- Electrical Machines Tutorial
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- Basic Concepts
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- Energy Stored in a Magnetic Field
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- Rotating Electrical Machines
- Faraday’s Laws of Electromagnetic Induction
- Concept of Induced EMF
- Fleming’s Left Hand and Right Hand Rules
- Transformers
- Electrical Transformer
- Construction of Transformer
- EMF Equation of Transformer
- Turns Ratio and Voltage Transformation Ratio
- Ideal and Practical Transformers
- Transformer on DC
- Losses in a Transformer
- Efficiency of Transformer
- Three-Phase Transformer
- Types of Transformers
- DC Machines
- Construction of DC Machines
- Types of DC Machines
- Working Principle of DC Generator
- EMF Equation of DC Generator
- Types of DC Generators
- Working Principle of DC Motor
- Back EMF in DC Motor
- Types of DC Motors
- Losses in DC Machines
- Applications of DC Machines
- Induction Motors
- Introduction to Induction Motor
- Single-Phase Induction Motor
- Three-Phase Induction Motor
- Construction of Three-Phase Induction Motor
- Three-Phase Induction Motor on Load
- Characteristics of 3-Phase Induction Motor
- Speed Regulation and Speed Control
- Methods of Starting 3-Phase Induction Motors
- Synchronous Machines
- Introduction to 3-Phase Synchronous Machines
- Construction of Synchronous Machine
- Working of 3-Phase Alternator
- Armature Reaction in Synchronous Machines
- Output Power of 3-Phase Alternator
- Losses and Efficiency of 3-Phase Alternator
- Working of 3-Phase Synchronous Motor
- Equivalent Circuit and Power Factor of Synchronous Motor
- Power Developed by Synchronous Motor
- Electrical Machines Resources
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- Electrical Machines - Discussion
Output Power of 3-Phase Alternator
Consider a three-phase alternator with a cylindrical rotor and is operating at a lagging power factor.
Let,
E = per phase induced EMF
V = per phase terminal voltage
Ia = per phase armature current
cos$\phi$ = power factor (lagging) of load
$\delta$ = power angle (angle between E and V)
Therefore, the output power of a three-phase alternator is given by,
$$\mathrm{\mathit{P_{0}}\:=\:3\mathit{VI_{a}cos\phi }\cdot \cdot \cdot (1)}$$
Approximate Output Power of 3-Phase Alternator
In a three-phase alternator, the resistance $R_{a}$ of the armature circuit is very small as compared to the synchronous reactance $X_{s}$ of the machine. Thus, we can neglect the armature resistance ($R_{a}$), after that we get approximate equivalent circuit of the alternator as shown in Figure-1. The phasor diagram of the circuit is also shown in Figure-1.
From the phasor diagram, we get,
$$\mathrm{\mathit{AB}\:=\:\mathit{I_{a}X_{s}cos\phi }\:=\:\mathit{E}\:\mathrm{sin\delta }}$$
$$\mathrm{\Rightarrow \mathit{I_{a}cos\phi }\:=\:\frac{\mathit{E\:\mathrm{sin\delta }}}{\mathit{X_{s}}}\cdot \cdot \cdot (2)}$$
Now, from equation (1) & (2), we get,
$$\mathrm{\mathit{P_{0}}\:=\:\frac{3\mathit{EV\:\mathrm{sin\delta }}}{\mathit{X_{s}}}\cdot \cdot \cdot (3)}$$
The expression in Equation-3 gives the approximate output power of a three-phase alternator.
When the alternator is operating at a constant speed with a constant field current, then Xs and E both are constant and therefore the terminal voltage V is also constant. Thus, from Equation-3, we can observe,
$$\mathrm{\mathit{P_{0}}\propto \:\mathrm{sin\delta }}$$
We know, when $\delta$ = 90°, then
$$\mathrm{\mathrm{sin\:90^{\circ}}\:=\:1}$$
Thus, the alternator supplies maximum power at $$ =90°, and it is given by,
$$\mathrm{\mathit{P_{max}}\:=\:\frac{3\mathit{EV}}{\mathit{X_{s}}}}\cdot \cdot \cdot (4)$$
The maximum power given by Equation-4 is called the static stability limit of the alternator.
Numerical Example
A 3-phase, 11 kV, 3 MVA star-connected alternator has a per phase synchronous reactance of 10 $\Omega$. Its excitation is such that the generated line EMF is 15 kV. When the alternator is connected to infinite busbars. Calculate the maximum output power of the alternator at the given excitation when armature resistance is neglected.
Solution
Given data,
$$\mathrm{\mathrm{Line\:voltage,}\mathit{V_{L}}\:=\:11\:kV\:=\:11000\:V}$$
$$\mathrm{\therefore\:\mathrm{Terminal\:voltage\:per\:phase,}\mathit{V}\:=\:\frac{11000}{\sqrt{3}}\:=\:6350.85\:V}$$
$$\mathrm{\mathrm{Generated\:line\:EMF}\:=\:15\:kV\:=\:15000\:V}$$
$$\mathrm{\therefore\:\mathrm{Generated\:EMF\:per\:phase,}\mathit{E}\:=\:\frac{15000}{\sqrt{3}}\:=\:8660.25\:V}$$
$$\mathrm{\mathrm{Synchronous\: reactance\: per\: phase,}\:\mathit{X_{s}}\:=\:10\:\Omega }$$
Therefore, the maximum power output of the alternator will be,
$$\mathrm{\mathit{P_{max}}\:=\:\frac{3\mathit{EV}}{\mathit{X_{s}}}\:=\:\frac{3\times 8660.25\times 6350.85}{10}}$$
$$\mathrm{\therefore \mathit{P_{max}}\:=\:16499\times 10^{3}W\:=\:16499\:\mathrm{kW}}$$