22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2013
Question 5 of 8: Heat Loss Through a Single-Glazed Window
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 5: Heat Loss Through a Single-Glazed Window (equal value)
Given. Single glass pane $\delta=10$ mm with $k_g\approx1.4\ \text{W/m}\cdot\text{K}$; inside air $T_i=25\ ^\circ\text{C}$, outside air $T_o=-15\ ^\circ\text{C}$; both surfaces exchange heat by natural convection only. No window dimension is printed, so a representative height $H=1.0$ m is assumed for the free-convection correlation (flagged below).
Find. The steady heat flux $q''$ through the window.
Figure 5 — Series thermal circuit: inside film (1/hᵢ), glass conduction (δ/k), outside film (1/hₒ). The glass resistance is negligible; the two convective films dominate.
Approach. The three resistances (inside film, glass, outside film) are in series, but the film coefficients depend on the (unknown) surface temperatures, so iterate: assume surface temperatures, evaluate each free-convection coefficient from the Churchill–Chu vertical-plate correlation, update the flux, and repeat to convergence.
Resistance network. With radiation negligible, $q''=(T_i-T_o)\big/\big(\tfrac{1}{h_i}+\tfrac{\delta}{k_g}+\tfrac{1}{h_o}\big)$. The glass term $\delta/k_g=0.010/1.4=0.0071\ \text{m}^2\text{K/W}$ is tiny beside the film terms ($\sim0.25$ each), so the answer is set by the two convection coefficients.
Free-convection coefficients (Churchill–Chu). For each vertical surface, $\mathrm{Ra}_H=g\beta\,\Delta T\,H^3/(\nu\alpha)$ and $\mathrm{Nu}=\big\{0.825+0.387\,\mathrm{Ra}^{1/6}/[1+(0.492/\mathrm{Pr})^{9/16}]^{8/27}\big\}^2$, then $h=\mathrm{Nu}\,k/H$. Converged film temperatures give$$h_i\approx3.95,\qquad h_o\approx4.10\ \text{W/m}^2\cdot\text{K}.$$
Converged surface temperatures and flux. Iteration settles at inner-surface $\approx4.9\ ^\circ\text{C}$ and outer-surface $\approx4.4\ ^\circ\text{C}$ (the glass carries almost no temperature drop), giving$$q''=\frac{25-(-15)}{\tfrac{1}{3.95}+0.0071+\tfrac{1}{4.10}}=\frac{40}{0.504}=\boxed{79\ \text{W/m}^2}.$$
Check (assumption)
The exam prints no window height, so $H$ must be assumed; a representative $H=1.0$ m and glass $k_g=1.4\ \text{W/m}\cdot\text{K}$ are used here. The answer is far less sensitive to that choice than it first appears: at $\mathrm{Ra}\sim10^9$ the Churchill–Chu bracket is dominated by its $\mathrm{Ra}^{1/6}$ term, so $\mathrm{Nu}\propto\mathrm{Ra}^{1/3}\propto H$ and $h=\mathrm{Nu}\,k/H$ is almost independent of height — only the constant 0.825 in the bracket leaves any residual dependence. Recomputing gives $q''=83.5\ \text{W/m}^2$ at $H=0.5$ m and $76.4\ \text{W/m}^2$ at $H=2.0$ m against $79.3\ \text{W/m}^2$ at $H=1.0$ m, i.e. only $+5\%$/$-4\%$ over a fourfold change in height. Report $q''\approx80\ \text{W/m}^2$ and state the assumed height, per exam Note 1.
