22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2013
Question 7 of 8: Thermocouple Radiation Error in a Duct
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — ideal-gas processes, gas power cycles, reheat Rankine and vapour-compression refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — natural-convection and internal-flow correlations, lumped-capacitance transients, radiation exchange, and the LMTD/ε–NTU heat-exchanger method. Freon-12 property data are read from the appendix supplied with the exam; steam properties from standard tables. This is an open-book, equal-value paper; candidates answer any five of the eight questions (three from one Part and two from the other) — all eight are worked here as a study resource.
Question 7: Thermocouple Radiation Error in a Duct (equal value)
Given. Bead reads $T_b=179\ ^\circ\text{C}=452.15$ K in a duct at $T_d=65\ ^\circ\text{C}=338.15$ K; $\varepsilon_b=0.7$, $\varepsilon_d=0.9$; bead 3 mm, duct $L=0.92$ m; the convective heat-transfer coefficient at the bead is $h=790\ \text{W/m}^2\cdot^\circ\text{C}$ (see note on the printed “790 W/m²”).
Find. The reading error $T_\text{gas}-T_b$ caused by radiation from the bead to the cooler duct wall.
Figure 6 — The bead gains heat by convection from the hot gas and loses it by radiation to the cooler duct wall, so it settles below the true gas temperature. Because the bead is tiny (D « L gives Fb-d ≈ 1), it radiates essentially as a small body enclosed by the duct.
Approach. Evaluate the configuration factor to confirm the bead behaves as a small body in a large enclosure, compute the net radiative loss from the bead, then a steady-state energy balance (convection in = radiation out) gives the temperature error directly.
Configuration factor. With a 3 mm bead and $L=0.92$ m, $F_{b\text{-}d}=0.92/\sqrt{D^2+0.92^2}\approx1$ for any duct diameter of order $L$ or smaller — the bead sees essentially the whole duct as its enclosure, so the net radiation reduces to the small-body result $q''_\text{rad}=\varepsilon_b\sigma(T_b^4-T_d^4)$.
Radiative loss from the bead.$$q''_\text{rad}=0.7(5.67\times10^{-8})(452.15^4-338.15^4)=\boxed{1140\ \text{W/m}^2}.$$
Steady-state energy balance → reading error. The bead is steady, so convection gain equals radiation loss: $h(T_\text{gas}-T_b)=q''_\text{rad}$. Hence$$T_\text{gas}-T_b=\frac{q''_\text{rad}}{h}=\frac{1140}{790}=\boxed{1.4\ ^\circ\text{C}},$$so the true gas temperature is about $180.4\ ^\circ\text{C}$ — the thermocouple under-reads by roughly $1.4\ ^\circ\text{C}$.
Check (source ambiguity)
Two items are not fully specified in the question: the duct diameter $D$ (needed for $F_{b\text{-}d}$) is left symbolic — taken as $F_{b\text{-}d}\approx1$ since $D\ll L$ for a slender duct — and the printed “790 W/m²” is read as the bead convection coefficient $h=790\ \text{W/m}^2\cdot^\circ\text{C}$, the quantity required to turn a radiative flux into a temperature error (this $h$ implies a bead Nusselt number of order 60, physically reasonable for gas flowing past a 3 mm sphere). The method — balance convection against radiation to the walls — is exact; only the numeric error scales inversely with $h$ (e.g. $h=395$ would double it to ≈ 2.9 °C).
