Transcript Slide 1
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HVAC
Module 1
METHODS OF HEAT TRANSFER
CONDUCTION
CONVECTION
RADIATION
ptD_transfer.ppt
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Overview of Thermal Physics Module:
1. Thermodynamic Systems:
Work, Heat, Internal Energy
0th, 1st and 2nd Law of Thermodynamics
2. Thermal Expansion
3. Heat Capacity, Latent Heat
4. Methods of Heat Transfer:
Conduction, Convection, Radiation
5. Ideal Gases, Kinetic Theory Model
6. Second Law of Thermodynamics
Entropy and Disorder
7. Heat Engines, Refrigerators
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METHODS OF HEAT TRANSFER
energy transfer (heat, Q)
due to a temperature difference, T
CONDUCTION
CONVECTION
RADIATION
Qnet
§17.7 p591
References: University
Physics 12th ed Young &
Freedman
Environment, TE
system, T
European heat wave, 2003 ~35 000 deaths in France
Live sheep trade
Sunday, October 26,
2003
Sheep to shore ... finally
The Labor Opposition will
pursue the Government
over the cost of the "ship
of death" saga, which
ended on Friday when
50,000 Australian sheep
at sea for three months
began being unloading in
Eritrea.
4
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Heat Conduction
Conduction is heat transfer by means of
molecular agitation within a material without
any motion of the material as a whole.
If one end of a metal rod is at a higher
temperature, then energy will be transferred
down the rod toward the colder end because
the higher speed particles will collide with the
slower ones with a net transfer of energy to
the slower ones.
Q
Q
TC
TH
conduction though glass
6
For conduction between two plane surfaces (eg
heat loss through the wall of a house) the rate of
heat transfer is
steady-state
Q
Q
T TC
kA H
t
L
TC
TH
A
energy
transferred
through
slab
Q
dQ
dT
k A
dt
dx
Q
Thermal conductivity k (W.m-1.K-1)
L
heat current H = dQ/dt
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Thermal Conduction through a uniform slab
steady-state
Q
T TC
kA H
t
L
dQ
dT
k A
dt
dx
TH
TC
0
temperature gradient
L
x
dT
dx
Material
diamond
Thermal conductivity
k (W.m-1.K-1)
2450
Cu
385
Al
205
Brick
0.2
Glass
0.8
Body fat
0.2
Water
0.6
Wood
0.2
Styrofoam
0.01
Air
0.024
Thermal conductivity, k
property of the material
kdiamond very high: perfect
heat sink, e.g. for high power
laser diodes
khuman low: core temp relatively
constant (37 oC)
kair very low: good insulator
* home insulation
* woolen clothing
* windows double glazing
Metals – good conductors:
electrons transfer energy from hot
to cold
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Heat Convection
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Convection is heat transfer by mass motion of a fluid such as
air or water when the heated fluid is caused to move away
from the source of heat, carrying energy with it.
Convection above a hot surface occurs because hot air
expands, becomes less dense and rises (natural or free).
Convection assisted by breeze, pump or fan – forced
convection.
Hot water is likewise less dense than cold water and rises,
causing convection currents which transport energy.
dQ
~ h A (T2 T1 )
dt
Convection coefficient, h
T between surface and air way from surface
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http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heatra.html
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Sea & Land Breezes, Monsoons
35 oC
20 oC
11 oC
What is the role of
heat capacity, c of water and soil?
http://iqsoft.co.in/hvac/module1.html
17 oC
RADIATION
Energy transferred by electromagnetic waves
All materials radiate thermal energy in
amounts determined by their temperature,
where the energy is carried by photons of
light in the infrared and visible portions of
the electromagnetic spectrum.
