I would like to ask about the conversion between the resistance of a PT100 thermistor and temperature
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I would like to ask the experienced professionals: regarding the conversion between the resistance value of a PT100 thermistor and temperature, for temperatures ranging from -200 to 500°, in a two-wire system, if the signal is sent to a temperature module that is designed to read only the thermistor signal, will the shielding wire used over a length of 80 meters between the thermistor and the module cause any interference with the temperature measurement? And if so, how significant would such interference be? Thank you to all the experienced drivers for their guidance.
Two-wire systems are subject to interference; a three-wire system should be used
Thermistors should be used in a three-wire or four-wire configuration; a two-wire setup will affect the accuracy of the measurements.
If PT100s are not affected by temperature changes, they are either three-wire or four-wire types; there is no two-wire version. Even if it is treated as a two-wire system, shorting the terminals only provides compensation for the resistance at the ends, not for the resistance along the wire.
The two-wire system cannot compensate for line resistance; using a three-wire system will suffice.
It is likely that the on-site environment is harsh or that it is for easier observation; that’s why the secondary meter is installed separately from the thermal resistor and not in the same location. The poster needn’t worry too much; the thermistor signal cable is installed as required with proper shielding and grounding, so 80 meters shouldn’t be a problem. We’ve used distances of 200 or 300 meters before; it’s no problem.
The experts upstairs have already answered your second question; I’ll answer your first one. According to the national standard JB/T 8622-1997 \"Technical Requirements and Calibration Tables for Industrial Platinum Resistance Thermometers,\" the resistance value of a Class A armored platinum resistance thermometer at 0°C is 100.00Ω； The allowable error is: ±(0.15℃ + 0.002|t|). Since at 0℃, 0.002|t| = 0, the allowable error is ±0.15℃, resulting in a corresponding resistance value of ±0.6Ω
I switched directly to an integrated transmitter, but I’m still very grateful for everyone’s answers. To measure the total resistance r of a 2*80 meter cable, it’s necessary to do so accurately using the 4-wire measurement method; when using the 2-wire method, the resistance of the multimeter probes must be taken into account. I’d like to learn how to carry out the 4-wire method from those with more experience