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THERMOCOUPLE LEAD RESISTANCE
Thermocouple lead length can affect instrument 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 .34 .85 .38 1.02 1.06 .58 7.00 1.47 1.26
12 .54 1.34 .61 1.65 1.65 .91 11.00 2.34 2.03
14 .87 2.15 .97 2.67 2.65 1.46 17.50 3.72 3.19
16 1.37 3.38 1.54 4.15 4.18 2.30 27.75 5.91 5.05
18 2.22 5.50 2.50 6.76 6.82 3.73 44.25 9.40 8.13
20 3.57 8.62 3.92 10.80 10.88 5.89 70.50 14.94 12.91
24 8.78 21.91 9.91 27.16 27.29 14.83 178.25 37.80 32.64
TABLE 2
Temperature Error in °F per 1000 feet of Leadwire
AWG Thermocouple Type:
No.JKTRSEBNC
10 .61 1.54 .69 1.84 1.91 1.04 12.60 2.65 2.27
12 .97 2.41 1.09 2.97 2.96 1.64 19.80 4.21 3.66
14 1.57 3.86 1.75 4.81 4.76 2.63 31.50 6.69 5.74
16 2.47 6.09 2.77 7.47 7.52 4.14 49.95 10.64 9.10
18 4.00 9.90 4.50 12.17 12.28 6.72 79.95 10.64 9.10
20 6.43 15.51 7.06 19.43 19.59 10.61 126.90 26.89 23.24
24 15.80 39.44 17.83 48.89 49.13 26.70 320.85 68.03 58.75
Example:
A 1/4 Din unit 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 = 2.47 (°F/1000 ft) from Table 2
Terr = 2.47 (°F/1000 ft) * 660 ft
Terr = 1.6 °F
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