Sloppy calibration

Calculus Level 4

The temperature curve of a resistance thermometer (PT100) is given a quadratic polynomial: $R(T) = R_0 (1 + a \cdot T + b \cdot T^2)$ with constants $R_0 = 100 \,\mathrm{\Omega}$ , $a = 4 \cdot 10^{-3} \frac{1}{^\circ \text{C}}$ and $b = -6 \cdot 10^{-7} \frac{1}{^\circ \text{C}^2}$ . The temperature $T$ is expressed in degrees Celsius.

Because he does not know the temperature curve, an engineer measures the resistance at the two temperatures $T = 0^\circ \text{C}$ and $T = 400^\circ \text{C}$ . For temperature range between he assumes a linear relationship $\tilde R(T) = R_0 (1 + c \cdot T)$ with $\tilde R(0^\circ \text{C}) = R(0^\circ \text{C})$ and $\tilde R(400^\circ \text{C}) = R(400^\circ \text{C})$ . He uses the self-calibrated curve for temperature measurements. What is the maximum error $\Delta T$ in the temperature measurement in the interval $0^\circ \text{C} \leq T \leq 400^\circ \text{C}$ ? Round the result to the nearest integer.

Hint: The graph is for illustrative purposes only and is not true to scale.

Bonus question: How does the temperature error change, if the upper temperatur point was $T_2 = 200^\circ \text{C}$ instead of $400^\circ \text{C}$ and the measurement range is reduced to $0^\circ \text{C} \leq T \leq 200^\circ\text{C}$ ?

6 °C 1 °C 24 °C 3 °C 12 °C

1 solution

Florian Wechslberger
Apr 16, 2018

$\begin{aligned} R(400\,^\circ)=\tilde R(400\,^\circ)~\Rightarrow~c=\frac{\tilde R(400\,^\circ C)/R_0-1}{T}=a+b\cdot 400\,^\circ C \end{aligned} \\~\\$

$\begin{aligned} \tilde T&=\frac{\tilde R/R_0-1}{c}\\ T&=\frac{-aR_0\pm\sqrt{a^2R_0^2-4bR_0(R_0-R)}}{2bR_0}\\ &=\frac{-a\pm\sqrt{a^2-4b+4bR/R_0}}{2b}~~,~T(R_0)\overset !=0~\Rightarrow +\\ \end{aligned} \\$

$\begin{aligned} \Delta T&=\tilde T-T=\frac{R/R_0-1}{c}+\frac{a-\sqrt{a^2-4b+4bR/R_0}}{2b}\\ \frac{\mathrm d\Delta T}{\mathrm dR}&=1/R_0c-\frac{4b/R_0}{4b\sqrt{a^2-4b+4bR/R_0}}\overset !=0\\ \Rightarrow c&=\sqrt{a^2-4b+4bR/R_0}\\ \Rightarrow 4bR/R_0&=c^2-a^2+4b\\ \Rightarrow~R/R_0&=(c^2-a^2)/4b+1\\ \Rightarrow~\Delta T&=\frac{c-a^2/c}{4b}+\frac{a-c}{2b}=\frac{-c-a^2/c+2a}{4b}\approx 6.383\,^\circ C\\ &\approx\boxed{6\,^\circ C} \end{aligned}$

Sloppy calibration

1 solution

0 pending reports