- Electrical Machines - Discussion
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- Power Developed by Synchronous Motor
- Equivalent Circuit and Power Factor of Synchronous Motor
- Working of 3-Phase Synchronous Motor
- Losses and Efficiency of 3-Phase Alternator
- Output Power of 3-Phase Alternator
- Armature Reaction in Synchronous Machines
- Working of 3-Phase Alternator
- Construction of Synchronous Machine
- Introduction to 3-Phase Synchronous Machines
- Methods of Starting 3-Phase Induction Motors
- Speed Regulation and Speed Control
- Characteristics of 3-Phase Induction Motor
- Three-Phase Induction Motor on Load
- Construction of Three-Phase Induction Motor
- Three-Phase Induction Motor
- Single-Phase Induction Motor
- Introduction to Induction Motor
- Applications of DC Machines
- Losses in DC Machines
- Types of DC Motors
- Back EMF in DC Motor
- Working Principle of DC Motor
- Types of DC Generators
- EMF Equation of DC Generator
- Working Principle of DC Generator
- Types of DC Machines
- Construction of DC Machines
- Types of Transformers
- Three-Phase Transformer
- Efficiency of Transformer
- Losses in a Transformer
- Transformer on DC
- Ideal and Practical Transformers
- Turns Ratio and Voltage Transformation Ratio
- EMF Equation of Transformer
- Construction of Transformer
- Electrical Transformer
- Fleming’s Left Hand and Right Hand Rules
- Concept of Induced EMF
- Faraday’s Laws of Electromagnetic Induction
- Rotating Electrical Machines
- Singly-Excited and Doubly Excited Systems
- Energy Stored in a Magnetic Field
- Electromechanical Energy Conversion
- Electrical Machines - Home
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Output Power of 3-Phase Alternator
Consider a three-phase alternator with a cypndrical 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,
$$mathrm{mathit{P_{0}}:=:3mathit{VI_{a}cosphi }cdot cdot cdot (1)}$$
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.
![Phasor](/electrical_machines/images/phasor.jpg)
From the phasor diagram, we get,
$$mathrm{mathit{AB}:=:mathit{I_{a}X_{s}cosphi }:=:mathit{E}:mathrm{sindelta }}$$
$$mathrm{Rightarrow mathit{I_{a}cosphi }:=:frac{mathit{E:mathrm{sindelta }}}{mathit{X_{s}}}cdot cdot cdot (2)}$$
Now, from equation (1) & (2), we get,
$$mathrm{mathit{P_{0}}:=:frac{3mathit{EV:mathrm{sindelta }}}{mathit{X_{s}}}cdot cdot cdot (3)}$$
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,
$$mathrm{mathit{P_{0}}propto :mathrm{sindelta }}$$
We know, when $delta$ = 90°, then
$$mathrm{mathrm{sin:90^{circ}}:=:1}$$
Thus, the alternator supppes maximum power at $$ =90°, and it is given by,
$$mathrm{mathit{P_{max}}:=:frac{3mathit{EV}}{mathit{X_{s}}}}cdot cdot cdot (4)$$
The maximum power given by Equation-4 is called the static stabipty pmit of the alternator.
Numerical Example
A 3-phase, 11 kV, 3 MVA star-connected alternator has a per phase synchronous reactance of 10 $Omega$. Its excitation is such that the generated pne EMF is 15 kV. When the alternator is connected to infinite busbars. Calculate the maximum output power of the alternator at the given excitation when armature resistance is neglected.
Solution
Given data,
$$mathrm{mathrm{Line:voltage,}mathit{V_{L}}:=:11:kV:=:11000:V}$$
$$mathrm{ herefore:mathrm{Terminal:voltage:per:phase,}mathit{V}:=:frac{11000}{sqrt{3}}:=:6350.85:V}$$
$$mathrm{mathrm{Generated:pne:EMF}:=:15:kV:=:15000:V}$$
$$mathrm{ herefore: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{3mathit{EV}}{mathit{X_{s}}}:=:frac{3 imes 8660.25 imes 6350.85}{10}}$$
$$mathrm{ herefore mathit{P_{max}}:=:16499 imes 10^{3}W:=:16499:mathrm{kW}}$$
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