Transfer
Content
New Era University – Virtual Learning Environment
College of Engineering & Technology
PHYSICS
271
Mechanisms of
Heat Transfer
4/15/16
OBJECTIVES:
At the end of this module, the student
should be able to:
1. Describe heat as a means of energy
transfer and other energy transfer
methods.
2. Calculate the rate of energy transfer
through
a given material and other quantities as
required in the given heat transfer
equations.
For use in Physics 271 at NEU-CET
2
4/15/16
MECHANISM OF HEAT
TRANSFER
SUB-TOPICS
•
Conduction
•
Convection
•
Radiation
•
Sample Problems
•
Assessment Questions
For use in Physics 271 at NEU-CET
3
HEAT TRANSFER
Conduction – Transfer of heat
from cooler to hotter object in
contact due to rapidly moving
molecules
Convection – Transfer of heat
due to entirely to the movement
of heated fluid
Radiation – Heat transfer due to
electromagnetic waves
<
^
>
For use in Physics 271 at NEU-CET
CONDUCTION
P = Q/t = kAΔT/
Q/t
Δx
T
2
A
T1
Δx
T1<T2
Where : Q = amount of heat (J)
t = time (s)
k= thermal conductivity
A= cross sectional area (m2)
ΔT=temp. Diff
Δx = thickness of material
<
^
>
For use in Physics 271 at NEU-CET
CONDUCTION
• Rate of thermal conduction depends on the
properties
of the substance being heated
•Conduction occurs only if there is difference in
temperature between two parts of the
conducting
medium
•Substances that are good thermal conductors
have
larger k
•The rate of change of temperature along the
material is called temperature gradient (ΔT/
Δx)
< ^ >
For use in Physics 271 at NEU-CET
•The ratio Δx/kA is called thermal resistance
CONDUCTION RPOBLEM
A copper bar 2 m long has a circular crosssection of radius 1 cm. One end is kept at
100°C and the other end is kept at 0 °C.
Surface of the bar is insulated so that there is
negligible heat loss. (kcu
=397W/m°C). Find
cu
a. Thermal resistance
b. Thermal gradient
c. Thermal current
d. Temperature of the bar, 0.25 m from hot end.
<
^
>
For use in Physics 271 at NEU-CET
CONDUCTION PROBLEM
For use in Physics 271 at NEU-CET
CONVECTION
P = Q/t = hAΔT
Where : Q = amount of heat (J)
t = time (s)
h= convection coefficient
A= cross sectional area
ΔT=temp. Diff
<
^
>
For use in Physics 271 at NEU-CET
NATURAL CONVECTION
For use in Physics 271 at NEU-CET
NATURAL CONVECTION
For use in Physics 271 at NEU-CET
NATURAL CONVECTION
For use in Physics 271 at NEU-CET
RADIATION
P = Q/t = σAeT4
Where : Q = amount of heat
t = time
σ = 5.67 x 10-8 W/m2k4
A= cross sectional area
e = emissivity
T=temp. In Kelvins
<
^
>
For use in Physics 271 at NEU-CET
RADIATION
For use in Physics 271 at NEU-CET
Radiation emitted by objects
•All objects that have a temperature greater
than 0 K emit radiation
•Hot objects emit more radiation that colder
objects
Q: How much radiation is being emitted by an
object, and at what wavelengths?
For use in Physics 271 at NEU-CET
Black Body Radiation
Black Body - any object that is a perfect emitter
and a perfect absorber of radiation -->>
object does not have to appear "black"
sun and earth's surface behave
approximately as black bodies
For use in Physics 271 at NEU-CET
Stefan-Boltzman Law
The Sefan-Boltzman law relates the total amount of
radiation emitted by an object to its temperature:
E=sT4
where:
E = total amount of radiation emitted by an object per square
meter (Watts m-2)
s is a constant called the Stefan-Boltzman constant = 5.67 x 10 -8
Watts m-2 K-4
T is the temperature of the object in K
Consider the earth and sun:Sun: T = 6000 K
so E = 5.67 x 10-8 Watts m-2 K-4 (6000 K)4 = 7.3 x 107
Watts m-2
this a lot of radiation!!!
