Chapter 15
Heat Transfer
Second Law of
Thermodynamics

Heat flows naturally from hot to
cold objects. Heat will not flow
spontaneously from cold object to
hot object.
Methods of Heat Transfer



Need to know the rate at which
energy is transferred
Need to know the mechanisms
responsible for the transfer
Methods include
• Conduction
• Convection
• Radiation
Conduction

The transfer can be viewed on an
atomic scale
• It is an exchange of energy between
microscopic particles by collisions
• Less energetic particles gain energy
during collisions with more energetic
particles

Rate of conduction depends upon the
characteristics of the substance
Conduction example




The molecules vibrate
about their equilibrium
positions
Particles near the stove
coil vibrate with larger
amplitudes
These collide with
adjacent molecules and
transfer some energy
Eventually, the energy
travels entirely through
the pan and its handle
Conduction, cont.

In general, metals are good
conductors
• They contain large numbers of electrons
that are relatively free to move through
the metal
• They can transport energy from one
region to another

Conduction can occur only if there is
a difference in temperature between
two parts of the conducting medium
Conduction, equation


The slab allows
energy to transfer
from the region of
higher temperature
to the region of
lower temperature
Rate of heat
transfer (L=x)
Q
T
 kA
t
L
Conduction, equation
explanation




A is the cross-sectional area
L = Δx is the thickness of the slab
or the length of a rod
Q is in Joules and t is in seconds
k is the thermal conductivity of the
material
• Good conductors have high k values
Good:
copper k=385
W/m·°C
and good
insulators
have
low k values
Intermediate: concrete k=0.8 W/m ·°C
Insulator: air k=0.024 W/m ·°C
Example
3mX15m concrete wall is 30 cm thick.
Outside surface –10 C, inside 20 C.
What rate does heat transfer through
it? (k=40x10-4 cal/cm·s·°C for concrete)
Example
What is equivalent thickness of glass
wool? (k=9.3x10-5 cal/cm·s·°C for glass
wool)
Glass wool is a much better
insulator than concrete!!
Example
30 cm concrete house wall with 2.0 cm
wood paneling. 10 C outside and 18C
inside.What is temperature @
wood/concrete interface?
(k=2.4x10-4 cal/cm·s·°C for wood)
Home Insulation

Substances are rated by their R
values
• R = L / k; L in inches,
• k in BTU·in/ft2·hour·°F


For multiple layers, the total R
value is the sum of the R values of
each layer
Wind increases the energy loss by
conduction in a home
Glass wool: k=0.27 BTU in/(ft^2 hour °F)
R=(1/k)·thickness = 3.7·thickness
Conduction and Insulation with
Multiple Materials


Each portion will have a specific
thickness and a specific thermal
conductivity
The rate of conduction through each
portion is equal
Example
7 ft2 sheet of material allows 80 BTU to
flow through it in 40 minutes when
temperature difference across it is
35F. What is R-value for the sheet?
Convection

Energy transferred by the movement
of a substance
• When the movement results from
differences in density, it is called natural
convection
• When the movement is forced by a fan
or a pump, it is called forced convection
Convection example



Air directly above
the flame is
warmed and
expands
The density of the
air decreases, and
it rises
The mass of air
warms the hand as
it moves by
Convection applications

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Boiling water
Radiators
Upwelling
Cooling automobile engines
Algal blooms in ponds and lakes
Convection Current Example




The radiator warms
the air in the lower
region of the room
The warm air is less
dense, so it rises to
the ceiling
The denser, cooler air
sinks
A continuous air
current pattern is set
up as shown
Heat Transfer by Convection
Q
 hAT
t
• T: temperature difference
• A: area of contact
• h: convection surface coefficient
Example
Heat transfer from glass to outside
through convection. Assume window
area to be 1 m2, glass temperature 5
°C and outside temperature –15°C.
For air in contact with solid surface
h=1.5-2.5 W/m2·°C. Amount of heat
lost in a day?
Radiation



Radiation does not require physical
contact
All objects radiate energy
continuously in the form of
electromagnetic waves due to
thermal vibrations of the molecules
Rate of radiation is given by Stefan’s
Boltzmann Law
Radiation example



The electromagnetic waves carry the
energy from the fire to the hands
No physical contact is necessary
Cannot be accounted for by conduction
or convection
Radiation equation
Q
4
 AeT
t
• The power is the rate of energy transfer,
in Watts
• σ = 5.6696 x 10-8 W/m2.K4
• A is the surface area of the object
• e is a constant called the emissivity

e varies from 0 to 1
• T is the temperature in Kelvins
Example
A square steel plate 10 cm on a side at
800 C. Find the rate of heat radiated.
In one hour?
Example
Temperature of a hot plate is doubled
from 100 C to 200 C. How is the rate
of heat radiated changed?