Page 300 - Industrial Power Engineering and Applications Handbook
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Protection of  electric motors  12/279
          The heat derived from this equation may be less than   i.e. 2.5 or 250%
        the  minimum  heat  (equation (12.5)) or even  more than   which  is even more than the maximum heat obtained from
        the maximurn heat (equation (1 2.4)) depending upon the   equation (12.4). Equation (12.6) is thus more appropriate for
        se\perity  and  the  phase  disposition  of  the  negative   a  protective  device  and  reflects  the  effect  of  a  negative
        component  with  reference  to  the  positive  component.   sequence component in a motor winding  more precisely.
        This can be  illustrated by  the following examples.
                                                        Rotor power
        Example 12.4
        Referring to Examples 12.2 and 12.3 above, the heat produced   The negative sequence voltage sets up a reverse rotating
        according to the empirical formula is as follows:   field and the slip of the rotor becomes '2 - S', compared
                                                       to the positive sequence slip S. The motor will thus operate
        Hes =  (1'  + 6 x 0.4')                        under the cumulative influence of these two slips, where
        1.e.  1.96 or 196%                             power  output P  can  be expressed by  (see also  Section
                                                       2.3).
        which is the same as that derived in Example 12.2.
        Corollary

        One can thus easily obtain the significance of the factor 6 to   where
        represent the status of the most affected winding of the motor   I,, = positive sequence current in the rotor circuit, and
        in the  event  of  a voltage  unbalance resulting in a negative   I,,  = negative sequence current in the rotor circuit
        sequence  current  component.  For  more  clarity,  consider
        equations (12.4) and (12.6) to ascertain the similarity in both
        these  equations.  Since both must  represent  the  maximum   Rotor heat
        heating effect
                                                       The unbalanced voltage will produce an additional rotor
         :.  I:  + 6I:  =  I:  + I,'  + 2. I,  x  I,   current at nearly twice the supply frequency. For example.
                                                       for a 2% slip, i.e. a slip of  1 Hz, the negative sequence
        or    51,'  = 21,  x  I"                       stator current, due to an unhalanced supply voltage, will
        or     I,  = 215  x  I,, Le. 0.4 I,            induce a rotor current at a frequency of (2f-  1) = 99 Hz
                                                       for a 50 Hz system. These high-frequency currents will
        Thus these two methods will yield the same result for a negative   produce  significant  skin  effects  in  the  rotor  bars  and
        sequence component of 40%. If the negative sequence current   cause high  eddy current  and hysteresis losses (Section
        I,  is lower than 40%, the heat produced from equation (12.6)   1.6.2(A-iv)). Total rotor heat may be represented by
        will be lower than the minimum heat obtained from equation
        (12.5). In contrast, for  a  negative sequence component of
        more than 40%, the situation is likely to be reversed, since   =(I;  + 3;")          (1 2.9)
        the heat produced as in equation (12.6) will  be higher than
        the  maximum  heat  produced  by  equation  (12.4). See  the   (refer to curve 4 of  Figure  12.3) and  cumulative rotor
        following example for more clarity.            current

        Example 12.5
        Consider a negative sequence component of  15% and 50%
        respectively.
                                                       Example 12.6
                                                       For  example  12.2,  the  rotor  heat  at  40%  stator  negative
        (a) For a 15°/0 negative sequence component:   sequence current
           From equation (12.5)                            = (1'  + 3 x 0.4')
           He,(min.)  = (1'+  0.15'  +  I x  0.15)     i.e.  1.48 or 148%
           i.e.  1.1725 or  117.25%
                                                       Refer to curve 4 of  Figure 12.3.
           and from equation (12.6)
           He, = (1'  + 6 x  0.15')
                                                       12.3  Fault conditions
           i.e. 1.135 or 113.5%
        which  is  even  less  than  the  minimum heat  obtained from   These are conditions in which overheating of the machine
        equation (12.5).                               may not trace back to its own thermal  curves as in  the
        (b) For a 50% negative sequence component:     first case. The  temperature  rise  may  now  be  adiabatic
                                                       (linear) and  not  exponential  and  hence  rapid.  Now  a
           From equation (1 2.4)                       normal  thermal  protection  device  may  not  be  able  to
           He,(max.)  = (1'  + 0.50'  + 2 x  1 x 0.5)   respond as in the previous case. Some conditions causing
                                                       overheating  may  not  recessarily  be  fault  conditions.
           i.e. 2.25 or 225%
                                                       Nevertheless, they may require fast tripping, and hence
           and from equation (12.6)                    are  classified  in  this  category  for  more  clarity.  Such
           Heq   (1'  + 6 x 0.5')                      conditions may be one or more of the following:
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