Heat Transfer-1 (color)

Materials Engineering Science
MESc. 5025
Instructor: Herve Marand
Chapter 6.
Heat Transfer

• There are three modes of heat transfer:
– Conduction
– Convection
– Radiation
Heat Transfer
• In its simplest form, the rate of heat transfer is equal to the product
of a driving force and a thermal conductance.
• Thermal conductances vary for each mode of heat transfer
• Conduction:
Conduction is accomplished via two mechanisms:
– Molecular Interactions:
Molecules in high energy levels (high T) impart energy to
adjacent molecules at lower energy levels.
This type of heat transfer necessitates the presence of molecules
of solid, liquid or gas and the existence of T gradient.

– Heat Transfer via Free Electrons:
This second conduction heat transfer mechanism is observed
primarily in metallic solids. The concentration of free electrons
varies considerably for metallic alloys and is very low for nonmetals.
The previous statement implies that the ability of solids
to conduct heat via this mechanism varies directly with the free
electron concentration (pure metals conduct heat the best). This
conduction mechanism also relies on the existence of a
temperature gradient.
• Theoretical Treatment of Heat Transfer by Conduction
A quantitative expression relating the rate of heat transfer, the
temperature gradient and the nature of the conducting medium
is attributed to Fourier (1822)
q x dT
= − k temperature
A dx gradient along x
Heat flux= heat thermal
flow rate per unit conductivity
area in direction x of material
A: area normal to
heat flow

• Units:
quantity S.I. unit non-S.I. unit
q Watt or J.s-1 Btu/hr
A m 2 ft 2
dT/dx K.m -1 °F.ft -1
k W. K -1 .m -1 Btu.hr -1 .ft -1 .°F -1
• Unit Conversion Factor:
1 Watt = 1 J.s -1 = 3.414 Btu.hr -1
1 Btu = 1054 Joules
• For isotropic materials and steady state conditions, the
Fourier equation can be generalized as:
q
A
= − k∇ T
k = k e + k i
- sign accounts for heat
transfer from hot to cold
molecular interactions
free electrons
⎛
⎜
⎜
∇ = ⎜
⎜
⎜
⎝
⎞
⎟
x
⎟
⎟
y⎟
⎟
z ⎠
is the
gradient
vector

• In most solids, the principal mode of thermal energy
assimilation is by increase in vibrational energy of the
atoms. Atoms in solids are constantly vibrating at very
high frequencies and with relatively small amplitudes. The
atomicvibrations of adjacent atoms are coupled through the
atomic bonding. These vibrations are coordinated in such
a way that traveling lattice waves are produced. (These
may be thought of as elastic waves or simply sound waves
(short wavelength and high frequency), which propagate
through a crystal at the velocity of sound. The vibrational
thermal energy for an ordered material consist of a series
of elastic waves (with a distribution of frequencies). Only
certain energies are allowed (quantization of the
vibrational energy). A single quantum of vibrational
energy is called a phonon. Note that phonon waves can
scatter free electrons during electronic conduction, thus
they control the temperature dependence of electronic
conduction

• In metals, the free electron mechanism of heat transport is
much more efficient than the phonon mechanism, because
phonons are more easily scattered than free electrons and
electrons ahave higher velocities.
• Thermal energy associated with phonons is transported in
the direction of their motion (net movement of phonons
from high to low temperature regions)
• Free electrons in hot regions have higher kinetic energy
tham these in cold regions. When they migrate to colder
regions, their kinetic energy is transferred to atoms as
vibrational energy (collision of electrons with phonons).
• Thermal conductivities of common pure metals or alloys
range from 20 to ca. 400 W.m -1 .K -1 . When impurities are
are present in a metal or alloying is used, the impurity
atoms act as scattering centers for the electrons, decreasing
the efficiency of electron motion (i.e. k e)

• In ceramics k e

• Polymers:
Thermal conductivity for most polymers is of the order of
0.3 W.m -1 .K -1 . Energy transfer is accomplished by the
vibrations and rotations of chain backbone and side chain
bonds.
The magnitude of the thermal conductivity increases with
the degree of crystallinity.
Polymers are used as low temperature thermal insulators
(porous polymershave enhanced insulative properties).
• Gases:
Thermal conductivity is low but increases with
temperature due to increased Brownian motion resulting in
a higher frequency of contact and increase in molecular
exchange rates. For dilute gases (low pressure, P

Kinetic Theory of Gases:
assumes ideal gas behavior.
Chapman-Erskog Theory:
accounts for pairwise interactions
based on the Lennard-Jones potential.
T: absolute temperature
m: molecular mass
M: molar mass
d: molecular diameter
k b: Boltzmann constant
Ω k, σ: related to Lennard-Jones potential
k =
k =
1
3 / 2 d 2
1.9891 ⋅10 −4
Ω k 2
k
3
T
b
m
T
M
see viscosity
section for
discussion on Ω k
Chapman-Erskog theory accounts better than the Kinetic
Theory for the temperature dependence of interactions.

