II
Natural Sciences Tripos Part II
MATERIALS SCIENCE
C17: Heat and Mass Transfer
Vapour Transport of ZnS under temperature gradient from the hot end (on the
right) at 1753K, condensing at the cold end (on the left) at 1473K
Name............................. College..........................
Dr R. V. Kumar
Easter Term 2014-15
2014 15
C17 - HEAT & MASS TRANSFER
Dr R.V. KUMAR
COURSE CONTENT (6 Lectures + 1 Examples Class)
Heat Transfer
 Basic concepts of heat transfer by conduction, convection and thermal
radiation and their relevance to metallurgical processes
 Heat conduction equation; convection and heat transfer calculations;
thermal resistance; heat transfer coefficient; selected dimensionless
groups; radiation from black and grey surfaces
 Case studies: Combined modes of heat transfer in (a) induction heating
(b) plasma spraying and ( c) stream shrouding in continuous casting
Mass Transfer
 Fluid flow and viscosity; mass transfer in metallurgical processes;
mass transfer coefficient and inter-phase mass transfer
 Case studies: Applications of mass transfer calculations to (a) gas
dissolution in molten metals and (b) metal refining reactions
Examples Class (one); Question Sheet (one)
Book-List
1. J.P. Holman, Heat Transfer, 8th Edition, McGraw-Hill, 1997 (Cf 12)
2. W.J. Beek, K.M.K. Muttzal, J.W. Van Heuven, Transport Phenomena,
John-Wiley & Sons, 1999 (Cf 13)
3. E.T Turkdogan, Fundamentals of Steelmaking, The Institute of
Materials, 1996, Ch. 2. (Dg 153)
4. G.H. Geiger and D.R. Poirier, Transport Phenomena in Metallurgy,
Addison-Wesley, 1973. (D 23)
1
HEAT TRANSFER - BASIC CONCEPTS
Heat transfer is energy transfer due to a temperature difference; Heat
transfer occurs whenever:
 there is a temperature gradient within a material
 there is temperature difference between a material and its surrounding
 two bodies at different temperatures are brought together
Everyday examples of heat transfer:
 naturally occurring such as solar energy transfer to earth
 specially engineered as in melting of materials using fuel or electrical
form of heating
Often the heat transfer is intentional and therefore equipment and
materials are designed to facilitate heat transfer. Sometimes heat transfer
is not desired, such as in minimising heat losses from a furnace, when
insulating materials are suitably provided.
Regardless of the mechanism or mode of heat transfer the driving
force for heat transfer is always temperature difference.
The Laws of Thermodynamics help us to evaluate the amount of heat
transferred (The First Law - Conservation of Energy) and the direction of
heat transfer (The Second Law - increase in Entropy for heat flow from
higher to lower temperature), but is not concerned with the rate at which a
heat transfer process takes place or with the manner in which it take
place.
The rate and the mode of heat transfer is the province of “heat transfer”
studies.
2
Modes of Heat Transfer
There are 3 modes of heat transfer: Conduction, Convection and
Thermal Radiation and the rate at which heat is transferred depends on
the mode of transfer.
These modes can be best illustrated by considering heat transfer between
two flat plates, each of area 1 m2, facing each other at a distance 5 cm
apart. The hot plate is maintained at a steady temperature of 80oC and the
cold plate at 20oC.
Space for sketching:
For a perfect vacuum between the plates, heat can only be transferred by
Thermal Radiation, which involves propagation of energy through space
at the speed of light, by electromagnetic phenomena similar to light and
radio wave propagation. Thermal radiation does not require the presence
of matter for its propagation, although it can be transferred through gases.
3
If a solid material such as copper is placed between the plates in perfect
contact, then heat is transferred by Conduction. In solids, heat
conduction involves transfer of energy by (a) vibrations through the
lattice structure of the solid and/or (b) free electrons. Good electrical
conductors like copper are also good heat conductors.
In liquids, heat conduction involves atomic vibration and free electron
movement as in solids, together with some translational movements of
ions/atoms.
In gases, where atoms/ molecules are far apart from each other, heat
conduction is governed by the rate at which the kinetic energy of gas
molecules can be transferred by random collisions, and therefore is a slow
process. In the above example, if solid copper is replaced by mineral
wool, which is essentially a composite of air pockets trapped in wool
fibres, heat conduction is dramatically reduced.
Although heat conduction through a gas is a slow process, if the plates
were separated by air alone, heat transfer would be higher than as
expected from conduction alone. This is because, air does not remain
stationary. Due to motion of fluid past the surfaces, heat transfer takes
place by Convection.
