Transcript Physics 207: Lecture 2 Notes

Lecture 27
Goals:
• Ch. 18
• Qualitatively understand 2nd Law of Thermodynamics
• Ch. 19
 Understand the relationship between work and heat in a
cycling process
 Follow the physics of basic heat engines and refrigerators.
 Recognize some practical applications in real devices.
 Know the limits of efficiency in a heat engine.
• Assignment
 HW11, Due Tues., May 5th
 HW12, Due Friday, May 9th
 For Thursday, Read through all of Chapter 20
Physics 207: Lecture 27, Pg 1
The need for something else: Entropy
V1
You have an ideal gas in a box of volume
V1. Suddenly you remove the
partition and the gas now occupies a
larger volume V2.
P
(1) How much work was done by the
system?
P
(2) What is the final temperature (T2)?
V2
(3) Can the partition be reinstalled with all
of the gas molecules back in V1?
Physics 207: Lecture 27, Pg 2
Free Expansion and Entropy
V1
You have an ideal gas in a box of
volume V1. Suddenly you remove
the partition and the gas now
occupies a larger volume V2.
P
(3) Can the partition be reinstalled with
all of the gas molecules back in V1
P
V2
(4) What is the minimum process
necessary to put it back?
Physics 207: Lecture 27, Pg 3
Free Expansion and Entropy
V1
You have an ideal gas in a box of
volume V1. Suddenly you remove
the partition and the gas now
occupies a larger volume V2.
P
(4) What is the minimum energy
process necessary to put it back?
P
Example processes:
V2
A. Adiabatic Compression followed by
Thermal Energy Transfer
B. Cooling to 0 K, Compression,
Heating back to original T
Physics 207: Lecture 27, Pg 4
Modeling entropy
 I have a two boxes. One with fifty pennies. The other has none.
I flip each penny and, if the coin toss yields heads it stays put. If
the toss is “tails” the penny moves to the next box.
 On average how many pennies will move to the empty box?
Physics 207: Lecture 27, Pg 6
Modeling entropy
 I have a two boxes, with 25 pennies in each. I flip each penny
and, if the coin toss yields heads it stays put. If the toss is “tails”
the penny moves to the next box.
 On average how many pennies will move to the other box?
 What are the chances that all of the pennies will wind up in
one box?
Physics 207: Lecture 27, Pg 7
2nd Law of Thermodynamics
 Second law: “The entropy of an isolated system never decreases. It
can only increase, or, in equilibrium, remain constant.”
Increasing
Entropy
Entropy measures the probability
that a macroscopic state will
occur or, equivalently, it
measures the amount of
disorder in a system
 The 2nd Law tells us how collisions move a system toward
equilibrium.
 Order turns into disorder and randomness.
 With time thermal energy will always transfer from the hotter to the
colder system, never from colder to hotter.
 The laws of probability dictate that a system will evolve towards the
most probable and most random macroscopic state
Physics 207: Lecture 27, Pg 8
Reversible vs Irreversible
 The following conditions should be met to make a process
perfectly reversible:
1. Any mechanical interactions taking place in the process
should be frictionless.
2. Any thermal interactions taking place in the process should
occur across infinitesimal temperature or pressure gradients
(i.e. the system should always be close to equilibrium.)
 Based on the above answers, which of the following processes
are not reversible?
1. Melting of ice in an insulated (adiabatic) ice-water mixture at
0°C.
2. Lowering a frictionless piston in a cylinder by placing a bag of
sand on top of the piston.
3. Lifting the piston described in the previous statement by
removing one grain of sand at a time.
4. Freezing water originally at 5°C.
Physics 207: Lecture 27, Pg 11
Reversible vs Irreversible
 The following conditions should be met to make a process
perfectly reversible:
1. Any mechanical interactions taking place in the process
should be frictionless.
2. Any thermal interactions taking place in the process should
occur across infinitesimal temperature or pressure gradients
(i.e. the system should always be close to equilibrium.)
 Based on the above answers, which of the following processes
are not reversible?
1. Melting of ice in an insulated (adiabatic) ice-water mixture at
0°C.
2. Lowering a frictionless piston in a cylinder by placing a bag of
sand on top of the piston.
3. Lifting the piston described in the previous statement by
removing one grain of sand at a time.
4. Freezing water originally at 5°C.
Physics 207: Lecture 27, Pg 12
Heat Engines and Refrigerators
 Heat Engine: Device that transforms heat into work ( Q  W)
 It requires two energy reservoirs at different temperatures
 An thermal energy reservoir is a part of the environment so
large with respect to the system that its temperature doesn’t
change as the system exchanges heat with the reservoir.
 All heat engines and refrigerators operate between two
energy reservoirs at different temperatures TH and TC.
Physics 207: Lecture 27, Pg 16
Heat Engines
For practical reasons, we would like an engine to do the maximum
amount of work with the minimum amount of fuel. We can
measure the performance of a heat engine in terms of its thermal
efficiency η (lowercase Greek eta), defined as
We can also write the thermal efficiency as
Physics 207: Lecture 27, Pg 17
Exercise Efficiency
 Consider two heat engines:
 Engine I:
 Requires Qin = 100 J of heat added to system to
get W=10 J of work (done on world in cycle)
 Engine II:
 To get W=10 J of work, Qout = 100 J of heat is
exhausted to the environment
 Compare
hI, the efficiency of engine I, to hII, the
efficiency of engine II.
Wcycle Qh  Qc
Qc
h

