Cambridge University Engineering Department
Engineering Tripos Part IIA
Module 3A6: Heat and Mass Transfer
Some Notes on Conduction
Dr. Christos Markides
cnm24@cam.ac.uk
Nomenclature:
T
Temperature [K]
Heat flow rate [W = kg.m2/s3]
Heat flow rate per unit area, or heat flux [W/m2 = kg/s3]
Heat flow rate per unit volume [W/m3 = kg/m.s3]
ρ
c
k
α
h
l
A
X
L
S
P
Density [kg/m3]
Specific heat capacity [J/kg.K]
Thermal conductivity [W/m.K]
Thermal diffusivity [m2/s]
Convective heat transfer coefficient [W/m2.K]
Characteristic length for heat transfer, relevant for Nusselt and Biot numbers [m]
Cross-sectional area available to conductive heat flow [m2]
Material thickness in planar geometries [m]
Cylinder length in cylindrical geometries [m]
Surface area available to convective heat flow [m2]
Circumferential length relevant to convective heat transfer [m]
Italics: Optional, background and explanatory text.
Bold: Important equations.
The Einstein Summation Convention is used throughout, with
.
“When I felt that the moment was right,
I went, and I did it___
And even when I found resistance, I remembered that:::
Sometimes we must go alone; and wait for the world to follow.
Now you must follow your own paths..__
When you think that the moment is right, even if you are unsure, or afraid_ seize it.
Go, and do it, and good luck.
I will be there if you need me_..”
CNM
Summary:
Usually, when both conduction and convection are possible, convection will play a stronger role than conduction.
The relative importance of the two heat transfer modes is described by the Nusselt number,
. Conduction
will only dominate at low Nu, which is far from being a realistic scenario in the majority of engineering applications
you will face. The most straightforward way of approaching conductive heat transfer in isolation is to consider solid
bodies, inside which there can be no fluid motion, and hence no convective heat transfer.
Assuming a constant (uniform and steady) thermal conductivity, you are solving a single equation,
that applies within (throughout the volume of) a solid material. Convective heat transfer, i.e. the
term
where T∞ is the bulk temperature of the fluid a long distance away from the solid surface, can only appear as a
boundary condition in such problems, with one exception that is mentioned below. Note that conduction, i.e. the
term that includes the thermal conductivity, is a process of diffusion of temperature (second order spatial
differential), in the same way that viscosity results in diffusion of momentum or velocity.
If the geometry of the solid body, and of the boundary conditions, exhibit symmetry about some axis or dimension,
in some coordinate system, the above equation reduces to one of,
In each case, you will never be asked to consider the full corresponding equation. Instead, you will consider
simplified cases, where one or more terms are zero.
Firstly, for steady problems the left hand side term is zero, and the temperature is only a function of one spatial
variable. The solution is not difficult; these are second order ordinary differential equations. You can integrate twice,
even if
is a known, but simple function of x, or r. You will obtain two constants of integration, and will require
two boundary conditions in order to fix them. In these problems, convection can only appear as a boundary
condition, that is, where the solid is in contact with the fluid the temperatures must be equal and the heat flow rates
must also match.
Secondly, if the unsteady term is re-introduced, you can be given a situation with no volumetric heating, but with
spatial temperature variations. The resulting solution involves the error function, is rather more complicated to
obtain, and you would probably either be supplied with it, or be guided through its derivation.
Finally, if the unsteady term is non-zero, but volumetric heating is allowed, you will almost invariably be told that the
lumped heat capacity model, or approximation, applies. This effectively means that you can ignore the final term on
the right hand side. To check that this assumption is adequate, you must verify that the Biot number,
, is
smaller than unity. The approximation assumes that the temperature throughout the solid is the same, and hence
that the temperature (of the whole solid) is only a function of time. The equations above simplify to first order
ordinary differential equations in time. The process of solving these is explained in detail in the Mathematics Data
Book, in the Differential Equations Section. There is now no distinction between the volume inside the solid and its
boundary. Hence, and exceptionally, convection can appear not only as a boundary condition, but as part of
.
The full transport equation for the conservation of energy/enthalpy is:
From left to right the terms represent: unsteadiness, convection, volumetric heating (such as due to external, e.g.
radiative, or internal, e.g. exothermic reaction, heating), work done with respect to gravity, conduction, and heating
due to shear/friction/viscosity. The derivation of this equation is beyond the scope of this document.
