Why heat and mass transfer?

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Heat transfer and mass transfer are kinetic processes that may occur and be studied separately or jointly. Studying them apart is simpler, but both processes are modeled by similar mathematical equations in the case of diffusion and convection (there is no mass-transfer similarity to heat radiation), and it is thus more efficient to consider them jointly. Besides, heat and mass transfer must be jointly considered in some cases like evaporative cooling and ablation.
Why heat and mass transfer ................................................................................................................................1
Fundamentals of heat transfer (what is it) ..........................................................................................................2
Thermodynamics of heat transfer .................................................................................................................. 3
Physical transport phenomena ....................................................................................................................... 5
Thermal conductivity ................................................................................................................................. 7
Heat equation ............................................................................................................................................... 10
Modelling space, time and equations........................................................................................................... 11
Case studies ................................................................................................................................................. 12
Nomenclature refresh................................................................................................................................... 12
Objectives of heat transfer (what for) ...............................................................................................................13
Relaxation time ............................................................................................................................................ 14
Conduction driven case (convection dominates) ..................................................................................... 15
Convection driven case (conduction dominates) ..................................................................................... 16
Heat flux ...................................................................................................................................................... 17
Temperature field......................................................................................................................................... 19
Dimensioning for thermal design ................................................................................................................ 20
Procedures (how it is done) ..............................................................................................................................21
Thermal design ............................................................................................................................................ 21
Thermal analysis .......................................................................................................................................... 21
Mathematical modelling .............................................................................................................................. 22
Modelling the geometry........................................................................................................................... 23
Modelling materials properties ................................................................................................................ 24
Modelling the heat equations ................................................................................................................... 25
Analysis of results........................................................................................................................................ 26
Modelling heat conduction ...............................................................................................................................27
Modelling mass diffusion .................................................................................................................................27
Modelling heat and mass convection ................................................................................................................27
Modelling thermal radiation .............................................................................................................................27
General equations of physico-chemical processes ...........................................................................................27
Books on Heat and Mass Transfer ....................................................................................................................27
Heat transfer and mass transfer are kinetic processes that may occur and be studied separately or jointly.
Studying them apart is simpler, but both processes are modelled by similar mathematical equations in the
case of diffusion and convection (there is no mass-transfer similarity to heat radiation), and it is thus more
efficient to consider them jointly. Besides, heat and mass transfer must be jointly considered in some cases
like evaporative cooling and ablation.
The usual way to make the best of both approaches is to first consider heat transfer without mass transfer,
and present at a later stage a briefing of similarities and differences between heat transfer and mass transfer,
with some specific examples of mass transfer applications. Following that procedure, we forget for the
Heat and mass transfer page 1
2. moment about mass transfer (dealt with separately under Mass Transfer), and concentrate on the simpler
problem of heat transfer.
There are complex problems where heat and mass transfer processes are combined with chemical reactions,
as in combustion; but many times the chemical process is so fast or so slow that it can be decoupled and
considered apart, as in the important diffusion-controlled combustion problems of gas-fuel jets, and
condensed fuels (drops and particles), which are covered under Combustion kinetics. Little is mentioned
here about heat transfer in the micrometric range and below, or about biomedical heat transfer (see Human
thermal comfort).
Heat transfer is the flow of thermal energy driven by thermal non-equilibrium (i.e. the effect of a non-
uniform temperature field), commonly measured as a heat flux (vector), i.e. the heat flow per unit time (and
usually unit normal area) at a control surface.
The aim here is to understand heat transfer modelling, but the actual goal of most heat transfer (modelling)
problems is to find the temperature field and heat fluxes in a material domain, given a previous knowledge
of the subject (general partial differential equations, PDE), and a set of particular constraints: boundary
conditions (BC), initial conditions (IC), distribution of sources or sinks (loads), etc. There are also many
cases where the interest is just to know when the heat-transfer process finishes, and in a few other cases the
goal is not in the direct problem (given the PDE+BC+IC, find the T-field) but on the inverse problem: given
the T-field and some aspects of PDE+BC+IC, find some missing parameters (identification problem), e.g.
finding the required dimensions or materials for a certain heat insulation or conduction goal.
Heat-transfer problems arise in many industrial and environmental processes, particularly in energy
utilization, thermal processing, and thermal control. Energy cannot be created or destroyed, but so-common
it is to use energy as synonymous of exergy, or the quality of energy, than it is commonly said that energy
utilization is concerned with energy generation from primary sources (e.g. fossil fuels, solar), to end-user
energy consumption (e.g. electricity and fuel consumption), through all possible intermediate steps of energy
valorisation, energy transportation, energy storage, and energy conversion processes. The purpose of thermal
processing is to force a temperature change in the system that enables or disables some material
transformation (e.g. food pasteurisation, cooking, steel tempering or annealing). The purpose of thermal
control is to regulate within fixed established bounds, or to control in time within a certain margin, the
temperature of a system to secure its correct functioning.
As a model problem, consider the thermal problem of heating a thin metallic rod by grasping it at one end
with our fingers for a while, until we withdraw our grip and let the rod cool down in air; we may want to
predict the evolution of the temperature at one end, or the heat flow through it, or the rod conductivity
needed to heat the opposite end to a given value. We may learn from this case study how difficult it is to
model the heating by our fingers, the extent of finger contact, the thermal convection through the air, etc. By
Heat and mass transfer page 2
3. the way, if this example seems irrelevant to engineering and science (nothing is irrelevant to science),
consider its similarity with the heat gains and losses during any temperature measurement with a typical
'long' thermometer (from the old mercury-in-glass type. to the modern shrouded thermocouple probe). A
more involved problem may be to find the temperature field and associated dimensional changes during
machining or cutting a material, where the final dimensions depend on the time-history of the temperature
Everybody has been always exposed to heat transfer problems in normal life (putting on coats and avoiding
winds in winter, wearing caps and looking for breezes in summer, adjusting cooking power, and so on), so
that certain experience can be assumed. However, the aim of studying a discipline is to understand it in
depth; e.g. to clearly distinguish thermal-conductivity effects from thermal-capacity effects, the relevance of
thermal radiation near room temperatures, and to be able to make sound predictions. Typical heat-transfer
devices like heat exchangers, condensers, boilers, solar collectors, heaters, furnaces, and so on, must be
considered in a heat-transfer course, but the emphasis must be on basic heat-transfer models, which are
universal, and not on the myriad of details of past and present equipment.
Heat transfer theory is based on thermodynamics, physical transport phenomena, physical and chemical
energy dissipation phenomena, space-time modelling, additional mathematical modelling, and experimental
Thermodynamics of heat transfer
Heat transfer is the relaxation process that tends to do away with temperature gradients (recall that ∇T→0 in
an isolated system), but systems are often kept out of equilibrium by imposed boundary conditions. Heat
transfer tends to change the local state according to the energy balance, which for a closed system is:
What is heat (≡heat flow)? Q≡∆E−W → Q =∆E|V,non-dis=∆H|p,non-dis (1)
i.e. heat, Q (i.e. the flow of thermal energy from the surroundings into the system, driven by thermal non-
equilibrium, not related to work or to the flow of matter), equals the increase in stored energy, ∆E, minus the
flow of work, W. For non-dissipative systems (i.e. without mechanical or electrical dissipation), heat equals
the internal energy change if the process is at constant volume, or the enthalpy change if the process is at
constant volume, both cases converging for a perfect substance model (PSM, i.e. constant thermal capacity)
to Q=mc∆T. However, it is worth to keep in mind that:
• Heat is the flow of thermal energy driven by thermal non-equilibrium, so that 'heat flow' is a
redundancy (i.e. a pleonasm, and the same for ‘work flow’). Heat must not be confused with stored
thermal energy, and moving a hot object from one place to another must not be called heat transfer.
But, in spite of all these remarks, it is common in normal parlance to say ‘heat flow’, to talk of ‘heat
content’, etc.
Heat and mass transfer page 3
4. • Heat is an energy flow, defined by (1) just for the case of mass-impervious systems (i.e.
Q≡Wadiab−W). When there are simultaneous energy and mass flows, heat flow must be considered at a
surface with no net mass flow.
• Heat input to a system, may not necessarily cause a temperature increase. In absence of work, a heat
input always increases internal energy (Q=∆E for W=0 in (1)), but this increment may be ‘sensible’
(i.e. noticeable as a temperature increase), or ‘latent’ (e.g. causing a phase change or other
endothermic reaction at constant temperature).
• A temperature increase in a closed system is not necessarily due to a heat input; it can be due to a
work input (e.g. ∆E=mc∆T=W for Q=0 in (1)), either with dissipation (e.g. internal stirring), or
without (isentropic compression).
• The First Law (1) shows that, for a steady state without work exchange, the heat loss by a system
must pass integrally to another system, i.e. for the interface, Q≡Qnet=Qin− Qout=0 for ∆E=W=0.
