Question 5 of 7: Thermocouple Temperature Measurement and Dynamic Response

Question 5: Thermocouple Temperature Measurement and Dynamic Response (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

a) A thermocouple's output voltage depends on the TEMPERATURE DIFFERENCE between the measuring (hot) junction and the reference (cold) junction, not on the measuring junction's absolute temperature alone — standard thermocouple voltage-vs-temperature tables are referenced to a 0°C cold junction, which classically was maintained physically by an ice bath. Cold junction compensation (CJC) corrects for whatever temperature the reference junction is ACTUALLY at. Without ice, this is done electronically: an independent local temperature sensor (a thermistor or IC sensor) measures the real (ambient) temperature at the point where the thermocouple wire meets the copper measurement circuitry, and the instrument adds a correction voltage — equal to what a thermocouple of the same type would generate between 0°C and that measured reference temperature — to the raw thermocouple EMF before converting it to a temperature reading, effectively synthesizing a virtual 0°C reference electronically.

b) A thermocouple junction is an inherently LOW-IMPEDANCE source — a simple junction of two dissimilar metal wires, typically only a few ohms to a few tens of ohms — and thermal (Johnson) noise voltage scales with the SQUARE ROOT of the source resistance, $v_n=\sqrt{4k_BTRB}$. Even though the useful thermocouple signal is small (tens of microvolts per °C), the intrinsic noise floor set by its own low resistance is smaller still, so the signal-to-noise ratio stays favourable despite the small absolute signal level; the low output impedance also makes a thermocouple relatively immune to capacitively-coupled interference compared with a high-impedance sensor.

c) Given. Small cylindrical thermocouple bead: length $L=2\ \text{cm}=0.02\ \text{m}$, diameter $D=0.5\ \text{cm}=0.005\ \text{m}$, density $\rho=5000\ \text{kg/m}^3$, specific heat capacity $c=350\ \text{J}/(\text{kg}\cdot\text{K})$, convective heat transfer coefficient $h=50\ \text{J}/(\text{s}\cdot\text{m}^2\cdot\text{K})$.

Find. The thermal time constant $\tau$ of the bead in moving air.

Approach. Treat the bead as a lumped thermal capacitance ($mc$) losing heat to the air by convection ($hA$); the resulting first-order energy balance gives $\tau=mc/(hA)$, so compute the bead's volume, mass, and full (lateral $+$ two end caps) surface area, then combine.

1. Volume and mass of the cylindrical bead. $$V=\dfrac{\pi}{4}D^2L=\dfrac{\pi}{4}(0.005)^2(0.02)=3.927\times10^{-7}\ \text{m}^3$$ $$m=\rho V=5000\times3.927\times10^{-7}=1.9635\times10^{-3}\ \text{kg}.$$
2. Total convective surface area (lateral $+$ two end caps). $$A=\pi DL+2\left(\dfrac{\pi}{4}D^2\right)=\pi(0.005)(0.02)+2\left(\dfrac{\pi}{4}(0.005)^2\right)=3.534\times10^{-4}\ \text{m}^2.$$
3. Lumped thermal time constant. An energy balance on the bead ($mc\,dT/dt=-hA(T-T_\infty)$) gives the classic first-order form with $$\tau=\dfrac{mc}{hA}=\dfrac{(1.9635\times10^{-3})(350)}{(50)(3.534\times10^{-4})}=\boxed{38.9\ \text{s}.}$$

d) Approach. A first-order sensor's response to a step input follows $\theta(t)=1-e^{-t/\tau}$; setting $\theta=0.95$ and solving for $t/\tau$ gives the universal (geometry-independent) multiple of $\tau$ needed for 95% of the ultimate response.

1. Solve for the time in $\tau$ units. $$1-e^{-t/\tau}=0.95 \;\Longrightarrow\; e^{-t/\tau}=0.05 \;\Longrightarrow\; \dfrac{t}{\tau}=-\ln(0.05)=\boxed{3.0\ \tau.}$$
2. Convert to seconds using part (c)'s $\tau$. $$t=3.0\times38.9\ \text{s}=\boxed{116.6\ \text{s} \approx 117\ \text{s}.}$$