Convection heat transfer is a mixture of conduction and bulk movement
of the fluid. Conduction takes place across a thin boundary layer of fluid
4
close to the surface. Fluid flow will arise by Natural Convection i.e.
density difference between air near the hot surface and air near the cool
surface. Hotter part of air will rise vertically upwards due to a buoyancy
effect and the cooled air near the cold plate will descend. The exact
amount of heat transferred will depend upon the properties of the fluid.
Also note that free radiation will occur irrespective of convection in
gases.
If air is replaced by water, the heat transfer increases by two orders of
magnitude in this example. If air at 20oC is physically blown past the hot
surface, the convective heat transfer is termed Forced Convection.
In reality, it is impossible to separate or isolate entirely one mode from
the others. However, for simplicity of analysis, we will consider each
mode separately.
5
CONDUCTION
Fourier’s First Law (for steady state heat conduction)
L
T1
T2
→Q
Q = -kA T/x, or
q = -k T/x = k (T1 - T2)/L
Q = Rate of heat transfer by conduction (W)
q = Rate of heat transfer per unit area (W m-2)
k = thermal conductivity of the material (W m-1 K -1)
A = surface area; L = Length
Typical values of k (in (W m-1 K -1)
diamond: 3000
metals: 390 (copper); 15 (stainless steel)
insulators: 1 (firebrick); 0.048 (mineral wool)
liquids: 0.6 (water); 0.07 (Freon 12)
gases: (at P =1 atm); 0.18 (hydrogen); 0.026 (air) - measuring “k” is a
good way of detecting hydrogen in air.
Heat conduction in a solid is influenced by the amount of heat transferred
by convection (natural or forced) at the surface.
Space for sketching:
Heat transfer rate to the surface of the
solid per unit area:
q = h(T - Ts), where
h = heat transfer coefficient (W m-2 K-1)
h = f (fluid property, velocity of fluid)
6
One-Dimensional Steady-State Conduction: For example, for
conduction through a solid slab or a cylinder maintained at T(x =0) = T1
and T(x=L) = T2, it can be shown that rate of heat flow:
Q (Watts) = qA = kA T/L, where T = T1 - T2 and A = surface area.
Using electrical analogy,
I (amp or C/s) = V(volts) / R(ohms),
the concept of thermal resistance can be developed. Therefore:
Q(W )  T ( K ) / Rth ( K / W )
where the Thermal Resistance Rth for the slab = L/Ak. Thermal resistance
can be defined for the various situations, (e.g. thermal resistance for
convection at a surface is 1/hA), and the corresponding electrical
analogue readily described.
7
Composite Media: Consider a simple series made up of two different
materials as will be described. Also consider convective heat transfer at
the two outer surfaces of the composite wall thus adding up to 4 thermal
components. At steady-state, the unidirectional heat flux through the four
parts of the entire circuit is constant. Thus:
Q  Ahi (Ti  T1 ) 
k1 A(T1  T2 ) k 2 A(T2  T3 )

 Aho (T3  To )
L1
L2
The problem is greatly simplified by making use of the concept of
thermal resistance. The four thermal resistance are in series and the
Rtotal = total resistance for the whole circuit is simply their sum = Ri + R12
+ R23 + Ro
And the heat flow is: Q = Ti - To/ Rtotal. We need to know the total
temperature drop across the system to calculate the heat flux which we
can use to determine the temperature at any position within the composite
wall.
Space for sketching:
8
The General Heat Conduction Equation:
It is a mathematical expression of the principle of energy conservation in
a solid substance, derived using a specified control volume of the material
in which heat transfer occurs by conduction only (similar in concept for
diffusion of matter). For simplicity, we will consider 1-D heat conduction
along x-direction, but also include the possibility of thermal energy
generation within the material itself.
Energy balance for a control volume: Thermal conduction into the left face +
thermal energy generated within control volume = thermal conduction out of the
right face + changes in internal energy of control volume, i.e. storage of thermal
energy, the heat conduction equation (Fourier’s 2nd Law) can be derived:
Terms: g = rate of thermal energy generation internally per unit volume;
 = density; c= specific heat; k = thermal conductivity of the material; these
three material properties are combined to define the “thermal diffusivity of the
material:  = k/c (in m2 s-1) and is associated with the diffusion of thermal
energy into the material during changes of temperature T with time t.
Space for Derivation:
9
Fourier’s 2nd law of heat conduction:
2T/x2 + g/k = (c/k) T/t = 1/ T/t
Typical values of  (m2 s-1) x 106 at 300K:
Copper - 114; Lead - 25; Steel - 12; Bricks - 0.5; Water - 0.13; Air - 0.3.
