Losses and Efficiency of 3-Phase Alternator

Losses in Three-Phase Alternator

The losses that occur in a three-phase alternator may be spanided into the following four categories −

Copper losses

Iron or core losses

Mechanical losses

Stray load losses

Read through this section to find out more about the types of losses that occur in a three-phase alternator.

Copper Losses

The copper losses occur in the armature winding and rotor winding of the alternator due to their resistance when currents flowing through them. Thus, these losses are also called I²R losses.

Iron or Core Losses

The iron or core losses occur in iron parts pke stator core and rotor core of the alternator. These losses consist of hysteresis loss and eddy current loss. The core losses occur because various iron parts of the alternator are subjected to the changing magnetic field.

Mechanical Losses

The mechanical losses occur in the moving or rotating parts pke rotor, shaft, bearings, etc. of the alternator. There are two main types of mechanical losses namely, frictional losses and windage losses. The frictional losses are due to friction of bearings in the alternator, while the windage losses are due to friction between the rotating parts of the alternator and the air inside the casing of the alternator.

Stray Load Losses

This category includes those losses in the alternator which cannot be easily determined. These losses are also called miscellaneous losses. The stray load losses may be because of the following reasons −

Distortion of main field flux due to armature reaction.

Non-uniform distribution of current over the area of cross-section of armature conductors.

In practical calculations, we take the stray load losses 1% of the full-load losses.

Note

The iron losses and mechanical losses together are called rotational losses because these losses occur in the alternator due to rotation of the rotor.

All these losses that occur in an alternator are converted into heat and result in the increase of the temperature and decrease in the efficiency of the alternator.

Efficiency of Three-Phase Alternator

The ratio of output power to input power of an alternator is called efficiency of the alternator. The efficiency is usually expressed in percentage.

$mathrm{mathrm{Efficiency,} : eta :=:frac{Output:Power}{Input:Power} imes 100\%:=:frac{Output:Power}{Output:Power+Losses} imes 100\%}$

Now, we will derive the expression for efficiency of a three-phase alternator. For that consider a three-phase alternator operating at a lagging power factor.

Let,

V= terminal voltage per phase

I_a = armature current per phase

cos $phi$ = load power factor (lagging)

Thus, the output power of a three-phase alternator is given by,

$$mathrm{mathit{P_{0}}:=:3:mathit{VI_{a}cos:phi }}$$

The losses in the alternator are,

$$mathrm{mathrm{Armature: copper: loss,}mathit{P_{cu}}:=:3:mathit{I_{mathit{a}}^{mathrm{2}}R_{a}}}$$

$$mathrm{mathrm{Field :winding: copper: loss}:=:mathit{V_{f}I_{f}}}$$

Where,V_f is the DC voltage across field winding and I_f is the DC field current.

$$mathrm{mathrm{Rotational: losses,}mathit{P_{r}}:=:mathrm{Core:losses:+:Mechanical:losses}}$$

$$mathrm{mathrm{Stray:load:losses}mathit{P_{s}}}$$

$$mathrm{ herefore mathrm{Total:losses:in:alternator,}mathit{P_{loss}}:=:3:mathit{I_{a}^{2}R_{a}:+:P_{r}:+:P_{s}:+:V_{f}I_{f}}}$$

Since, the speed of rotation of the rotor is constant so the rotational losses are constant. The field winding copper losses are also constant. If we assume stray-load losses to be constant. Then, we have,

$$mathrm{mathrm{Total:constant:losses,}mathit{P_{c}}:=:mathit{P_{r}:+:P_{s}:+:V_{f}I_{f}}}$$

$$mathrm{ herefore:mathrm{Variable:losses} :=:mathrm{3}mathit{I_{a}^{mathrm{2}}R_{a}}}$$

Hence, the efficiency of the alternator is given by,

$$mathrm{eta :=:frac{mathit{P_{0}}}{mathit{P_{0}+mathrm{Losses}}}:=:frac{3mathit{VI_{a}cosphi }}{3mathit{VI_{a}cosphi :+:mathrm{3}mathit{I_{a}^{mathrm{2}}R_{a}}:+P_{c}}}cdot cdot cdot (1)}$$

Equation-1 can be used to determine the efficiency of a three-phase alternator.

Condition for Maximum Efficiency

The efficiency of an alternator will be maximum when the variable losses are equal to the constant losses, i.e.,

$$mathrm{mathit{P_{c}}:=:3:mathit{I_{a}^{mathrm{2}}R_{a}}cdot cdot cdot (2)}$$

In practice, the maximum efficiency of an alternator usually occurs at about 85% of the full rated load.

Numerical Example

A three-phase alternator has a per phase terminal voltage of 230 V, and the per phase armature current of 14.4 A. The resistance of the armature circuit of the alternator is 0.5Ω, and the constant losses are 200 watts. Calculate the efficiency and maximum efficiency of the alternator if it is supppng a load at 0.8 lagging power factor.

Solution

$$mathrm{mathrm{Efficiency,}eta :=:frac{3mathit{VI_{a}cosphi }}{3mathit{VI_{a}cosphi :+:mathrm{3}mathit{I_{a}^{mathrm{2}}R_{a}}:+P_{c}}}}$$

$$mathrm{Rightarrow:eta :=:frac{3 imes 230 imes 14.4 imes 0.8}{left ( 3 imes 230 imes 14.4 imes 0.8 ight ):+:left ( 3 imes 14.4^{2} imes 0.5 ight ):+:200}}$$

$$mathrm{ herefore eta :=:0.9395:=:93.95\%}$$

For maximum efficiency of the alternator,

$$mathrm{mathit{P_{c}}:=:3:mathit{I_{a}^{mathrm{2}}R_{a}}}$$

$$mathrm{ herefore eta_{max} :=:frac{3mathit{VI_{a}cosphi }}{3mathit{VI_{a}cosphi :+:mathrm{2}mathit{P_{c}}}}:=:frac{3 imes 230 imes 14.4 imes 0.8}{left ( 3 imes 230 imes 14.4 imes 0.8 ight ):+:left (2 imes 200 ight )}}$$

$$mathrm{ herefore eta_{max} :=:0.9521:=:95.21\%}$$