Efficiency, rating, and applications of dynamos

EFFICIENCY, RATING, AND APPLICATIONS OF DYNAMOSBY:

MA. JANINE LOSABIO

JEMAR PALERMO

CAMILLE ALEXIS ROXAS

POWER LOSSES IN DYNAMOS:

A dynamo is a machine that converts energy from one form to another. In the generator, it is from mechanical energy to electrical energy; in the motor, it is from electrical energy to mechanical energy. When this conversion takes place at a uniform rate, that is, when the energy received by the machine per unit time and the energy delivered by the machine in the same unit of time are constant, then, it is proper to say that a dynamo converts power from one form to another.

The power received by a dynamo is called the input; in a generator it is mechanical power, while in a motor it is electrical power. The power delivered by a dynamo is called its output; in a generator, it is electrical power, while in a motor it is mechanical power. Although mechanical power is usually expressed in horsepower ( 1 hp = 33 00 ft-lb of energy per min ), it is customary to change the mechanical unit to its equivalent electrical unit of watts when making calculations involving power losses and efficiency ( 1 hp = 746 watts ).

The power input to a dynamo is always more than its power output; that is, a generator or motor cannot convert all the power it receives into useful output. This implies, that some of the power input is used to perform functions that do not show up as power output.

The difference between the power input to machine and its power output is called power loss because it is unavailable to drive a mechanical load in a motor or to supply electrical power in a generator.

This power loss always produces heating in the dynamo; therefore, the greater the power loss, as a percentage of the power input, the hotter will the machine tend to become. If this loss should reach an excessive value, the temperature rise might be high enough to cause failure.

TWO GENERAL CLASSIFICATIONS OF POWER LOSSES IN ELECTRIC

MACHINES

1. Those that are caused by the rotation of the armature (rotational losses)

2. Those that result from a current flow in the various parts of the machine (electrical losses)

Rotational losses will vary only if the speed changes, while the electrical losses are affected by the current values through the various electrical elements such as the armature winding, the fields, and the brush contacts.

ROTATIONAL LOSSES MAY BE DIVIDED INTO FIVE PARTS:

1. bearing friction2. brush friction3. wind friction (windage)4. hysteresis5. eddy currents

HYSTERESIS LOSS Takes place in the revolving armature core because the magnetic

polarity in the iron changes in step with the changing positions of the magnetic material under the various poles.

This hysteresis loss is magnetic in character but results only because the armature core is turning; it may therefore be properly classified as rotational loss.

In generators, whose speeds are usually constant, and in shunt motors that operate at substantially constant speed, the hysteresis loss is fairly constant because of the flux density in the core iron changes very little with the load.

In series and compound motors, the hysteresis loss does vary somewhat with the changes in load because both the speed and flux density are affected by the load.

EDDY CURRENT This result because the generated voltages in

the iron near the outside surface are greater than those closer to the center of the shaft because of the higher speed; the difference in potential then causes the currents to flow in the iron. Since the eddy currents have paths mostly parallel to the shaft, the logical way to minimize their magnitude is to introduce high resistances in the form of air spaces in direct line to such paths.

Eddy currents are electromagnetic in character but result only because of the rotation of the armature core; these, too, may therefore be regarded primarily as a rotational loss. Its value in watts depends upon the core flux density, the speed of rotation, and the thickness of the laminations; it is independent of the quality of the magnetic iron.

The eddy current loss is affected in much the same way by the load as is hysteresis; if the speed and flux densities remain constant, the eddy current loss does not change; if they vary, as they must in series and compound motors, the eddy current loss will change.

COPPER LOSS Always occur when there is current flow through the

various copper circuits. One of them takes place in the armature winding and is equal to IA

2RA. There will be another loss at the brush contacts between the copper commutator and the carbon brushes; this loss depends upon the brush contact voltage drop and the armature current IA. In low voltage machines, (115 to 230 volts) the brush drop EB varies between 1 and 3 volts, while in the higher voltage machines (550 to 1100 volts) EB may be as much as 6 volts. From this it may be seen that the armature resistance copper loss is approximately proportional to the square of the load, while the brush contact loss is nearly directly proportional to the load.

VARIOUS FIELD COPPER LOSSES

1. shunt field2. series field3. interpole field4. compensating – winding field

FACTORS THAT RESULT STRAY LOAD LOSS

1. the distortion of the flux because of armature reaction

2. lack of uniform division of the current in the armature winding through the various paths and through individual conductors of large cross sectional area

3. short circuit currents in the coils undergoing commutation

EFFICIENCY OF DIRECT CURRENT GENERATORS

The efficiency of a dc generator is the ratio of the electrical power output ET X IL to the mechanical power input, converted to watts.

