Generator Tutorial

Copyright © of Alstom Grid UK Limited

1. For the system below, provide the following:

a. C.T. and V.T. ratios b. Single line connections for complete protection c. Relay ANSI numbers

140MVA132/13.8kV

140MVA13.8kV

40 Ohms

EXC

10MVA13.8/6.6kV

132kV


2. Calculate the size and rating of transformer and loading resistor for

the distribution transformer earthing of the generator below.

The Generator is 75,000 kVA, 11.5kV at 60Hz having surge protective capacitors at its terminals. The distribution transformer is to have 110V secondary winding NOTE 1) The rated primary voltage of the distribution transformer should

be approximately √3 times the generator line-neutral voltage, in order to avoid excessive magnetizing inrush when an earth fault occurs.

2) Overvoltages due to field forcing are not considered. 3) Copper losses in the distribution transformer are not

considered. 4) The distribution transformer rating can be reduced by the

following factors depending on its required operational time.

0.25uF

0.25uF

0.25uF

R

0.32uF

0.32uF

0.32uF

0.06uF

0.004uF

0.06uF

0.004uF0.06uF

0.004uF


1 Minute – 4.7 5 Minutes – 2.8 30 Minutes – 1.8 1 Hour – 1.6 2 Hours – 1.4 For this example a 1 minute rating has been considered. Note It is preferable to disconnect the generator from the system and remove excitation immediately upon the occurrence of a fault in order to confine the damage as much as possible. On small systems, some operators prefer to have the earth fault relay sound an alarm, giving the system operator a chance to make provision for the loss of the generator.


3. If the generator shown in question 2 is Resistance earthed,

calculate the ohmic size of resistor to limit the earth fault current to 200A and the percentage of stator winding protected by an overcurrent relay in the generator neutral.

Effective primary setting of relay = 15% Current Transformer ratio = 250/1


Solutions

1.

For Faults here BusBar Protection operates and initiates relay which trips generator

132kV

625/0.58

6000/20

1000/1

20/1

6000/20

6000/20

6000/20

6000/20

200/1

64A

EXC

32

50N

20/1

87G-T

87G

20/1

4646

20/1

40

U.T.

87

625/1

625/1

625/1

140MVA132/13.8kV

6000/20

13.8kV/110V

20/1REF

450/120/1

51

625/1

450/1

50N

B.B.

87

51 50i

950/0.851000/1

1000/1

51S

50N

10MVA13.8/6.6kV


Relay References: 40 Field Failure 46 Negative Phase Sequence 51 IDMT O/C 51I Interlocked O/C 64A Negative Biased Earth Fault 50N Instantaneous Earth Fault 50Nref Restricted Earth Fault

87UT Unit Transformer Differential 87BB Busbar Protection 87GT Overall Differential 32 Reverse Power Protection 51S Standby Earth Fault

Generator – Transformer Overall differential Connections

Considering the fault shown above. If the infeed from the external system is very small the maximum bias will be obtained by connecting the overall differential between the generator and unit transformer C.T.s

Op

87GT

BiasBias


Connections as shown above are used where infeed from the system is large in comparison with the generator contribution.

Op

87GT

Bias Bias


2. The zero-sequence capacitance per phase of each circuit

component is listed below and totaled:

Generator winding 0.32 µF Generator Surge capacitor 0.25 µF Generator Leads 0.06 µF Power Transformer 0.004 µF Total 0.634 µF The capacitive reactance can be found by

fCπ21

Therefore, 610634.06021

−××××π

= 4184 Ohms

The per phase charging current is as follows :-

So, 41846620 = 1.58A

The steady state 3∅ capacitive kVA developed is

358.135.11

×× =31.5kVA

NOTE: During a phase to earth fault on a high impedance earthed system the fault current is approximately 3x steady state the capacitive charging current (i.e. 4.74A) To keep transient overvoltages to a minimum the loading resistor should also dissipate 31.5kVA i.e. equal to the charging kVA as shown below:-


Choosing a transformer of 11.5kV primary and 110V secondary the open circuit secondary voltage will be 110/√3 = 63.5V The current rating of the resistor will be

5.6331500 = 496A and the

resistance 496

5.63 = 0.128 Ohms

During an earth fault the transformer dissipates:- AkV 74.4

35.11

× =31.5kVA

However, the transformer rating is a function of its rated voltage and rated current, which are 11.5kV and 4.74A respectively. Therefore:- Transformer continuous rating = AkV 74.45.11 × = 54.5kVA In this example the transformer need only be rated for 1 minute and hence it can be de-rated accordingly:- Actual transformer rating =

7.45.54 kVA = 11.6kVA

VTRANSIENT

Neutral to Earth

Faulted Phase

Resistor kW Charging kVA

(% EØ/N PEAK)

Unfaulted Phase

400

.2 .4 .6 .8 1.0 2.0

300

200

100


3. The degree of the generator winding protected may be

determined as follows:

If V = normal phase-neutral voltage. R = Ohmic value of earthing resistance = Ω==

− 332006.6 kV

IenVp

IA = Primary fault setting of relay in amps. IG = Generator full load current IP = Primary fault setting of relay in percent of generator full

load current. IR = Rating of earthing resistance in percent of generator full

load current. Then percentage of winding unprotected = 100×

×V

RIA

= %5.18100

660033)25015.0(

=×××

and percentage of winding protected = 1001 ×⎟⎠

⎞⎜⎝

⎛ ×−

VRIA

= 100)1875.01( ×− = 81.25%

or using ratings in terms of generator full load current

= 100×=G

R IR

VI

= 100

5.113100075

336600

×

××

= 1003756200

×

= 5.3%


And percentage of winding protected = 1001 ×−R

P

II

= 1003.53756

5.371 ×−

= 100

3.5995.01 ×−

= 1001875.01 ×− = 81.25%

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Generator Tutorial