• The Second Law teaches that heat always flows from the hotter system towards the colder one. Even
when we want to extract heat from a cold system like in refrigeration, we must procure a colder
working substance for heat to flow down the temperature gradient to the working fluid (later to be
compressed to a higher temperature than that of the heat sink, to finally dispose of the thermal energy
again by letting heat to flow down the temperature gradient to the ambient).
In Thermodynamics, sometimes one refers to heat in an isothermal process, but this is a formal limit for
small gradients and large periods. Here, in Heat Transfer, the interest is not in heat flow, Q, but on heat-
flow-rate, Q =dQ/dt, that should be named just heat rate, because the 'flow' characteristic is inherent to the
concept of heat, contrary for instance to the concept of mass, to which two possible 'speeds' can be ascribed:
mass rate of change, and mass flow rate. Heat rate, thence, is energy flow rate at constant volume, or
enthalpy flow rate at constant pressure:
dQ dT
Q ≡
What is heat flux (≡heat flow rate)? = mc ≡ KA∆T (2)
dt dt PSM,non-dis
where the global heat transfer coefficient K (associated to a bounding area A and the average temperature
jump ∆T between the system and the surroundings), is defined by (2); the inverse of K is named global heat
resistance coefficient M≡1/K. Notice that this is the recommended nomenclature under the SI, with G=KA
being the global transmittance and R=1/G the global resistance, although U has been used a lot instead of K,
and R instead of M.
In most heat-transfer problems, it is undesirable to ascribe a single average temperature to the system, and
thus a local formulation must be used, defining the heat flow-rate density (or simply heat flux) as q ≡ dQ dA
. According to the corresponding physical transport phenomena explained below, heat flux can be related to
temperature difference between the system and the environment in the classical three modes of conduction,
convection, and radiation:
Heat and mass transfer page 4
5. conduction q =−k ∇T
What is heat flux density (≈heat flux)? K T convection q ≡ h (T − T∞ ) (3)
q =∆

radiation=q εσ (T 4 − T04 )
These three heat-flux models can also be viewed as: heat transfer within materials (conduction), heat transfer
within fluids (convection), and heat transfer through empty space (radiation).
Notice that heat (related to a path integral in a closed control volume in thermodynamics) has the positive
sign when it enters the system, but heat flux, related to a control area, cannot be ascribed a definite sign until
we select 'our side'. For heat conduction, (3) has a vector form, stating that heat flux is a vector field aligned
with the temperature-gradient field, and having opposite sense. For convection and radiation, however, (3)
has a scalar form, and, although a vector form can be forged multiplying by the unit normal vector to the
surface, this commonly-used scalar form suggest that, in typical heat transfer problems, convection and
radiation are only boundary conditions and not field equations as for conduction (when a heat-transfer
problem requires solving field variables in a moving fluid, it is studied under Fluid Mechanics). Notice also

that heat conduction involves field variables: a scalar field for T and a vector field for q , with the associated
differential equations relating each other (because only short-range interactions are involved), which are
partial differential equations because time and several spatial coordinates are related.
Another important point in (3) is the non-linear temperature-dependence of radiation, what forces to use
absolute values for temperature in any equation with radiation effects. Conduction and convection problems
are usually linear in temperature (if k and h are T-independent), and it is common practice working in
degrees Celsius instead of absolute temperatures.
Finally notice that (1) and (2) correspond to the First Law (energy conservation), and (3) incorporates the
Second-Law consequence of heat flowing downwards in the T-field (from hot to cold).
Physical transport phenomena
Heat flow is traditionally considered to take place in three different basic modes (sometimes superposed):
conduction, convection, radiation.
• Conduction is the transport of thermal energy in solids and non-moving fluids due to short-range
atomic interactions, supplemented with the free-electron flow in metals, modelled by the so-called

Fourier's law (1822), q =−k ∇T , where k is the so-called thermal conductivity coefficient (see
below). Notice that Fourier's law has a local character (heat flux proportional to local temperature
gradient, independent of the rest of the T-field), what naturally leads to differential equations. Notice
also that Fourier's law implies an infinite speed of propagation for temperature gradients (thermal
waves), which is nonsense; thermal conduction waves propagate at the speed of sound in the
medium, as any other phenomena small perturbation. In crystalline solids, packets of quantised
energy called phonons serve to explain thermal conduction (as photons do in electromagnetic
Heat and mass transfer page 5
6. • Convection, in the restricted sense used in most Heat Transfer books, is the transport of thermal
energy between a solid surface (at wall temperature T) and a moving fluid (at a far-enough
temperature T∞), modelled by a thermal convection coefficient h as in the second line of (3), named
Newton's law (1701); in this sense, heat convection is just heat conduction at the fluid interface in a
solid, whereas in the more general sense used in Fluid Mechanics, thermal convection is the
combined energy transport and heat diffusion flux at every point in the fluid. Notice, however, that
what goes along a hot-water insulated pipe is not heat and there is no heat-transfer involved; it is
thermal energy being convected, without thermal gradients. Related to fluid flow, but through porous
media is percolation; a special case concerns heat transfer in biological tissue by blood perfusion (i.e.
the flow of blood by permeation through tissues: skin, muscle, fat, bone, and organs, from arteries to
capillaries and veins); the cardiovascular system is the key system by which heat is distributed
throughout the body, from body core to limbs and head.
• Radiation is the transport of thermal energy by far electromagnetic coupling, modelled from the basic
black-body theory (fourth-power-law of thermal emission), Mbb=σT4, named Stefan-Boltzmann law
(proposed by Jozef Steafan in 1879 and deduced by his student Ludwig Bolzmann in 1884), σ being
a universal constant σ=5.67·10-8 W.m-2.K-4, modified for real surfaces by introducing the emissivity
factor, ε (0<ε<1). Radiation is emitted as a result of the motion of electric charges in atoms and
molecules (by thermal vibrations or external forces), and radiation is absorbed by matter increasing
the atomic motion of electric particles. Notice that (3) only applies to radiation heat transfer when the
surface absorptance to radiation coming from the environment at temperature T0 is equal to surface
emissivity at temperature T, what is usually not the case if both temperatures are of different orders
of magnitude. If absorption at a surface is not total, part of the incident radiation may be reflected or
transmitted behind (scattered, in general). Heat transfer by thermal radiation is not only important at
high temperatures; even at room temperatures, T≈300 K, the equivalent linear coefficient of heat
transfer is K=4σT3=6.1 W/(m2·K), comparable to a typical natural convective coefficient in air of
h=10 W/(m2·K).
The three heat-transfer modes above-mentioned, often appear at the same time on a thermal problem, but
seldom with the same importance, what allows for simple one-mode analysis in many instances. A combined
case that appears in many cases is the heat flow by convection from one fluid to another fluid separated by
an intermediate solid wall (single, as in pipes, or double, as in modern window panes); in such cases, dealing
with an overall heat-transfer coefficient, K, is very helpful (one has just to apply (2)); e.g. typical values for a
modern house may be: K=0.5 W/(m2·K) for walls, K=0.6 W/(m2·K) for the roof, K=0.7 W/(m2·K) for the
floor, and K=3 W/(m2·K) for a double-pane window.
Notice finally that isothermal surfaces are usually assumed in convection- and radiation- heat-transfer
problems, and that the temperature field is only solved in heat conduction problems (except when big
computation codes are used to solve the whole fluid mechanics problem, namely in CFD).
Heat and mass transfer page 6
7. Thermal conductivity
The thermal conductivity, k, of a given isotropic material at given conditions, is the proportionality constant

defined by Fourier's law, q =−k ∇T . For non-isotropic materials, k is no-longer a scalar magnitude but a
tensor. Representative k-values are presented in Table 1.
Table 1. Representative thermal conductivities.
k [W/(m·K)] Comments
Order of 10 (good In metals, Lorentz's law (1881), k/(σT)=constant, relates thermal
magnitude for conductors) conductivity, k, to electrical conductivity, σ, when electron-gas conduction
solids 1 (bad is dominant; however, good electrical conductivity is not synonymous of
conductors) good thermal conductivity: mercury only has k≈9 W/(m·K) being a good
electrical conductor, and diamond has k≈1000 W/(m·K) being a good
electrical insulator.
Phonon theory indicates that electrical insulators have k(T< k(T>>TD)∝1/T, whereas pure metals have an electronic contribution (on top
of that of phonons) of k(T<>TD)≈constant.
Some values for metal alloys are in Table 2 below.
Ceramics and polymers are, in general poor conductors (see Solid data).
Aluminium 200 Very pure aluminium may reach k=237 W/(m·K) at 288 K, decreases to
k=220 W/(m·K) at 800 K; going down, k=50 W/(m·K) at 100 K, increasing
to a maximum of k=25∙103 W/(m·K) at 10 K and then decreasing towards
zero proportionally to T, with k=4∙103 W/(m·K) at 1 K).
Duralumin (4.4%Cu, 1%Mg, 0.75%Mn, 0.4%Si) has k=174 W/(m·K),
increasing to k=188 W/(m·K) at 500 K.
