Heat Loss Through a Single-Pane Window (Natural Convection Both Sides): 17-Phys-B6 Applied Thermodynamics and Heat Transfer, December 2013, Q5

Question 5: Heat Loss Through a Single-Pane Window (Natural Convection Both Sides)

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. Single glass pane, $\delta=10$ mm, indoor air $25\,{}^{\circ}\text{C}$, outdoor air $-15\,{}^{\circ}\text{C}$, both surfaces cooled/heated by natural (free) convection only, radiation neglected. Glass thermal conductivity $k=1.4$ W/m·K (typical soda-lime window glass, not printed in the source). The window height needed to evaluate the Rayleigh number is not given in the source; a representative residential window height $L=1$ m is assumed (flagged below).

Given / assumed data
Quantity	Symbol	Value
Indoor air temperature	$T_i$	$25\,{}^{\circ}\text{C}$
Outdoor air temperature	$T_o$	$-15\,{}^{\circ}\text{C}$
Glass thickness	$\delta$	10 mm
Glass conductivity	$k$	1.4 W/m·K (assumed)
Window height	$L$	1 m (assumed)

Find. The heat-transfer rate per unit area $q''$ through the window.

Series thermal-resistance network for the window: indoor natural convection, glass conduction, outdoor natural convection.

Approach. Model the window as three resistances in series (indoor film, glass, outdoor film) and solve for the two unknown glass surface temperatures $T_{s,i}$, $T_{s,o}$ by requiring the same $q''$ through all three, with $h_i$ and $h_o$ each evaluated from the Churchill–Chu vertical-plate natural-convection correlation at their own film temperature (the coefficients depend on the very surface temperatures being solved for, so the system is solved iteratively/numerically rather than by hand).

Governing equations. With the same heat flux through all three resistances, $$q'' = h_i(T_i-T_{s,i}) = \frac{k}{\delta}(T_{s,i}-T_{s,o}) = h_o(T_{s,o}-T_o)$$ where each convection coefficient comes from the Churchill–Chu correlation for a heated (or cooled) vertical plate, evaluated at its own film temperature $T_f=(T_{surface}+T_{air})/2$: $$\text{Ra}_L=\frac{g\beta|T_s-T_\infty|L^3}{\nu\alpha}, \qquad \text{Nu}_L=\left\{0.825+\frac{0.387\,\text{Ra}_L^{1/6}}{\big[1+(0.492/\text{Pr})^{9/16}\big]^{8/27}}\right\}^2, \qquad h=\frac{\text{Nu}_L\,k_{air}}{L}$$
Numerical (iterative) solution. Solving the three coupled equations simultaneously (air properties evaluated at each film temperature) converges to $$T_{s,i} = 4.96\,{}^{\circ}\text{C}, \qquad T_{s,o} = 4.38\,{}^{\circ}\text{C}$$ $$h_i = 4.03\text{ W/m}^2\cdot\text{K}\ (\text{Ra}_L\approx2.25\times10^9), \qquad h_o = 4.17\text{ W/m}^2\cdot\text{K}\ (\text{Ra}_L\approx3.06\times10^9)$$ Both films land in the turbulent natural-convection range for a 1 m-tall plate at this $\Delta T$, and (as expected for a single glazing) the two film coefficients are close in magnitude since both sides see a similar temperature difference from the glass to the ambient air.
Total resistance and heat flux. The glass conduction resistance is tiny next to the two air films: $$R''_{tot} = \frac{1}{h_i}+\frac{\delta}{k}+\frac{1}{h_o} = \frac{1}{4.03}+\frac{0.010}{1.4}+\frac{1}{4.17} = 0.2481+0.0071+0.2398 = 0.4955\text{ m}^2\cdot\text{K/W}$$ $$\boxed{q'' = \frac{T_i-T_o}{R''_{tot}} = \frac{25-(-15)}{0.4955} = 80.7\text{ W/m}^2}$$ (check: $h_i(T_i-T_{s,i})=4.03\times20.04=80.7$ and $h_o(T_{s,o}-T_o)=4.17\times19.38=80.7$ — all three legs agree.)

Check: The source does not print a window height; $L=1$ m is assumed as a typical residential dimension (per the exam's own "state your assumptions" instruction). Both films are in the turbulent range ($\text{Ra}_L>10^9$), where the Churchill–Chu correlation tends to $\text{Nu}_L\propto\text{Ra}_L^{1/3}\propto L$, so $h$ is almost independent of height: re-solving the same system gives $q''=85.0$ W/m² for $L=0.5$ m and $77.8$ W/m² for $L=2$ m (within about $\pm5\%$ of the 80.7 W/m² found here). The conclusion — single glazing dominated by the two air films, glass conduction negligible — does not depend on the choice.

Question 5 — results
Quantity	Value
Inside glass surface temperature	$4.96\,{}^{\circ}\text{C}$
Outside glass surface temperature	$4.38\,{}^{\circ}\text{C}$
Indoor film coefficient $h_i$	4.03 W/m²·K
Outdoor film coefficient $h_o$	4.17 W/m²·K
Heat flux $q''$	80.7 W/m²

Topic. Series conduction/natural-convection resistance network for a single-pane window.

Key relations. Churchill–Chu correlation for a vertical plate (all Rayleigh numbers); series thermal resistance $R''_{tot}=1/h_i+\delta/k+1/h_o$; $q''=\Delta T_{overall}/R''_{tot}$.

Why this works. Because $h_i$ and $h_o$ each depend on the unknown surface temperature they are trying to predict, the three-equation system is inherently implicit; solving it self-consistently (rather than guessing a surface temperature once) is what a full analysis requires, and is exactly why a real design table or an iterative/numerical solve is standard practice, not a shortcut.

Common pitfall. Assuming the glass conduction resistance dominates (it is almost 35× smaller than either air film here) and ignoring the natural-convection correlation's own $\Delta T$-dependence, which would understate how much of the overall temperature drop actually occurs at the air films rather than across the glass itself.

Source. F. P. Incropera and D. P. DeWitt, Fundamentals of Heat and Mass Transfer, 7th ed., Ch. 9 (free/natural convection, vertical-plate correlations).

Question 5 of 8: Heat Loss Through a Single-Pane Window (Natural Convection Both Sides)

Question 5: Heat Loss Through a Single-Pane Window (Natural Convection Both Sides)