# Dc generators : power losses, generator data and ratings, magnetohydrodynamic (mhd) generation and summary of dc generators .

19–9 POWER LOSSES

All machines suffer power losses in the form of heat. Electrical machines, in particu- lar, suffer two kinds of power losses: copper losses (also known as I2R losses) and stray power losses; see Figure 19–36.

This phenomenon is sometimes explained by comparing the machine to a leaky water pipe, with one end of the pipe representing the input, and the other end the output. Such an illustration, depicting the losses on a motor, is represented in Figure 21–23. This analogy should make the fact clear that in all machines

Power input = Power output + All losses

As an exercise to further illustrate this point, make a sketch analogous to the generator shown in Figure 19–36.

Mechanical losses are caused by friction in bearings and brushes and by friction of the air, called windage. Windage loss varies with speed but is independent of load current. Core losses consist of hysteresis and eddy current losses. Hysteresis loss is due to molecule friction in the iron of the armature because of continual reversal of magnetization. Hysteresis loss increases with increase in flux density and increase in speed. Eddy current losses are due to small circulating, or eddying, currents.

Copper losses are a result of heat production by currents in the armature and field circuits of the machine. In other words:

Total I2R losses = I2R of armature + I2R of field(s)

EXAMPLE 19–2

Given: A shunt generator with an armature resistance of 1 ohm, delivering 9.5 amperes to a load at a terminal voltage of 120 volts; the shunt field has a resistance of 240 ohms, as shown in Figure 19–37.

Find:

a. Total I2R losses

b. The total emf generated

Efficiency of a Machine

Conversion of energy always incurs losses. In other words, the power output of a machine is always less than its power input; and the difference between the two represents the losses.

Pout = Pin - Losses

The ratio between the power output and the power input is described as the efficiency

of the machine. Efficiency is generally expressed as a percentage, as follows:

EXAMPLE 19–3

Given: The generator of Example 19–2, Figure 19–37, requiring a mechanical input of 2 horsepower.

Find:

a. The efficiency of the machine

b. The stray power losses

Solution

19–10 GENERATOR DATA AND RATINGS

The full-load rating of a generator is a statement of conditions that provide efficient operation without exceeding safe limits of speed and temperature. These data, which are supplied on the nameplate of the machine, include speed, voltage, power output (kW) or current output, and allowable temperature rise.

If operated at very low current output, efficiency is low. At a speed that is too low, air circulation is poor and overheating can result. Higher than normal current over a long time raises temperature and can damage insulation and burn the commutator and brushes. Standard voltages for larger DC generators are 125, 250, 275, and 600 volts.

The temperature rise allowed in a machine is a rise above 40°C, which is taken as a standard surrounding temperature. For example, the temperature of Class A insulation (enamel, oil-impregnated paper, and cotton) should not exceed 105°C. A machine with this insulation is rated 50°C temperature rise. This rating allows the average, or surface, temperature of a coil to be 90°C (40 1 50) while allowing for hot spots in the center of the coil to be 15°C higher than 90°C, which is the 105°C specified limit.

19–11 MAGNETOHYDRODYNAMIC (MHD) GENERATION

MHD generation, or energy conversion, uses the principles already discussed but in a different manner. Instead of moving a wire carrying electrons through a magnetic field, a stream of electrons and ions is made to move through a magnetic field. The high-speed electrons and positive ions are deflected in opposite directions by the magnetic field. Electrons

collect at one plate and positive ions at another. The excess negative and positive charges on the plates cause them to be at a different electrical potential, like the terminals of a battery. Think of the superheater in Figure 19–38 as a place for heating a gas, either by burning it or by applying heat from an external source. The temperature of the superheated gas is so high that gas molecules ionize, forming a cloud of electrons and positive ions.

The ionized gas blows through a nozzle, entering the magnetic field area at a tempera- ture of about 2,000°C and a velocity of over 1,500 miles per hour. As shown in Figure 19–38, the magnetic lines of force are directed at right angles to the page, as from an N pole above the page to an S pole behind the page. Just as electrons in a wire in an ordinary generator are forced to move in a direction relative to the magnetic field and to their direction of motion, electrons in this device are forced upward toward the anode.

