Hach 900 MAX - Manual (Page 79)

Section 5
Page 77
8990options.fm Conductivity Probe
5.5.4 Conductivity Probe Calibration
1. From the Main Menu, select OPTIONS > ADVANCED OPTIONS >
CALIBRATION > CONDUCTIVITY.
2. Clean and dry the probe.
3. Place the sensor and thermometer in the calibration solution
(Cat. No. 3230). The temperature sensor is located in the middle of the
sensor body allowing the probe to be completely submerged in
the solution.
4. Allow the sensor to stabilize in the solution about 10 minutes to ensure
that the probe and the solution are the same temperature.
5. Enter the temperature correction factor or enter zero for no correction
factor.
Note: The temperature correction factor is used to compensate for the effects of
temperature on the conductivity readings at the point of installation. The
conductivity of a solution is temperature sensitive. Therefore the actual
conductivity of the solution will change with the temperature. Each site may have a
different correction factor depending on the major constituent of the flow stream.
This is not used for calibration and has no effect on the calibration of the sensor.
Below are some examples of compensation factors of various liquids.
• 0.96%/°C 5% Sulfuric Acid
• 1.88%/°C Dilute Ammonia
• 1.91%/°C ‘Typical’ Wastewater
• 1.97%/°C Potassium Chloride
• 2.12%/°C Salt (Sodium Chloride)
• 2.84%/°C 98% Sulfuric Acid
• 4.55%/°C Ultra-pure Water
6. With the sensor still in the calibration solution, press any key. Wait for the
sensor to stabilize. Calculate the actual conductivity of the calibration
solution. If using the KCl solution provided by the manufacturer, make
your selection from the
Table 12 on page 78 . If using a solution other than
1.0 mS @ 25 °C KCl available from manufacturer, calculate the
conductivity of the solution using temperature correction factors. See the
following example.
Example: The KCl calibration solution is 1.0 mS at 25°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