Instruction manual

MODBUS System Set Up
34 025-9209
A
VI
N 256
S
MM
SS
ML
HL
HL
L
=
×
×
−






×
−
−






+
The sample rate is returned in 10ths of a second. The default sample rate is 10, or 1 sample
per second. The time period (in seconds) over which the accumulation took place can be
calculated as:
PN
R
10
=
×
÷
Once the scaled average value and time period are known, the total can be calculated by
multiplying the average value by the time period. If the time units for the sensor are
something other than seconds, a conversion factor will also have to be included.
T
AP
SecondsPerSensorTimeUnits
=
×
Here is an example. We will use the same sensor that we used in our discussion of analog
input configuration. It measures water flow and outputs a current of 4mA for a flow of 0
gal/min and a current of 20mA for a flow of 100 gal/min. When the accumulator is read, we
get a value of 12345678, number of samples of 66666 and sample rate of 20. Therefore, we
know the following:
V = 12345678 M
H
= 100 gal/min
N = 66666 samples S
L
= 4 mA
R = 20 S
H
= 20 mA
M
L
= 0 gal/min I = 20 mA
Plugging these into the equation for the scaled average value results in the following
equation.
A
12345678 20mA
65432 256
4mA
100gal / min 0gal / min
20mA 4mA
0gal / min=
×
×
−






×
−
−






+
()
A 14.74mA 4mA
6.25gal / min
1mA
0gal / min=−×






+
A 67.13
g
al / min
=
The time period over which the value was accumulated is:
P 65432 20 10
=
×
÷
P 130864Sec
=
Finally, the total scaled accumulated value is: