Page 74 - Electric Machinery Fundamentals
P. 74

5U   ELECTRIC MACH1NERY FUNDAMENTALS

                                         S =  VI                       (1- 62)

           The units of apparent power are volt-amperes (VA), where  I VA =  1 V X  I A. As
           with  reactive  power, apparent power is given a distinctive set of units to  avoid
           confusing it with real and reactive power.

           Alternative Forms of the Power Equations

           If a load has a constant impedance, then Ohm's law caD be used to derive alterna-
           tive expressions for the real, reactive, and apparent powers supplied to the load.
           Since the magnitude of the voltage across the load is given by
                                         V= [Z                         (1-63)
           substituting Equation (1-63) into Equations (1- 60) to  (1- 62) produces equations
           for real, reactive, and apparent power expressed in terms of current and impedance:
                                       p  =  / 2ZCOS e                 (1-64)
                                       Q = / 2Z sin e                  (1-65)

                                       S =  / 'Z                       (1- 66)
           where Izi s the magnitude of the load impedance Z.
                   i
                Since the impedance of the load Z can be expressed as
                             Z = R + jX  = IZI cos e + jlZI in e
                                                      s
           we see from this equation that R  = IZI cos e and X = Izi sin e, so the real and
           reactive powers of a load can also be expressed as

                                          P =  [' R                    (1- 67)
                                         Q =  [ 2X                     (1-68)
           where R is the resistance and X is the reactance of load Z.


           Complex Power
           For simplicity in  computer calculations, real and  reactive  power are sometimes
           represented together as a complex power S, where

                                       s =  P + j Q                    (1-69)
           The complex power S supplied to a load can be calculated from the equation
                                        S =  VI'                       (1-70)

           where the asterisk represents the complex conjugate operator.
                To understand this equation, let's suppose that the voltage applied to a load
           is V  = V L  IX  and  the current through the load is I  = / L  [3.  Then the complex
           power supplied to the load is
   69   70   71   72   73   74   75   76   77   78   79