CHAPTER 7
DC MACHINES

7.1 INTRODUCTION

An electric machine is a link between electrical and mechanical systems. The conversion of electrical to mechanical systems is based on transferring voltage and current electrical quantities to mechanical quantities as torque and angular speed and vice versa. Generators and motors can be used interchangeably. Feeding a generator from the mains will convert it to a motor, while driving a motor with a prime mover will convert it to a generator. Generators or motors can be AC or DC machines.

DC and AC machines are subject to electromechanical energy conversion, which satisfies Lenz's, Faraday's or Lorentz's fundamental laws. We present herein a simple illustration of two principles on which such machines are developed:

  • conductor moving in a uniform magnetic field
  • current-carrying conductor in a uniform magnetic field

7.2 CONDUCTOR MOVING IN A UNIFORM MAGNETIC FIELD

The movement of a conductor in a magnetic field gives induced voltage in a conductor in the magnetic field, as shown in Figure 7.1 [1],

Diagram shows conductor of length L placed along y axis moving with velocity v along x axis in magnetic field B.

Figure 7.1 Conductor moving in a magnetic field.

(7.1)numbered Display Equation

where B is the flux density in Tesla or Weber/m2, l is the conductor length (m), and V is the velocity of its movement in the magnetic field (ms−1).

7.3 CURRENT-CARRYING CONDUCTOR IN A UNIFORM MAGNETIC FIELD

When a current-carrying conductor is in a magnetic field of density B Tesla, as shown in Figure 7.2 [1], the conductor experiences a force F given by

Diagram shows wire carrying current I placed perpendicular to magnetic lines between two magnet bars and force F directed perpendicular both to field lines and wire.

Figure 7.2 Current-carrying conductor in a magnetic field.

A tuning force is impacted on the conductor, causing it to rotate in the magnetic field; this is also due to Lorentz's law.

Of all the rotating machines defined so far, synchronous, induction, and DC machines all use these fundamental principles to operate. To generalize these two principles, machines are designed based on two major components, namely a rotating (moving) part called the rotor and a stationary part called the stator.

The windings are made of copper, which allows current to flow through or allows for induced voltage based on the two principles. In general, the armature winding used to denote rotor and field winding is also used to denote stator winding for same class of machines.

Each of the machines by principle follows the concept—either electric-power conversion to mechanical torque, referred to as motor action, or mechanical-power conversion to electrical power, known as generator action.

7.4 DC-MACHINE CONSTRUCTION AND NAMEPLATE PARAMETERS

DC machines are built with salient magnetic poles on the stator and have commutating poles between the main poles. The coils are situated in the interpolar region with contact through carbon brushes. The windings are based on loop configuration and allow for pairs of poles in the wave winding. The brush holder holds the machines against the commutator surface by springs to maintain a constant pressure and smooth running speed.

A general schematic of a DC machine is shown in Figure 7.3. It is important to note that compensating winding is in the main poles, whereas the interpole is inserted in the commutating poles, which allows for reversal of current. So, we can claim here that a DC machine can be modeled in three configurations based on the location of the shunt or series winding of the interconnected windings of rotor and stator. Electric motors are described as shunt, series, or compound wound-type configuration depending on exciting current.

Diagram shows DC machine with labels of stator, rotor or armature, armature winding, field winding, pole, and air gap. Schematic diagram shows shunt field, communicating field, armature, series field, and compensating winding.

Figure 7.3 Front elevation and schematic diagram of DC machine.

7.5 DC MACHINE PERTINENT NAMEPLATE PARAMETERS

In specifying machines for practical applications, it is important that the parameters are clearly spelled out to ensure easy equipment selection and accurate execution of tasks. Consequently, manufacturers usually include a nameplate on the body of the machine that spells out the machine's important parameters. Important parameters of DC machines that may be found on the nameplate include:

  • armature and shunt field current (amps)
  • armature resistance Ra (ohm)
  • horsepower rating (HP)
  • interpole winding and brushes resistance (ohm)
  • machine type by winding (series, shunt, or compound wound)
  • nominal power rating P (kW)
  • number of poles P
  • rotating losses
  • speed N (rpm)
  • terminal voltage V (volts)

7.6 DEVELOPMENT AND CONFIGURATION OF EQUIVALENT CIRCUITS OF DC MACHINES

From the principles of a rotating conductor, as detailed in Section 7.3, DC-machine armature rotates in a magnetic field produced by the stator magnetic poles. By applying Lenz's law, the voltage induced in the armature winding is given by:

where l is the length of conductor in the slot of the armature, φ is the flux per pole ωm is the mechanical speed, r is the distance of conductor from , and A is the area per pole .

