Problem 1: The trouble with DC electrical systems

GEOS 24705 / ENST 24705 / ENSC21100 Problem set #12 Due: Tuesday May 13 Problem 1: The trouble with DC electrical systems In the previous problem set you read about Edison s first electrical company and his 1882 Pearl Street generation station that operated at 110 V direct (DC) voltage, and saw how small his service area was. Edison was forced to keep the area small because he needed to minimize his resistive losses without making his copper transmission wires prohibitively heavy and expensive. (Note that wiring was one area of technology that Edison didn t have to invent from scratch. Copper wire manufacturing already existed: people had needed copper wires since the invention of the telegraph by Samuel Morse of Morse code in the 1830s. But telegraph wires could be very thin, as they carried only tiny amounts of electrical power in sending telegraph signals. Edison s business model of sending large amounts power down a wire for lighting was a new thing.) In the last problem set you should have derived the expression for resistive losses in a wire. That came from combining the definition of power carried by a current P = I V and the voltage drop produced by a resistance V = I R. If a current I is flowing down a wire (determined by other constraints) then any resistance of the wire itself produces a (hopefully small) voltage drop along the wire V wire = I R wire, and a resulting power loss P wire = I V wire = I 2 R wire. There don t seem to be any descriptions of Edison s wiring network (at least none readily available), but you can make some assumptions to estimate what his wire thickness was. A. First, consider how Edison s losses would scale with the various factors he could control in designing his system. Edison s system design can set a) the resistance of his wires, b) the maximum total power that is transmitted down each wire, and c) the voltage V that his generator produces. Combine the equations above to derive a simple equation that relates P wire to these factors (R wire, P, and V). B. Now manipulate the equation in 1) to write down an expression for the fractional loss of power to line losses: P wire /P. C. What was the total power P Edison transmitted along a single line? Here you have to make some assumptions. The Pearl Street station had 6 generators connected to steam engines capable of putting out ~ 100 kw of power each (over 100 horsepower). When Edison first started operations, he was serving 85 different customers. As a first guess we can assume that each customer had an individual power line. D. For a reality check, let s consider the actual lights that Edison was powering, and think of the resistance of the lightbulbs themselves. Although his first customers only had about 400 lights, the power station was sized for a larger number: newspaper reports claimed that it could light 7200 lamps. How much power would an 1880s lightbulb draw? Does that sound reasonable to you? Assuming that Edison keeps his line losses manageable (and therefore has a relatively small voltage drop in the lines), then nearly his entire 110

V voltage drop will occur over the filament of the lightbulb. What is the resistance of the lightbulb filament? (If you have power in Watts and voltage in Volts, then your answer for resistance will come out in Ohms). E. If Edison wanted to keep his line losses to 10% or less (P wire /P < 0.1), how low must the resistance R wire of each line be? Compare the resistance of the line to that of the lightbulb. F. From your map of the last problem set, estimate a maximum length of line in Edison s first distribution network. We ll assume that all Edison s main power lines are of this length. G. Using the formula we derived in class for the resistance of a wire, and resistivity of copper (1.68 x 10-8 ohm- meters), what is the cross- sectional area of each of Edison s lines, with your assumptions? What is the diameter of each line? (Because of your guess about the number of lines, this answer may be different from Edison s true wire diameter, but your answer should be reasonable to a factor of several.) H. (Optional) You d expect that a brand- new technology would be expensive costs always come down over time. Edison sold power from the Pearl Street Station for 24 cents per kwh. That doesn t sound terribly expensive, but remember this was 1882. Look up the inflation since 1882 and give Edison s price in current dollars. In previous problem sets you looked up the retail electricity rate. By what factor have electricity costs dropped? Problem 2. AC electrical systems Because Edison s standard lost out to Tesla s, your household electricity is not DC but instead is alternating current (and voltage). Tesla s AC generating systems produce voltage that rises and falls in a sine wave, as shown below. VOLTAGE That means that each generator connected to our grid drives an alternating current that sloshes back and forth from the power plant through the transmission lines. It may seem a strange concept, but it is possible to transmit power down a line even if the charges simply oscillate back and forth without net motion. Purely resistive devices like incandescent lightbulbs, or toasters don t care whether they are powered with alternating or direct current. An incandescent lightbulb will light no matter what direction the current is flowing through it, and will therefore light (and consume power) even if the current is changing direction.