Anisotropicity of the Thermal Conductivity
• In general, in crystalline materials, the simplified Fourier
equation is not rigorously valid asthe thermal conductivity
may be different in different directions in the crystal.
• The scalar k must be replaced by the tensor k (3x3). The
Tensor is symmetrical (Onsager’s principle)
˜
k =
˜
k =
⎛
⎜
⎜
⎝
k11 k21 k31
k12 k22 k32
k13 k23 k33
⎛ k1 0 0 ⎞
⎜ ⎟
0 k2 0
⎜
⎟
⎝ 0 0 k3⎠
⎞
⎟
⎟
⎠
where: -k ij = rate of heat transfer per
unit cross section parallel to the axis
defined by “i” from a temperature
gradient along the “j”-axis
if the crystal axes are matching the
laboratory axes
Note: that q and ∇ T
will generally not be parallel unless the crystal
is cubic (i.e. k1=k2=k3)

Example Calculation #1
• Steam is transported through a 11/2 “ mild steel pipe. The inside
and outside pipe wall temperatures are 205°F and 195°F,
respectively. OD = 1.9” ; ID = 1.5”; k = 24.8 Btu.hr -1 .ft -1 .°F -1
– 1) Find the heat loss for 10 feet of pipe
– 2) FInd the heat flux based upon inside and outside surface
areas.
• Heat is transferred radially. The applicable scalar form of the
Fourier rate equation is:
qr = −kA
For the steady state case, qr is constant
dT ⎫
dr ⎬
A = 2 rL
⎪
⇒ qr = −k( 2 rL)
⎪ ⎭
dT
dr
rodr
To
qr ∫ = −2 kL ∫ dT ⇒ qr ln
r
ro ⎛ ⎞
⎜ ⎟ = −2 kL( To − Ti )
⎝ ⎠
r i
T i
r i

( )
qr = 2 kL Ti − To ln r ⎛ o ⎞
⎜ ⎟
⎝ ⎠
q r
A i
qr
A o
=
=
r i
65, 000
⋅ 1.5 ⎛ ⎞
⎜ ⎟ ⋅10
⎝ 12 ⎠
65, 000
⋅ 1.9 ⎛ ⎞
⎜ ⎟ ⋅10
⎝ 12 ⎠
= 2 ⋅ 3.14 ⋅ 24.8 ⋅ 10 ⋅10
ln 1.9 ⎛ ⎞
⎜ ⎟
⎝ 1.5 ⎠
The heat flux is q/A (rate of heat loss per unit area)
= 16, 700 Btu/hr.ft 2
= 13,150 Btu/hr.ft 2
= 65, 000 Btu/hr

Example Calculation #2
• For steady state heat conduction through a wall with
dimensions and surface temperatures shown below, express:
T 0
x=0
– 1) the heat flow rate for a constant thermal conductivity k.
– 2) the heat flow rate, when the thermal conductivity of the
wall material varies linearly with temperature k = k 0(1+βT).
x
x=L
T L
• 1)
• 2)
q x dx = −kAdT
L
q x ∫ dx = −kA ∫ dT
0
T L
T 0
( )
q x L = −kA T L − T 0
q x = kA
L T ( L − T0) q x = −k0 ( 1 + T)A
dT
dx
q x = k0 A
L 1 + 2 T0
⎡
⎢ + TL
⎣
( )
⎤
⎥ ⎦
TL − T0
[ ]

Convection
• Convection Heat Transfer: involves the exchange of heat
between a fluid and a surface or an interface.
• Two kinds of convection exist:
– Forced convection in which fluid motion past a surface
is caused by an external agency
– Free convection in which density changes in the fluid
which result from energy exchange lead to a natural
(i.e.unassisted) fluid motion.
• Newton (1701) first expressed the basic rate equation for
convective heat transfer (Newton’s law of cooling).
rate of convective
heat transfer (Btu/hr)
q = hA( Tsurf − T ) fluid temperature
driving force
convective heat
transfer coefficient
(Btu.hr -1 .ft -2 .°F -1 )
area normal to the
direction of heat flow

• Regardless of the flow phenomenon involved, it is known that
directly adjacent to the interface, the energy transfer mechanism is
that of conduction. The surface conductive layers of fluid control
the heat transfer rate and determine the value of “h”, which is
justifiably often called the “film coefficient”. The only difficulty in
describing convective phenomena lies in the evaluation of “h”.
• If we write : q = KΔT where K is the thermal conductance.
K convection = hA
K conduction hollow cylinder =
K conduction plane wall = kA
L
2 kL
ln r ⎫
⎪
⎪
⎪
⎪
⎬ Btu / hr.°F
⎛ o ⎞ ⎪
⎜ ⎟
⎝ ri ⎠ ⎪
⎪
⎪
⎭ ⎪

• Range of values for “h” allows the convective mechanism
to be understood
– mechanism h (Btu.hr -1 .ft -2 .°F -1 )
condensing water vapor 1,000 to 20,000
boiling water 500 to 5,000
forced convection (H 2O) 50 to 3,000
forced convection (air) 5 to 100
free convection 1 to 10

Example Calculations #3
Given the steel pipe with conditions described in Example #1, with
10°F air surrounding the pipe and 210°F steam flowing on the
inside, evaluate the convective heat transfer coefficients on each of
the pipe surfaces and indicate the operating mechanisms.
1) q r/A i = 16,700 Btu.hr -1 .ft -2
T steam - T surface = 210 -205°F
h = 16,700 / 5 = 3,340 Btu.hr -1 .ft -2 .°F -1
Steam is condensing on the inside of the pipe
2) q r/A o = 13,150 Btu.hr -1 .ft -2
T surface - T air = 195 -10°F
h = 13,150 / 185 = 71 Btu.hr -1 .ft -2 .°F -1
Air must be in forced convection

Thermal Radiation
• Heat transfer by radiation requires no medium for
propagation. Radiant exchange between surfaces is in fact
maximum when no material occupies the intervening
space. This type of energy transfer by radiation is an
electromagnetic phenomenon and the exact nature still not
completely understood.
• A perfectly emitting or absorbing body is called a black
body. The rate at which it emits radiant energy is given by
the Stephan-Boltzmann law of Thermal radiation:
radiant emission in Btu/hr
q
A
= T 4
Absolute
Temperature
area of emitting surface in ft 2 Stephan-Boltzmann constant