Question 10.3: Calculate power factor versus current angle for saliency rat...
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.
Equation(10.26) is the basis for the power factor calculation. The calculation is carried out for saliency ratio values of 5, 10, and 50.
\cos \varphi=\frac{\left(L_{ d }-L_{ q }\right) i_{ d } i_{ q }}{\sqrt{\left(L_{ d } i_{ d }\right)^{2}+\left(L_{ q } i_{ q }\right)^{2}} \sqrt{i_{ d }^{2}+i_{ q }^{2}}} (10.26)
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} } }Electric current angle changes with changes in the id and iq components. The diagram in Figure10.4 shows that current angle is linked to the current components by means ofEquation(10.28).
u_{ d }=R_{ s } i_{ d }+\frac{ d \psi_{ d }}{ d t}-\omega_{ r } \psi_{ q } (10.4)
\kappa=\arctan \sqrt{\frac{L_{ d }}{L_{ q }}} (10.28)
k = 90° – arctan \frac { i_{d}}{i_{q}}or more simply
k = arccos \frac { i_{d}}{i_{s}}If id changes from 0.9 to zero, the current angle will move from 25.8° to 90°. For example, with a saliency ratio of 10 at id =0.2, the iq component will be as follows.
i_{q} = \sqrt { i^{2}_{s} + i^{2}_{d} } = \sqrt {1^{2} – 0.2^{2} } = 0.979At this same saliency ratio of 10 and with Lq =0.2 and Ld =2, the calculation of power factor for the current angle given by this expression,
k = 90° – arctan \frac { 0.2 }{0.979} = 90° – 11.54 ° = 78.45°results in
cos φ_{10} = \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 {(2 – 0.2) 0.2 ⋅ 0.979}{ \sqrt { (2⋅ 0.2)^{2} + (0.2 ⋅ 0.979)^{2} } ⋅ 1 } = \frac { 0.352}{0.445} = 0.79So,the maximum value results at this current angle
k = arctan \sqrt { \frac { L_{d}}{L_{q}} } =arctan√10= arctan 3.16 = 72.45°and from Equation(10.29) will reach the value
\cos \varphi _{\max }=\frac{\frac{L_{d}}{L_{q}}-1 }{\frac{L_{d}}{L_{q}}+1 } (10.29)
cos φ_{max10} = \frac { \frac {L_{d}}{L_{q}} – 1 } { \frac {L_{d}}{L_{q}} + 1 } = \frac {10 -1 }{10 + 1 }=\frac {9}{11} = 0. 818At the electric current angle of 72.45°, which corresponds to current components id =0.3 and iq =0.954, the power factor reaches its maximum of 0.818. This can be confirmed
using Equation(10.27).
Figure10.7 illustrates for the range of the current angles κ between 25° and 90° and for saliency ratios 5, 10, and 50.
For a saliency ratio of Ld /Lq=5, the maximum power factor value will be reached at the following current angle.
Figure10.7 Power factor of a SynRM as a function of current angle at different saliency ratios with Lq=0.2pu
The maximum power factor will be
cos φ_{max5} = \frac { \frac {L_{d}}{L_{q}} – 1 } { \frac {L_{d}}{L_{q}} + 1 } = \frac {5 -1 }{5 + 1 } = \frac {4}{6} ≈0. 67For a saliency ratio of Ld/Lq=50, the maximum power factor value will be reached at
k_{50} = arctan \sqrt { \frac { L_{d}}{L_{q}} } = arctan \sqrt {50} = arctan 7.07 = 81.95°and will be
cos φ_{max50} = \frac { \frac {L_{d}}{L_{q}} – 1 } { \frac {L_{d}}{L_{q}} + 1 } = \frac {50 -1 }{50 + 1 } = 0.96For Ld/Lq=10, the current angle becomes κ=72.7°, and the maximum power factor will be cos φmax=0.82. A large saliency ratio leads to a higher operating current angle, and therefore a good power factor.