Iron and steel 50 Pure iron (ferrite) has k=80 W/(m·K).
Cast iron (96%Fe, 4%C) has k=40 W/(m·K).
Mild-carbon steel with <0.4%C have k=52 W/(m·K), and decreases with
carbon content: k=42 W/(m·K) for carbon steel with 1%C, k=32 W/(m·K)
for carbon steel with 1.5%C, etc.
Conductivity decreases with temperature in carbon steels (e.g. from 50
W/(m·K) at 300 K to 30 W/(m·K) at 1000 K).
Conductivity increases with temperature in stainless steels (e.g. from 18
W/(m·K) at 300 K to 25 W/(m·K) at 1000 K). Going down, k=9 W/(m·K)
at 100 K, k=0.7 W/(m·K) at 10 K). Conductivity decreases with alloying
from k=26 W/(m·K) to k=15 W/(m·K).
Order of 1 (inorganic) Liquids are, in general poor conductors (see Liquid data), with k decreasing
magnitude for 0.1 (organic) with temperature, in general.
liquids Liquid metals are good conductors: mercury has k=9 W/(m·K), and molten
sodium has k=60 W/(m·K). The Na-K eutectic alloy (a room-temperature
liquid with 22%wt Na) has k=23 W/(m·K) at 100 ºC.
Heat and mass transfer page 7
8. Thermal oils are not good on thermal conductivity (k=0.1 W/(m·K)) but on
thermal stability (can work up to 600 K).
Water 0.6 Thermal conductivity of saturated water grows from k=0.57 W/(m·K) at
273 K to k=0.69 W/(m·K) at about 400 K, and then decreases to k=0.64
W/(m·K) at 500 K, k=0.50 W/(m·K) at 600 K, and k=0.24 W/(m·K)
towards the critical point at 647 K.
Water properties may be used as a first approximation for many natural
aqueous solutions (milk, wine, beer, vinegar, seawater, urine, fruit juices,
Order of 10−2 Gases are very poor thermal conductors (see Gas data); hydrogen, with
magnitude for k=0.17 W/(m·K), and helium, with k=0.14 W/(m·K), are, by far the best
gases conductors.
According to the simplest generalised transport theory in gases, thermal
diffusivity, mass diffusivity and kinematic viscosity of gases have the same
values (a=Di=ν ≈10-5 m2/s). With this theory, gas thermal conductivity
increases with the square root of temperature, and do not change with
pressure; in reality, a linear dependence with temperature better fits
experimental results.
Air 0.024 Thermal conductivity of foamed materials cannot be below that of air,
unless the bubbles are air-tight and the foaming agent used CO2 (k=0.015
W/(m·K)), R134a (CF3CH2F, k=0.014 W/(m·K)), or so.
Measuring k is based on measuring heat flux against temperature gradient in steady-state set-ups, or on
measuring thermal diffusivity in transient processes. Measuring k in liquids is very difficult because test
sizes are restricted to avoid heat convection, and it is even harder to measure k in gases because, not only
convection must be avoided, but thermal radiation too. Many times, conductivity-values at different
temperatures are needed, but notice that measuring k always implies a temperature difference to generate the
heat flow, which cannot be too small without compromising uncertainty in the measure, thus all k-values are
more or less averaged values..
A summary of the different approaches followed to measure thermal conductivities is presented below,
quoting the formulation to be used, which is covered in detail separately under Heat Conduction modelling.
Steady state methods to measure thermal conductivity of materials (most accurate, but slow and expensive):
• Planar geometry. A slab (or a long isolated bar for good conductors) is kept between a guarded heater
at one side of the sample (e.g. an electrical heater mat insulated on the other side, or in a symmetric
configuration), and a guarded cooler on the other side of the sample (e.g. cooling water, or a
thermoelectric refrigerator). If the heat flow is unidirectional, the conductivity is given by
= k ( Q A ) / ( ∆T L ) , but, as it is very difficult to measure Q without losses, a reference-k material is
used to measure k by comparison. Most accurate values for poor conductors are obtained in a
Heat and mass transfer page 8
9. multilayer set-up like H-T-R1-T-S-T-R2-T-C, where H stands for the heater, T stands for an
aluminium disc with embedded thermocouples, R1 and R2 stand for two reference-material discs
(preferably with a similar conductivity as the sample), S stands for the sample, and C for the cooler.
Very thin materials like films and foils are tested by stacking several sample layers together.
• Cylindrical geometry. A cylindrical sample with a central heater and a cooler on the outside is used.
The conductivity is given
= by k ( Q L ) / ( 2π∆T ln ( Rext Rint ) ) , but, as above, a reference-k
material can be used to measure k by comparison.
Unsteady methods to measure thermal conductivity of materials (not so accurate, but quicker (half a minute
instead of half an hour), cheaper, and yield thermal diffusivity directly):
• Planar geometry. The most used variant of this method is the quasi-steady state method (Fitch-1935).
A disc-shape sample (a thin slab may do) is sandwiched between a planar wall of a metal container
with a well-stirred bath at a fixed temperature, Tw, and a small copper disc of mass mCu, and area ACu,
well-insulated on all sides except that in contact with the sample. Assuming linear quasi-steady heat
transfer through the sample, its conductivity is obtained by the copper energy-balance,
k= ( ( ))
− ( mCu cCu L ( ACu t ) ) / ln (TCu (t ) − Tw ) (TCu,initial − Tw ) ; accuracy can be enhanced by
optimisation of sample and copper thicknesses. Another planar method is based on the time lag
between two thermocouples on each side of a thin sample, when a short light-pulse is shined on one
side; this is known as flash method, and, although already introduced in the 1950s, still lacks
accuracy; the thermal diffusivity of the sample is usually estimated as a=0.14L2/t1/2, where L is
sample thickness, and t1/2 is the time it takes for the rear thermocouple to reach half of the maximal
temperature increase (temperature falls afterwards, due to heat losses, axially and laterally).
• Cylindrical geometry (also known as line heat source method). The method is based on the
temperature rise at radius R within the sample, after a centred line heater of given power is switched
on. The most used variant of this method is the singular case with R=0, i.e. when temperature rise is
measured just at the axis, where the heater is also located; a line-heat-source probe holds both a
heater and a thermometer, either inside a narrow tube, or on the outside of a tube or a rod (internal
placement makes the probe more robust, but external placement yields more accurate data). The
conductivity of the sample is obtained by fitting the straight portion of the T(t) versus lnt plot, and
using k = ( Q ( 4π ) ) / ( dT d( ln t ) ) (the curved initial and final portions should be discarded). The
sample diameter D must be large enough, D2/(at)>10, and probe diameter d narrow enough, say
d/D<10, and even so, uncertainties are typically around 5%.
• Spherical geometry (also known as thermistor method). The method is based on measuring the
temperature rise of a thermistor of radius R, encapsulated in a nearly-spherical bead, embedded
within the sample, and used as a point source of constant power, Q . The thermal diffusivity of the
sample material is obtained from Q = ( )
4π Rk + 4 R 2 π k ρ c t (TR (t ) − T∞ ) , i.e. by linear fitting of the
thermistor temperature versus the inverse of the square root of time. This method required calibration
with a medium of known thermal conductivity to find the effective bead radius, R, and even so,
uncertainties are typically around 20%, mainly due to heat losses through the connectors..
Heat and mass transfer page 9
10. Heat equation
The heat-transfer equation is the energy balance for heat conduction through an infinitesimal non-moving
volume. To deduce it, we start from the energy balance (2) applied to a system of finite volume:
dH ∂T  
Q →
= ∫ ρc − ∫ q ⋅ ndA + ∫ φ dV
dV = (4)
dt p V
∂t A V
where φ is some energy release rate per unit volume (e.g. by nuclear or chemical reactions), sometimes
written as q gen . Equation (4) can be read as "the time-increment of enthalpy within de volume is due to the
heat input through the frontier plus the energy dissipation in the interior", the minus sign coming from the

choice of n as the normal outwards vector. When the Gauss-Ostrogradski theorem of vector calculus is used
to transform the area-integral to the volume-integral, and (4) is applied to an infinitesimal volume within the
system, one gets:
∂T   
∫ ρc
dV = − ∫ q ⋅ ndA + ∫ φ dV = − ∫ ∇ ⋅ q dV + ∫ φ dV

∂T 
V →0
 → ρ c = −∇ ⋅ q + φ (5)
Finally, considering Fourier's law (3) and constant material properties (density ρ, thermal capacity c,
conductivity k, and their combination, the thermal diffusivity:
a≡ (6)
one gets the so-called heat equation:
∂T ∂T φ
ρc =k ∇ 2T + φ , or =a∇ 2T + (7)
∂t ∂t ρc
The heat equation (7) is the most well-known parabolic partial-differential equation (PDE) in theoretical
physics, φ being a non-homogeneous term. The heat equation is also known as diffusion equation, and it has
solutions that evolve exponentially with time to the steady state. At steady state, the heat equation becomes
an elliptic PDE named Poison's equation, which, without the non-homogeneous term, becomes Laplace
equation, Ñ2T =0. Besides parabolic and elliptic PDE, the third type is the hyperbolic PDE
∂2ψ/∂t2−c2∂2ψ/∂x2=0, typical of wave-like phenomena.
