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- (PDF) Energy Conversion - Electric Machinery book Fitzgerald solution | emre erkan - preserbelleodo.ga
- Fundamentals of Electrical Machines.pdf
- Electric Machinery Fundamentals Solutions Manual
- Electric Machinery Fundamentals

Electric Machinery. Sixth Edition. A. E. Fitzgerald. Late Vice President for Academic Affairs and Dean of the Faculty. Northeastern University. Charles Kingsley, Jr. Arthur E. Fitzgerald, Charles Kingsley, Jr., and Stephen D. Umans, Electric Machinery, Sixth Edition, McGraw-Hill, 1. (Three-Phase Circuits) 2. ( Magnetic. Energy Conversion - Electric Machinery book Fitzgerald solution. Emre Erkan. 1 PROBLEM SOLUTIONS: Chapter 1 Problem Part (a): lc lc Rc = = =0 A/Wb.

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Electric Machinery Fitzgerald 7th Edition PDF Electric Machinery Fitzgerald 7th Edition Book PDF Extra Link Book Description: This seventh edition of Fitzgerald . He has taught electric machines, power electronics, and electric drive systems for There has been widespread use of electric machines in industries and. Stephen J. Chapman received a B.S. in Electrical Engineering from Louisiana. State University () and an M.S.E. in Electrical Engineering from the Univer-.

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It sounds like an 8-pole motor supplied by 60 Hz. Reversing two phases is the procedure for reversing the direction of a three-phase induction motor. For a slip of 0. Since the rotor is 5 wavelengths long, the armature winding will link 10 poles of flux with 10 turns per pole. Chapter 5 Problem 5. Reduced to Unchanged part c: Unchanged part d: Increased to The windings are orthogonal and hence the mutual inductance is zero.

Thus, the equivalent inductance is simply equal to the phase self-inductance. Problem 5. See plot below. Minimum current will when the motor is operating at unity power factor.

Interpolating the given data shows that this corresponds to a value of around V line-to-line. Using generator convention for current part c: It looks like an inductor. It was underexcited, absorbing reactive power. It increased. The answers are the same. The rotor angle is From the solution to Problem 5. The generator base impedance is 1. Thus part a: Chapter 6 Problem 6. Problem 6. Therefore the motor has 6 poles.

Because this is an induction machine, the car in this case will never reach synchronous speed. For a slip of 6. Hence the magnitude of the armature voltage must vary linearly with electrical frequency and hence the desired speed.

The electrical frequency of the rotor currents is equal to the slip frequency and hence will remain constant. Since the torque is proportional to the square of the voltage, the torque-speed characteristic will simply be reduced by a factor of 4.

Four poles.

Counterclockwise part c: Four poles part e: The induced voltage will therefore be at a frequency of 25 Hz. R1 will decrease by a factor of 1. R2 will decrease by a factor of 3. All values will decrease by a factor of 3. The following table of results was obtained from a MATLAB script which implemented the equivalent-circuit equations assuming the core-loss resistance Rc is in parallel with the magnetizing reactance.

The impedance of the feeder must be added in series with the armature resistance R1 and leakage reactance X1.

Hence 0. Thus the rotational loss is 3. Similarly, the input power to the equivalent circuit is Thus, the core losses are equal to The equivalent circuit is solved in the normal fashion.

For ease of calculation, the core loss can be accounted for by a resistor connected at the equivalent-circuit terminals based upon the results of part a , this corresponds to a resistance of 1. The shaft input power is equal to the negative of the shaft power calculated from the equivalent circuit plus the rotational loss power. The electrical output power is equal to the negative of the input power to the equivalent circuit.

Basically the same calculation as part b. The impedance of the feeder must be added to armature impedance of the induction motor. Taking the ratio of Eqs. As in the solution to Problem 6. The parameters are calculated following exactly the procedure found in Example 6.

The results are: Therefore 1. From the solution to Problem 6. If the rotor resistance is doubled, the motor impedance will be the same if the slip is also doubled. The torque will equal to full-load torque.

With no external resistance, the slope of the torque-speed characteristic is thus 5. The slope of the desired torque-speed charac- teristic is 4. The total loss is approximately equal to the sum of the open-circuit loss and the short-circuit loss kW.

We will again solve this problem with the load connected to the high-voltage side but referred to the low-voltage side. The load power is equal to 0. The voltage rating is V: Hence as an auto- transformer operating with a load at 0.

The voltage rating is 78 kV: The referred load voltage Vload and current Iload will be in phase and can be assumed to be the phase reference. Thus we can write the phasor equation for the sending-end voltage as: Chapter 3 Problem 3.

Problem 3. Under the assumption that all electrical transients have died out, the solution will be the same as that for Problem 3. Four poles part b: Thus, part a: The battery will supply only the energy dissipated in the coil. The required force is equal to N the mass of the slab times the acceleration due to gravity, 9. Note that as the coil moves upward in the slot, the energy associated with the leakage flux associated withing the coil itself remains constant while the energy in the leakage flux above the coil changes.

Hence to use the energy method to calculate the force on the coil it is necessary only to consider the energy in the leakage flux above the slot. The curve of spring force versus angle is plotted as a straight line on the plot of part d. The intersection with each curve of magnetic force versus angle gives the equilibrium angle for that value of current.

It will not rotate. Substituting the given expressions for the coil currents gives: The flux linkages of the voice coil can be calculate in two steps. For a fixed voltage, the dynamic equations become: Following the derivation of Example 3. Chapter 4 Problem 4. Problem 4. It is an induction motor. It sounds like an 8-pole motor supplied by 60 Hz.

Reversing two phases is the procedure for reversing the direction of a three-phase induction motor. For a slip of 0. Since the rotor is 5 wavelengths long, the armature winding will link 10 poles of flux with 10 turns per pole. Chapter 5 Problem 5. Reduced to Unchanged part c: Unchanged part d: Increased to The windings are orthogonal and hence the mutual inductance is zero. Thus, the equivalent inductance is simply equal to the phase self-inductance.

Problem 5. See plot below. Minimum current will when the motor is operating at unity power factor. Interpolating the given data shows that this corresponds to a value of around V line-to-line.

Using generator convention for current part c: It looks like an inductor.

It was underexcited, absorbing reactive power. It increased. The answers are the same. The rotor angle is From the solution to Problem 5. The generator base impedance is 1. Thus part a: Chapter 6 Problem 6. Problem 6. Therefore the motor has 6 poles. Because this is an induction machine, the car in this case will never reach synchronous speed. For a slip of 6.

Hence the magnitude of the armature voltage must vary linearly with electrical frequency and hence the desired speed. The electrical frequency of the rotor currents is equal to the slip frequency and hence will remain constant. Since the torque is proportional to the square of the voltage, the torque-speed characteristic will simply be reduced by a factor of 4. Four poles.

Counterclockwise part c: Such a transformer is commonly called an iron- core transformer. Most transformers are of this type.

The following discussion is concerned almost wholly with iron-core transformers. Types of cores for power transformer both types are constructed from thin laminations electrically isolated from each other minimize eddy currents 16 Core Form: a simple rectangular laminated piece of steel with the transformer windings wrapped around two sides of the rectangle.

Shell Form: a three legged laminated core with the windings wrapped around the centre leg. Figure 5. The primary and secondary windings are wrapped one on top of the other with the low-voltage winding innermost, due to 2 purposes: 1. It simplifies the problem of insulating the high-voltage winding from the core. It results in much less leakage flux 17 1.

Distribution transformers are generally used in power distribution and transmission systems. Power transformers are used in electronic circuits and come in many different types and applications.