The anode may be negatively charged already because of the previous accumulation of electrons. As more electrons come roaring by, the magnetic field deflects them. Their high kinetic energy slams them onto the anode, building up the negative charge still more. Positive ions are deflected in the opposite direction. They collide with the cathode and pick up electrons from it; thus the cathode maintains a positive charge (electron deficiency). Anode and cathode connect to the useful external circuit.

When and if MHD converters can be built to produce huge amounts of electrical energy, efficiency much higher than that for conventional steam turbine generator plants is possible. The size of the converter may be relatively small; Westinghouse has built an MHD generator that produces 2.5 kW from a unit about the size of this book.

Serious problems must yet be overcome. Extremely high temperatures are needed for the gases to ionize sufficiently to stay ionized rather than to recombine. Heat sources capable of producing the extremely high temperature must be developed. New materials must be found to resist critically high temperatures, or better use must be made of materials now available.

SUMMARY

• DC armatures carry coils in a series-parallel arrangement. There are two or more parallel paths though the armature, and each path consists of several coils in series.

• The magnetic field of the generator is usually supplied by current from the generator itself. This process is called self-excitation.

• Shunt field coils, in parallel with the load circuit, produce field magnetism that decreases slightly as the current output of the generator increases.

• Series field coils, in series with the load circuit, produce field magnetism that increases in proportion to the output current.

• Compound generators have both series and shunt fields.

• DC generators that are a part of specialized motor-control circuits are usually separately excited.

• Armature reaction, which is field distortion due to armature current, causes commutation difficulties that can be corrected by interpoles. It also causes field weakening, which can be corrected by compensating windings.

Achievement Review

1. A four-pole generator field has a flux density of 75,000 lines per square inch.

Each pole face is 5 inches square. Calculate the total flux.

2. Name three types of field coil arrangements for self-excited generators. Which gives the steadiest voltage output?

3. Name the three factors of greatest influence in determining the emf produced in a generator armature.

4. What is armature reaction? Name the two main disagreeable effects of armature reaction. How can both of these effects be corrected?

5. After standing idle for several weeks, an engine-driven DC compound generator is started up. It runs well but produces no voltage. Suggest reasonable causes and remedies.

6. If the efficiency of the generator is the only consideration, should the armature resistance be high or low? Should the shunt field resistance be high or low? Should the series field coil be high or low resistance?

7. Draw typical load–voltage characteristic curves for

a. A shunt generator

b. An undercompounded generator

c. An overcompounded generator

d. A series generator

8. Give three different reasons why the terminal voltage of a shunt generator decreases with an increase in load current.

9. A shunt generator is rated 125 volts, 25 kilowatts; armature resistance is 0.08 ohm; shunt field circuit resistance is 25 ohms. Determine

a. The induced emf in the armature at rated load

b. The watts loss in the armature

c. The watts loss in the shunt field circuit

d. The total power generated in the armature

10. A 10-kilowatt, 120-volt DC generator has an output of 120 volts at rated load.

With no load, a voltmeter across the output terminals reads 110 volts. Determine the voltage regulation. Is this machine a shunt, cumulative compound, or differential compound generator? How can you tell?

11. A 12-kilowatt, 240-volt, 1,500-revolutions-per-minute shunt generator has an armature resistance of 0.2 ohm and a shunt field resistance of 160 ohms. The stray power losses are 900 watts. Assuming shunt field current is constant, calculate

a. The efficiency at rated load

b. The efficiency at half-rated load

12. Explain how interpoles accomplish their purpose in a DC generator. State the polarity rule for interpoles.

13. A 10-kilowatt, 230-volt, long-shunt compound DC generator has efficiency 5 82%, armature resistance 5 0.15 ohm, series field 5 0.1 ohm, shunt field 5 100 ohms. At rated load, calculate

a. The armature current

b. The voltage across the brushes

c. The generated emf

d. The total copper losses

e. The horsepower of the prime mover

14. A separately excited 6-kilowatt generator has a terminal voltage of 135 volts at no load. At full load, the terminal voltage is 120 volts with speed and field excitation unchanged. Armature resistance 5 0.25 ohm. Find

a. The amount of voltage decrease caused by armature reaction

b. The voltage regulation

15. Complete the internal and external connections for the compound generator illustrated in the sketch below. This generator is to be connected as a cumulative compound long-shunt machine. The interpole field windings are to be a part of the armature circuit terminating at the connection points A1 and A2. Be sure the connections for all main field poles and interpoles are correct so that the proper polarities will be obtained.