Question 10.2: Calculate the per-unit electromagnetic torque versus load an...

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.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Equation (10.15) can be rearranged in terms of  per-unit electromagnetic torque. See also Equation(10.1). The result can be the basis for diagramming Te=f(δs) with saliency ratio as  a parameter.

P_{\text {out }}=\omega_{ s } T_{ e }=u_{ s }^{2} \frac{L_{ d }-L_{ q }}{2 L_{ d } L_{ q } \omega_{ s }} \sin 2 \delta_{ s }       (10.15)

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)

T_{e} = \frac {p_{out}}{w_{s}} = \frac {u^{2}_{s} L_{d} – L_{q}}{w_{s} 2 L_{d}L_{q}w_{s}} sin 2δ_{s} = \frac { u^{2}_{s} 1- \frac{l_{q}}{L_{d}}}{w^{2}_{s} 2 L_{q} } sin 2δ_{s} = 1 \frac { 1- \frac {1}{L_{d}/l_{q}}}{2L_{q} } sin 2δ_{s}

For a saliency ratio Ld/Lq=10 and a load angle of 20°

T_{e} = 1 \frac { 1- \frac {1}{L_{d}/l_{q}}}{2L_{q} } sin 2δ_{s} = \frac { 1- \frac{1}{10}}{2⋅0.27} sin 2⋅20° =1.666 sin 40° = 1.07

 

T_{emax10} = 1 \frac { 1- \frac {1}{L_{d}/l_{q}}}{2L_{q} } sin 2δ_{s} = \frac { 1- \frac{1}{10}}{2⋅0.27} sin 2⋅45° =1.666

Therefore, the maximum value for κ=45° is 1.67. Using the same approach, the calculated per-unit torques versus load angles for saliency ratios of  Ld/Lq=5,10, and 50 are illustrated in Figure10.5. The maximum values for a saliency ratio of 5 and 50 are as follows.

T_{emax5} = 1 \frac { 1- \frac {1}{L_{d}/l_{q}}}{2L_{q} } sin 2δ_{s} = \frac { 1- \frac{1}{5}}{2⋅0.27} sin 2⋅45° =1.48

 

T_{emax50} = 1 \frac { 1- \frac {1}{L_{d}/l_{q}}}{2L_{q} } sin 2δ_{s} = \frac { 1- \frac{1}{50}}{2⋅0.27} sin 2⋅45° =1.81

Figure10.5 Torque production capacity of a SynRM at different saliency ratios. The q-axis per-unit inductance is constant at Lq=0.27.

Maximum torque is produced when the load angle is δs=π/4. Figure10.5 illustrates how the torque production capacity of a SynRM changes with load angle for the saliency ratios of 5,10, and 50. The q-axis inductance is held constantat Lq=0.27pu.

10.1

Related Answered Questions