Cover a Pt100 probe with your hand and it warms in seconds; let go and still air needs tens of minutes to carry the heat back out. Cooling follows an exponential with a time constant of about 10 minutes for a 6 mm probe in still air, and a reading is settled only after about 5 time constants. Log the value every 30 seconds and judge the channel only if the trace jumps or freezes.
A bench Pt100 that is still reading warm minutes after your hand has left it is not faulty. The hand heats the probe by direct contact, which is fast, and the room can only take that heat back out through a film of still air, which is slow. Both directions follow one law, Newton's law of cooling; the difference is the coupling on each side, about twenty times stronger under the palm than in the air. So a warm-up that finishes in half a minute has a matching cool-down that runs tens of minutes, roughly 50 minutes for a typical 6 mm probe. The slow smooth fall is physics. A wiring or channel fault looks different: the value freezes, jumps, or settles far from ambient. This page gives the rule, the numbers, and a logging test that tells them apart.
Why the rise is fast and the fall is slow
Under the palm, heat crosses into the sheath by conduction through pressed contact. Skin pressed on metal couples in the hundreds of W/(m²·K). Around the probe, heat leaves by natural convection through still air, where standard heat-transfer references put the film coefficient at 5 to 25 W/(m²·K). Same probe, same temperature gap, couplings tens of times apart.
The rise looks normal because the contact path does the heating. The fall looks wrong because weak air convection is the only path left. The asymmetry belongs to the coupling, not the probe: warmed from still air, the same probe climbs exactly as slowly as it cools. Pull it out of a warm water bath instead and it settles in a minute or two, because water convects far harder than air.
The rule the cooling curve follows
Newton's law of cooling: the gap between probe and room shrinks by the same fraction in every equal slice of time.
T(t) = T_room + (T_start − T_room) × e^(−t/τ)
τ is the time constant, the thermal mass over the coupling: τ = mc/(hA), with m the probe's mass in kg, c its specific heat in J/(kg·K), h the film coefficient in W/(m²·K) and A the surface in contact with the air in m². For a slender sheathed probe this collapses to τ = ρ × c × d ÷ (4h): sheath diameter on top, air film underneath. Halve the diameter and the time constant halves with it.
The percentages come straight from the exponential. One time constant leaves 37% of the gap, three leave 5%, five leave 0.7%. Sensor-response guidance says to allow 5 to 10 time constants for a reading to reach equilibrium, so five is the working minimum for a settled value.
Worked example: a 6 mm Pt100 after a 30-second hand-warm
The inputs, and where each comes from:
- Probe: 6 mm outer diameter, 100 mm long, stainless-steel sheath, the common bench construction. No dimensions were given for the probe being tested, so these are typical values for the class; other sheath sizes scale as shown below.
- Stainless steel: density 8000 kg/m³, specific heat 500 J/(kg·K), standard reference values.
- Air film: h = 10 W/(m²·K), mid-range for natural convection in still air.
- Release gap: skin under a pressed palm sits near 33 °C on a 20 °C bench, so the probe starts 13 K above ambient.
- Test interval: 30 s, the sampling used in the log below.
The rule is τ = mc/(hA). The arithmetic:
- Volume: V = π × (0.003 m)² × 0.100 m = 2.83 × 10⁻⁶ m³
- Mass: m = 8000 kg/m³ × 2.83 × 10⁻⁶ m³ = 0.0226 kg, about 23 g
- Surface: A = π × 0.006 m × 0.100 m + 2 × π × (0.003 m)² = 0.00189 m² + 0.00006 m² = 0.00194 m²
- Lumped-model check: h × (d/2) ÷ k = 10 W/(m²·K) × 0.003 m ÷ 15 W/(m·K) = 0.002, far under the 0.1 limit for treating the sheath as one mass at one temperature
- Time constant: τ = (0.0226 kg × 500 J/(kg·K)) ÷ (10 W/(m²·K) × 0.00194 m²) = 11.3 J/K ÷ 0.0194 W/K = 583 s, about 10 minutes
Result, starting from a 13 K gap at the moment the hand leaves:
- 10 min: 13 K × 0.37 = 4.8 K left, so the display shows about 24.8 °C
- 20 min: 4.8 K × 0.37 = 1.8 K left, about 21.8 °C
- 30 min: 0.7 K left, about 20.7 °C
- 50 min: 13 K × 0.007 = 0.09 K left, settled to within a tenth of a degree
The warming side closes the loop. A 30-second warm-up through the same mass and surface needs a coupling of mc/(τA) = 11.3 J/K ÷ (30 s × 0.00194 m²) = about 194 W/(m²·K), twenty times the air film. That twenty-to-one ratio is the whole effect.
Decision: with a 6 mm sheath and h = 10 W/(m²·K), wait 50 minutes before judging the settled value. At that point, a display reading more than about 1 K above the ambient reference recorded before the test points to a nearby heat source, an insulating mount, or a fault, and waiting longer fixes none of those three. Construction moves the constant, not the method.
What moves the time constant
| Change | Effect on τ | Numbers to expect |
|---|---|---|
| Sheath diameter | Proportional, τ = ρcd/(4h) | 3 mm: about 5 min; 6 mm: about 10 min; 8 mm: about 13 min, all at h = 10 W/(m²·K) |
| Airflow | Raises h, shortens τ | Forced air runs from tens to a few hundred W/(m²·K) |
| Mounting | An air void insulates and stretches the response a lot; metal contact adds a conduction path | A probe sitting in a loose thermowell can take several times the bare-probe time |
| Element and build | Construction spreads response times | One test series on weather-station platinum probes found a fourfold spread between designs, and thermocouples respond faster than RTDs of similar size |
Airflow is the lever you control on a bench: a fan, a doorway draft or a fume hood cuts the time constant several-fold. The coefficient estimates worked through for estimating heat-transfer coefficients in forced convection ovens transfer to any moving-air case.
Log every 30 seconds and read the shape
- Let the probe sit untouched until the display holds steady for a few minutes, and record that value. It is the ambient reference for everything that follows.
- Press your palm over the sheath for a fixed 30 seconds, then let go without moving the probe. Keep the interval identical on every run so the traces compare.
- Log the displayed temperature every 30 seconds until the display stops moving, at least five time constants.
- Divide each gap by the one before it. A healthy trace shrinks by a similar fraction at every step, about 37% per time constant.
- Run the whole cycle a second time. Two smooth, similar curves close the case; the probe is following physics.
If the log comes off a trend on the HMI instead of a stopwatch, check how many samples it actually kept before you trust the shape: a trend can record fewer samples than its trigger asks for, and that missing-samples fault has its own write-up at Why an HMI trend chart records fewer samples than the trigger.
Trace shapes and the decision
| Trace shape | What it points to | Next move |
|---|---|---|
| Smooth decay, similar fraction at every step | Normal thermal lag | Wait out the five time constants and change nothing |
| Value freezes while the probe is still warm | Stuck channel, scaling or update path | Compare the input's raw or diagnostic value with the displayed one |
| Jumps or steps in the decay | Loose termination or unstable input | Isolate the RTD circuit per your site's procedure, then check every termination |
| Settles smoothly well above the reference | Heat source nearby, bench equipment, a cabinet, sun, or a shallow-immersion mount | Move the probe to free air and rerun the log |
| Warm probe reads negative | Lost sign on a bipolar RTD input | Follow the fix in Fix an HMI showing lost sign on bipolar RTD temperatures |
Isolate the RTD input circuit before you loosen a single terminal. The signal side is low voltage; what shares the panel with it may not be.
If this test comes up regularly, a small shop-floor screen beside the bench showing the expected curve and the five logging steps as a static slide answers the same question for the next person; ours plays slides offline from a card.
Recommendation: Log the displayed temperature every 30 seconds through one full cool-down and call it healthy physics when the gap shrinks by a similar fraction at every step. Take the wiring or the channel configuration apart only when the trace freezes, jumps, or sits more than about 1 K above the ambient reference recorded before the test once five time constants have passed.
Questions
How do I get the time constant from my own cooling trace?
Pick two points on the smooth part of the decay, separated by a known interval. Divide the time between them by the natural log of the gap ratio. An 8 K gap that falls to 3 K in 10 minutes gives 10 ÷ ln(2.67), about 10 minutes. Five time constants from release is your settle point, so about 50 minutes for this probe.
Can the input's filter be blamed for part of the lag?
Input smoothing adds its own delay on top of the probe's, and the two stack. Where the RTD input's documentation offers a filter or damping setting, run the bench test with the fastest setting so you measure the probe alone, then restore smoothing for a noisy line. A filter averages electrical noise; it cannot pull heat out of the sheath any faster.
Why does the settled value sit slightly above the room thermometer?
Two reasons. The measuring current warms the sensing element a little, typically a few tenths of a degree or less at the milliamp-level excitation RTD inputs use. And the probe settles to its immediate surroundings, so a bench near a drive cabinet or a lamp reads warmer than a wall thermostat. Record your ambient reference with the same probe, untouched, and both effects are absorbed.
Planning information, not legal or engineering advice. Rules differ by place: check codes and permits with your local authority and your licensed trades.