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Copyright: © 2015
|Pages: 26

DOI: 10.4018/978-1-4666-8441-6.ch008

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TopThe losses in rotating electric machines may be classified as

*1.*Electric losses

*2.*Rotational losses

The electric losses in d.c. machines (motor as well generator) include I^{2} R losses in the field circuits and the armature circuits, while rotational losses include windage, friction plus brush friction losses and stray load losses.

The stray load losses result from the load and cannot be measured directly. These include eddy current and hysteresis losses.

The rotational losses are also called no-load losses and can be determined by no-load tests.

Efficiency is defined as the ratio of the useful output to the input, and the expression preferred for calculations of efficiency as:

If the losses are know for a given output, the input is the sum of the losses and the output and the efficiency can be calculated to a high degree of accuracy even if there are small errors in the losses. The brush contact loss is usually taken as 2Ia. An example will illustrate the calculations of efficiency.

The following data apply to a 100 kw, 250V, 6-pole 900 rpm shunt compound generator. No-load rotational losses = 3840 W. Armature resistance 0.012 Ω; Series field resistance = 0.004 Ω shunt field current = 2.60 A.

Assume a stray load loss of 1% of the output and calculate rated load efficiency.

Since the generator is long shunt compound hence the total resistance (not including brushes) is the sum of the resistance of the armature, series field and commentating pole field windings.

The armature current is the sum of the load current and the shunt field current or

The losses are as follows.

No. load rotational losses = 3840 Armature circuit copper loss = (402.6)The input at rated is therefore, 100,000 + 9535 = 109,535 watts and the efficiency is

A 10 kw, 230 V, 1750 rpm shunt generator was run light as a motor to determine its rotational losses at its rated load. The applied voltage was 245 V and the armature current drawn was 2A. The field resistance of the generator was 230Ω and the armature circuit resistance was 0.2 Ω. Calculate

*1.*The rotational losses at full load.

*2.*The full load armature circuit loss and the field loss.

*3.*The generator efficiency at ½ and rated load.

Therefore loss

.The field loss =

* Note that copper/electric losses are proportional square of armature current hence if the load decreases by ½ the loss will decrease by ¼.

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