Stephen J. Electric machinery fundamentals / Stephen Chapman. - 4th ed. p. em. Includes index. ISBN I. E lectric machinery. I. T itle. T K *n.d. strionalerispi, VANT. ELECTRIC. MACHINERY. FUNDAMENTALS. FIFTH EDITION. Electric Machinery. Fundamentals. Fifth Edition. · **. Micror, EN. Unfortunately, the situation in sinusoidal ac circuits is more complex, be- cause there can be a phase difference between the ac voltage and the ac currenl.
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copyrighl page. Library of Congress CataJoging-in-PubIication Data. Chapman, Stephen 1. Electric machinery fundamentals 1 Stephen 1. Chapman. -5th ed. Chapman. Stephen J. Electric machinery fundamentals / Stephen Chapman. - 4th ed. p. em. Includes index. ISBN I. Electric machinery. I. Title. This edition of Electric Machinery Fundamentals makes selected use of MATLAB to enhance a student's learning experience where appropriate. For example.
Author: Stephen Chapman, Stephen J. Chegg Solution Manuals are written by vetted Chegg Control Theory experts, and rated by students - so you know you're getting high quality answers. Solutions Manuals are available for thousands of the most popular college and high school textbooks in subjects such as Math, Science Physics , Chemistry , Biology , Engineering Mechanical , Electrical , Civil , Business and more. It's easier to figure out tough problems faster using Chegg Study. Unlike static PDF Electric Machinery Fundamentals solution manuals or printed answer keys, our experts show you how to solve each problem step-by-step.
A wire is shown in Figure P which is carrying 5. Calculate the magnitude and direction of the force induced on the wire. The wire is shown in Figure P is moving in the presence of a magnetic field. With the information given in the figure, determine the magnitude and direction of the induced voltage in the wire. Repeat Problem for the wire in Figure P The core shown in Figure P is made of a steel whose magnetization curve is shown in Figure P How much flux is produced in the core by the currents specified?
What is the relative permeability of this core under these conditions? Was the assumption in Problem that the relative permeability was equal to a good assumption for these conditions? Is it a good assumption in general? It is not very good in general. Its depth is 8 cm, and there are turns on the center leg. The remaining dimensions are shown in the figure. The core is composed of a steel having the magnetization curve shown in Figure c.
Answer the following questions about this core: Is it twice the current in part a? Similarly, the magnetizing intensity H required to produce a flux density of 0. Similarly, the magnetizing intensity H required to produce a flux density of 1. A two-legged magnetic core with an air gap is shown in Figure P The depth of the core is 5 cm, the length of the air gap in the core is 0.
The magnetization curve of the core material is shown in Figure P Assume a 5 percent increase in effective air-gap area to account for fringing. How much current is required to produce an air-gap flux density of 0.
What are the flux densities of the four sides of the core at that current? What is the total flux present in the air gap? An air-gap flux density of 0. Bag 0. A transformer core with an effective mean path length of 10 in has a turn coil wrapped around one leg. Its cross-sectional area is 0. If current of 0. What is the flux density? Sketch the voltage present at the terminals of the coil. This voltage will be the same polarity as the direction shown on the core, so it will be positive.
Figure P shows the core of a simple dc motor. The magnetization curve for the metal in this core is given by Figure c and d. Assume that the cross-sectional area of each air gap is 18 cm2 and that the width of each air gap is 0. The effective diameter of the rotor core is 4 cm. The relative permeability of this core is shown below: This is a design problem, and the answer presented here is not unique. Other values could be selected for the flux density in part a , and other numbers of turns could be selected in part c.
These other answers are also correct if the proper steps were followed, and if the choices were reasonable. Notice that the saturation effects become significant for higher flux densities. Does the load consume reactive power from the source or supply it to the source? Figure P shows a simple single-phase ac power system with three loads. The current in Loads 1 and 2 is the same as before.
Since less reactive power has to be supplied by the source, the total current flow decreases. Demonstrate that Equation can be derived from Equation using simple trigonometric identities: A linear machine has a magnetic flux density of 0.
What is the initial current flow? What is the efficiency of the machine under these circumstances? A linear machine has the following characteristics: What is its final steady-state speed?
What is the new steady-state speed of the bar? Reducing the flux density B of the machine increases the steady-state speed, and reducing the battery voltage VB decreases the stead-state speed of the machine. Both of these speed control methods work for real dc machines as well as for linear machines.
