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STX Signal Transmitter Installation and Operation ... - Kistler-Morse

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Chapter 3. St<strong>and</strong>-Alone <strong>STX</strong> Analog Calibration <strong>and</strong> Setup<br />

Calibration by Adding a<br />

Known Quantity of Material<br />

This calibration method does not require<br />

the vessel to be completely empty. The<br />

principle follows.<br />

Zero [point (1) in Figure 3-3] is set to the low<br />

current output (4 mA or 0 mA). A known<br />

quantity of material, representing at least<br />

25% of the vessel’s total capacity, is added to<br />

the vessel. Span [point (2) in Figure 3-3] is<br />

set to a current output proportional to the<br />

change in weight. Points 1 <strong>and</strong> 2 define the<br />

dashed straight line. The slope of the line is<br />

called the Scale Factor, which is calculated<br />

internally. Zero [point (3) in Figure 3-3] is<br />

then set to a current output proportional to<br />

the estimated amount of material in the<br />

vessel, adjusting the ‘location’ of the line to<br />

the solid line.<br />

The calibration is reasonably accurate,<br />

because it is based on the known difference<br />

between the currents (based on the difference<br />

in weights) <strong>and</strong> counts. However, if the<br />

estimated weight is incorrect, the actual<br />

‘location’ of the line is incorrect, resulting<br />

in errors in output. The greater the error<br />

in the estimated weight, the greater the<br />

resulting error.<br />

The accuracy of the calibration improves<br />

the greater the known quantity of material<br />

added during the calibration procedure. For<br />

example, adding 50% of the vessel’s total<br />

capacity results in greater accuracy than<br />

adding 25% of the total capacity.<br />

Example: Operating in 4-20 mA mode, maximum<br />

vessel live load = 100,000 lbs, current live load =<br />

25,000 lbs.<br />

1. Input Zero point as 4 mA.<br />

2. Add 50,000 lbs to vessel.<br />

3. Change in weight as percentage of maximum<br />

live load<br />

= 50,000/100,000 = 50%,<br />

corresponding to 12 mA from Table 3-1.<br />

4. Input Span point as 12 mA.<br />

5. Estimated current live load as percentage of<br />

maximum live load<br />

= (25,000 + 50,000)/100,000 = 75%,<br />

corresponding to 16 mA from Table 3-1.<br />

6. Input Zero point as 16 mA.<br />

Analog calibration is complete.<br />

0-20 mA Mode 4-20 mA Mode<br />

Current % of Current % of<br />

Output Maximum Output Maximum<br />

(mA) Live Load (mA) Live Load<br />

0 0 4 0<br />

1 5 5 6.25<br />

2 10 6 12.5<br />

3 15 7 18.75<br />

4 20 8 25<br />

5 25 9 31.25<br />

6 30 10 37.5<br />

7 35 11 43.75<br />

8 40 12 50<br />

9 45 13 56.25<br />

10 50 14 62.5<br />

11 55 15 68.75<br />

12 60 16 75<br />

13 65 17 81.25<br />

14 70 18 87.5<br />

15 75 19 93.75<br />

16 80 20 100<br />

17 85<br />

18 90<br />

19 95<br />

20 100<br />

Table 3-1. Proportional Current Outputs<br />

Indicated<br />

Current<br />

Current Proportional<br />

to Estimated<br />

Actual Load<br />

Corrected Calibration Line, Same Slope<br />

Zero<br />

(3)<br />

Current Proportional<br />

to Change in Load (2)<br />

Add<br />

Known Weight<br />

4 mA (4-20 mA mode)<br />

or 0 mA (0-20 mA mode)<br />

Calculated<br />

Zero Counts<br />

(1)<br />

Zero<br />

Span<br />

Slope =<br />

Change in Current<br />

Change in Counts<br />

Calculated<br />

Span Counts<br />

Figure 3-3. Calibration by Adding a Known Quantity of Material<br />

Weight<br />

<strong>Signal</strong><br />

in Counts<br />

3-3

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