Transcript 投影片 1 - National Cheng Kung University

Part Three: Thermodynamics
Steam turbine converts the energy of high-pressure steam to mechanical energy and then electricity.
World energy consumption (2008): ~ 151012 (15 trillion / tera) watts., mostly from fossil fuels
Applications of the laws of thermodynamics:
• Combustion engines.
• Sun  Heat flows on Earth  Climate.
• Global warming.
• Big bang: Heat flow in the universe.
List of countries by electricity consumption
Average power per
capita (watts per
person)
Rank
Country
Electricity consumption
Year of Data
(MW·h/yr)
Source
—
World
16,830,000,000
2005
CIA Est.[3] 6,464,750,000 2005
297
2005
CIA[1]
298,213,000
1,460
1,315,844,000 2009
277
1
United States
3,816,000,000
Population
As of
2005
2
China
3,640,000,000
2009
[4]
—
European Union[5]
2,820,000,000
2004
CIA Est.
459,387,000
2005
700
3
Russia
985,200,000
2007
CIA Est.
143,202,000
2005
785
4
Japan
974,200,000
2005
CIA
128,085,000
2005
868
5
Germany
593,400,000
2007
CIA Est.
82,329,758
2009 (CIA Est.) 822.22
6
Canada
540,200,000
2005
CIA
32,268,000
2005
7
India
488,500,000
2005
CIA
1,103,371,000 2005
50.5
8
France
451,500,000
2005
CIA
60,496,000
2005
851
9
South Korea
368,600,000
2007
CIA
47,817,000
2005
879
10
Brazil
368,500,000
2005
CIA
186,405,000
2005
226
11
United Kingdom
348,700,000
2005
CIA
59,668,000
2005
667
12
Italy
307,100,000
2005
CIA
58,093,000
2005
603
13
Spain
243,000,000
2005
CIA
43,064,000
2005
644
14
South Africa
241,400,000
2007
CIA
47,432,000
2005
581
15
Taiwan (Republic of China) 221,000,000
2006
CIA
22,894,384
2005
1,101
16
Australia
219,800,000
2005
CIA
20,155,000
2005
1,244
17
Mexico
183,300,000
2005
CIA
107,029,000
2005
195
1,910
Part Three: Thermodynamics
16. Temperature and Heat
17. The Thermal Behavior of Matter
18. Heat, Work, and the First Law of Thermodynamics
19. The Second Law of Thermodynamics
16. Temperature & Heat
1.
2.
3.
4.
Heat , Temperature & Thermodynamic Equilibrium
Heat Capacity & Specific Heat
Heat Transfer
Thermal Energy Balance
How does this photo reveal heat loss from the house?
And how can you tell that the car was recently driven?
IR photo:
engine & brakes hot
Studies of thermal properties:
•
Thermodynamics: Relations between macroscopic properties.
•
Statistical mechanics: Atomic description.
16.1. Heat , Temperature & Thermodynamic Equilibrium
Thermodynamic equilibrium:
State at which macroscopic properties of system remains unchanged over time.
Examples of macroscopic properties:
L, V, P, , , …
2 systems are in thermal contact if heating
one of them changes the other.
Otherwise, they are thermally insulated.
A,B in eqm
B,C in eqm

Two systems have the same temperature
A,C in eqm
 they are in thermodynamic equilibrium
0th law of thermodynamics:
2 systems in thermodynamic equilibrium with a 3rd system are themselves in equilibrium.
Gas Thermometers & the Kelvin Scale
Constant volume gas thermometer T  P
Kelvin scale:
P = 0  0 K = absolute zero
Triple point of water  273.16 K
Triple point: T at which solid, liquid &
gas phases co-exist in equilibrium
Mercury fixed at this level
by adjusting h  P  T.
All gases behave similarly as P  0.
Temperature Scales
Celsius scale ( C ) :
Melting point of ice
at P = 1 atm  TC = 0 C.
Boiling point of water at P = 1 atm  TC = 100 C.
 Triple point of water = 0.01C
TC  T  273.15
 TC  T 
Fahrenheit scale ( F ) :
Melting point of ice
at P = 1 atm  TF = 32 F.
Boiling point of water at P = 1 atm  TF = 212 F.
TF 
180
TC  32
100
Rankine scale ( R ) :
0R  0 K
TR  TF
9



T


T
F
C 

5


Supplement
Conditions for
thermodynamic equilibrium
Isolated
ideal gas
Fixed, thermally
conducting partition
P, V, n, T
PV=nRT
P jVj = nj R Tj
P1 , V1 ,
n1 , T1
P2 , V2 ,
n2 , T2
Movable, thermally
conducting partition
P1 , V1 ,
n1 , T1
P2 , V2 ,
n2 , T2
Porous, movable,
thermally conducting
partition
P1 , V1 ,
n1 , T1
T1 = T2
j = 1, 2
(Local eqm.)
P jVj = nj R Tj
j = 1, 2
T1 = T2 & P1 = P2
P2 , V2 ,
n2 , T2
Same as no partition
P jVj = nj R Tj
j = 1, 2
T 1 = T2 , P1 = P2 &  1 = 2
=n/V
Heat & Temperature
A match will burn your finger, but doesn’t provide much heat.

Heat ~ amount
Temperature ~ intensity
Brief history of the theory of heat:
1. Heat is a fluid (caloric theory: 1770s) that flows from hot to cold bodies.
2. B.Thompson, or Count Rumford, (late 1790s): unlimited amount of
heat can be produced in the boring of canon  heat is not conserved.
3. J.Joule (1840s): Heat is a form of energy.
Heat is energy transferred from high to low temperature regions.
16.2. Heat Capacity & Specific Heat
Heat capacity C of a body :
Q  C T
Q = heat transferred to body.
C  J / K
Specific heat c = heat capacity per unit mass
Q  m c T
c  J /  kg K 
1 calorie (15C cal) = heat needed to raise 1 g of water from 14.5C to 15.5C.
1 BTU (59F) = heat needed to raise 1 lb of water from 58.5F to 59.5F.
1 cal  thermochemical   4.184 J
1 BTU  1055 J
1 kcal  4 BTU
c = c(P,V) for gases  cP , cV .
Example 16.1. Waiting to Shower
The temperature in the water heater has dropped to 18C.
If the heater holds 150 kg of water, how much energy will it take to bring it up to 50C?
If the energy is supplied by a 5.0 kW electric heating element, how long will that take?
Q  m c T
 150 kg  4184 J / kg K 50C 18C   20 MJ
Q
20 MJ
t 

P
5.0 kW
20 106 J

 4000 s  1 h 7 m
5.0 103 J / s
The Equilibrium Temperature
Heat flows from hot to cold objects until a common equilibrium temperature is reached.
For 2 objects insulated from their surroundings:
Q1  Q2  0  m1 c1 T1  m2 c2 T2
When the equilibrium temperature T is reached:
m1 c1 T  T1   m2 c2 T  T2   0

T
m1 c1 T1  m2 c2 T2
m1 c1  m2 c2
GOT IT? 16.1.
A hot rock with mass 250 g is dropped into an equal mass of pool water.
Which temperature changes more?
crock = 0.20 cal / g C
cwater = 1.0 cal / g C
Explain.
 Trock changes more
Example 16.2. Cooling Down
An aluminum frying pan of mass 1.5 kg is at 180C,
when it was plunged into a sink containing 8.0 kg of water at 20C.
Assuming none of the water boils & no heat is lost to the environment,
find the equilibrium temperature of the water & pan.
T
m1 c1 T1  m2 c2 T2
m1 c1  m2 c2
1.5 kg  900 J / kg K 180 C   8.0 kg  4184 J / kg K  20 C 


