Consider the questions below that relate
to the lab. Once you have explored the issues, you will write-up
a report following the guidelines provided. We will be looking
for a write-up that pulls the information together and does
not simply answer each of the given questions in order
of presentation. Your formal write-up will be done outside
of the lab period. Take notes of your work, and then after
the lab, organize these notes into your write-up.
Please keep in mind the following guidelines
when taking part in group work:
The formal write up for this lab should be
from four to five pages long (counting the title page and
your description of the group process) and should be typed.
The text should be double spaced. Be sure that every
member of the group signs the lab report. By signing your
name, you are indicating that you contributed significantly
toward the completion of that report.
(on separate lines).
Your grade will be determined by the content
and mechanics of your final report. The only way that we have
to determine whether or not you understand the concepts discussed
in class is through the way in which you communicate your
understanding of the concepts. The primary goal of this exercise
is for you to demonstrate, through the process of a formal
write-up, that you can effectively communicate complex ideas.
We will pay close attention to the following aspects of your
write-up:
The University Writing Center is located
in Hamiton Smith, room 7. Their phone number is 862-3272.
We highly recommend that you take full advantage of the resources
that the Writing Center offers. Please make sure that you
contact the people at the writing center and make an appointment
with them at least two days prior to the time that you would
like to meet with them. The sooner that you meet with them
in the writing process the better. For example, it would be
best to meet with them to show them your outline before the
first draft is started.
One of the common problems
that earth scientists face is the estimation of the amount
of a particular material within the subsurface (ex: water
or oil). The total material within a geological formation
is estimated by using information on the concentration of
the material (amount/volume) and the volume of the region
containing the material. This module will look at an example
of this problem by estimating the volume of oil in a petroleum
reservoir. The analysis will focus on the evaluation of the
gross resource potential and on making decisions about possible
development.
In the petroleum industry,
decisions on reservoir development are based largely on financial
considerations. We need to know how much oil is in a given
volume of rock as well as the total volume of oil in the reservoir.
The amount of oil in a given volume of rock depends on many
things. It is important to recognize that not all of the oil
is recoverable (extractable), but we are still interested
in the total amount of oil in place. We will leave the worries
about extraction to the engineers.
We will first examine how water or oil becomes
trapped within geologic formations. Rock formations are formed
from a composite of solid materials, and they often have empty
spaces (or voids) within them. (See Figure 1-1) These void
spaces can be filled with water or oil. For a given volume
of a rock formation, a certain amount is composed of solid
material and the rest is composed of void space. The ratio
of the void space and the total volume (volume of void space
divided by the total volume) is called the porosity. The value
of the porosity can be different for different positions within
the formation. (see Figure 1-2)
Figure 1-2: Porosities
can vary for different types of rocks. (figure taken
from Bear,1972)
Different kinds of fluids can accumulate within
the void spaces of a rock formation. For example, both oil
and water can be trapped. Because the two fluid do not mix
and the density of oil is less than water, the oil will
tend to move upwards within the formation. In some formations,
natural "traps" form (See Figure 1-3) . We will focus on
a particular trap called a "convex trap". As it moves upward,
the oil may become trapped by a layer of rock that has an
extremely small porosity (such as clay).
These trapped volumes of oil, or reservoirs,
are the sources of oil that the engineers want to extract.
The decision to drill for the oil is based on many factors.
It is not always obvious where these traps occur, and it
is not always clear how much oil is available to extract.
We will concentrate on this later problem. We will attempt
to estimate how much oil can be found within a given reservoir.
Figure 1-3: Natural traps can form where oil collects.
We will focus on the trap in (a). (Jacob Bear, 1972)
Figure 1-4: Water and oil can become
trapped within some formations. (Figure taken from Bear,
1972)
2. Test Case
First let's consider a simple example. If
we do not have a lot of information about the geological
site, an approach we may take is to approximate the petroleum
reservoir as a simple geometry (e.g., a cylinder) and use
an average porosity. First, we assume that the formation
is roughly cylindrical in shape and has a uniform (or isotropic)
porosity.
The total oil in place, M(oil), is then simply
estimated by
In general, the porosity in the reservoir
sandstone can vary with the depth, and the geometry of the
reservoir is non-trivial. We will need a more general way
to approximate the total volume of the oil, Moil. Building
upon the idea above, the reservoir can be split into a set
of horizontal disks. Here, the porosity can change in the
vertical direction, so we denote the porosity as s(z). Then
the total oil in place can be approximated by
If the geometry is not cylindrical and the
radius changes with respect to z, the vertical direction,
we can include the changing radius in our equation as follows:
CLASS EXERCISE
Objective: Calculate the volume of oil in
a petroleum reservior.