For motors, the design would obviously be different depending on whether the current is AC or DC. You should be able to guess from the class discussion of DC motors that alternating current might actually allow a very simple motor, since it could solve the problem of the loop that flopped back and forth when connected to a battery. Instead of having to switch the connections, you could just switch the direction of the current exactly as an AC generator already does. AC can therefore have some advantages in making simple electric motors. In the U.S. electrical system, all generators must have the same frequency of oscillation, 60 times per second (60 Hz), AND all must be perfectly synchronized. They need not produce the same voltage, though, since voltages can be easily transformed. Transmission lines are at high voltage, but before electricity reaches your house the voltage is down- converted to 110 V. But what does that specified standard of 110 V really mean? Voltage in an AC system varies from 0 to some maximum amplitude Vmax back down through 0 and then to a negative Vmax. The specified voltage is some single value that must represent that whole pattern. How to pick that representative value? We obviously can t use the average voltage, since that s zero. We could use the maximum voltage V max, or the average absolute voltage V average that you just calculated in problem 2. But that s not what is used. Instead, voltage is given as the root- mean- squared voltage: the square root of the average of the square of the voltage: V rms = V 2 (t) (where the horizontal bar represents a time average). In this problem you ll understand why that was chosen. A. Determine Vmax, the peak voltage you get on your household electrical system, from your known V rms. For people with calculus, integrate the sine wave to find Vrms. For others, use the formula V rms = V max / 2 The diagram below describes a simple system, with current flowing from an applied voltage V to ground through a resistive load R. (That resistive load could be a lightbulb, for example). As mentioned above, the power dissipated in the resistive load is the product of current and voltage: P = I ΔV. The current that flows in a resistive system is proportional to the voltage drop (Ohm s law: ΔV= I R). B. Write expression for the power dissipated in the resistive load (Above you used an expression in terms of I and R, now also write that in terms of V and R).

C. Your answers in B should give you insight: Why do we use V rms? What s so special about squaring the voltage? D. (Optional) Now that you understand AC voltage you can use your methods of Problem 1 to understand modern lightbulbs, which have tungsten filaments (with resistivity 5.6 x 10-8 ohm- meters). What is the resistance of the filament in a 100 W lightbulb? That is, what resistance will give you 100 W of power dissipation, when connected to your 110 V household electricity? To give you that power consumption, the lightbulb companies choose the appropriate filament design. Make a reasonable assumption for the length of the filament, and then estimate the thickness of its filament. Problem 3. DC and AC motors and generators All motors and generators are fundamentally made possible by the interactions between moving charges and magnetic fields that we discussed in class. We saw that any moving charge will produce its own magnetic field (Ampere s Law). That means that any moving charge will also interact with an external magnetic field, leading to some kind of force. In class we wrote down the equation for the Lorentz force: F = qv x B (where q is the charge, v the velocity, and B the magnetic field; the bold quantities means these have some direction; and the x operator means that the force depends on the component of the velocity that is perpendicular to the magnetic field, and that the force itself is in a 3 rd direction. (See the right hand rule diagram:) In class we said that both motors and generators involved a wire loop, a magnetic field, an electrical current, and a force. I gave these definitions: An electrical motor: if you flow current through a current loop in the presence of a magnetic field, it will rotate and do work (Ampere s Law) An electrical generator: if you apply work to rotate a current loop in the presence of a magnetic field, it will cause a current to flow (Faraday s Law) But both of these laws can be readily explained just using the Lorentz force. In this problem you ll do that using the animations at http://www.animations.physics.unsw.edu.au/jw/electricmotors.html Note: you know that motors and generators must operate on similar principles because every motor can also be a generator. If you flow a current through the wire loop you can impose a force; if you apply force to rotate the wire loop, you can produced a current. (This is analogous to our rule in thermodynamics that every heat engine if run in reverse can be a heat pump.) Another note: we could also describe motors and generators in terms of voltages instead of currents. The only way a current flows is that there is a voltage difference across the two ends of the wire. In the case of a motor, you apply a voltage difference to drive a current flow that than makes the motor turn; in the case of a generator, by turning the generator you create a voltage difference that drives a current flow. (In the optional problem you can consider how much current flows in a motor.)