Boundary and Initial Conditions:
The heat conduction equation requires specifying one initial condition
(e.g. distribution of temperature at t =0) and 2 boundary conditions (for
each co-ordinate), which specify the thermal conditions at the boundary
surfaces.
Initial condition: T(x, t=0) = Ti
For example, at a boundary surface, one of the following may be
specified:
 the temperature (boundary condition of the 1st kind)
 the heat flux (boundary condition of the 2nd kind)
 heat transfer by convection into the surrounding fluid at a specified
temperature (boundary condition of the 3rd kind)
Examples:
B.C. of the 1st Kind: A surface maintained at a constant temperature;
T (0, t) = To at the interface between a wax candle and the flame; or the
liquid/solid interface in a welding pool (moving boundary)
10
B.C. of the 2nd Kind: A fixed value of heat flux at a boundary, i.e. q is
constant; -q/k = dT/dx (x=0, t) = constant; (slope of T vs x is constant at
x=0)
Under special condition, if dT/dx = 0 i.e. q =0, then it is called
“homogenous boundary condition of the 2nd kind” and the boundary is
described as an “adiabatic boundary” This situation may be approximated
in sand mould casting of a liquid metal where the thermal conductivity of
sand is very low compared with that of the metal and all the heat flow is
in the vertical (y) direction. Heat transfer into the insulating mould is
considered to be zero: dT/dx (x =0, t) = 0.
B.C. of the 3rd Kind: Convective heat transfer into a fluid at a fixed
temperature T ; Therefore heat balance at the boundary is achieved when
heat transfer by conduction = heat transfer by convection;
q = -k dT/dx (x =0) = h (Ts - T)
11
Heat Transfer Solutions
One-D heat conduction equation in the radial direction (cylindrical coordinates) 2T/r2 + 1/r(T/r) + g /k = 1/ (T/t)
One-D heat conduction equation in the radial direction (spherical coordinates) 1/r2 (r2 T/r)/r + g /k = 1/ (T/t)
An example of how cylindrical co-ordinates can be used to solve steady
state radial flow:
Long Hollow Cylinder: For steady radial flow of heat through the wall
of a hollow cylinder as shown below, cylindrical co-ordinates are
convenient. Since in this case the temperature depends only on the radial
co-ordinate, the equation is:
1 d (rdT / dr )
0
r
dr
The boundary conditions are:
T (r1) = T1 and
It can be shown that the thermal resistance is Rcyl 
Space for sketch:
12
T (r2) = T2
ln( r2 / r1 )
2kL
Space for Derivation:
13
Solution for Unsteady-State Heat Conduction Problems: Let us
consider heat transfer by conduction in solids, in which the temperature
varies not only with position in space, but also undergoes a continuous
change with time at any position described by the partial differential
equation:
 T =  2 T
t
 x2
The initial condition:
T(x, 0) = Ti (uniform)
The boundary conditions: T(0, t) = To and T(, t) = Ti
T
To
Ti
x
For a flat semi-infinite plate or slab, the form is identical to the diffusion
equation, and the solution is of the same form:
T(x,t) = To + (Ti - To) erf (x/ 2 t)
It can be shown that when
x = 4 (t)0.5, T = Ti which imply that no penetration of heat at such
values of x and t for a given .
At x = (t)0.5, T = ½ (To + Ti). The following table illustrates these
points clearly:
Copper (= 114 x 10-6 m2/s)
Glass ((=0.6x10-6 m2/s)
Time, t
1s
1hr
1s
1hr
Distance, x
43 mm
2.6 m
3.1 mm
0.19 m
14
Dimensionless Parameters:
The expression x/(t)0.5, is a useful expression for estimating the extent of
heat penetration by conduction. The inverse of the square of this
parameter is defined as the
Fourier Number (Fo) = t/L2
Another useful dimensionless number called Biot Number (Bi) is defined
as the ratio of convective heat transfer and conductive heat transfer at the
surface of a solid:
Bi = hL/k
(where h = convective heat transfer coefficient at the surface, k = thermal
conductivity of the solid and L = characteristic length of conduction)
In the semi-infinite slab problem discussed above, the temperature at the
surface is maintained at a constant value, a situation that can be described
as akin to h → .
For a situation, where k is very large in relation to h i.e. k → .
Therefore, Bi → 0. For heat transfer situations, where the Biot number is
small (Bi ≤ 0.1), the heat transfer is limited only by convection.
