Transcript Lecture 1 – Putting Safety Into Perspective

Heat Transfer
Mechanisms
Dr. AA
Department of Chemical Engineering
University Teknology Malaysia
Conduction
Dr. AA
Department of Chemical Engineering
University Teknology Malaysia
Heat Conduction
Key Question:
How does heat pass
through different
materials?
Heat Transfer
• The science of how heat flows is called heat
transfer.
• There are three ways heat transfer works:
conduction, convection, and radiation.
• Heat flow depends on the temperature
difference.
Thermal Equilibrium
• Two bodies are in
thermal equilibrium with
each other when they
have the same
temperature.
• In nature, heat always
flows from hot to cold
until thermal equilibrium
is reached.
Heat Conduction
• Conduction is the transfer of heat through
materials by the direct contact of matter.
• Dense metals like copper and aluminum are
very good thermal conductors.
Heat Conduction
• A thermal insulator is a material that conducts
heat poorly.
• Heat flows very slowly through the plastic so
that the temperature of your hand does not rise
very much.
Heat Conduction
• The ability to conduct
heat often depends
more on the structure of
a material than on the
material itself.
– Solid glass is a thermal
conductor when it is
formed into a beaker or
cup.
– When glass is spun into
fine fibers, the trapped
air makes a thermal
insulator.
Thermal Conductivity
• The thermal conductivity of a material describes
how well the material conducts heat.
Thermal Conductivity
• Heat conduction in
solids and liquids
works by
transferring energy
through bonds
between atoms or
molecules.
Heat Conduction Equation
Thermal conductivity
(watts/moC)
Heat flow
(watts)
Area of cross section (m2)
PH = k A (T2 -T1)
L
Temperature
difference (oC)
Length (m)
Variables for conduction
Convection
Dr. AA
Department of Chemical Engineering
University Teknology Malaysia
Convection
Key Question:
Can moving matter
carry thermal energy?
Convection
• Convection is the transfer
of heat by the motion of
liquids and gases.
– Convection in a gas occurs
because gas expands when
heated.
– Convection occurs because
currents flow when hot gas
rises and cool gas sink.
– Convection in liquids also
occurs because of
differences in density.
Convection
• When the flow of gas or
liquid comes from
differences in density
and temperature, it is
called free convection.
• When the flow of gas or
liquid is circulated by
pumps or fans it is
called forced
convection.
Convection
• Convection depends on
speed.
• Motion increases heat
transfer by convection in
all fluids.
Convection
• Convection depends on
surface area.
• If the surface contacting
the fluid is increased,
the rate of heat transfer
also increases.
• Almost all devices made
for convection have fins
for this purpose.
Forced Convection
• Both free and forced convection help to
heat houses and cool car engines.
Heat Convection Equation
Heat transfer coefficient
(watts/m2oC)
Heat flow
(watts)
Area contacting fluids (m2)
PH = h A (T2 -T1)
Temperature
difference (oC)
Radiation
Dr. AA
Department of Chemical Engineering
University Teknology Malaysia
Radiant Heat
• We do not see the
thermal radiation
because it occurs at
infrared wavelengths
invisible to the human
eye.
• Objects glow different
colors at different
temperatures.
26.3 Radiant Heat
• A rock at room
temperature does not
“glow”.
• The curve for 20°C
does not extend into
visible wavelengths.
• As objects heat up
they start to give off
visible light, or glow.
• At 600°C objects glow
dull red, like the
burner on an electric
stove.
Radiant Heat
• As the temperature rises, thermal
radiation produces shorterwavelength, higher energy light.
• At 1,000°C the color is yelloworange, turning to white at
1,500°C.
• If you carefully watch a bulb on a
dimmer switch, you see its color
change as the filament gets
hotter.
• The bright white light from a bulb
is thermal radiation from an
extremely hot filament, near
2,600°C.
Radiant Heat
• The graph of power
versus wavelength
for a perfect
blackbody is called
the blackbody
spectrum.
Radiation striking a solid surface has one of three
fates:
1.
2.
3.
Absorption absorptivity (a)
Transmission transmissivity
(t)
Reflection reflectivity (z)
How are these properties related ?
a + t +z=1
Two special cases require definition:
If all of the energy is either reflected or absorbed (no
transmitted radiation), we define the body as
Opaque a + z = 1
If all of the energy striking a surface is absorbed, we define
the body as
Black body a = 1
For heat transfer calculations, we often assume that the
properties a, t, and r are independent of wavelength.
When this assumption is made we say that we have gray
surfaces.
Let us return to the subject of radiation
emitted by a surface.
Total emissive power is defined as the total amount of
energy leaving the surface per unit time per unit area:
W = energy/area-time
[Btu/hr-ft2 or W/m2]
Note: Emissive power is a function of wavelength. The
important wavelengths for heat transfer are 0.5 -50 µm. For
temperatures above 1500°F, the important wavelength
range is between 0.5 and 5 µm . In our analysis, we will use
the average values over all wavelengths.
Emissivity
The emissivity is the ratio of the emissive power of a
surface compared to the
maximum emissive power.
 =
W
WBlack Body
How does the emissivity relate to the
absorptivity (a) at thermal equilibrium?
=a
Although this strictly applies at thermal equilibrium, we
normally assume that it applies at all temperatures.
Stefan-Boltzman Law
Finally, we must ask how the emissive power of a black body
is related to temperature. The answer is provided by the
Stefan Boltzman Law.
W = sT4
where s = 0.1714 x 10-8 Btu/hr-ft2-°R4
constant)
= 5.676 x 10-8 W/m2-K4
(Stefan-Boltzman
For an object that is not a black body (i.e., not a perfect
radiator), we can write the following expression:
W = sT4
T is absolute temperature
Heat Transfer Equation
To calculate the heat transfer rate by radiation, we must
include terms for
energy output and energy received from the surroundings.
Energy
output:
 sT
Energy
input:
4
s
s aT
4

