Electric machinery fitzgerald 5th edition pdf free download

A V kW 0. At full load, the efficiency is 90 percent. The armature resistance is 0. A V, hp, Hz, eight-pole, Y-connected synchronous motor has a rated power factor of 0. 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.

Therefore, the stator apparent power limit is The maximum E A is V or 1. 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 phase angle 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 4. A V, 60 Hz, three-phase Y-connected synchronous motor has a synchronous reactance of 1. A V kVA 0. Ignore all losses. Its efficiency at full load is 96 percent. What is the phase current of the motor at rated conditions?

Answer the following questions about the machine of Problem Is it consuming reactive power from or supplying reactive power to the power system? Is the machine operating within its ratings under these circumstances? It is operating within its rating limits. A V three-phase six-pole Hz induction motor is running at a slip of 3. Answer the questions in Problem for a V three-phase two-pole Hz induction motor running at a slip of 0.

A kW, V, Hz, two-pole induction motor has a slip of 5 percent when operating a full-load conditions. At full-load conditions, the friction and windage losses are W, and the core losses are W. A V four-pole Hz Y-connected wound-rotor induction motor is rated at 30 hp. The power consumed by the Thevenin equivalent circuit must be the same as the power consumed by the original circuit.

For the motor in Problem , what is the slip at the pullout torque? What is the pullout torque of this motor? SOLUTION The slip at pullout torque is found by calculating the Thevenin equivalent of the input circuit from the rotor back to the power supply, and then using that with the rotor circuit model. For the motor of Problem , how much additional resistance referred to the stator circuit would it be necessary to add to the rotor circuit to make the maximum torque occur at starting conditions when the shaft is not moving?

Plot the torque-speed characteristic of this motor with the additional resistance inserted. The resulting torque-speed characteristic is: If the motor in Problem is to be operated on a Hz power system, what must be done to its supply voltage? What will the equivalent circuit component values be at 50 Hz?

Answer the questions in Problem for operation at 50 Hz with a slip of 0. Calculate the slip and the electrical frequency of the rotor at no-load and full-load conditions. What is the speed regulation of this motor [Equation ]? What is the rotor copper loss in this motor? The power crossing the air gap of a 60 Hz, four-pole induction motor is 25 kW, and the power converted from electrical to mechanical form in the motor is Figure a shows a simple circuit consisting of a voltage source, a resistor, and two reactances.

Find the Thevenin equivalent voltage and impedance of this circuit at the terminals. Figure P shows a simple circuit consisting of a voltage source, two resistors, and two reactances in parallel with each other. If the resistor RL is allowed to vary but all the other components are constant, at what value of RL will the maximum possible power be supplied to it? Prove your answer. Use this result to derive the expression for the pullout torque [Equation ].

A V Hz four-pole Y-connected induction motor is rated at 25 hp. For the motor in Problem , what is the pullout torque? What is the slip at the pullout torque? What is the rotor speed at the pullout torque? If the motor in Problem is to be driven from a V Hz power supply, what will the pullout torque be? What will the slip be at pullout? The resulting equivalent circuit is shown below.

At what slip does Pout equal the rated power of the machine? An appropriate program is shown below. It follows the calculations performed for Problem , but repeats them at many values of slip, and then plots the results.

It produces an output power of 75 kW at 3. Why is this so? A V six-pole Y-connected hp design class B induction motor is tested in the laboratory, with the following results: No load: V, Plot the torque-speed characteristic of the motor in Problem What is the starting torque of this motor?

A V four-pole hp Hz Y-connected three-phase induction motor develops its full-load induced torque at 3. The two curves are plotted below. As you can see, only the 0. The Thevenin equivalent of the input circuit was calculated in part a.

The easiest way to find the line current or armature current at starting is to get the equivalent impedance Z F of the rotor circuit in parallel with jX M at starting conditions, and then calculate the starting current as the phase voltage divided by the sum of the series impedances, as shown below. Answer the following questions about the motor in Problem What will the voltage be at the motor end of the transmission line during starting? Note that the terminal voltage sagged by about 9.

Note that this voltage sagged by 6. In this chapter, we learned that a step-down autotransformer could be used to reduce the starting current drawn by an induction motor. While this technique works, an autotransformer is relatively expensive. A earlier in this chapter. Answer the following questions about this type of starter. At start-up, the motor develops 1. This motor is to be started with an autotransformer reduced voltage starter. A wound-rotor induction motor is operating at rated voltage and frequency with its slip rings shorted and with a load of about 25 percent of the rated value for the machine.

