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        be the losses and the more economical will be the design.   -  (2.873   C2 cm) x (1' x 12 x 2.54 cm)
                                                                      x
        But we cannot reduce the thickness and increase the diameter
                                                                              x
        of the enclosure indefinitely. A very large diameter may also   IT x [(37.46 - 0.43) 2.54 cm] (0.43 x 2.54 cm)
        become  unsuitable to  match  at  the  generator  end.  The   = 0.271 x   Rift
        generator terminal dimensions, in fact, decide the size of the
        enclosure. For the generator rating under consideration, we   and
        are taking the outside diameter of the enclosure as 1500 mm
        (it should  be considered according to  the site  requirement   %c105*  = RdcPO  [l +   (lo5 - 20)1
        and adjusted for the support insulators selected). The terminal   = 0.271 x 10-6[1 + 3.96 x   x 851
        spacing of the generator between phases will also determine
        the centre spacing of the IPB. In our calculations we assume   = 0.271 x   [I ,33661
        it to be around 68 inches (which may almost be the case in
        practice).                                            = 0.3622 x   Wft
        Enclosure's approximate area of  -  20 000 x 0.98   Skin effect ratio, from Figure 31.17 for
                                -
        cross-section                400
                                = 49 square inches
        (assuming the enclosure's current to be around 98% of  the   = 371.54
        conductor's current                                    0.43
                                                       and t/d  = -
        Enclosure's outer diameter = 1500 mm                  37.46
                                                             = 0.011
                           D = 59 inches
        :.   enclosure thickness  t c- 0.27 inch (Figure 31.16)   :.  Skin effect ratio = 1.03
                                                       :.  W, (20 000)'  x 0.3622  1 0-6 x  1.03
                                                                           x
                                                           =
                         [49  = n(D - t) . t = IT(=  59)tl
                                                           = 149.22 Wlft
        :.  exact area of  cross-section  =   (59' - (59 - 0.54)')   2 By the enclosure
                                4
                              = 49.79 square inches    (a) By current
                                                                 /,)'
        Enclosure surface area per foot run A,  = x x 59 x  12   W,  = (0.98  . RaC
                                = 2223.12 square inches   The current through the continuous enclosure is assumed to
                                                       be at  about 98% of  that of  the main conductor. The actual
        Checking the suitability  of the above sections by   current may  be less than this  as a  result of  low enclosure
        heat balancing                                 thickness, which is considered to be less than the depth of
                                                       penetration 4. But it may not cause a high field in the space,
        Heat generated (In the outdoor part)           leading  to  heating of  steel  structures,  beams,  columns or
        1 By the conductor                             other equipment  installed in the vicinity  or step and touch
                                                       voltages higher than permissible. This is a subject for further
             WI  =  I:  R,,  W/ft                      investigation and experiments need be conducted  to establish
                                                       that the field in the space is within reasonable limits. Otherwise
                =  I,'   Rdc x skin effect ratio       the enclosure thickness  may  need to  be increased slightly
                                                       and the diameter reduced accordingly.
                                                                   (2.873 x lo4) (1 x 12 x 2.54)
                                                       Now,  R,m  =
                                                                 IT x (59 - .27) x 2.54 x 0.27 x 2.54
                                                               = 0.273 x   aft
                                                       and RdcgSt = 0.273 x   [l  + 3.96 x   x (95- 20)]
                                                               = 0.273 x  10-6[1 .297]
                                                               = 0.354 x 10-6 aft
                                                       and skin effect ratio from Figure 31.1 7
                                                       for  jw  or 375.8
                                                                 0.27
                                                                  59
                                                       and   tld = - 0.0046
                                                                     =
                                                       *We  have  the  conductor  to  be operating  at  its  maximum
                                                       permissible temperature,  although  it  may  be  operating  at
                                                       less than this and generating less heat than assumed.
                                                       tHere  also  we  have  assumed  the  maximum  operating
                                                       temperature although the enclosure may be operating at less
         Figure 31.16  IPB conductor and enclosure for Example 31.1   than this.
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