GGR124H1 Lecture : Bullet point notes of topics covered in lecture

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25 Jan 2011

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N course revolves around the macro study of cities in the contemporary and historical context. N north american urban layout is based on the greek and roman models. N the city is a system: urban system. N the micro approach (the individual city) consists of studying the internal relationships between the various elements of cities. N the latter view regroups the social, economical end infrastructural factors that interact with each other. N urbanization concepts will cover the methodology of urban geographical study. N urban growth and evolution will delve into the progress of cities. N the purposes of urban centres and their subsectors will be evaluated. N housing is the prime component in urban social development. N city governance determines how it functions in relation to all other levels of government. N urban: permanent settlement of a concentrated, dense population.

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3. 3 Stratosphere-troposphere exchange

The rate of exchange of air between the troposphere and thestratosphere is critical for determining the potential of variouspollutants emitted from the surface to reach the stratosphere andaffect the stratospheric ozone layer. One of the first estimates ofthis rate was made in the 1960s using measurements of strontium-90(90Sr) in the stratosphere. Strontium-90 is a radioactive isotope(half-life 28 years) produced in nuclear explosions. It has nonatural sources. Large amounts of 90Sr were injected into thestratosphere in the 1950s by above-ground nuclear tests. Thesetests were banned by international treaty in 1962. Following thetest ban the stratospheric concentrations of 90Sr began to decreaseas 90Sr was transferred to the troposphere. In the troposphere,90Sr is removed by wet deposition with a lifetime of 10 days (bycontrast there is no rain, and hence no wet deposition, in thestratosphere). An intensive stratospheric measurement network wasoperated in the 1960s to monitor the decay of 90Sr in thestratosphere. We interpret these observations here using a 2-boxmodel for stratosphere-troposphere exchange with transfer rateconstants kTS and kST (yr-1) between the tropospheric andstratospheric reservoirs. The reservoirs are assumed to beindividually well-mixed.

Let mS and mT represent the masses of 90Sr in the stratosphere andin the troposphere respectively. Observations of the decrease inthe stratospheric inventory for the period 1963-1967 can be fittedto an exponential mS(t) = mS(0)exp(-kt) where k = 0.77 yr-1.

1. Write mass balance equations for mS and mT in the 1963-1967period.

2. Assuming that transfer of 90Sr from the troposphere to thestratosphere is negligible (we will verify this assumption later),show that the residence time of air in the stratosphere is tS =1/kST = 1.3 years.

3. Let m'T and m'S represent the total masses of air in thetroposphere and the stratosphere, respectively. Show that theresidence time of air in the troposphere is tT = tS (m'T/m'S) = 7.4years. Conclude as to the validity of your assumption in question2.

4. Hydrochlorofluorocarbons (HCFCs) have been adopted asreplacement products for the chlorofluorocarbons (CFCs), which werebanned by the Montreal protocol because of their harmful effect onthe ozone layer. In contrast to the CFCs, the HCFCs can be oxidizedin the troposphere, and the oxidation products washed out byprecipitation, so that most of the HCFCs do not penetrate into thestratosphere to destroy ozone. Two common HCFCs have trade namesHCFC-123 and HCFC-124; their lifetimes against oxidation in thetroposphere are 1.4 years and 5.9 years, respectively. There are noother sinks for these species in the troposphere. Using our 2-boxmodel, determine what fractions of the emitted HCFC-123 andHCFC-124 penetrate the stratosphere.

In a circuit containing only resistors, the basic (though notnecessarily explicit) assumption is that the current reaches itssteady-state value instantly. This is not the case for a circuitcontaining inductors. Due to a fundamental property of an inductorto mitigate any "externally imposed" change in current, the currentin such a circuit changes gradually when a switch is closed oropened.
Consider a series circuit containing a resistor of resistance andan inductor of inductance connected to a source of emf andnegligible internal resistance. The wires (including the ones thatmake up the inductor) are also assumed to have negligibleresistance.
Let us start by analyzing the process that takes place after switchis closed (switch remains open). In our further analysis, lowercaseletters will denote the instantaneous values of various quantities,whereas capital letters will denote the maximum values of therespective quantities.
Note that at any time during the process, Kirchhoff's loop ruleholds and is, indeed, helpful:
.
Part A
Immediately after the switch is closed, what is the current in thecircuit?
Hint A.1
How to approach the problem
Hint not displayed
ANSWER:

zero
Answer not displayed
Part B
Immediately after the switch is closed, what is the voltage acrossthe resistor?
Hint B.1
Ohm's law
Hint not displayed
ANSWER:

