ISA T1000 PLUS User Manual page 72

DOC. MIE91093 Rev. 1.34 Page 72 of 145
ZB = (K1+K2)*ZN = 12,72 Ohm.
. Choose the parameters corresponding to point A. As Z = V/I,
taking V as the reference, on the X axis the I angle is – 90°,
or 270° ; on the –X axis it is 90°. Now, as we want to move
from the point with 0 impedance down to A, it is apparent that
the test is best performed by setting a constant value for the
current, say IT, and modifying the voltage. The zero
impedance point corresponds to V = 0; the point A
corresponds to the voltage VA = ZA * IT.
In our instance, if we choose It = 5 A, then VA = 5.78 V. In
conclusion, the test of point A is conducted by:
- Setting the current at IT = 5 A;
- Setting the I angle at 90°;
- Increasing the voltage from 0 until the relay trips.
Next point to be tested is B: how will we perform the test? As
per point A, we can set the starting current at 5 A; but, what
about the starting voltage? As you can see in the diagram, you
have to start from a voltage VS that must be HIGHER than
VS = ZB * IT = 63.6 V,
Then DECREASE the voltage rather than increasing it, until the
relay trips. In our instance, starting from 80 V and decreasing
it would do.
Next, we want to test point C: how to do it? One could
compute the corresponding angle, and perform the test as per
A or B; however, minor errors in the angle would cause big
errors; C could also be missed. In this situation, the way is to
compute the current and voltage corresponding to ZC, and
modify the phase angle between them. In our instance:
ZC = ZA + (ZB-ZA)/2 = 6.94 Ohm;
IT = 5 A; VC = 34.68 V.
The test is performed applying these current and voltage at
the angle of 0°, and then increasing the angle until the relay
trips. Note that if you use the same current and voltage, start
from 180° and decrease the angle, you get the symmetric
point C'.