and radiant exchange in the interstices (small voids between solids) .

A.2 Metallic Solids

In metallic solids, k is dependent upon (1) transport by lattice waves (vibration) and (2)

free electron drift.

Phonon Contribution -- here the lattice wave mechanism is the same as in nonmetallic

solids.

Electron drift contribution -- this is independent of the lattice wave mechanism and

parallel network theory applies, hence

(3)

where

kp = thermal conductivity contribution due to lattice waves ( phonons )

ke = thermal conductivity contribution due to electron drift

Relative magnitudes -

Pure metals: ke - 30 kp

(pure metals have k 1 to 2 orders of magnitude larger than nonmetallic solids ) .

alloys (disordered molecular structure):

Here, the mean free paths of phonons and electrons are about the same length. Thus,

kp and ke are of the same order of magnitude.

Ways to measure ke and kp:

(a) In a strong magnetic field, ke = 0 and the measured k = kp

(b) Use of Wiedemann-Franz law -- in a metallic solid,

ke = Lo T (4)

where Lo = Lorentz Constant

- electrical conductivity

T = absolute temperature

Substitute (4) into (3) to obtain the phonon contribution,

Kp = K - Lo T (5)

A.3 Alloys

Alloys constitute special cases of metallic solids. Types of alloys are:

(a) continuous solid solutions

(b) eutectics - treat by mass-ratio mixture law

(c) limited solid solutions

(d) intermetallic compounds

4.4 Semiconductors

Semiconductors have very few valence electrons. Consequently, thermal energy transport

is by phonons with slight supplementation by electron transport.

at low temperature -- kp >> ke

at high temperature -- kp ≅ ke or ke > kp

A 5 Phase Transitions in Solids

Crystalline, polymorphous solids may exist in more than one structural (molecular lattice)

form. The transition from one structural form to another result from rearrangement of the atoms

in the molecular lattice due to change of state parameters, i.e., pressure and temperature.

Example: Sulfur - rhombic form below 368.5 K

monoclinic form above 368.5 K

The change from the rhombic to the monoclinic form results in a sharp decrease in k. Many other

solids such as petroleum based waxes, paraffins, etc., exhibit phase transition prior to melting.

B. Liquids

The molecular structure in the liquid state is less ordered than that in the solid state; this

is a result of the molecular bonding being weaker in the liquid than in the solid state. Also, liquid

motion renders thermal conductivity measurements more difficult than those for solids due to

convective currents increasing the thermal energy transport.

Melting of a solid results in a reduction of thermal conductivity due to the degeneration

of molecular order; transport by lattice waves is thereby impeded. An abrupt change in k usually

occurs just as the solid melts -- this has been confirmed for water, sulfur and many metals. An

exception to this is bismuth which has a higher speed of sound in the liquid than in the sol id

phase.

B.1 Ordinary Liquids

Correlation of Liquid Thermal Conductivirty With Other Properties

The difficulties in measuring k of liquids has resulted in many attempts to predict k from

values of more easily measured properties. These center on:

(1) acoustic velocity

(2) properties other than acoustic velocity.

The most successful has been the former which, based upon kinetic theory, has resulted in

where: γ = ratio of specific heats, cp/cv, given by 1 + 2/f where f is

the number of molecular degrees of freedom

ρ = mass density

M = molecular weight

R = universal gas constant

Cliq = speed of sound in the liquid

No = number of molecules per mole

The most difficult term in this equation to obtain is cliq which is related to (l) the acoustic velocity

in the gaseous state of this liquid and (2) the molecular volume.

Equation (6) represents the best that can be done using kinetic Theory. it holds

reasonably we1l for many liquids at 1 atmosphere pressure and over a range of temperature.

However, it overestimates kliq at high pressures badly.

B Liquid Metals

In liquid metals, free electron drift is the major mechanism for thermal energy transport.

Compared with other liquids, this free electron contribution results in very high conductivity -- l0

to .l000 times that for ordinary liquids.

C Gases

The transition to the gas phase results in very large distances between molecules; as a

result the volume available per molecule at standard pressure and temperature is on the order of

1000 times the volume of the molecule. Free movement or randomness of motion results. This

randomness allows use of statistical mechanics and kinetic theory. Kinetic theory leads to a

simple expression for k,

ADVANCED HEAT TRANSFER

MCEG 5223

Unit 5. Variation of Properties With Temperature and Pressure

0biectives - The objectives of this unit are (1) to depict the general effects of temperature and

pressure upon the thermal conductivities of various types of materials in solid,

liquid and gaseous states,and (2) to present a technique of modifying thermal

conductivity data of gases to account for very high pressures.

THE VARIATION OF THERMAL CONDUCTIVITY WITH TEMPTRATURE

The following is a highly simplified summary of temperature effect upon thermal

conductivity.

A. Thermal Conductivity of Solids

A.1 Crystalline Materials

The thermal conductivity of the solid phase of a metal of fixed composition

is primarily dependent only upon temperature.

