{"id":318132,"date":"2023-11-17T08:55:57","date_gmt":"2023-11-17T08:55:57","guid":{"rendered":"\/forum\/forums\/reply\/318132\/"},"modified":"2023-11-17T08:55:57","modified_gmt":"2023-11-17T08:55:57","slug":"318132","status":"publish","type":"reply","link":"https:\/\/innovationspace.ansys.com\/forum\/forums\/reply\/318132\/","title":{"rendered":"Reply To: How to simulation no- load and full load PM generator"},"content":{"rendered":"

<p style=”text-align: justify;”>Hellow,<\/p><p style=”text-align: justify;”>You can follow below steps.<\/p><p style=”text-align: justify;”>For no load simulation of Generator :<\/strong><\/p><\/p>\n

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  1. Cnahge the ecxitation type to Current and you can give current as Zero (0 A).<\/li>\n
  2. Run the simulation. We can find out here no load induced voltage, no load core loss and cogging torque.<\/li>\n<\/ol>\n

    <p style=”text-align: justify;”>For Full load Simulation of Generator:<\/strong><\/p><\/p>\n

      <li style=”text-align: justify;”>Change the ecxitation type as voltage, enter the winding resistance and end winding leakage inducthance values.<\/li>\n

      <li style=”text-align: justify;”>we can use voltage experssion which given below.<\/li>\n

      <li style=”text-align: justify;”>1. Solution:<\/strong>Va= Vpeak*sin(2*pi*f*Time-delta),Vb= Vpeak*sin(2*pi*f*Time-120deg-delta),Vc= Vpeak*sin(2*pi*f*Time-240deg-delta). Condition: <\/strong>no load induced voltage of phase a has to be zero at 0 sec. we can do parametric setup on delta variable or if you know the delta (load angle) you can enter the value in degrees.This type of setup will not consider mechanical transients, just speed is enough. (or) <\/li>\n

      <li style=”text-align: justify;”>2. Solution:<\/strong>If you want to consider mechanical transients, your approuch is correct. But Moment of inertia is required and load torque is 4.7746 Nm as per the input provided by you. Va= Vpeak*sin(2*pi*f*Time),Vb= Vpeak*sin(2*pi*f*Time-120deg),Vc= Vpeak*sin(2*pi*f*Time-240deg).(or)<\/li>\n

      <li style=”text-align: justify;”>3. Solution:<\/strong>we can get load resistance from per phase power and per phase current. consider your machine has 3 phase and connected in star. Then Rl=(per phase power)\/(per phase current)^2= 10000\/(48)^2 =4.34027 ohm. Enter the (4.34027+winding resistance) in resistance tab in excitation window. Va=0, Vb=0 and Vc=0.<\/li>\n<\/ol>\n

      <p> <\/p><p>Hope it will help you.<\/p><p> <\/p><p>Thank you,<\/p><p>Harinath Vudumula.<\/p><\/p>\n","protected":false},"template":"","class_list":["post-318132","reply","type-reply","status-publish","hentry"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t