Download March 30 -- April 1 - Department of Physics | Oregon State

Presentation outline:
• Some semantics -- the meaning of the word „alternative”
• List of those we can call “alternatives”
• If we are interested in “alternatives”, why should we also
know much about energy sources that are not “alternatives”?
I. e., about fossil fuels, and how they are used?
• Major fossil fuel resources, how long will they last, and
their distribution over the globe.
• What are fossil fuels used for? Major ways of using them.
• Heat engines – the main tools of coverting fossil fuel
energy to other usable forms.
• Types of heat engines.
•Some thermodynamics: First Law, and then a longer
introduction to the Second Law.
The title of this course is Energy Alternatives
Let’s first precisely define what it means. Take the Webster
Definition and look up Alternative. We find:
Adjective:
1: offering or expressing a choice <several alternative plans>
2: different from the usual or conventional: as
● a: existing or functioning outside the established cultural, social,
or economic system <an alternative newspaper> <alternative lifestyles>
b: of, relating to, or being rock music that is regarded as an alternative
to conventional rock and is typically influenced by punk rock, hard rock,
hip-hop, or folk music
c: of or relating to alternative medicine <alternative therapies>
Noun:
1 a: a proposition or situation offering a choice between two or more things
only one of which may be chosen
b: an opportunity for deciding between
two or more courses or propositions
2 a: one of two or more things, courses, or propositions to be chosen
● b: something which can be chosen instead <the only alternative to
intervention>
3: alternative rock music
Clearly, the highlighted are the most appropriate Energy Alternatives.
In short: generally, the term Energy Alternatives refers to resources
that can be chosen instead of the established methods of energy
production.
Traditional fuels & resources;
Energy alternatives:
• Coal (since early 1700s);
• Oil (since mid-XIX Century);
• Natural gas (as above);
• Hydropower (many millennia!);
• Nuclear fission (since 1950s).
• Solar energy (direct usage);
• Wind (solar, too! – indirectly);
• Bio-fuels (again, solar!);
• Hydropower (one more solar!);
• Nuclear (returning to favors);
• Ocean waves;
• Tides;
• Geothermal energy;
• ……. (probably a few items can
be still added).
Extracting energy
from the first three
involves burning
“Traditional” methods – we don’t like them (why?).
Think green:
They are our enemy! We want to eliminate it!
Well – and keep in mind what the greatest military
leaders in history always used to say:
Rule Number One
for a victorious
campaign:
Know your enemy!
Learn about all
its weaknesses
and strengths!
Fossil fuels – basic facts and numbers:
Major – global resources:
• Coal:
997,748 million short tons (4,416 BBOE; 2005)
• Oil: 1,119 to 1,317 billion barrels (2005-2007)
• Natural gas: 6,183 - 6,381 trillion cubic feet (1,161 BBOE; 2005-2007)
Minor (or not yet fully exploited):
• Tar sands (contain “bitumen”, a form of heavy oil): 1.7 trillion(!) BBOE;
• Oil shales (as above) 411 gigatons, or 2.8 to 3.3 trillion(!) BBOE;
• Methane hydride – (resources unknown, by some believed very large).
BBOE = Billion Barrels of Oil Equivalent
Energy conversion – a convenient program
Flows (daily production) during 2006
Oil: 84 million barrels per day;
Gas: 19 million barrels oil equivalent per day {MBOED}
Coal: 29 million barrels oil equivalent per day MBOED
How long will those resources last?
Years of production left, due the most optimistic reserve estimates
(Oil & Gas Journal, World Oil)
Oil: 43 years
Natural Gas: 167 years
Coal: 417 years
(FYI, not for any longer discussion in class)
The distribution of coal, oil and gas deposits by country, shown using colors
Red – largest resources; Black – smallest resources
COAL:
GAS:
OIL:
TOTAL:
How are fossil fuels used? We just burn them,
that’s all! But in many different ways:
• Simple combustion;
• To generate heat needed in many types
of industrial processes, e.g., smelting,
chemical synthesis, ….
• In heat engines, using various types
of combustion, propelling cars, trucks,
railway engines, planes, ships, …
• In heat engines, to generate mechanical energy, and then electric power;
Major heat engine types:
• Steam engines (in historical context, mostly).
• Internal combustion engines
• Steam turbines and gas turbines.
Steam engine and internal combustion engine animations
Another Web site with engine animations
Steam Turbine
Animation
Conclusions: Heat (or, rather thermal energy)
can be transformed into mechanical energy.
