- #1

Martin Harris

- 103

- 6

- Homework Statement
- Induction Motor calculation

- Relevant Equations
- $$Psn (stator nominal power) = Pelectrical (electrical input power)$$

$$Psn (stator nominal power) = \frac {Pn} {efficiency}$$

$$sn(nominal slip) = \frac {Ljr} {Pelectromagnetic} $$

$$ Ωn (nominal speed) = Ωs * (1-sn)$$

$$Mn(nominal torque) = \frac {Pn} {Ωn}$$

$$ILs (stator line current) = \frac {Psn} {sqrt(3) * uls * cosφ}$$

Given the AC Induction (asynchronous motor) in 3 phases:

Requirements:

a) Power loss calculation, Power loss balance.

b) sn (nominal slip) = ? Ωn (Nominal speed) = ? Mn (nominal torque) = ?

c) ILs (current through the statoric line) = ?

a) LFe= Lmechanical = $$ \frac {1.5} {100} * 5000 W = 75 W$$

Hence, LFe = Lmechanical = 75W (Iron loss = Mechanical loss = 75W)

Lv = $$ \frac {1} {100} * 5000 W = 50 W$$

Hence Lv = 50W (ventilation loss on the rotor)Psn (stator nominal power) = Pelectrical (electrical input power) = $$\frac {Pn} {0.9} = 5555.5555555555555555555555555556 W$$

i) $$ Ljs + Ljr = Psn - (Pn+Lfe + Lmechanical +Lv) = 5555.555555555555556 W - 5200W$$

$$ Ljs + Ljr = 355.5555555555555555555555555556W $$

ii) $$Ljr = \frac {2} {3} * Ljs$$

from i) and ii)

$$Ljs = 213.33333333333333333333333333336 W $$

$$Ljr = 142.22222222222222222222222222224 W $$

$$ Pelectromagnetic (electromagnetic power) = Psn(input electrical power) - Ljs - LFe $$

$$ Pelectromagnetic= 5555.55555555555555556 W - 213.33333333333333333333336 W - 75W $$

$$ Pelectromagnetic = 5267.2222222222222222222222222222 W$$

$$Ltotal (totalloss) = Ljs+LFe+Ljr+Lmechanical+Lv = 555.5555555555555555556 W$$

$$Ltotal (totalloss) = 555.5555555555555555555555555556 W$$

b) $$sn(nominal slip) = \frac {Ljr} {Pelectromagnetic}$$

$$sn= \frac {142.22222222222222222222222222224 W} {5267.2222222222222222222222222222 W } $$

$$sn(nominal slip) = 0.02700137116337939035966670182471 [-] $$

$$ Ωn (nominal speed) = Ωs * (1-sn)$$

where Ωs is the (synchronization speed)

$$Ωs = \frac {2*π*fstator} {p} = \frac {2*π*50Hz} {2} =157.07963267948966192313216916398 [rad/s] $$

$$Ωn (nominal speed) = 157.079632679489661923132 [rad/s] * (1-0.027001371163379390359)$$

$$Ωn = 152.838267215303462845785704096[rad/s]$$

$$ns (syncrhonization speed) [rpm] = \frac {60*fstator} {2} = 1500 [rpm]$$

$$nn (nominal speed) [rpm] = ns*(1-sn) = 1500 [rpm] * (1-0.02700137116337939035966670)$$

$$nn (nominal speed) [rpm] = 1459.4979432549309144604999472629 [rpm] $$

$$Mn(nominal torque) = \frac {Pn} {Ωn} = \frac {5000W} {152.83826721530346284578570409637 [rad/s]}$$

$$Mn(nominal torque) = 32.714320118249532323572617030565 [Nm] $$

c) $$Psn (electrical power input) = sqrt(3) * uls * ILs * cosφ$$

Hence

The current through the stator line:

$$ILs = \frac {Psn} {sqrt(3) * uls * cosφ} = \frac {5555.5555555555555555555555555556 W} {sqrt(3) * 220V * 0.88}$$

$$ILs = 16.567673013935541910845637640667 [A]$$

I would be more than grateful if someone can confirm these calculations.

Many thanks!

