# Power Flow Diagram and Losses of Induction Motor

Digital ElectronicsElectronElectronics & Electrical

The 3-phase input power fed to the stator of a 3-phase induction motor is given by,

$$\mathrm{ЁЭСГ_{ЁЭСЦЁЭСа} = \sqrt{3} ЁЭСЙ_ЁЭР┐ЁЭР╝_ЁЭР┐ cos \varphi_i = 3 ЁЭСЙ_{ЁЭСаЁЭСЭтДО} ЁЭР╝_{ЁЭСаЁЭСЭтДО} cos \varphi_i}$$

Where,

• VL = Line voltage
• IL = Line current
• Vsph = Stator phase voltage
• Isph = Stator phase current
• cos \varphii = input power factor

## Stator Losses

• Stator copper losses or I2R losses in the stator winding resistances, which are given as follows −

$$\mathrm{ЁЭСГ_{ЁЭСаЁЭСРЁЭСв} = 3 ЁЭР╝_{ЁЭСаЁЭСЭтДО}^{2} ЁЭСЕ_{ЁЭСаЁЭСЭтДО}}$$

• Hysteresis and eddy current losses in the stator core, which are known as stator-core losses and are given by −

$$\mathrm{ЁЭСГ_{ЁЭСаЁЭР╢} = ЁЭСГ_{ЁЭСатДО} + ЁЭСГ_{ЁЭСаЁЭСТ}}$$

Therefore, the power output of the stator will be,

$$\mathrm{ЁЭСГ_{ЁЭСЬЁЭСа} = ЁЭСГ_{ЁЭСЦЁЭСа}− ЁЭСГ_{ЁЭСаЁЭСРЁЭСв} − ЁЭСГ_{ЁЭСаЁЭР╢}}$$

This output power of the stator (Pos) is transferred to the rotor of the machine across the air-gap between the stator and the rotor. It is also known as air-gap power (Pg) of the machine. Hence,

$$\mathrm{Power\:output\:of\:stator (ЁЭСГ_{ЁЭСЬЁЭСа})\:=\: Air\:gap\:power (ЁЭСГ_ЁЭСФ)\:=\:Input\:power\:to\:rotor (ЁЭСГ_{ЁЭСЦЁЭСЯ})}$$

## Rotor Losses

• Rotor copper losses or I^2R losses in the rotor resistance, which are given by,
• $$\mathrm{ЁЭСГ_{ЁЭСЯЁЭСРЁЭСв} = 3 ЁЭР╝_2^2 ЁЭСЕ_2}$$

• Rotor core losses or hysteresis and eddy current losses, which are given as follows −
• $$\mathrm{ЁЭСГ_{ЁЭСЯЁЭР╢} = ЁЭСГ_{ЁЭСЯтДО} + ЁЭСГ_{ЁЭСЯЁЭСТ}}$$

• Friction and windage losses (Pfw)
• Stray load losses (Pmisc) consisting of all the losses which are not included in the above losses like losses due to harmonic fields.

## Mechanical Power Developed (Pm)

If the rotor copper losses are subtracted from the rotor input power (Pg or Pir), then the remaining power is converted from electrical power to mechanical power. It is known as developed mechanical power (Pm).

$$\mathrm{Developed\:mechanical\:power,\: ЁЭСГ_ЁЭСЪ = ЁЭСГ_{ЁЭСЦЁЭСЯ} − ЁЭСГ_{ЁЭСЯЁЭСРЁЭСв}}$$

$$\mathrm{⇒ ЁЭСГ_ЁЭСЪ = ЁЭСГ_ЁЭСФ − ЁЭСГ_{ЁЭСЯЁЭСРЁЭСв} = ЁЭСГ_ЁЭСФ − (3ЁЭР╝_2^2ЁЭСЕ_2)}$$

Therefore, the output power of the motor is given by,

$$\mathrm{ЁЭСГ_ЁЭСЬ = ЁЭСГ_ЁЭСЪ − ЁЭСГ_{ЁЭСУЁЭСд} − ЁЭСГ_{ЁЭСЪЁЭСЦЁЭСаЁЭСР}}$$

The power Po is known as useful power or shaft power.

## Rotational Losses

At the starting and during the acceleration, the rotor core losses are high and decreases with the increase in the speed of the motor. The friction and windage losses are zero at start and increases with the increases in the speed.

Consequently, the sum of core losses and friction and windage losses is approximately constant with varying speed of the motor. Hence, these losses may lumped together and are called as rotational losses and given as follows −

$$\mathrm{ЁЭСГ{ЁЭСЯЁЭСЬЁЭСбЁЭСОЁЭСбЁЭСЦЁЭСЬЁЭСЫЁЭСОЁЭСЩ\:ЁЭСЩЁЭСЬЁЭСаЁЭСаЁЭСТЁЭСа}= ЁЭСГ_ЁЭР╢ + ЁЭСГ_{ЁЭСУЁЭСд} + ЁЭСГ_{ЁЭСЪЁЭСЦЁЭСаЁЭСР}}$$

Then, the output power of the motor is given by,

$$\mathrm{ЁЭСГ_ЁЭСЬ = ЁЭСГ_ЁЭСЪ − ЁЭСГ{ЁЭСЯЁЭСЬЁЭСбЁЭСОЁЭСбЁЭСЦЁЭСЬЁЭСЫЁЭСОЁЭСЩ\:ЁЭСЩЁЭСЬЁЭСаЁЭСаЁЭСТЁЭСа}= ЁЭСГ_ЁЭСЬ = ЁЭСГ_ЁЭСЪ − ЁЭСГ_ЁЭР╢ − ЁЭСГ_{ЁЭСУЁЭСд} − ЁЭСГ_{ЁЭСЪЁЭСЦЁЭСаЁЭСР}}$$

The power diagram of the induction motor is shown in the figure below. 