A more general heat equation takes account also of the effect of relative motion between the material system

and the coordinate system with a velocity v , which, in a fix reference frame (Eulerian reference frame) takes
the form:
Heat and mass transfer page 10
11. ∂T φ 
= a∇ 2T + − ∇ ⋅ (Tv ) (8)
∂t ρc
although we will only consider such motions when analysing moving heat sources in a stationary solid (to
change to a reference frame moving with the source), of application to machining, grinding, cutting, sliding,
welding, heat treatment, and so many materials processing. The most general heat equation (e.g. to be used
in computational fluid dynamics (CFD), must include in the energy release term φ, viscous dissipation and
the dilatation work due to the time-variation of pressure along a fluid line, if any, although both are usually
negligible energy contributions. To solve the heat equation, besides the parameters explicitly appearing in it
(a, φ, ρ, c...), appropriate bounding conditions for the variables are required, i.e. initial conditions for time
and boundary conditions for the space variables.
Modelling space, time and equations
Space-time modelling may refer to the consideration of continuous or discrete processes in space and time,
but space-time modelling in heat transfer usually refers to the consideration of processes as steady or non-
steady, zero-dimensional, one-dimensional, two-dimensional or three-dimensional geometry, planar,
cylindrical or spherical, etc. So important this modellization is, that heat transfer books, and in particular the
heat conduction part, is usually divided in different chapters for the different space-time models: steady one-
dimensional conduction, unsteady one-dimensional conduction, steady two-dimensional conduction, etc.
As in many other engineering problems, the steady state solution is usually analysed first, in heat-transfer
problems, leaving transient effects for a more advanced phase, but many undesirable events may occur
during transients. Here in this context, thermal shock (e.g. breaking a glass by pouring hot water) and local
overheating (e.g. charring shoes and cloth before getting warm), can be mentioned.
Every step in problem-solving may have an associated mathematical modelling, from geometrical definition
and materials properties, to results and conclusions. We want now to consider the main mathematical tools
used to formulate and solve heat-transfer problems, traditionally divided into classical analytical methods
(partial differential equations in a continuum, developed in the 19th century), and modern numerical
methods (discrete set of algebraic equations applied to small elements of the system, developed in the late
20th century, like the finite element method FEM, the finite difference method FDM, or the boundary
element method BEM. In both cases there are imposed bounding conditions (BC), which, in heat transfer,
are the initial conditions (heat equation is 1st order in time) and boundary conditions (heat equation is 2nd
order in space variables); the latter are usually classified as:
• First kind boundary conditions, when the temperature is known at the boundary.
• Second kind boundary conditions, when the temperature-gradient is known at the boundary. It often
happens that, on the whole boundary of a closed system, conditions of a first kind apply to some
parts, and of the second kind to the others.
• Third kind boundary conditions, when the temperature-gradient at the boundary is a known function
of the temperature there.
Heat and mass transfer page 11
12. • Fourth kind boundary conditions, when neither the temperature-gradient nor the temperature-gradient
are known at the boundary, but are functions of other boundary conditions in the problem (i.e. the
boundary at hand is an intermediate boundary in a much larger system).
In practice, all real problems are of the third and fourth kind, because it is really difficult to force a constant
temperature at a boundary (thermostatic baths and blocks are often used for the purpose), and even more
difficult to guarantee a constant heat flux.
By the way, continuous field theory (as used in Fluid Mechanics, Elasticity, or Electromagnetism) is a
simplifying recourse to modelling the influence of many discrete microscopic particles on other many-
particles systems (i.e. the goal of Thermodynamics). Classical field theory started with Newton's Law of
Gravitation, followed with Euler's Law of fluid motion (later expanded to Navier-Stokes equation), and
Fourier's heat equation, and peaked with Maxwell's electromagnetism equations. But Fourier was the first to
solve a multi-dimensional field equation (a PDE), inventing Fourier's series and separation of variables to
solve the heat equation (he did it to win a prize offered by the French Academy).
Case studies
To better illustrate heat transfer theory and applications, we are considering the following two simple
practical examples to throw light on the different approaches to solve heat transfer problems:
• Cooling-down of a hot sphere in a water bath (sphere-cooling, for short), where a glass sphere with
D=1 cm in diameter, is taken out of a bath at T1=100 ºC (e.g. boiling water) and submerged in a
bath of ambient water at T∞=15 ºC with an estimated convective coefficient of h=500 W/(m2∙K).
We take for glass k=1 W/(m∙K), ρ=2500 kg/m3 and c=800 J/(kg·K). This is a one-dimensional,
spherical-symmetry, unsteady problem of practical relevance in materials processing.
• Heating-up of a rod in ambient air by an energy source at one end (rod-heated-at-one-end, for
short), where an aluminium rod of length L=0.1 m and diameter D=0.01 m, is being heated at one
end with Q 0 =10 W (from an inserted electrical heater), while being exposed to a ambient air at
T∞=15 ºC with an estimated convective coefficient h=20 W/(m2∙K). We take for aluminium k=200
W/(m∙K), ρ=2700 kg/m3 and c=900 J/(kg·K), though k may vary from 120 W/(m∙K) in the typical
aerospace alloy (Al-7075) to 220 W/(m∙K), in pure aluminium (Al-1100). This is a quasi-one-
dimensional unsteady problem, which has a non-trivial steady state temperature profile. The name
rod usually refers to centimetric-size elements; for much smaller rods, the word spine is more
common, and the word beam for much larger elements.
These case studies may seem too stereotyped, but they are relevant to heat transfer practice, and they allow
for comparison of practical numerical approaches with simple analytical exact solutions.
Nomenclature refresh
A science is a set of concepts and their relations. Good notation makes concepts more clear, and helps in the
developments. Unfortunately, heat transfer notation is not universally followed.
Heat and mass transfer page 12
13. ISO 31-4 & 31-6 here
Heat flux , dQ/dt Φ Q
Heat flux density , dQ/(dtdA) φ or q q
Thermal conductivity, dQ/(dtdA∆T) λ or k k
Thermal conductance (or heat conductance), dQ/(dt∆T), G=1/R G G
Thermal conductance coefficient (or heat transfer coeff.), dQ/(dtdA∆T) K K (old U)
Thermal resistance (or heat resistance), R=1/G R R
Thermal insulation coefficient (or heat insulatio coeff.), M=1/K M M
Thermal diffusivity, k/(ρcp) a or α a
Radiant energy, [J] Q or W (not used)
Radiant energy density, [J/m ] w (not used)
Radiant energy flux, [W] Φ or P Φ
Incident radiations:
- Irradiance (or incident (radiant energy) flux, direct plus diffuse), [W/m2] E E (old i)
- Absorptance (at an opaque surface, or through transmitting media) α α
Emerging radiations:
- Exitance (or emerging (radiant energy) flux), [W/m2] M M
- Radiant intensity (or emerging radiant energy along a direction), [W/sr] I (not used)
- Radiance (or emerging radiant energy flux along a direction), [W/m ∙sr] L (not used)
2 c)
- Emittance, [W/m ], is not defined; only exitance. Planck law is M(λ) (only defined for M
for a blackbody. (exitance=emission+reflexion+transmission) a blackbody)
- Emisivity = real emittance / blackbody emittance ε ε
a) The desire to use a common symbol for all kind of fluxes is praiseworthy, but we still keep here to the
traditional symbol Q for the heat flux, usually simplified to Q in most heat-transfer texts when there is no
possible confusion between heat and heat flux.
b) In Heat Transfer, q and Q (instead of the here-used q and Q ) may be used for heat flux density and heat
flux, respectively, to simplify notation, but q and Q are preferable when a Heat Transfer course is
combined with Thermodynamics, to avoid symbol overriding.
c) Planck law: M λ = , gives the spectral exitance of a blackbody.
5   c2  
λ exp   − 1
  λT  
Heat transfer theory may be used to compute heating/cooling times in heat transfer problems, or to compute
temperature fields and heat fluxes, or to compute required dimensions or properties for heat insulation or
conduction. In some special cases, the goal is to find the value of a parameter in a thermal problem that
produces branching solutions, as in the onset of Bénard-Marangoni convection when heating a thin liquid
layer from below (bifurcation analysis).
Heat-transfer problems may arise in typical thermal applications, like heating and cooling; e.g., the
defrosting problem in refrigeration and air conditioning, due to thermal insulation of the ice layer (frosty ice
Heat and mass transfer page 13
14. conductivity can be as low as that of wood), admits several solutions, all of them controlled by heat transfer.