Eq. (20)

Per cent efficiency = watts output x 100 watts input

Since watts input = watts output + watts losses,Per cent efficiency = watts output

x 100 watts output + watts losses

Eq. (21)

Another way to write the equation for more accurate calculations when slide rule is used is

Eq. (22)

Per cent efficiency = 1 – watts losses x 100 watts output + watts losses

EXAMPLE 1: A 5 KW generator has a total loss of 700 watts when operating

at full load. Calculate the per cent efficiency.

Per cent efficiency = watts output x 100 watts input = 5000 watts x 100 5000 watts + 700 watts = 87.7 %

EXAMPLE 2. A 50 KW generator has full load efficiency of 90.5 per

cent. Determine the total losses.

90.5 = 50, 000 x 100 50, 000 + losses Losses = 50, 000 x 100 - 50, 000 90.5 Losses = 5250 watts

TWO METHODS OF DETERMINING THE EFFICIENCY OF A GENERATOR:

1. By directly measuring the total power output and the total power input

2. By making certain necessary tests from which the various power losses are determined

A calibrated motor is one the outputs of which are known for all values of input (the calibrated motor output equals generator input). Ordinarily, direct efficiency tests are both difficult to perform, especially upon large machines, and somewhat inaccurate. There is a considerable waste of power when actual loads must be applied. For these reasons, it is usually customary to determine the efficiency by the so called conventional method; by which the various power losses are determined by calculation.

To illustrate why the conventional efficiency determination is more accurate than one measured directly, assume an efficiency of 90 per cent. Under this condition, the losses will be about 11 per cent of the output [losses = (output/efficiency) – output]. Thus an error of even as much as 5 per cent in calculating the losses will result in an error in the overall efficiency of less than 0.6 per cent. An error of 5 per cent in making measurements by the direct method would yield worthless results.

EFFICIENCY OF DIRECT CURRENT MOTORS

The efficiency of a dc motor is the ratio of the mechanical power output, converted to watts, to the electrical power input. As percentage, this statement may be written in equation form

Per cent efficiency = hp output x 746 x 100 eq. (23)

watts input

Since watts input = (hp input x 746) + watts losses, eq. (24)

Per cent efficiency = 1- hp output x 746 x 100 eq. 25

(hp output x 746) + watts losses

In making the test for the rotational loss, the impressed voltage across the armature must equal the terminal emf minus the brush contact and armature resistance drops at full load.

EA = (ET – EB) – IARA

IMPORTANCE OF EFFICIENCY A dynamo that operates at a comparatively

high efficiency losses little power; on the other hand, one that operates at a low efficiency losses much power. Since all power losses are converted into heat, it follows that the temperature rise of a machine is affected very definitely by the efficiency. If the temperature to which a generator or motor rises exceeds well-established practices, it tends to cause insulation failure and eventual breakdown.

If a machine has a low efficiency and the temperature rise tends to become excessive, it usually becomes necessary to use cooling fans mounted on the shaft; such fans require additional power, so that the efficiency becomes still lower.

Another factor of importance is the cost of operation. The energy cost of the losses in a dynamo, as a percentage of the total operating cost, goes up as the efficiency becomes less. Therefore, in a machine of a given output, the energy charge increases as the efficiency drops because the operator pays for input, which is the sum of the output and the losses.

RATING OF GENERATORS AND MOTORS

Generators are rated in terms of kilowatt output at a given speed and voltage. motors are rated in terms of horsepower output at a certain speed and voltage.

In the operation of any dc generator or motor, it is well to remember that there are two danger signals: temperature rise and commutator sparking.

SELECTION OF GENERATORS AND

MOTORSWhen a dc generator or motor is selected

for a given application, there are many great factors that must be borne in mind if the proper choice is to be made.

1. What should be the rating in kilowatts or horsepower, voltage, and speed?

2. The type of machine whether shunt, series, or compound.

For a generator, it is also necessary to know such things as the degree of compounding, the way it is to be mounted, the kind of prime mover it will have, where it will be located, the type of control it will require, whether or not it will be paralleled with other machines, the general service conditions, and whether it is belted, geared, or coupled.

In this case of a motor, selection should be made on the basis of such factors as the following: starting torque requirements, overload possibilities, type of service (such as continuous duty or intermittent duty), the regulation, whether or not the motor will be reversed, type of speed control, how it is to be mounted (i.e., on the floor, side wall, or ceiling, horizontal or vertical), the surrounding conditions (whether damp or hot), the altitude, and the kind of driver (i.e., belt, coupling, gear, etc.)

An improper installation may mean poor service, inefficiency, high cost for energy and fixed charges, early breakdown, and much loss of time, as well as other more or less serious matters.