Consequently, we can recast Equation 7.3 as

(7.4)numbered Display Equation

This leads to computation of terminal voltage as a function of the number of turns connected in series with the armature, given as

(7.5)numbered Display Equation

where a is the number of parallel paths and N is the total number of turns in the winding.

(7.6)numbered Display Equation

(7.7)numbered Display Equation

Therefore, we observe that Ea is a function of φ and ωm, given represented by a constant k for a specific machine.

So we may now write

(7.8)numbered Display Equation

where k varies with each machine configuration depending on N, P, π, and a.

Assume

numbered Display Equation

then

(7.9)numbered Display Equation

Ea is now fully defined as voltage in generator operation and Ea is given as back emf in motor operation.

Typically, we develop the equivalent circuit (Figure 7.4) as follows under steady-state operations. As a generator (DC machine) we write

Circuit diagram shows voltage v sub(f) applied across resistor R sub(f) and terminal voltage v sub(t) across armature resistance R sub(a) with voltage E sub(a) and current I sub(a).

Figure 7.4 DC machine equivalent circuit.

(7.10)numbered Display Equation

and

(7.11)numbered Display Equation

or simply

(7.12)numbered Display Equation

Combining these two equations

(7.13)numbered Display Equation

where ka depends on number of conductor and pole pairs, i.e.,

(7.14)numbered Display Equation

Based on the second fundamental equation of machine operation (Equation 7.2), torque is developed based on resulting force on the conductor by Lorentz's equation. Torque is produced by armature winding with current in a magnetic-field winding of the stator.

The force armature conductor is calculated as

(7.15)numbered Display Equation

(7.16)numbered Display Equation

(7.17)numbered Display Equation

which gives the average torque

(7.18)numbered Display Equation

where ka is a design constant that depends on .

Therefore, in the case of motor operation:

  • The terminal voltage generated is greater than emf and the electromagnetic torque produces rotation against a load.
  • Alternatively, when the terminal voltage Vt is less than the generated voltage in the case of a generator, the electromagnetic torque opposes the torque applied to the shaft by the prime mover.

Simply, in the case of a motor

numbered Display Equation

where Tωm is the developed power within the mechanical system. The converse is also true for a motor or generator operation. So, we can write

(7.19)numbered Display Equation

where Ka, Wm, φ, and Ia are already defined.

Example 1

A shunt generator delivers 450A at 230 V and the resistance of the shunt field and armature are 50 ohm and 0.03 ohm, respectively. Calculate the generated emf.

Solution:

Current through shunt field winding

numbered Display Equation

Load current

numbered Display Equation

Armature current

numbered Display Equation

Armature voltage drop

numbered Display Equation

Example 2

A four-pole DC machine has an armature of radius 12.5 cm, effective length of 25 cm. The poles cover 75 percent of the armature periphery. The armature windings consist of thirty-three coils each having seven turns. The coils are accommodated in thirty-three slots. The average flux density under each pole is 0.75T. If the armature is wound, determine:

  1. armature constant ka
  2. induced-armature voltage when the armature rotates at 1,000 rpm
  3. current and electromagnetic torque developed when the armature is 400A
  4. power developed by the armature

If the armature is wave wound, repeat (a)–(d) above.

Solution:

numbered Display Equation

pole area

numbered Display Equation

or

numbered Display Equation

wave wound

numbered Display Equation

7.7 CLASSIFICATION OF DC MACHINES

The field current and armature current can be interconnected in many ways to provide various performance characteristics for DC-machine configuration.