Any temperature-measure involves some heat transfer problem. Besides, many other heat-transfer problems
come from non-thermal pursuits; for instance:
• Cooling electronic equipment. Microprocessor computing power is limited by the difficulty to
evacuate the energy dissipation (a Pentium 4 CPU at 2 GHz in 0.18 µm technology must dissipate
76 W in an environment at 40 ºC without surpassing 70 ºC). Most electronics failures are due to
overheating by improper ventilation or fan malfunction. Bipolar junctions in silicon wafers fail to
keep the energy gap between valence and conduction electrons above some 400 K, but at any
working temperature there is some dopant diffusion and bond-material creep, causing some
random failures, with an event-rate doubling every 10 ºC increase; depending on the reliability
demanded, bipolar junctions are usually limited to work at 90..100 ºC. Electrical powers up to a
few watts can usually be dissipated by natural convection to ambient air, and up to few hundred
watts by forcing air with fans (cheap, but noisy and wasteful), and liquid cooling is usually needed
beyond 1 kW systems.
• Cooling rubbing parts. In a mechanical transmission, the oil loses its lubricating capacity if
overheated; in a hydraulic coupling or converter, the fluid leaks under the pressure created. In an
electric motor, overheating causes deterioration of the insulation. In an overheated internal-
combustion engine, the pistons may seize in the cylinders.
• Lamp design. The size of an incandescent lamp is governed by heat transfer (the filament needs a
bulb to keep away from oxygen, but the bulb-size is large in glass bulbs to avoid glass softening by
high temperatures, and small in halogen lamps to be hot enough to maintain the halogen cycle
inside (to avoid deposition at a cold bulb), in spite of the little size for the electrical resistance R
that must be fed at low voltage for the same power P=VI=V2/R).
• Materials processing like casting, welding, hot shaping, crystal growth, etc. Materials machining is
limited by the difficulty to evacuate the energy dissipation. And not only engineering materials:
food processing and cooking, dish washing, cloth washing, drying and ironing, and many other
house-hold tasks, are dominated by heat transfer.
• Energy conversion devices, like solar collectors, combustors, nuclear reactors, etc.
• Environmental sciences like meteorology, oceanography, pollutant dispersion, forest fires, urban
planning, building, etc.
Relaxation time
The two usual limits for thermal interaction in Thermodynamics are the isothermal process and the adiabatic
process, the former corresponding to the limit of very slow heat transfer due to an infinitesimal temperature
difference along an infinite time, and the latter corresponding to a quick process with negligible heat
transfer. In Heat Transfer, however, we are interested on finite-time process, and a basic question is to know
the thermal inertia of the system, i.e. how long the heating or cooling process takes, usually with the
intention to modify it, either to make the system more permeable to heat, more insulating, or more
'capacitive', to retard a periodic cooling/heating wave.
Heat and mass transfer page 14
15. When the heat flow can be imposed, as when heating water with a submerged electrical resistor, the
minimum time required is obtained from the energy balance, dH / dt = Q = mc∆T / ∆t ; e.g. to heat 1 kg of
water from 15 ºC to 95 ºC with a 1000 W heater, the minimum time is ∆t= mc∆T / Q =1·4200·(95-15)/
1000=336 s (the actual value in practice depends on the way of heating, the geometry of the vessel, and the
way temperature is measured; typically, 30% more time is required with highly efficient types of heaters like
a microwave oven or induction heaters, and up to 100% more time with inefficient heaters as external
electrical resistors).
For the case where the heat flux is not imposed but a temperature gradient is imposed, an order-of-
magnitude analysis of the energy balance, dH / dt = Q → mc∆T/∆t=KA∆T, shows that the relaxation time is
of the order ∆t=mc/(KA), and, depending on the dominant heat-transfer mode in K, two extreme cases can be
considered: solids in well-stirred fluids (convection dominates, and evolution is driven by conduction), and
highly conducting solids (convection-driven case).
Conduction driven case (convection dominates)
Problems where the thermal conductance from a solid system to a surrounding fluid, K=h, is much larger
than the thermal conductance within the solid, K≈k/L, i.e. where Bi≡hL/k→∞ (Bi is a non-dimensional
parameter called Biot number). In this case, the boundary condition imposes a constant temperature on the
body surface, and the time ∆t it takes for heat to penetrate to the centre of the body, of characteristic length L
(volume divided by surface), i.e. the relaxation time may be guessed from (2-7):
∆H mc∆T mc∆T ρ L3c ρ cL2 L2
∆t ≈ ≈ ≈ ≈ = = (9)
Q KA∆T kA ∆T kL2 1 k a
where the ∆T from initial to final states of the system has been assumed to be of the same
magnitude of the representative ∆T from the system to the surroundings (although the former is not uniform
and the latter is not constant but decreases with time). Thus, the time it takes for the centre to reached a mid-
temperature representative of the forcing is L2/a, i.e. increases with the square of the size, decreases with
thermal diffusivity, and is independent of temperature.
• Exercise 1. Make an estimation of the time for the 1 cm glass-ball to cool down, in our sphere-
cooling problem stated above.
Solution. In this case Bi=hL/k=500·(0.01/6)/0.6=1.5>1, large enough, and the time it takes for the
centre to reach a representative temperature of the heating process (e.g. a mid-temperature between
the initial and the final, say 60 ºC), is ∆t=ρcL2/k=2500·800·(0.01/6)2/1=6 s, where the characteristic
length of a spherical object, L=V/A=(πD3/6)/(πD2)=D/6, has been used.
Could this model be applied to the cooling down of a hot potato in air?
Notice that this convection-dominated model is not applicable to our rod-heated-at-one-end problem,
since, for it Bi=hL/k=20·(0.01/6)/200=10-4<<1 (neither to the hot potato in air.
• Exercise 2. Make an estimation of the time it takes to boil an egg.
Heat and mass transfer page 15
16. Solution. Assuming an egg at 5 ºC (from the fridge) is suddenly put into boiling water at 100 ºC (a
bad practice since the jiggling might break the shell against the walls; it is better to keep it below
boiling), and neglecting chemical energy changes, the time it takes for the centre to reach mid
temperatures (say 50 ºC), is ∆t=ρcL2/k=1000·4000·(0.04/6)2/0.6=300 s (i,e, about 5 minutes), where
the characteristic length of a nearly spherical object, L=D/6 has been used, with an egg diameter of
D=4 cm, and thermal properties of water (most food, as the human body itself, have a water content
of some 70%). This is a convection-dominated problem since the convective coefficient is of order
h=1000 W/(m2·K) and thus Bi=hD/k=1000·(0.06/6)/0.6=17>>1.
Three levels of egg boiling may be distinguished: suck-egg (also named egg from the shell, or oeuf à
la coque), boiled for about two minutes, what leaves it semi-liquid throughout; soft-boiled egg,
boiled for 3 to 5 minutes, what leaves a barely solid outer white, a milky inner white and a warm
yolk, to eat with a spoon from the shell; hard-cooked egg, boiled for 10 to 15 minutes, what leaves a
solid to be peeled and consumed apart. When heating an egg, the globular-folded aminoacids in
albumin (a sol dispersion) start to unfold and stick, becoming less fluid and less transparent, forming
a gel (a porous network of interconnected solid fibres spanning the all the volume of a liquid
medium). Egg white starts to coagulate at about 60 ºC and ends at 65 ºC, when proteins denaturize;
the yolk has different proteins and more fat, and coagulates from 65 ºC to 70 ºC.
It is important to be aware of the huge increase in heat transfer rate that even a small fluid convection may
bring. If air inside a room of size L=3 m were to be heated with a radiator just by heat conduction (without
air motion), a time ∆t=L2/a=32/10−5=106 s (i,e, about 10 days!) would be required, whereas in the real case,
the heat-up time may be estimated as ∆t=L/v, where v is an average speed of the natural convection due to
the draught caused by the heated air close to the radiator; if we estimate the air velocity in its vicinity from
the draught pressure balance, ∆ρg∆z=(1/2)ρv2max, with ∆ρ=ρα∆T=ρ∆T/T, we get vmax=(2g∆z∆T/T)1/2=0.8
m/s for a typical radiator height of ∆z=1 m, heating some 10 K the surrounding air at 300 K; if we take as
room-average speed v=vmaxδ/L=0.8·0.1/3=0.03 m/s (having assumed a draught thickness of δ=0.1 m), the
heating time (by natural convection) is now ∆t=L/v=3/0.03=102 s, not a bad guess (103 s is a more realistic
order of magnitude, but nothing comparable to the 106 s we got in the diffusion case).
Convection driven case (conduction dominates)
Problems where the thermal transmittance within the system, K=k/L, is much larger than the thermal
transmittance to the surroundings, K=h, (i.e. where Bi≡hL/k→0). In this case, the temperature within the
body can be assumed spatially uniform and the relaxation time may be guessed from (2-7):
∆H mc∆T mc ρ L3c ρ cL
∆t ≈ ≈ ≈ ≈ = (10)
Q KA∆T hA hL2 h
where, as before, the ∆T from initial to final states of the system has been assumed to be of the same
magnitude of the representative ∆T from the system to the surroundings. The difference now is that the
convective coefficient with the environment is not a material property but depends a lot on the motion
outside, and that now the relaxation time is directly proportional to the size L and not its square function.
