Induction motor calculation problems

In summary, the given parameters for this theoretical induction motor include resistance through the stator and rotor, stator and rotor inductance, magnetic inductance, frequency, number of pairs of poles, slip, and stator and phase voltage. Using these parameters, the stator current, angle, rotor current, magnetizing current, stator power, joule stator loss, electromagnetic power, mechanical power, electromagnetic torque, and shaft torque were calculated. However, there may be issues with the parameters provided, such as a very small magnetic inductance, resulting in unrealistic values for the stator current and power losses. It is also noted that the electromagnetic torque and shaft torque may not be identical due to friction.
  • #1
Martin Harris
103
6
Homework Statement
Given the 3-phased induction machine working as a motor in AC, please see the following diagram attached below, and its parameters.
Relevant Equations
Kirchoff's Law
Ohm's Law
Given the following input parameters:
ParameterValue
Rs (Resistance through stator)1.4 Ω
Rr (Resistance through stator)0.7 Ω
Ls (stator inductance) = Lr (rotor inductance)0.002 H
xs = xr = 2*π*f*Ls0.6283i Ω
Lm(magnetic inductance)0.01 H
xm = 2*π*f*Lm3.1415iΩ
f (frequency)50 Hz
p (number of pairs of poles)3
s (slip)0.04
Vsl (Stator line voltage) - star configuration ; Vsp(Stator phase voltage)380 V ; 220V
zs = Rs +xs(1.4 + 0.6283i)Ω
zm = xm because Rw = 0 (please see diagram attached below)3.1415iΩ
zr= Rr/s +xr(17.5 + 0.6283i)Ω

Please note that everything that has an underline below is treated as a phasor.

Following the main s (stator) branch, the m (magnetizing) and r (rotor) sub-branches are in parallel as they can be seen in the attached diagram.

a) (stator current) = ?
$$\underline {Is} =? $$
$$\underline {Is} = \frac {Vsl/sqrt(3)} {z_ab} (Eq1)$$
$$z_{ab} = z_s + z_{parallel} (Eq2)$$
$$z_s = R_s+x_s = (1.4 + 0.6283i) Ω (Eq3)$$
According to Microsoft Math Solver:
$$z_{parallel} = \frac {zm*zr} {zm+zr} =0.5389+3.0254i (Eq4)$$
Substituting Eq3 and Eq4 into Eq2, yields:
$$z_{ab} =(1.9389+3.6537i ) Ω $$
Hence substituting z_ab and Vsl/sqrt(3) into Eq1
$$\underline {Is} =\frac {380/sqrt(3)} {(1.9389+3.6537i ) Ω} $$
$$\underline {Is} =(24.8634−46.8523i) A $$
$$ I_s =53.04080 A $$

b) φ =? (angle)
$$φ = arctan \frac {46.8523} {24.8634} $$
$$φ = 62.04 degrees $$

C) rotor current = ? magnetizing current =?
$$\underline {Ir} = \frac {\underline {Vparallel}} {z_{r}/s} (Eq5)$$
$$\underline {Vparallel} =\underline {Is}*z_{parallel} $$
$$\underline {Vparallel} =(24.8634−46.8523i) A * (0.5389+3.0254i)Ω$$
$$\underline {Vparallel} =155.1458+49.9730i$$
$$V_{parallel} = 162.9954 V $$

Substituting back into Eq5 yields:
$$\underline {Ir} = \frac {155.1458+49.9730i} {17.5 + 0.6283i }$$
$$\underline {Ir} = 8.95645348+2.53403773i$$
$$I_r = 9.3079 A $$

$$\underline {Im} = \frac {\underline {Vparallel}} {z_{m}} (Eq6)$$
$$\underline {Im} = \frac {155.1458+49.9730i} {3.1415i} $$
$$\underline {Im} = 15.9073−49.3858i $$
$$I_r = 51.8844 A $$

d) Stator Power = ?
$$\underline {P_s} = 3* \frac {U_{s}} {sqrt(3)} * I_{s}* cos (φ) $$
$$\underline {P_s} = 3* \frac {380V} {sqrt(3)} *53.04080 *cos (62.04 degrees)$$
$$\underline {P_s} = 16367.8966 W$$

e) Joule stator loss = ? Joule rotor loss = ?
$$L_{js} = 3*R_s* I_s^2 $$
$$L_{js} = 3*1.4 Ω* 53.04080^2 A^2 $$
$$L_{js} = 11815.9711 W $$

$$L_{jr} = 3*R_r* I_r^2 $$
$$L_{jr} = 3*0.7 Ω* 9.3079^2 A^2 $$
$$L_{jr} = 181.9377 W $$

$$L_{mech} = (1.5/100) * P_s = 24.5684 W $$
$$L_{ventilation} = (1/100) * P_s = 16.3789 W $$

f) Electromagnetic Power = ? Mechanical Power =? Assuming negligible L_Fe (Iron losses ~= 0)
$$P_{elm} = P_s - L_{js} = 16367.8966 W - 11815.9711 W $$
$$P_{elm} = 4551.9255 W $$

$$P_{mech} = P_{elm} - L_{jr} = P_{elm} * (1-s) $$
$$P_{mech} = 4551.9255 W -181.9377 W$$
$$P_{mech} = 4369.9878 W $$

g)Electromagnetic torque =? Shaft torque = ?