15
Newtonian Heating or Cooling (Lumped Systems)
When a flat plate is cooled from an initial uniform temperature of Ti by a
fluid at temperature Tf, the temperature distribution, which varies with
time, as shown below, can be described by the transient heat conduction
solution. For Bi ≤ 0.1, the temperature gradients within the plate are
negligible, and the temperature of the whole plate varies only as a
function of time. Such a process of cooling (or heating) is called
“Newtonian”.
Space for sketching:
Rate of Heat lost by convection during cooling = Rate of decrease of
internal energy
hA(T - Tf) = - VCP dT/dt
where V = volume of the plate and A = area of the plate exposed to the
fluid. By rearranging and integrating, it can be derived that
T  Tf
Ti  T f
 exp(
 hAt
)
VC p
In deriving the above expression, although A and V are specified, no
geometrical restrictions have been imposed; also note that the solution
does not contain the conductivity of the material.
16
CONVECTION
We have seen how convection can provide a possible boundary condition
for conduction problems in the form of a heat transfer coefficient, h. In
dealing with heat transfer by convection, i.e. energy transport between
fluids and surfaces, we are mainly concerned with obtaining the value of
h, for various systems and investigating how the heat transfer coefficient
varies as a function of the fluid’s properties, such as thermal conductivity,
viscosity, density and specific heat, the system geometry, the flow
velocity and the temperature differences.
17
Fluid motion is the distinguishing feature of heat transfer by convection.
In free convection heat transfer, the fluid motion arises solely due to a
local fluid buoyancy difference caused by temperature difference.
In forced convection heat transfer, the fluid motion is caused by
imposition of a pressure difference by using a pump or a fan.
Flow in pipes, ducts, nozzles ad heat exchangers etc., are examples of
internal flow, where the flow is bounded by solid surfaces apart from an
inlet and outlet. If the fluid experiences only one solid boundary, such as
fluid flow past a flat plate or a cylinder, the flow is deemed an external
flow.
The fluid exerts a drag force on the surface as a result of viscous action,
which affects the convective heat transfer. These effects must be
considered to analyse heat transfer.
Forced Convection (External Flow): Consider flow of a fluid over a flat
plate;
Space for sketching:
When fluid particles make contact with the surface of the plate, they
assume zero velocity, which act to retard motion of molecules in the
adjoining fluid layer, resulting in a viscous effect, until at a distance y =
, from the surface when the effect is negligible. And further at x = 0, at
the leading edge,  = 0 (no viscous effects) and at x =L,  = max
(maximum viscous effects)
18
These viscous forces are described in terms of shear stress  between the
fluid layers. For many fluids, the shear stress is proportional to the
velocity normal gradient:
 =  du/dy
The constant of proportionality is called as the dynamic viscosity, 
(N.s m-2) or (Pa.s); fluids which satisfy this equation are called
Newtonian fluid. The region affected by the viscous force is called the
“boundary layer” and  = boundary layer thickness.
When viscous forces are dominant, the flow tends to be streamlined
(called laminar) and the fluids flow along flow lines which do not touch
each other, i.e. there is no cross flow. When inertial forces dominate over
viscous forces, the regularity is disturbed, and a transition to turbulent
flow is observed, where regions of fluid can move to and fro across
streamline. For the above situation, Re is defined as a function of x, f(x):
Rex =  u x / (at location x)
Often simplified to Rex = u x /
where  is the kinematic viscosity = / = dynamic viscosity/density
For fluid flow, Reynolds number (Re), which is dimensionless number
represents the ratio of inertial force / viscous force: Re = uL/ (where u =
velocity of the fluid, L = characteristic dimension of the flow geometry.
Reynolds number (Re) provides a criteria for determining whether the
flow is laminar (streamlined) or turbulent (erratic). In general, high
velocity, high fluid density and low fluid viscosity promote turbulent
flow.
Space for sketching:
Laminar Flow
Turbulent Flow
19
Flat Plate: The start of transition from laminar to turbulent flow over a
flat plate has been found to be at Rex > 5 x 10 5.
This is satisfied at x = xc - the critical distance from the leading edge.
Space for sketching:
20
Dimensionless Parameters: Since a large number of variables are
involved in calculating the values of heat transfer coefficient (and for
mass transfer coefficients as we shall see later), the coefficients are
frequently reported as correlation’s of dimensionless numbers. We have
already considered the dimensionless numbers Fo and Bi and defined
their physical significance.