For most loads, the induced torque will decrease. This reduces the phase current and line current in the motor and on the secondary side of the transformer by a factor of 0. However, the current on the primary of the autotransformer will be reduced by another factor of 0. When it is necessary to stop an induction motor very rapidly, many induction motor controllers reverse the direction of rotation of the magnetic fields by switching any two stator leads.

When the direction of rotation of the magnetic fields is reversed, the motor develops an induced torque opposite to the current direction of rotation, so it quickly stops and tries to start turning in the opposite direction. If power is removed from the stator circuit at the moment when the rotor speed goes through zero, then the motor has been stopped very rapidly. This technique for rapidly stopping an induction motor is called plugging.

The motor of Problem is running at rated conditions and is to be stopped by plugging. Calculate the torque-speed characteristic for this induction motor, and compare it to the torque- speed characteristic for the single-cage design in Problem How do the curves differ? Explain the differences. As a result, the impedance of the rotor is calculated as the parallel combination of these two current paths.

Also, recall that rotor reactance varies with rotor frequency. We must apply this equation to calculate the rotor impedance at any slip, and then divide the resulting reactance by the slip to get to the equivalent impedance at locked-rotor conditions the reactance at locked-rotor conditions is the term that goes into the torque equation. What is the power flowing into or out of the machine? When the loop goes beyond the pole faces, eind will momentarily fall to 0 V, and the current flow will momentarily reverse.

Therefore, the average current flow over a complete cycle will be somewhat less than 5. Refer to the simple two-pole eight-coil machine shown in Figure P Consider the internal resistance of the machine in determining the current flow.

There are 16 conductors in this machine, and about 12 of them are under the pole faces at any given time. A dc machine has 8 poles and a rated current of A.

How much current will flow in each path at rated conditions if the armature is a simplex lap-wound, b duplex lap-wound, c simplex wave-wound? How many parallel current paths will there be in the armature of an pole machine if the armature is a simplex lap-wound, b duplex wave-wound, c triplex lap-wound, d quadruplex wave-wound?

An eight-pole, kW, V dc generator has a duplex lap-wound armature which has 64 coils with 10 turns per coil. How wide must each one be? Since it is duplex-wound, each brush must be wide enough to stretch across 2 complete commutator segments.

Since there are 16 parallel paths through the machine, the armature resistance of the generator is 0. Figure P shows a small two-pole dc motor with eight rotor coils and 10 turns per coil. The flux per pole in this machine is 0. Ignore any internal resistance in the motor. If K is known, then the speed of the motor can be found. Refer to the machine winding shown in Figure P How wide should they be?

At the time shown, those windings are 1, 2, 9, and Each brush should be two commutator segments wide, since this is a duplex winding. Of that number, an average of 14 of them would paths, which produces 7 conductors per path.

Therefore, there are 28 conductors divided among 4 parallel Describe in detail the winding of the machine shown in Figure P If a positive voltage is applied to the brush under the north pole face, which way will this motor rotate? If a positive voltage is applied to the brush under the North pole face, the rotor will rotate in a counterclockwise direction. Column 1 contains field current in amps, and column 2 contains the internal generated voltage EA in volts.

In Problems through , assume that the motor described above can be connected in shunt. The equivalent circuit of the shunt motor is shown in Figure P Assuming no armature reaction, what is the speed of the motor at full load?

What is the speed regulation of the motor? Assume no armature reaction, as in the previous problem. How does it compare to the result for Problem ? What is the starting current of this machine if it is started by connecting it directly to the power supply VT?

How does this starting current compare to the full-load current of the motor? This much current is extremely likely to damage the motor. Assume that the armature reaction increases linearly with Plot the torque-speed characteristic of this motor assuming no armature reaction, and again assuming a increases in armature current.

For the separately excited motor of Problem a What is the maximum no-load speed attainable by varying both V A and Radj? Neglect armature effects in this problem. Both curves are plotted on the same scale to facilitate comparison.

How does the new speed compared to the full-load speed The motor is connected cumulatively compounded and is operating at full load. What will the new speed calculated in Problem ? For Problem , the motor is now connected differentially compounded as shown in Figure P The motor is now connected differentially compounded. Its magnetization curve is The core losses are W, and the mechanical losses are W at full load.