1. Suppose that a sandstone formation looks
like a stack of disks (the height of each disk is 15m):
What is the total volume of the formation?
2. Assuming a constant porosity of 0.1 in
the reservoir, calculate the volume of oil in the reservoir.
3. Now suppose the formation has the following
configuration:
r(i) = 10m is the inner radius and r(o) is
the outer radius of the reservoir (the same as problem 1).
Again, using a porosity of 0.1, calculate the total volume
of oil in the reservior.
4. Draw a side view of the oil reservoir in
problem 3.
5. Suppose that the dome-shaped reservoir
can be described using two parabolas.
The inside parabola is z=20-0.089r2 and the
outside parabola is z=45-0.018r2
a) Calculate the "important" z values. (hint:
look at the figure and divide it into sections)
b) Set up column one of your spreadsheet to
hold the different z values.
c) Set up column two to be the inside radius
at a particular z value.
d) Set up column three to be the outside radius
at a particular z value.
e) Set up column four to hold the porosities
at a given z value.
f) Use the last column to calculate the volume
of oil in each given slice.
GAMMA RAY BURSTS
Application of Trigonometric
Functions in Astronomy
This module examines a
mysterious phenomenon in astronomy and how trigonometric
functions are part of the language used to describe the
phenomenon. Back in the 1950's, a campaign was waged to
stop the testing of nuclear weapons in the atmosphere. It
was discovered that these nuclear tests, several of which
had already taken place, created serious health problems
in people around the world. Both the United States and USSR
were paranoid about each other performing nuclear testing
and believed that the other country might even go so far
as to test nuclear weapons on the dark side of the moon
where it would be harder to detect. Because of this concern,
both countries sent up satellites to detect the energetic
photons, known as gamma rays, that would be released by
such a bomb.
Our satellites did,
in fact, see explosions of gamma rays. However, it did not
take long to realize that these were not nuclear bomb explosions;
they had the wrong distribution of energy and they did not
behave like a nuclear bomb. But these explosions (that is,
short transient events) were considered interesting by a
few people who had access to these data obtained from military
satellites. However, because it was military data, they
were not allowed to talk about it for five years after it
was first observed. The first paper concerning these explosions,
known as Gamma Ray Bursts or GRBs for short, appeared in
1971.
Between 1971 and 1991
there were occasional random observations of GRBs by scientific
satellites that were in space for other reasons. In particular,
the Pioneer Venus orbiter had a small detector. The Interplanetary
Network, established in 1979, as well as the Solar Maximum
Mission that was designed and built here at UNH, were also
able to detect GRBs. By 1991, about 500 of these bursts
had been seen, but because the detectors were small, only
the brightest bursts could be detected.
In the beginning there
were many theories about what kind of process might create
such an explosion. It might have been supernovae or flare
stars; however, by 1991 the most well-accepted theory was
that these were events associated with neutron stars and
the reconnection of magnetic fields on these stars. Researchers
were confident of this theory because similar events had
been seen on flare stars, so it was a very plausible explanation.
No theory is without
its consequences, and we now need to look at one of the
consequences to better understand the theory. These neutron
stars were capable of releasing a certain amount of energy.
This amount was small enough that if these explosions happened
in distant galaxies we would not be able to see them since
they would be too faint. But if they happened in our
galaxy, we would see them. This means that if we plotted
their location in the sky, we would find the bursts occurring
along the galactic plane. This did not agree with the data
that had been taken so far. These data showed that the gamma
ray bursts were distributed uniformly across the sky (called
an isotropic distribution); this is not what we would expect
from the neutron star theory. But this could be understood
even if we accept the neutron star theory. To see this,
consider the stars that are visible to the naked eye. If
we look at stars in Durham, we see only the brightest (i.e.,
the closest) ones because of the light pollution, and these
appear to be be evenly distributed (isotropic) in the sky.
However, if we go to a mountain in Maine in the middle of
nowhere, we are able to see the fainter stars in the galactic
plane, and the distribution of stars is no longer isotropic.
Scientists were convinced
that if they could do the equivalent of going to a mountaintop
in Maine they would see the galactic distribution. In April
1991 they did just that: they launched the Compton Gamma
Ray Observatory with one instrument aboard whose only job
was to look for GRBs. This detector was much bigger than
previous ones and could see much fainter bursts. What did
they see? Not a galactic distribution but an isotropic distribution.
