Manual

11
THERMOCOUPLE LEAD RESISTANCE
Thermcouple lead length can affect instrument accuracy since the size (gauge) and the length
of the wire affect lead resistance.
To determine the temperature error resulting from the lead length resistance, use the following
equation:
Terr = TLe * L where; TLe = value from appropriate table below
L = length of leadwire in thousands of feet
TABLE 1
Temperature error in °C per 1000 feet of Leadwire
AWG Thermocouple Type:
No. J K T R S E B N C
10 .68 1.71 .76 2.05 2.12 1.15 14.00 2.94 2.53
12 1.08 2.68 1.21 3.30 3.29 1.82 22.00 4.68 4.07
14 1.74 4.29 1.95 5.34 5.29 2.92 35.00 7.44 6.37
16 2.74 6.76 3.08 8.30 8.35 4.60 55.50 11.82 10.11
18 4.44 11.00 5.00 13.52 13.65 7.47 88.50 18.80 16.26
20 7.14 17.24 7.84 21.59 21.76 11.78 141.00 29.88 25.82
24 17.56 43.82 19.82 54.32 54.59 29.67 356.50 75.59 65.27
TABLE 2
Temperature Error in °F per 1000 feet of Leadwire
AWG Thermcouple Type:
No. J K T R S E B N C
10 1.22 3.07 1.37 3.68 3.81 2.07 25.20 5.30 4.55
12 1.94 4.82 2.18 5.93 5.93 3.27 39.60 8.42 7.32
14 3.13 7.73 3.51 9.61 9.53 5.25 63.00 13.38 11.47
16 4.93 12.18 5.54 14.93 15.04 8.28 99.90 21.28 18.20
18 7.99 19.80 9.00 24.34 24.56 13.44 159.30 33.85 29.27
20 12.85 31.02 14.12 38.86 39.18 21.21 253.80 53.79 46.48
24 31.61 78.88 35.67 97.77 98.26 53.40 641.70 136.07 117.49
Example:
An MRC is to be located in a control room 660 feet away from the process. Using 16 AWG,
type J thermocouple, how much error is induced?
Terr = Tle * L
TLe = 4.93 (°F/1000 ft) from Table 2
Terr = 4.93 (°F/1000 ft) * 660 ft
Terr = 3.3 °F