. The ideal transformer will not be treated during lecture, but the textbook includes a section that covers the topic well. Read that section. Be sure that you are reading the section on the ideal transformer. This material is testable. (There is nothing to submit for this problem.)
2(a). A transformer has a primary-to-secondary turns ratio of 100:1. The voltage on the primary side is 27.7 kv.a.c. at 60.0 Hz (when ECE 3254 is treating electric power topics, you can always assume 60.0 Hz unless stated otherwise). A load is connected across the secondary winding. Assuming that the transformer is ideal, what voltage is developed to the load? Briefly show your work. Sketch a diagram.
2(b). The load of part a is a 2.00-ohm load. How much current does the load draw from the transformer’s secondary winding? [Answer: 138.5 amps. Show your work.]
2(c). How much current must be supplied to the transformer’s primary winding?
2(d). How much power does thesecondary winding deliver to the load? In view of the first law of thermodynamics and that the transformer is assumed to be ideal, how much power is drawn by the transformer’s primary winding? (The obvious answer is the correct answer. This is not a trick question.)
2(e). Show that your answer to part d is consistent with your answers to parts b and c.
2(f). The transformer’s primary winding is supplied by a pair of wires from the electric utility. Show that, with the 2.00-ohm load connected across the secondary winding, the primary winding appears to the utility as it were a 20.0-kilohm resistor.
2(g). Briefly explain why, if the 2.00-ohm load is removed from the secondary, leaving an open circuit, the primary—though indeed connected to the utility—appears to the utility as it were an open circuit. (In a practical, nonideal transformer, the primary winding unfortunately does draw some current even when the secondary is open-circuited. The attendant power is dissipated via magnetic fields as heat in the transformer’s iron core. Fortunately however, manufacturers of electric transformers have learned how to limit open-circuit currents to a small, nuisance level.)
2(h). (In your homework paper, a much abbreviated problem statement suffices.) In an industrial plant —when equipment is moved, hoists are operated, forklifts are driven, assembly lines are retooled, maintenance is carried out, and so on—electrical accidents occasionally happen. Electrical conduits can be damaged and other mishaps can occur, suddenly short-circuiting a transformer’s secondary winding. To limit danger to personnel and hazard to property to the extent possible, the electrical engineer plans in advance for such accidents. When a transformer’s secondary winding is short-circuited, the concept of the ideal transformer fails. An engineer probably cannot usefully model a transformer as an ideal transformer in the short-circuit case. In the short-circuit case, rather, the transformer’s primary winding appears to be not as it were a short circuit but approximately as it were an inductor. Suppose that the transformer primary winding appears to be a 0.500-H inductor when the secondary is shorted. How much current does the transformer’s primary winding draw? (There is no part i.)
2(j). Referring to part h, express the voltage across and current entering the transformer’s primary winding in the time domain in the short-circuit case. That is, write expressions for v(t) and i(t). It is recommended to work in phasors and then to change to the time domain at the end of your work, but don’t forget to adjust for rms.
2(k). Observe that, in the case of parts h and j, a circuit breaker will open, whether (preferably) the industrial plant’s circuit breaker or (preferably not) the utility’s circuit breaker. Therefore, the shortcircuit current will flow only for a fraction of a second. (There is nothing to submit for part k. In practical industrial applications, the main circuit breakers can cost tens of thousands of dollars each, but that is less than the cost of the fires and explosions that must otherwise sporadically occur; for though care is taken, it is impossible to avoid 100.0 percent of electrical mishaps in industrial facilities. The expensive circuit breakers thus do a necessary job and are simply part of the budgeted cost of an industrial plant’s electrical installation. It is interesting to note that the main circuit breaker itself is normally mounted in a steel cabinet in a locked room where forklifts and such are unlikely to run into it! Or else it is in a fenced utility yard like the yard east of Durham Hall.) (There is no part l.)
2(m). Returning to parts a through f, in which a 2.00-ohm load is connected across the transformer’s secondary winding, suppose that the electric utility supplies the transformer’s primary winding by a pair of wires (two wires) suspended from utility poles (note: this is a simplification; in reality, the electric utility distributes power in three phases and uses some other engineering techniques that slightly complicate the matter). The wires are 50.0 miles long and each has a resistance of 1.50 ohms per mile. How much power is lost in the utility’s wires?
2(n). Referring to part m, compute the system’s efficiency. (There is no part o.)
2(p). If no transformer were available, the utility would have to supply power at 277 v.a.c. over its 50.0-mile pair of wires. Suppose that no transformer were available and that the utility did indeed supply power at 277 v.a.c. Suppose that the industrial plant’s 2.00-ohm load were directly connected across the utility’s pair of wires. Repeat parts m and n in this case.
2(q). The cost of an industrial plant’s main transformer can run to the hundreds of thousands of dollars. Seemingly, this cost could be saved if the electric utility and the plant only used the same voltage. A voltage like 27.7 kv is too high to handle safely inside a plant (this is why the utility suspends wires at such voltages in the air from tall poles, for current can arc through air at about 4 kv per cm), but it would seen that the utility could safely use 277 v. This is indeed true: the utility could safely use 277 v. In view of part p, briefly, why doesn’t the utility use 277 v.?
3. Consider again problem 2, parts a through d. Suppose that the electric utility supplied 27.7 kv.d.c. rather than 27.7 kv.a.c. How much power would the transformer’s secondary winding deliver to the load in this case? (Answer: zero. This is a trick question. Transformers work according to Faraday’s law, a law of magnetics, by which a time-varying magnetic flux in an iron core induces a voltage in a coil wound about that core. According to Faraday, the voltage is proportional not only to the flux’s strength but also to the frequency at which the flux varies. At d.c., the frequency is zero. Therefore, at d.c., zero voltage and thus zero power are delivered. Transformers do not work at d.c. Interestingly, transformers work somewhat better at 60 Hz than at 50 Hz, whereas 60 Hz has still proved to be a practical frequency—not too fast—for the mechanical bearings, rotors and shafts of four-pole electric motors; so, electrically in terms of frequency, we probably have it a little better in North America than they do in the old world. There is nothing to submit for this problem but its material is testable, of course.)
4(a). In view of the lecture of April 13, show that the power delivered to a balanced three-phase, wyeconnected load is P = (√3)IV, where I and V are respectively the line current and the voltage between one line and another (rather than the voltage between a line and neutral). Sketch a suitable schematic and phasor diagram.
4(b). If the voltage from a line to neutral is 277 v.a.c., then compute the voltage between one line and another. [Answer: 480 v.a.c.]
4(c). Suppose that the current is 10.0 amps a.c. in each line. Compute the power delivered to the threephase load of part a.
4(d). How much power is delivered to each of the load’s three phases? [Answer: one third of the total is delivered to each. Your paper should state how much power this is.]