Dear P310'ers,
Our first test is Thursday, October 1, during
class in our
usual classroom. By then you will be in pretty good shape for the test
because you will have done and understood the homework problems. That
said,
I list here key points that are likely to occur to me when I
composed this year's Exam I. You might use this as a kind of
check
list for your
own preparation. Feel free to contact me if you have questions.
(brabson@indiana.edu,
IU Tel: 855-8707, Home Tel: 332-6507) Ben Brabson
Energy Units: There are lots of different units for energy including the standard SI unit, J (Joule). For example, 1 Kcal = 4186 Joules is the mechanical or heat energy necessary to raise the temperature of 1 kg of H2O by 10C, and 1 Btu = 1055 Joules is the energy to raise the temperature of 1 pound of water by 10F. A Joule is also the work to lift an object weighing 1 Newton by 1 meter and a jelly doughnut has about 106 Joules of chemical energy. A very convenient energy unit for discussions about individual molecules and atoms is the electron-volt = 1 ev = 1.6 x 10-19 Joules, the work to move an electron through a potential difference of 1 volt.
Work: W = F o d = Fd cos(Q) defines what we mean by work. Positive work in Joules is done when F in Newtons is in the same direction as d in meters.
Fgravity = mg where g = 9.8 N/kg on the surface of the Earth. We often round g to 10 N/kg for convenience.
Potential energy, PE, can be written as: PE = mgh , where h = vertical height. You store energy as potential energy either by pulling two attractive objects apart (lifting a book away from the Earth) or by pushing two repulsive objects together (pushing two positive charges together into a nucleus). Only changes in potential energy are important. Where you choose to put zero potential energy is up to you. For gravitation problems zero is often taken to be at floor level. For electrostatic problems considering forces between a couple of charges, Q1 and Q2, zero potential energy is usually taken to be at r = infinity giving U = kQ1Q2/r. Then, a system is a "bound" system when its total energy (Etotal = KE + U) is negative.
Kinetic energy, KE, energy of motion, KE = 1/2mv2,can
also be viewed as a way to store energy. A rotating flywheel has
kinetic energy.
Energy conservation for a closed system says that: Etotal(now)
= Etotal(later).
For objects in an approximately
constant gravitational field (near the surface of the Earth, for
example where g = 9.8 N/kg)
ignoring
heat and friction, energy conservation says:
mgh1 + 1/2mv12 = mgh2
+ 1/2mv22
Power = Energy/time = Work/time in Joules/second
or watts. (Though it
makes
sense
to ask how many Joules of energy were consumed in a day, it does not
make
sense to ask how many watts were consumed in a day. If you find
yourself wondering how to convert Joules to watts be sure to carefully
figure out why the question doesn't make sense.
Chemical Energy & Bonding: An attractive force between atoms in a molecule. Covalent bonding takes place when shared electrons between positive ions provide the electrostatic attractive interaction. Ionic bonding is an extreme form of covalent bonding where an electron from one atom spends most of its time around a neighboring atom. Then, an attractive force holds the two ionized atoms together. Van der Waals (weak) bonds exist between dipoles, neutral atoms or molecules that are polarized (non-symmetric charge arrangement). The polarization can either be induced or permanent. In each case the potential energy of a bound system is negative. The binding energy is the work required to separate the components of the molecule. [We didn't discuss this explicitely in class. Chapter 29 in Physics, 5th Edition by Doug Giancoli gives a lovely 5-page description of these bonding processes.]
T2 = doubling time for an exponential. T2
= (ln 2)/k = 0.693/k, where k is the growth constant. For small
values, k is
approximately equal to the fractional increase per year or, times 100,
the percent
interest
per year paid by a bank, R. Approximately, T2
= 70 / (%/yr),
the so-called "Rule of 70." For large values of R, we must calculate k
= ln(1+R/100).
A production plot shows the quantity of a resource extracted
each year plotted versus time. A consumption
plot does the same
for
the quantity/year of a
resource consumed. The area
under such a
curve,
of course, represents the total quantity produced or consumed.
The production curve shapes we discussed were 1.)
constant, 2.) linearly increasing/decreasing, 3.) exponential, and 4.)