  1. Field poles in DC machines are configured by four methods (as shown in Figure 7.5):
    1. series-field winding (connected in series)
    2. shunt-field winding (connected in parallel)
    3. short-shunt compound winding
    4. long-shunt compound winding
Image described by caption

Figure 7.5 DC-machine configurations. (a) Series DC machine (b) Shunt DC machine (c) Compound (short shunt) (d) Compound (long shunt).

The compound-machine configuration combines series and shunt winding and these two windings have different performance. The rheostat helps control the field current to achieve the desired performance.

Given these configurations, we now develop analysis for DC generators using the circuit equivalent.

Consider the machine to be separately excited (Figure 7.6), then the relevant equations are:

Circuit diagram shows current I1 flows through dc source, variable resistor, and inductor in first loop and current I sub(L) flows through armature winding and load. Voltage V taken across load.

Figure 7.6 Separately excited DC machine.

numbered Display Equation

and

numbered Display Equation

where

numbered Display Equation

These equations define the termed or external Vt of the separately excited DC generator.

The relationship between Vt and It for different operating conditions is shown in Figure 7.7 [2].

Terminal voltage versus terminal current graph shows horizontal line depicting terminal characteristics with no armature reaction, decreasing line depicting terminal characteristics with armature reaction, and increasing line containing operating point depicting load characteristics.

Figure 7.7 Terminal and load characteristics of separately excited generator.

From the terminal characteristics of a separately excited DC generator, there is a voltage drop due to armature reaction (ΔVAR) from the machine internal characteristics to external characteristics. The operating point is the intersection between the generator's external characteristics and the load characteristics. The operating point gives the operating values of load current and terminal voltage.

Example 1

A shunt generator delivers 450A at 230 V and the resistance of the shunt field and armature are 50 ohm and 0.03 ohm, respectively. Calculate the generated emf.

Solution:

Current through shunt field winding = Ish = 230/50 = 4.6A

Load current

numbered Display Equation

Armature current

numbered Display Equation

Armature voltage drop

numbered Display Equation

Example 2

A long-shunt compound generator delivers a load current of 50A at 500 V and has armature, series-field, and shunt-field resistances of 0.05 ohm, 0.03 ohm, and 250 ohm, respectively. Calculate the generated voltage and the armature current. Allow 1 V per brush for contact drop.

Solution:

numbered Display Equation

Current through armature and series winding = 50 + 2 = 52A

Voltage drop on series field winding = 52 × 0.03 = 1.56V

Armature voltage drop = IaRa = 52 × 0.05 = 2.6V

Drop at brushes = 2 × 1 = 2V

Now,

numbered Display Equation

Example 3

A short-shunt compound generator delivers a load current of 30A at 220 V and has armature, series-field, and shunt-field resistance of 0.05 ohm, 0.3 ohm, and 200 ohm, respectively. Calculate the induced emf and armature current. Allow 1.0 V per brush for contact drop.

Solution:

numbered Display Equation

Example 4

A 12 kW, 100 V, 1,000 rpm DC shunt generator has armature resistance Ra = 0.1 ohm, shunt-field winding resistance Rf = 80 ohm, and Nf = 1, 200 turns/pole. Rated field current is 1A. Find Vt at full load if Ea is 98 V.

If armature resistance is at full load of 0.06A, determine full load at Vt.

Solution:

Given that

numbered Display Equation

Example 5

Determine the field current required to make the terminal voltage of Vt = 100V at full load.

Solution:

numbered Display Equation

so

numbered Display Equation

7.8 VOLTAGE REGULATION

(7.20)numbered Display Equation

Vt, NL is the no-load terminal voltage, and is the full-load terminal voltage.

Since Ea is equal to Vt, NL,

(7.21)numbered Display Equation

7.9 POWER COMPUTATION FOR DC MACHINES

For DC-machine operation, DC current must be provided. Use

(7.22)numbered Display Equation

at armature reaction

(7.23)numbered Display Equation

where

(7.24)numbered Display Equation

Finally, the total resistance

(7.25)numbered Display Equation

This is used to determine the performance of the machine.

