# Induction motor calculations

• Engineering
Martin Harris
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:
 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!

Last edited:

Your calculation, it seems to me, is o.k. I have only the following remark:
If according to Steinmetz schematic diagram of an induction motor Rr*(1-sn)/sn*Ir^2=Pn then Rr*Ir^2=Ljr=sn*(Pn+Ljr) and sn=Ljr/(Pn+Ljr)
Now if Pinput-Ljs=Pn+Ljr then sn is o.k. However, Pinput=Pn/0.9=5555.56 and Ljs=213.33 then Pinput-Ljs=5342.2 and Pn+Ljr=5000+142.2=5142.2 W .In this case sn=0.02766.

#### Attachments

• Equivalent Steinmetz Gap Power.jpg
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Martin Harris
Usually, in IEC World, 220 V it is the line-to-neutral voltage and the line-to-line is 380 V [now according to IEC std 60038 has to be 400/231 V]

Martin Harris
Dr.D
It is pretty hard to justify so many decimal places when the input values are not nearly so precisely given.

Martin Harris
Martin Harris
Your calculation, it seems to me, is o.k. I have only the following remark:
If according to Steinmetz schematic diagram of an induction motor Rr*(1-sn)/sn*Ir^2=Pn then Rr*Ir^2=Ljr=sn*(Pn+Ljr) and sn=Ljr/(Pn+Ljr)
Now if Pinput-Ljs=Pn+Ljr then sn is o.k. However, Pinput=Pn/0.9=5555.56 and Ljs=213.33 then Pinput-Ljs=5342.2 and Pn+Ljr=5000+142.2=5142.2 W .In this case sn=0.02766.
Thanks for the reply it's much appreciated.
I was told $$\frac {Ljr} {Ljs} = \frac {2} {3}$$
I was just given the following formula:
$$sn = \frac {Ljr} {Pelectromagnetic}$$
where sn = nominal slip, Ljr = 142.2222 W and Pelectromagnetic = Psn - Ljs - LFe = 5555.5555W - 213.3333W - 75W = 5267.2217 W

Hence I get sn (nominal slip) = 0.02700
I don't get where I am comitting the mistake.

Martin Harris
Usually, in IEC World, 220 V it is the line-to-neutral voltage and the line-to-line is 380 V [now according to IEC std 60038 has to be 400/231 V]
Yes, indeed, I was told that, but asked to use the stator line voltage at 220V and frequency at 50 Hz, but I get what you mean, that's right.

Martin Harris
It is pretty hard to justify so many decimal places when the input values are not nearly so precisely given.
Indeed, though the result won't change by a lot, even if I trim down the decimals.