Examples on Generator Protection

Generator Protection Part 13
Examples on Generator Protection
Example 1 :The neutral point of a 11 kV alternator is earthed through a resistance of 12Ω, the relay is set top operate when there is out of balance current of 0.8 A. The C.T.s have a ratio of 200/5. What percentage of the winding is protected against earth faults. What must be the minimum value of earthing resistance required to give 90% of protection to each phase ?
Solution : The given value are,
               VL = 11 kV              R = 12 Ω              C.T. ratio = 2000/5
                io = relay current = 0.8 A
...              Io = minimum operating line current (C.T. primary)
                     = iox (2000/5) = (0.8 x 2000)/5 
                     = 320 A
                  V = line to neutral voltage = VL/√3
                      = (11x103)/√3 = 6350.8529 V
...                 % Winding unprotected = (RIo/V) x 100 = (12x320/6350.829) x 100
                      = 60.46%
...                %Winding protected = 100 - 60.46 = 39.53%
       Thus with R = 12 only 39.53 % winding is protected
       It is necessary to give 90% protection.
...              % Winding unprotected = 100 - 90 = 10%
...              10% = (RIo/V) x 100 
...               10 = (Rx320/6350.8529) x 100 
...               R = 1.9846 Ω
       This is the minimum value of resistance to give 90% protection to the largest machine.
Example 2 : A 50 MVA, 3 phase, 33 kV synchronous generator is protected by the Merz-Price protection using 1000/5 ratio C.T.s. It is provided with restricted earth fault protection with the earthing resistance of 7.5 Ω . Calculate the percentage of winding unprotected in each phase against earth faults if the minimum operating current of the relay is 0.5 A.
Solution : The given values are,
                   VL= 33 kV      C.T. ratio = 1000/5      R = 7.5Ω
                    io = 0.5 A = relay current 
...                  Io = minimum operating current (primary)
                        = io x (1000/5) = (05 x 1000)/5
                        = 100 A
                     V = VL/√3 = (33x103)/√3
                         = 19052.55 V
...                   % Winding unprotected = (RIo/V) x 100
                         = (7.5x100/19052.55) x 100
                         = 3.936%
Example 3 : A 13.2 kV, 3 phase, 100 MW at 0.8 p.f. lag, alternator has reactance of 0.2 p.u. If it is equipped with a circulating current differential protection set to operate at least at 500 A fault current, determine the magnitude of the neutral grounding resistance that leaves the 10% of the winding unprotected.
Solution : The given values are,
                  VL = 13.2 kV       = 0.8       P = 100 MW           X = 0.2 p.u.
Now           P = √3 VL ILcos
...               100 x 106 = √3 x 13.2 x 103 x IL x 0.8
...                IL = 5467.33 A = I = full load current
       The p.u. reactance is given by,
       p.u. X = (IX)/V     where X = reactance per phase
...              0.2 = 5467.33X/(13.2x103 /√3)   where V = VL/√3
...              X = 0.2787 Ω per phase
       % of unprotected winding = 10%
...      Reactance of unprotected winding = (10/100) x 0.2787
                = 0.02787 Ω
       Voltage induced in 10% of unprotected winding
                 = (10/100) x V = (10/100) x (13.2x103 /√3) = 762.1023 V
Let this voltage be      v = 762.1023 V
               Z = √(r2 + x2)
where      Z = Impedance offered to the fault
                r = Resistance in neutral 
                x = Reactance of 10% of winding
Now         v = Voltage induced in 10%winding
                   = 762.1023 V
                i = Fault current = 500 A
...             √(r2 + x2) = 762.1023/500
...             √(r2 + (0.02787)2)= 1.5242
...               r2 +(0.02787)2= 2.3232
...               r2 = 2.3224
...                r = 1.524
       This is the required resistance in neutral earthing.

Example 4: An alternator stator winding protected by a percentage differential relay is shown in the Fig. 1. The relay has 15% slope of characteristics (I1 - I2 ) against (I1 + I2 /2). The high resistance ground fault has occurred near the grounded neutral end of the generator winding while the generator is carrying load. Thwe currents flowing at each end of the generator winding are also shown. Assuming C.T. ratio to be 500/5 A, will the the relay operate to trip the circuit breaker ?
Fig. 1

Solution : From the given current at two ends, let us calculate C.T. secondary current at two ends,
                    i1 = (300 + j 0) x (5/500) = 3 A
and               i2 = (340 + j 0) x (5/500) = 3.4 A
      The direction of currents are shown in the Fig. 2.
Fig. 2

      The current flowing through the relay coil is i1 - i2.
                   i1 - i2 = 3 - 3.4 = -0.4 A
While          (i1 + i2 )/2 = (3 + 3.4)/2 = 3.2 A
       From the characteristics of 15 % slope, corresponding to (i1 + i2 )/2 the out of balance current required is,
                   i1 - i2 = Slope x (i1 + i2 )/2
                   = 0.15 x 3.2
                    = 0.48 A
       This is shown in the Fig.3. Thus - must be more than 0.48 A i.e. above the line to operate the relay but actual point is located below the line in negative torque region. Hence the relay will not operate.
Fig. 3

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