Losses in DC Machines

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In DC machines (generator or motor), the losses may be classified into three categories namely,

• Copper losses
• Iron or core losses
• Mechanical losses

All these losses appear as heat and hence raise the temperature of the machine. They also reduce the efficiency of the machine.

Copper Losses

In dc machines, the losses that occur due to resistance of the various windings of the machine are called copper losses. The copper losses are also known as I²R losses because these losses occur due to current flowing through the resistance of the windings.

The major copper losses that occur in dc machines are as,

$$\mathrm{\mathrm{Armature\:copper\:loss}\:=\:\mathit{I_{a}^{\mathrm{2}}R_{a}}}$$

$$\mathrm{\mathrm{Series\:field\:copper\:loss}\:=\:\mathit{I_{se}^{\mathrm{2}}R_{se}}}$$

$$\mathrm{\mathrm{Shunt\:field\:copper\:loss}\:=\:\mathit{I_{sh}^{\mathrm{2}}R_{sh}}}$$

In dc machines, there is also a brush contact loss due to brush contact resistance. In practical calculation, this loss is generally included in armature copper loss.

Iron Losses

The iron losses occur in core of the armature of a DC machine due to rotation of the armature in the magnetic field. Because these losses occur in core of the armature, these are also called core losses.

There are two types iron or core losses namely hysteresis loss and eddy current loss.

Hysteresis Loss

The core loss that occurs in core of the armature of a dc machine due to magnetic field reversal in the armature core when it passes under the successive magnetic poles of different polarity is called hysteresis loss. The hysteresis loss is given by the following empirical formula,

$$\mathrm{\mathrm{Hysteresis\:loss,}\mathit{P_{h}}\:=\:\mathit{k_{h}B_{max}^{\mathrm{1.6}}fV}}$$

Where, $\mathit{k_{h}}$ is the Steinmetzs hysteresis coefficient, $\mathit{B_{max}}$ the maximum flux density,f is the frequency of magnetic reversal, and V is the volume of armature core.

The hysteresis loss in dc machines can be reduced by making the armature core of such materials that have a low value of Steinmetzs hysteresis coefficient like silicon steel.

Eddy Current Loss

When the armature of a DC machine rotates in the magnetic field of the poles, an EMF is induced in core of the armature which circulates eddy currents in it. The power loss due to these eddy currents is known as eddy current loss. The eddy current loss is given by,

$$\mathrm{\mathrm{Eddy\:current\:loss,}\mathit{P_{e}}\:=\:\mathit{k_{e}B_{max}^{\mathrm{2}}f^{\mathrm{2}}t^{\mathrm{2}}V}}$$

Where,$\mathit{K_{e}}$ is a constant of proportionality, and tis the thickness of lamination.

From the expression for eddy current loss it is clear that the eddy current loss depends upon the square of thickness of lamination. Therefore, to reduce this loss, the armature core is built up of thin laminations that are insulated from each other by a thin layer of varnish.

Mechanical Losses

The power losses due to friction and windage in a dc machine are known as mechanical losses. In a dc machine, the friction loss occurs in form of bearing friction, brush friction, etc. while the windage loss occurs due to air friction of rotating armature.

The mechanical losses depend upon the speed of the machine. But these losses are practically constant for a given speed.

Note− Iron or core losses and mechanical losses together are known as stray losses.

Constant and Variable Losses

In DC machines, we may group the above discussed losses in the following two categories −

• Constant Losses
• Variable Losses

Those losses in a DC machine that remain constant at all loads are called constant losses. These losses include − iron losses, shunt field copper loss, and mechanical losses.

Those losses in a DC machine that vary with load are known as variable losses. The variable losses in a DC machine are − armature copper loss and series field copper loss.

Total losses in a DC machine = Constant losses + Variable losses

Note − Iron losses and mechanical losses together are known as stray losses, i.e.

Strey Losses = Iron losses + Mechanical losses

Losses in a Transformer

The power losses in a transformer are of two types −

• Iron or Core Losses
• Copper Losses

Iron or Core Losses

The irons losses consist of hysteresis and eddy current losses and occur in the core of the transformer due to alternating flux. The iron losses of the transformer can be determined by the open-circuit test.

$$\mathrm{\text{Hysteresis Loss,} \:P_{h} \: = \: K_{h} \: B_{max}^{1.6} \:fV \: \text{ Watts }}$$

$$\mathrm{\text{Eddy Current Loss,} \:P_{e} \: = \: K_{e} \: B_{max}^{2} \:f^{2}t^{2}V \: \text{ Watts }}$$

Also,

$$\mathrm{\text{Iron or core Losses, } \:P_{i} \: = \: P_{h} \: + \: P_{e} \: = \: \text{ Constant Losses}}$$

The hysteresis losses can be minimised using silicon steel whereas the eddy current losses can be reduced using core made up of thin laminations.

