Honeywell 9782 Series - Manual (Page 128)

Conductivity/Resistivity Analyzer/Controller
9782 Series Conductivity/Resistivity Analyzer/Controller - Operator’s Manual 7/99 10-12
10.8 Entering Values for Lead Resistance Compensation (Wide Range
Only)
Introduction
If you use standard Honeywell cell lead lengths of 7 or 20 feet connected directly to the
Analyzer/Controller, no compensation for lead resistance is necessary. Similarly, if a junction
box is used to extend the leads up to 150 feet, no compensation is required. However, if longer
leads are used (greater than 150 feet), signal quality can be adversely affected unless you enter
information that will permit the 9782 to compensate for lead resistance.
If you use a single wire gauge (12, 14, 16, or 18 AWG) in a length up to 1500 feet, simply
specify the gauge and length as described in Table 10-9.
If mixed wired gauges are used, or lead length or wire gauge are not within the stated ranges, the
9782 can still perform the compensation. However, you must first calculate the lead resistance,
then put it in terms of the available settings for AWG gauge and length.
The resistance of each available gauge choice (in copper wire) is:
12 AWG = 1.6 ohms per 1000 feet
14 AWG = 2.5 ohms per 1000 feet
16 AWG = 4.0 ohms per 1000 feet
18 AWG = 6.4 ohms per 1000 feet
For example, suppose each lead between the cell and Analyzer/Controller consists of 500 feet of
12 gauge wire and 1000 feet of 18 gauge wire.
1000 ft of 18 AWG 500 ft. of 12 AWG
9782
Honeywell
Junction Conductivity Cell
Box
Figure 10-1 Example of a Conductivity Loop
Because there are two different types of wire used in each lead to the cell in this example, the
total lead resistance is calculated as follows:
(2 x 0.5 x 1.6) + (2 x 1 x 6.4) = 14.4 ohms
Since the 9782 only allows entry of one wire gauge type, we allow for the worst-case condition
by dividing the total resistance by the resistance per thousand feet of the higher resistance gauge
wire. In our example this would be:
14.4 ohms
÷
6.4 ohms per thousand feet of 18 AWG wire = 2,250 feet
The length to enter is one-half this number, 1125 feet, because the 9782 Analyzer/Controller
already accounts for the fact that there is always a pair of conductor wires in the system loop.
Therefore, in our example we would use the procedure in Table 10-9, and specify the wire gauge
as 18 AWG and the length as 1125 feet.