It was a shock! This may mean that the source of the gamma
ray bursts is much closer (like the bright visible stars)
or much further away (since the distribution of galaxies
is also isotropic). Either way, the neutron star theory
was in trouble.
Before researchers could
accept a conclusion that destroyed a good theory, the conclusion
needed to be made quantitative; they needed to characterize
mathematically how closely the distribution approximated
an isotropic distribution. One way to do this is to find
the average distance from the galactic plane (averaged over
all bursts observed). Isotropic distributions and galactic
distributions will have very different values for this distance,
so this will provide a measure of deviation from isotropy.
CLASS EXERCISE
In this laboratory exercise,
you will first calculate the average distance from the galactic
plane for an isotropic distribution (this is the theory),
and you will then calculate the average value for several
different data sets. You will be asked to judge which data
sets are consistent with isotropy and which are not.
A. The first step
is to state the question more precisely mathematically.
1. We begin
by imagining that the sky is like a globe seen from
the inside. The height of a burst above the galactic
plane (which is analogous to the equator on a globe)
would be x = rsinq
where q
is the angle that corresponds
to latitude (i.e., north and south). Verify
that this is the correct formula for the height using
the sketch.
2. Because we cannot calculate the radius,
r, we will look only at the value of sin (q
). Now imagine that we have a galactic distribution
and an isotropic dietibution, both of which are symmetric
about the galactic equator. What is the average distance
from the plane if distances below are negative and distances
above are positive? (Given that any symmetric distribution
will give the average value of q
= 0, we look at the average value of sin2q
this gives different values for different distributions
since there is no cancellation of positive and negative
values.)
B. To find the throeretical value for the
average, we must calculate the average value of sin2q
for an isotropic distribution. How do we do this?
1. We will start with a similar example: finding
the average value of a die thrown. Suppose that I throw
a six sided die. For one throw, what number should I expect
to get? We can calculate this nmumber by adding all possible
outcomes multiplied by the probability for each outcome:
2. We now see that
our job is to calculate the average value of sin2q
. That is, we must calculate the summation of sin2q
multiplied by the probability
of obtaining q for
all values of q
. We now can calculate that probability.
Instead of looking
at a uniform distribution on the sphere, let’s begin
with a uniform distribution on a line. Imagine something
(e.g., M& s) being uniformly distributed along a line
one meter in length. What fraction of the M& s can be
found in the first 1/2 meter? What fraction is in the middle
1/3 meter? What fraction is in the last 1/4 meter? (Draw
a picture)
3. Now imagine that
we have the same M& s evenly distributed on a one meter
by one meter square table. What fraction is in the upper
quadrant? What fraction is in the lower 1/3 by 1/3 meter.
How do you know? What is your general rule for finding how
much stuff is in a given area? (Draw a picture)
4. Now we need to extend
these general ideas to even distributions of M& s on
a sphere. Look at the sheet with galactic coordinates. If
you look at the area between the 60-degree contour and the
90-degree contour and compare that to the area between the
0-degree contour and the 30-degree contour, are they the
same? If not, which is bigger?
5. It is assumed that the
area on a globe between latitude q
and q+Dq
is proportional to cosq.
Give arguments to support this based on the plot.
6. We are now to the point
where we know that the probability to be at an angle q
is proportional to cosq
for an isotropic distribution. But this might be multiplied
by a constant. We must determine that constant. We get the
constant by requiring that the probability of being somewhere
on the globe is one. That is,
Total probability = Sum
of the probabilities for each possible value = 1
Does this this hold for
tossing a die?
7. Now, for locations on
the globe we have that
This leads to an expression
for F,
where F is the normalization
constant you want to determine. Now we would like to sum
over all values of q,
but we cannot because there are an infinite number of them.
Instead, to calculate this sum, evaluate cosq
for 20 evenly spaced values of q
between -p/2
and p/2
and find F. Repeat this for 50 values, then 100 values.
What value do you find? Do you think you would find something
much different if you could sum over all these values?
8. Now we are ready to
find the average value of sin2q:
Again, calculate this sum
for 20, 50, and 100 values of q
(evenly spaced from -p/2
to p/2).
Does it seem to be converging to an answer? What is it?
9. Now take each of the
data sets and, using a spreadsheet, and do the following:
a. Load the spreadsheet
called grb.xcl.
b. Calculate
the average value of the three columns and compare it
to your answer for an isotropic distribution. Which
data sets appear to be isotropic?
(Note that none
of the data sets agree with isotropy exactly. This is
where error analysis becomes essential. Error analysis
would tell us how large a deviation we would expect from
the istropic distribution value given the error in angles
and the number of bursts included in the average. But
this is a subject for another time.)
