Question 10.4: Figure10.8 shows power factor versus load angle for a SynRM ...
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.
The RMS load angle Equation(10.1) can be written in per-unit form as follows.
P=3U^{2}_{sph}\frac{L_{d}-L_{q}}{2\omega _{s}L_{d}L_{q}}\sin 2\delta _{s}=3U^{2}_{sph}\frac{1-\frac{L_{q}}{L_{d}} }{2\omega _{s}L_{q}}=\sin 2\delta _{s} (10.1)
P = u^{2} \frac { L_{d} – L_{q} } { 2 w_{s} L_{d} L_{q} } sin 2δ_{s} = u^{2}_{spu} \frac { 1- \frac {L_{q}}{L_{d} } }{2w_{s} L_{q} } sin 2δ_{s}
At the Ld /Lq =50 saliency ratio with a load angle of δs=10°,per-unit power is
p = 1^{2} \frac { 1- \frac {0.2}{10 } }{2⋅ 1 ⋅ 0.2 } sin (2 ⋅ 10° ) = 0.83This expression can be simplified by assuming there is no leakage. With Ψs=Ψm=1, Ψmd =Ψmcosδs=0.98. The current id needed to excite the machine is as follows.
i_{d} = \frac {Ψ_{md}}{L_{md} } = \frac {0.98}{10} = 0.098Correspondingly, q-axis current is
i_{q} = \sqrt { i^{2}_{spu} + i^{2}_{dpu} } = \sqrt {1^{2} – 0.098^{2} } = 0.995So the electric current angle becomes κ=arcos(0.995/.098)=84.4°, and the power factor is
cos φ = \frac {(L_{d} – L_{q}) i_{d}i_{q}}{ \sqrt { (L_{d}i_{d})^{2} + (L_{q}i_{q})^{2} } \sqrt { i^{2}_{d} + i^{2}_{q} } } = \frac {(10 – 0.2) 0.098 ⋅ 0.995}{ \sqrt { (10⋅ 0.098)^{2} + (0.2 ⋅ 0.995)^{2} } ⋅ \sqrt { 1} } = 0.955