Earth: T = 288K
so E = 5.67 x 10-8 Watts m-2 K-4 (288 K)4 = 390 Watts m-2
Q: If you double the temperature of an object, how much
For use in Physics 271 at NEU-CET
more radiation will it emit?
QUESTIONS FOR THOUGHT:
1. Why does it get colder on clear nights
versus
cloudy nights?
2. Is the atmosphere a blackbody? Why
or why
not?
3. Which would have the greatest effect
on the
earth's greenhouse effect: Removing
all of the
CO2 from the atmosphere or removing
all of the
water vapor? Explain.
use in Physics 271 at NEU-CET
4. Explain whyForan
increase in cloud cover
RADIATION
A student is trying to decide what to
wear. The surroundings (his bedroom)
are at 20°C. If the skin temperature of
the unclothed student is 35°C, what is
the net energy loss from his body in
10.0 minutes by radiation? Assume that
the emissivity of the skin is 0.900 and
that the surface area of the student is
1.50 m22.
<
^
>
For use in Physics 271 at NEU-CET
Given: T1 = 20 °C +273 = 293K
T2 = 35°C +273 = 308K
A = 1.50 m2
t = 10 minutes
e = 0.900
This is heat transfer by radiation and the formula
to recall is
4
SolvingQ/t
for =
Q,eδAT
Q = eδAT4 t
= (0.900)(5.67 x 10-8 W/(m 2K 4 )( 1.50 m2)
[(308K)4-(293K)] 4] (10minx60s/1min)
Q = 74,820.95 J or 7.5x104 J
To visualize the amount of heat released, let us convert
this to calories:
74,820.95 J x ( 1 cal / 4.186 J) = 17,874.1 cal
= 17.87 kcal
ASSESSMENT:
1. The process of heat transfer in which heat is transmitted by actual
mass motion of the molecules from one place to another is called
a. convection
b. conduction
c. radiation
d. all of these
2. A thermos bottle is a double walled pyrex glass with silver walls. Silver
is a very
good reflector with low emissivity. The silver walls minimize energy
transfer by
a. convection
b. conduction
3. It is a mechanism of heat transfer that requires no material medium
c. radiation
d. all of these
for their
passage.
a. convection
b. conduction
c. radiation
d. all of these
4. They are considered poorest conductors because molecules are far
apart.
a. metals
b. gas
c. liquid
d. solid
5. It is a transfer of heat due entirely to the gross movement of the
heated fluid.
a. convection
b. conduction
c. radiation
d. all of these
For use in Physics 271 at NEU-CET
ASSESSMENT:
6. A copper bar, 2m long. One end is kept at 0 C ° and the other end is
kept at 100
C °. The rate of change of temperature along the bar is called
a. thermal resistance
b. temperature gradient
c. thermal current
d. thermal conductivity
7. In #6, the copper bar (k=397W/m∙ °C) has a circular cross-section of
radius 1 cm.
Surface of the bar is insulated so that there is negligible heat loss. Its
thermal
current is
a. 63 W
b. 6.3 W
8. Which device/instrument is NOT governed by the principle of heat
c. 0.63 W
d. 630 W
transfer ?
a. metal foil covering of the satellites
b. thermos bottle
c. leather
shoes
d. halogen cooktop stove
9. Which
is NOT
a perfect blackbody
a. sun
c. earth’s atmosphere
b. earth’s surface
d. moon
10. It is a perfect emitter and perfect absorber of radiation
a. solid objects
b. liquid objects
c. black body
d. gaseous elements
For use in Physics 271 at NEU-CET
Assessment:
These questions are of low and medium level of
cognitive complexity to measure the learning of the
students after the discussion. When you score
below 9 points, you should review this topic and
understand the concept of mechanisms of heat
transfer
before
proceeding
to
solve
questions/exercises of higher level of cognitive
complexity.