CHEMICAL REACTION RATES
Application of Graphical Methods and
Differentiation in Chemistry
Objective: In this module, you will
be given realistic data for the concentrations of reactant
in solution at various times. You must then determine a
useful functional form by graphical methods using a spreadsheet
program, and you will have to use a trial and error process
or intuition for this step. Having used the data to obtain
a functional form, you will analytically differentiate the
model to produce a differential rate equation. The differential
equation is more transparently linked to the physical processes
involved. In the more advanced analysis, you will start
from solutions of a differential equation and then work
backwards to derive the differential equation. You will
construct your own technique for solving differential equations.
Reaction Rates and Mechanisms
Quantitative study of the rates of change
of concentrations of species in chemical reactions is a
powerful tool in understanding the mechanism of a reaction
-- that is, what is actually happening during a reaction.
For instance, one step in the series of reactions resulting
in the depletion of ozone (O3) in the stratosphere
is:
But how do we know that the ozone (O3)
does not just fall apart on its own? In other words, how
do we know that the following reaction does not take place:
Does O subsequently combine with Cl? What
really happens is called the mechanism of a reaction.
Monitoring the rates of reaction can help. If both Cl and
O3 must be present simultaneously for the reaction
to occur, then the frequency of reaction will depend on
the amount present of both species.
More precisely, the frequency of reaction
will be proportional to the concentration of both species.
That is:
Frequency ~ [Cl] and Frequency ~ [O3]
Or,
Frequency ~ [Cl] [O3]
The more frequent the reaction, the more rapidly
O3 will be depleted. Thus,
(Notice that we have replaced the proportionality
sign with an equality by inserting a constant, k.) The minus
sign is used to recognize the fact that the O3
concentration is decreasing with time. This Equation is
called the rate law for the reaction.
On the other hand, if the ozone had decayed
through an encounter with another ozone molecule, then the
rate law would have been:
The differential rate law is clearly linked
to the mechanism of the reaction. In other words, if we
can observe the amount of O2, then we can try to see if
the change in its concentration is consistent with one of
the equations above.
Rate Data
In the real world, the rate law must be inferred
from experimental data. Typically, one could observe the
ozone concentration as a function of time. For simplicity,
letıs assume that [Cl], the concentration of chlorine, is
large and, therefore, changes little as time progresses.
Then the rate law becomes:
where kı = k [Cl] is a constant.
Solving the differential equation yields:
This is the exponential decay law.
However, attempting to fit such a function
can be a risky procedure. A much safer method is to plot
the data points and draw some simple curve through them.
Preferably, a straight line will provide a reasonably good
fit to the data.
Taking logarithms of both sides of the equation
above, we arrive at:
A plot of (ln[O3]t)
versus t is a straight line with intercept (ln[O3]t=0)
and slope (k').
CLASS EXERCISE
Given the following data for the concentration
of X in mole/liter ([X]t=0) vs. time in seconds,
find the differential rate law:
Steps:
1. Using the data given above, graph [X] vs.
t.
2. Graph - (change in) [X]/Dt for each value
of t.
3. Find the best straight-line that fits the
data. That is, find the functions of [X] and t such that
f([X]) vs. g(t) is a straight line. (Hint: Try [X]n
vs. t and ln[X] vs. t)
4. You have now found a simple function of
[X] that, when plotted against t, gives approximately a
straight-line graph. Rewrite the equation of the straight-line
graph with [X] as the subject. Then, using linear regression,
determine the intercept and slope.
5. What is the slope of the line?
6. Now differentiate the straight line equation
to obtain the differential rate law.
Example: If you find the natural logarithm
of [X] -- ln[X] -- to be linear in t, then f([X]) = ln[X]
and g(t) = t. The equation is: ln[X] = constant - m t.
Differentiate this equation to get the rate law (don't forget
that [X] is a function of t.)
7. What does the number m tell you? (Give
examples of what it means to have a larger and a smaller
value of m.)
8. Open the spreadsheet, rate.xcl. Examine
the column of data and approximate the value for a rate
law describing the reaction. (Go through the same steps
as above.)
ADVANCED EXERCISE
Find the integrated form for the rate law
(Equation 1) above without assuming [Cl] is constant. Hint:
Try differentiating ln([O3]/[Cl]), where [O3]
= [O3]t=0 - x and [Cl] = [Cl]t=0
- x. Then remember what happens if [Cl] is constant. Once
you get to the differential rate law, backtrack to see how
to solve such equations in the future.