Ampere s law and the Lorentz force In class we drew on the blackboard one example of Ampere s law the idea that any current (a line of moving charges) generates a magnetic field around it. And we said that this property meant that two wires with currents could generate a force on each other, so that the wires would be pushed together or apart (depending on the direction of the currents). But this basically IS a particular application of the Lorentz force law. The moving charge in wire 2 (moving with velocity v) experiences the magnetic field B that is generated by the moving charges in wire 1, and vice versa. A Draw two diagrams showing B, v, and F, one for a charge in wire 1 (being affected by wire 2), and one for the opposite case. Show that Ampere s Law is entirely consistent with the Lorentz force law. (I believe though that when I drew the example on the board I got the force direction backwards, so be careful with your right hand rule!) DC motors A DC motor is one that is driven by a constant voltage difference. The most common application of DC motors is use with a battery as the power source: any motor driven by a battery is a DC motor. DC motors are common in your life: they drive the fans in your laptop, cordless power drills, the propellers on a drone, your childhood Dancing Elmo toy. A DC motor isn t actually powered by a steady current flow: in class we showed how a simple direct current can turn a wire loop only partway before it stalls. To produce continual rotational motion, a DC motor uses some mechanical device to switch the current back and forth. In this problem you ll look at the animation at http://www.animations.physics.unsw.edu.au/jw/electricmotors.html#dcmotors to understand the simple DC motor. In this animation the battery terminals are just shown as lines; the longer is the positive terminal, so current flows from this terminal to the other one. Note that I don't think the text of this page is great, but the animations are very good. B Play the animation, then freeze it at a few different parts of the cycle (at least 4) and print out the diagram for each snapshot of the cycle. Pick one segment of the wire to think about (one of the two sides of the loop). For each of your printouts, draw arrows showing the direction of 1) the magnetic field B, 2) the current in your segment of the loop (that is, the direction of motion v of the moving charges), 3) the resulting force F, and 4) the instantaneous motion of the segment of the wire loop.

C Now for each snapshot of part B, partition the total force F on your wire segment into two parts, one part in the direction of motion, and another perpendicular to it. The first part helps push the loop around; the second part just tries to stretch the loop. Draw on your diagram an arrow that shows that part of the force (if any) that is in the direction of motion. This bit is the force that turns the motor. You saw in class that when the loop is tilted vertical, there is no turning force on the motor. D Explain how the turning force you deduce relates to the graph on the right of the animation. E For practical usage, people don't want motors that deliver jerky, intermittent force. You d like a motor whose force is fairly constant. The simplest way to achieve that is to build a motor that is essentially several DC motors in one, with their coils offset. Then the total force is smoothed out. The pictures below show a multi- coil DC motor, and the turning force from each coil. Show that the total force delivered is then smoother. You can just choose a few sample points on the graph and add up the force from each coil, and plot that sum (or those who want can do the math explicitly).