For use in Physics 271 at NEU-CET
College of Engineering & Technology
PHYSICS
271
Mechanisms of
Heat Transfer
4/15/16
OBJECTIVES:
At the end of this module, the student
should be able to:
1. Describe heat as a means of energy
transfer and other energy transfer
methods.
2. Calculate the rate of energy transfer
through
a given material and other quantities as
required in the given heat transfer
equations.
For use in Physics 271 at NEU-CET
2
4/15/16
MECHANISM OF HEAT
TRANSFER
SUB-TOPICS
•
Conduction
•
Convection
•
Radiation
•
Sample Problems
•
Assessment Questions
For use in Physics 271 at NEU-CET
3
HEAT TRANSFER
Conduction – Transfer of heat
from cooler to hotter object in
contact due to rapidly moving
molecules
Convection – Transfer of heat
due to entirely to the movement
of heated fluid
Radiation – Heat transfer due to
electromagnetic waves
<
^
>
For use in Physics 271 at NEU-CET
CONDUCTION
P = Q/t = kAΔT/
Q/t
Δx
T
2
A
T1
Δx
T1<T2
Where : Q = amount of heat (J)
t = time (s)
k= thermal conductivity
A= cross sectional area (m2)
ΔT=temp. Diff
Δx = thickness of material
<
^
>
For use in Physics 271 at NEU-CET
CONDUCTION
• Rate of thermal conduction depends on the
properties
of the substance being heated
•Conduction occurs only if there is difference in
temperature between two parts of the
conducting
medium
•Substances that are good thermal conductors
have
larger k
•The rate of change of temperature along the
material is called temperature gradient (ΔT/
Δx)
< ^ >
For use in Physics 271 at NEU-CET
•The ratio Δx/kA is called thermal resistance
CONDUCTION RPOBLEM
A copper bar 2 m long has a circular crosssection of radius 1 cm. One end is kept at
100°C and the other end is kept at 0 °C.
Surface of the bar is insulated so that there is
negligible heat loss. (kcu
=397W/m°C). Find
cu
a. Thermal resistance
b. Thermal gradient
c. Thermal current
d. Temperature of the bar, 0.25 m from hot end.
<
^
>
For use in Physics 271 at NEU-CET
CONDUCTION PROBLEM
For use in Physics 271 at NEU-CET
CONVECTION
P = Q/t = hAΔT
Where : Q = amount of heat (J)
t = time (s)
h= convection coefficient
A= cross sectional area
ΔT=temp. Diff
<
^
>
For use in Physics 271 at NEU-CET
NATURAL CONVECTION
For use in Physics 271 at NEU-CET
NATURAL CONVECTION
For use in Physics 271 at NEU-CET
NATURAL CONVECTION
For use in Physics 271 at NEU-CET
RADIATION
P = Q/t = σAeT4
Where : Q = amount of heat
t = time
σ = 5.67 x 10-8 W/m2k4
A= cross sectional area
e = emissivity
T=temp. In Kelvins
<
^
>
For use in Physics 271 at NEU-CET
RADIATION
For use in Physics 271 at NEU-CET
Radiation emitted by objects
•All objects that have a temperature greater
than 0 K emit radiation
•Hot objects emit more radiation that colder
objects
Q: How much radiation is being emitted by an
object, and at what wavelengths?
For use in Physics 271 at NEU-CET
Black Body Radiation
Black Body - any object that is a perfect emitter
and a perfect absorber of radiation -->>
object does not have to appear "black"
sun and earth's surface behave
approximately as black bodies
For use in Physics 271 at NEU-CET
Stefan-Boltzman Law
The Sefan-Boltzman law relates the total amount of
radiation emitted by an object to its temperature:
E=sT4
where:
E = total amount of radiation emitted by an object per square
meter (Watts m-2)
s is a constant called the Stefan-Boltzman constant = 5.67 x 10 -8
Watts m-2 K-4
T is the temperature of the object in K
Consider the earth and sun:Sun: T = 6000 K
so E = 5.67 x 10-8 Watts m-2 K-4 (6000 K)4 = 7.3 x 107
Watts m-2
this a lot of radiation!!!