If all the variables can be identified, which are sufficient to specify a
given heat (or mass) transfer situation, then it is possible to obtain
pertinent “Dimensionless Groups” without any reference to the basic
differential equations at all. There is in fact a systematic way of
determining the logical groupings of the pertinent variables in the form of
dimensionless parameters by using a technique called the Buckingham pi
theorem. The dimensionless numbers themselves have important
physical significance. The correlation between the different
dimensionless groups for a given heat transfer situation will allow us to
calculate the value of h, the heat transfer coefficient.
21
Space for deriving Dimensionless Parameters:
22
Thermal Boundary Layer: If the fluid stream has a temperature
different from solid surface, then a thermal boundary layer analogous to
the concept of the velocity boundary layer will also develop. Consider a
fluid at a uniform temperature T at the leading edge flowing past a flat
plate which is maintained at a constant temperature Ts:
Space for sketching:
At x = 0, the T (y) = T. At the surface of the plate, the fluid particles
achieve the same temperature as the plate’s surface (thermal equilibrium
is assumed). Temperature gradients (T/y) which is a function of x,
develop as shown in the figure above. The region of the fluid in which the
temperature gradients exist is referred to as the thermal boundary layer
(t).
At any distance x, from the leading edge, the local heat flux at the surface
of the plate (y=0) is given by:
qx = -kf (T/y)y=0
where, kf is thermal conductivity of the fluid. Since there is no fluid
motion at the surface of the flat plate, the Fourier’s First law can be used.
The heat flux at the surface can also be written in terms of heat transfer
coefficient, h.
qx = hx (Ts - T)
Therefore, the local heat transfer coefficient hx = -kf (T/y)y=0 / (Ts - T)
A good estimate of hx can be shown  kf/t. The mean heat transfer
coefficient, for the whole surface can be obtained by integrating the
above equation over the distance x = 0 to x = L along the flat surface.
h = 1/L  hx dx
23
Heat Transfer with Laminar Forced Convection over a Flat Plate:
From the above discussions it is clear that convective heat transfer over a
surface involves both momentum transfer coupled with energy transfer.
For any given condition of velocity profile, it is the relative magnitude of
/ which determines the magnitude of convective heat transfer.
Correlations between three important dimensionless (empirically
obtained) numbers can be used to estimate the value of heat transfer
coefficient.
Reynolds number (Rex ) = inertial forces/viscous forces = u x /
Prandtl number (Pr) = momentum diffusivity /thermal diffusivity = /
Nusselt number (Nux) = total heat transfer/conductive heat transfer =
hxx/kf
In general for forced convection: Nu = f (Re, Pr).
Also note that all the fluid properties, e.g. ,  and kf are evaluated at a
mean boundary layer temperature, called the film temperature given by:
Tfilm = 0.5(Ts + T).
A local value (at any x) for a given parameter is defined by using a subscript: as hx,
Rex, Nux etc., while the average value (integrated over all x and then divided by L)
over the surface is defined by dropping the subscript as h, Re or Nu respectively.
24
The correlation for laminar flow over a flat plate is given by: (for 0.6
= Pr = 50; Re < 5 x 105)
local Nusselt number: (Nux) = 0.332 Pr1/3 Rex
(Nux) = hxx/kf
Nu (for average h over the plate length, L) = 0.664 Pr1/3 Re = hL/kf
Where Re = ReL
For turbulent flow over a flat plate: 5 x 105 < Re < 107; 0.6 < Pr < 60
Nu (for average h over the plate length, L) = 0.037 Pr1/3 Re0.8
For heat transfer to a sphere of dia, d (laminar flow): 2 < Re < 200
Nu = 2 + 0.6 Pr1/3 Red
25
FREE CONVECTION HEAT TRANSFER:
In forced convection, the fluid flows as a result of an external force
resulting in a known velocity distribution which can be used in the
energy/momentum equation. Convection may be induced by difference
in density across the flow resulting in buoyancy. Consider a hot vertical
plate placed in a body of fluid at rest. (Space for sketching):
Temperature variation within the fluid will generate a
density gradient, which will give rise to convective
motion due to buoyancy forces. The momentum and the
energy equations still contain velocities terms, but these
are not knowable and hence the situation is more
complex.
Generally, the rate of mixing is lower than in forced convection, and
therefore the h values are lower. Heat transfer by free convection is a very
important mode of heat transfer in several practical situations such as: in
electrical and electronic engineering applications in the cooling of
electrical components; in the heating of a room by central heating system
with radiators.
Typical Values of Heat Transfer Coefficient, h (W m-2 K-1)
Natural Convection: Gases: 2-30 Liquids: 50-1000
Forced Convection: Gases: 6-600 Water: 200-17000 Boiling Water: 3000 - 60000
Condensation: 6000 - 120000
An exact evaluation of h for free convection is very complicated.
