Page 312 - Practical Power System and Protective Relays Commissioning

306 Practical Power System and Protective Relays Commissioning

I set 5 0:2 I N HV side
500 3 10 3 1
5 0:2 3 p ffiffiffi 3
3 3 500 600
5 0:1924 A
5 192:24 mA

From the CT magnitude curve check the value of I m at V S 5 180
V-I mag 5 45 mA.
The total operating current of the relay 5 I Rv 1 I RI 1 I CT HV side mag 1 I CT
LV side mag 1 I R CT neutral mag and total operating current 5 I total .
VA 1
I RV 5 5 5 5:55 mA
V 180
I RI 5 30 mA
I total 5 5:55 1 30 1 45 1 45 1 45 5 170:55 mA

Then we need a shunt resistor calculated as follows:
I of R shunt 5 192.24 170.55 5 21.69 mA
180
R shunt 5 5 8219:17 Ω
0:02169

I total 5 I RV 1 I RI 1 I CT HV mag 1 I CT LV mag 1 I CT neutral mag 1 I R shunt
5 5:55 1 30 1 45 1 45 1 45 1 21:69
5 192:24 mA

Subchapter 19.6

Generator Protection Setting

19.6.1 INTRODUCTION
As generators are a very costly component of the power system they should
have a good protection system. The complexity of this protection system
varies depending on the size of the units. The settings of these protections
are set at different values depending on the utilities setting polices. Here we
provide some examples of generator protection settings.

19.6.2 WORKED EXAMPLES

Worked Example 1 (as shown in Fig. 19.6.1)
Power station data:
Generator unit (G)