《电机学-同步电机》作业思路.pdf

    5‐21 400kW6300V(Y )cosϕN=0.8 ψ =60 E =7400V X X 0 0 d q           E U cosθ=+I X ⎧⎪ 0 d d ⎨   U sinθ I X ⎪⎩ q q   U UNϕ 6300 / 3 3637.31(V)                 ϕ acosϕ acos 0.8 36.87°  N N   θ ψ =−ϕ 60=°−36.87° 23.13=°  0 N       P 3U I cosϕ  =  N N N N P 400k N Iϕ I N 45.821(A)   3UN cosϕN 3 ×6300 ×0.8   I I sinψ 45.821sin 60° 39.6825(A) ⎧⎪d ϕ 0 ⎨   I I cosψ 45.821cos 60° 22.9107(A) ⎪⎩q ψ 0   ⎧ E0 −U cosθ 7400 −3637.31×cos 23.13° X d 102.188(Ω) ⎪⎪ I d 39.6825 ⎨   ⎪ U sinθ 3637.31×sin 23.13° Ω X q 62.364( ) ⎪ I 22.9107 ⎩ q     ⎧ * * * *