Power Flow The input power to a three-phase induction machine is given by P i n = 3 – √ V L L I L cos θ = 3 V 1 I 1 cos θ P i n =...

Power Flow

The input power to a three-phase induction machine is given by
${P}_{in}=\sqrt{3}{V}_{LL}{I}_{L}\mathrm{cos}\theta =3{V}_{1}{I}_{1}\mathrm{cos}\theta$
Output power can be found by subtracting the losses from the input power

Losses

1. Stator Copper Loss. The stator resistive losses
${P}_{SCL}=3{I}_{1}^{2}{R}_{1}$
2. Rotor Joule Loss. The rotor resistive losses. This is often called rotor copper loss, but since the rotor conductors are aluminum, rotor joule loss is the more correct terminology.
${P}_{RCL}=3{I}_{2}^{2}{R}_{2}$
3. Core Loss, or Iron Loss. The losses due to eddy current and hysteresis losses in the laminations. This can be calculated using the resistor ${R}_{c}$. At times, core losses are grouped with friction and windage and stray loss as rotational losses.

Rotor Power

The power transferred to the rotor is called the "Airgap Power". Consider the equivalent circuit below (the core loss resistance has been removed and core losses grouped into rotational loss).
From the above circuit, it can be seen that the total power transfer to the rotor is given by
${P}_{gap}=\frac{3{I}_{2}^{2}{R}_{2}}{s}$
To find the power converted to the mechanical system the rotor joule loss must be subtracted from the total rotor power
$\begin{array}{rl}{P}_{conv}& ={P}_{gap}-{P}_{RCL}\\ {P}_{conv}& =\frac{3{I}_{2}^{2}{R}_{2}}{s}-3{I}_{2}^{2}{R}_{2}\\ {P}_{conv}& =3{I}_{2}^{2}{R}_{2}\frac{1-s}{s}\end{array}$
From the above equations, it can be seen that power converted to the mechanical system is a function of the airgap power and slip:
${P}_{conv}=\left(1-s\right){P}_{gap}$
Final output power may be obtained by subtracting the rotational loss from ${P}_{conv}$.
${P}_{out}={P}_{conv}-{P}_{rotational}$

Torque

As with all rotating mechanical systems in steady state, torque can be found from the power and mechanical speed
$\tau =\frac{P}{\omega }$
In the case of an induction machine, the electromagnetic torque generated by the machine can be found using
$\begin{array}{rl}\tau & =\frac{{P}_{conv}}{{\omega }_{m}}\\ \tau & =\frac{\left(1-s\right){P}_{gap}}{\left(1-s\right){\omega }_{s}}\\ \tau & =\frac{{P}_{gap}}{{\omega }_{s}}\end{array}$
Writing the torque in terms of the rotor current:
$\tau =\frac{3{I}_{2}^{2}{R}_{2}}{s{\omega }_{s}}$
Finally, to find the available shaft torque after rotational losses, the output power must be used.

Name

BASIC ELECTRICAL,12,BATTERIES,4,CIRCUIT THEORIES,9,CONTROL SYSTEMS,3,DC MOTOR,1,DIGITAL ELECTRONICS,1,DISTRIBUTED GENERATION,2,DISTRIBUTION,6,ELECTRICAL DRIVES,1,ELECTRICAL LAWS,8,ELECTRONICS DEVICES,2,General,7,GENERATION,3,GENERATOR,1,HIGH VOLTAGE,4,ILLUMINATION,1,INDUCTION MOTOR,7,MATERIALS,1,MEASUREMENT,1,MOTOR,1,POWER ELECTRONICS,2,PROJECTS ON INDUCTION MOTOR,1,PROTECTION,1,SMART GRID,3,SWITCHGEAR,4,SYNCHRONOUS MOTOR,1,TRANSFORMER,6,TRANSMISSION,4,
ltr
item
Electrical for Us: Power & Torque of Induction Motor
Power & Torque of Induction Motor
https://2.bp.blogspot.com/-04jTMT79Ntk/WEpmKB0OlXI/AAAAAAAAAgg/5RCH0AEhrtUwZ4-zRQuawIyfiU5xW2yVgCLcB/s1600/High-Voltage-Explosion-Proof-AC-Induction-Motor-Output-Power-Massive-Torque-Induction-Motor-Aluminium-Frame-Sf1-15-YE2-132S-4-11-Kw.jpg
https://2.bp.blogspot.com/-04jTMT79Ntk/WEpmKB0OlXI/AAAAAAAAAgg/5RCH0AEhrtUwZ4-zRQuawIyfiU5xW2yVgCLcB/s72-c/High-Voltage-Explosion-Proof-AC-Induction-Motor-Output-Power-Massive-Torque-Induction-Motor-Aluminium-Frame-Sf1-15-YE2-132S-4-11-Kw.jpg
Electrical for Us
http://www.sanjaysah.com.np/2016/12/power-torque-of-induction-motor.html
http://www.sanjaysah.com.np/
http://www.sanjaysah.com.np/
http://www.sanjaysah.com.np/2016/12/power-torque-of-induction-motor.html
true
3851448774078769448
UTF-8