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The results of the tests are shown below. A single-phase power system is shown in Figure P The load on the transformer is 90 kW at 0. When travelers from the USA and Canada visit Europe, they encounter a different power distribution system. It has turns of wire on the V side and turns of wire on the V side. Sketch the magnetization current that would flow in the transformer.
What is the rms amplitude of the magnetization current? What percentage of full-load current is the magnetization current? The MATLAB program shown below calculates the flux level at each time, the corresponding magnetization current, and the rms value of the magnetization current.
The rms magnetization current is 0. This is true because the peak flux is higher for the 50 Hz waveform, driving the core further into saturation. What is the voltage regulation of the transformer? What is its voltage regulation under these conditions? Find its efficiency. SOLUTION a The open-circuit test was performed on the low-voltage side of the transformer, so it can be used to directly find the components of the excitation branch relative to the low-voltage side.
The magnetizing impedance is j80 per unit. Repeat this process for power factors of 0. A three-phase transformer bank is to handle kVA and have a The kVA must be It is supplied with power directly from a large constant-voltage bus. SOLUTION From the short-circuit information, it is possible to determine the per-phase impedance of the transformer bank referred to the high-voltage side.
Referred to the primary side of one of the transformers, the load on each transformer is equivalent to kVA at V and 0. Calculate all of the transformer impedances referred to the low-voltage side. An autotransformer is used to connect a It must be capable of handling kVA.
There are three phases, connected Y-Y with their neutrals solidly grounded. Two phases of a A farmer along the road has a V feeder supplying kW at 0. The single-phase loads are distributed evenly among the three phases.
Also find the real and reactive powers supplied by each transformer. Assume the transformers are ideal. From the figure, it is obvious that the secondary voltage across the transformer is V, and the secondary current in each transformer is A. The primary voltages and currents are given by the transformer turns ratios to be V and What are the transmission losses of the system? What are the transmission losses of the system now? Note: The transformers may be assumed to be ideal. Also, the transmission losses in the system are reduced.
Consider the transformer to be ideal, and assume that all insulation can handle V. Answer the questions of Problem for this transformer. Autotransformers are normally only used when there is a small difference between the two voltage levels. SOLUTION The impedance of a transformer can be found by shorting the secondary winding and determining the ratio of the voltage to the current of its primary winding. SOLUTION a The equivalent of this three-phase transformer bank can be found just like the equivalent circuit of a single-phase transformer if we work on a per-phase bases.
The open-circuit test data on the low-voltage side can be used to find the excitation branch impedances referred to the secondary side of the transformer bank. What is the efficiency of the transformer at rated conditions and unity power factor? What is the voltage regulation at those conditions?
The core losses in resistor RC would be V 2 1. Figure P shows a power system consisting of a three-phase V Hz generator supplying two loads through a transmission line with a pair of transformers at either end. What is the power factor of the generator? With the switch closed? What is the effect of adding Load 2 to the system? The power system can be divided into three regions by the two transformers.
This problem is a good example of the advantages of power factor correction in power systems. We must right a new function halfwave3 to simulate the output of a three-phase half-wave rectifier. This output is just the largest voltage of v A t , v B t , and vC t at any particular time. It is identical to the one in the textbook. This answer agrees with the analytical solution above.
We must right a new function fullwave3 to simulate the output of a three-phase half-wave rectifier. Explain the operation of the circuit shown in Figure P What would happen in this circuit if switch S1 were closed? Therefore, a voltage oriented positive-to-negative as shown will be applied to the SCR and the control circuit on each half cycle. The process starts over in the next half cycle. Therefore, less power is supplied to the load.
What is the rms voltage on the load under these conditions? Note: Problem is significantly harder for many students, since it involves solving a differential equation with a forcing function. This problem should only be assigned if the class has the mathematical sophistication to handle it. However, capacitor C charges up through resistor R, and when the voltage vC t builds up to the breakover voltage of D1, the DIAC will start to conduct.
When it turns ON, the voltage across the SCR will drop to 0, and the full source voltage vS t will be applied to the load, producing a current flow through the load. The SCR continues to conduct until the current through it falls below IH, which happens at the very end of the half cycle. At the end of the half cycle, the voltage on the capacitor is again essentially 0 volts, and the whole process is ready to start over again at the beginning of the next half cycle. This calculation is much harder than in the examples in the book, because in the previous problems the source was a simple DC voltage source, while here the voltage source is sinusoidal.
However, the principles are identical. To determine the voltage vC t on the capacitor, we can write a Kirchhoff's Current Law equation at the node above the capacitor and solve the resulting equation for vC t.
One problem with the circuit shown in Figure P is that it is very sensitive to variations in the input voltage v ac t. For example, suppose the peak value of the input voltage were to decrease.