Hubbert
(~normal dist.). Presented with such a curve, you should be able to
estimate
the total resource by making an approximate estimate of the area under
the curve.
A semilog plot of an exponential function is straight line. Extrapolating an exponential is easier done on a semilog plot. The energy consumption per year in the US has grown essentially exponentially for 140 years. (Do you expect this to continue for the next 140 years?)
Exponential behavior occurs when dN ~ Ndt. When discussing population growth, N is the number of people and dN the number of babies arriving in time dt. That is, growth by a certain fraction (dN/N) per time interval (dt) leads to exponential growth because the differential equation, dN/dt = kN has the solution: N = N0 ekt where k = growth constant. When a production curve follows an exponential (as US oil consumption did from 1890-1970, then the total resource used, QT is given by the integral of the exponential: QT = (N0/k)(ekT - 1) and the resource used in time T equals T = (1/k)*ln[(kQT/N0) + 1].
Hubbert model: N = NM exp(-z2/2)
where NM = QT/(sqrt(2p)*s)
and z = (t - TM)/s.
The shape of the distribution is that of a normal distribution with
mean, NM and standard deviation, s. For
Hubbert
any three parameters are sufficient to determine the entire production
model. A convenient table in McFarland, Hunt, & Campbell gives the
area under the center of a normal curve for a number of values of
z. This table serves to relate the area under a normal curve to
its parameters.
Resources and Reserves: For a particular energy comodity, reserves estimate the
amount in the ground that can be recovered at competitive prices. A
much less well determined number, resources estimate the
total amount of a a comodity including that part yet
to
be discovered. A large energy unit used when discussing reserves and
resources is a quintillion Btu = 1 Q
= 1000 quads = 1018
Btu = 1.055 x 1021 Joules. Also,
recall our discussions about gas (methane)
hydrates
where a very rough estimate of the resource is 400 Q, none of which has
been successfully recovered.
Roughly, world resources are:
US
World
Gas
1
Q
10 Q
Oil
0.6
Q
7-12 Q
Uranium-235
1
Q
2-3 Q
Coal
40 Q
200 Q
(the dominant fossil fuel resource)
Fusion
(Ocean deuterium: 1010 Q)
Methane
Hydrates
(rough guess ~400 Q)
Oxidation or burning of a hydrocabon:
C + O2 ---> CO2 + 95 Kcal/mol of
carbon
2H2 + O2 ---> 2H2O + 68
Kcal/mol of hydrogen
Specific chemical energy content of many fuels lies between 20 and 45 MJ/kg. For example, Coal contains ~29 MJ/kg, Methanol about 20 MJ/kg, Natural gas about 43 MJ/kg...
Photosynthesis: CH2O
is simplest of the carbohydrates, Cm(H20)n
where m,n
are integers.
An example of
photosynthesis of this simplest carbohydrate is:
gsun + CO2 + H2O
---> C(H20) + O2 Where gsun
is a photon of energy from the sun.
THERMODYNAMICS:
Definition of temperature: For an ideal monatomic gas, temperature,
T is defined by: 3/2kT = <KE>molecule,
where
k = Boltzmann's constant = 1.38 x 10-23 Joules/K.
Total internal energy, U, is defined by the sum of the kinetic energies of the individual molecules in the gas (assuming no potential energy between molecules). Be careful to distinguish the idea of temperature from the idea of internal energy.
Pressure = Force/Area P[N/m2] = F[N] / A[m2]
Specific Heat: When heating
a substance, the quantity of heat needed is
proportional
to the mass and to the change in temperature. The constant of
proportionality
is called the specific heat, c.
Q = c m DT,
where c = 1 Kcal/(kg*K) for
water, for example.
Specific heat for gases, cp, cv:
dQ = cpndT
(constant
pressure)
dQ = cvndT
(constant
volume)
First Law of Thermodynamics: dQ =
dU
+ dW
where dW = p dV is the work done by the
system,
dQ the heat added to the
system and dU the change in the internal
energy.
