Radiation Error in a Thermocouple Reading: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2013, Q7

Question 7: Radiation Error in a Thermocouple Reading

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.

Given. Bead reading (its own equilibrium temperature) $T_{th}=179\,{}^{\circ}\text{C}$; duct wall temperature $T_w=65\,{}^{\circ}\text{C}$; bead emissivity $\varepsilon_b=0.7$; duct emissivity $\varepsilon_d=0.9$; duct 0.92 m long, no duct diameter $D$ printed anywhere in the source. Convective "rate of 790 W/m$^2$" — read, per the exam's own licence to state assumptions, as the convective heat-transfer coefficient $h_c=790\text{ W/m}^2\cdot\text{K}$ with its temperature-unit suffix dropped in print, as the paper also does in Question 6's "J/k C" specific heat, since a bare heat flux value alone cannot be combined with $F_{b-d}$, $\varepsilon_b$, $\varepsilon_d$ to solve for anything — only a coefficient can.

Given data
Quantity	Symbol	Value
Indicated (bead) temperature	$T_{th}$	$179\,{}^{\circ}\text{C}$ (452.15 K)
Duct wall temperature	$T_w$	$65\,{}^{\circ}\text{C}$ (338.15 K)
Bead emissivity	$\varepsilon_b$	0.7
Duct emissivity	$\varepsilon_d$	0.9
Convection coefficient (read)	$h_c$	790 W/m²·K
Duct length	$L$	0.92 m

Find. The error in the thermocouple reading, $T_{gas}-T_{th}$.

Approach. At steady state the bead's convective heat gain from the (hotter) gas balances its radiative heat loss to the (cooler) duct wall. The bead is a very small sphere compared with the duct's inner surface for any physically sensible duct diameter, so in the two-gray-surface radiation network the duct-emissivity term vanishes; with no duct diameter printed, the configuration factor $F_{b-d}=L/\sqrt{D^2+L^2}$ is taken as unity (its limit for $D\ll L$); compute the radiative loss from the KNOWN bead and wall temperatures, then back out the true gas temperature from the convective balance using $h_c$.

Radiative loss from the bead to the duct wall. For a small sphere (area $A_b$) radiating to a much larger surrounding surface (area $A_d\gg A_b$) with view factor $F_{b-d}\to1$ (assumed above), the two-surface gray-enclosure network reduces to $$q''_{rad} = \varepsilon_b\,\sigma\left(T_{th}^4-T_w^4\right)$$ — independent of the duct's own emissivity $\varepsilon_d$, since the $(1-\varepsilon_d)/(\varepsilon_d A_d)$ term is negligible whenever $A_b\ll A_d$ (true for a 3 mm bead in any real duct, regardless of the unprinted diameter $D$). Numerically, $$q''_{rad} = 0.7\times5.670\times10^{-8}\times\left(452.15^4-338.15^4\right) = 1140.0\text{ W/m}^2$$
True gas temperature from the convective balance. At steady state the convective heat gained from the gas equals the radiative heat lost to the wall, $$h_c\left(T_{gas}-T_{th}\right) = q''_{rad}$$ $$T_{gas}-T_{th} = \frac{q''_{rad}}{h_c} = \frac{1140.0}{790}$$ $$\boxed{\text{Error} = T_{gas}-T_{th} = 1.44\,{}^{\circ}\text{C}, \qquad T_{gas}=180.4\,{}^{\circ}\text{C}}$$

Check: No duct diameter $D$ is printed anywhere in the source, so the given $F_{b-d}$ formula cannot be evaluated numerically; $F_{b-d}=1$ is adopted, per the exam's own instruction to state assumptions. The duct-emissivity term is negligible for ANY diameter ($A_b\ll A_d$), but the view factor is not: if a diameter were specified and $F_{b-d}$ applied to the bead–wall exchange (the open ends contributing nothing), the error scales with $F_{b-d}$ — $D=0.1$ m gives $F=0.994$ and 1.43°C, $D=0.3$ m gives $F=0.951$ and 1.37°C, $D=0.5$ m gives $F=0.879$ and 1.27°C. The 1.44°C found here is therefore the upper (slender-duct) bound.

Question 7 — results
Quantity	Value
Radiative heat loss from bead	1140.0 W/m²
Error in reading ($T_{gas}-T_{th}$)	$1.44\,{}^{\circ}\text{C}$
True gas temperature	$180.4\,{}^{\circ}\text{C}$

Topic. Radiation error in gas-temperature measurement by an immersed-thermocouple probe.

Key relations. Small-object-in-large-enclosure radiation exchange $q''_{rad}=\varepsilon_b\sigma(T_1^4-T_2^4)$; steady-state bead energy balance $h_c(T_{gas}-T_{th})=q''_{rad}$; general two-gray-surface network $Q_{1\to2}=\sigma(T_1^4-T_2^4)/\big[(1-\varepsilon_1)/(\varepsilon_1A_1)+1/(A_1F_{12})+(1-\varepsilon_2)/(\varepsilon_2A_2)\big]$.

Why this works. A thermocouple immersed in a duct where the wall runs colder than the gas always under-reads, because it settles at the temperature where convective heating from the gas exactly balances radiative cooling to the wall — it can never show the true gas temperature unless $h_c\to\infty$ or the wall and gas are at the same temperature. The duct's own emissivity genuinely drops out of the answer here because the bead's surface area is many orders of magnitude smaller than the duct wall's, so essentially all of the wall behaves as a near-blackbody cavity from the bead's point of view regardless of the wall material's actual emissivity.

Common pitfall. Including the duct emissivity $\varepsilon_d$ in the radiative exchange formula when the small-object limit already makes it irrelevant (a common over-application of the general two-surface network); and reporting the error with the wrong sign, since a wall colder than the gas always makes the reading LOW, not high.

Source. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 13 (radiation exchange between surfaces, view factors, small-object approximation).

Question 7 of 8: Radiation Error in a Thermocouple Reading

Question 7: Radiation Error in a Thermocouple Reading