Page 174 - Practical Power System and Protective Relays Commissioning

174 Practical Power System and Protective Relays Commissioning

For example, a 240 MVA SGT has a name plate rating of 20%;
20 3 100
Rating on 100 MVA 5
240
5 8:33%
2. MVA
Fault rating and fault in-feed are expressed in MVA. This is related to
percent impedance by the following:
Percent impedance 5 10; 000=MVA if the base is 100 MVA
For example, if the fault infeed at a particular substation is 35,000
MVA, then the source impedance, expressed as a percentage on 100
MVA base, will be 0.286.
Similarly, if it is calculated that the impedance to a particular fault is
2% (on 100 MVA), then the fault infeed will be 5000 MVA.
To derive the fault current (I f ), either:
a. Use I f 5 fault infeed (MVA)/(kV 3 1.732) kA
b. Or I f 5 10,000/(kV 3 1.732 3 Z%) kA
3. Ohmic impedance
To convert percent impedance to primary ohmic impedance:

Taking Z% 5 percent impedance on 100 MVA
10 4
Z% 5
MVA
V2
MVA 5
Z
4
10 3 Z
Z% 5
V 2
2
Then, primary impedance (Ω) 5 Z% 3 (kV /10,000)
For example: if Z% 5 2, and kV 5 400, then Z prim 5 32 Ω.
2
Conversely, Z% 5 Z prim (10,000/kV ).
To convert primary ohmic impedance to secondary ohmic impedance,
this can be done by using the following impedance conversion factor
This depends upon the VT and CT ratios, as follows:
Z sec 5 Z prim 3 ðCT ratio=VT ratioÞ
For example, for 400 kV, with 2000/1 CTs:

Z sec 5 Z prim 3 ð2000 3 110=396; 000Þ
General notes for calculating the impedance to fault and fault current
are described here. To calculate the fault current, it is first necessary to
draw out the network involved. From this the impedance network is