Question 10.1: Draw a space-vector diagram of a SynRM, which has the follow...
Draw a space-vector diagram of a SynRM, which has the following per unit parameters. Lmd =3, Lmq =0.2,Lsσ =0.1, us =1, is =1, ωs =1, Ψs =1, Ψmd =0.95, Ψmq =0.18, and Ψsσ =0.1. Calculate cosφ, Ld /Lq , the electric current components, load angle, current angle, and pu electromagnetic torque and power.
The SynRM space-vector diagram was shown in Figure10.4. The space vectors comprised-axis and q-axis components.
The per unit values of the magnetizing flux linkage for the d- and q-components can be expressed as follows.
cos δ_{m} = \frac {Ψ_{md}}{Ψ_{m} } = \frac {0.95}{0.967} = 0.983 → δ_{m} = 10.7°
The angle between Ψs and Ψm can be calculated using the cosine law.
(L_{sσ}i_{s})^{2} = Ψ^{2}_{s} + Ψ^{2}_{m} – 2Ψ_{s}Ψ_{m} cos(δ_{s} – δ_{m} )cos (δ_{s} – δ_{m}) = \frac { Ψ^{2}_{s} + Ψ^{2}_{m} – (l_{sσ}i_{s})^{2}}{2Ψ_{s}Ψ_{m}} = \frac {1^{2} + 0.967^{2} – (0.1 ⋅1)^{2}} { 2⋅ 1 ⋅ 0.967 } = 0.995 → (δ_{s} – δ_{m}) = 5.5°
(δ_{s} – δ_{m}) = 5.5° → δ_{s} = 5.5° + δ_{m} = 5.5° + 10.7°
δ_{s} = 16.2°
The electric current per-unit components according to Equations(10.12) and (10.13) are
i_{ d }=\frac{1}{L_{ d }} \psi_{ s } \cos \delta_{ s } (10.12)
\dot{i}_{ q }=\frac{1}{L_{ q }} \psi_{ s } \sin \delta_{ s } (10.13)
i_{d} = \frac {1} { L_{d}}Ψ_{spu} cos δ_{s} = \frac {1}{3.1}1⋅cos 16.2° = 0.31i_{q} = \frac {1} { L_{q}}Ψ_{s} sin δ_{s} = \frac {1}{0.31}1⋅sin 16.2° = 0.9
Per-unit torque as a function of load angle, calculated with space vectors is
T_{e} = Ψ^{2}_{s} (\frac {L_{d} – L_{q}}{L_{d} L_{q}}) \frac{1}{2} sin 2δ_{s} = 1^{2} \frac {3.1 – 0.3}{2⋅0.3 ⋅3.1} sin 2⋅ 16.2° = 0.81Neglecting losses, the per-unit output power Pout becomes
P_{out} = w_{s} T_{e} = u^{2}_{s} \frac {L_{d} – L_{q}}{2L_{d}L_{q} w_{s}} sin 2δ_{s} = 1^{2} \frac {3.1 – 0.3}{2⋅ 0.3 ⋅ 3.1 ⋅1} sin 2⋅16.2° = 0.81Because the q-axis is perpendicular to the d-axis, us is perpendicular to Ψs . Therefore, the angle between the voltage and q-axis is the load angle δs . The angle between is and the q-axis must be added to get the phase angle φ between the voltage and current.
sin^{-1} \frac {i_{d}}{i_{s}} = sin^{-1} \frac {0.309}{1} = 18°φ = 16.18° + 18° = 34.18° → cos φ = 0.827
The electric current angle is shown at the base of the Figure10.4 diagram.
κ + φ = 90° +δ_{s} → k = 90° + δs – φ = 90° + 16.18° – 34.18° = 72°Expressed more simply, the statement becomes
k = arccos \frac {i_{d} }{i_{s}} = arccos \frac {0.309}{1} = 72°Per-unit torque can be checked using the electric current angle.
t_{e} = \frac {1}{2} i^{2}_{s} (L_{d} – L_{q} ) sin 2k = \frac{1}{2} ⋅ 1^{2} (3.1 -0.3 ) sin 2⋅ 72 = 0.82This result is close to the 0.81 calculated above (with in a small rounding error). The saliency ratio is Ld/Lq =3.1/0.3=10.33.
