Short Version : 19. 2 nd Law of Thermodynamics
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Transcript of Short Version : 19. 2 nd Law of Thermodynamics
Short Version : 19. 2nd Law of Thermodynamics
19.1. Reversibility & Irreversibility
Block slowed down by friction:
irreversibleBouncing ball:
reversible
Examples of irreversible processes:
• Beating an egg, blending yolk & white
• Cups of cold & hot water in contact Spontaneous process:
order disorder
( statistically more probable )
19.2. The 2nd Law of Thermodynamics
Heat engine extracts work from heat reservoirs.
• gasoline & diesel engines
• fossil-fueled & nuclear power plants
• jet engines
Perfect heat engine: coverts heat to work directly.
Heat dumped
2nd law of thermodynamics ( Kelvin-Planck version ):There is no perfect heat engine.
No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.
EfficiencyNet WorkHeat Input
e
h c
h
Q QeQ
1 c
h
(any engine)
(Simple engine)
0U W Q (any cycle)
Hero Engine
Stirling Engine
Carnot Engine
1. isothermal expansion: T = Th , W1 = Qh > 0
2. Adiabatic expansion: Th Tc, W2 > 0
3. isothermal compression: T = Tc , W3 = Qc < 0
Adiabatic compression : Tc Th , W4 = W2 < 0
Ideal gas:
Adiabatic processes:
1 1h B c CT V T V
1 1c D h AT V T V
C B
D A
V VV V
1 1c ccarnot
h h
Q TeQ T
3 ln2
Bh AB h
A
VQ Q n R TV
ln Dc CD c
C
VQ Q n R TV
AB: Heat abs.
C D: Heat rejected:
B C: Work done BC BC c hW U nR T T
D A: Work done
32
DA DA
h c
W U
nR T T
Engines, Refrigerators, & the 2nd Law
Carnot’s theorem:
1.All Carnot engines operating between temperatures Th & Tc have the same efficiency.
2.No other engine operating between Th & Tc can have a greater efficiency.
Refrigerator: extracts heat from cool reservoir into a hot one.
work required
2nd law of thermodynamics ( Clausius version ):
There is no perfect refrigerator.
perfect refrigerator: moves heat from cool to hot reservoir without work being done on it.
No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.
Perfect refrigerator Perfect heat engine
Clausius Kelvin-Planck
Carnot engine is most efficient
eCarnot = thermodynamic efficiency
eCarnot erev > eirrev
Carnot refrigerator, e = 60%
Hypothetical engine, e = 70%
19.3. Applications of the 2nd LawPower plant
fossil-fuel : Th = 650 K
Nuclear : Th = 570 K
Tc = 310 K
3101650fossil
KeK
52 %
3101570nuclear
KeK
46 %
Actual values:
efossil ~ 40 % enuclear ~ 34 % ecar ~ 20 %
Prob 54 & 55
Heat source
Boiler
Turbine
Generator
Electricity
Condenser
Waste water
Cooling water
Application: Combined-Cycle Power Plant
Turbine engines: high Th ( 1000K 2000K ) & Tc ( 800 K ) … not efficient.
Steam engines : Tc ~ ambient 300K.
Combined-cycle : Th ( 1000K 2000K ) & Tc ( 300 K ) … e ~ 60%
Example 19.2. Combined-Cycle Power Plant
The gas turbine in a combined-cycle power plant operates at 1450 C.
Its waste heat at 500 C is the input for a conventional steam cycle, with its condenser at 8 C.
Find e of the combined-cycle, & compare it with those of the individual components.
273 81273 1450combinede
84 %
273 5001273 1450turbinee
55 %
273 81273 500steame
64 %
Refrigerators
Coefficient of performance (COP) for refrigerators :
cQCOPW
c
h c
QQ Q
c
h c
TT T
COP is high if Th Tc .
Max. theoretical value (Carnot)
1st law
W = 0 ( COP = ) for moving Q when Th = Tc .
Example 19.3. Home Freezer
A typical home freezer operates between Tc = 18C to Th = 30 C.
What’s its maximum possible COP?
With this COP, how much electrical energy would it take to freeze 500 g of water initially at 0
C?
5.31 273 18
30 18
c
h c
TCOP
T T
Q L m 334 / 0.5kJ kg kg 167 kJ Table 17.1
QWCOP
167
5.31kJ
31 kJ
2nd law: only a fraction of Q can become W in heat engines. a little W can move a lot of Q in refrigerators.
