If possible, the exact level of the tank should be
known and should be as large as possible (> 70 %).
Too low a level reduces the accuracy of the upper poi
nt (corresponds to the
full tank). An ammeter
must be connected to the current
pick-off at the electronic insert.
Let us presume that the level was determined for
90 %. Now the current value that corresponds to
the level of 90 % must be determined. The upper
current value can be adju
sted with the +/– keys.
The + key increases the value, the – key reduces the value.
The following must also be considered:
1. The lower current value (= empty tank, 0 %) is 4 mA.
2. The upper current value (= full tank, 100 %) is 20 mA.
3. This results in a measuring range of
16 mA for a change from 0 % to 100 %,
i.e. 0.16 mA increase in
the current for every 1 % increase in the level.
4. For a 90 % level, this is 90
% x 0.16 mA/% which equals 14.4 mA. This must be added to the
4 mA to achieve the current value to be set: 14
.4 mA + 4 mA = 18.4 mA. (You can also take
the upper current value and then subtract 10 % x 0.16 mA/% = 1.6 mA from 20 mA.)
If possible, the exact level of the tank should beknown and should be as large as possible (> 70 %).Too low a level reduces the accuracy of the upper point (corresponds to thefull tank). An ammetermust be connected to the currentpick-off at the electronic insert.Let us presume that the level was determined for90 %. Now the current value that corresponds tothe level of 90 % must be determined. The uppercurrent value can be adjusted with the +/– keys.The + key increases the value, the – key reduces the value.The following must also be considered:1. The lower current value (= empty tank, 0 %) is 4 mA.2. The upper current value (= full tank, 100 %) is 20 mA.3. This results in a measuring range of16 mA for a change from 0 % to 100 %,i.e. 0.16 mA increase inthe current for every 1 % increase in the level.4. For a 90 % level, this is 90% x 0.16 mA/% which equals 14.4 mA. This must be added to the4 mA to achieve the current value to be set: 14.4 mA + 4 mA = 18.4 mA. (You can also takethe upper current value and then subtract 10 % x 0.16 mA/% = 1.6 mA from 20 mA.)
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