Heat and mass transfer page 16
17. • Exercise 3. Make an estimation of the time it takes for the rod to reach steady state, in our rod-
heated-at-one-end problem.
Solution. For a rod with lateral convection, Bi=hL/k=20·(0.01/4)/200=2.5·10-4<<1, where the and
thence the time it takes to heat up (for the centre to reach a mid temperature between the initial and
the steady state), is ∆t=ρcD/h=2700·900·(0.01)/20=1200 s (i,e, some 20 minutes), where the
characteristic length of a cylindrical object, L=V/A=(πD2/4)/(πD)=D/4 has been used, instead of the
axial length, which is of little relevance in the transient heat up.
• Exercise 4. Make an estimation of the time it takes for a hot potato to cool down.
Solution. Assuming a baked potato is brought out of an oven at 100 ºC (mind that even after 1 hour in
an oven at 250 ºC, food can only reach some 100 ºC while it is moist), and left in air at 20 ºC, the
time it takes for the centre to reach a representative temperature of the cooling process (e.g. a mid-
temperature between the initial and the final, say 60 ºC), is ∆t=ρcL/h=1000·4000·(0.06/6)/10=4000 s
(i,e, more than 1 hour!), where the characteristic length of a nearly spherical object, L=D/6 has been
used, with a potato diameter of D=6 cm, the thermal properties of water (most food, as the human
body itself, have a water content of some 70%), and a typical value of the convective coefficient in
calm air of h=10 W/(m2·K). We have neglected chemical energy changes, but the result obtained is
longer than experience tells (say, over half an hour), mainly due to radiation losses (wrapping a hot
potato in aluminium paper, a very poor emitter, keeps it hot for nearly one hour). Notice that potatoes
should never be forced-cooling by water immersion (unlike most other vegetables) because they
would get moist. Moreover, potato skin should be pierced before baking to let vapour escaping.
Potatoes loss weight on baking (oven and microwave): from 10% the large ones to 25% the small
ones. Can you now predict how long will it take for a frozen turkey of 5 kg to thaw when taken out of
the freezer? (Beware of the phase change.)
Heat flux
Thermodynamics is concerned with heat accounting Q=∆E−W, but Heat Transfer is focussed on rates, so
that one question is to know the relaxation time for a given amount of heat (see before), and another question
is the flow-rate function considered here (yet, a third one would be to find the materials and geometry that
satisfy a prescribed flow-rate or relaxation time; see below).
Typical heat-flux problems mostly consist on finding the heat flux corresponding to prescribed temperatures
at the boundaries of a given geometry and known material. In many circumstances, the simple steady state,
one-dimensional planar geometry, may be a good approximation, and in other cases the uniform-temperature
approximation is acceptable; for more complicated problems, the heat-flux problem must be found
concurrently with the temperature field (next point). A couple of simple applications follow.
• Exercise 5. Find the heat flux through a composite wall, with application to the bottom of a freezer
boat with -30 ºC cargo temperature, 1 mm thick stainless steel lining sheet, 10 cm thick polyurethane
insulation, 1 cm thick air layer, and 1 cm thick steel of hull in contact with water at 10 ºC.
Heat and mass transfer page 17
18. Solution. Assuming steady state in one-dimensional planar geometry, the heat flux must be the same
through every layer, so that for the combined sandwich:
T2 − T1 T −T T −T 1
q = K ∆T = k12 = k23 3 2 = ... = n 1 ⇒ K= (11)
L12 L23 L L
∑ ki ∑ ki
i i
where the interfaces are named 1,2,…,n; i.e., the overall thermal transmittance is given by (11), best
recalled in terms of their inverse, the thermal resistance coefficient R=1/K, what establishes that the
overall resistance is the sum of the partial resistances R=∑Ri=∑Li/ki (electrical analogy of a circuit
with resistances in series). For the numerical example here:
Tn − T1 (10 + 273) − (−30 + 273) 40
=q = = = 9.7 W/m 2 (12)
Li 0.01 0.01 0.1 0.001 0.0002 + 0.42 + 3.7 + 0
∑ k 52 + 0.024 + 0.027 + 20
where thermal conductivity values for steel, air, polyurethane, and stainless steel of (52, 0.024, 0.027,
and 20) W/(m∙K), respectively, have been assumed.
• Exercise 6. Find the heat transfer rate during the cooling down of the glass ball in our sphere-cooling
problem stated above.
Solution. If uniform temperature inside the ball could be assumed, the energy balance would yield an
ordinary differential equation, easily integrated:
dH dT T − T∞  hA 
Q= → mc = −hA (T − T∞ ) ⇒
= KA∆T  = exp  − t (13)
dt dt T0 − T∞  mc 
showing an exponential temperature decrease and a corresponding exponentially decreasing heat
flow-rate, that for the numerical values of the glass ball cooling in water gives
T − T∞  hA   6h   6 ⋅ 500   t 
=exp  − t  =exp  − t  =exp  − t  =exp  −  (14)
T0 − T∞  mc   ρ Dc   2500 ⋅ 0.01 ⋅ 800   7 s
 hA   t 
− hA (T − T∞ ) =
Q = − hπ D 2 (T0 − T∞ ) exp  − − (13 W ) exp  −
t =  (15)
 mc   7 s
i.e., the water bath starts getting 13 W from the ball, but at a decreasing rate insomuch that
 = ρ cV ∆T= 2500 ⋅ 800 ⋅ (π 0.013 /6) ⋅ (100 − 15)= 89 J (in effect, (13 W)·(7 s)≈90 J).
∫ Qdt
Exercise 7. Dew on window panes
Heat and mass transfer page 18
19. Temperature field
Although in many thermal problems (and even in some heat transfer problems) temperature is assumed
uniform in a system (as in the hot-potato problem just explained, and in many heat convection and radiation
problems), temperature is never uniform in practice (equilibrium systems are just limit models; we live in a
non-equilibrium world). The heat equation (4), with the appropriate initial and boundary conditions, can
provide a spatial temperature distribution at every time.
Many times, particularly in thermal control problems, only the extreme temperatures are sought. Living
beings have very short temperature ranges (human cannot support body temperatures outside T=(310±4) K,
and typical electrical batteries deteriorate outside T=(300±30) K. Refractory materials are those material able
to withstand high temperatures (say >1500 K), without failure by fusion or decomposition; they are usually
classified according to chemical behaviour, as acid refractories (fireclay), neutral refractories (coal, graphite,
refractory metals and metal carbides), and basic refractories (metal oxides).
Two examples of temperature field computations are presented below:
• Exercise 8. When temperature at a surface of a semi-infinite solid (of constant properties, initially at
T∞) is suddenly brought to T0), the disturbance propagates to a depth x in a time t such that the
temperature profile (one-dimensional because of the initial and boundary conditions), and heat flux
density, are given respectively by:
∂ 2T 1 ∂ T η≡ ∂ 2T ∂T
− = 0  2 at
→ + 2η = 0 → =
T c1 + c2 erf (η )
∂ x a ∂t
∂η 2
T ( x, t ) − T0  x  (T∞ − T0 ) exp  − x 2 
= erf   q ( x, t ) = −k   (16)
T∞ − T0  2 at  π at  4at 
a being the thermal diffusivity of the material. Notice the high idealisation of the statement: semi-
infinite extent (in one dimension, infinite in the other two), and infinitely quick temperature jump,
what makes the problem of little practical use, but notice the simplicity of the result too (practical
heat-transfer problems usually end up with a myriad of numbers difficult to crunch, or with a
multiplicity of partial graphics difficult to integrate).
• Exercise 9. Find the steady temperature profile in our rod-heated-at-one-end problem, with a
prescribed temperature value at one end, instead of the constant heat source.
Solution. We can consider this problem a one-dimensional heat conduction problem axially, with
internal heat sinks to account for the actual lateral heat losses by convection (i.e. with
φAdx=−hpdx(T−T∞), p being the perimeter and A the cross-section area), and apply (7) to an
infinitesimal slice:
∂T d 2T hp
ρ cAd=
x kAdx∇ T − hpdx (T − T∞ ) 
2 t →∞

= 0 − (T − T∞ ) (17)
∂t dx 2 kA
Heat and mass transfer page 19
20. and, imposing the boundary conditions:
d 2T hp 
0 = 2 − (T − T∞ )   T ( x) − T∞ cosh  m ( L − x ) 
dx kA   T −T = cosh ( mL )
  0 ∞
=T x =0 T0 ⇒  (18)
  sinh  m ( L − x ) 
dT =
  Q ( x ) hpkA ( T − T )
0 = −kA cosh ( mL )
∞ 0
dx x = L  

where m ≡ hp /(kA) , h being the convective coefficient, p the perimeter of the rod cross-section
(πD if circular) , A the cross-section area (πD2/4 if circular), and L the rod length.
