2
$\begingroup$

In the experiment for calibrating a platinum resistance thermometer, we are always approximating the resistance of the platinum thermometer by $$R(T) \approx R_0 (1+\alpha T),$$ taking the reference at $0^o C$ and using Celsius scale.

We measure temperature by a thermometer of least count $0.5^o C$, and resistance of the platinum thermometer by a Carey Foster bridge which is sensitive up to $0.005\Omega$.

We work in the range of $0^o C-100^o C$.

Qusetion: How can we justify (without saying that, experimentally it is observed...) the usage of the approximation for $R(T)$ for this temperature range given the precision of our apparatus?


What I’m basically asking is the justification for ignoring the higher order terms in the Taylor expansion of $R(T)$ at $T_0$ for ranges as large as $100^o C$?

$\endgroup$
  • $\begingroup$ Well, you could look at the tables of resistance vs temperature for your particular resistor and see how linear it is. $\endgroup$ – Jon Custer Sep 20 at 13:59
  • $\begingroup$ I asked on theoretical grounds... $\endgroup$ – Atom Sep 20 at 14:00
  • $\begingroup$ Well, the theory is in the Taylor expansion. The check if it is reasonable is the actual data for you particular material. Deriving resistivity vs temperature for an arbitrary material across a large temperature range is, well, quite difficult. $\endgroup$ – Jon Custer Sep 20 at 14:02
  • $\begingroup$ @JonCuster So the range of validity comes purely from experiment? $\endgroup$ – Atom Sep 20 at 14:04
  • $\begingroup$ Indeed. You might be interested in tsapps.nist.gov/publication/get_pdf.cfm?pub_id=904572 - it would suggest that the range should be based on the particular calibration requested. $\endgroup$ – Jon Custer Sep 20 at 14:07
2
$\begingroup$

The response of a platinum resistance thermometer (PRT) is better approximated with the so-called Callendar-Van Dusen equation (this is defined in the standard IEC EN 60751):

$R(t) = \begin{cases} R_0\left[1+At+Bt^2+C(t-100\,{}^\circ\mathrm{C})t^3\right]& t<0 \\[5mm] R_0(1+At+Bt^2)& t\geqslant 0 \end{cases}$

where $t$ is the Celsius temperature and the nominal values of the coefficients are (a calibration can give more accurate values)

$\begin{aligned} &R_0 = 100\,\Omega \\ &A = 3.9083\times 10^{-3}/{}^\circ\mathrm{C}, \\ &B = -5.775\times 10^{-7}/{}^\circ\mathrm{C}^2, \\ &C = -4.183\times 10^{-12}/{}^\circ\mathrm{C}^4, \end{aligned}$

For $t$ from $0\,{}^\circ \mathrm{C}$ to $100\,{}^\circ \mathrm{C}$, the response is

$R(t) = R_0(1+At+Bt^2)$

If you use instead the linear approximation $ R(t) = R_0(1+\alpha t)$, you will measure the temperature $t'$ such that (assuming that $R_0$ is the same for the two equations; otherwise, modify accordingly)

$R_0(1+At+Bt^2) = R_0(1+\alpha t')$

The difference $\Delta t = t'-t$ is the measurement error that you get from using the linear approximation. Substituting $t' = t+\Delta t$ in the above equation and simplifying, you get

$At+Bt^2 = \alpha(t+\Delta t)$

from which

$\Delta t = \dfrac{1}{\alpha}\left[(A-\alpha)t+B t^2\right].$

The above equation represents a parabola, and substituting the value of $\alpha$ that you use you can find the maximum error along the whole measurement range. For instance, if you use $\alpha = 3.85\times 10^{-3}/{}^\circ \mathrm{C}$, the nominal average temperature coefficient from $0\,{}^\circ \mathrm{C}$ to $100\,{}^\circ \mathrm{C}$ (according to the EU standards, for a US PRT you may get $\alpha = 3.92\times 10^{-3}/{}^\circ \mathrm{C}$), the error $\Delta t$ is maximum at $t = 50{}^\circ \mathrm{C}$ and $(\Delta t)_\mathrm{max}\approx 0.38{}^\circ \mathrm{C}$. This specific case is represented in the figure below (own picture):

Pt100 transfer characteristic

Depending on your application and the accuracy class of the PRT, you can then evaluate whether you can consider the above error negligible or not.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.