Page 321 - Industrial Power Engineering and Applications Handbook
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121300  Industrial Power Engineering and Applications Handbook
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                                               Speed (rpm) -               N, = 980




                                 Figure 12.32  Determining the duration of starting current



        accelerating torque curve into three sections, A,  B and C as
        shown, to ascertain the magnitude of the starting current and   :.   I,,  =
        acceleration time for the different sections.
                                                            = 682% of  I,
        Section A
                                                       (b)  T,  = 48%
        (a)   I,  = 750% of I,                            or 215 mkg
             /b = 730% of  I,
                                                       and starting time t,E  = 7.71  seconds.
                                                       Consider the safe time as 8.8 seconds.
        :.   la"=
                                                       Section C
              = 740% Of  1,
                                                       (a)   I,  = 630% of  I,
        (b)   & = 82% of  T,                                1,  = 100% of  lr
        or     - 450 x 974   o,82                      :.   I,=   v-
                   980                                          ~
              = 367 rnkg                                     = 450% Of  I,
                          1866 x 333.3                 (b)  & = 80% or
        and starting time  t,A =
                           375 x 367
                                                             = 358 mkg.
                          4.52 seconds
                                                       and starting time t,C  = 4.36 seconds.
        Assume the safe time to be 5.2 seconds (considering a safety   Consider the safe time as 5.0 seconds.
        factor of nearly 15%)                          Plot the starting current and thermal curve of the motor as in
                                                       Figure 12.33.
        Section E                                      Considering TJ  = 94% and p.f. = 0.85, then
                                                                  450
        (a)  lb = 730% of  I,                          I, =   x  2.64 x 0.85 x 0.94   Amp. (Max.) at 80% of 3.3 kV
            1,  = 630% of  I,                            = 123.2 A
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