Derive(10.37). idref=ψsmaxref/√2La

Question

Derive (10.37).

[latex] i_{dref} = \frac { Ψ_{smaxref}}{ \sqrt{2} L_{d} } [/latex]

√

Accepted Answer

Differentiating the torque equation

[latex] T_{e} = \frac {3}{2} p ( \frac { L_{d} }{L_{q} } - 1 ) i_{d} \sqrt { Ψ^{2}_{s} - ( L_{d} i_{d} ) ^{2} } [/latex]

with respect to i_d, results in

[latex] \frac { ∂T_{e} }{ ∂i_{d} } = \frac {3}{2} p( \frac { L_{d} }{L_{q} } - 1 ) [\sqrt { Ψ^{2}_{s} - ( L_{d} i_{d} ) ^{2} } + i_{d} \frac { -2L^{2}_{d} i_{d} } { 2 \sqrt { Ψ^{2}_{s} - ( L_{d} i_{d} ) ^{2} } } ] [/latex]

The zero for the differential can be found by setting the bracketed term to zero.

[latex] [\sqrt { Ψ^{2}_{s} - ( L_{d} i_{d} ) ^{2} } + i_{d} \frac { -2L^{2}_{d} i_{d} } { 2 \sqrt { Ψ^{2}_{s} - ( L_{d} i_{d} ) ^{2} } } ] = 0 [/latex]

This results in

[latex] [ \frac { Ψ^{2}_{s} - 2( L_{d} i_{d} ) ^{2} }{ \sqrt { Ψ^{2}_{s} - ( L_{d} i_{d} ) ^{2} } } ] =0 [/latex]

The numerator becomes 0 when

[latex] i_{d} = \frac { Ψ_{s} }{ \sqrt {2} L_{d} } [/latex]

Question 10.5: Derive(10.37). idref=ψsmaxref/√2La

Related Answered Questions

Figure10.8 shows power factor versus load angle for a SynRM motor drive at rated voltage and speed and saliency ratios of 5, 10, and 50. Check the operating point power factor versus load angle for the saliency ratio of 50 with Lq=0.2 pu at the rated voltage, current, and frequency. ...

Verified Answer:

Calculate power factor versus current angle for saliency ratios between 5 and 50 and Lq=0.2 at rated voltage, current, and frequency. Calculate the current angle at which the power factor reaches its maximum value and the maximum power factor at that current angle. ...

Verified Answer:

Calculate the per-unit electromagnetic torque versus load angle for a saliency ratio between 5 and 50. Lq=0.27 at rated voltage and frequency. ...

Verified Answer:

Verified Answer: