Chapter 1
The Heat Equation

1  Some Examples of Conduction Heat Transfer Systems

Here we provide a few selected examples of practical situations involving conduction heat transfer processes. Examples are selected somewhat arbitrarily and partly reflect my own experience.
Perishable foodstuffs can be preserved by freezing. The food is frozen by refrigeration at sufficiently low temperature. Since the freezing rate affects the desirable characteristics of the food, food engineers are interested in controlling the freezing process in order to obtain optimal food characteristics such as flavor and nutritional value.
Thermal energy is internally generated in microelectronic devices by the electric current passing through them. The resulting heat must be dissipated to prevent malfunction of the device. Microelectronic engineers want to determine conduction heat transfer rates inside microelectronic packages in order to design them reliably.
Pressurized water nuclear reactors produce energy by radioactive disintegration of ceramic fissile material contained inside fuel rods. The generated energy is conducted through the cladding of the fuel pin and transferred onto a primary water circuit for subsequent transfer onto a steam circuit. Nuclear engineers are interested in controlling the rate of conduction heat transfer through the cladding in order to prevent its potentially catastrophic failure.
Domestic heating requirements raise steeply during the winter in northern latitudes. Ideally, we would like to maintain a comfortable warm interior at the lowest possible price. Construction and material engineers are interested in using construction materials that produce the lowest rates the conduction heat transfer.
A common processing method using in the metallurgical industry is continuous casting. Steel, copper and aluminum are routinely produced using this technology. The process is used to convert refined liquid metal into solidified ingot. Liquid metal is poured into a chilled reciprocating mold on one end and the (partially solidified) ingot is extracted on the other end. For the metal to freeze heat must be conducted through the solidified shell in contact with the mold and though the mold wall. Continuous casting engineers are interested in controlling the rate of solidification in order to avoid metallurgical defects or catastrophic breakouts.
In the manufacture of jet engines, turbine disks are mill annealed by heating and maintaing them at a selected temperature and subsequent air cooling. The treatment leads to optimal metallurgical structure and properties in the finished component. Heat treating engineers are interested in controlling the rates of heating and cooling during heat treatment in order to optimize the resulting component properties.
Metal components can be joined or heat treated by using high density energy sources. Appropriate sources are laser or electron beams. As the beam impinges on the joint it produces localized melting of the material. Heat dissipation by conduction into the surrounding material produces rapid solidification of the weld. Welding engineers are interested in controlling the size and shape of the molten pool in order to obtain optimal properties in the finished joint.
Last but not least is the following theory which is currently being considered by forensic engineers investigating the catastrophic collapse of the WTC towers following last years terrorist attacks in NYC. It is suggested that the insulating material layers covering the underlying steel structure were suboptimal and were unable to impede the conduction of the heat generated by the burning jet fuel. As a result, the steel beams reached temperatures in the creeping regime, became unable to bear the weight of the structure and catastrophic collapse ensued.
The list above could easily be made much larger.
Another good reason to undertake the study of conduction heat transfer is that the fundamental notion involved (energy conservation) has important analogues in other physical systems of considerable interest. Specifically, the principle of mass conservation is used to formulate and solve problems in diffusional mass transfer and the principles of conservation of mass and momentum are the foundation of fluid mechanics. Much of the intuition and insight acquired from a study of conduction heat transfer can be utilized to advantage when studying diffusion and fluid flow.
Furthermore, the mathematics of heat conduction has also important applications in the study of Brownian motion, probability theory and in the theory of financial investments.

2  Thermal Energy Transport in Solids: Heat Flux

Temperature is a measure of the degree of hotness at a point in a material and is a macroscopic manifestation of the potential and kinetic energies of the constituent molecules. When temperature differences exist between two distinct points in a medium the result is the transport of thermal energy downhill the temperature gradient. The rate of energy transport has the units of energy per unit time (J/s = W). The amount of thermal energy crossing a unit area per unit time while flowing in the direction of decreasing temperature is the heat flux vector
q = - k T .
Here, q is the heat flux (W/m2), T is the temperature (K), is the gradient operator and k is the thermal conductivity of the medium (W/m K). The term T is thus the temperature gradient vector.
The thermal conductivity is the rate of thermal energy transfer per unit area and per unit temperature gradient. Thermal energy is transported within a solid by the electrons and the phonons (lattice vibrations) inside the material. The transport of energy is hindered by the presence of imperfections or by any kind of scattering sites.
Thermal conductivities of solids at room temperature vary from 0.1 W/m K for good insulators (e.g. asbestos) up to 400 W/m K for good conductors (e.g. silver). The conductivity changes mildly with temperature except at very low temperatures where it can acquire very large values. For instance, pure copper at 10 K has a conductivity of about 20,000 W/m K.
Note that if there is macroscopic transport of matter (e.g. fluid flow) inside the body, the mass flow makes an additional contribution to the transport of energy (convective heat transfer). This contribution is disregarded when studying conduction heat transfer.

3  The Heat Equation

The heat equation is a differential statement of thermal energy balance. The thermal energy content E (J) of a material body can change only if (i) energy is removed/added through its bounding surface or (ii) energy is generated or absorbed within the body.
Consider a small volume element V inside a material undergoing heat conduction processes. The rate of change of thermal energy per unit volume [H\dot] (in W/m3) at a point inside the volume is

H
 
= H

t
where H is the enthalpy or heat content per unit volume of material in J/m3. From thermodynamics, the following relationship expresses the accumulated enthalpy in a material resulting from increasing its temperature from T1 to T2,
H =
T2

T1 
rCp dT
where r and Cp are, respectively the density (in kg/m3) and specific heat (in J/kg K) of the material. More complex H-T relationships are also possible.
Introducing the above H-T relationship, the rate of change of thermal energy then becomes

H
 
= rCp T

t
Now, the net outflow of thermal energy through the bounding surface A of volume V is given by the divergence theorem as
-


A 
q ·n dA = -


V 
·q dV
Finally, the rate of internal heat generation/absorption at a point in the body (W/m3) is assumed given as g(r,t).
The thermal energy balance equation for the volume V is



V 
H

t
dV =


V 
[- ·q + g(r,t) ] dV
However, since the integrals are equal the arguments are also equal and the most general form of the differential thermal energy balance equation is
H

t
= - ·q + g(r,t)
and is called the heat equation.
More restrictive forms of the heat equation are
rCp T

t
= ·k T + g(r,t)
for a simple isotropic material and
T

t
= a2 T
for constant thermal properties and no heat generation. Here
a = k

rCp
is the thermal diffusivity of the material (in m2/s) and measures the speed of penetration into the body of an applied thermal load at its surface. Values of a range from 0.1 ×10-6 for cork to 300 ×10-6 for potassium.
The following are commonly used forms of the heat equation
rCp T

t
=

x
(k T

x
) +

y
(k T

y
) +

z
(k T

z
) + g
in rectangular Cartesian coordinates (x,y,z),
rCp T

t
= 1

r

r
(k r T

r
) + 1

r2

f
(k T

f
) +

z
(k T

z
) + g
in cylindrical coordinates (r,f,z),
rCp T

t
= 1

r2

r
(k r2 T

r
) + 1

r2 sinq

q
(k sinq T

q
) + 1

r2 sin2 q

f
(k T

f
) + g
in spherical coordinates (r,f,q), and
rCp T

t
= 1

a
[

u1
(k a

a12
T

u1
) +

u2
(k a

a22
T

u2
) +

u3
(k a

a32
T

u3
)] + g
in general orthogonal curvilinear coordinates (u1,u2,u3), where a1,a2,a3 are the scale factors and a = a1a2a3.

Exercise Consider the problem of one-dimensional transient conduction heat transfer inside a rod of cross sectional area A where energy is conducted along the direction of the axis of the rod (x-axis). At some arbitrary interior point of the rod x introduce a control volume of size Dx. Perform an energy balance over the control volume (i.e. accumulation = input - output) and derive the one-dimensional heat equation.

4  Boundary Conditions

Boundary conditions are the representation of the thermal energy balance at the bounding surface of the material. They measure heat exchange interactions between the material and its surroundings in W/m2. Common heat exchange mechanisms are convection
qconv = h (T - T)
where h is the heat transfer coefficient (W/m2 K) and T is the bulk temperature of the surrounding environment; radiation
qrad = es(T4 - Tr4)
where e, s and Tr are, respectively the emissivity of the material (-), the Stefan-Boltzmann constant ( = 5.729 ×10-8 W/m2 K4) and the temperature of the far-field reflecting surfaces; and heat supply from an external energy source, qsup.
The energy balance equation at the surface of the material is then
qn + qsup = qconv + qrad
Boundary conditions can sometimes be expressed in linear form. For example, if the temperature at the surface is specified one obtains boundary conditions of the first kind. Alternatively, if the heat flux at the surface is given, boundary conditions of the second kind result. Next, if a (linear) relationship between surface heat flux and temperature is specified, for example,
- k T

n
= h (T - T)
we talk about boundary conditions of the third kind. Finally, at the surface of contact of two solids an interface boundary condition can be stated in terms of an interfacial thermal conductance.

5  Moving Solids

Moving material carries with it its thermal energy by mass flow. If the material is moving along the x direction with velocity ux, the amount of thermal energy transported per unit area per unit time is
qmass = rCp T ux
and the total heat flux (conduction plus mass transport) is
qx = - k T

x
+ rCp T ux
For a material containing a (known) velocity field u = (ux,uy,uz) the energy balance equation becomes
rCp [ T

t
+ u ·T ] = k 2 T + g.
This is sometimes called the convection-diffusion equation.