1.5 kg 900 J / kg K   8.0 kg  4184 J / kg K 
 26 C
16.3. Heat Transfer
Common heat-transfer mechanisms:
• Conduction
• Convection
• Radiation
Conduction
Conduction: heat transfer through direct physical contact.
Mechanism: molecular collision.
Heat flow H , [ H ] = watt :
H
dQ
dt
 k A
Thermal conductivity k ,
[ k ] = W / mK
T
x
conductor
insulator
Example 16.3. Warming a Lake
A lake with flat bottom & steep sides has surface area 1.5 km2 & is 8.0 m deep.
The surface water is at 30C; the bottom, 4.0C.
What is the rate of heat conduction through the lake?
Assume T decreases uniformly from surface to bottom.
H  k A
T
30C  4.0C
   0.61 W / m  K  1.5 106 m2 
 3.0 MW
x
8.0 m
Power of sunlight is ~ 1 kW / m2 .
H  k A
T
x
applies only when T = const over each (planar) surface
For complicated surface, use
H  k A
dT
dx
Prob. 72 & 78.
Composite slab:
H must be the same in both slabs to
prevent accumulated heat at interface
H  k1 A
T T
T2  T1
 k2 A 3 2
x1
x2
R
Thermal resistance :

T1  T2  H R1
T2  T3  H R2

H 
H
T
R
T1  T3
R1  R2

x
kA
[R]=K/W
T2  T1
T T
 3 2
R1
R2
Resistance in series
GOT IT? 16.2.
Rank order the 3 temperature differences.
H, A, x same for all three
 k T = const
T1  T3  T2
Insulating properties of building materials are described by the R-factor ( R-value ) .
R  R A
x T

A
H
k
R   m2 K / W
U.S.
R   ft 2  F  h / BTU
1 ft 2  F  h / BTU  0.176 m2 K / W
H  k A
T
A
dT

  T
R
R
dx
= thermal resistance of a slab of unit area
Example 16.4. Cost of Oil
The walls of a house consist of plaster ( R = 0.17 ), R-11 fiberglass
insulation, plywood (R = 0.65 ), and cedar shingles (R = 0.55 ).
The roof is the same except it uses R-30 fiberglass insulation.
In winter, average T outdoor is 20 F, while the house is at 70 F.
The house’s furnace produces 100,000 BTU for every gallon of oil,
which costs $2.20 per gallon.
How much is the monthly cost?
R wall  0.17  11  0.65  0.55  12.37
Arect  2  36 ft  28 ft   10 ft   1280 ft 2
1
Agable  2    28 ft   14 ft  tan 30   226 ft 2
2
R roof  0.17  30  0.65  0.55  31.37
 14 ft 
2
Aroof  2   36 ft   
  1164 ft
 cos 30 
Awall  Arect  Agable  1506 ft 2
 1

H wall  
BTU / h / ft 2 / F  1506 ft 2   70F  20F   6073 BTU / h
 12.37

 1

H roof  
BTU / h / ft 2 / F  1164 ft 2   70F  20F   1853 BTU / h
 31.37

Q   6073 1853 BTU / h 24 h / d 30 d / month  5.7 MBTU
Cost  5.7 MBTU 10 gal / MBTU
$ 2.20 / gal 
 $126
Convection
Convection = heat transfer by fluid motion
T      rises
Convection cells in liquid film between glass plates
(Rayleigh-Bénard convection, Benard cells)
Examples:
• Boiling water.
• Heating a house.
• Sun heating earth  Climate, storms.
• Earth mantle  continental drift
• Generation of B in stars & planets.
H convec  T
Radiation
Glow of a stove burner  it loses energy by radiation
Stefan-Boltzmann law for radiated power:
P
 e T4
A
 = Stefan-Boltzmann constant = 5.67108 W / m2 K4.
A = area of emitting surface.
0 < e < 1 is the emissivity ( effectiveness in emitting radiation ).
e = 1  perfect emitter & absorber ( black body ).
Black objects are good emitters & absorbers.
Shiny objects are poor emitters & absorbers.
Stefan-Boltzmann law :
P
 e T4
A
Wien‘s displacement law : max = b / T
b  2.898  103 mK
 P  T4  Radiation dominates at high T.
Wavelength of peak radiation becomes shorter as T increases.
Sun ~ visible light.
Near room T ~ infrared.
RT 