Making the usual assumption that  = a, and multiplying by
area yields:

q = sA T  T
4
s
4

)
This is the expression for an object totally enclosed by
surroundings at T∞.
View factors
View Factor
Previously, we found that for a body totally enclosed by its
surroundings, the net rate of heat transfer by thermal
radiation is given by the following expression:
q = sA(Ts4 - T24)
The equation for q given above is one of the most
important and commonly used results, however, it does not
cover all situations.
View Factor
Suppose we have two surfaces at temperature T1 and T2, but
both are finite in area, and neither surface is completely
enclosed by the other.
An example might be the floor and ceiling of a room. Only a
fraction of the energy leaving the ceiling strikes the floor and
vice versa.
To account for this incomplete exchange of energy, we
define the view factor, F12:
F12 = fraction of energy leaving A1 reaching A2
View Factor
The calculation of view factors is a straightforward exercise in calculus
as shown in the figure on the preceding page. For each point on the
surface A1, we consider rays of thermal energy emanating out equally
in all directions. The fraction of these rays (actually, the total solid
angle) which strikes A2 gives the fraction of energy reaching that
surface. Integrating over all points on surface A1 and averaging gives
the view factor F12.
The following relationship is true:
A1F12 = A2F21
What is the energy transfer rate
from 1 to 2 and vice versa?
q1->2 =
s AF T
q2->1 =
s A2 F T
4
1 12 1
4
21 2
Analytical Expression of View Factor
Case 1: Differential surface parallel to a finite rectangular surface
L1
L2
D
1  X
Y
Y
X 
1
1
F12 =
tan
+
tan


2
2
2
2
2  1+ X
1+ X
1+ Y
1+ Y 
where X=L1/D and Y=L2/D
Analytical Expression of View Factor
b
a
c
Case 2: Plane circular surface
with common central normal
1+ B + C 
2
F12 =
3
1+ B
2B
2
+ C ) 4B C
2
where B = b/a and C = c/a
3
2
2
Analytical Expression of View Factor
Case 3: Plane element A1 to sphere of radius r2;
normal to centre of element passes through
centre of sphere
 r2 
F12 =  
h
2