Protective Elements
Example
Let us assume the following impedance values (each given in polar coordinates, i. e. as norm [vector length] and
angle):
• ·
X'
d
= 3.6 Ω ∠ 90°
• ·
X
T
= 2.04 Ω ∠ 90°
• ·
Z
S
= 8.9 Ω ∠ 85°
This gives the following:
• ·
Z
=
X
T
+
Z
S
≈ 10.9 Ω ∠ 86°
Therefore one gets the following:
• · Mho Pos.Imp.Reach = |
Z
| = 10.9 Ω
• · Mho Pos.Imp.Angle = tan
-1
(
Z
) = 86°
• · Mho Offs.Imp.Reach =
X'
d
= 3.6 Ω
• · Mho Offs.Imp.Angle = −90° = 270°
For the calculation of the blinder distances we again neglect the R-parts, i. e. we set |
Z
S
| ≈
X
S
.
With δ
S
=120° as stability limit we get the following:
• · Blinder B = ½⋅(
X'
d
+
X
T
+
Z
S
)⋅tan(90° − ½δ
S
) = ½(3.6 Ω + 2.04 Ω + 8.9 Ω)⋅tan(90° − ½⋅120°) ≈ 4.2 Ω
With δ
A
=240° (to get the same stability limit 120° for power swings during motoring of the generator):
• · Blinder A = ½⋅(
X'
d
+
X
T
+
Z
S
)⋅tan(90° − ½δ
O
) = ½(3.6 Ω + 2.04 Ω + 8.9 Ω)⋅tan(90° − ½⋅240°) ≈ −4.2 Ω
With
f
S,max
= 2.0 Hz it follows:
To Next Page