sunTsun
TRT
.502 m  5778K 
300 K
 9.66 m
GOT IT? 16.3.
Name the dominant form of heat transfer from
Radiation
(a)
a red-hot stove burner with nothing on it.
conduction
(b)
a burner in direct contact with a pan of water.
(c)
the bottom to the top of the water in the pan once it boils.
convection
Example 16.5. Sun’s Temperature
The sun radiates energy at the rate P = 3.91026 W, & its radius is 7.0 108 m.
Treating it as a blackbody ( e = 1 ), find its surface temperature.
P  e  AT 4
 = 5.67108 W / m2 K4
1/4


P

T 
 e   4  R2  



3.9 1026 W


2
8
2
4
8
5.67

10
W
/
m
K
4


7.0

10
m






 5.8 103 K
1/4






Conceptual Example 15.1. Energy-Saving Windows
Why do double-pane windows reduce heat loss greatly compared with
single-paned windows?
Why is a window’s R-factor higher if the spacing between panes is small?
And why do the best windows have “low-E” coatings?
Thermal conductivity (see Table 16.2):
Glass k ~ 0.8 W/mK
Air
k ~ 0.026 W/mK
 Layer of air reduces heat loss greatly & increases the R-factor .
This is so unless air layer is so thick that convection current develops.
“low-E” means low emissivity, which reduces energy loss by radiation.
Making the Connection
Compare the for a single pane window made from 3.0-mm-thick glass
with that of a double-pane window make from the same glass with a
5.0-mm air gap between panes.
x
R 
k
R single
Rdouble
Glass
Air
3.0  103 m

0.8 W / m  K
k ~ 0.8 W/mK
k ~ 0.026 W/mK
R
R x

A kA
 0.004 m2  K / W
3.0  103 m
5.0  103 m
 2

 0.8 W / m  K   A  0.026 W / m  K   A
R double  0.2 m2  K / W
R double  50  Rsingle

0.2 2
m  K /W
A
16.4. Thermal Energy Balance
A house in thermal-energy balance.
System with fixed rate of energy input
tends toward an energy- balanced state
due to negative feedback.
Heat from furnace balances
losses thru roofs & walls
Example 16.6. Hot Water
A poorly insulated electric water heater loses heat by conduction at the rate of
120 W for each C difference between the water & its surrounding.
It’s heated by a 2.5 kW heating element & is located in a basement kept at 15 C.
What’s the water temperature if the heating element operates continuously.
T=?
Conductive
heat loss
H  120 W / C  T 15C   2.5 kW
2.5 103 W
T  15C 
 36C
120 W / C
Electrical
energy in
Heating
element
Example 16.7. Solar Greenhouse
A solar greenhouse has 300 ft2 of opaque R-30 walls,
& 250 ft2 of R-1.8 double-pane glass that admits solar energy at the rate of 40 BTU / h / ft2.
Find the greenhouse temperature on a day when outdoor temperature is 15 F.
H
T A T

R
R
 300 ft  T
2
H wall 
30 ft  F h / BTU
2
 250 ft  T
2
H glass 
1.8 ft  F h / BTU
2
 10 BTU / h / F  T
 139 BTU / h / F  T
H sun   40 BTU / h / ft 2  250 ft 2 
104 BTU / h
 67 F
T 
149 BTU / h / F
 104 BTU / h
 Hwall  H glass
T  15 F  67 F  82 F
Application: Greenhouse Effect & Global Warming
Average power from sun :
Total power from sun :
S  960 W / m2
HS  S  R2E
Power radiated (peak at IR) from Earth :
H E  e   4  RE2  T 4
HS  HE
e 1
1/4

C.f.  T   15 C


960 W / m 2

T 
  5.67 108 W / m 2 K 4   4 


 255 K  18 C
natural greenhouse effect
Mars: none
Greenhouse gases: H2O, CO2 , CH4 , …
passes incoming sunlight, absorbs outgoing IR.
Venus: huge
CO2 increased by 36%
0.6 C increase during 20th century.
1.5 C – 6 C increase by 2100.