Appendices

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: (Note: the analyzer accounts for the fact that there is always a pair of conductor wires in the system loop.)

(0.5 x 6.4) + (0.2 x 10.2) = 5.24 ohms

Since the analyzer 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:

5.24 ohms ⎟ 10.2 ohms per thousand feet of 22 AWG wire = 514 feet

Therefore, in our example we would use the procedure in Table 6-5, and specify the wire gauge as 20 AWG and the length as 514 feet. (20 AWG wire simulates 22 AWG coax)

January 2009

UDA2182 Universal Dual Analyzer Product Manual

203

Page 215
Image 215
Honeywell manual January UDA2182 Universal Dual Analyzer Product Manual 203