Question

Thermocouple system has a time constant of 8 sec. It is used to measure temperature of a surface varying sinusoidally between 400\(^\circ\)C and 450\(^\circ\)C with a periodic time of 60 sec. The thermocouple will indicate the maximum temperature of

• A. 450.0\(^\circ\)C
• B. 44(4)2\(^\circ\)C
• C. 425.0\(^\circ\)C
• D. 405.0\(^\circ\)C

Hint:

First-Order System Response. For sinusoidal input \(A_{in\sin(\omega t)\), output amplitude \(A_{out = M \times A_{in\), where \(M = 1/\sqrt{1+(\omega\tau)^2\). Output also has phase lag. Max indicated value = Mean Value + \(A_{out\).

Answer 1

The Correct Option is B

The thermocouple system is modeled as a first-order system with time constant \(\tau = 8\) sec
The input temperature \(T_{in}(t)\) varies sinusoidally
Mean temperature \(T_{mean} = \frac{400 + 450}{2} = 425^\circ\)C
True amplitude of variation \(A_{true} = \frac{450 - 400}{2} = 25^\circ\)C
Periodic time \(T_{period} = 60\) sec
Angular frequency \(\omega = \frac{2\pi}{T_{period}} = \frac{2\pi}{60} = \frac{\pi}{30}\) rad/s
The amplitude of the indicated temperature variation (\(A_{indicated}\)) is attenuated relative to the true amplitude (\(A_{true}\)) due to the system's time constant
The attenuation factor (magnitude ratio M) is: $$ M = \frac{A_{indicated}}{A_{true}} = \frac{1}{\sqrt{1 + (\omega \tau)^2}} $$ Calculate \(\omega \tau\): $$ \omega \tau = \left(\frac{\pi}{30}\right) \times 8 = \frac{8\pi}{30} = \frac{4\pi}{15} $$ $$ \omega \tau \approx \frac{4 \times (3)14159}{15} \approx \frac{1(2)566}{15} \approx 0
8377 $$ Calculate M: $$ M = \frac{1}{\sqrt{1 + (0
8377)^2}} \approx \frac{1}{\sqrt{1 + 0
7018}} = \frac{1}{\sqrt{(1)7018}} \approx \frac{1}{(1)3045} \approx 0
7665 $$ Calculate the indicated amplitude: $$ A_{indicated} = M \times A_{true} \approx 0
7665 \times 25^\circ\text{C} \approx 19
16^\circ\text{C} $$ The maximum temperature indicated by the thermocouple is the mean temperature plus the indicated amplitude: $$ T_{max, indicated} = T_{mean} + A_{indicated} \approx 425^\circ\text{C} + 19
16^\circ\text{C} \approx 44(4)16^\circ\text{C} $$ This is closest to 44(4)2\(^\circ\)C