Assume that the mechanical losses vary as the cube of the speed of the motor and that the core losses are constant. The magnetization curve expressed in terms of Neglect rotational losses. Column 1 contains magnetomotive force in ampere-turns, and column 2 contains the internal generated voltage EA in volts.

The motor described above is connected in shunt. Note that is curve is plotted on the same scale as the shunt motor in Problem Derive the shape of its torque- speed characteristic. A series motor is now constructed from this machine by leaving the shunt field out entirely. Derive the torque-speed characteristic of the resulting motor.

To make a practical series motor out of this machine, it would be necessary to include 20 to 30 series turns instead of The motor is to start An automatic starter circuit is to be designed for a shunt motor rated at 20 hp, V, and 75 A.

The with no more than percent of its rated armature current, and as soon as the current falls to rated value, a starting resistor stage is to be cut out.

How many stages of starting resistance are needed, and how big should each one be? The maximum desired starting current is 2. The three stages of starting resistance can be found from the resistance in the circuit at each state during starting. Armature reaction may be ignored in this machine. What is the output torque of the motor? What would happen to the motor if its field circuit were to open?

Ignoring armature reaction, what would the final steady-state speed of the motor be under those conditions? The magnetization curve for a separately excited dc generator is shown in Figure P Its field circuit is rated at 5A. The current is The minimum possible field current occurs and minimum speed and field current.

The machine in Problem is reconnected as a shunt generator and is shown in Figure P The magnetization curve and the field resistance line are plotted below. As you can see, they intersect at a terminal voltage of V. This program created the plot shown above. Note that there are actually two places where the difference between the E A and VT lines is 3.

The code shown in bold face below prevents the program from reporting that first unstable point. Tell user. As shown in the figure below, there is a difference of 7.

The program to create this plot is identical to the one shown above, except that the gap between E A and VT is 7. The resulting terminal voltage is about V. Note that the armature reaction reduces the terminal voltage for any given load current relative to a generator without armature reaction.

The point where the distance between the reaction. E A and VT curves is exactly 4. The new point where the distance between the E A and VT curves is exactly 4. Note that decreasing the field resistance of the shunt generator increases the terminal voltage. Its equivalent circuit is shown in Figure P Answer the following questions about this machine, assuming no armature reaction. Compare it to the terminal characteristics of the shunt dc generators in Problem d. If the machine described in Problem is reconnected as a differentially compounded dc generator, what will its terminal characteristic look like?

Derive it in the same fashion as in Problem Click here to download:- Electric machinery 6th edition fitzgerald. This site uses Akismet to reduce spam. Learn how your comment data is processed. Fitzgerald Textbook Solutions. Find interactive solution manuals to the most popular college math, physics, science, and engineering textbooks. Fitzgerald,Arthur Eugene Fitzgerald,Charles Kingsley,Alexander Kusko Book Resume: The exciting new sixth edition of "Electric Machinery" has been extensively updated while retaining the emphasis on fundamental principles and physical understanding that has been the outstanding feature of this classic book.

Fitzgerald, Charles Kingsley, Jr. Umans — Details — Trove It is hard to understand even when you have the solution. Electric Machinery, Sixth Edition - A.

Arthur Eugene , Subjects Electric machinery. Magnetic Circuits and Magnetic Materials. Fundamentals of electric circuits 6th - WordPress. A First Course In Optimization Theory Solution Manual This seventh edition of Fitzgerald and Kingsley's Electric Machinery by Stephen Umans was developed recognizing the strength of this classic text since its first edition has been the emphasis on building an understanding of the fundamental physical principles underlying the performance of electric machines.

This is just one of the solutions for you to be successful. As understood, execution does not suggest that you have astounding points. Solutions Manual to accompany Electric Machinery This is a straight forward, easy to understand book about three-phase electric systems and machinery.

One of the best I have seen. Also, it is a great price. Another perk is that it appears that the author has put many electronic supplements i. The author writes from a practical industry perspective. We will be glad if you go back us more. Fitzgerald, Charles Kingsley Jr.

To make this manual easier to use, it has been made self-contained. Both the original problem statement and the problem solution are given for each problem in the book. It begins with a review of the fundamentals of circuit theory and electromagnetics and then introduces the concept of electromechanical energy conversion.

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