zero
Answer not displayed
Part C
Immediately after the switch is closed, what is the voltage acrossthe inductor?
Hint C.1
A formula for voltage across an inductor
Hint not displayed
ANSWER:

zero
Answer not displayed
Part D
Shortly after the switch is closed, what is the direction of thecurrent in the circuit?
ANSWER:
clockwise
counterclockwise
There is no current because the inductor does not allow the currentto increase from its initial zero value.
Answer not displayed
Part E
Shortly after the switch is closed, what is the direction of theinduced EMF in the inductor?
ANSWER:
clockwise
counterclockwise
There is no induced EMF because the initial value of the current iszero.
Answer not displayed
Part F
Eventually, the process approaches a steady state. What is thecurrent in the circuit in the steady state?
ANSWER:

Answer not displayed
Part G
What is the voltage across the inductor in the steady state?
Hint G.1
Induced EMF and current
Hint not displayed
ANSWER:
zero

Answer not displayed
Part H
What is the voltage across the resistor in the steady state?
ANSWER:
zero

Answer not displayed
Part I
Now that we have a feel for the state of the circuit in its steadystate, let us obtain the expression for the current in the circuitas a function of time. Note that we can use the loop rule (goingaround counterclockwise):
.
Note as well that and . Using these equations, we can get, aftersome rearranging of the variables and making the subsitution,
.
Integrating both sides of this equation yields
.
Use this last expression to obtain an expression for . Rememberthat and that .
Express your answer in terms of , , and . You may or may not needall these variables. Use the notation exp(x) for .
ANSWER:
=
Answer not displayed
Part J
Just as in the case of R-C circuits, the steady state here is neveractually reached: The exponential functions approach their limitsasymptotically as . However, it usually does not take very long forthe value of to get very close to its presumed limiting value. Thenext several questions illustrate this point.
Note that the quantity has dimensions of time and is called thetime constant (you may recall similar terminology applied to R-Ccircuits). The time constant is often denoted by . Using , one canwrite the expression

as
.
Find the ratio of the current at time to the maximum current.
Express your answer numerically, using three significantfigures.
ANSWER:
=
Answer not displayed
Part K
Find the time it takes the current to reach 99.999% of its maximumvalue.
Express your answer numerically, in units of . Use threesignificant figures.
ANSWER:
=
Answer not displayed

Part L
Find the time it takes the current to reach 99.999% of its maximumvalue. Assume that ohms and millihenrys.
Express your answer in seconds, using three significantfigures.
ANSWER:
=
Answer not displayed
seconds
Part M
Find the time it takes the current to reach 99.999% of its maximumvalue. Assume that ohms and henrys.
Express your answer in seconds, using three significantfigures.
ANSWER:
=
Answer not displayed
seconds
Part N
What fraction of the maximum value will be reached by the currentone minute after the switch is closed? Again, assume that ohms andhenrys.
Use three significant figures in your answer.
ANSWER:
=
Answer not displayed
Now consider a different situation. After switch has been closedfor a long time, it is opened; simultaneously, switch is closed, asshown in the figure. This effectively removes the battery from thecircuit.
The questions below refer to the time immediately after switch isopened and switch is closed.
Part O
What is the direction of the current in the circuit?
ANSWER:
clockwise
counterclockwise
The current is zero because there is no EMF in the circuit.
Answer not displayed
Part P
What is happening to the magnitude of the current?
ANSWER:
The current is increasing.
The current is decreasing.
The current remains constant.
Answer not displayed
Part Q
What is the direction of the EMF in the inductor?
ANSWER:
clockwise
counterclockwise
The EMF is zero because the current is zero.
The EMF is zero because the current is constant.
Answer not displayed
Part R
Which end of the inductor has higher voltage (i.e., to which end ofthe inductor should the positive terminal of a voltmeter beconnected in order to yield a positive reading)?
Hint R.1
How to approach the problem
Hint not displayed
ANSWER:
left
right
The potentials of both ends are the same.
The answer depends on the magnitude of the time constant.
Answer not displayed
Part S
For this circuit, Kirchhoff's loop rule gives
.
Note that , since the current is decreasing. Use this equation toobtain an expression for .
Hint S.1
Initial and final conditions
Hint not displayed
Hint S.2
Let us cheat a bit
Hint not displayed
Express your answer in terms of , , and . Use exp(x) for .
ANSWER:
=
Answer not displayed

GGR124H1 Lecture : Bullet point notes of topics covered in lecture

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