And there is a range of heat engines that
can be harnessed to perform many useful tasks.
Unfortunately… The reality is not
so brilliant as one might think.
There is one annoying “troublemaker”
that adds much gloom to the picture.
The name of that troublemaker is
The Second Law of Thermodynamics
So – even more unpleasant news:
we have to go back to physics!
If we want to know what the 2nd Law is about, we have to
know first what the 1st Law of Thermodynamics says, right?
About the First Law of Thermodynamics:
SYSTEM:
A system: a single body, or
more bodies that in contact
with one another.
U:
There is a physical quantity called the
INTERNAL THERMAL ENERGY of a
system – or “internal energy” in short.
Conventionally, it is denoted as U .
Energy may be added to the system, thus increasing its U
(we call such a process “heating”).
-- or –
Energy may be taken away from the system, thus lowering its U
(we call such a process “cooling”).
There is a temperature, at which no more
energy can be removed from the system.
We call it the absolute zero. Its value is:
T = - 492.3 ˚F, or in the SI system, T = 0 K.
Now, QQQ (Quick Quiz Question):
Energy is added to a system. Its temperature:
(a)
(b)
(c)
(d)
Increases
Decreases
Does not decrease
Does not increse
1 K = 1.8 ºF
Water freezes
at 273.15 K
Boils at 373.15 K
Celsius Scale:
Water freezes
at 0 ºC, boils at
100 ºC.
So, the increment
in the SI scale and
in Celsius scale is
the same: 1ºC= 1 K.
Now, the First Law: essentially, it’s the Energy
Conservation Law, but expressed in a way
specifically applying to thermal phenomena:
U  Q  W
The total
change in
the system
internal
energy
The change due
to transfer of
heat (heat flowing
in or out from
another system)
The change due to
mechanical work
done ON the system,
or the work delivered
BY the system (then - )
IMPORTANT! A common misconception is to confuse HEAT with
the INTERNAL ENERGY. Internal energy is the amount of energy
contained by the system. Heat is the energy that flows in or out
from/to a warmer/cooler body which is in contact with the system.
This
slide
is not
for
going
through
it in
detail in
class,
but
rather
for you
to read
before
or after
the
class.
This
also aplies to
the
next
slide.
The First Law was an easy part. But in order to explain what
the Second Law talks about, we have to introduce the notion
of ENTROPY.
Entropy is widely regarded as one of the most difficult concepts
in university physics curriculum. It’s a parameter that characterizes the thermal state of a system. Other state parameters are
the internal energy U, volume V, the amount of substance (usually expressed as the number of moles N – a mole consists of
6.0221023 molecules of a given substance – who can tell why
such an “exotic” number?), the temperature T, and pressure p.
They are all “intuitively clear”, am I right?
In contrast, entropy, conventionally denoted as S, is an abstract
function. Its mathematical definition is not particularly difficult:
T
dQ
dQ
Differenti ally : dS 
; hence : S (T )  
T
T
T 0
However, for a student it may not be a straightforward thing to
understand its physical meaning, and “what it is good for”.
4 U to read on your own before the class
Entropy is an even greater challenge for an instructor, than for
a student – I mean, doing a “quality work” when teaching this
topic. Dr. Tom has been teaching thermal physics at OSU for
more then ten years, and he knows that trying to tell everything
relevant about entropy in the course of a single class hour would
not be a “quality job”. Rather, in the thermal physics classes he
teaches he spends several hours, introducing the entropy in a
systematic manner, step by step. Entropy is not a good topic
for being taught in a “crash-course” fashion.
This course is not a thermal physics course, and entropy
is “just a small episode”. We can only talk about that briefly.
Therefore, this presentation is limited to some basic facts that
I am asking you to accept without proof.
Here we define the entropy as it is done in classical thermodynamics, which is a macroscopic theory. In statistical thermodynamics, which is a microscopic approach, one uses a different definition – in terms of thermal disorder: S  kB  ln 
where Ω is “the measure of disorder”. Both definitions are equivalent, as can be shown – however, the latter is not particularly
useful for analyzing the performance of thermal engines, and
therefore we will use the “classical definition”
Entropy – important facts “in a nutshell”:
The entropy of a thermally isolated system (meaning: no heat
can be transferred in or out) may only increase or remain
constant in time, but it cannot decrease. In other words:
 dS 
 0.
 