Thermal radiation wavelength ranges:
IR
~ 100 - 0. 8 m
Visible ~ 0.8 - 0.4 m 800 – 400 nm
UV
~0.4 - 0.1 m
For exam: more detail than in the textbook
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Ludwig Boltzmann
(1844-1906)
All objects above absolute zero
emit radiant energy and the rate
of emission increases and the peak
wavelength decreases as the
temperature of object increases
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Thermography
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Absorption & Stefan-Boltzmann Law
Incident radiation (INTENSITY I energy passing through a square metre
every second
Iinc = P / A Iabs = a Iinc
Power absorbed by surface of an object
Qabs
dQabs
Pabs
Aa Ts 4
dt
• Surface Area, A
• Absorption coefficient, a = 0 to 1
• Stefan-Boltzmann constant
σ = 5.67 x 10-8 W.m-2.K-4
A, a
Ts
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Emission & Stefan-Boltzmann Law
Power radiated from the surface of an object
dQrad
Prad
Ae T 4
dt
A, e, T
• Surface Area, A
• Emissivity, e = 0 to 1
• Stefan-Boltzmann constant
σ = 5.67 x 10-8 W.m-2.K-4
Pnet Prad Pabs
Pnet > 0 net heat transfer out of system
Qrad
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A blackbody absorbs all the radiation incident upon it and
emits the max possible radiation at all wavelengths
(e = a = 1)
A graybody is a surface that absorbs a certain proportion of
the energy of a blackbody, the constant being constant over
the entire band of wavelengths
(0 e = a < 1)
emissivity
e
absorption coefficient (absorptivity) a
Stefan-Boltzmann constant = 5.6710-8 W.m-2.K-4
Wien’s Displacement Law
peak
b
T
Wien constant
b = 2.89810-3 m.K
Blackbody: absorbs
ALL the EMR radiation
falling on it & emits the
max possible energy
over all wavelengths
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Wien’s Displacement Law
peak
b
T
Wien constant:
b = 2.89810-3 m.K
Blackbody radiation curves show different
peak wavelengths at various temperatures
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Stefan-Boltzmann constant, = 5.67 x 10-8 W.m-2.K-4
*
*
*
*
*
*
*
*
emissivity, e = 0 to 1 Blackbody, e = 1
Absorption coefficient, a = 0 to 1
At a temperature T a = e all wavelengths
T > 700 oC visible radiation (dull red ~ 800 oC
white ~ 2000 oC)
Black surface (e ~ 1) – good emitter / absorber
Polished surface (e ~ 0.01) –
poor emitter / absorber, good reflector
Hot stars – blue
Cool stars - red
Water (e ~ 0.96)
Earth (e ~ 0.3)
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Sun (6000 K - hot!)
Earth (300 K - cold!)
Visible radiation
Infrared radiation
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Sun & Photosynthesis
What about life on
Earth if the Sun was
hotter or colder ?
TSun ~ 6000 K
peak ~ 480 nm (blue/green)
UV (ionization of molecules) ~ 9%
Visible (excite molecules)
~ 49 %
IR (warm)
~ 42%
0.1% of radiant energy captured by
chlorophyll of plants
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Emissivity, e – the nature of the surface
Summer clothing: white reflects radiant
energy better than black.
Wrap an ice-cube in black cloth
and another in aluminium foil and
place both in the sunshine. What
will happen?
Why is the pupil of the eye black?
e ~ 0.8
e ~ 0.4
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Problem D1:
Estimate the Sun’s temperature
Prad Ae T 4
Assume e = 1
Distance from Sun to Earth: RSE = 1.5 x 1011 m
Radius of the Sun: RS = 6.9 x 108 m
Solar radiation at Earth’s surface: I = 103 W.m-2
= 5.67 x 10-8 W.m-2.K-4
RS
Solution
Power radiated by Sun
RSE
Prad = I A = I 4pRSE2 = (103)(4p)(1.5x1011)2 W = 2.83 x 1026 W
Surface area of the Sun, ASun = 4pRS2 = 5.98 x 1018 m2
T 4 = Prad / (ASun e ) T = 5.4 x 103 K
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Problem D.2 Estimate the Earth’s surface Temperature TE
(assume NO atmosphere)
Solar constant Io= 1360 W.m-2
Earth albedo (reflectivity) aE = 0.3
RSE
Earth
Adisk = pRE2
Power absorbed by Earth:
Pabs = (1-aE) AdiskIo
Power radiated by Earth:
Prad = AE TE4
Pabs = Prad TE = 255 K = -18 oC
e=1
radiation emitted from
the surface of a sphere
AE = 4pRE2
What is natural greenhouse effect?