MODULE FOUR. ANTENNAE DESIGN
Application of Logarithms in Wireless Communication
Systems
This module involves modeling the wireless
communication channel in free space. ìFree spaceî
means that you are far away from objects that would reflect
or scatter radio waves, such as the ground or buildings.
Thus, the calculations presented here can be applied to
communication between satellites in space or between a satellite
and an earth station. Our objective in this exercise is
to calculate the power that will be received by an antenna
for a specified transmit power.
Basic Concept: The Antenna
An antenna is a transducer: It converts
electrical energy to electromagnetic energy, and vice versa.
For most applications, we are interested in
determining the power available at the terminals of the
receiving antenna (Power Out). The received power will be
a function of the power radiated or transmitted (Power In),
the distance between the two antennae (R), and other factors
(e.g., terrain, foliage, atmospheric refraction, buildings).
Because we are assuming free-space propagation, we will
be ignoring these other factors.
Consider the case of an isotropic antenna
in free space. Isotropic means that radiation is equal in
all directions. It should be noted that isotropic antennae
are impossible to build, yet they are used as a reference
in defining the performance characteristics of real antennae.
Suppose that a power of Pt Watts
is fed to our hypothetical isotropic transmitting antenna
(Pt = Power In). If we sum all of the power that
is being radiated by the antenna, then, by the conservation
of energy, it must equal the power that we are putting into
the antenna. We can sum the power leaving (i.e., being radiated
by) the antenna by integrating (summing) the Poynting vector
(P), a measure of pewer density with units of Wattrs/m2,
over any closed surface surrounding the antenna.
Mathematically, this sum would be of the form:
Since radiation is radial and equal in all
directions, the intergral is equal to |P| times the
surface area of the sphere, |P|4(pi)R2.
We know how much power is being put into the antenna and
can solve the equation for P,
This gives us power density as a function
of the power in, Pt, and the distance from the
antenna, R. The maximum power that can be collected by a
receiving antenna, Pr, is given by Pr
= |P|Ae, where Ae is the Effective
Aperture of the receiving antenna. You can think of the
antenna as a bucket that collects electromagnetic energy,
with Ae being the size of that bucket. For any
antenna the aperature is given by:
where G is the gain of the antenna. The gain
of an antenna represents its ability to radiate more energy
in one direction than another. For the isotropic antenna
we are considering here, G=1 since power is radiated equally
in all directions. Using G=1 in the above equations, we
can calculate the power received:
Converting to decibels
Why Bother? All calculations could
be performed without converting to decibels, but the calculations
are much easier, and less prone to errors, if decibels are
used. Decibels are based upon logarithms; this enables the
summing of terms rather than multiplication.
[Recall: log(a) + log(b) = log(ab) and
log(a) - log(b) = log(a/b)]
In determining the received signal power we
need to multiply and divide many terms; if we can use addition
and subtraction rather than multiplication and division,
then our job will be much easier.
To convert our equation to decibels, first
substitute for all known constants in the Friis equation:
This can be simplified further to:
Now take 10 x log10 of both sides to get the
ratio of the received power to the transmitted power (Pr/Pt)
in decibels, or dB (this is the way to express a power ratio
in decibels):
This represents the electromagnetic path
gain between two isotropic transmitting and receiving
antennae in decibels (dB). Note that a negative gain in
dB represents a loss.
Accounting for antenna gains, and finding
the total received power
Because the gain of an antenna is expressed
in dB with respect to an isotropic antenna (dBi), we can
readily incorporate antenna gain into our calculations.
To keep our units consistent, we need to express our input
power in decibels. This is typically expressed in dBm, in
which the input power is referenced to a milliwatt (1 milliwatt
= 0.001 Watt). Keep in mind that decibels always represent
a ratio: dBm = 10log(Pt/0.001).
(Power Out)dBm = (Power In)dBm
+ (Transmitter Antenna Gain)dBi
+
(Path Gain)dB
+ (Receiver Antenna Gain)dBi
CLASS EXERCISE
Problem
Determine the received power (in Watts) if
the transmitter and receiver are 10 km apart operating at
a frequency of 125 MHz with 4 Watts transmitted power, and
both the transmitting and receiving antennae are half-wave
dipoles (Antenna Gain = 2.15 dBi for a half-wave dipole).
Method of Solution
1. Find the transmitted power in dBm.
2. Determine the Path Gain in dB.
3. Sum the gains and the transmitted power
to determine the received power in dB:
4. Convert the received power, Pr,
from decibels to Watts.
