A 3-phase induction motor with 4 poles is connected to a voltage source with an amplitude of 209 Vrms and a frequency of 120 Hz. The circuit has negligible line impedance and you can assume the simplified circuit model is valid. The motor has Rs = 2.3 LaTeX: \OmegaΩ, R'r = 0.7 LaTeX: \OmegaΩ, Xs = 1.3 LaTeX: \OmegaΩ, X'r = 1.8 LaTeX: \OmegaΩ , and Xm = 76 LaTeX: \OmegaΩ. In addition, the motor is spinning at a speed of 2,464 rpm. What is the output torque at the specified speed in Nm?

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abdullahjawaid1 abdullahjawaid1 · Answer 1 · 2019-12-25T18:11:49+01:00

Answer:

T = 25.41 Nm

Explanation:

Calculating Nsync (Synchronous Speed):

[tex]Nsync = 120f/P[/tex]

[tex]Nsync = 120 x 120 / 4\\Nsync = 3600 rpm[/tex]

[tex]Wsync = 3600 * 2\pi /180\\Wsync = 377 rad/s[/tex]

Calculating s (Slip):

[tex]s = (Nsync - Nm) / Nsync \\

[tex]s = (3600-2464)/3600\\s = 0.3156[/tex]

Calculating Vth (Thevenin Voltage):

[tex]Vth = Vph (Xm / \sqrt{Rs^{2} + (Xs+Xm)^{2} })\\Vth = 209 (76 / \sqrt{(2.3)^{2} + (1.3 + 76)^{2} }\\Vth = 205.39 V[/tex]

Calculating Rth (Thevenin Resistance):

[tex]Rth = Rs (Xm/Xs + Xm)^{2} \\Rth = 2.3 (76/1.3 + 76)^{2} \\Rth = 2.22 ohm[/tex]

Calculating Xth (Thevenin Reactance):

Xth = Xs = 1.3 ohm

Calculating Torque:

[tex]T = (3Vth^{2}Rr/s) / (Wsync[(Rth+Rr/s)^{2} + (Xth + Xr)^{2}])\\T = (3*205.39*0.7/0.3156) / 377[(2.22+0.7/0.315)^{2} + (1.3+1.8)^{2}][/tex]

T = 280699 / 377 [19.69 + 9.61]

T = 25.41 Nm