Parameter | Value |

Pn (Nominal Power) = Pmechanical (output power at the shaft) | 5 kW = 5000W |

uls (Voltage through the stator line) | 220 V |

fstator (stator frequency) | 50 Hz |

p (Number of pole pairs) | 2 |

LFe (Iron loss) = Lmechanical (mechanical loss) | LFe = Lmechanical 1.5% Pn |

Lv (Ventilation loss on the rotor) | Lv = 1%*Pn |

LJr (Rotor Joule loss) = (2/3) * LJs (Stator Joule loss) | #To be calculated below# |

η (efficiency of the induction motor) | 0.9 |

cosφ (power factor) | 0.88 |

Requirements:

a) Power loss calculation, Power loss balance.

b) sn (nominal slip) = ? Ωn (Nominal speed) = ? Mn (nominal torque) = ?

c) ILs (current through the statoric line) = ?

**Attempt at a solution:**a) LFe= Lmechanical = $$ \frac {1.5} {100} * 5000 W = 75 W$$

Hence, LFe = Lmechanical = 75W (Iron loss = Mechanical loss = 75W)

Lv = $$ \frac {1} {100} * 5000 W = 50 W$$

Hence Lv = 50W (ventilation loss on the rotor)Psn (stator nominal power) = Pelectrical (electrical input power) = $$\frac {Pn} {0.9} = 5555.5555555555555555555555555556 W$$

i) $$ Ljs + Ljr = Psn - (Pn+Lfe + Lmechanical +Lv) = 5555.555555555555556 W - 5200W$$

$$ Ljs + Ljr = 355.5555555555555555555555555556W $$

ii) $$Ljr = \frac {2} {3} * Ljs$$

from i) and ii)

$$Ljs = 213.33333333333333333333333333336 W $$

$$Ljr = 142.22222222222222222222222222224 W $$

$$ Pelectromagnetic (electromagnetic power) = Psn(input electrical power) - Ljs - LFe $$

$$ Pelectromagnetic= 5555.55555555555555556 W - 213.33333333333333333333336 W - 75W $$

$$ Pelectromagnetic = 5267.2222222222222222222222222222 W$$

$$Ltotal (totalloss) = Ljs+LFe+Ljr+Lmechanical+Lv = 555.5555555555555555556 W$$

$$Ltotal (totalloss) = 555.5555555555555555555555555556 W$$

b) $$sn(nominal slip) = \frac {Ljr} {Pelectromagnetic}$$

$$sn= \frac {142.22222222222222222222222222224 W} {5267.2222222222222222222222222222 W } $$

$$sn(nominal slip) = 0.02700137116337939035966670182471 [-] $$

$$ Ωn (nominal speed) = Ωs * (1-sn)$$

where Ωs is the (synchronization speed)

$$Ωs = \frac {2*π*fstator} {p} = \frac {2*π*50Hz} {2} =157.07963267948966192313216916398 [rad/s] $$

$$Ωn (nominal speed) = 157.079632679489661923132 [rad/s] * (1-0.027001371163379390359)$$

$$Ωn = 152.838267215303462845785704096[rad/s]$$

$$ns (syncrhonization speed) [rpm] = \frac {60*fstator} {2} = 1500 [rpm]$$

$$nn (nominal speed) [rpm] = ns*(1-sn) = 1500 [rpm] * (1-0.02700137116337939035966670)$$

$$nn (nominal speed) [rpm] = 1459.4979432549309144604999472629 [rpm] $$

$$Mn(nominal torque) = \frac {Pn} {Ωn} = \frac {5000W} {152.83826721530346284578570409637 [rad/s]}$$

$$Mn(nominal torque) = 32.714320118249532323572617030565 [Nm] $$

c) $$Psn (electrical power input) = sqrt(3) * uls * ILs * cosφ$$

Hence

The current through the stator line:

$$ILs = \frac {Psn} {sqrt(3) * uls * cosφ} = \frac {5555.5555555555555555555555555556 W} {sqrt(3) * 220V * 0.88}$$

$$ILs = 16.567673013935541910845637640667 [A]$$

I would be more than grateful if someone can confirm these calculations.

Many thanks!

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