DC generators The next animation on the page is a DC generator. But the animation looks exactly like that of the motor! That s because generators are just the converse of motors. In a generator, you re applying mechanical force to turn the coil, and the movement of the wires in a magnetic field somehow induces a current to flow. You can think of this process as also governed by the Lorentz force. The metal wire is full of little charges (think: electrons). So your velocity v in this case is now the motion of the wire itself. That motion, in a magnetic field, produces a force on the charges in the wire (let s call it an electromotive force in this case). The charges will start to flow down the wire if that force is in the direction of the wire itself. F Repeat what you did in parts B and C above. Print out 4 (or more) snapshots of the generator cycle, and on each draw arrows showing the directions of 1) the magnetic field B, 2) the direction of movement of your wire segment v, 3) the resulting force F, and 4) the component of that force that is in the direction of the wire. G Show and explain how the resulting electromotive force pushing current around the wire is given by the graph at the right of this animation. Notes: applying an electromotive force basically means applying a voltage. The sign convention here is that if the charges are being pushed toward the negative terminal, then the force (voltage) is positive. AC generators Scroll down the page til you get to the animation of the AC generator (which here is called an alternator ). Note that it looks exactly like the DC generator, except for the rings. So we don t need to repeat the previous problem. Just H Explain the difference between the rings in the DC and AC generators, and how the change in the rings affects the voltages produced by the generators that is, why in the AC generator, the voltage switches between positive and negative. In this animation the generator produces an alternating voltage that would slosh current back and forth. But that alternation would also result in alternating power production: since P = I*V, when there is no voltage and no current then the generator is momentarily producing no electrical power. That seems like a bad design. If the generator is being spun by an engine or turbine that produces a steady constant turning force, you d want to ensure that that mechanical push can be translated into a steady production of electrical power. And you d like to be able to deliver a steady electrical power to your customers, rather than going from maximum to zero power once every 60 seconds. To achieve that steadiness, a commercial AC generator is builts as three generators in one to smooth out the power production (just as a DC motor is several motors in one to smooth out its force). There are three different coils (not connected), each producing a different voltage and current. Those voltages are then transmitted to the customers by three different wires. Look at all transmission lines they always have some repetition of multiples of three.

3- phase generation. Image from koehler.me.uk 3- phase transmission. Image from Launceton College Why this rule of three? Because it turns out that three generators can combine perfectly to produce exactly what you d want. First, you want no NET movement of charge sloshing back and forth across the country, so you want the voltages in the three power lines to perfectly sum to zero. If charges are moving in one direction in one wire, they should be moving the opposite direction in the others wires to compensate. And, you want absolutely steady production of power. Since P = I*V and the current I is proportional to voltage (V = I*R à I = V/R), then power is proportional to I 2. (We used this same relationship to think about resistive heating in the wires, but in this case we re thinking about the total power delivered to all the customers. In each case power is proportional to the square of the current.) The diagram below shows the three different voltages from a modern AC generator. It also shows (at bottom) those voltages squared, i.e. the power in each transmission line.

I Demonstrate that the voltages sum to zero (and therefore there is no net current in the transmission lines). For those who are comfortable with sines & cosines, consider the sum of V 1 =sin(wt), V 2 =sin(wt + f 2 ), and V 3 =sin(wt + f 3 ) and prove that it equals zero. (You need standard trig identities but nothing fancier than that). For those who are not comfortable, pick four locations on the top figure above, and at each location measure V 1, V 2, and V 3 by hand and add them to show the sum is constant at all times. J Demonstrate that the sum of the V 2 s is constant (and therefore that power transmission is constant). As before, for those comfortable with sine and cosines, show analytically that V 1 2 + V 2 2 + V 3 2 = constant. For those not comfortable, pick four locations along the figure and measure and add by hand. K Can you get the same properties if you use only two phases? Discuss. Problem 4: DC motors (the problem is optional but do the reading) I asked earlier, when you connect a DC motor to a battery, how much current would flow? The answer to that question is a little counter- intuitive, but it s deep and interesting and governs why electric motors are particularly useful for certain applications. First, think about what might happen if the motor loop wasn t allowed to turn at all if you put some kind of brake on it. Connecting the loop to a battery would then be just like placing a wire across the battery terminals. Because the wire has low resistance, a huge amount of current would flow (set by Ohm s law, DV battery = I R wire, so I = DV battery /R wire ). The battery would lose all its energy very quickly, since the power flowing out of the battery must be P = I * DV battery = DV battery 2 /R wire. And, since the only place for that power to go is into the wire as heat, you d probably melt your motor. But you know that motors don t actually melt if you allow them to turn. If you went to lab, you saw that when you connected a wire to a battery with a magnet nearby, and allowed it to turn, the wire turned but didn t heat up. In some way, allowing the motor to turn must reduce the current that flows through it! That s odd, and fascinating. What the DC motor actually does in the process of turning is to create a kind of back voltage (call it V e ) proportional to how fast it is turning, and in the opposite direction. (Why does it do this? Because a motor is also a generator, so turning the loop makes a counter- acting voltage.) This back voltage means that in a motor that is turning, the total voltage drop that pushes current around the wire is no longer V battery but a smaller number, V battery - V e. (I m dropping the D notation here just for clarity.) The resulting current that flows is then only:

