Explanation: Te(av)= 1/8 Im2(Ld-Lq)sin2δ+1/2 IfImMmaxsinδ = Asinδ+Bsin2δ. 12. An electromechanical energy conversion device has cylindrical stator but salient pole rotor. Rotor is not excited. If δ is the angle between stator field and rotor long axis, then average torque developed is proportional to (A and B are constants)
A. Asinδ
B. Asin2δ
C. Asinδ + Bsin2δ
D. zero
Answer:A
Explanation: Te(av)= 1/8 Im2(Ld-Lq)sin2δ+1/2 IfImMmaxsinδ = Asinδ+Bsin2δ. 12. An electromechanical energy conversion device has cylindrical stator but salient pole rotor. Rotor is not excited. If δ is the angle between stator field and rotor long axis, then average torque developed is proportional to (A and B are constants)
A. Asinδ B. Asin2δ C. Asinδ + Bsin2δ D. zero
Explanation: We know, Te= 1/8 Im2(Ld-Lq)sin2δ+1/2 IfImMmaxsinδ, rotor is not excited implies If=0, and thus Te= 1/8 Im2(Ld-Lq)sin2δ = Asin2δ 13. The self and mutual inductances of a doubly excited magnetic system are Ls=0.6+0.20cos2θr H and Lr=0.75+0.30cos2θr H and Msr=0.8cosθr H. For a stationary rotor at an angular position of θr=60°,what is the magnitude of torque when the currents is=20A DC and ir=10A DC?