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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
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Power Developed by Synchronous Motor
In this chapter, we will derive the expression for mechanical power developed (Pm) by a three-phase synchronous motor. Here, we will neglect the armature resistance Ra of the synchronous motor. Then, the armature copper loss will be zero and hence the mechanical power developed by the motor is equal to the input power (Pin) to the motor, i.e.,
$$mathrm{mathit{P_{m}}:=:mathit{P_{in}}}$$
Now, consider an under-excited (i.e.,Eb<V) three-phase synchronous motor having zero armature resistance (i.e.,Ra = 0), and is driving a mechanical load.
![Mechanicalpower](/electrical_machines/images/mechanicalpower.jpg)
The phasor diagram of one phase of this synchronous motor is shown in the figure. Because the motor is under-excited, thus it will operate at a lagging power factor, say (cos $phi$). From the phasor diagram, it is clear that $mathit{E_{r}}:=:I_{a}X_{s}$ and the armature current per phase $I_{a}$ lags the resultant EMF $mathit{E_{r}}$ by an angle of 90°.
Therefore, the input power per phase to the motor is given by,
$$mathrm{mathit{P_{in}}:=mathit{VI_{a}}:cos:phi :cdot cdot cdot (1)}$$
Since $mathit{P_{m}}$ is equal to $mathit{P_{in}}$, therefore,
$$mathrm{mathit{P_{m}}:=mathit{VI_{a}}:cos:phi :cdot cdot cdot (2)}$$
From the phasor diagram, we have,
$$mathrm{mathit{AB}:=mathit{I_{a}X_{s}}:cos:phi :=:mathit{E}_{mathit{b}}:sin:delta}$$
$$mathrm{ herefore mathit{I_{a}:cosphi :=:frac{E_{b}:sindelta }{mathit{X_{s}}}}:cdot cdot cdot (3)}$$
Using equations (2) & (3), we obtain,
$$mathit{P_{m}:=:frac{mathit{VE_{b}}sindelta }{X_{s}}}cdot cdot cdot (4)$$
This is the expression for mechanical power developed (Pm) per phase by the synchronous motor.
For 3-phases of the motor, the developed mechanical power is given by,
$$mathit{P_{m}:=:frac{3mathit{VE_{b}}sindelta }{X_{s}}}cdot cdot cdot (5)$$
Also, from the equations (4) & (5), it is clear that the mechanical power developed will be maximum when power angle ($delta$= 90°) electrical. Therefore,
For per phase,
$$mathit{P_{max}:=:frac{mathit{VE_{b}}}{X_{s}}}cdot cdot cdot (6)$$
For 3-phase,
$$mathit{P_{max}:=:frac{mathit{3VE_{b}}}{X_{s}}}cdot cdot cdot (7)$$
Important Points
The following important points may be noted about the mechanical power developed by a three-phase synchronous motor −
The mechanical power developed by the synchronous motor increases with the increase in power angle ($delta$) and vice-versa.
If power angle ($delta$) is zero, then the synchronous motor cannot develop mechanical power.
When the field excitation of the synchronous motor is reduced to zero, i.e., ($E_{b}$ = 0), the mechanical power developed by the motor is also zero, i.e. the motor will come to a stop.
Numerical Example
A 3-phase, 4000 kW, 3.3 kV, 200 RPM, 50 Hz synchronous motor has per phase synchronous reactance of 1.5 $Omega$. At full-load, the power angle is 22° electrical. If the generated back EMF per phase is 1.7 kV, calculate the mechanical power developed. What will be the maximum mechanical power developed?
Solution
Given data,
Voltage per phase, $mathit{V}:=:frac{3.3}{sqrt{3}}:=:1.9:kV$
Back EMF per phase,$mathit{E_{b}}:=:1.7:kV$
Synchronous reactance,$X_{s}:=:1.5Omega $
Power angle,$delta :=:22^{^{circ}}$
Therefore, the mechanical power developed by the motor is,
$$mathrm{mathit{P_{m}}:=:frac{3:mathit{VE_{b}:sindelta} }{mathit{X_{s}}}:=:frac{3 imes 1.9 imes 1.7 imes mathrm{sin}:22^{circ}}{1.5}}$$
$$mathit{ herefore P_{m}}:=:2.42 imes 10^{6}:W:=:2.42:MW$$
The mechanical power developed will be maximum when ($delta$ =90°),
$$mathit{P_{max}}:=:frac{3mathit{VE_{b}}}{X_{s}}:=:frac{3 imes 1.9 imes 1.7}{1.5}$$
$$mathit{ herefore P_{max}}:=:6.46:mathrm{MW}$$
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