7.9.1 Loss Calculation

(7.26)numbered Display Equation

(7.27)numbered Display Equation

(7.28)numbered Display Equation

where

numbered Display Equation

7.9.2 Torque Computation

Assume a DC machine has Va, Ia, Ea, Kmax, then

numbered Display Equation

This can be written as function of losses

(7.29)numbered Display Equation

and the corresponding torque

(7.30)numbered Display Equation

(7.31)numbered Display Equation

(7.32)numbered Display Equation

(7.33)numbered Display Equation

where

numbered Display Equation

so that

(7.34)numbered Display Equation

nm is in rad/s.

Using the loss formula, we can determine power flow and efficiency of the machine.

7.10 POWER FLOW AND EFFICIENCY

From the Figure 7.8a and Figure 7.8b, we can compute efficiency

Top flow diagram shows shunt field, armature circuit, and rotational losses between input power from mains and shaft power to load and bottom flow diagram shows losses between input power from prime mover and output power.

Figure 7.8 Losses and energy flow in DC generators and motors. (a) Energyflow in DC motor (b) Energy flow in DC generators.

numbered Display Equation

where

numbered Display Equation

(7.35)numbered Display Equation

(7.36)numbered Display Equation

Copper loss depends on armature-winding loss (I2aRa) or field-winding losses (I2fRf) when Ia and If are armature- and field-winding currents, respectively, Ra and Rf are armature and field resistances, respectively, brush loss Pb = VbIa, and Vb is the voltage drop across brushes, which is typically about 2 V.

7.10.1 Core Loss

The core loss is derived from the eddy-current and hysteresis losses, as discussed in Chapter 2. Core loss depends on the value of the magnetic flux density B and the speed of rotation of the machine. Mechanical losses can be given as a function of friction and windage. Stray losses account for approximately 1 percent of full-load loss. Power losses in DC machines are given or empirically computed.

Example 1

A shunt generator delivers 195A at terminal voltage 250 V. The armature resistance and shunt-field resistance are 0.02Ω and 50Ω, respectively. The iron and frictional losses equal 950 W. Find:

  1. emf generated
  2. copper loss
  3. output of the prime mover
  4. commercial, mechanical, and electrical efficiencies

Solution:

numbered Display Equation

7.11 DC MOTORS

Under a steady-state condition, the interaction between voltage and current for DC machines is given as

(7.37)numbered Display Equation

(7.38)numbered Display Equation

where we further suggest, as discussed in Section 7.12, that for a motor

(7.39)numbered Display Equation

from which it can be further shown that

(7.40)numbered Display Equation

So, as usual

(7.41)numbered Display Equation

Different configurations can be represented as follows:

  1. shunt motors, whose field current can be controlled by use of variable resistance in series with the field winding
  2. series motors, for which it is difficult to control the armature and field winding independently, so we treat flux as largely dependent on the armature current and field current
  3. compound motors, where the effect of armature current on the flux depends on different issues (short and long winding)

7.12 COMPUTATION OF SPEED OF DC MOTORS

From the fundamental expression

numbered Display Equation

(7.42)numbered Display Equation

which means that speed, ωm, depends on the applied voltage Vt, armature current Ia, armature resistance Ra, and flux φ per pole.

The schematic diagram (as seen in Figure 7.9) of each DC-motor configuration with respect to control options is given as follows:

Diagram shows R sub(f) connected parallel to R sub(a) in part (a), field winding connected in series to R sub(a) in part (b), and R sub(f) connected parallel to R sub(a) on one side and series filed connected on other side in part (c).

Figure 7.9 Configuration types of DC motors. Configuration (a) shunt DC motor (b) Series DC motor (c) Compound motor.

For configuration (a), shunt motor

(7.43)numbered Display Equation

(7.44)numbered Display Equation

So, we can write

(7.45)numbered Display Equation

For configuration (b), series DC motor

(7.46)numbered Display Equation

(7.47)numbered Display Equation

So, we can write

(7.48)numbered Display Equation

For configuration (c), compound motor

(7.49)numbered Display Equation

(7.50)numbered Display Equation

numbered Display Equation

But

(7.51)numbered Display Equation

Then,

(7.52)numbered Display Equation

(7.53)numbered Display Equation

The following figures namely Figures 7.10, 7.11, and 7.12 show the torque/armature current characteristics, torque/speed characteristics and speed/power characteristics respectively of the dc motor.