Dimensioning for thermal design
The goal of most heat transfer modelling is to find the temperature field and heat fluxes in a material
domain, given a set of constraints: general heat equation (e.g. set as a partial differential equation, PDE),
boundary conditions (BC), initial conditions (IC), distribution of sources or sinks (SS), etc. In a few cases
the goal is not in the direct problem (given the PDE+BC+IC+SS, find the temperature field), but on the
inverse problem: given the T-field and some aspects of PDE+BC+IC+SS, find some missing parameters
remaining (identification problem).
Perhaps the very simplified, yet very important, problem of one-dimensional steady heat transfer between
two bodies, separated by a solid layer, can make more clear the several different goals in heat transfer: heat
fluxes, T-fields, material characterisation, and dimensioning:
•= Q kA (T1 − T2 ) / L , i.e. find the heat flux for a given set-up and T-field.
• T=  / ( kA) ) , i.e. find the temperature corresponding to a given heat flux and set-up.
T2 + QL
Notice that our thermal sense (part of the touch sense) works more along balancing the heat
flux than measuring the contact temperature, what depends on thermal conductivity of the
object; that is why Galileo masterly stated that we should ascribe the same temperature to
different objects in a room, like wood, metal, or stone, contrary to our sense feeling.
• k QL
=  / ( A∆T ) , i.e. find an appropriate material that allows a prescribed heat flux with a
given T-field in a given geometry.
•= L kA (T1 − T2 ) / Q , i.e. find the thickness of insulation to achieve a certain heat flux with a
given T-field in a prescribed geometry.
Other typical example of thermal design follows.
• Exercise 10. Find the minimum conductivity for a pot handle of length L=0.2 m and A=1 cm2 cross-
section, to avoid hand-burning (assume Tburn=45 ºC) when holding the handle up to the middle while
the end at the pot is at boiling-water temperature.
Solution. We start assuming that the hand is not modifying the thermal problem; i.e., we want to find
when we have Tburn at L/2. A first analysis shows that a key point in the thermal problem is missing:
Heat and mass transfer page 20
21. what causes temperature to fall along the handle? The answer is, of course, heat losses to ambient air
by convection, which should be modelled. Assuming ambient air at T∞=20 ºC and a convective
coefficient of h=10 W/(m2∙K), one may establish the desired relation from (18):
Tburn − T∞ cosh ( mL / 2 ) 45 − 20 cosh ( mL / 2 ) hp 2
= → = → = L constant (19)
T0 − T∞ cosh ( mL ) 100 − 20 cosh ( mL ) kA
where m ≡ hp /(kA) has been substituted, to reach the conclusion that the allowed conductivity
increases with h, p and L (of course, when more convection or longer handle, more conductive
handles can be allowed), and decreases with A (the larger the cross-section, the most insulating the
handle material must be). Notice that, for a given area, larger perimeter handlers are best. Assuming a
square solid handle, the above constant has a value of 6.12, m=12.4 m, and the maximum allowable
handle conductivity is k=26 W/(m∙K), i.e. a stainless-steel handle can be allowed (from Thermal data
of solids, k=16..26 W/(m∙K), depending on the type). In most cases, however, non-metal handles are
implemented, a good reason being that the user tends to hold the handle much closer to the pot root,
to decrease the force moment.
Thermal design
Design is an intricate multidisciplinary top-down activity (see Thermal Systems, for an overview). Thermal
design, in heat-transfer problems, aims at providing a suitable configuration (materials, components,
geometry, arrangement...), amongst different possibilities, trying to optimise the cost/benefit. For instance, a
thermal designer may be asked to provide solutions to keep a computer CPU dissipating 70 W without
becoming hotter than 70 ºC; amongst the different possibilities, the most common one nowadays is to leave
some free-room nearby and blow air with a fan (with the associated noise and dissipation increase), but
using a heat-pipe to efficiently-connect the internal chip with an external ample sink is already taking over
(e.g. in laptops); high-power-dissipation devices may demand liquid cooling loops or even phase change
loops (which might be expandable in some cases, similar to animal sweating).
Thermal design requires a broad knowledge of the subject (and related subjects), and is left to a later stage in
training, except for simple 'design' problems where the configuration is already given and only a parameter
of the configuration is to be optimised. The most common endeavour for beginners is to solve well-defined
thermal problems, i.e. to perform some heat transfer analysis to find temperatures, heat fluxes, or relaxation
Thermal analysis
To solve a heat-transfer problem in practice, to find the temperature field and heat fluxes, like for any other
engineering task, there are not magic recipes, but sound understanding of the subject matter. The practitioner
should not compile a set of graphics, tables and formulas, much less the student. On the contrary, they
should master the principles of heat transfer, and have an idea of the different tools available.
Heat and mass transfer page 21
22. Several steps are usually taken to solve a heat-transfer problem:
1. Mathematical modelling of the physical problem. This is the most creative phase in solving a
problem. Sometimes, physical analogies help to build the mathematical model. The electrical
analogy consists on taking thermal resistances as resistors, thermal capacities as capacitors, heat
sources as intensity sources, the thermal environment as electrical ground, and applying
Kirchhoff's law to the network. It is important, however, to keep in mind that analogies are just
analogies, not identities, and care is needed to avoid stretching them beyond their applicability.
2. Mathematical solution of the mathematical problem. Although it is just a mathematical burden,
engineers must be aware of the available methods of solution, and their pros and cons, in order to
direct the previous idealisation towards feasible, available, affordable, efficient and solvable
problems. The two basic approaches are:
• Analytical solutions, which gives a whole and concise parametric solution, but only in
extremely idealised problems (only of academic interest or to check numerical simulations).
• Numerical solutions, which gives particular solutions to any practical problem, but without an
overview of the influence of the parameters (several particular cases must be solved to have
an idea of the influences).
3. Analysis of the results (analytical or numerical) and physical interpretation. In some
circumstances, particularly with new or complicated problems, some experimental tests, where
the temperature field and heat fluxes are metered in an instrumented sample, are required to
provide evidence of the goodness of the mathematical modelling.
In actual practice, heat-transfer problems are solved numerically by using a large commercial computer
package, usually an integrated fluid-thermal-structural CFD-package, or at least with inputs and outputs
compatible with main commercial packages for mechanical and structural analysis.
Mathematical modelling
The mathematical modelling is the idealisation of the physical problem until a well-defined set of
(mathematical) constraints, representing the main features, is established. Mathematical modelling is
required not only in analytical work but also in actual heat-transfer practice, where a large commercial
computer package is used; the user has to identify and approximate the actual geometry of the system, has to
select the most appropriate terms from the list of supplementary effects in the PDE, must approximate the
boundary conditions according to specific package procedures, and, most important of all, the user has to
give knowledgeable feed-back on possible weaknesses and improvements, since heat-transfer analysis, as
any other engineering activity, is an iterative process that must be refined as needed; effort proportional to
expected utility (a common error of beginners, both at school and at work, is to spend too much effort and
time pursuing very precise numerical solutions to 'what if' preliminary problems that are discarded soon
afterwards, or even before being finished!).
Heat and mass transfer page 22
23. Mathematical modelling is the most creative part in the whole process of solving heat-transfer problems.
Modelling usually implies approximating the geometry, materials properties, and the heat transfer equations.
Modelling the geometry
In thermal problems, the first task is to identify the system under study. On one side, the geometry is
idealised, assuming perfect planar, cylindrical or spherical surfaces, or a set of points and a given
interpolation function. Besides the edges or boundaries (which are usually fixed, as in Fig. 1, except in some
special cases like the Stefan problem of moving phase-change), further information is needed to know if the
region or domain of interest lies inside, outside, or in between boundaries. Additionally, several numerical
methods of solving heat-transfer problems, make use of a subdivision of the domain in small sub-domains
called elements, and procedures are needed to carry out an automatic meshing and the associated numbering.
Location procedures are also needed to know to which element a given point belongs, which are the
neighbour elements, and so on.
Fig. 1. The space-time domain is divided in the spatial domain or boundary, D (that may be one-, two- or
three-dimensional, and is usually assumed independent of time in thermal problems, D(t)=D(t0)),
and the time domain (that is one-dimensional, with a clear start, t=t0, and a clear bias, t>t0).
The most complicated case occurs when boundary conditions are imposed on free-moving boundaries, i.e.
surfaces with a priori unknown locations which separate geometric regions with different characteristics, as
in heat-transfer problems with phase change; e.g. freezing of liquids or moist solids, casting, or
polymerisation. This type of moving-boundary-value problems is known as Stefan problem, because Jozef
Stefan was the first, in 1890, to analyse and solve it, when studying the rate of ice formation on freezing
water, although a similar problem was first stated in 1831 in a paper by Lamé and Clapeyron. Phase-change
materials are very efficient thermal-energy stores, either to accommodate heat input to heat output, or even
to get rid of large amounts of thermal energy by ablation. In the normal case of phase-change accumulators,
only the solid/liquid phase-change is considered, and with some buffering space to avoid large pressure
build-up; this void fraction, plus the usual metal mesh used to increase thermal conductance, makes thermal
modelling complicated.