Heat Pumps
ccooling
QCOPW
c
h c
QQ Q
in
out in
TT T
Heat pump as AC :
hheating
QCOPW
h
h c
QQ Q
in
in out
TT T
Heat pump as heater :
Ground temp ~ 10C year round (US)
Heat pump: moves heat from Tc to Th .
19.4. Entropy & Energy Quality
Energy quality measures the versatility of different energy forms.
2nd law:Energy of higher quality can be converted completely into lower quality form.
But not vice versa.
Entropy
lukewarm: can’t do W,
Carnot cycle (reversible processes):
c c
h h
Q TQ T
0c h
c h
Q QT T
Qh = heat absorbed
Qc = heat rejected
Qh , Qc = heat absorbed
0C
d QT
C = any closed path
2
1
d QST
S = entropy [ S ] = J / K
Irreversible processes can’t be represented by a path.
Entropy
lukewarm: can’t do W,
Carnot cycle (reversible processes):
c c
h h
Q TQ T
0c h
c h
Q QT T
Qh = heat absorbed
Qc = heat rejected
Qh , Qc = heat absorbed
0C
d QT
C = any closed path
2
1
d QST
S = entropy [ S ] = J / K
Irreversible processes can’t be represented by a path.
C = Carnot cycle
Contour = sum of Carnot cycles.
Entropy change is path-independent.
( S is a thermodynamic variable )
S = 0 over any closed path
S21 + S12= 0
S21 = S21
Entropy in Carnot CycleIdeal gas:
ln Bh AB h
A
VQ Q n R TV
ln Dc CD c
C
VQ Q n R TV
Adiabatic processes:1 1
h B c CT V T V
1 1c D h AT V T V
C B
D A
V VV V
ABC ABQ Q
ADC DCQ Q
Heat absorbed:
Heat rejected:
ln Bh
A
Vn R TV
ln Cc
D
Vn R TV
ABABC
h
QST
ln B
A
Vn RV
DCADC
c
QST
ln C
D
Vn RV
ABCS
Irreversible Heat Transfer
Cold & hot water can be mixed reversibly
using extra heat baths.
f
c
T
c T
dQST
1
cQT
f
h
T
h T
dQST
2
hQT
c hS S S 1 2
c hQ QT T
1 2
c cQ QT T
1 2
1 1c T T
Q
0
0S Actual mixing, irreversible processes
reversible processes
T1 = some medium T.
T2 = some medium T.
2 1T T
Adiabatic Free Expansion
. exp. 0adW 0vacuump
Adiabatic Qad.exp. = 0
. exp. 0adU
. exp. 0adT
S can be calculated by any reversible
process between the same states.
dQST
1 dQT
QT
2
1
lnVn RV
0
WT
2
1
1 V
V
n R T dVT V
p = const.Can’t do work
degraded.
isothermal
Entropy & Availability of Work
Before adiabatic expansion, gas can do work isothermally
2
1
lnVW Q n R TV
After adiabatic expansion, gas cannot do work, while its entropy increases by
QST
2
1
lnVn RV
unavailableE T S
In a general irreversible process minunavailableE T S
Coolest T in system
Example 19.4. Loss of
A 2.0 L cylinder contains 5.0 mol of compressed gas at 300 K.
If the cylinder is discharged into a 150 L vacuum chamber & its temperature remains at 300 K,
how much energy becomes unavailable to do work?
2
1
lnVn R TV
minunavailableE T S
150 25.0 8.314 / / 300 ln2.0L Lmol J K mol K
L
54 kJ
A Statistical Interpretation of Entropy
Gas of 2 distinguishable molecules occupying 2 sides of a box
Microstates Macrostates probability of macrostate
1/4
2 ¼ = ½ 1/4
Gas of 4 distinguishable molecules occupying 2 sides of a box
Microstates Macrostates probability of macrostate
1/16 = 0.06
4 1/16 = ¼ =0.25
4 1/16 = ¼ =0.25
1/16 = 0.06
6 1/16 = 3/8 = 0.38
Gas of 100 molecules Gas of 1023 molecules
Equal distribution of molecules
lnBS k Statistical definition of entropy : # of micro states
Entropy & the 2nd Law of Thermodynamics
2nd Law of Thermodynamics : 0S in any closed system
S can decrease in an open system by outside work on it.
However, S 0 for combined system.
S 0 in the universe
Universe tends to disorder
Life ?