Earth: T = 288K
so E = 5.67 x 10-8 Watts m-2 K-4 (288 K)4 = 390 Watts m-2
Q: If you double the temperature of an object, how much
For use in Physics 271 at NEU-CET
more radiation will it emit?
QUESTIONS FOR THOUGHT:
1. Why does it get colder on clear nights
versus
cloudy nights?
2. Is the atmosphere a blackbody? Why
or why
not?
3. Which would have the greatest effect
on the
earth's greenhouse effect: Removing
all of the
CO2 from the atmosphere or removing
all of the
water vapor? Explain.
use in Physics 271 at NEU-CET
4. Explain whyForan
increase in cloud cover
RADIATION
A student is trying to decide what to
wear. The surroundings (his bedroom)
are at 20°C. If the skin temperature of
the unclothed student is 35°C, what is
the net energy loss from his body in
10.0 minutes by radiation? Assume that
the emissivity of the skin is 0.900 and
that the surface area of the student is
1.50 m22.
<
^
>
For use in Physics 271 at NEU-CET
Given: T1 = 20 °C +273 = 293K
T2 = 35°C +273 = 308K
A = 1.50 m2
t = 10 minutes
e = 0.900
This is heat transfer by radiation and the formula
to recall is
4
SolvingQ/t
for =
Q,eδAT
Q = eδAT4 t
= (0.900)(5.67 x 10-8 W/(m 2K 4 )( 1.50 m2)
[(308K)4-(293K)] 4] (10minx60s/1min)
Q = 74,820.95 J or 7.5x104 J
To visualize the amount of heat released, let us convert
this to calories:
74,820.95 J x ( 1 cal / 4.186 J) = 17,874.1 cal
= 17.87 kcal
ASSESSMENT:
1. The process of heat transfer in which heat is transmitted by actual
mass motion of the molecules from one place to another is called
a. convection
b. conduction
c. radiation
d. all of these
2. A thermos bottle is a double walled pyrex glass with silver walls. Silver
is a very
good reflector with low emissivity. The silver walls minimize energy
transfer by
a. convection
b. conduction
3. It is a mechanism of heat transfer that requires no material medium
c. radiation
d. all of these
for their
passage.
a. convection
b. conduction
c. radiation
d. all of these
4. They are considered poorest conductors because molecules are far
apart.
a. metals
b. gas
c. liquid
d. solid
5. It is a transfer of heat due entirely to the gross movement of the
heated fluid.
a. convection
b. conduction
c. radiation
d. all of these
For use in Physics 271 at NEU-CET
ASSESSMENT:
6. A copper bar, 2m long. One end is kept at 0 C ° and the other end is
kept at 100
C °. The rate of change of temperature along the bar is called
a. thermal resistance
b. temperature gradient
c. thermal current
d. thermal conductivity
7. In #6, the copper bar (k=397W/m∙ °C) has a circular cross-section of
radius 1 cm.
Surface of the bar is insulated so that there is negligible heat loss. Its
thermal
current is
a. 63 W
b. 6.3 W
8. Which device/instrument is NOT governed by the principle of heat
c. 0.63 W
d. 630 W
transfer ?
a. metal foil covering of the satellites
b. thermos bottle
c. leather
shoes
d. halogen cooktop stove
9. Which
is NOT
a perfect blackbody
a. sun
c. earth’s atmosphere
b. earth’s surface
d. moon
10. It is a perfect emitter and perfect absorber of radiation
a. solid objects
b. liquid objects
c. black body
d. gaseous elements
For use in Physics 271 at NEU-CET
Assessment:
These questions are of low and medium level of
cognitive complexity to measure the learning of the
students after the discussion. When you score
below 9 points, you should review this topic and
understand the concept of mechanisms of heat
transfer
before
proceeding
to
solve
questions/exercises of higher level of cognitive
complexity.
For use in Physics 271 at NEU-CET