Note that, by using the principle of conservation of mass (Equation 4a, below), for any scalar quantity φ,
where the Material Derivative is defined as
Then, and given that
.
, we can re-write Equation 1 either in terms of internal energy or enthalpy:
This last one looks similar to, and indeed results from the same “master” (Boltzmann) conservation equation as the
familiar transport equations for the conservation of mass and momentum:
On the left hand side we have unsteadiness and advection, and on the right hand side buoyancy/gravity, pressure
forces and shear/friction/viscosity. Note that, the left hand sides of all conservation equations contain unsteady and
advective (or convective in the energy equation) terms, while the right hand sides contain surface terms, such as
pressure, viscous stresses and heat flux, and external terms, such as buoyancy/gravity and volumetric heating.
We can make one further simplification. If we multiply Equation 4b by ui and subtract this from Equation 3b we get:
We can bring temperature into the equation, in place of the internal energy, by use of the relations
,
and Fourier’s Law of Heat Conduction,
,
where
is a fluid property called the coefficient of volumetric thermal expansion.
For ideal gases,
, so that:
Finally, the term involving the material derivative of pressure is usually completely ignored, unless the physical
process involves significant compressions and expansions, such as for example in internal combustion engines. In this
course it can be completely neglected, resulting in,
that is even more similar to Equation 4b, which was the governing equation for the conservation of momentum.
Purely conductive heat transfer within a medium is best considered when that medium is a solid. This is because fluid
motion, and hence convective heat transfer, cannot exist within solids. Usually, when both are present, convection
will play a stronger role than conduction. A non-dimensional number that describes this relative importance of the
two heat transfer modes is the Nusselt number:
It is important to point out that in the Nusselt number, k is the thermal conductivity of the fluid, as opposed to the
Biot number, where k is the thermal conductivity of an adjacent solid. Also, the Biot number is mainly a ratio of
temperature differences, or spatial temperature gradients, rather than of heat flow rates. Conduction will only
dominate at low Nu, when k is high relative to h.
In solids, all velocities, the pressure and density are constants, hence Equations 3a, 3b and 5 become,
where the term on the left hand side is the unsteadiness, and the first and second terms on the right hand side are
the volumetric heat generation and conductive heat transfer respectively.
The unsteady term can be written in terms of the specific heat capacity of the solid and we may assume that the
heat conduction term can be approximated by Fourier’s Law of Heat Conduction,
:
Equation 10 is the full governing equation for physical situations of heat transfer in which convective heat transfer
can be completely ignored; that is, in solid materials whence there is no possibility of heat transfer by fluid flow.
Heat conduction is taken into account by the final term on the right hand side. In this course, you will deal with solid
materials that have constant (in both space and time) thermal conductivity, such that:
Geometrical symmetry of the solid and symmetry in the boundary conditions allow us to consider only a single
spatial dimension. The full, unsteady heat equation in one (Cartesian) dimension for constant thermal conductivity
reduces to:
Furthermore, in this course you will only deal with one-dimensional steady conductive heat transfer:
1. With no volumetric heat generation and constant thermal conductivity:
2. With constant (i.e. uniform and steady) volumetric heat generation and constant thermal conductivity:
Note that there is now only one independent variable, x, so that the partial derivative becomes an ordinary one.
In these steady problems, as long as there are temperature variations in the solid, and the conduction term (last
term on the right hand side of Equation 11) is non-zero,
is a generation (or loss) throughout the volume of the
solid and not a boundary condition. Even in the case of finite heat generation, you will only encounter situations of
constant, that is both uniform (in space) and steady (in time),
. The heat generation term can then be, but not
necessarily so, from radiative heat transfer or energy release from exothermic chemical reactions.
The unsteady problems you might face will not involve any heat generation if the conduction term is non-zero.
Conversely, you will only be presented with a situation of unsteady heat transfer involving finite, but always
constant, heat generation, if you can ignore all spatial temperature variations in the solid, such that the conduction
term is zero. This is called the ‘lumped heat capacitance’ model. In these unsteady, lumped heat capacitance
problems,
appears as a generation (or loss) term, either within the volume of the solid as before, or as a
boundary condition, in which case it can be due to convective heat transfer from fluid motion at the wall of the solid.