Transformers The turns ratio of the transformer is What are its voltage regulation and efficiency? Since no particular equivalent circuit was specified, we are using the approximate equivalent circuit referred to the primary side. The secondary voltage and current are What is its voltage regulation? The results of the tests are shown below.
The resulting equivalent circuit is 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? The open-circuit test performed on the low-voltage side of the transformer yielded the following data: 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 program to calculate the secondary voltage of the transformer as a function of load is shown below: A three-phase transformer bank is to handle kVA and have a The kVA must be The ratings for each transformer in the bank for each connection are given below: 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. It is much easier to solve problems of this sort in the per-unit system, as we shall see in the next problem.
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? 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. The transformer connection is shown below: Answer the questions of Problem for this transformer. Autotransformers are normally only used when there is a small difference between the two voltage levels.
Prove the following statement: 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. For the transformer connected as an ordinary transformer, the impedance referred to the primary N C is: The open-circuit test was performed on the low-voltage side of this transformer bank, and the following data were recorded: 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.
Since the low-voltage side of the transformer is Y-connected, the per-phase open-circuit quantities are: When it is tested as a conventional transformer, the following values are measured on the primary V side of the transformer: What is the efficiency of the transformer at rated conditions and unity power factor?
What is the voltage regulation at those conditions? VSC 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.
The base impedances of each region will be: This problem is a good example of the advantages of power factor correction in power systems. Introduction to Power Electronics 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. The function is shown below: It is identical to the one in the textbook.
This answer agrees with the analytical solution above. Over that interval, the output voltage is: 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? 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.
Solving for B1 and B2 yields: It reaches a voltage of 30 V at a time of 3. This problem could also have been solved using Laplace Transforms, if desired. 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. The above discussion assumes that transformer T3 is never in either state long enough for it to saturate. Figure P shows a relaxation oscillator with the following parameters: 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 The component values in this circuit are: 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: 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. AC Machinery Fundamentals The simple loop is rotating in a uniform magnetic field shown in Figure has the following characteristics: 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? Synchronous Generators 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?
The resulting per-phase equivalent circuit is shown below: 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. The voltage regulation would be: Note that the maximum current will be A in any case.
A phasor diagram representing the situation at lagging power factor is shown below: 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. This generator is operating in parallel with a large power system infinite bus. How much reserve power or torque does this generator have at full load? Sketch the corresponding phasor diagram. Assume I F is still unchanged. The loads supplied by the two generators consist of kW at 0.
That action will increase the terminal voltages of the system without changing the power sharing between the generators. Three physically identical synchronous generators are operating in parallel. They are all rated for a full load of 3 MW at 0. The no-load frequency of generator A is 61 Hz, and its speed droop is 3. The no-load frequency of generator B is The no-load frequency of generator C is At what load does one of the generators exceed its ratings?
Which generator exceeds its ratings first? Why or why not? This plot reveals that there are power sharing problems both for high loads and for low loads. Generator B is the first to exceed its ratings as load increases. Its rated power is reached at a total load of 6. On the other hand, Generator C gets into trouble as the total load is reduced.
When the total load drops to 2. A paper mill has installed three steam generators boilers to provide process steam and also to use some its waste products as an energy source. Since there is extra capacity, the mill has installed three 5-MW turbine generators to take advantage of the situation.
Each generator is a V kVA 0. Generators 1 and 2 have a characteristic power-frequency slope sP of 2. At what frequency does this limit occur? How much power does each generator supply at that point? Generator 3 will be the first machine to reach that limit.
Generator 3 will supply this power at a frequency of 5. The field currents of the three generators must then be adjusted to get them supplying a power factor of 0. At that point, each generator will be supplying its rated real and reactive power.
Its armature resistance RA is 0. The core losses of this generator at rated conditions are 7 kW, and the friction and windage losses are 8 kW. The open-circuit and short-circuit characteristics are shown in Figure P Column 1 contains field current in amps, and column 2 contains short-circuit terminal current in amps. It is 0. The open circuit voltage at 0. This is a straight line, so we can determine its value by comparing the ratio of the air-gap voltage to the short-circuit current at any given field current.
Each of these files are organized in two columns, where the first column is field current and the second column is either open-circuit terminal voltage or short-circuit current. A program to read these files and calculate and plot X S is shown below. The saturated synchronous reactance at rated conditions was found to be 0.