Empirical Correlations used to evaluate h, require the introduction of
another dimensionless parameter called Grashof number (Gr):
Gr = buoyancy force / viscous force = g (Ts - T)L3 / 2
For free convection, the Nusselt number is correlated as: Nu = f(Gr, Pr)
As with forced convection, all fluid properties are evaluated at the film
temperature: Tfilm = 0.5(Ts + T).
26
Natural or Free Convection over a Vertical Plate (for Laminar condition)
Nu = (GrL/4)-0.25 0.902 Pr
(0.861 +Pr)0.25
4
(10 < GrLPr < 109)
Natural or Free Convection over a Vertical Plate (for Turbulent condition)
Nu = 0.0246 GrL2/5 Pr7/15 (1 + 0.494 Pr2/3)-2/5
(109 < GrLPr < 1012)
Space for sketching:
Hot Vertical Plate
Cold vertical plate
27
THERMAL RADIATION
Thermal radiation is electromagnetic energy in transport which can travel
through empty space and originates from a body because of its
temperature. The rate at which energy is emitted depends on the
substance itself, its surface condition and its surface temperature. Thus
radiation heat transfer is the result of interchange between various
surfaces or bodies. The emission is in the form of electromagnetic waves
travelling in straight lines with wavelengths   0.1 to 100 m, covering
a significant part of UV, all of the visible and most of the IR spectrum.
A perfect emitter or absorber against which the radiation properties are
compared is called a “black body” (a hypothetical ideal surface). The
radiation emitted by a black body varies continuously with  and
increases with temperature. The maximum emissions are achieved at
shorter wavelengths as temperature is increased.
Space for sketching:
The total rate of emission at any particular temperature is found by
summing the radiation over all wavelengths (area under the curve of E vs
)
eb =  T4 (Stefan-Boltzmann Law), where
 = Stefan-Boltzmann Constant = 5.67 x 10-8 W m-2 K-4
28
Any real body emitting thermal radiation has a total emitting power
which is some fraction of the emissive power of a black radiator, which is
defined as the emissivity of the material ():
e = eb
The emissivity of a typical real body varies with wavelength and the
variation can be quite irregular. Frequently, it is assumed that the
emissivity of a surface does not vary with wavelength. Such a surface is
known as a grey surface.
Any incoming radiation can be absorbed, reflected or transmitted by a
given material. If we define the total irradiation, G, of a material as the
flux of thermal radiant energy incoming to the surface and the total
radiosity, J as the total flux of radiant energy leaving the surface of the
body (energy emitted + energy reflected), then:
G = G + G + G
or  +  +  = 1
where  = absorptivity (fraction of G absorbed) = 1 for a black body;  =
reflectivity (fraction of G reflected); and  = transmittivity (fraction of G
transmitted) = 0 for opaque materials.
J = e + G
Space for sketching:
Imagine a region in space completely filled with black body radiation. A
real body 1 emitting radiation at a rate e1 is placed in this region; the net
rate of energy transferred from the body is
q = e1 - 1eb = energy emitted - energy absorbed = 1eb - 1eb
If the body is in thermal equilibrium with the black body radiation, then q
= 0, and 1 = 1
Thus the emissivity and absorptivity of any body at thermal equilibrium
are equal.
29
A cavity with a small aperture is the closest approximation in practice to a
black body. Since radiation entering the cavity is subject to repeated
partial absorption and reflection, the effective absorptivity of the cavity to
the external radiation is unity. (space for sketching):
The energy emitted from any point on the walls of the enclosure is eb;
when this radiation strikes other parts of the enclosure, a certain fraction
equal to eb is reflected. A second reflection equals (eb), while the
third one is (eb). Since the hole is assumed to be very small,
radiation that can escape is neglected, and thus the radiation is therefore
made up of infinite number of such reflections.
Therefore, e (hole) = eb ( 1 +  + 2 + 3 + ……)
or e (hole) = eb ( 1/1-)
Since  = 1 - , and  =  (thermal equilibrium), e (hole) = eb.
Radiation from a cavity is therefore very nearly black. If the walls are
maintained at temperature T, the emissive power from the cavity is = T4.
A radiation standard can be constructed with heavy metal internal walls
that are made rough and oxidised to increase emissive power. The walls
can be heated electrically, while the exterior is well insulated.
