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41-748P
For example, from Figure 8 page 15, an output of 0.6
per unit kW on 30# hydrogen pressure curve is -0.4
per unit reactive kVA. Therefore,
(
)
(
kVA
=
0.6
C
= 0.721 per unit
θ
and,
=
Tan 1
–
Converting to the impedance curve:
2
Vt
Z
=
----------------- -
(
)c
kVA
|Z| = 1.39 per unit
Since the angle remains the same, the impedance
plot conversion is:
∠
1.39
33.6°
, as shown in Figure 9, page
Z
=
–
20.
Continue this process with other points from curves
such as those of figure 8 until a complete curve like
that of figure 9 is obtained. After plotting the steady-
state stability limit and the machine capability curves
on the R-X diagram, plot the impedance unit circle
between the stability limit and the capability curve.
(Note in Figure 9 the relay circle cannot be plotted
within the 60# -V
= 0.95 curve, since the machine is
T
beyond the steady-state stability limit for these condi-
tions.) This plot defines the desired reach Z
radius R of the relay circle. Then use the following
procedure to select tap settings.
1000 kV
Z base
=
--------------------------------------- -
(
where
Z
= one per unit primary ohms/as seen from
base
the relay
kV
= rated phase-to-phase voltage of the
machine
kVA = rated kVA of the machine
R
= the current transformer ratio
C
R
= the potential transformer ratio
V
The actual settings, Z
and Z
A
Z
= (Z
per unit) X (Z
A
A
10
2
2
)
– (
)
+
0.4
0.4
–
------- -
=
–
33.6°
0.6
2
1.0
=
-------------- -
0.721
A
2
(
)
R C
)R V
kVA
, are:
C
)
base
Z
= (Z
C
C
(2R-Z
where R = radius of circle in per unit.
The tap-plate settings are made according to
equations:
(
Z A or Z C
where:
T
= compensator tap value
S
= autotransformer primary tap
M
= autotransformer secondary tap value
(M is a per-unit value determined by taking the sum
of the values between the L and the R leads. The
sign is positive when L is above R and acts to lower
the Z setting. The sign is negative when R is above L
and acts to raise the Z setting).
The following procedure should be followed to obtain
an optimum setting of the relay:
1. Select the lowest tap S which give a product
of 18.6S
product of 6S
2. Select a value of M that will most nearly
make it equal to:
TS
----- - 1.
M
=
and
Z
If the sign is negative, then the M taps are connected
with the R lead above the L lead to raise the setting.
(2)
5.3 SAMPLE CALCULATIONS
Assume that a KLF relay is to be applied to the fol-
lowing machine:
3-phase, 60 hertz, 3600 rpm, 18 kV, rated at 0.9 pf,
183,500 kVA at 45#H
R
= 1400/1
C
If the recommended setting from Figure 9 is used:
Z
per unit =
A
Z
per unit =
C
(The relay circle in Figure 9 was obtained by trial an
error using a compass to get the desired radius and
(3)
offset.)
KLF Loss-of-Field Relay
per unit) X (Z
) =
base
) X (Z
)
A
base
TS
)
=
------------- -
±
1
M
greater than desired Z
A
greater than desired Z
C
–
.
2
R
= 150/1
V
1.68
2R - Z
= 2(0.94) -1.68 = 0.20
A
(4)
(5)
and a
A
.
C
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