Then the time that it takes capacitor C1 to charge up to the breakover voltage of the DIAC will increase, and the SCR will be triggered later in each half cycle. Therefore, the rms voltage supplied to the load will be reduced both by the lower peak voltage and by the later firing. This same effect happens in the opposite direction if v ac t increases. How could this circuit be modified to reduce its sensitivity to variations in input voltage?
SOLUTION If the voltage charging the capacitor could be made constant or nearly so, then the feedback effect would be stopped and the circuit would be less sensitive to voltage variations. This diode holds the voltage across the RC circuit constant, so that the capacitor charging time is not much affected by changes in the power supply voltage.
Explain the operation of the circuit shown in Figure P, and sketch the output voltage from the circuit. When that happens, the circuit looks like: Since the top of the transformer is now grounded, a voltage VDC appears across the upper winding as shown. This voltage induces a corresponding voltage on the lower half of the winding, charging capacitor C1 up to a voltage of 2VDC, as shown.
Now, suppose that a pulse is applied to transformer T2. At that time, the circuit looks like: Now the voltages on the transformer are reversed, charging capacitor C1 up to a voltage of 2VDC in the opposite direction. The output voltage from this circuit would be roughly a square wave, except that capacitor C2 filters it somewhat. Note that vC t and vD t look the same during the rising portion of the cycle.
In the circuit in Figure P, T1 is an autotransformer with the tap exactly in the center of its winding. Explain the operation of this circuit. Assuming that the load is inductive, sketch the voltage and current applied to the load. What is the purpose of SCR 2? What is the purpose of D2? This chopper circuit arrangement is known as a Jones circuit. When that happens, current will flow from the power supply through SCR1 and the bottom portion of transformer T1 to the load.
At that time, a voltage will be applied to the bottom part of the transformer which is positive at the top of the winding with respect to the bottom of the winding. This current causes C to be charged with a voltage that is positive at its bottom with respect to its top. This condition is shown in the figure above. Now, assume that SCR2 is triggered. Current then flow through the capacitor, SCR2, and the load as shown below.
This current charges C with a voltage of the opposite polarity, as shown. SCR2 will cut off when the capacitor is fully charged. Alternately, it will be cut off by the voltage across the capacitor if SCR1 is triggered before it would otherwise cut off. Diode D2 in this circuit is a free-wheeling diode, which allows the current in the load to continue flowing for a short time after SCR1 turns off.
A series-capacitor forced commutation chopper circuit supplying a purely resistive load is shown in Figure P What causes it to turn off? Assume that three time constants must pass before the capacitor is discharged. This happens when the capacitor charges up to a high enough voltage to decrease the current below IH. The capacitor discharges through resistor R. It can be considered to be completely discharged after three time constants. These more complex circuits provide special paths to quickly discharge the capacitor so that the circuit can be fired again soon.
A parallel-capacitor forced commutation chopper circuit supplying a purely resistive load is shown in Figure P Assume that three time constants must pass before the capacitor is charged. If we assume that it takes 3 time constants to fully charge the capacitor, then the time until SCR1 can be turned off is 0. Note that this is not a very realistic assumption.
If the power to the load must be turned on and off rapidly, this circuit could not do the job. These more complex circuits provide special paths to quickly charge the capacitor so that the circuit can be turned off quickly after it is turned on.
Figure P shows a single-phase rectifier-inverter circuit. Explain how this circuit functions. What are the purposes of C1 and C2? What controls the output frequency of the inverter? Therefore, this circuit is a current source inverter.
The rectifier and filter together produce an approximately constant dc voltage and current across the two SCRs and diodes at the right of the figure. The applied voltage is positive at the top of the figure with respect to the bottom of the figure. To understand the behavior of the inverter portion of this circuit, we will step through its operation. Then the voltage V will appear across the load positive-to-negative as shown in Figure a.
At the same time, capacitor C1 will charge to V volts through diode D3, and capacitor C2 will charge to V volts through diode D2.
Then a voltage of V volts will appear across the load positive-to-negative as shown in Figure b. At the same time, capacitor C1 will charge to V volts with the opposite polarity from before, and capacitor C2 will charge to V volts with the opposite polarity from before.
The cycle continues in this fashion. Capacitors C1 and C2 are called commutating capacitors. The output frequency of this rectifier-inverter circuit is controlled by the rates at which the SCRs are triggered. The resulting voltage and current waveforms assuming a resistive load are shown below.
A simple full-wave ac phase angle voltage controller is shown in Figure P However, capacitor C charges up through resistor R, and when the voltage vC t builds up to the breakover voltage of D1, the PNPN diode will start to conduct.
