Page 994 - Industrial Power Engineering and Applications Handbook
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An  isolated phase bus system  311939

            10           I                                            3.133 x  IO-"  x  IO"
                \        :
                                                                 2n
            09           I                              ...  Spz,, = -     1  x50
                                                                = 1.26 cm or  12.6 mm
            08                                          and  6pp8s 12.6 1/12.16
                                                                =
           to/                                                  = 14.01 mm
          0                                              Since most of the current will tlow through SP, ;I  thicker
           $06
          %                                             conductor will only add to the bulk and cost of the tube
          c                           penetration)      without  proportionately  raising  its  current-carrying
           5 0.5
           k                                            capacity.  A  greater  thickness  does  not  assist  in  heat
           2                                            dissipation, as the heat  is dissipated more quickly from
           $04               lie = 0.368                the outside than  the inside surface of a body.
          c
          _m
          $03                                           31.4.2  Thickness of enclosure
                                                        For a near-total  shielding of  the  field produced  by  the
            02
                                                        main  conductors  (i.e. for  I,  - I,  to  be  very  low), it  is
                                                        essential to have the thickness of the enclosure as near to
            01                                          6,  as  possible.  But  this  may  prove  to  be  a  costly
                                                        proposition. In addition, a higher induced current in the
             0                                          enclosure  will  also mean  higher  losses. This  has  been
               0        1 6P       26P       36P        established by computing the cost of  the enclosure and
                         Depth from surface t  +
                                                        capitalizing the cost of losses for minimum losses in the
         Figure 31.11  Skin effect in tubular isolated conductors in   enclosure
         terms of  depth of  penetration Sp
                                                          Thickness of tubular conductor (I j   = rrI2
                                                             Depth of penetration   )
         aluminium, grades EIE-M and E-91E, we have worked                         t  =  IT  6,   (31.6)
         out the likely values of 6,  at an operating temperature of   or              -
         85°C a\ follows:
                                                        But this will  also prove to be a costly proposition.
         1 For grade EIE-M                               A smaller surface area may require forced cooling and
                                                        an extra cost for the same while a larger enclosure would
            pzo = 2.873 x  IO-"  a cm                   mean a higher cost for the enclosure itself. Considerations
                                                        of cost will thus determine a larger enclosure or a forced
            azo = 4.03 x  IO-?  per  "C                 cooling. Two factors have been considered relevant here:
         :.   px4 = plo  [I  + 4.03 x  IO-'  x (85-ZO)]
                                                         Loss factor: This is a function of  losses  (W, in  the
                                                                                             j
               = p30 x  1.26195 R cm                     main conductor (W,) and the enclosure (W?). A curvc
         Therefore depth ot penetration at 20°C          as illustrated in Figure 3 1.12 can be established between
                                                          the  losses  versus  thickness  (t) of  the  enclosure  by
                                                         experiments.
                                                         Optimization factor: This is a function of the cost of
                                                         enclosure  for  different  thickness.  1,  the  cost  of  the
              = 1.207 cm or  12.07 inm                    cooling  system  (if  cooling  is  considered  necessary)
                                                          and  the  capitalized  cost  of  the  losses  at  different
         and at 85°C
                                                          thickness t. A curve as shown in Figure 3 1.13.  rather
         GpgT  =  12.07.\/1.26195                         similar  to  that  in  Figure  31.12,  can  be  established
                                                          theoretically between the total cost of the IPB system
             = 13.56 mm                                   versus t.
         (The suffix refers to the temperature.)           The above two curves will help optimize the thick-
                                                          ness, t, of  the enclosure for a total  rninirriurri ~osl oP
         2 For grade E-91E                                the system. Enclosure losses may not be lowest at this
               pZn = 3. I 33 x  I OF  cm                  thickness as shown  in  Figure 31.12, but  they  would
                                                          maintain  the temperature rise of the enclosure within
               a2,] = 3.63 x  lo-'  per  "C               limits. The magnetic field in the space. being already
          :.   ps5 = pzo  ( 1  + 3.63 x  IO-'  x 65)      very low, would require no other measure. Moreover.
                                                          a  small  field  in  the  space  may  causc  only  a  small
                  = p20x I  236 !2 cm                     amount of  heat  in  nearby  structures,  which  may  be
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