Determine how DFIM behaviour changes as rotor voltage is reduced or increased by 1% unit from the no-load slip value of 0.25 pu. The per-unit rotor resistance is Rr=0.01, and the leakage inductance is Lrσ=0.1.

Question

Determine how DFIM behaviour changes as rotor voltage is reduced or increased by 1% unit from the no-load slip value of 0.25 pu. The per-unit rotor resistance is R_r=0.01, and the leakage inductance is L_rσ=0.1.

Accepted Answer

The base slip and rotor voltage at DFIM no-load are s₀=u'_r=s₀u_s=0.25.
Vector direction can be assumed according to Figure11.53. Then, rotor voltage can be dropped to u'_r=0.24. Based on Equation(11.106), rotor current will be

[latex] i'_{r} = \frac { u'_{r} - s_{0}u_{s} } { R'_{r} + jsw_{s} L'_{rσ} } = \frac {0.24e^{j3π/2} - 0.25 ⋅1 e^{j3π/2} }{0.01 + j0.025 ⋅ 1 ⋅ 0.1 } = \frac {-0.01 e^{j3π/2}}{0.01 + j0.025 } = -0.37e^{j3π/2-0.37jπ} [/latex]

Figure11.54 Vector diagrams of machine behaviour for small voltage changes from the no-load value to a value nearer to the 0.25 base slip.
Increasing rotor voltage to u'_r=0.26 in the original vector direction results in the
following.

[latex] i'_{r} = \frac { u'_{r} - s_{0}u_{s} } { R'_{r} + jsw_{s} L'_{rσ} } = \frac {0.26e^{j3π/2} - 0.25 ⋅1 e^{j3π/2} }{0.01 + j0.25 ⋅ 1 ⋅ 0.1 } = \frac {0.01 e^{j3π/2}}{0.01 + j0.025 } = -0.37e^{j3π/2-0.37jπ} [/latex]

The voltage that produces current in the rotor winding is the difference between u'_r0=s₀u_sand the real rotor voltage u'_r. The rotor winding angular frequency is dictated by the base
slip frequency resulting in s₀w_s Figure11.54 illustrates the vector diagrams of the two example scenarios.

Question 11.4: Determine how DFIM behaviour changes as rotor voltage is red...

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