Does Newton's Law of Cooling Predict Exchanger Settling?

Newton's law of cooling separates two effects that are often confused: input power sets the equilibrium temperature, while the ratio of thermal conductance to heat capacity sets the first-order rate. For a linear lumped system, changing heater power does not change the rate constant k. It changes the temperature excursion. A different fitted k indicates that conductance, heat capacity, boundary conditions, sensor dynamics, or the validity of the one-state model changed.

Thermal model and identifiable parameters

For one lumped temperature T, ambient temperature Ta, applied electrical power P, effective power fraction eta, thermal conductance G, and heat capacity C, the energy balance is:

C dT/dt = eta P - G(T - Ta)

The term G here means the complete effective conductance from the modeled thermal mass to its surroundings. In the simple convection form, G = hA. Combining h and A avoids trying to identify two quantities that the temperature record cannot separate.

For constant power and constant coefficients, the solution is:

T(t) = Tinf + (T0 - Tinf) exp(-kt)
k = G/C
Tinf = Ta + eta P/G
tau = 1/k

A single transient identifies k = G/C, not G and C separately. The steady-state power tests add another relationship. If the slope of steady temperature rise versus applied power is S = d(Tinf-Ta)/dP, then:

S = eta/G
G/eta = 1/S
C/eta = 1/(kS)

Individual values of G and C require either a known eta or an independent measurement of one parameter. Under the explicit assumption that all measured electrical power reaches the modeled lump and all heat rejection is represented by G, set eta = 1, giving G = 1/S and C = 1/(kS).

Check 1: First-order model validity

Check 1 uses the recorded 12 W transient sampled every 200 ms. Fit the complete heating record to the first-order equation, then calculate the residual R(t) = Tinf - T(t). For a valid single-state model, ln|R(t)| forms a straight line over the useful portion of the record, with slope -k.

1. Estimate T0 from the samples immediately before or at the power step.
2. Fit Tinf and k together rather than fixing Tinf from a short record.
3. Plot residuals against time. Expect small, pattern-free temperature errors.
4. Plot ln|Tinf-T(t)| against time. Expect one approximately constant slope.

A curved semi-log plot, a fast initial slope followed by a slow tail, or patterned residuals means one exponential is not representing the full exchanger. A heat exchanger commonly contains several thermal masses: heater, hot-side structure, interface, cold-side structure, fluid or load, and thermistor. Those nodes can produce multiple time constants even when every material property is constant. In that branch, fit a multi-exponential or physical multi-node model before extrapolating to other powers.

If the first-order residual checks pass, continue to Check 2. If they fail only during the first few samples, examine thermistor response, acquisition filtering, and the exact time of the power step before rejecting the thermal model.

Check 2: Steady-state power linearity

Check 2 uses the steady cold-side temperatures recorded at 2 W, 4 W, 6 W, and 8 W. Record the ambient temperature associated with each test and calculate DeltaT = Tinf - Ta. Regress DeltaT against measured electrical power.

DeltaT = intercept + S P

Expect a straight line if effective conductance and delivered-power fraction remain constant. An intercept near zero is expected when Ta represents the true thermal boundary temperature and the measurement has no offset. A nonzero intercept directs attention to ambient drift, thermistor offset, preheating, unmeasured background power, or an omitted thermal boundary.

Use actual power calculated from measured heater voltage and current at each operating point. A power-supply setting alone does not account for heater-resistance changes or wiring voltage drop. If the four points are linear, retain their fitted slope S and continue to Check 3. If they curve systematically, treat G, eta, or both as temperature-dependent and do not reduce the complete range to one constant.

Check 3: Rate-constant invariance

Check 3 determines whether k can be reused at other power levels. The fitted value from the 12 W record is . Under the explicit assumption that this value represents the dominant first-order thermal mode, the corresponding time constant is:

The large separation between the 200 ms sampling interval and a time constant means the record is densely sampled. More samples do not replace a sufficiently long observation period; the late portion of the response is what separates the asymptote from the exponential rate.

Fit k independently at additional power levels whenever transient records are available. Expect statistically similar values when G and C are constant. If k varies, use k = G/C to structure the diagnosis: conductance changed, effective thermal mass changed, or the fitted first-order pole shifted because several modes contribute differently at different temperatures.

Power itself is not an argument of k in the linear model. A fitted empirical function k(P) may interpolate tested data, but it conceals the physical dependency. Associate the rate with measured temperature or operating condition only after repeated transients demonstrate that dependency.

Check 4: Boundary and load consistency

Check 4 compares everything that can alter the heat path between tests. The cold-side response belongs to the complete assembly and connected load, not only to the nominal heat-exchanger body. Changing contact pressure, flow, mounting, insulation, fluid inventory, attached mass, or ambient conditions changes the parameters being identified.

If load and boundary conditions match and both linearity checks pass, continue to parameter estimation. If they do not match, create a separate parameter set for each configuration; no temperature-only calculation can predict an unmeasured change in attached thermal mass or conductance.

Parameter estimation from the recorded tests

The existing data can identify the combinations needed for prediction without direct access to surface area or mass. It cannot separate h from A, nor separate G and C unless eta is known.

1. For each 2 W, 4 W, 6 W, and 8 W test, pair the final cold-side temperature with its measured ambient temperature.
2. Fit DeltaT = S P, or include an intercept while diagnosing offsets. Report the slope in kelvins per watt.
3. Calculate G/eta = 1/S. Its unit is watts per kelvin.
4. Use the 12 W transient fit to obtain k. The recorded fit gives .
5. Calculate C/eta = 1/(kS). Its unit is joules per kelvin.
6. If an independent power-loss test establishes eta, calculate G = eta/S and C = eta/(kS).

Fit all steady-state and transient records jointly when possible. A joint fit uses one shared k and one shared power-to-temperature slope while allowing each run its own initial and ambient temperature. This directly tests the claim that the same physical parameters apply at every power.

Settling-time prediction and final verification

Steady state is approached asymptotically, so “time to steady state” requires a tolerance. For a fractional completion f, measured from the initial temperature toward the final temperature:

t = -ln(1-f)/k

With , and assuming that rate applies at the target operating condition, 63.2% completion takes one time constant or ; 95% takes 2.996 tau, about ; and 99% takes 4.605 tau, about . These are fractional settling times and therefore do not change with power in the linear model.

For an absolute temperature band epsilon around the final value, use:

t = (1/k) ln(|Tinf - T0|/epsilon)

Absolute-band settling time can increase with power because a larger temperature excursion must decay into the same band, even while k remains constant.

1. Check 1: Predict the target equilibrium from Tinf = Ta + SP. Expect the measured final cold-side temperature to fall within the regression's established error band.
2. Check 2: Predict the full trace from T(t) = Tinf + (T0-Tinf)exp(-kt). Expect residuals without a time-dependent curve.
3. Check 3: Compare the time to the selected fractional completion. Expect the same normalized settling time at each power when k is invariant.
4. Check 4: Repeat one tested power from the same initial and boundary conditions. Expect both the fitted Tinf and k to repeat before using the model for an untested power.

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