Figure P shows a three-phase full-wave rectifier circuit supplying power to a dc load. The circuit uses SCRs instead of diodes as the rectifying elements. At what phase angle should the SCRs be triggered in order to operate this way? Sketch or plot the output voltage for this case.
The sketch of output voltage is reproduced below, and the ripple is 4. The following table shows which SCRs must conduct in what order to create the output voltage shown below. The times are expressed as multiples of the period T of the input waveforms, and the firing angle is in degrees relative to time zero.
These waveforms are shown below. The inputs to this function are the current phase angle in degrees, the offset angle of the waveform in degrees, and the firing angle in degrees.
What could be done to reduce the harmonic content of the output voltage? It would be easy to modify the function to use any arbitrary dc voltage, if desired. After the voltages are generated, function vout will be used to calculate vout t and the frequency spectrum of v out t. Finally, the program will plot v in t , v x t and v y t , v out t , and the spectrum of v out t.
Note that in order to have a valid spectrum, we need to create several cycles of the 60 Hz output waveform, and we need to sample the data at a fairly high frequency.
This problem creates 4 cycles of vout t and samples all data at a 20, Hz rate. Declare arrays. There are two plots here, one showing the entire spectrum, and the other one showing the close-in frequencies those under Hz , which will have the most effect on machinery. Note that there is a sharp peak at 50 Hz, which is there desired frequency, but there are also strong contaminating signals at about Hz and Hz.
If necessary, these components could be filtered out using a low-pass filter. This increase in sidelobe frequency has two major advantages: it makes the harmonics easier to filter, and it also makes it less necessary to filter them at all. Since large machines have their own internal inductances, they form natural low-pass filters.
If the contaminating sidelobes are at high enough frequencies, they will never affect the operation of the machine. Thus, it is a good idea to design PWM modulators with a high frequency reference signal and rapid switching.
Calculate the current that would flow through the resistor. How does this number compare to the amount of electric power being generated by the loop? This machine is acting as a generator, converting mechanical power into electrical power. Develop a table showing the speed of magnetic field rotation in ac machines of 2, 4, 6, 8, 10, 12, and 14 poles operating at frequencies of 50, 60, and Hz.
A three-phase four-pole winding is installed in 12 slots on a stator. There are 40 turns of wire in each slot of the windings. The flux per pole in the machine is 0. Since this is a four-pole machine, there are two sets of coils 4 slots associated with each phase. A three-phase Y-connected Hz two-pole synchronous machine has a stator with turns of wire per phase. What rotor flux would be required to produce a terminal line-to-line voltage of 6 kV?
What happens to the resulting net magnetic field? If an ac machine has the rotor and stator magnetic fields shown in Figure P, what is the direction of the induced torque in the machine? Is the machine acting as a motor or generator? The machine is acting as a generator. In the early days of ac motor development, machine designers had great difficulty controlling the core losses hysteresis and eddy currents in machines.
They had not yet developed steels with low hysteresis, and were not making laminations as thin as the ones used today.
To help control these losses, early ac motors in the USA were run from a 25 Hz ac power supply, while lighting systems were run from a separate 60 Hz ac power supply. What was the fastest rotational speed available to these early motors? At a location in Europe, it is necessary to supply kW of Hz power. The only power sources available operate at 50 Hz. It is decided to generate the power by means of a motor-generator set consisting of a synchronous motor driving a synchronous generator.
How many poles should each of the two machines have in order to convert Hz power to Hz power? A V kVA 0. At 60 Hz, its friction and windage losses are 24 kW, and its core losses are 18 kW. The field circuit has a dc voltage of V, and the maximum I F is 10 A.
Column 1 contains field current in amps, and column 2 contains open-circuit terminal voltage in volts. It is 4. The required torque is PIN Assume that the field current of the generator in Problem has been adjusted to a value of 4. What happens to the phasor diagram for the generator? Assume that the field current of the generator in Problem is adjusted to achieve rated voltage V at full load conditions in each of the questions below. Note that the maximum current will be A in any case.
Assume that the field current of the generator in Problem has been adjusted so that it supplies rated voltage when loaded with rated current at unity power factor.
You may ignore the effects of R A when answering these questions. This generator is a very long way from that limit. Since sin This machine can also be paralleled with the normal power supply a very large power system if desired. Show both by means of house diagrams and by means of phasor diagrams what happens to the generator. How much reactive power does the generator supply now?
Show this behavior both with phasor diagrams and with house diagrams. The generator must have the same voltage as the power system. The phase sequence of the oncoming generator must be the same as the phase sequence of the power system.
The frequency of the oncoming generator should be slightly higher than the frequency of the running system. The circuit breaker connecting the two systems together should be shut when the above conditions are met and the generator is in phase with the power system.
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