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Radiation Combined with Convection: At very high temperatures, as
radiant energy depends upon the fourth power of temperature, can
completely dominate over convection. In many practical situations at
intermediate temperatures, both modes of energy transport must be
considered. The total heat flow is the sum of the convective and the
radiant heat flow. It is often convenient to use the total heat transfer
coefficient,
ht = h + hr
where the radiant heat transfer coefficient
hr = q / (T1 - T2) = 1  (T14- T24) / (T1 - T2)
corresponding to a situation where surface 1 is surrounded by or exposed
only to some fluid at a bulk temperature T2.
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MASS TRANSFER
Mass Transfer Coefficient:
Analogous to heat transfer coefficient, the mass transfer coefficient can
be defined for transport of mass at the interface of two phases as flux of
component A divided by concentration difference:
k M  J M (C Ao  C )
Mass transfer coefficient has both diffusional and bulk flow contribution.
In analogy with Heat Transfer, we can define Dimensionless Numbers to
evaluate mass transfer coefficients for different situations.
The Sherwood Number (Sh) has the same or similar functional
dependence on the Reynolds (Re) and Schmidt (Sc) Numbers as the
Nusselt does to Reynolds and Prandtl Numbers.
Sherwood Number: Shx = kM x/D = Total mass transfer/Diffusive mass
transfer
Schmidt Number: Sc = /D = Momentum diffusivity / Mass diffusivity
Sh = f (Re, Sc) from the functional dependence, we can calculate the
value of kM
Space for sketching:
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Laminar flow along a flat plate:
Sh = 0.323 (Re)1/2 (Sc)1/3
for Re < 5 x 105
Laminar flow through a circular pipe (diameter = d):
Sh = 1.86(Sc. Re)0.8
for Re < 5 x 105
Stokes flow (Re < 1) to solid sphere: Sh = 1 + (Sc.Re)1/3
Forced convection around a solid Sphere:
Sh = 2 + (Re)1/2 (Sc)1/3
Space for an example:
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for 2 < Re < 200
A small droplet of fluid moving through another fluid with Stokes
velocity:
kM = (4DUb / d)1/2
(d = diameter of the particle; D = diffusivity of the species being
transported; Ub = Stokes law terminal velocity)
For a small particle moving through a fluid, the terminal velocity arises
due to force balance between buoyancy and viscous drag force:
Ub = d2 g  / 18
 = density difference;  = dynamic viscosity of the fluid through which
the particle is moving
Example:
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C17 - HEAT & MASS TRANSFER
Examples Class (RVK)
1. The campaign life of continuously operating furnaces is limited by the fastest
eroding section, which frequently is due to attack on the refractory walls by molten
slag. One solution is to freeze a layer of the slag on the walls which protects the
wall from the molten slag. This can be accomplished by placing water-cooled
copper plates behind the refractory brick.
Calculate the heat flux required and the thickness of refractory with a 0.05 m thick
slag layer for a slag at 1473 K having a natural convection heat transfer coefficient
of 100 W m-2 K-1 with the wall of the furnace. The freezing point of slag is 1373 K.
Assume that the heat transfer coefficient remains unaffected when the solid slag
layer is formed. The copper cooling plate behind the brick is 0.02 m thick, the
cooling water is at 323K and that the water heat transfer coefficient for forced
convection is 104 W m-2 K-1. Assume rectangular geometry. Also given the
following data:
Material
Thermal Conductivity(k) W m-1 K-1
Water
Copper
Refractory Brick
Slag (solid or liquid)
0.55
320
2
4
2. A steel slab 0.2 m thick x 1.0 m wide x 3.0 m long has been just withdrawn from a
re-heating furnace (see attached diagram). Assume that temperature in the slab is
uniform at 1273K. It is now moving along a roller system at a speed of 0.5 m/s, as
illustrated. Assuming an ambient air temperature of 298K and ignoring radiation
and natural convection effects, calculate: (a) the average convective heat transfer
coefficient from the top surface; (b) the local heat transfer coefficient in the middle
of the top surface; (c) the initial rate of heat loss from the top surface; (d) the
temperature at the mid-point of the surface and at the middle of the ingot after one
hour of cooling. State all the assumptions you need to make to solve this part.
Data for steel: k = 38 W m-1 K-1;  = 7200 Kg m-3; Cp = 400 J Kg-1 K-1;
Data for air at film temperature of 785.5K:  = 0.465 Kg m-3; Cp = 1088 J Kg-1
K-1; k = 0.0568 W m-1 K-1;  = 7.9 x 10-5 m2 s-1; Prandtl number (Pr) = 0.688.
3. Show, using Fourier’s laws of heat conduction, that the Nusselt number has a
limiting value of 2 for steady state heat transfer from a spherical particle to a
motionless fluid (Also valid for very low values of Reynold’s number).