 1
Qh
Qh
Qh
Physics 207: Lecture 27, Pg 18
Exercise Efficiency
 Compare
hI, the efficiency of engine I, to hII, the efficiency of
engine II.
 Engine I:
 Requires Qin = 100 J of heat added to system to get
W=10 J of work (done on world in cycle)
 h = 10 / 100 = 0.10
 Engine II:
 To get W=10 J of work, Qout = 100 J of heat is exhausted
to the environment
 Qin = W+ Qout = 100 J + 10 J = 110 J
 h = 10 / 110 = 0.09
Wcycle Qh  Qc
Qc
h

 1
Qh
Qh
Qh
Physics 207: Lecture 27, Pg 19
Refrigerator (Heat pump)
 Device that uses work to transfer heat from a colder object to a
hotter object.
K
QCold
WIn

What you get
What you pay
Physics 207: Lecture 27, Pg 20
The best thermal engine ever, the Carnot engine
 A perfectly reversible engine (a Carnot engine) can be
operated either as a heat engine or a refrigerator between the
same two energy reservoirs, by reversing the cycle and with no
other changes.
 A Carnot cycle for a gas engine
consists of two isothermal processes
and two adiabatic processes
 A Carnot engine has max. thermal
efficiency, compared with any other
engine operating between TH and TC
hCarnot  1 
TCold
THot
 A Carnot refrigerator has a
maximum coefficient of performance,
compared with any other refrigerator
operating between TH and TC.
K Carnot 
TCold
THot TCold
Physics 207: Lecture 27, Pg 21
The Carnot Engine