isolated
 dt system
This is the Second Law of Thermodynamics –
or, rather one of its many formulations. There
are many other formulations that one can find in the literature,
but they are all equivalent.
One funny fact: the shortest of all those formulations states:
It is not possible to build a Perpetual Motion Machine of the Second Kind
Q: Who derived the Second Law, and how?
A: It has not been “derived” mathematically. It is an EMPIRICAL LAW,
based on zillions of experimental results and observations. What
scientist only did, they “digested” all that information and formulated
The conclusion in the form of a law of physics.
What is the “Perpetual Motion Machine of the Second Kind”? When the Energy
Conservation Law was formulated, it became clear that building a purely mechanical
perpetual motion device was not possible. But some “inventors” did not give up!
They said: Well, we accept that work cannot be created out of nothing. But note that
that oceans are almost infinite reservoirs of thermal energy. Let’s convert this energy
to work – such a machine would not violate the Energy Conservation Law!
Entropy in a nutshell, cont. – situations in which the
entropy in a thermally isolated system increases:
The entropy must increase when an irreversible process occurs
in the system. Example:
T1
T2
T1
T2
T1 > T2
1. Two bodies of different temperatures
in an isolated system.
Tf
Tf
T2 < Tf < T1
3. After a while, equilibrium is reached. Now the
two bodies have the same temperature.
2. Bodies brought into thermal contact.
Heat flows.
We will work this example in
class on the blackboard, and
we will show that the entropy
indeed increases in such a
process.
The calculations below will be done in class on the board – this slide is
only for you to read before the class, or later when preparing for a test.
Suppose that the internal energy U of each body in
the preceding slide is proportion al to its temperatu re,
so that : U  c  T (pretty normal behavior at room T ).
Since no work is involved, U is only due to heat
input/outp ut Q, so that Q  U  c  T
Let' s calculate the change in the body' s entropy S
when it tempera ture changes from T ' to T ' ' :
dQ
c  dT
dT
 T '' 
S  

c 
c  ln  
T T' T
T
 T' 
T'
T'
T ''
T ''
T ''
The calculations below will be done in class on the board – this slide is
only for you to read before the class, or later when preparing for a test.
OK., we said that when the two bodies are broght
into contact, heat flows until an equilibriu m is
establishe d, and both bodies have the same temperatu re
Tf . So, the entropy of the warmer body (1) changes by :
 Tf
dT
S1  c  
 c  ln 
T
 T1
T1

 (note : since T f  T1 , S1  0).

The entropy of the cooler body (2) will change by :
Tf
 Tf
dT
S 2  c  
 c  ln 
T
 T2
T2

 (here, S 2  0).

The net change in the system entropy is then :
Tf
S net
 Tf
 S1  S 2  c  ln 
 T1

 Tf
  c  ln 

 T2
 Tf 2 


  c  ln 
 T1  T2 



We still
don’t know
Tf -- but it is
easy to find
it (see next
slide).
The calculations below will be done in class on the board – this slide is
only for you to read before the class, or later when preparing for a test.
Since for both bodies U  c  T , we find :
For Body 1 : U1  c  T f  T1 
For Body 2 : U 2  c  T f  T2 
The net change is U1  U 2 , and it must be zero
because of the energy conservati on. So, we get :
c  T f  T1   c  T f  T2   c  2T f  T1  T2   0
T1  T2
which yields : T f 
2
This result we plug into the expression for S net :
S net
 T f2 
 T1  T2 2 
  c  ln 
 c  ln 

 T T 
 4T1  T2 
 1 2
Now we need only to
check the sign of the
ln function.
The calculations below will be done in class on the board – this slide is
only for you to read before the class, or later when preparing for a test.
 T1  T2 2 
 T12  T22  2T1  T2 
 T12  T22  2T1  T2  4T1  T2 
ln 
  ln 
  ln 

