SERVICE MANUAL R5888B
QUADRAMHO Chapter 1
Page 13 of 27
19.2 Data
Line to be protected: 10km, 60kV, single circuit between MAY 60kV and JUNE
60kV substations.
Note: Sample calculations will be presented for MAY relay only.
Special requirement: Line protection must be able to clear ground faults with up to
40 ohms fault resistance.
Assumptions: 150kV fault level is infinite at each bulk supply point.
Transformers have an
X
/
R
ratio equivalent to 87°.
Line data:
Positive sequence series impedance (Z
1
) = 0.5 /70° ohm/km
= (0.171 + j0.47) ohm/km
Zero sequence series impedance (Z
0
) = 2.0 /68° ohm/km
= (0.749 + j1.85) ohm/km
Voltage transformer ratio = 60,000/110 volt
Current transformer ratio = 200/5A
Line length = 10km
Relay selected: Quadramho type B, with quadrilateral ground fault
impedance characteristics. This is because the line is short
and a high resistive coverage is required for ground faults.
Scheme to be used: Permissive underreach transfer tripping scheme which
requires only a single signalling channel. Alternatively the
Zone 1 extension scheme may be selected if a signalling
channel is not available.
19.3 Calculation of maximum and minimum source impedance at MAY 60kV
a) Minimum source impedance is when both 50MVA transformers at MAY are
switched in and the 50MVA transformer at APRIL is switched in.
Positive and zero sequence impedances of each 50MVA transformer:
= 0.125 x
60
2
= 9.0
/87° = (0.47 + j8.99) ohms
50
Positive sequence impedance of each APRIL – MAY line:
= 30 x (0.171 + 0.47) = (5.13 + j14.1) ohms
Zero sequence impedance of each APRIL – MAY line:
= 30 x (0.749 + j1.85) = (22.5 + j55.5) ohms
Positive sequence impedance of APRIL – MAY line in parallel plus the APRIL
transformer impedance:
=
(
5.13 + j14.1)
+ (0.47 + j8.99) = (3.04 + j16.04) ohms
2
Zero sequence impedance of APRIL – MAY lines in parallel plus APRIL
transformer impedance:
=
(22.5 + j55.5)
+ (0.47 + j8.99) = (11.7 + j36.74) ohms
2
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