Page 135 - Handbook of Electrical Engineering
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INDUCTION MOTORS     117

                  Table 5.7.  Per-unit resistances and starting-to-full-load current ratio for HV four-pole motors

                  Rated power   Slip (pu)     R 1       R 20       R 21     R c     I s /I n
                  (kW)
                      630        0.00828    0.00809    0.00688   0.0285    39.01    5.84
                      800        0.00932    0.00804    0.00764   0.0288    45.16    5.45
                     1,100       0.01050    0.00780    0.00844   0.0287    52.88    5.00
                     1,500       0.01120    0.00742    0.00889   0.0280    59.20    4.66
                     2,500       0.01120    0.00650    0.00878   0.0256    65.29    4.35
                     5,000       0.00895    0.00495    0.00713   0.0207    62.59    4.40
                     6,300       0.00785    0.00441    0.00633   0.0189    59.01    4.53
                     8,000       0.00667    0.00386    0.00545   0.0169    54.24    4.71
                    11,000       0.00515    0.00308    0.00450   0.0143    53.06    5.02


                   Table 5.8.  Per-unit reactances and starting-to-full-load torque ratio for HV four-pole
                   motors
                   Rated power   Slip (pu)    X 1      X 20      X 21     X M     T s /T n
                   (kW)
                       630        0.00828    0.109    0.120     0.0594    3.213    0.934
                       800        0.00932    0.126    0.112     0.0546    3.403    0.828
                      1,100       0.01050    0.147    0.104     0.0501    3.635    0.697
                      1,500       0.01120    0.165    0.0996    0.0474    3.834    0.593
                      2,500       0.01120    0.182    0.0976    0.0460    4.085    0.473
                      5,000       0.00895    0.177    0.106     0.0498    4.242    0.391
                      6,300       0.00785    0.173    0.111     0.0528    4.243    0.377
                      8,000       0.00667    0.155    0.119     0.0570    4.217    0.365
                     11,000       0.00515    0.135    0.134     0.0647    4.145    0.350


           At standstill the slip is 1, therefore the equivalent impedance is,

                                                    (R 22 + jX 22 )(R 33 + jX 33 )
                                Z 231 = R 231 + jX 231 =                                   (5.7)
                                                    R 22 + jX 22 + R 33 + jX 33
           At full-load the slip is s, therefore the equivalent impedance is,

                                                   (R 22 /s + jX 22 )(R 33 /s + jX 33 )
                             Z 230 = R 230 /s + jX 230 =                                   (5.8)
                                                   R 22 /s + jX 22 + R 33 /s + jX 33

                 Taking the real and imaginary parts of each equation separately yields the four equations
           required for the solution. The given values are R 230 , X 230 , R 231 , X 231 and the full-load slip s.The
           solution is the set of values R 22 , X 22 , R 33 ,and X 33 .

                 The iterative solution can be carried out by one of various algorithms, for example New-
           ton’s approximation to find roots, steepest descent to find a minimum quadratic error, rough search,
           successive substitution. Newton’s method in four dimensions works reasonably well, although insta-
           bility can set in if the incremental changes are allowed to be too large. Hence some ‘deceleration’ is
           required to stabilise the algorithm. The method of successive substitution is more efficient, but also
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