• Exercise 11. Find the time for a liquefied-nitrogen-gas pool, 4 mm thick, to vaporise when suddenly
spread over ground.
Solution. The problem of spreading and vaporisation of cryogenic liquids, when there is a spillage
over ground or water, is similar to the problem of water pouring over a very hot plate. Initially, the
temperature jump is so large that there is a violent vaporisation at the contact surface, with formation
of a thin (say tenths of a millimetre) vapour layer in between that isolates the liquid from the solid.
Even with this vapour resistance, the solid starts to cool down, until the temperature jump is not
enough to generate the vapour layer, which collapses and brings the liquid directly in contact with the
Heat and mass transfer page 23
24. solid, increasing very much the solid cooling-rate, and changing the vaporisation from film boiling to
nucleate boiling (see Heat transfer with phase change). This phenomenon was first described by J.G.
Leidenfrost, in 1756, and is named after him. If we here disregard the initial vapour layer, and
consider a uniform liquid layer of initial thickness L, vaporising at a rate controlled by the heat flux
being supplied from the ground, which is modelled as a semi-infinite solid with a fixed temperature-
jump at the surface (see Similarity solutions in Heat conduction), the energy balance gives:
m vap hLV dL
q0 = = −ρ hLV (20)
A dt
ρ and hLV being the density and vaporisation enthalpy of the liquid, whereas the heat flux is (Case 1
from Table 6 in Heat conduction):
q0 = k (21)
π at
k and a being the thermal conductivity and diffusivity of the solid, and ∆T the constant temperature
jump form the liquid to the solid far away.
The solution is then:
dL k ∆T 2k ∆T t  L ρh 
= − → L= L0 −
 ⇒ t0= π a  0 LV  (22)
dt ρ hLV π at ρ hLV πa  2k ∆T 
L0 being the initial layer thickness and t0 being the time for the whole layer to vaporise (when
L(t)=0). Substituting numerical values for liquid methane (as an approximation to LNG mixture,
from Liquid property data), a=k/(ρc)=0.18/(423·3480)=0.12·10-6 m2/s, ρ=423 kg/m3, hLV=510 kJ/kg,
k=0.18 W/(m·K), c=3480 J/(kg·K), with L0=4 mm and ∆T=T0−Tb=288-112=176 K, we finally have
t0=70 s, i.e. about one minute.
One should keep in mind that real applications usually have complex geometry, with different materials (e.g.
thermal problems in electronic boards), and the fact that a good modelling should only retain key thermal
elements with approximated shapes, as major heat dissipaters with box or cylindrical shapes, and most
sensitive items (e.g. oscillators, batteries). Most of the times, the geometry, material and boundary
conditions are such that real 3D problems can be modelled as 2D or even 1D, with immense effort-saving.
Modelling materials properties
Once the system is defined, its materials properties must be idealised, because density, thermal conductivity,
thermal capacity, and so on, depend on the base materials, their impurity contents, actual temperatures, etc.
(see Table 1 above.) Most of the times, materials properties are modelled as uniform in space and constant in
time, for each material, but, whether this model is appropriate, or even the right selection of the constant-
property values, requires insight.
Heat and mass transfer page 24
25. Unless experimentally measured, thermal conductivities from generic materials may have uncertainties of
some 10%. Most metals in practice are really alloys, and thermal conductivities of alloys are usually much
lower than those of the components, as shown in Table 2; it is good to keep in mind that conductivities for
pure iron, mild steel, and stainless steel, are (80, 50, 15) W/(m·K), respectively. Besides, many common
materials (like graphite, wood, holed bricks, reinforced concrete), are highly anisotropic, with directional
heat conductivities, particularly all modern composite materials. And measuring k is not simple at all: in
fluids, avoiding convection is difficult; in metals, minimising thermal-contact resistance is difficult; in
insulators, minimising heat losses relative to the small heat flows implied is difficult; the most accurate
procedures to find k are based on measuring thermal diffusivity a=k/(ρc) in transient experiments.
Table 2. Thermal conductivities of some typical alloys and its elements.
Alloy k [W/(m·K)] k [W/(m·K)] k [W/(m·K)]
of alloy of element of element
Alu-bronze C-95400 59 393 (Cu) 220 (Al)
(10% Al, >83% Cu, 4% Fe, 2% Ni)
Mild steel G-10400 51 (at 15 ºC) 80 (Fe) 2000 (C, diamond)
(99% Fe, 0.4% C) 25 (at 800 ºC) 2000 (C, graphite, parallel)
6 (C, graphite, perpend.)
2 (C, graphite amorphous)
Stainless steel S-30400 16 (at 15 ºC) 80 (Fe) 66 (Cr)
(18.20% Cr, 8..10% Ni) 21 (at 500 ºC) 90 (Ni)
Unless experimentally measured, convective coefficients computed from generic correlations may have
uncertainties of some 10%, whereas those taken from 'typical value' tabulations are just coarse orders of
magnitude, e.g. when it is said that typical h-values for natural convection in air are 5..20 W/(m2·K) and one
assumes h=10 W/(m2·K).
Unless experimentally measured on the spot, absorptance coefficients and emissivities of a given surface can
have great uncertainties, which in the case of metallic surfaces may be double or half, due to minute changes
in surface finishing and weathering.
Modelling the heat equations
The equations defining a heat-transfer problem, in systems where thermal conduction is the only heat-
transfer mechanism in the interior, are the heat equation (5), and its bounding conditions (initial and
boundary conditions). In systems with internal convection, the above equations must be solved concurrently
with the fluid mechanics equations of Navier-Stokes. In systems with internal radiation, very complicated
integral-differential equations appear when one considers spectral absorptances and multidirectional
dispersions. Here we restrict the rest of the analysis to conductive systems, with convective and/or radiative
effects entering only as boundary conditions.
Heat and mass transfer page 25
26. There are a number of commercial packages for numerical solutions of PDE (like NASTRAN), applicable in
principle to thermal, structural, fluid and electrical problems. However, in practice, the thermal problem may
be highly non-linear (particularly if radiation is important) and it may be inconvenient to use the same
discretization or even the same problem for thermal and structural analysis (in many cases the number of
nodes and elements is 1 to 2 orders of magnitude larger for structural than for thermal analysis) To use these
commercial packages, the user first makes use of a pre-processor (included in the package or dedicated ones
like MSC/Patran or SCRC/Ideas) to draws the geometry or to import it as a CAD-file, to defines the
materials (from a pre-loaded list or entering its properties), and to indicates a mesh type and size, what,
together with and the specification of the particular boundary conditions (what is usually the hardest task),
completes the input to the solver. After some time (always longer than expected) the solver produces a huge
amount of information (output from the solver) that must (always) first be checked out for validity, before
any further analysis. The user needs a post-processor (included in the package or a dedicated one like
MSC/Patran or SCRC/Ideas) to interpret the results.
Perhaps the key point to remember when actually doing the mathematical modelling of thermal problems is
that it is nonsense to start demanding great accuracy in the solution when there is not such accuracy in the
input parameters and constraints. Without specific experimental tests, there are big uncertainties even in
materials properties, like thermal conductivity of metal alloys, entrance and blocking effects in convection,
and particularly in thermo-optical properties.
Analysis of results
The analysis of the results may be quite different in the case of a closed analytical solution than for the case
of a numerical solution. In the last case, the interpretation of the numerical solution to judge its validity,
accuracy and sensitivity to input parameters can be quite involved. The direct solution usually gives just the
set of values of the function at the nodes, what is difficult to grasp for humans in raw format (a list of
numbers or, for regular meshes, a matrix). Some basic post-processing tools are needed for:
• Visualization of the function by graphic display upon the geometry or at user-selected cuttings.
Unfortunately many commercial routines, besides the obvious geometry overlay, only present the
function values as a linear sequence of node values and don't allow the user to select cuts.
Additional capabilities as contour mapping and pseudo-colour mapping are most welcome.
• Computation of function derivatives (and visualization). Some times only the function is
computed, and the user is interested in some special derivatives of the function, as when heat
fluxes are needed, besides temperatures.
• Feedback on the meshing, refining it if there are large gradients, or large residues in the overall
thermal balance. It is without saying that the user should do all the initial trials (what usually
takes the largest share of the effort) with a coarse mesh, to shorten the feedback period.
• Precision and sensitivity analysis by running some trivial cases (e.g. relaxing some boundary
condition) and by running 'what-if' type of trials, changing some material property, boundary
condition and even the geometry.
Heat and mass transfer page 26
27. A global checking that the detailed solution verifies the global energy equation gives confidence in 'black
box' outputs and serves to quantify the order of magnitude of the approximation.
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