Torque versus armature current graph shows concave up increasing curve depicting series motor, almost linearly increasing curve depicting shunt motor, and curve with intermediate slope depicting compound motor.

Figure 7.10 Torque/armature-current characteristics.

Speed versus torque graph shows concave down decreasing curve representing DC shunt motor and concave up decreasing curve depicting DC series motor.

Figure 7.11 Torque/speed characteristics of a DC-motor configuration.

Speed versus power graph shows concave down decreasing curve representing DC shunt motor and concave up decreasing curve depicting DC series motor.

Figure 7.12 Speed/power characteristics of a DC-motor configuration.

7.12.1 Connection Description

A shunt machine is a constant-speed machine with low-speed regulation. Speed can be controlled by field flux.

For a series motor, recall the equation for Te given as

(7.54)numbered Display Equation

If Ra ≈ 0, Te is a function of flux φ, terminal voltage will be

(7.55)numbered Display Equation

from which

(7.56)numbered Display Equation

Also

numbered Display Equation

(7.57)numbered Display Equation

(7.58)numbered Display Equation

Thus, the speed/power relationship is as shown in Figure 7.13.

Speed versus power graph shows concave down decreasing curve representing DC shunt motor and concave up decreasing curve depicting DC series motor.

Figure 7.13 Speed–power relationship.

The relationship between speed and torque for each class of DC-machine configuration operating as a motor is as shown in Figure 7.14.

Speed versus torque graph shows concave up curve depicting series motor, concave down curves depicting shunt motor, induction motor, and synchronous machine, and s-shaped curve depicting compound motor.

Figure 7.14 Speed-torque characteristics of electric machines.

Example 1

A 440 V shunt motor has armature resistance of 0.8 ohm and field resistance of 200 ohm. Determine the back emf when giving an output of 7.46 kW at 85 percent efficiency.

Solution:

numbered Display Equation

Now,

numbered Display Equation

Example 2

The following data is for a DC shunt motor:

numbered Display Equation

At no load, armature resistance = 0.5Ω, speed = 1,000 rpm, armature current = 4A, torque = 10Nm.

  1. Find the armature current and the speed of the motor.
  2. If the motor is required to develop 10 hp at 1,200 rpm, compute the required value of the external series resistance in the field circuit.

Solution:

Assume field current

numbered Display Equation

with machine on load

numbered Display Equation

We can now obtain

numbered Display Equation

Since

numbered Display Equation

we can recompute

numbered Display Equation

with corresponding speed

numbered Display Equation

For 10 Hp at 1,200 rpm,

numbered Display Equation

So, given

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or

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Example 3

The input to a 230 V DC shunt motor is 11 kW. Calculate:

  1. torque developed
  2. efficiency
  3. speed at this load

The particulars of the motor are: no-load current is 5A, no-load speed is 1,150 rpm, armature resistance is 0.5 ohm, and shunt-field resistance is 110 ohm.

Solution:

No-load input = 220 × 5 = 1,100 W

numbered Display Equation

with 11 kW input,

numbered Display Equation

7.13 DC-MACHINE SPEED-CONTROL METHODS

Three methods are used mainly for DC machine control:

  1. field control
  2. armature-resistance control
  3. solid-state rectifier control

7.13.1 Field-Control Method

From

(7.59)numbered Display Equation

Vt is kept constant so ωm depends on If only. φ decreases while speed increases and since If is directly proportional to φ, it follows that by varying If, the machine's speed can be varied. One way to do this is to insert a variable resistance in the field circuit to control If. This is one of the simplest but most efficient methods of DC-machine speed control.

7.13.2 Armature-Resistance Control Method

Terminal voltage and field current are maintained consistent at their rated values. Motor speed is easily controlled by varying armature resistance through an external rheostat. This control method is simple and inexpensive but has the disadvantage of low efficiency.

