
- Electrical Machines - Home
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- DC Machines
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- Induction Motors
- Introduction to Induction Motor
- Single-Phase Induction Motor
- 3-Phase Induction Motor
- Construction of 3-Phase Induction Motor
- 3-Phase Induction Motor on Load
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- 3-Phase Induction Motor Fault Types
- Synchronous Machines
- Introduction to 3-Phase Synchronous Machines
- Construction of Synchronous Machine
- Working of 3-Phase Alternator
- Armature Reaction in Synchronous Machines
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- Losses and Efficiency of an Alternator
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- Output Power of Synchronous Motor
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- V Curves & Inverted V Curves of Synchronous Motor
- Torque in Synchronous Motor
- Construction of 3-Phase Synchronous Motor
- Synchronous Motor
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- Power Flow in Synchronous Motor
- Types of Faults in Alternator
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- Solid State Motor Starters
- Characteristics of Single-Phase Motor
- Types of AC Generators
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- Distribution Factor
- Electrical Machines Basic Terms
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- Stator and Rotor in Electrical Machines
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- Discussion
Losses in a Transformer
The following power losses may occur in a practical transformer −
Iron Loss or Core Loss
Copper Loss or I2R Loss
Stray Loss
Dielectric Loss
In a transformer, these power losses appear in the form of heat and cause two major problems −
Increases the temperature of the transformer.
Reduces the efficiency of the transformer.
Iron Loss or Core Loss
Iron loss occurs in the magnetic core of the transformer due to flow of alternating magnetic flux through it. For this reason, the iron loss is also called core loss. We generally use the symbol ($\mathit{P_{i}}$) to represent the iron loss. The iron loss consists of hysteresis loss ($\mathit{P_{h}}$) and eddy current loss ($\mathit{P_{e}}$). Thus, the iron loss is given by the sum of the hysteresis loss and eddy current loss, i.e.
$$\mathrm{\mathrm{Iron\:loss,}\mathit{P_{i}}\:=\:\mathrm{Hysteresies\:loss(\mathit{P_{h}})}\:+\:\mathrm{Eddy\:current\:loss(\mathit{P_{e}})}}$$
The hysteresis loss and eddy current loss (or iron loss) are determined by performing the open-circuit test on the transformer.
The empirical formulae for the hysteresis loss and eddy current loss are given by,
$$\mathrm{\mathit{P_{h}}\:=\:\mathit{k_{h}f\:B_{m}^{x}}\:\cdot \cdot \cdot (1)}$$
$$\mathrm{\mathit{P_{e}}\:=\:\mathit{ke\:B_{m}^{\mathrm{2}}\:f^{\mathrm{2}}t^{\mathrm{2}}}\:\cdot \cdot \cdot (2)}$$
Where,
The exponent of Bm, i.e. "x" is called the Steinmetzs constant. Depending on the properties of the core material, its value is ranging from 1.5 to 2.5.
kh is a proportionality constant whose value depends upon the volume and quality of the material of core.
ke is a proportionality constant which depend on the volume and resistivity of material of the core.
f is the frequency of the alternating flux in the core.
Bm is the maximum flux density in the core.
t is the thickness of each core lamination.
Therefore, the total iron loss or core loss can also be written as,
$$\mathrm{\mathit{P_{i}}\:=\:\mathit{k_{h}f\:B_{m}^{x}}\:+\:\mathit{ke\:B_{m}^{\mathrm{2}}\:f^{\mathrm{2}}t^{\mathrm{2}}}\:\cdot \cdot \cdot (3)}$$
Since the input voltage to the transformer is approximately equal to the induced voltage in the primary winding, i.e.
$$\mathrm{\mathit{V_{\mathrm{1}}}\:=\:\mathit{E_{\mathrm{1}}}\:=\:4.44\:\mathit{f\phi _{m}N_{\mathrm{1}}}}$$
$$\mathrm{\Rightarrow \mathit{V_{\mathrm{1}}}\:=\:4.44\:\mathit{f\:B_{m}AN_{\mathrm{1}}}}$$
Where, A is the cross-sectional area of the transformer core, N1 is the number of turns in the primary winding and f is the supply frequency.
$$\mathrm{\therefore \mathit{B_{m}}\:=\:\frac{\mathit{V_{\mathrm{1}}}}{4.44\mathit{fAN_{\mathrm{1}}}}\:\cdot \cdot \cdot (4)}$$
Hence, from equations (1) & (4), we get,
$$\mathrm{\mathit{P_{h}}\:=\:\mathit{k_{h}f}\left ( \frac{\mathit{V_{\mathrm{1}}}}{4.44\mathit{fAN_{\mathrm{1}}}} \right )^{x}}$$
$$\mathrm{\Rightarrow \mathit{P_{h}}\:=\:\mathit{k_{h}f}\left ( \frac{\mathrm{1}}{4.44\mathit{AN_{\mathrm{1}}}} \right )^{x}\cdot \left ( \frac{\mathit{V_{\mathrm{1}}}}{\mathit{f}} \right )^{x}}$$
$$\mathrm{\Rightarrow \mathit{P_{h}}\:=\:\mathit{k_{h}}\left ( \frac{\mathrm{1}}{4.44\mathit{AN_{\mathrm{1}}}} \right )^{x}\cdot \mathit{V_{\mathrm{1}}^{x}}\:\mathit{f^{(\mathrm{1}-x)}}\:\cdot \cdot \cdot (5)}$$
Thus, Equation (5) shows that the hysteresis loss depends upon both input voltage and supply frequency.
Again, from equations (2) & (4), we get,
$$\mathrm{\mathit{P_{e}}\:=\:\mathit{k_{e}f^{\mathrm{2}}t^{\mathrm{2}}}\left ( \frac{\mathit{V_{\mathrm{1}}}}{4.44\mathit{fAN_{\mathrm{1}}}} \right )^{\mathrm{2}}}$$
$$\mathrm{\Rightarrow \mathit{P_{e}}\:=\:\mathit{k_{e}\left ( \frac{\mathit{V_{\mathrm{1}}}}{\mathrm{4.44}\mathit{AN_{\mathrm{1}}}} \right )^{\mathrm{2}}\mathit{t^{\mathrm{2}}}\:\cdot \cdot \cdot \mathrm{(6)}}}$$
Hence, from equation (6), we can conclude that the eddy current loss in the transformer is proportional to the square of the input voltage and is independent of the supply frequency.
Therefore, the total core loss can also be written as,
$$\mathrm{\mathit{P_{i}}\:=\:\mathit{k_{h}\left ( \frac{\mathrm{1}}{\mathrm{4.44}\mathit{AN_{\mathrm{1}}}} \right )^{\mathrm{2}}\cdot \mathit{V_{\mathrm{1}}^{\mathit{x}}f^{(\mathrm{1-x})}}\:+\:\mathit{k_{e}}\left ( \frac{V_{\mathrm{1}}}{\mathrm{4.44}\mathit{AN_{\mathrm{1}}}} \right )^{\mathrm{2}}\mathit{t^{\mathrm{2}}}\:\cdot \cdot \cdot \left ( \mathrm{7} \right )}}$$
In practice, transformers are connected to an electric supply of constant frequency and constant voltage, thus, both f and Bm are constant. Therefore, the core or iron loss is practically remains constant at all loads.
We can reduce the hysteresis loss by using steel of high silicon content to construct the core of transformer while the eddy current loss can be minimized by using core of thin laminations instead of solid core. The open-circuit test is performed on a transformer to determine the iron or core loss.
Copper Loss or I2R Loss
Power loss in a transformer that occurs in both the primary and secondary windings due to their Ohmic resistance is called copper loss or I2R loss. We usually represent the copper loss by PC. Therefore, the total copper loss in a transformer is the sum of power loss in the primary winding and power loss in the secondary winding, i.e.,
$$\mathrm{\mathit{P_{c}}\:=\:\mathrm{Copper\:loss\:in\:primary\:+\:Copper\:loss\:in\:secondary}}$$
$$\mathrm{\Rightarrow \mathit{P_{c}}\:=\:\mathit{I_{\mathrm{1}}^{\mathrm{2}}}\mathit{R_{\mathrm{1}}}\:+\:\mathit{I_{\mathrm{2}}^{\mathrm{2}}}\mathit{R_{\mathrm{2}}}\:\cdot \cdot \cdot (8)}$$
Since,
$$\mathrm{\mathit{I_{\mathrm{1}}}\mathit{N_{\mathrm{1}}}\:=\:\mathit{I_{\mathrm{2}}}\mathit{N_{\mathrm{2}}}}$$
$$\mathrm{\Rightarrow \mathit{I_{\mathrm{1}}}\:=\:\left ( \frac{\mathit{N_{\mathrm{2}}}}{\mathit{N_{\mathrm{1}}}} \right )\mathit{I_{\mathrm{2}}}\:\cdot \cdot \cdot (9)}$$
$$\mathrm{\therefore \mathit{P_{c}}\:=\:\left [ \left ( \frac{\mathit{N_{\mathrm{2}}}}{\mathit{N_{\mathrm{1}}}} \right )I_{\mathrm{2}} \right ]^{\mathrm{2}}\:\mathit{R_{\mathrm{1}}}\:+\:\mathit{I_{\mathrm{2}}^{\mathrm{2}}}\mathit{R_{\mathrm{2}}}\:=\:\left [ \left ( \frac{\mathit{N_{\mathrm{2}}}}{\mathit{N_{\mathrm{1}}}} \right )^{\mathrm{2}}\mathit{R_{\mathrm{1}}}\:+\:\mathit{R_{\mathrm{2}}} \right ]\mathit{I_{\mathrm{2}}^{\mathrm{2}}}\:\cdot \cdot \cdot (10)}$$
From Equation (10), it is clear that the copper loss in a transformer varies as the square of the load current. For this reason, the copper loss is also referred as "variable loss" because in practice a transformer is subjected to variable load and hence has variable load current.
We conduct the "short-circuit test" on the transformer to determine the value of its copper loss. In a practical transformer, the copper loss accounts for about 90% of the total power loss in the transformer.
Stray Loss
In practical transformer, a fraction of the total flux follows a path through air and this flux is called leakage flux. This leakage flux produces eddy currents in the conducting or metallic parts like tank of the transformer. These eddy currents cause power loss, which is known as stray loss.
Dielectric Loss
The power loss occurs in insulating materials like oil, solid insulation of the transformer, etc. is known as dielectric loss. The dielectric loss is significant only in transformers working on high voltages.
Although, in practice, the stray loss and dielectric loss are very small, constant and may be neglected.
From the above discussion, we found that a transformer has some losses which are constant and some other are variable. Thus, we may categorize losses in a transformer in two types namely constant losses and variable losses.
Therefore, the total losses in a transformer are the sum of constant losses and variable losses, i.e.,
Total losses in transformer = Constant losses + Variable losses