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Optional device installation
Example:
The KCl calibration solution is 1.0 mS. at 25°C. The temperature correction factor for KCl
is 1.97%/°C. If the actual temperature of the KCl at the time of calibration is 18.4 °C, then
the solution has a conductivity value of 0.870 mS.
a.
Find the difference between the labeled temperature and the actual temperature
of the calibration solution at the time of calibration.
25 °C – 18.4 °C = 6.6 °C
b.
Multiply the difference (6.6) by the correction factor per °C (1.97% or 0.0197).
6.6 °C x 0.0197/°C = 0.13002
c.
If the calibration temperature is lower than the labeled value, then subtract that
value from the standard (1.0 mS) to get the actual value to be used for calibration.
1.0 mS - (correction factor) 0.13002 = 0.86998 mS
d.
If the calibration temperature is higher than the labeled value, then add that value
to the standard (1.0 mS) to get the actual value to be used for calibration.
7.
Using the value that was calculated in step 6, enter the conductivity of the solution
then press
ACCEPT
.
Calibrating the conductivity temperature
Necessary when logging temperature only.
1.
Place the probe in a liquid. Wait for the temperature reading to stabilize.
2.
Enter the actual temperature of the liquid (the current reading is shown
for reference). Temperature calibration is complete.
Goodnal STP ST041 Pre Treatment (Sigma 900 MAX All Weather Refrigerated Sampler) Vendor Manual
Q-Pulse Id VM373
Active 29/10/2013
Page 88 of 122
