Page 295 - Handbook of Electrical Engineering
P. 295

FAULT CALCULATIONS AND STABILITY STUDIES     281

           The contribution from the lower voltage motors is found as follows.

           The average ratio of starting current to rated current (I s /I n ) for the lower voltage motors may be
           assumedtobe,
                                      I s
                                        = 6.5 pu for lower voltage motors
                                     I n
           Their sub-transient impedance Z      is
                                      lm
                                              1.0
                                        Z    lm  =  = 0.153 pu(at S lm )
                                              6.5
                 Typical lower voltage motors have a starting power factor of between 0.25 and 0.35 lagging,
           hence assume 0.35.

           The sub-transient impedance Z      of the motors becomes,
                                     lm
                                          Z    lm  = 0.054 + j0.143 pu

                 The equivalent impedance Z td of the distribution transformers can be found by the method in
           sub-section 11.5.2.5 or taken as,

                                        Z td = 0.0 + j0.055 pu(at S td )

           Again assume that the air-gap emf E      is 1.0 pu.
                                          lm
           The contribution I      from the lower voltage motors is,
                           lm

                                                      E lm
                                     I
                                      lm  =
                                            Z
                                              lm       Z td
                                                +                S base
                                            S lm  2K td (S lm + S ls )
                                                      1.0S lm
                                                                                         (11.3)
                                                            j0.06
                                        =
                                            0.054 + j0.143 +      S base
                                                            K td2
           Where

                                                      S ls
                                    K td2 = 2K td 1.0 +
                                                     S lm
           The total sub-transient symmetrical rms fault current I      is,
                                                         frms

                                            I      = I + I      + I                       (11.4)
                                             frms  g   hm    lm
           11.6.1 Calculate the Sub-Transient Peak Fault Current Contributions

           Many power system networks can be reduced to a simple series-connected circuit containing a resis-
           tance R and an inductance L, for the purpose of calculating the transient fault current. Furthermore a
   290   291   292   293   294   295   296   297   298   299   300