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4. For corrosive and/or high temperature environments, metallic parts are often
coated with refractory materials by plasma spraying. As shown in the schematic
diagram, an arc is struck between the water-cooled anode and cathode, and
nitrogen is injected down the annulus between the electrodes. The gas can be
heated up to 30,000 K. The particles to be sprayed are injected into the jet through
the anode, and they melt in their passage through the flame and solidify on the
substrate. Calculate the maximum size of alumina particles that can be heated to its
melting point (2300 K) in a 10,000 K plasma jet, assuming the particles enter the
jet at the anode and remain in the jet until they hit the substrate. Ignore radiation
from the non-polar gas employed.
Given:
the distance between the electrode and the substrate is 0.3 m;
gas velocity is 6 m s-1 ;
initial particle temperature is 298 K;
k (nitrogen) = 0.08 W m-1 K-1; k (alumina) = 2 W m-1 K-1;
Cp (alumina) = 1.25 J g-1 K-1;
density (alumina) = 4000 Kg m-3.
(Hint: Assume very small Reynold’s number)
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37
C17 - HEAT & MASS TRANSFER
Question Sheet (RVK)
1. A water-cooled copper lance, used for example to inject oxygen into a basic
oxygen steelmaking furnace, has inner and outer diameters of 0.16 and 0.2 m
respectively. Assume water flowing inside has a heat transfer coefficient of 15000
W m-2 K-1, and the furnace gases outside have an average coefficient of 400 W m-2
K-1. Determine the % of the total resistance that each phase contributes and the
temperature at each interfaces given that the hot gas is at 2073K and the water at
273K. The copper lance usually has slag splashed onto it, which can adhere to the
lance. What happens to the total resistance when 5 mm of slag freezes on the lance
and how does this change the lance temperature profile? Given, thermal
conductivity, k (W m-1 K-1): copper - 320; slag - 4.
2. Copper shot is made by dropping molten metal droplets at 1473K into water at
300K. Assuming the droplets to be spheres with diameter of 5 mm, calculate the
total time for the droplets to cool to 373K, given the following data:
For copper: Freezing point: 1356 K; Latent heat of fusion = 204 kJ Kg-1; Cp
(liquid) = 502 J Kg-1 K-1; Cp (solid) = 376 J Kg-1 K-1; k(liquid) = 260 W m-1 K-1;
k(solid) = 340 W m-1 K-1; (liquid) = 8480 Kg m-3; (solid) = 8970 Kg m-3;
Quench data for water: h (1473 - 873K) = 80 W m-2 K-1; h(873 - 373K) = 400 W
m-2 K-1;
3. Define “Biot Number” for a solid plate, stating the significance of this
dimensionless number in heat transfer.
How much Calorie (kcal) per day will an Emperor Penguin be generating in order
to maintain its body temperature at 35oC, while standing stationary in the Antarctic
weather condition with an average temperature of -25oC, given the following data:
Average thermal conductivity of the Penguin,
Heat transfer coefficient,
1 Calorie = 1 kcal = 4.2 kJ
k:
h:
1.0 W m-1 K-1
1 W m-2 K-1
You can model the Penguin as a cuboid of dimensions: 1 m height x 0.2 m width x
0.2 m depth.
Calculate the surface temperature of the Penguin and the temperature profile within
the body, under a stormy weather when the value of the heat transfer coefficient
changes from 1 to 30 W m-2 K-1, assuming that the rate of heat generated is
unaffected by the weather and that this rate is uniformly generated within the
volume of the body. What is the maximum body temperature under these
conditions?
(Tripos 2010)
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4. (a) Show that a cavity with a small aperture can be approximated to a black body
for thermal radiation.
(b) Describe how a practical black body radiation standard may be constructed.
(c) Show that emissivity and absorptivity of any body at thermal equilibrium are
equal.
(d) The mass-transfer coefficient KM can be used to describe the rate of dissolution of
an oxygen bubble rising in a liquid. Estimate the ratio of KM values for a given initial
radius bubble in two cases:
(i) for a bubble in molten glass at 1500K with a kinematic viscosity, ν = 104
Pa.s
(ii) for a bubble in molten copper at 1500K with ν = 10-2 Pa.s
(Tripos 2010)
Answers:
1. gas: 94%; copper: 2.6% and water: 3.1%; T = 374K and 329K; gas: 64.5%; slag:
32.5; copper: 1.9% and water: 2.2%; T = 930, 350 and 313K.
2. 56 seconds
3. 1086 Cal/day; The maximum body temperature is at the centre, T (x=0) = -16.4oC
4. 10-3
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