General rule l: k for a pure metal usually decreases with increasing temperature; alloying

elements tend to reverse this trend.

Example

Similar results hold for copper, magnesium, nickel and aluminum with various alloying agents.

(Notice that pure aluminum. does not exhibit simple or monotonical1y decreasing k with

temperature). As stated in unit 1, the thermal conductivity of a metal can be well represented

over a wide temperature range by

where θ = t - to, and ko is the thermal conductivity at to. For many engineering applications the

range of temperature change is relatively small, say not more than a few hundred degrees, and

equation (13) can be replaced by the linear expression

4.2 Amorphous Materials

Some of the amorphous materials are homogenous and some are not. The thermal

conductivity of a non-homogeneous material is usually markedly dependent upon the apparent

bulk density,

The total volume includes the void volume, such as that caused by air pockets within the overall

boundaries of the material.

General rule 2: For a non homogeneous amorphous material, k usually increases with

increasing bulk density. In addition to the variation with density, k for these materials exhibit

strong temperature dependence.

General rule 3: For an amorphous material, homogeneous or non homogeneous, k usually

increases with increasing temperature.

B. Thermal Conductivity of Liquids

For liquids k is seen to depend strongly upon temperature. These values are relatively

insensitive to pressure.

General rule 4: Thermal conductivities of most liquids decrease with increasing

temperature.

There are numerous exceptions to this, a notable exception to general rule 4 is water

which exhibits increasing k up to about160oC and decreasing k there-after. Water has the highest

thermal conductivity of all common liquids (except liquid metals).

C. Thermal Conductivity of Gases

For high pressure (i.e., on the order of the critical pressure or greater), the effect of

pressure is significant. Modifications for high pressure will be discussed in the following section.

Two of the most important gases for engineering applications are air and steam -- no distinction

is made in our study between a gas and a vapor.

General rule 5. Thermal conductivities of gases increase with increasing temperature.

THE VARIATION OF THERMAL CONDUCTIVITY WITH PRESSURE

A. Solids

A.l Metallic solids - Thermal conductivity decreases linearly with increasing

pressure. The effect is very small -- negligibly so for most engineering applications. (Pressure

distorts crystalline lattice and thereby

induces added resistance to phonon and electron transport).

A.2 Amorphous Solids - Thermal conductivity increases with increasing pressure due

to increased molecular contact. Again, the effect is usually negligib1e.

B. Liquids

Thermal conductivity increases with increasing pressure, again due to increased molecular

contact. Here too, the effect is usually small. Reported experimental values indicate an increase

in k by a factor of 2 over the range from 0 to 12000 kg/cm2 (0 to l.l6 x l04 atm). Engineering

applications usually involve pressures up to l00 atmospheres, not l0,000 atmospheres. At design

pressures exceeding l000 atmospheres, it is wise to investigate the effect of pressure upon the

thermal conductivity of a liquid.

C. Gases

At moderate pressures, k data are essentially independent of pressure. At very low

pressures where the medium is not a true continuum, the mean free path length is influenced by

container size. Thus, the conductivity becomes pressure dependent at very 1ow pressures.

At high pressure, say above the critical pressure, the viscosity and the specific heat at

constant volume both increase with pressure, thus the thermal conductivity increases with

increasing pressure. One way of estimating the effect of high pressure upon the thermal

conductivity of a gas is by use of the generalized conductivity chart. As a general rule, whenever

the compressibility factor Z in the equation of state

deviates significantly from unity, then the effect of pressure upon the viscosity and the thermal

conductivity of the gas will be significant.

It is always preferable to use actual thermal conductivity data at high pressures for a

specific gas, whenever such data are available. In the absence of specific gas data the use of the

generalized correlation chart is suitable for most gases, excluding steam and air. Its use with air

would result in significant over correction ; steam behaves erratically and this precludes

generalization.

ADVANCED HEAT TRANSFER

MCEG 5223

Unit 6. Analyses of Steady-State 0ne Dimensional Conduction in Slabs, Tubes, Rods and Pin-Type Fins

Obiective - The objectives of this unit are to (l) begin a study of analysis of steady-state

conduction problems by solution of the appropriate coordinate form of the

previously derived conduction equation, (2) to obtain solutions for the

temperature distribution in several simple geometrical configurations and

(3) to apply the results to obtain the heat flux for practical design problems.

CONDUCTION-EQUATION SOLUTIONS FOR STEADY CONDITIONS

A. Slab

The simplest geometrical configuration is the single-layer slab as shown in the following

figure. Further, the easiest boundary conditions for us to handle are steady, uniform temperatures

t1 and t2 on the two y-z faces of the slab.