Again, ωm changes depend on Vt. It is easy to compute voltage regulation or speed regulation using speed regulation

(7.60)numbered Display Equation

The method of control here is applicable to shunt-, series-, and compound-motor configurations.

7.13.3 Solid-State Rectifier Control

Silicon-controlled rectifiers (SCRs) are one of the solid-state devices widely used in the control of DC motors. The SCR is used to rectify the voltage and control the armature current of the DC machine by adjusting the firing angle with the application of the appropriate trigger voltage to the gate of the SCR. The firing angle of the SCR determines the amount of DC current supplied to the motor's armature. Thus, by controlling the armature current, the machine is also controlled. This is consistent with the equation

(7.61)numbered Display Equation

For a DC-series motor it can be shown that

numbered Display Equation

from which we can see that

(7.62)numbered Display Equation

Thus, increasing the armature current results in a decrease in motor speed, which can be achieved using the SCR.

7.14 WARD LEONARD SYSTEM

For a solid-state control rectifier using a DC machine, the speed-control system is called the Ward Leonard system. It is capable of controlling a DC motor from zero to maximum or maximum to zero.

The configuration of the Ward Leonard system is shown in Figure 7.15 [3]. AC source (induction or synchronous machine) serves as the prime mover for a DC generator. The generator in turn supplies Vt, which supplies a DC motor for operation.

Schematic diagram shows three phase line, motor M2, switch S, variable resistor R, generator G, generator E, and motor M1 with voltage V across it.

Figure 7.15 Ward Leonard system control scheme.

Varying the field current of the DC machine, which varies the speed of the motor, can consequently control the armature voltage in the motor. That is, the motor speed can be reduced or increased by the motor field current with changes in flux. Further, if the current of the generator is reversed, the polarity of the generator-armature voltage will reverse as well. These reversals also change or cause the motor direction of rotation to change. The speed variation occurs as the field current changes back and forth in the opposite direction for the Ward Leonard system.

Illustrative Problems and Examples

  1. Calculate the flux in a four-pole dynamo with 722 armature conductors generating 500 V when running at 1,000 rpm, when the armature is (a) lap wound (b) wave wound.

    Solution:

    The generated emf is given by

    numbered Display Equation
    1. For lap wound C = 2P
      numbered Display Equation

      For lap wound,

      numbered Display Equation

      Now,

      numbered Display Equation
    2. For wave winding A = 2
      numbered Display Equation
  2. A DC generator generates an emf of 520 V. It has 2,000 armature conductors, flux per pole of 0.13 Wb, speed of 1,200 rpm, and the armature winding has four parallel paths. Find the number of poles.

    Solution:

    numbered Display Equation

    For four-parallel path

    numbered Display Equation
  3. When driven at 1,000 rpm with a flux per pole of 0.02 Wb, a DC generator has an emf of 200 V. If the speed is increased to 1,100 rpm and at the same time the flux per pole is reduced to 0.019 Wb per pole, what is the induced emf?

    Solution:

    numbered Display Equation
  4. A 230 V shunt motor has an armature resistance of 0.26Ω. Assuming a 2 V brush drop, calculate the back emf when the armature current is 36A.

    Solution:

    numbered Display Equation

7.15 CHAPTER SUMMARY

We have presented the concepts and fundamental principles of DC machines in this chapter. The internal structures were briefly discussed. Also, the various configurations and their important characteristics were discussed. We distinguished between DC generators and DC motors and discussed methods of speed control.

DC motors have found wide applications in control systems. As a rotating machine with an inertial mass, it is of interest to us as it applies to the microgrid. The behavior of DC machines as a generator or motor—for instance on a DC microgrid—should be of practical interest. The student is encouraged to learn these principles as an integral part of the knowledge relative to energy conversion in the smart grid and microgrid. Further, this chapter provided principles for studying the performance of DC-machine applications.

EXERCISES

  1. A DC series motor operates at 750 rpm with a line current of 80A from the 230 V mains. Its armature circuit resistance is 0.14Ω and its field resistance is 0.11Ω. Assuming that the flux corresponding to a current of 20A is 40 percent of that corresponding to a current of 80A, find the motor speed at a line current of 20A at 230 V.