From the open-circuit characteristic, the required field current would be 0. What is the voltage regulation of this generator at the rated current and power factor?
If this generator is operating at the rated conditions and the load is suddenly removed, what will the terminal voltage be? What are the electrical losses in this generator at rated conditions? If this machine is operating at rated conditions, what input torque must be applied to the shaft of this generator? Express your answer both in newton-meters and in pound-feet. The input power to this generator is equal to the output power plus losses.
Assume that the generator field current is adjusted to supply V under rated conditions. What is the static stability limit of this generator? You may ignore R A to make this calculation easier.
How close is the full-load condition of this generator to the static stability limit? Normal generators would have more margin than this. If the field current and the magnitude of the load current are held constant, how will the terminal voltage change as the load power factor varies from 0.
Make a plot of the terminal voltage versus the impedance angle of the load being supplied by this generator. Also, the magnitude of the armature current is constant. Assume that the generator is connected to a V infinite bus, and that its field current has been adjusted so that it is supplying rated power and power factor to the bus. You may ignore the armature resistance R A when answering the following questions.
The reactive power supplied would increase as shown below. A MVA A three-phase Y-connected synchronous generator is rated MVA, Its synchronous reactance is 0. The phase voltage is Note that this is not quite true, if the armature resistance R A is included, since R A does not scale with frequency in the same fashion as the other terms.
Two identical kVA V synchronous generators are connected in parallel to supply a load. The prime movers of the two generators happen to have different speed droop characteristics. When the field currents of the two generators are equal, one delivers A at 0. A generating station for a power system consists of four MVA kV 0. Three of these generators are each supplying a steady 75 MW at a frequency of 60 Hz, while the fourth generator called the swing generator handles all incremental load changes on the system while maintaining the system's frequency at 60 Hz.
Suppose that you were an engineer planning a new electric co-generation facility for a plant with excess process steam. You have a choice of either two 10 MW turbine-generators or a single 20 MW turbine generator.
What would be the advantages and disadvantages of each choice? If two 10 MW generators are chosen, one of them could go down for maintenance and some power could still be generated. A MVA three-phase Open-circuit test Field current, A Line voltage, kV Short-circuit test Field current, A Armature current, A The armature resistance is 0. Express the answer both in ohms per phase and in per-unit.
The extrapolated air-gap voltage at this point is The generator is connected in parallel with a Hz If the internal generated voltage E A is decreased by 5 percent, what will the new armature current I A be? Synchronous Motors A V, 60 Hz, four-pole synchronous motor draws 50 A from the line at unity power factor and full load.
Assuming that the motor is lossless, answer the following questions: Express the answer both in newton-meters and in pound-feet. Explain your answer, using phasor diagrams. This increase in E A changes the angle of the current I A , eventually causing it to reach a power factor of 0.
A V, 60 Hz, hp 0. Ignore its friction, windage, and core losses for the purposes of this problem. How near is this value to the maximum possible induced torque of the motor for this field current setting? The induced torque is POUT A V hp 0. The open-circuit characteristic of this motor is shown in Figure P Answer the following questions about the motor, assuming that it is being supplied by an infinite bus.
What would the new power factor be? How much reactive power is being consumed or supplied by the motor? This voltage would require a field current of 4. Since the power supplied by the motor does not change when I F is changed, this quantity will be a constant. The new power factor is cos 3. At a power factor of 0. The internal generated voltage at 0. Plot the V-curves I A versus I F for the synchronous motor of Problem at no-load, half-load, and full- load conditions.
It may simplify the calculations required by this problem. Also, you may assume that R A is negligible for this calculation. Note that we are assuming that R A is negligible in each case. The flattening visible to the right of the V-curves is due to magnetic saturation in the machine. If a Hz synchronous motor is to be operated at 50 Hz, will its synchronous reactance be the same as at 60 Hz, or will it change?
Think about the derivation of X S. The armature reaction voltage is caused by the armature magnetic field B S , and the amount of voltage is directly proportional to the speed with which the magnetic field sweeps over the stator surface. The higher the frequency, the faster B S sweeps over the stator, and the higher the armature reaction voltage Estat is.
Therefore, the armature reaction voltage is directly proportional to frequency. Similarly, the reactance of the armature self-inductance is directly proportional to frequency, so the total synchronous reactance X S is directly proportional to frequency.