I = (V battery - V e )/R wire And the power lost to resistive heating is also correspondingly smaller: P heat = I *(V battery - V e ) But the total power lost from the battery must still be P battery = I * V battery How can those two quantities (P battery and P heat ) be different? Because some of the total power goes to work! The work that the motor does is the difference between the total power lost by the battery and the part of that power that goes to resistive heating: P work = P battery P heat. Or P work = I * V e When does the motor do maximum work? As the motor spins faster, V e increases, which would tend to increase the work, but I decreases, which would tend to decrease the work. To answer the question, you have plug in the expression for I: P work = V e *(V battery - V e )/R wire = [- V e 2 + (V battery *V e ) ] * 1/R wire Check the logic of this quadratic equation. Remember that the back- voltage V e is proportional to how fast the motor is turning. If the motor stops (V e =0), it can (obviously!) do no work. But if the motor is going so fast that V e = V battery, it again can do no work. Somewhere between 0 speed and this top speed, the motor must do its maximum work. Now, how do you set how fast the motor goes, and how much work it does? The answer is, you don t get to tell the motor what to do. You have only two things you can control: what voltage you apply to the motor what load you connect to the motor Think of the load as a weight you re asking the motor to lift. It s analogous to the problem of the Newcomen engine and water pump. If you attach too heavy a load to the motor, it can t turn and doesn t do work. If you attach no load to the motor, it has no way to do work and spins uselessly at its maximum rate. You get the maximum work from the motor if you match the load and voltage just right (just as the Newcomen engine designers had to match each engine to its pump.) What happens if you overload the motor? You should know from your problem sets that if you overloaded the Newcomen engine, it would just stop. The cylinder would heat and cool but the piston would not move. That s wasteful, but not dangerous. Overloading a DC motor is however dangerous. The overloaded motor also stops, but now you have a bigger problem, because as the motor slows, V e goes down and the current rises. That is, as the motor slows, the battery delivers MORE power, but LESS of that power goes to doing work. Instead, the extra power is dumped as heat into the wires. If you overload a DC motor, it can melt.

Problem 4 (Optional) A. At what applied voltage does a DC motor do maximum work? Discuss. B. Make a power- speed curve for a DC motor: draw a graph showing the variation of the power output P work vs. the rotations per time at which the motor turns (call it w). The back voltage V e is proportional to that speed, so let V e = w /k, where k is some constant determined by the motor design. You can use the result above that the maximum power comes at V e = V battery /2. C. Make a torque- speed curve for a DC motor: draw a graph showing torque vs. w. We haven t discussed torque yet; we ll talk about it in the next class. Torque (t) is a kind of turning force that is the force exerted times a lever arm distance; it has the units of energy. The power output is then equal to torque x rotation rate: P work = t * w. D Your answer in C is related to why Silicon Valley moguls buy $100,000 electric Tesla sports cars. Explain and discuss.