Copper Losses

Copper losses occur in the primary and secondary windings of the transformer due to their resistance. These can be determined by short circuit test.

$$\mathrm{\text{Copper Losses, } \:P_{cu} \: = \: I_{1}^{2}R_{1} \: + \:I_{2}^{2}R_{2}}$$

Losses in Rotating AC Machines

The losses occur in rotating AC machines are also same as those are in DC machines. These losses can be classified into two categories as −

Fixed or Constant Losses

• Stator iron loss
• Friction and windage loss

Variable Losses

• Stator Copper Loss
• Rotor Copper Loss

Electric Machine Efficiency

The efficiency of an electric machine is defined as the ratio of the output power to the input power, i.e.

$$\mathrm{\text{Efficiency, } \: \eta \: = \: \frac{\text{Output Power }\:(P_{0})}{\text{Input Power}\:(P_{i})}}$$

$$\mathrm{\because\: \text{Input Power } \: = \: \text{Output Power } \: + \: \text{ Losses}}$$

$$\mathrm{\therefore \: \text{ Efficiency, } \: \eta \: = \: \frac{\text{Output Power}\:(P_{0})}{\text{Output Power }\:(P_{0}) \: + \: \text{ Losses}} \: = \: \left(1 \: + \: \frac{\text{Output Power }\:(p_{0})}{\text{ Losses}}\right)}$$

Numerical Example #1

The armature resistance of a compound long shunt DC motor is 0.0858 Ω. It has shunt and series field resistances of 60 Ω and 0.06 Ω respectively. The motor draws a total current of 100 A. If the shunt field current and series field current are 2 A, determine the total cu loss of the motor.

Solution

Armature Current,

$$\mathrm{I_{a} \: = \: I_{r} \: + \: I_{sh} \: = \: 100 \: + \: 2 \: = \: 102}$$

Therefore, Armature Cu Loss,

$$\mathrm{= \: I_{a}^{2}R_{a} \: = \: 102^{2} \: \times \: 0.0858 \: = \:892.66 \: W}$$

Series Field Cu Loss,

$$\mathrm{= \: I_{se}^{2}R_{se} \: = \: I_{a}^{2}R_{se} \: = \: 102^{2} \: \times \: 0.06 \: = \: 624.24 \: W}$$

Shunt Field Cu Loss,

$$\mathrm{= \: I_{sh}^{2}R_{sh} \: = \: 2^{2} \: \times \: 60 \: = \: 240 \: W}$$

∴ Total Cu Losses,

$$\mathrm{P_{cu} \: = \: I_{a}^{2}R_{a} \: + \: I_{se}^{2}R_{se} \: + \: I_{sh}^{2}R_{sh}}$$

⇒ Total Cu Losses,

$$\mathrm{P_{cu} \: = \: 892.66 \: + \: 624.24 \: + \: 240 \: = \: 1756.9 \: W}$$

Numerical Example #2

A power transformer has a core material for which hysteresis coefficient is 120 J/m³and eddy current-loss coefficient is 250. Its volume is 10000 cm³ and the maximum flux density is 1.18 Wb/m² . The core is built up of thin laminations of thickness 8 mm. What is the total iron/core loss in watts, if the frequency of the alternating current is 50 Hz?

Solution

The hysteresis power loss is given by,

$$\mathrm{P_{h} \: = \: K_{h}B_{max}^{1.6} \: fV}$$

$$\mathrm{= \: 120 \: \times \: (1.18)^{1.6} \: \times \: 50 \: \times \: 10000 \: \times \: 10^{-6}}$$

$$\mathrm{= \: 78.19 \: W}$$

And, the eddy current loss is given by,

$$\mathrm{P_{e} \: = \: K_{e}B_{max}^{2} \: f^{2}t^{2} \: V}$$

$$\mathrm{= \: 250 \: \times \: (1.18)^{2} \: \times \: 50^{2} \: \times \: (8 \: \times \: 10^{-3})^{2} \: \times \: 10000 \: \times \: 10^{-6}}$$

$$\mathrm{= \: 0.557 \: W}$$

Therefore,

Total core losses = $\mathrm{78.19 \: + \: 0.557 \: = \: 78.747 \: W}$

Numerical Example #3

In a 25 kVA transformer, the iron loss is 250 W and full load copper loss is 400 W. Find the efficiency at full load at 0.8 power factor lagging.

Solution

Full Load Output,

$$\mathrm{P_{0} \: = \: 25 \: \times \: 0.8 \: = \: 20 \: kW}$$

Total Full Load Losses,

$$\mathrm{= \: 250 \: + \: 400 \: = \: 650 \: W \: = \: 0.65 \: kW}$$

Full Load Input Power,

$$\mathrm{P_{i} \: = \: 20 \: + \: 0.65 \: = \: 20.65 \: kW}$$

Therefore, Full Load Efficiency,

$$\mathrm{\eta \: = \: \frac{P_{0}}{P_{i}} \: \times \: 100 \: = \: \frac{20}{20.65} \: \times \: 100 \: = \: 96.85\%}$$