1
Question
A 3-phase, 50 Hz salient-pole synchronous motor has a per phase direct-axis reactance Xd of 0.8 p.u and quadrature axis reactance Xq of 0.6 p.u. Resistance of machine is negligible. It is drawing full load current at 0.8 p.f. lagging. When the terminal voltage is 1 p.u. per phase induced voltage in p.u. is_______
- 0.8
- 0.8
Open in App
Solution
The correct option is A 0.8
For a sailent pole synchronous motor,
E′f=Vt−jIaXq−IaRa
Vt=1 p.u.
Here, Ra=0
Xq=0.6 p.u.
Ia=1 p.u.
ϕ=−36.86∘
E′f=1∠0−1∠−36.86∘×0.6∠90∘
E′f∠δ=1−0.6∠53.13∘
=1−0.36−j0.48
E′f∠δ=0.64−j0.48
E′f∠δ=0.8∠−36.86∘
as δ=−36.86∘
and ϕ=−36.86
ψ=δ−ϕ
ψ=0∘
So, Id=Ia sin ψ=0 p.u.
Ef=E′f−Id(Xd−Xq)
Ef=E′f=0.8 p.u.
For a sailent pole synchronous motor,
E′f=Vt−jIaXq−IaRa
Vt=1 p.u.
Here, Ra=0
Xq=0.6 p.u.
Ia=1 p.u.
ϕ=−36.86∘
E′f=1∠0−1∠−36.86∘×0.6∠90∘
E′f∠δ=1−0.6∠53.13∘
=1−0.36−j0.48
E′f∠δ=0.64−j0.48
E′f∠δ=0.8∠−36.86∘
as δ=−36.86∘
and ϕ=−36.86
ψ=δ−ϕ
ψ=0∘
So, Id=Ia sin ψ=0 p.u.
Ef=E′f−Id(Xd−Xq)
Ef=E′f=0.8 p.u.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
