Page 512 - Industrial Power Engineering and Applications Handbook
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            10 x 103                                  2 Relay current setting
        I, = 7                                        Considering a ground circuit resistance of, say, 2 R:
            %I3 x 3.3
                                                            3.3 x 1000
          = 1750 A                                     :.  I,  =
                                                              45  x 2
        The fault  level of the system,               where  /g  is the ground fault current
                  100
        I,,  = 1750 x - (equation (13.5))                  = 952.6 A (say 950 A)
                  10.8
          = 16.20 kA                                   Let us consider a setting of, say, 30% of  Ig:
                                                       ...   = 0.3 x 950
        (Assume a lower value of  x; (12 - 1.2 = 10.8%) to be on the   /,f
        safe side.)                                        = 285 A
        Consider CTs with a ratio of  2000/1 A and having R,,  = 7 R   Referring to Figure 15.28(a) the number of  CTs that will fall
        and the magnetizing chracteristics as in Figure 15.31. Consider   in parallel,
        a lead resistance from CT terminals to the relay to be 0.5 R
        per lead.                                      N=6
        :.  Total lead resistance, R, = 2 x 0.5        and Im corresponds to the relay voltage setting of 65 V from
                                                       the curve of  Figure 15.31 = 15 mA.
                            = 1 R  (presuming this to  be at   From equation (15.1)
                               the operating temperature)
                                                          285 = 2000 (6 x 0.015 + ISt)
        Fault current in terms of the secondary,
                                                       :.   I*t  =  ~  285  - 6 x 0.015
                    1
        is,  = 16200 x - 8.1 A                                 2000
                       =
                   2000                                      = 0.1425  - 0.09
        1 Relay voltage setting (stability limit)
                                                             = 0.0525 A
        V,, = isc(RCT + R,) (considering the resistances at the operating
            temperature)                               Therefore the relay can be set between 5-7.5%  of  1 A.
          = 8.1 (7 + 1)                                3 Stabilizing resistance
                                                       Total desired relay circuit impedance
          = 64.8 V
        say, 65 V  or nearest higher setting available on the relay.
        :.  Minimum kpv,  V,  = 2 x 65
                                                         -   65   - 1238R
                       = 130 V                             0.0525
                                                       Relay resistance
                                                       R   - VA  -  =  L
                                                        r-
                                                           I:;   (0.0525)'
                                                         = 363 R
                            Knee point
                                                       :.  Required stabilizing resistance
                                                       R,t  =I238 - 363 = 875 R
                 /I                                    where V,,  = is, x R, (considering that there are no other feeds
                                                       4 Peak voltage across the relay circuit

                                                       v,=  2
                                                                   J
                                                               &
                                                                                              (1 5.2)
                                                                      m
                                                                     to the generator internal fault from other
                                                                     sources)
                                                              = 8.1 x  1238
                      I
                      I
                      I
                           I I
                           I
                                                              = 10,027.8 V
                                                       :.   V,  =  2&,,/130(10,027.8  -130)
                     15    I,   1.5/,                         = 3208 V
                         Excitation current I,,,  (mA) +
                     (Ampere-turns converfed into Amps.)   which is a marginal case. It is, however, advisable to provide
                          RCT  = 7 Cl                  a non-linear resistance.
                     Core material - CRGO silicon steel
                                                       5 Specification for class PS CTs
        Figure 15.31  Assumed magnetizing characteristic of 2000/1
        A class PS CTs                                 CTR = 2000/1
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