Transducer Output
section of the
Transducer Test Screen
should read 1000 * 0.998
=
998.00 Watts
The
Accuracy
displayed in the
Transducer Output
section would be equal to the
following,
(998.0 - 999.98 / 999.98) * 100 = % accuracy or
-0.198
%
If this were a 0.2 % transducer, then the firmware would compare the accuracy values
between the Setting Screen and the Test Screen and would display
PASS
in the
Transducer Output
section of the test screen.
Note: All of the calculations are very similar when testing VAR 1 1/2
Element transducers.
The primary difference is replacing the COS function with the SIN
function. For example, let us assume that the test angle for the VAR transducer is 30
degrees. This is an important point, since if they were in phase, the SIN of 0
°
is 0, thus
the Var contribution is 0 at the in-phase condition. Only rotating the phase angle 30
degrees do we create the measured output VAR's would be,
120.00 * 4.166 * SIN 30 + 120.00 * 4.166 * SIN 30 = 499.92 VARs
3.8.10.7.4
Watt/VAR 2 Element
This transducer is normally used in three-phase, three wire delta application, which
requires 2 voltages and 2 currents to test. Normally, the PT's and CT's are connected to
A and C phases. The MPRT will automatically select two voltage and current channels,
V1
,
V3
,
I1
and
I3
(in the event that there is no V3/I3 channel, then V2 and I2 will be
used). The test will initially start at the default voltage value, that is set in the
Default
Setting Screen
, for each Voltage output. Thus, for a default voltage setting of 120 volts,
V1 will be set to 120 Volts at an angle of 0
°
, and V3 will be 120 Volts at 300
°
(delta
connected PT's). This assumes that the default phase angle is 0 - 360 Degrees lag, and
not +/- 180 degrees. If the +/- 180 degree phase angle option is used, then V3 will be at
+60
°
. I1 and I3 will be phase shifted 30
°
with their respective voltages, or I1 at 30
°
lag
and I3 at 270
°
lag (or + 90
°
). When the user inputs the
MAX.
Watts value in the
Transducer Setting Screen
, the firmware can calculate the required test currents for full
scale value.
The formula required to calculate Watts for 2 Element transducer is,
V1 *
* I1 * (COS 30
°
+
∅
) + V3 *
* I3 * (COS 30
°
-
∅
) = Total Watts
Where
∅
is the incremental angular change between V1 and I1 and V3 and I 3.
Example: The default voltage is 120.00 Volts AC, and the user inputs 1000 Watts as
Max. Value. The current required for full scale output from the transducer is,
120
*
* I1 * COS 30
°
+ 120 *
* I3 * COS 30
°
= 1000 Watts
I1 = 500 Watts/(120 Volts *
* COS 30
°
) , or I1 = 500/180.00
Since I1 = I3, then I1 and I3 will be 2.7777 Amperes each
-
145
www
. ElectricalPartManuals
. com