6  Anisotropic Media

In anisotropic materials thermal energy flows at different rates along different directions. This is taken into account by assigning a second order tensor character to the thermal conductivity. The x-component of the heat flux vector qx becomes
qx = - ( k11 T

x
+ k12 T

y
+k13 T

z
)
and similarly for the other two directions. For orthotropic materials heat still travels at different rates along different directions but the heat flux along any direction is driven only by the temperature gradient along that direction, i.e. kij = 0 when i j. The heat equation then becomes
rCp T

t
=

x
(kx T

x
) +

y
(ky T

y
) +

z
(kz T

z
) + g.

7  Conduction in Composite Media

Composites are formed by bringing together into close contact distinct materials. The material properties of the assembly change discontinuously at the contact surface. Also, the mechanical conditions of the contact at the interface affect the rate at which energy flows across it. In general, the analysis of heat transfer at an interphase interface characterized by its normal n and separating the two materials 1 and 2 requires specification of two conditions: a) the heat flux balance
(k1 T1) ·n = (k2 T2) ·n
and the relationship between the temperatures on both sides of the interface T1 and T2 usually in terms of the contact conductance h, i.e.
(- k1 T1) ·n = h (T1 - T2)
The so called perfect thermal contact condition simply states that at the interface h and T1 = T2. The above equations must be added to the stated initial and boundary conditions of the particular problem at hand.

8  Fundamental Solutions of Heat Conduction Problems

The heat equation is used to produce a mathematical representation of a physical system undergoing heat conduction processes. The mathematical formulation requires statement of the differential thermal energy balance equation as well as the statement of appropriate boundary and initial conditions. The solution of the problem consists of determining the function T(r,t) which satisfies the differential equation subject to the stated boundary and initial conditions. This is the direct heat conduction problem. Direct problems are usually well posed (i.e. they have unique solutions stable under small perturbations in the input data.) Solutions to direct problems can be found by analytical techniques for the simplest cases (linear problems) or using numerical methods in more complex situations.
The one-dimensional heat equation for a material with constant thermal diffusivity a = 1 is
T

t
= 2 T

x2
The following are three well known and useful simple solutions of the above. They will be used a great deal subsequently.
First is the source solution:
u1(x,t) = exp(-x2/4t)




4 pt
Next is the derived source solution:
u2(x,t) = xexp(-x2/4t)

t


4 pt
Finally, there is the complementary error function solution:
u3(x,t) = erfc( x




4t
)
Inverse heat conduction problems can also be formulated. In this case the objective is the determination of the boundary conditions required to produce a given temperature distribution inside the material. Inverse problems are ill-posed (i.e. they often have multiple solutions.) Inverse problems are invariably solved using numerical techniques.

Exercise Show that the three solutions given above are indeed solutions of the heat equation.

9  Lumped Parameter Formulation

A lumped parameter formulation is an approximation which facilitates the solution of heat transfer problems. The key assumption is the neglect of temperature gradients inside the body of volume V and surface area A so that its temperature is only a function of time. However, this assumption clearly amounts to the neglect of the heat conduction processes inside the material and should be used with caution.
This approximation is valid if
Bi = h L

k
<< 1
where L = V/A is a characteristic length and Bi, the Biot number is a measure of the ratio of internal and external resistances to heat flow.
If the only mechanism for energy exchange with the surroundings is convection through the bounding surface, the differential thermal energy balance equation (in W) becomes
rCp V dT

dt
= - h A (T - T)
this is a first order ordinary differential equation which is easily solved.

Exercise A 150 micrometer diameter steel sphere (r = 7,700 kg/m3, Cp = 460 J/kg K, k = 25 W/m K) is quenched from a temperature of 1200 K using an air jet (h = 10,000 W/m2 K) at room temperature (T = 300 K). Calculate the value of Bi for this system and, if possible, use the lumped parameter model to estimate the time it takes for the temperature of the sphere to reach 325 K.

10  Units and Conversion Factors




Quantity SI Units English Units
E J BTU
E/t W BTU/hr
H J/m3 BTU/ft3
H/t W/m3 BTU/hr ft3
q W/m2 BTU/hr ft2
k W/m K BTU/hr ft oF
r kg/m3 lb/ft3
Cp J/kg K BTU/lb oF
a m2/s ft2/hr
h W/m2 K BTU/hr ft2o F
s 5.729 ×10-8 W/m2 K4 0.173 ×10-8 BTU/hr ft2o R4



English Units SI Units
1 BTU 1,055 J
1 BTU/hr 0.29307 W
1 BTU/ft3 37,260 J/m3
1 BTU/hr ft3 10.3497 W/m3
1 BTU/hr ft2 3.154591 W/m2
1 BTU/hr ft oF 1.7307 W/m K
1 lb/ft3 16.02 kg/m3
1 BTU/lb oF 4,186.8 J/kg K
1 ft2/hr 2.5807 ×10-5 m2/s
1 BTU/hr ft2oF 5.6782 W/m2 K



Exercise Verify all the units and unit conversions given in the tables above.



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On 23 Sep 2009, 17:12.

Updated: 2009-09-23, 17:12