Induction motor calculation problems

[Martin Harris]

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:

Parameter	Value
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*Ls	0.6283i Ω
Lm(magnetic inductance)	0.01 H
xm = 2*π*f*Lm	3.1415iΩ
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 I_m+I_r > I_s (main current).

Was expecting I_s (main stator current) = I_m+I_r (as the 2 sub-branches run in parallel)

Furthermore the Joule Stator Loss seems to be enormous L_js =11.81 kW given that the stator Power was calculated as P_s =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!

 

[Gordianus]

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

 

[Martin Harris]

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.

 

[Gordianus]

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.

 

[Babadag]

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.

 

[Babadag]

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]

 

[Martin Harris]

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.