4
T

T
4
T

T
4
T

T
1
2
1
2




 1 2 
 4T1  T2  T1  T2 2 
 T1  T2 2 
 ln 
  ln 1 

4T1  T2
4T1  T2 



The second term in the function' s argument is always
a positive number, so that the argument is always  1,
meaning that the logarithm is always positive - therefore ,
the entropy in the process considered always increases, Q. E. D.
In contrast,
in reversible
processes
the system
entropy does
not change
Example of a reversible process
There is a cylinder, made of a thermally insulating
material, with a heavy piston, also insulating. Initially, the gas is compressed and exerts an upward
force on the piston larger than it weight. When
released, the piston starts mowing up. The gas
pressure gradually drops. The piston oscillates
up and down, in analogy to a mass on a spring.
The process is reversible: first, the gas energy U
is converted into potential energy of the piston,
and then the sequence is reversed, going through
the same stages as when moving up. The gas first delivers, then
absorbs work, but its entropy remains constant during the process.
Another example of an irreversible process: two
different gases in a container:
The container is divided
into two chambers of
equal volumes V by a
partition, and each gas
occupies a separate
chamber
The partition opens, the
gases spontaneously
mix, and now the gas
mixture occupies a
volume of 2V.
Plugging the hole will
not return the system
to the initial state. The
process is irreversible!
One more thing we need: the Ideal Gas Laws
pV  NRT called the equation of state; R  8.314 J/K  mole
3
U  NRT the internal energy of a monoatomic gas;
2
R  8.314 J/K  mole is the so - called " gas constant".
There are several processes that were given special names.
We will soon need to use two of them:
• Isothermal expansion/compression: the gas temperature is kept
constant throughout the process, thus pV = const.
• Adiabatic expansion/compression: no heat flows in or out during
process, meaning that Q = 0, and consequently, since S = Q / T ,
there is no change in the entropy. Therefore, the process is often
called “isoentropic” expansion/compression.
Ideal gas processes
Carnot
Cycle:
(or Carnot
Engine, if
you prefer
“Engine”):
P-V diagram:
Start looking
HERE and
travel
clockwise
Carnot Engine animation
Another one
3
Ideal gas : pV  NRT ; Internal energy : U  NRT ;
2
dQ
Entropy : dS 
, so that at constant T : Q  T  S
T
Carnot Engine operation
First stroke, A→B : Isothermal expansion at high T :
“Hot
source”:
At point A
At point B
T = Th = const., so that U = const. All heat absorbed converted to work!!
WAB  QAB  Th  S AB
Ideal gas : pV  NRT ; U 
3
dQ
NRT ; Entropy : dS 
; at constant T : Q  T  S
2
T
Carnot Engine operation (2)
Second stroke, B→C : Adiabatic expansion, no heat input, S = 0 :
Thermally
insulating
shield
At point B
At point C
Now the gas still delivers work, but uses its internal energy for that:
WB C
3
 U B  U C  NR  (Th  Tc )
2
Ideal gas : pV  NRT ; U 
3
dQ
NRT ; Entropy : dS 
; at constant T : Q  T  S
2
T
Carnot Engine operation (3)
Third stroke, C→D : Isothermal compression at low temperature (Tc):
“Heat
sink”:
At point C
At point D
T = Tc = const., so that U = const. All work absorbed is dumped
in the form of heat to the “heat sink”!!
WC D  QC D  Tc  SC D
It’s a negative
value, keep in
mind!
Ideal gas : pV  NRT ; U 
3
dQ
NRT ; Entropy : dS 
; at constant T : Q  T  S
2
T
Carnot Engine operation (4)
Fourth stroke, D→A : Adiabatic compression, no heat input, S = 0 :
At point D
Back at point A
The gas still absorbs work, but energy cannot get out in the form of heat,
so all work goes to increasing the internal energy, thus heating up the gas:
WD A
3
 U D  U A  NR  (Tc  Th )
2
A negative
value, note.
Let’s collect all results from the above analysis in a table:
Now, calculating the Carnot Engine efficiency will be a piece of cake!
Column 2 yields:
QA B QC  D

0
Th
TC
(continued:)
QA B QC  D

0
Th
TC
(from the last slide)
Wnet  QA B  QC  D (last column in the table)
From the first :
QC  D
TC
  QA B
Th
Plug into the second :
Wnet
 TC 
TC
 QA B  QA B  QA B  1  
Th
 Th 
Hence, the ratio of the work output to the heat input is :
 TC 
Wnet
   1  
QA B
 Th 
This is which we call the
“thermodynamic efficiency” – even though
a better term would be “inefficiency” :o)))