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Greenhouse Effect
Earth’s albedo (reflectivity)
aE = 0.3
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American Journal of Physics: Feb 1983
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Selective surfaces
Emissivity, e – the nature of the surface
Value of e is temperature and wavelength dependent
e
Selective
surface
used in
solar
collectors
1
0
short
Good emitter / absorber at short
wavelengths
long
Poor emitter / absorber
at long wavelengths
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1
Black paint
1
Black skin
e
e
White paint
0
White skin
0
short
long
short
long
Black skin absorbs heat more readily than white skin but
also radiates heat more readily - the heat balance
favours black skin in the tropics
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Problem D.3
Steel plates are placed on a non-conducting opaque surface,
normal to incident solar radiation (direct + diffuse) of
750 W.m-2.
Neglecting convection, calculate the equilibrium temperature
T for a polished steel plate (e = 0.07) and a dull steel plate (e
= 0.8). Assume graybody behaviour.
Calculate the effect of coating a steel plate with a selective
surface
e = 0.92 < 2 m
short
e = 0.10
> 2 m
long
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Solution
Identify / Setup
Prad Ae T 4
Pabs = a I A I = 750 W.m-2
e=a
Equilibrium: Pabs = Prad
Execute
a I A e I A Ae T
0.25
I
T
T 339 K = 66 oC
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The equilibrium temperature is
the same for both surfaces –
the equilibrium temperature is
unaffected by the area and
nature of the surface if the
emissivity (and absorptivity)
remain constant over the range
of temperature & wavelength.
Assume Sun a blackbody at 5800 K
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Absorbed: virtually all the incident
radiation lies at < 2 m thus
Pabs = e I A = (0.92)(750)A = 690A
Radiated: most of the radiation will
be at wavelengths > 2 µm
e
short
Prad = e A T4 = 0.1 A T4
long
0.92
Equil Pabs = Prad
0.10
T = 591 K = 317
oC
250
oC
hotter
2 m
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Problem D.4
An igloo is a hemispherical enclosure built of ice. Elmo’s igloo has an inner
radius of 2.55 mm and the thickness of the ice is 0.30 m. This thickness
can be considered small compared to the radius. Heat leaks out of the
igloo at a rate determined by the thermal conductivity of ice,
kice = 1.67 W.m-1.K-1.
At what rate must thermal energy be generated inside the igloo to maintain
a steady air temperature inside the igloo at 6.5 oC when the outside
temperture is -40 oC?
Ignore all thermal energy losses by conduction through the ground, or any
heat transfer by radiation or convection or leaks.
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Solution
Identify / Setup
thickness t = 0.30 m
-40 oC
6.5
oC
kice = 1.67 W.m-1.K-1
radius r = 2.55 m
The rate of energy production must be equal to the rate of
loss of thermal energy by conduction through the
hemispherical ice wall.
Rate of energy transfer by conduction
dQ
dT
k A
dt
dx
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dQ/dt = ? W
k = 1.67 W.m-1.K-1
dT = {6.5 – (– 40)} oC = 46.5 oC
thickness of ice, dx = 0.30 m
area, A = surface area of hemisphere = (4 p R2) / 2 = 2 p R2
Because the thickness of the ice is much smaller than either the inside or outside
radius, it does not matter which radius is used – taking the average radius
R = (2.55 + 0.15) m = 2.70 m
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Execute
dQ
dT
k A
dt
dx
dQ/dt = – (1.67)(2p)(2.70)2(46.5)/0.30
dQ/dt = – 1.2 104 W
Evaluate
sensible value
units
significant figures
did I answer the question ?
W
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Problem D.5
Suppose a human could live for 120 min unclothed
in air at 8 oC.
How long could they live in water at 8 oC?
Problem D.5
Suppose a human could live for 120 min unclothed in air at 8 oC. How
long could they live in water at 8 oC?