A V kW 0. The rotational losses are also to be ignored. A V Y-connected synchronous motor is drawing 40 A at unity power factor from a V power system. The field current flowing under these conditions is 2. Assume a linear open-circuit characteristic. Therefore, I A2 0. A synchronous machine has a synchronous reactance of 2.
How much power P is this machine consuming from or supplying to the electrical system? How much reactive power Q is this machine consuming from or supplying to the electrical system? Figure P shows a synchronous motor phasor diagram for a motor operating at a leading power factor with no RA. This generator is supplying power to a V kW 0. The synchronous generator is adjusted to have a terminal voltage of V when the motor is drawing the rated power at unity power factor.
What is its new value? To make finding the new conditions easier, we will make the angle of the phasor E A, g the reference during the following calculations. The resulting phasor diagram is shown below. Note that an increase in machine flux has increased the reactive power supplied by the motor and also raised the terminal voltage of the system.
This is consistent with what we learned about reactive power sharing in Chapter 5. A V, kW, Hz, four-pole, Y-connected synchronous motor has a rated power factor of 0. At full load, the efficiency is 91 percent.
The armature resistance is 0. Find the following quantities for this machine when it is operating at full load: The Y-connected synchronous motor whose nameplate is shown in Figure has a per-unit synchronous reactance of 0.
Is it the armature current or the field current that limits the reactive power output? Express the answer both in newton- meters and in pound-feet. We will have to check each one separately, and limit the reactive power to the lesser of the two limits. The stator apparent power limit defines a maximum safe stator current.
This limit is the same as the rated input power for this motor, since the motor is rated at unity power factor.
The maximum E A is V or 1. A V three-phase Y-connected synchronous motor has a synchronous reactance of 1. A V, kVA, 0. Ignore all losses. A hp V 0. Its efficiency at full load is 89 percent. What is the phase current of the motor at rated conditions? To simplify this part of the problem, we will ignore R A. Answer the following questions about the machine of Problem Is it consuming reactive power from or supplying reactive power to the power system?
With the remaining rotor arms, a motor can produce sufficient torque to begin spinning the rotor, and a generator can provide useful power to an external circuit. Two or more fixed brushes connect to the external circuit, either a source of current for a motor or a load for a generator.
Commutator segments are connected to the coils of the armature, with the number of coils and commutator segments depending on the speed and voltage of the machine. Large motors may have hundreds of segments. Each conducting segment of the commutator is insulated from adjacent segments. Mica was used on early machines and is still used on large machines. Many other insulating materials are used to insulate smaller machines; plastics allow quick manufacture of an insulator, for example.
The segments are held onto the shaft using a dovetail shape on the edges or underside of each segment. Insulating wedges around the perimeter of each segment are pressed so that the commutator maintains its mechanical stability throughout its normal operating range.
In small appliance and tool motors the segments are typically crimped permanently in place and cannot be removed. When the motor fails it is discarded and replaced. On large industrial machines say, from several kilowatts to thousands of kilowatts in rating it is economical to replace individual damaged segments, and so the end-wedge can be unscrewed and individual segments removed and replaced.
Replacing the copper and mica segments is commonly referred to as "refilling". Refillable dovetailed commutators are the most common construction of larger industrial type commutators, but refillable commutators may also be constructed using external bands made of fiberglass glass banded construction or forged steel rings external steel shrink ring type construction and internal steel shrink ring type construction.
Disposable, molded type commutators commonly found in smaller DC motors are becoming increasingly more common in larger electric motors. Molded type commutators are not repairable and must be replaced if damaged.
In addition to the commonly used heat, torque, and tonnage methods of seasoning commutators, some high performance commutator applications require a more expensive, specific "spin seasoning" process or over-speed spin-testing to guarantee stability of the individual segments and prevent premature wear of the carbon brushes. Such requirements are common with traction, military, aerospace, nuclear, mining, and high speed applications where premature failure can lead to serious negative consequences.
Friction between the segments and the brushes eventually causes wear to both surfaces. Carbon brushes, being made of a softer material, wear faster and may be designed to be replaced easily without dismantling the machine. Older copper brushes caused more wear to the commutator, causing deep grooving and notching of the surface over time. The commutator on small motors say, less than a kilowatt rating is not designed to be repaired through the life of the device. On large industrial equipment, the commutator may be re-surfaced with abrasives, or the rotor may be removed from the frame, mounted in a large metal lathe , and the commutator resurfaced by cutting it down to a smaller diameter.
