Thermocouple time constant and frequency response

For a first-order system with unit sensitivity, a step from $T_i = 25\ ^\circ$C to $T_\infty = 200\ ^\circ$C gives

$$T(t) = T_\infty - (T_\infty - T_i)e^{-t/\tau} = 200 - 175\,e^{-t/2}$$

(a) Time to indicate 190 $^\circ$C

$$190 = 200 - 175e^{-t/2} \Rightarrow e^{-t/2} = \frac{10}{175} \Rightarrow t = -2\ln(0.05714) = 5.72\ \text{s}$$

(b) Reading after 5 s

$$T = 200 - 175e^{-5/2} = 200 - 175(0.0821) = 185.6\ ^\circ\text{C}$$

(c) Time for 99% of the step

$$e^{-t/\tau} = 0.01 \Rightarrow t = \tau\ln 100 = 2(4.605) = 9.21\ \text{s}$$

(d) Sinusoidal bath temperature at 0.1 Hz

$\omega = 2\pi(0.1) = 0.6283$ rad/s, so $\omega\tau = 1.2566$.

$$M = \frac{1}{\sqrt{1 + (\omega\tau)^2}} = \frac{1}{\sqrt{1 + 1.579}} = 0.623$$ $$\phi = -\tan^{-1}(\omega\tau) = -\tan^{-1}(1.2566) = -51.5^\circ$$

The thermocouple shows only 62.3% of the true amplitude and lags by $51.5^\circ$ (time lag $= 51.5/360 \times 10 = 1.43$ s).

Answer: (a) 5.72 s; (b) 185.6 $^\circ$C; (c) 9.21 s; (d) amplitude ratio 0.623, phase lag $51.5^\circ$.