Solution
Identify / Setup
Thermal conductivities kair = 0.024 W.m-1.K-1 kwater = 0.6 W.m-1.K-1
Execute
Q
T
k A
t
x
tW k A
t A kW
Q
T
kW A
tW
x
k
tW t A A
kW
&
Q
T
k A A
t A
x
0.024
(120)
min 4.8 min
0.6
Evaluate
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Problem D.6
Why do droplets of water dance over the very hot pan ?
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Why do droplets of
water dance over
the very hot pan ?
Water at the bottom of the drops is evaporated and
provides insulation against further evaporation.
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Problem D.7
An aluminium pot contains water that is kept steadily boiling (100 ºC). The
bottom surface of the pot, which is 12 mm thick and 1.5104 mm2 in area,
is maintained at a temperature of 102 °C by an electric heating unit. Find
the rate at which heat is transferred through the bottom surface. Compare
this with a copper based pot. The thermal conductivities for aluminium and
copper are
kAl = 235 W.m-1.K-1
and
kCu = 401 W.m-1.K-1
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Solution
Identify / Setup
Base area
A = 1.5x104 mm2
= 1.5x10-2 m2
TC = 100 oC
dQ
dT
k A
dt
dx
TH = 102 oC
Base thickness
L = 12 mm = 12x10-3 m
kAl = 235 W.m-1.K-1
Execute
kCu = 401 W.m-1.K-1
Al, dQ/dt = 5.9x102 W
dT/dx = (TH – TC) / L
dQ/dt = ? W
Cu, dQ/dt = 1.0x103 W
Cu pots ~ 2 times more efficient
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“Body Heat”
convection
Heat lost by
is very important for humans
For a naked person, h ~ 7.1 W.m-2.K-1
Assume the person’s surface area is 1.5 m2, skin
temperature of 33 oC and surrounding temperature 29 oC.
A = 1.5 m2
T = 33 oC
TE = 29 oC
dQ/dtconvection = h A T = (7.1)(1.5)(4) W = 43 W
If there is a breeze, convection losses are greater – wind
chill factor
Viscosity of fluids slows natural convection near a stationary surface by producing
a boundary layer which has about the same insulating values of 10 mm plywood.
For the naked person, estimate the net rate of energy
radiated.
A = 1.5 m2 T = 33 oC = (33 + 273) K = 306 K
TE = (29 + 273) K = 302 K
assume e = 1 = 5.67x10-8 W.m-2.K-4
Pradiation = e A (T 4 – TE4) = 39 W
dQ/dtloss = dQ/dtradiation + dQ/dtconvection = 39 W + 43 W = 82 W
Ans. similar to the rate at which heat is generated by the
body when resting
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CONVECTION
Why not heat the
water at the top?
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46
CONVECTION
Forced convection
Warm air rises above the
ground
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47
CONVECTION
Why do we get this
pollution haze?
Temperature inversion
prevents air rising and the
dispersing the pollution
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48
CONVECTION
In bathrooms, the
heater is often near
the ceiling. Problem ?
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Why are the cooling
coils at the top ?
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RADIATION
f peak bf T
Wien constant
bf = 2.83 kB/h s-1.K
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RADIATION
Why are pipes in solar panels painted black ?
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51
RADIATION
Highly reflecting metal foil keeps
inside temperature low.
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52
RADIATION
Lemur at left is nocturnal,
so the dark fur poses no
disadvantage in absorbing
excessive sunlight.
Lemur at the right is active
during the day; it points
its belly toward the sun on
cold mornings.
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Think about
* Two different materials at the same temperature have different emissivities.
Which one glows the brightest?
* Some animals have hair which is composed of solid tubular strands, while
others have hollow, air-filled tubes. Where would one more likely find the latter
animal: In cold climates, or warm?
* Steel reinforcement bars add stability to concrete walls. Do they also enhance
the insulating value of concrete?
* Wind chill factor
* Drapes hung close to a cold window
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* Clothing
* Should you lower the blinds and draw the curtains on a hot day?
* When one steps from a shower on a cold morning, why does
the tile floor seem so much colder than the air?
* Place a wooden spoon and a metal spoon in the freezer. Which will
cool faster? After several hours, what would they feel like?
* Why do people become "flushed" when overheated?