A 3-phase, 415 V, 6-pole, 50 Hz, star-connected slip-ring induction motor has a total stator and rotor reactance of 1.5 Ω referred to the stator. The machine drives pure inertia load; the moment of inertia of the rotor and load being 11 kg m^2 . Direct on-line starting is used and the rotor

Question

A 3-phase, 415 V, 6-pole, 50 Hz, star-connected slip-ring induction motor has a total stator and rotor reactance of 1.5 [latex] \Omega[/latex] referred to the stator. The machine drives pure inertia load; the moment of inertia of the rotor and load being [latex] 11 kg m^{2}[/latex]. Direct on-line starting is used and the rotor circuit resistance is adjusted so that the load is brought to 0.96 of the synchronous speed from rest in shortest possible time. Neglecting stator losses, compute the acceleration time and the value of the rotor resistance referred to the stator.

Accepted Answer

s = 1 – 0.96 = 0.04

Substituting in Eq. (9.101) [latex] \left(s_{\max , T} ight)_{ opt }=\sqrt{\frac{\left(s_{1}-s_{2} ight)^{2}}{2 \ln \left(s_{1} / s_{2} ight)}}[/latex]

[latex] \left(s_{\max , T} ight)_{ opt }=\left[\frac{1-(0.04)^{2}}{2 \ln (1 / 0.04)} ight]^{1 / 2}[/latex]

= 0.394

From Eqs (9.105) [latex] \left(R_{2}^{\prime} ight)_{ opt } =\left(X_{1}+X_{2}^{\prime} ight)\left(s_{\max , T} ight)_{ opt }[/latex]

and (9.103) [latex] t_{A}(\min )=\frac{J \omega_{s}}{2 T_{\max }}\left[\frac{1-s^{2}}{2\left(s_{\max , T} ight)_{ opt }}+\left(s_{\max , T} ight)_{ opt } \ln \frac{1}{s} ight][/latex]

[latex] \left(R_{2}^{\prime} ight)_{ opt }=1.5 imes 0.394=0.591 \Omega[/latex]

[latex] t_{A}(\min )=\frac{J \omega_{s}}{2 T_{\max }}\left[\frac{1-s^{2}}{2\left(s_{\max , T} ight)_{ opt }}+\left(s_{\max , T} ight)_{ opt } \ln \frac{1}{s} ight][/latex]

From Eq. (9.94) [latex] T_{\max }=\frac{3}{\omega_{s}} \cdot \frac{0.5 V_{T H}^{2}}{\left(X_{1}+X_{2}^{\prime} ight)}[/latex] , assuming [latex] V_{TH} = V[/latex]

[latex] T_{\max }=\frac{3}{2 \pi imes 1000} \cdot \frac{0.5(415 / \sqrt{3})^{2}}{1.5}=548.2[/latex]

[latex] =\frac{11(2 \pi imes 1000 / 60)}{2 imes 548.2}\left[\frac{1-(0.04)^{2}}{2 imes 0.394}+0.394 imes \ln \left(\frac{1}{0.04} ight) ight][/latex]

= 2.66 s

Question 9.19: A 3-phase, 415 V, 6-pole, 50 Hz, star-connected slip-ring in...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions