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GGRA03H3 Lecture 7: Megacities Lecture 7

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yellowdolphin818
20 Apr 2011
School
UTSC
Department
Geography
Course
GGRA03H3
Professor
Andre Sorensen

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Document Summary

Megacities ggra03 lecture 7: why are megacities important, many cities are rapidly growing and soon will pass the 10 million thresholds. 10 million and above = megacities (current definition) Special issues of very large cities 2: cost of living, land cost, business cost goes up in larger cities, advantages: business much more efficient in larger cities because you have more. If you invest in transportation then you can reduce transport cost suppliers, huge supply of labour and advertising firms. If you are a company it"s good to be in a large scale economy and having a bigger selection of employees (deep pool of talent) and the input and market that you need www. notesolution. com. 2/3 of all development are illegal (building houses, factories) without permit therefore impossible for government to make a plan cause no one will follow it) poor cities.

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Related Questions

Situations arise in engineering when we want to control the flow rate of a solution. This solution could be pumped into a holding tank, distillation column, heat exchanger, reactor, etc. This means that the flow rate becomes a function of time. When we turn on a pump, the flow rate is (for the most part) constant. The problem is that we may want the flow rate to be very fast at first, and then very slow towards the end of the process. This is especially true in certain chemical reactors where the concentration of reactants is always important. Control systems that could handle this type of situation are very expensive. Pumps (relatively speaking) are cheap. So, having a bank of pumps will usually solve the problem. The pumps are simply turned on or off to regulate the flow. Mathematically, this amounts to taking the volumetric flow rate function and finding a piecewise-linear approximation. Suppose that, because of the application we are running, we need a flow rate given by the function ,where t is in hours and v(t) is in 100 cubic feet per hour (e.g., v(0)=1 which means that at time t=0, we need a flow rate of 1*100=100 cubic feet per hour). Find a linear approximation and a piecewise-linear approximation using first two lines and then three lines. Think about what using more lines would do to the approximation. You will need to find suitable values of t to be able to find your approximation. As a further exercise, try to find values of t for your approximation that keeps the error (i.e., the difference between the approximation and the real function v(t) ) less than 0.1.Hint: to help with this, ask yourself this question: where would more lines be useful - where the function changes a lot or where the function changes a little?]

Consider a satellite of mass ms in a circular earth orbit. Anysatellite orbiting in the upper reaches of our atmosphere willexperience a small drag force, Fd, opposite to itsdirection of motion. This drag force will cause the satellite tospiral inward toward smaller orbital radii. In the process, theorbital velocity of the satellite actually increases! This problemwill investigate how a drag force can cause a body in orbit tospeed up.

The situation is sketched in Fig. 1. The diagram shows separateforce and velocity diagrams for the satellite. The mass of theearth is denoted me, while the mass of the satelliteis ms. The tangential velocity of the satellite isvt and the small radial velocity is acquires because itis spiraling inwards is vr. The size of vr isgreatly exaggerated in the diagram, and we can take vr<< vt and use vt as an adequateapproximation to the magnitude of the total velocity, v. We willcall θ the ‘angle of de- scent’.

Turning to the force diagram, we see that the total force actingon the satellite is the vector sum of F⃗d andF⃗g. Again, the drag force is much weaker than thegravitational force, so Fd << Fg . Wecan use this to our advantage by just considering the effect ofFg along the line of descent. We can see from thediagram that Fg has a component along the line ofdescent given by Fg sin(θ).

(a) First show how we arrive at the following two basicequations governing this system, and carefully describe thephysical meaning of the quantity msad inwords:

Fgsinθ−Fd = msad (1)

Fg = (ms*vt2)/r= G(msme)/r2 (2)

(b) We can find out more about ad and its relation tothe various forces and velocities by powerful

use of the chain rule for differentiation:

Explain Eq. 3 in your own words. Where did vr comefrom and why is the ≃ sign there?

ad = dv/dt = (dv/dr)*(dr/dt) =vr(dv/dr) ≃ vr(dvt/dr) (3)

(c) We can now make use of Eqs. 2 and 3 to show that:

msad = (1/2)Fg sinθ

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