We begin with the general conduction equation, assuming constant thermal conductivity. This is,

for the cartesian coordinate system,

For the problem of interest, the following assumptions/conditions hold:

(1) steady-state

(2) body very large in y- and z-directions

(3) no internal conversion/generation

and the conduction-equation reduces to

Notice that the temperature is a function of x on1y, hence the ordinary second derivative. To

effect

a solution of (l), we need 2 boundary conditions on t(x) -- one for each order of the highest order

derivative with respect to each independent variable (x in this case). These are:

The general solution to (1) is

Applying boundary condition (2),

and (4) becomes

Applying boundary condition (3)

or,

and (5) becomes

Rearranging

which is the expression for the temperature at any position in the slab. (Question why doesnt this

expression contain any thermal properties?)

To obtain the heat flux through the slab, we apply Fouriers Law

Differentiating equation (7),

and substituting (9) into (8) yields

or,

Notice the analogy between equation (10.a) and Ohms law:

Q - I (current)

t - V (potential voltage)

l/kA - R (resistance)

From this, it is understandable that l/kA is often termed the thermal resistance

Considering the case of the composite slab as shown in the adjacent figure, it is clear that

we can apply equation (l) to each of the two slabs separately. The

boundary conditions on slab (l) are

The solution for the temperature in the

first slab is clearly

and

In like fashion, the boundary conditions for slab (2) are

and we readily obtain

From equations (13) and (15)

which add to yield (since Q is the same for both layers)

or,

Clear1y, this can be extended to any number of material layers.

In like fashion, when the heat flux at the outer faces is convective, Newton's law of

cooling gives

at the left-hand and right-hand convective surfaces, respectively. In this equation, t1 and t5 are the

fluid temperatures at the sides of the body, but sufficiently far away from the solid wall to be

thermally undisturbed.

Solving (19) for (t1–t2), for example,

and clearly 1/hA, is also a thermal resistance. Using equations (l6), (17) and (20),

or,

where t1 is the undisturbed fluid temperature on the left side and t4 is the right-hand side surface

temperature. Extensions to additional layers and convective boundaries on both sides is obvious.

A generalized form of (21) is

where Rth is a thermal resistance, conductive or convective

B. Tube

For the tube (hollow cylinder), the cylindrical coordinate system is more convenient. The

general conduction equation in cylindrical coordinates is:

Under assumptions/conditions

(l) no azimuthal (-direction) temperature variation

(2) no axial (z-direction) temperature variation

(3) no internal energy generation or conversion

(4) steady-state

the conduction equation reduces to

Again, as in the case of the slab, we need two boundary conditions to completely specify the

problem. These are:

Note that equation (24) is a second order, ordinary differential equation with non-

constant coefficients. Hence, the solution is slightly more involved than the usual second order

equation with constant coefficients. In this case, we note that

Thus, we require the solution of

Subject to the boundary conditions of equations (25) and (26). Performing successive indefinite

integrations,

Applying the two boundary conditions to evaluate C1 and C2 in a manner similar to what we did

for the slab (home work problem)

To determine the heat flux we apply Fourier's law

where A is the area (a function of r) perpendicular to r. Thus, A = 2rL and from equation (30),

and

Substituting into (31),

Thus, we see that the thermal resistance of a single cylindrical layer is

For a composite on multi-layer tube, such as that which results from adding one or more

layers of insulation to a pipe, the extension parallels that for the composite slab (or multi-layered

plane wall).

For the two-layer cylinder with inside and outside surface temperatures t i and to

prescribed,

Also, if the inner, the outer, or both surfaces are exposed to a fluid convective boundary

condition, the thermal resistance of each of these is

which can be easily shown by application of Newton's law of cooling. Here, of course, the r and

the h must be for the particular surface under consideration.

Example: Suppose the inner surface of a two-layered (insulated) tube is specified to be t1. This is

frequently possible to determine when the inner fluid is liquid--often the liquid temperature is

essentially the same as the inner pipe surface. The outer surface is exposed to ambient air at t f.

The

pipe has conductivity ka while that of the insulation kb.

The heat flux is then

The extension of this approach to any number of layers and convective resistance

inside, outside, or on both of these surfaces is obvious.

CRITICAL THICKNESS OF INSULATION

Read the critical thickness section of the text. In this application, it is assumed that the tube wall

outer surface temperature is known (for thin metal tubes, this is often nearly the same as the

inner tube surface temperature. Then, the heat flux through the insulation (radii r i and ro) and to a

surrounding convective fluid (surface coefficient ho) is

where k is the thermal conductivity of the insulating material, ti is the inner surface insulating

material temperature and tf is the surrounding fluid temperature.

Now by examination of the two terms in the denominator of (38a), we see that an

increase in ro increases the first term but it decreases the second. The question is "what value of

outer radius ro maximizes the heat transfer rate?" To answer this, we note that the numerator is a

constant for a given application. Thus we need to determine the maximum value of the

denominator. Using the usual condition for an extremal, we differentiate the denominator and

then equate the derivative with respect to the independent variable of concern, ro, to zero. Thus,

zero. Thus,

Hence,

EXTENDED SURFACE HEAT TRANSFER