Carnot showed that the thermal efficiency of a Carnot
engine is:
Tcold
hCarnotcycle  1
Thot
 All real engines are less efficient than the Carnot engine
because they operate irreversibly due to the path and
friction as they complete a cycle in a brief time period.
Physics 207: Lecture 27, Pg 22
Problem
 You can vary the efficiency of a
Carnot engine by varying the
temperature of the cold
reservoir while maintaining the
hot reservoir at constant
temperature.
Which curve that best represents
the efficiency of such an
engine as a function of the
temperature of the cold
reservoir?
Tcold
 Carnotcycle  1
Thot
Temp of cold reservoir
Physics 207: Lecture 27, Pg 23
Other cyclic processes: Turbines
 A turbine is a mechanical device that extracts thermal energy
from pressurized steam or gas, and converts it into useful
mechanical work. 90% of the world electricity is produced by
steam turbines.
 Steam turbines &jet engines use a Brayton cycle
Physics 207: Lecture 27, Pg 26
Steam Turbine in Madison
 MG&E, the electric power plan in Madison, boils water to
produce high pressure steam at 400°C. The steam spins the
turbine as it expands, and the turbine spins the generator. The
steam is then condensed back to water in a Monona-lake-watercooled heat exchanger, down to 20°C.
 Carnot Efficiency?
hCarnot  1 
TCold
THot
 1
 0.56
293K
Physics673
207: Lecture
K 27, Pg 27
The Sterling Cycle
 Return of a 1800’s thermodynamic cycle
Isothermal
expansion
Isothermal
compression
SRS Solar System (~27% eff.)
Physics 207: Lecture 27, Pg 28
Sterling cycles
 1 Q, V constant  2 Isothermal expansion ( Won system < 0 ) 
1
3 Q, V constant  4 Q out, Isothermal compression ( Won sys> 0)
1 Q1 = nR CV (TH - TC)
2 Won2 = -nR TH ln (Vb / Va)= -Q2
3 Q3 = nR CV (TC - TH)
4 Won4 = -nR TL ln (Va / Vb)= -Q4
1
Gas
2
QCold = - (Q3 + Q4 )
T=TH
QHot = (Q1 + Q2 )
h = 1 – QCold / QHot
Gas
Gas
T=TH
T=TC
4
Gas
T=TC
3
P
1 2
x
4
start
Va
3
Vb
TH
TC
V
Physics 207: Lecture 27, Pg 29
Power from ocean thermal gradients… oceans
contain large amounts of energy
Carnot Cycle Efficiency
hCarnot = 1 - Qc/Qh = 1 - Tc/Th
See: http://www.nrel.gov/otec/what.html
Physics 207: Lecture 27, Pg 30
Ocean Conversion Efficiency
hCarnot = 1 - Tc/Th = 1 – 275 K/300 K
= 0.083 (even before internal losses
and assuming a REAL cycle)
Still: “This potential is estimated to be about 1013 watts of base load
power generation, according to some experts. The cold, deep seawater
used in the OTEC process is also rich in nutrients, and it can be used to
culture both marine organisms and plant life near the shore or on land.”
“Energy conversion efficiencies as high as 97% were achieved.”
See: http://www.nrel.gov/otec/what.html
So h =1-Qc/Qh is always correct but
hCarnot =1-Tc/Th only reflects a Carnot cycle
Physics 207: Lecture 27, Pg 31
Internal combustion engine: gasoline engine
 A gasoline engine utilizes the Otto cycle, in which fuel and air
are mixed before entering the combustion chamber and are
then ignited by a spark plug.
Otto Cycle
(Adiabats)
Physics 207: Lecture 27, Pg 32
Internal combustion engine: Diesel engine
 A Diesel engine uses compression ignition, a process by which
fuel is injected after the air is compressed in the combustion
chamber causing the fuel to self-ignite.
Physics 207: Lecture 27, Pg 33
Thermal cycle alternatives
 Fuel Cell Efficiency (from wikipedia)
Fuel cells do not operate on a thermal cycle. As such, they are not
constrained, as combustion engines are, in the same way by thermodynamic
limits, such as Carnot cycle efficiency. The laws of thermodynamics also hold
for chemical processes (Gibbs free energy) like fuel cells, but the maximum
theoretical efficiency is higher (83% efficient at 298K ) than the Otto cycle
thermal efficiency (60% for compression ratio of 10 and specific heat ratio of
1.4).
 Comparing limits imposed by thermodynamics is not a good predictor of
practically achievable efficiencies
 The tank-to-wheel efficiency of a fuel cell vehicle is about 45% at low loads
and shows average values of about 36%. The comparable value for a Diesel
vehicle is 22%.
 Honda Clarity
(now leased in CA and gets
~70 mpg equivalent)
This does not include H2
production & distribution
Physics 207: Lecture 27, Pg 34
Fuel Cell Structure
Physics 207: Lecture 27, Pg 35
Problem-Solving Strategy: Heat-Engine Problems
Physics 207: Lecture 27, Pg 36
Going full cycle
 1 mole of an ideal gas and PV= nRT  T = PV/nR
T1 = 8300 0.100 / 8.3 = 100 K
T2 = 24900 0.100 / 8.3 = 300 K
T3 = 24900 0.200 / 8.3 = 600 K T4 = 8300 0.200 / 8.3 = 200 K
(Wnet = 16600*0.100 = 1660 J)
12
P
DEth= 1.5 nR DT = 1.5x8.3x200 = 2490 J
2
3
24900
Wby=0
Qin=2490 J
QH=2490 J
N/m2
23
DEth= 1.5 nR DT = 1.5x8.3x300 = 3740 J
Wby=2490 J Qin=3740 J QH= 6230 J
4
1
8300
34
N/m2
DEth = 1.5 nR DT = -1.5x8.3x400 = -4980 J
Wby=0
Qin=-4980 J QC=4980 J
41
100
200 V
DEth = 1.5 nR DT = -1.5x8.3x100 = -1250 J
liters
liters
Wby=-830 J Qin=-1240 J QC= 2070 J
QH(total)= 8720 J QC(total)= 7060 J h =1660 / 8720 =0.19 (very low)
Physics 207: Lecture 27, Pg 37
Lecture 27
• Assignment
 HW11, Due Tues., May 5th
 HW12, Due Friday, May 9th
 For Thursday, Read through all of Chapter 20
Physics 207: Lecture 27, Pg 39