Hence, in this course will only deal with unsteady heat transfer:
1. In one-dimension with no heat generation and constant thermal conductivity:
2. In lumped heat capacity situations without spatial temperature variations in the solid, and with constant (i.e.
uniform and steady) heat generation:
Summarising, for steady conductive heat transfer scenarios in one dimension, with constant thermal conductivity,
and without volumetric heating, from Equation 12, and noting that there is now only one independent variable, x, so
that the partial derivative becomes an ordinary one,
and with volumetric heating being present in addition:
For unsteady conduction in one dimension, with constant thermal conductivity, and without volumetric heating,
from Equation 12:
Finally, for unsteady heat transfer, with heat generation, in lumped heat capacity solids without spatial temperature
variations in the solid:
However, note that the final case is not really a conduction problem, and as such will not be dealt with. It suffices to
mention the Biot number; a non-dimensional number that relates temperature gradients, or differences, in the solid
(numerator), to those in the adjacent fluid (denominator), which must be small for the lumped model approximation
to be valid:
These equations are for a one-dimensional Cartesian co-ordinate system; we could have written equations
equivalent to Equations 15 to 17 in one-dimensional cylindrical or even spherical coordinates. To do this we would
have to replace
with one of (see Mathematics Data Book, Vector Calculus Section),
We consider first steady conduction in a solid of planar geometrical symmetry, without heat generation and with
constant thermal conductivity, i.e. Equation 15. We are after the spatial temperature distribution T(x). The starting
point for the planar solution is to integrate Equation 15 twice,
where it is clear that d/k is the temperature at x = 0, i.e. d = kT(x = 0).
This linear solution for the spatial temperature profile T(x) applies within a solid, planar slab with boundary
conditions exhibiting planar symmetry. In fact, the geometry and temperature boundary conditions can also be:
1. Cylindrically symmetric, with logarithmic temperature profiles.
2. Spherically symmetric, with hyperbolic temperature profiles (see Mathematics Data Book, Vector Calculus
Section).
In problems with cylindrical symmetry, Equation 15 becomes,
whereas for geometries with spherical symmetry, Equation 15 becomes,
Integration of Equation 20 twice gives:
Integration of Equation 21 twice gives:
The constants in these equations are fixed by the boundary conditions.
The heat flow rate due to this temperature distribution is given by Fourier’s Law in one-dimension, which is either
in planar coordinates, or
in cylindrical and spherical coordinates (see
Mathematics Data Book, Vector Calculus Section). The heat flux is either
, or
respectively.
Hence, in the planar case and from Equation 19a, or equivalently after differentiating Equation 19b,
and we realise that the left hand side is equal to the (negative of the) heat flux
. It is evident that the heat flux is
the same at all x (constant). Now, because this geometry exhibits planar symmetry, with A being independent of x,
the heat flow rate is also the same at all x (constant).
We now return to Equation 19b for the planar case, and after substituting for c and d, and re-arranging,
or, in terms of the heat flow rate,
If then a boundary condition is given such that the temperature x = X is known,
Note that ΔT = T(x = X) – T(x = 0). Returning to the temperature distribution inside the solid, i.e. Equation 25, we can
replace with this temperature difference:
One can use Equation 25 when
is known, or Equation 28 when ΔT is known.
We now consider the cylindrical case, and Equation 22a:
The heat flux is given by,
that is not independent of r (constant). However, the heat flow rate through an area A = 2πrL is,
that is independent of r (constant), similar to the planar geometry.
We can substitute this back into Equation 22b for the temperature distribution:
We can use a boundary condition for the temperature T(r = R) at r = R to eliminate d’,
and hence,
A simple re-arrangement provides the heat flow rate,
and if a further boundary condition is given such that the temperature r = R’ is known:
Note that ΔT = T(x = R’) – T(x = R).
Returning to the temperature distribution inside the solid, i.e. Equation 33, we can replace
with this temperature
difference:
Finally, consider the spherical case, and Equation 23a:
The heat flux is given by,
that is again not independent of r (constant). However, the heat flow rate through an area A = 4πr2 is,
that is independent of r (constant), similar to the planar and cylindrical geometries.
We can substitute this back into Equation 23b for the temperature distribution:
We can use a boundary condition for the temperature T(r = R) at r = R to eliminate d’,
and hence,
A simple re-arrangement provides the heat flow rate,
and if a further boundary condition is given such that the temperature r = R’ is known:
Note that ΔT = T(x = R’) – T(x = R).