$$M_{electromagnetic} = \frac {P_{elm}} {Ω_s} $$
$$Ω_s = \frac {2*π*f} {p} = 104.7197 rad/s $$
$$M_{electromagnetic} = \frac {4551.9255 W} {104.7197 rad/s } $$
$$M_{electromagnetic} = 43.4677 Nm$$

$$M_{shaft} = \frac {P_{mech}} {Ω} $$
$$Ω = Ω_s*(1-s) = 100.5309 rad/s $$
$$M_{shaft} = \frac {4369.9878 W} {100.5309 rad/s } $$
$$M_{shaft} = 43.4691Nm$$

I think I am doing something wrong because Kirchoff's Law, doesen't seem to apply such that the the sum of the currents sub-branches Im+Ir > Is (main current).

Was expecting Is (main stator current) = Im+Ir (as the 2 sub-branches run in parallel)

Furthermore the Joule Stator Loss seems to be enormous Ljs =11.81 kW given that the stator Power was calculated as Ps =16.367 kW.

For sure I did something wrong, and I still didn't realize it, and that's why I need your help. I would be more than grateful if someone could check my calculations. Many thanks!
 

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  • #2
Once again, I'm reading the Forum with my phone and my eyes aren't what they used to be. I'll check later with a large PC screen. Lm is too small
 
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  • #3
Gordianus said:
Once again, I'm reading the Forum with my phone and my eyes aren't what they used to be. I'll check later with a large PC screen. Lm is too small
Right, okay. Thanks a lot!
That was the given Lm = 0.01 H

I just find it weird that Kirchoff's Law, doesen't seem to apply such that the the sum of the currents sub-branches Im+Ir > Is (main current).

Furthermore the Joule Stator Loss seems to be enormous Ljs =11.81 kW given that the stator Power was calculated as Ps =16.367 kW.
 
  • #4
I checked your calculations and, at first glance, found no mistakes.
I insist, Lm is too small. This motor has no core!
No wonder the stator current is huge and the power losses shoot through the roof.
Let's assume the motor has no load (s=0). If you repeat the calculation will find the stator current is still too large. This is crazy. A good motor with no load should have a low stator current.
 
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  • #5
Your calculation-it seems to me- is correct [except some calculation insignificant errors since in my calculation Im+Ir=Is, indeed]. But Pmech has to be less-you have to subtract the ventilation and mechanical losses.
 
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  • #6
I agree with Gordianus: the motor data it is not for a real induction motor [efficacity 24%? and stator copper losses of 72%?]. However it is only a theoretical exercise, I agree]
 
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  • #7
I wonder if the electromagnetic torque should be exactly the same (identical) to the shaft torque.
I guess that's feasible if we assume negligible friction.
 

1. What is an induction motor?

An induction motor is a type of electric motor that converts electrical energy into mechanical energy. It works by using electromagnetic induction to create a rotating magnetic field, which then drives the motor's rotor to turn and produce mechanical output.

2. What are some common problems with induction motors?

Some common problems with induction motors include overheating, bearing failure, and electrical faults. These issues can be caused by factors such as excessive load, poor maintenance, or electrical imbalances.

3. How do you calculate the power of an induction motor?

The power of an induction motor can be calculated by multiplying its rated voltage by its rated current and power factor. The formula is P = V x I x PF, where P is power in watts, V is voltage in volts, I is current in amps, and PF is power factor (typically between 0.8 and 0.95).

4. What is the slip of an induction motor?

The slip of an induction motor refers to the difference between the synchronous speed (the speed of the rotating magnetic field) and the actual speed of the motor's rotor. It is expressed as a percentage and is used to determine the motor's torque and efficiency.

5. How can you troubleshoot induction motor problems?

To troubleshoot induction motor problems, it is important to first identify the symptoms and potential causes. This can involve checking for overheating, inspecting the bearings and electrical connections, and testing the motor's voltage and current. It may also be necessary to consult a professional or refer to the manufacturer's guidelines for specific troubleshooting steps.

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