The largest of equipment can include a lathe turning attachment directly over the commutator. A tiny 5-segment commutator less than 2 mm in diameter, on a direct-current motor in a toy radio control ZipZaps car.
Brush construction[ edit ] Various types of copper and carbon brushes.
However, these hard metal brushes tended to scratch and groove the smooth commutator segments, eventually requiring resurfacing of the commutator. As the copper brushes wore away, the dust and pieces of the brush could wedge between commutator segments, shorting them and reducing the efficiency of the device.
Fine copper wire mesh or gauze provided better surface contact with less segment wear, but gauze brushes were more expensive than strip or wire copper brushes. Modern rotating machines with commutators almost exclusively use carbon brushes, which may have copper powder mixed in to improve conductivity. Metallic copper brushes can be found in toy or very small motors, such as the one illustrated above, and some motors which only operate very intermittently, such as automotive starter motors.
Motors and generators suffer from a phenomenon known as 'armature reaction', one of the effects of which is to change the position at which the current reversal through the windings should ideally take place as the loading varies. Early machines had the brushes mounted on a ring that was provided with a handle. During operation, it was necessary to adjust the position of the brush ring to adjust the commutation to minimise the sparking at the brushes.
This process was known as 'rocking the brushes'. Various developments took place to automate the process of adjusting the commutation and minimizing the sparking at the brushes. One of these was the development of 'high resistance brushes', or brushes made from a mixture of copper powder and carbon.
Also, the high resistance brush was not constructed like a brush but in the form of a carbon block with a curved face to match the shape of the commutator. The high resistance or carbon brush is made large enough that it is significantly wider than the insulating segment that it spans and on large machines may often span two insulating segments.
The result of this is that as the commutator segment passes from under the brush, the current passing to it ramps down more smoothly than had been the case with pure copper brushes where the contact broke suddenly. Similarly the segment coming into contact with the brush has a similar ramping up of the current. Thus, although the current passing through the brush was more or less constant, the instantaneous current passing to the two commutator segments was proportional to the relative area in contact with the brush.
The introduction of the carbon brush had convenient side effects. Carbon brushes tend to wear more evenly than copper brushes, and the soft carbon causes far less damage to the commutator segments. There is less sparking with carbon as compared to copper, and as the carbon wears away, the higher resistance of carbon results in fewer problems from the dust collecting on the commutator segments.
The ratio of copper to carbon can be changed for a particular purpose. Brushes with higher copper content perform better with very low voltages and high current, while brushes with a higher carbon content are better for high voltage and low current. High copper content brushes typically carry to amperes per square inch of contact surface, while higher carbon content only carries 40 to 70 amperes per square inch.
The higher resistance of carbon also results in a greater voltage drop of 0. As the brush and commutator wear down, the spring steadily pushes the brush downwards towards the commutator. Eventually the brush wears small and thin enough that steady contact is no longer possible or it is no longer securely held in the brush holder, and so the brush must be replaced. It is common for a flexible power cable to be directly attached to the brush, because current flowing through the support spring would cause heating, which may lead to a loss of metal temper and a loss of the spring tension.
When a commutated motor or generator uses more power than a single brush is capable of conducting, an assembly of several brush holders is mounted in parallel across the surface of the very large commutator. This parallel holder distributes current evenly across all the brushes, and permits a careful operator to remove a bad brush and replace it with a new one, even as the machine continues to spin fully powered and under load. High power, high current commutated equipment is now uncommon, due to the less complex design of alternating current generators that permits a low current, high voltage spinning field coil to energize high current fixed-position stator coils.
This permits the use of very small singular brushes in the alternator design. In this instance, the rotating contacts are continuous rings, called slip rings , and no switching happens. Modern devices using carbon brushes usually have a maintenance-free design that requires no adjustment throughout the life of the device, using a fixed-position brush holder slot and a combined brush-spring-cable assembly that fits into the slot. The worn brush is pulled out and a new brush inserted.
Brush contact angle[ edit ] Different types of brushes have different brush contact angles  Commutator and brush assembly of a traction motor ; the copper bars can be seen with lighter insulation strips between the bars. Each dark grey carbon brush has a short flexible copper jumper lead attached. Parts of the motor field winding, in red, can be seen to the right of the commutator. The different brush types make contact with the commutator in different ways.