For an engine where it must return to its original condition after each
cycle, dU = 0 in one cycle, so dQ = dW for a heat engine, dQHot
= dQCold + dW. (Be sure to be able to draw the diagram
for a heat engine.) This leads immediately to the definition of
efficiency
for a heat engine:
real efficiency
= 1
- QCold/QHot
Here, both QCold and QHot are measured in energy
units such as Joules or KCal.
Entropy change is defined by dS = dQ/T = change in entropy,
and
entropy is a measure of the disorder of a system in units of Joules/K.
Second Law of Thermodynamics: The entropy for an
isoloated
system must always remain the same or increase. This has as its
consequence
that the maximum efficiency is given by:
maximum efficiency
= emax
=
1 - Tcold/Thot
Coefficients of performance (COP) for heat pumps, AC, and Refrigerators = "good stuff"/ "what you pay"
Engines:
dU = 0 for a full cycle of an engine
Work done in a cycle = work inside the loop in the
pV diagram.
It is useful to be
able to calculate DW, DQ,
DU
for a full cycle of an engine.
Ideal Gas:
pV = nRT , R = 8.31 J/mol/K = 1.99 cal/mol/K
U
= 3/2 NkT = nRT and <1/2mv2> = 3/2 kT
(monatomic gas)
dU
= 3/2 NkdT = 3/2 nRdT (monatomic gas)
pVg
= TVg-1 = constant
(adiabatic)
For a
monatomic
gas: cp
= 5 cal/mol/K and cv
= 3 cal/mol/K.
For a diatomic gas:
cp = 7 cal/mol/K
and cv = 5 cal/mol/K.
Conduction and convection:
Q/t = (kA)dT/dx where Q in Joules is the heat transfered
through a thickness dx of
material with coefficient of
conductance, k when the temperature difference across the material is
dT.
Q/t = (AdT)/R where R
= R1 +
R2 + R3... and
R = dx/k
for materials and
R = 1/h for dead air layer
near various surfaces:
h = 0.18 (dT)1/4
for vertical surfaces...
h = 0.13 (dT)1/4
for ceiling surfaces...
h = 0.25 (dT)1/4
for floor surfaces...
Electromagnetic Radiation: For periodic wave motion we can write: v = f l, or velocity = frequency x wavelength. The velocity of light in vacuum is a constant, v = 3.00 x 108 m/s.
Intensity [Watts/m2] =
Power[Watts]/Area[m2],
For example solar intensity at the top of the atmposhere = I0
= 1350 Watts/m2.
Stefan's Law: The Power emitted
by a black body at temperature T is given by: P =
sAT4,
s
= 5.67 x 10-8 Watts/m2/K4.
Wein's Law: The maximum
wavelength
emitted by a body at temperature T is given by the equation:
l[m]T[K]
= 2.9 x 10-3 m K
Ideas from Articles and Readings:
1.) Jatropha beans as a potential source of biofuel
2.) Driving on Biomass, John Ohlrogge et al.
3.) BP "Giant' Oil Discovery
4.) How Much Coal
5.) Minnesota Ecologist Pushes Prairie Biofuels
6.) MIT Airconditioning
7.) Leaping the Efficiency Gap
8.) Sandia Laboratory's Stirling Energy System for collecting solar
energy
***************************** Test I will not include the
following
items *************
The greenhouse effect is
illustrated
by the atmosphere which is largely tranparent to visible light but
absorbing
of infrared light because of molecules like CO2, H2O,
CH4, CFC's...
Emissivity in black body radiation:
An ideal
emitter
(e = 1) is also an ideal absorber (a
= 1). It absorbs all of the radiation that hits it and is called
a black body.
Selective Surfaces: In general,
the emissivity, e(l) and the absorptivity, a(l),
are equal at the same wavelength, l. This
does
not restrict their values at different wavelengths. A surface
that
has a large
absorptivity (and emissivity) in the visible and a small absorptivity
(and emissivity) in the infrared, for example, is called a selective
surface.
It absorbs well in the visible but does not emit well in the infrared.
Therefore, it can reach higher temperatures when sitting in the sun
than
does a surface with a large absorptivity (and emissivity) in the
infrared
like a hot black road. Such surfaces are useful in solar
collectors
where one wants to reach the highest possible temperatures.