Calculating Power Losses in an AC Induction Motor

• Engineering
• Martin Harris
In summary: The frequency of the rotor is, in this case, frotor=sn*fstator].In summary, the AC Induction (asynchronous motor) in 3 phases has a nominal power of 5 kW, a stator voltage of 220V, a stator frequency of 50 Hz, and 2 pole pairs. The iron and mechanical losses are both 75W, while the ventilation loss on the rotor is 50W. The efficiency of the motor is 0.9 and the power factor is 0.88. Using these values, the power loss balance can be calculated, and the nominal slip is approximately 0.027. The nominal speed is 1459.4979 rpm and the
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
It is pretty hard to justify so many decimal places when the input values are not nearly so precisely given.

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.

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.

Dr.D said:
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.

By-the -way, in my opinion the power losses Lfe and Ljs may be extracted from rated stator power and ventilation and friction [mechanical losses] from Pgap [then Pgap=Lv+Lmech+Pn+Ljr] neglecting Lfe for rotor where the frequence is very low [sn*fstator] and the iron losses depend on B[magnetic flux density] and frotor.

Martin Harris

1. What is an induction motor?

An induction motor is a type of AC motor that uses electromagnetic induction to convert electrical energy into mechanical energy. It is widely used in various industrial and commercial applications due to its simplicity, reliability, and cost-effectiveness.

2. How is the efficiency of an induction motor calculated?

The efficiency of an induction motor is calculated by dividing the output power by the input power. Output power is the mechanical power produced by the motor, while input power is the electrical power supplied to the motor. The efficiency of an induction motor typically ranges from 85% to 95%.

3. What are the main factors that affect the performance of an induction motor?

The main factors that affect the performance of an induction motor include the number of poles, frequency of the power supply, voltage, and load. The number of poles determines the speed of the motor, while the frequency and voltage affect the motor's torque and power output. The load on the motor also affects its speed and efficiency.

4. How do you calculate the starting torque of an induction motor?

The starting torque of an induction motor can be calculated using the formula T = (3V^2 * R2) / (s * R1), where V is the supply voltage, R2 is the rotor resistance, R1 is the stator resistance, and s is the slip of the motor. The slip is the difference between the synchronous speed and the actual speed of the motor.

5. What is the significance of slip in induction motor calculations?

Slip is a crucial factor in induction motor calculations as it determines the motor's speed and torque. A higher slip means a lower speed and a higher torque, while a lower slip results in a higher speed and a lower torque. Slip is also used to calculate the efficiency and power factor of an induction motor.

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