Returning to the temperature distribution inside the solid, i.e. Equation 41, we can replace
with this temperature
difference:
Equations 25, 33 and 41 give the temperature distribution within the solid for the various geometries, if the heat
flow rate is known. If instead the temperature difference is known, we can substitute ΔT with from Equations 28,
36 and 44 respectively.
Temperature,
or
Planar
Cylindrical
Spherical
In terms of
In terms of ΔT
Table 1. Temperature distribution for steady conduction, without volumetric heat generation and constant thermal
conductivity, in various geometries exhibiting one-dimensional symmetry.
Equations 27, 35 and 43 are convenient in that they all resemble Ohm’s Law,
, with the equivalent of
current being heat flow rate, the equivalent of potential/voltage difference being temperature difference and the
equivalent of resistance being:
Planar
Cylindrical
Spherical
Heat flow rate,
Conductive (Solid)
Thermal Resistance
Table 2. Heat flow rate and conductive resistance for one-dimensional conduction in various geometries.
As we have seen, these equations apply within a solid material, but the solutions, i.e. the spatial temperature
distributions and temporal evolution thereof, also depend on the boundary conditions. In unsteady problems, the
solutions also depend on the initial conditions. Occasionally, you will deal with problems where heat flows through
two, or more, solids with different properties, in thermal/physical contact with each other. At the boundaries of
each solid material, i.e. the contact surfaces, the temperatures must match, as must the heat flow rates. Finally, if
there is a contact thermal resistance Rc, the heat flow rates must still match, but the temperatures will be slightly
different. The contact resistance will always be supplied to you, and you can use that to obtain the temperature
difference between the solids. The heat flowing out of one solid must always be equal to the heat flowing into the
adjacent solid; otherwise heat would be accumulating in the interface and the problem would be unsteady.
Mathematically, at the interface x = X” or r = R” with finite contact thermal resistance,
and the heat flow rates can be obtained from Table 2.
It is evident from Table 2 that the thicker we make a solid the more resistance to conductive heat flow it offers.
Consider the cylindrical and spherical geometries with a fixed inner radius. If we want to decrease the heat flowing
into or out of the cylinder or the sphere, pure considerations of conduction would lead us towards making the wall
as thick as possible, such that the outer radius is as large as possible. What this does not take into account is that as
the outer radius increases, the area available for convective heat transfer increases, acting in opposition to our goal
of reducing heat flow rate.
Here we must realise that the total resistance to heat flow comes from a conductive resistance, in series with a
convective resistance. Convective heat transfer is given by,
where Rout is the outer radius of the cylinder or sphere and
is temperature in the fluid far away from the solid.
So, the thermal resistance due to convection between T(r = Rout) and
and S =
4πRout2
is
. Since S = 2πRoutL for the cylinder,
for the sphere, the convective resistance decreases as Rout increases.
Ultimately, we would like to the critical value of Rout that total resistance with respect to while Rin is kept constant:
Cylindrical
Spherical
Conductive (Solid)
Thermal Resistance
Convective (Fluid)
Thermal Resistance
Total
Thermal Resistance, RTOT
Table 3. Conductive, convective and total thermal resistances in cylindrical and spherical geometries.
Consider now steady conduction in a solid of planar geometrical symmetry, with heat generation and with constant
thermal conductivity, i.e. Equation 16. We are after the spatial temperature distribution T(x). The starting point for
the planar solution is to integrate Equation 16,
twice,
where it is clear that d/k is the temperature at x = 0, i.e. d = kT(x = 0).
This quadratic solution for the spatial temperature profile T(x) applies within a solid, planar slab with boundary
conditions exhibiting planar symmetry. In fact, the geometry and temperature boundary conditions can also be:
1. Cylindrically symmetric, with logarithmic temperature profiles.
2. Spherically symmetric, with hyperbolic temperature profiles (see Mathematics Data Book, Vector Calculus
Section).
In problems with cylindrical symmetry, Equation 16 becomes,
whereas for geometries with spherical symmetry, Equation 16 becomes,
Integration of Equation 49 twice gives:
Integration of Equation 50 twice gives:
It can be seen that these problems can be easily extended for non-uniform but still steady
only in space, as long as the form of
or
, that is varying in
is given, and the integration of
Equations 16, 47, 49, 50, 51a and 52a is still possible.
The constants in the above equations are fixed by the boundary conditions, as before.
Furthermore, the heat flow rate due to these temperature distributions is again given by Fourier’s Law in onedimension, which is either
in planar coordinates, or
in cylindrical and spherical
coordinates (see Mathematics Data Book, Vector Calculus Section). The heat flux is either
respectively.
