Recent content by Martin Harris

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

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...

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.

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

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...

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 *...

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

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

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=...

$$Vstator_{peak} = 120√2 *√2 = 240 V$$ $$Istator_{peak} = i_{q} (current on q axis)= 20*√2 = 28.2842 A $$ $$ω{15} = Ω_{15} * p = 2π * 15 = 94.2477 rad/s$$ $$ω{15} = Ω_{15} * ψ_{pm}$$ 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...

Hi Thanks a lot! What if Vsl = 380 V in Delta connection? Is = 380V / Z_ab (overall impedance) = phase current, right? In that case I don't have to / sqrt(3) to get the phase current, as I already have it calculated

Given the same circuit schematic as above What I don't understand is...say for example we have the Voltage on the stator line Vsl= 380 V Then if the connection is delta, Voltage on the stator line = Voltage of the stator phase = 380V, right? Vsl = Vsph = 380V When I calculate the electrical...

Hi! Thanks a lot, so active power P =√3*Vsl*Is*cos (phi) ? In my notes I wrote it as P =3*Vsl*Is*cos (phi) ; even if it is delta connection as I thought Is = phase current Can we say that the Joule Rotor Losses = 3*R_r (rotor resistance) * Ir2 (rotor current squared)? And that Joule Stator...

Given that the connection is delta So if we want to calculate the active Power P, can we say P = 3* Vsp (voltage stator phase) * I(stator current) * cos (phi) ? P = 3 * 220V * 21.6395 A * cos (145 degrees) ?

Thanks a lot for the confirmation! In this case if the connection is delta, what would happen then? V stator line = V stator phase = 220V Since we calculated the phase currents above. I think in this case (delta connection) the active Power P = 3*V stator phase *I phase*cos(phi)