  2. A DC shunt machine is connected across a 500 V supply. The total armature winding resistance is 0.05Ω and the field-circuit resistance is 100Ω. Calculate the ratio of speed as a generator to speed as a motor for the line current to be 200A.

  3. A shunt motor draws 26.5A from a 120 V supply when it drives a load at 25 rps. The resistance of the field circuit is 120Ω and the armature-winding resistance is 0.2Ω. What is the electromagnetic torque developed?

  4. When a motor is operating under load, the armature takes 8,600 W and its current is 38A. If the armature-circuit resistance, including brushes, is 0.4Ω, what horsepower is developed by the motor?

  5. A 240 V, 20HP shunt motor has a resistance of 0.18Ω for the armature circuit, not including brushes (allow for a brush drop of 2 V), and 240Ω for the field circuit. The rated speed is 1,200 rpm and the full-load efficiency is 0.87. Calculate:

    1. torque
    2. stray losses
    3. rotational losses
  6. The resistance of the armature circuit of a 240 V shunt motor is 0.1Ω, including brushes. When the field rheostat is adjusted so that the resistance of the shunt field circuit is 120Ω, the current taken by the motor is 82A and the speed is 900 rpm. Neglecting armature reaction, calculate:

    1. current
    2. speed if the load torque is increased by 50 percent
  7. A 110 V shunt generator has full-load current of 100A, shunt-field resistance of 55Ω, and constant losses of 500 W.

    1. If full-load efficiency is 88 percent, find armature resistance.
    2. Assuming voltage to be constant at 110 V, calculate the efficiency at half full load and at 50 percent overload.
    3. Find the current.
  8. A DC-series motor on full load takes 50A from 230 V DC mains. The total resistance of the motor is 0.22Ω. If the iron and friction losses together amount to 5 percent of the input, calculate the power delivered by the motor shaft. Total voltage drop due to the brush contact is 2 V.

  9. A two-pole, DC shunt motor operating from a 200 V supply takes a full-load current of 35A, the no-load current being 2A. The field resistance is 500Ω and the armature has a resistance of 0.6Ω. Calculate the efficiency of the motor on full load. Take the brush drop as being equal to 1.5 V per brush arm.

BIBLIOGRAPHY

  1. R. Ramshaw and R. G. van Heeswijk, Energy Conversion: Electric Motors and Generators, Saunders College Publishing, Philadelphia, 1990.
  2. E. H. Smith, Mechanical Engineers Reference Book, 12th ed., Butterworth Heinemann Ltd., Oxford, 1994.
  3. R. K. Rajput, A Textbook of Electrical Machines, Laxmi Publications Ltd., New Delhi, 2006.
  4. H. Saadat, Power System Analysis, 3rd ed., PSA Publishing, 2010.
  5. A. E. Fitzgerald, C. Kingsley, and S. D. Umans, Fitzgerald & Kingsley's Electric Machinery, 4th ed., McGraw-Hill, New York, 1983.
  6. C. A. Gross, Power System Analysis, John Wiley & Sons, New York, 1979.
  7. S. J. Chapman, Electric Machinery Fundamentals, 5th ed., McGraw Hill, New York, 2012.
  8. T. Gönen, Electrical Machines with MATLAB, 2nd ed., Taylor & Francis Group, Boca Raton, FL, 2011.
  9. W. L. Matsch and J. D. Morgan, Electromagnetics and Electromechanical Machines, 3rd ed., John Wiley & Sons, Hoboken, NJ, 1986.
  10. M. S. Sarma, Electric Machines: Steady-State Theory and Dynamic Performance, WMC Brown Publishers, Dubuque, IA, 1985.
  11. G. R. Slemon, Electric Machines and Drives, Addison-Wesley Publishing, New York, 1992.
  12. P. C. Sen, Principles of Electric Machines and Power Electronics, 2nd ed., John Wiley & Sons, Hoboken, NJ, 1997.
  13. Electronics Tutorials, “Magnetic Hysteresis,” www.electronics-tutorials.ws/electromagnetism/magnetic-hysteresis.html.
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