Engineering Torque calculation of a motor with permanent magnets

[Martin Harris]

Homework Statement

Given the synchronous machine working as a motor with Permanent Magnets with the following characteristics
Vsn = 120√2
Isn = 20 A
Ls = 75 mH
Rs = 0.1 Ω
p = 1 (1 pair of poles)
E0 = 119 V (at 15 rps)
e0 (peak) =119√2 = 168.3 V

Calculate the torque (M = ?) when the base speed of the synchronous motor is doubled.

Relevant Equations

$$M = 3/2 * p *ψ_{pm} * i_q$$

$$Vstator_{peak} = 120√2 *√2 = 240 V$$
$$Istator_{peak} = i_{q} (current on q axis)= 20*√2 = 28.2842 A $$
$$ω{15} = Ω_{15} * p = 2π * 15 = 94.2477 rad/s$$
$$ω{15} = Ω_{15} * ψ_{pm}$$
Hence
$$ψ_{pm} = e0/ω15 = 168.3V/94.2477 rad/s => ψ_{pm} = 1.79 Wb$$
$$M = 3/2 * p * ψ_{pm} * iq $$

Hence

$$ M = 3/2 * 1 * 1.79 Wb *28.2842 A = 75.9432 Nm $$
Now I am being asked to calculate the torque M2 by doubling the base speed ω_b which I calculated and obtainted 85.8068 rad/s

So ω = 2*ω_b = 171.6136 rad/s

M2 = ? When ω = 2*ω_b and Vs = V_sn (Stator Voltage = Nominal Stator Voltage)

I have this following formula for torque:
$$M = 3/2 * p * ψ_{pm} * i_{q} $$
But don't know how to imply that the base speed is doubled in order to get M2, any guidance would be more than appreciated!

 

[Baluncore]

But don't know how to imply that the base speed is doubled in order to get M2, any guidance would be more than appreciated!

How might you double the base speed ?
Base speed, RPM = Hz * 60 / p ;
If p = 1 it cannot be reduced any further. Therefore the AC supply frequency must be doubled.
The current will be limited by the increased impedance of the windings. The torque is proportional to the current.

 

[Martin Harris]

Thanks a lot for your reply, it's much appreciated!

I am asked to calculate the electromagnetic torque for a synchronous motor with permanent magnets if the base speed is doubled.
I calculated the base speed = 85.8068 rad/s = 819 rpm
So double the base speed would be = 171.6136 rad/s= 1638.78916638031 rpm

But no idea how to implement this to calculate the new electromagnetic torque for a speed that would be double than the base one.

 

[Baluncore]

But no idea how to implement this to calculate the new electromagnetic torque for a speed that would be double than the base one.

You know the new supply frequency, the motor winding inductance and resistance, which gives you;
impedance; Zs = Rs + XLs·j ;
How much current will flow through the windings?

 

[Martin Harris]

The new supply frequency would be 13.7*2 Hz = 27.4 Hz (double the initial frequency)
Impedance Zs = (0.1 + 2*π* 27.4 Hz * 75*10^-3H*i)Ω
Zs = (0.1+12.9119 i)Ω
Is = Vs/Zs
$$Is = \frac {120√2 V} {(0.1+12.9119 i)Ω} $$
Is = (0.101786415−13.142560069i)A
Is = 13.14295 A

 

[Baluncore]

The product of the motor current by the PM field generates torque.

 

[Martin Harris]

Right, but what would be the value of the PM field? I got initially 1.79 Wb but for E0 = 119V at 15 rps and hence e0 (peak) = 168.3 V

 

[Baluncore]

Right, but what would be the value of the PM field?

You can find the constant field for the PM by doing the same calculations for the original base speed. Things do tend to be ratiometric so you may not need to go that far back.

 

[Martin Harris]

Right, so then the constant field for the PM would be the same as before? Namely:
$$ψ_{pm} = e0/ω_{at 15 rps} = 168.3V/94.2477 rad/s => ψ_{pm} = 1.79 Wb$$
?
I find it weird that after I doubled the base speed in Post #5, the current decreased from 20 Amps to 13.14 Amps..

So now if I do 13.14A * 1.79 Wb = 23.52 Nm, which means the torque decreased as the speed is now double the base speed for the PMSM

 

[Baluncore]

I find it weird that after I doubled the base speed in Post #5, the current decreased from 20 Amps to 13.14 Amps..

The real winding resistance did not change, only the reactance increased with frequency.

 

[Martin Harris]

Indeed, I did that calculation in Post #5, and since the reactance increased, the impedance increased, and since the voltage stood the same, then the current dropped to 13.14 A.

So, the PM field stays at 1.79 Wb? As it was initially? If so...then the new electromagnetic torque = 23.52 Nm when doubling the base speed, as compared to 75.94 Nm at normal base speed without any multiplier.

 

[Baluncore]

So, the PM field stays at 1.79 Wb? As it was initially?

How could it change.

If so...then the new electromagnetic torque = 23.52 Nm when doubling the base speed, as compared to 75.94 Nm at normal base speed without any multiplier.

Normal / Double = 20 / 13.14 = 75.9432 N·m / ?

 

[Martin Harris]

How could it change.

Was thinking that it is 1.79 for e0 at 15 rps, but has nothing to do with the base speed I suppose.

Normal / Double = 20 / 13.14 = 75.9432 N·m / ?

Well, I'm not sure if my calculation on the new current and new electromagnetic torque is correct.
Normal / Double = 20 / 13.14 = 75.9432 N·m / 13.14*1.79 so

Normal / Double = 20A / 13.14A = 75.9432 N·m / 23.52 N·m

 

[Martin Harris]

Is that (Post #13) correct? I am confused now, I was actually thinking that I have to use ω = 2*ωb, for the new electromagnetic torque calculation.

 

[Martin Harris]

Would the PM flux stay constant if I double the base speed of the PMSM (motor) ?

I am given E0 (15rps) = 119V
hence e0(15rps) = 119 * sqrt(2) V = 168.2914 V and the ω15 (rps) = 94.2477 rad/s

But the base speed was calculated ωb = 85.8068 rad/s, which is different from ω15 (rps)

So now I'm asked to calculate the new torque M2 if ω=2*ωb and Us = Usn = 120*sqrt(2)So ω=2*ωb = 2*85.8068 rad/s = 171.6136 rad/s
But I have no idea what happens with the PM Flux if the base speed is doubled, because the initial PM flux was calculated as: Ypm = e0(at 15rps) / ω(at 15rps) = 1.79 Wb, should I use the same PM Flux for the new torque calculation when the base speed is doubled? If the answer to this question is yes, then the new torque will decrease yielding 23.52 N·m by doubling the base speed ωb.

 

[Martin Harris]

The new supply frequency would be 13.7*2 Hz = 27.4 Hz (double the initial frequency)
Impedance Zs = (0.1 + 2*π* 27.4 Hz * 75*10^-3H*i)Ω
Zs = (0.1+12.9119 i)Ω
Is = Vs/Zs
$$Is = \frac {120√2 V} {(0.1+12.9119 i)Ω} $$
Is = (0.101786415−13.142560069i)A
Is = 13.14295 A

I'm a bit puzzled because I don't use the new supply frequency anywhere to get the new torque at double the base speed.

For the new torque I multiplied the new current Is*Ymp (which was used as the same as the previous one) = 13.14295 A * 1.79 Wb