7SR18 Applications Guide
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Example calculation
Case 1: Source X/R = 50 FL = 14 kA Case 2: Source X/R = 17.5 FL = 40 kA
I
F
= 14kA, X/R = 50 I
F
= 40kA, X/R = 17.5
Z
S
= 132,000 / (√3 x 14,000) = 5.443 Ohms. Z
S
= 132,000 / (√3 x 40,000) = 1.905 Ohms.
As the busbar X/R is known for both cases the X and R components of the source impedance may be found.
X
S
=Cos(Tan
-1
(1/X/R ))xZ
S
=-j5.442 Ohms X
S
=Cos(Tan
-1
(1/X/R ))xZ
S
=-j1.9046 Ohms
R
S
=Sin(Tan
-1
(1/X/R))xZ
S
=0.109 Ohms R
S
=Sin(Tan
-1
(1/X/R))xZ
S
= 0.0381 Ohms
Cable Impedance = 0.39 – j1.277 Ohms Cable Impedance = 0.39 – j1.277 Ohms
Total Impedance = 0.499 – j6.719 Ohms Total Impedance = 0.4281 – j3.1816 Ohms
The X/R for external fault = 13.46 The X/R for external fault = 8.91
Z
T
= √(0.499
2
+ 6.719
2
) = 6.737 Ohms Z
T
= √(0.4281
2
+ 3.1816
2
) = 3.210 Ohms
External Fault Level = 132kV / (root 3 x Z
T
) External Fault Level = 132kV / (root 3 x Z
T
)
= 11,312 A = 23,741 A
Both Cases should be considered when arriving at the CT minimum e.m.f. requirements
The X/R= ranges from 8.91 to 13.46. The through-fault level ranges from 11.312 A to 23.741 A.
As the bias break point is being set to 0.5 xIN the following CT formula is applicable:
201 ≤××=
X
forRsIVk FM
CT Requirements:
Case 1: Source X/R=50 FL=14 kA Case 2: Source X/R=17.5 FL=40 kA
Vk ≥ 1 x 11312/1000 x (R
LL
+ R
PH
+ R
CT
) Vk ≥ 1 x 23741/1000 x (R
LL
+ R
PH
+ R
CT
)
From the above, Case 2 requirements are more onerous and should be used to calculate the Vk minimum
required.
CT’s for Substation A: CT’s for Substation B:
Vk ≥ 1 x 23741/1000 x (1.95+0.05+5) Vk ≥ 1 x 23741/1000 x (3.5+0.05+5)
Vk ≥ 167 volts Vk ≥ 203 volts
The above figures are recommended for the relay, however the safety margin of 20% may be used if CT core size
makes fitting the CT into the switchgear chamber difficult. In the above example the absolute lower limits would
be:
Vk ≥ 167 / 1.2 = 140 volts Vk ≥ 203 / 1.2 = 170 volts
This 20% reduction is attributable to the fact the CT formulae were based on the saturation emf (e
sat
). The e
sat
of a
CT is always ≥ 120 % of the CT knee-point voltage. The relay would still remain stable for these CT knee-point
voltages as there are other safety margins built into the formulae.
These additional safety margins are:
• The CT requirements was based on three-phase fault levels, therefore only the single core run between
the relay and CT needed to be considered as the lead burden. The formulae used included the full lead
loop resistance.
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