, or
Hence, in the planar case and from Equation 47, or equivalently after differentiating Equation 48,
and we realise that the left hand side is equal to the (negative of the) heat flux
. It is evident that both the heat
flux and the heat flow rate are not independent of x. We can use Equations 48 and 53, along with boundary
conditions that will be supplied, to find the values of the constants c and d.
We now consider the cylindrical case, and Equation 51a:
The heat flux is given by,
that is not independent of r. The heat flow rate through an area A = 2πrL is,
that is also not independent of r. We can use Equations 51b and 55, along with boundary conditions that will be
supplied, to find the values of the constants c and d.
Finally, consider the spherical case, and Equation 52a:
The heat flux is given by,
that is again not independent of r (constant). The heat flow rate through an area A = 4πr2 is,
that is also not independent of r. We can use Equations 52b and 58, along with boundary conditions that will be
supplied, to find the values of the constants c and d.
The solution of the unsteady heat conduction equation with temperature variations in the solid is only relevant
when the Biot number is not small enough for one to employ the lumped heat capacity approximation. The
governing equation of temperature becomes,
where
is the thermal diffusivity. Note that, the final form of this equation can be compared directly to the
momentum equation,
The second order spatial differential term in Equation 17, which includes the thermal diffusivity, that is the same
term that includes the thermal conductivity, describes a process of diffusion of temperature, in the same way that
the viscous stress term, which includes kinematic viscosity and second order spatial differentials of velocity, results
in diffusion of momentum or velocity.
You will not be asked to solve Equation 17, although the solution, at least in the most simple of cases, is in fact much
simpler to obtain than you might think. It is contained in your lecture notes and it is good at least be able to follow it
and to remember its form, because a tricky question might be set that guides you through the solution. This could
involve something like, asking you to verify that,
is a solution of Equation 17, after which you might be asked what the heat flow rate, or heat flux, is. For this you will
be given the differential of the error function and you will have to differentiate Equation 59 with respect to x, and
use Fourier’s Law,
.
Finally, we will derive various equations for heat transfer from finned surfaces.
h δS [T(x) - T∞]
Q(x+δx)
Q(x)
T o , Qo
L
x
Figure 1. Problem set-up for heat transfer from fins.
We assume radial symmetry, even though the circumference, P, and cross-section, A, of the fin can have an arbitrary
shape. Still, the circumference and cross-section do not vary axially. We also assume temperature variations only in
the axial direction, that is that the fin is longer than it is thick.
The heat flow rate being conducted through the fin is,
at x, and,
at x+δx, where:
The heat lost by convection must be the difference between
and
:
Substituting the terms from Equations 60 and 61:
Using Equation 62b:
We can define a temperature difference,
, and a new constant,
:
Equation 66 is a simple second order ordinary differential equation. There is no particular integral, since there is no
term on the right hand side. The two roots of the auxiliary equation are
, giving a complimentary function,
which is the general solution. We return to using our original variable that was T rather than θ:
We can also obtain an equation for the heat flow rate/heat flux through the fin, simply by differentiating Equation 68
with respect to x, and substituting the result into Equation 60:
All that remains is to remove the constants by using the boundary conditions, whether these are expressed as
temperature boundary conditions, or heat flow rate/heat flux boundary conditions.
The first condition is an isothermal one at the root of the fin, x = 0, with
or, but not as likely, a constant heat input one, with
,
,
The second boundary condition will come from one of four different situations at the tip, x = L, of the fin. There are
four cases.
For example, if we specify an isothermal boundary condition, with a certain temperature
:
We now have two equations, Equations 70 and 71, and two unknowns, A and B, that we can evaluate before
substituting them back into Equation 68 for T(x) or Equation 69 for
.
Another situation is to be given a heat flow rate, or heat flux boundary condition. A typical situation is to have an
insulated/adiabatic tip:
Again, there are two equations, Equations 70 and 72, and two unknowns, A and B, that can be evaluated before
substituting them back into Equation 68 for T(x) or Equation 69 for
.
The other cases are: (i) convection at the tip, with
, and (ii) a very long fin such that
. The solutions are all in the lecture notes. Note the appearance of the non-dimensional group,
that is nothing other than the square root of the Biot number, with A/P a characteristic lengthscale for heat transfer
in the fin.
CNM
22 of February, 2008
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