Question 8 of 8: Instantaneous heat gain through a west-facing double-glazed window

Problem 8: Instantaneous heat gain through a west-facing double-glazed window (20 points)

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. A 2 m² vertical insulating glass unit facing due west.

Find. The instantaneous heat gain, solar plus conduction, at that instant.

Approach. Locate the sun, get the clear-day irradiance on the window from the ASHRAE clear-sky model, convert it to a solar heat gain factor for reference glass, apply a shading coefficient for the actual glazing, and add the conduction term from a resistance network built on the given inside film coefficient.

1. Solar position. July 21 gives a declination $\delta = 20.6^{\circ}$ and 18:00 solar time gives an hour angle $H = 15(18-12) = 90^{\circ}$. The altitude follows from $$\sin\beta=\cos L\cos\delta\cos H+\sin L\sin\delta=\sin 40^{\circ}\sin 20.6^{\circ}=0.2262 \Rightarrow \beta=13.1^{\circ}$$ and the azimuth from $$\cos\phi=\frac{\sin\beta\sin L-\sin\delta}{\cos\beta\cos L}\Rightarrow \phi=106.1^{\circ}\ \text{west of south}$$
2. Incidence angle on the glass. The wall azimuth of a west window is $\psi = 90^{\circ}$, so the wall solar azimuth is $\gamma=\phi-\psi=16.1^{\circ}$ and $$\cos\theta=\cos\beta\cos\gamma=0.9740(0.9609)=0.9359\ \Rightarrow\ \boxed{\theta=20.6^{\circ}}$$ Because $|\gamma| < 90^{\circ}$ the glass is in direct sun, which is the check that a west window at 6 p.m. really does see beam radiation — it is at its worst hour of the day.
3. Clear-sky irradiance. With the ASHRAE July constants $A = 1085\ \text{W/m}^2$, $B = 0.207$, $C = 0.136$ and a ground reflectance of 0.2, $$G_{ND}=\frac{A}{e^{B/\sin\beta}}=\frac{1085}{e^{0.207/0.2262}}=434\ \text{W/m}^2$$ $$G_D=G_{ND}\cos\theta=407,\quad G_d=CG_{ND}F_{ws}=29.5,\quad G_R=G_{ND}(C+\sin\beta)\rho_g F_{wg}=15.7\ \text{W/m}^2$$ $$G_t = 407+29.5+15.7=452\ \text{W/m}^2$$ with the vertical-surface view factors $F_{ws}=F_{wg}=0.5$.
4. Solar heat gain factor for reference glass. The SHGF is what 3 mm double-strength sheet glass would transmit plus the inward-flowing part of what it absorbs. With the ASHRAE angular polynomials at $\cos\theta = 0.9359$, $\tau_b = 0.876$ and $\alpha_b = 0.047$, the diffuse values $\tau_d = 0.799$ and $\alpha_d = 0.054$, and an inward-flowing fraction $N = h_i/(h_i+h_o) = 8/(8+22.7) = 0.261$: $$\text{SHGF}=G_D(\tau_b+N\alpha_b)+(G_d+G_R)(\tau_d+N\alpha_d)=\boxed{398\ \text{W/m}^2}$$ which is 126 Btu/h·ft² — consistent with the tabulated ASHRAE value for a west exposure at 40°N in July.
5. Shading coefficient of the actual unit. The window is not reference glass: it is a 6 mm gray heat-absorbing outer light with a clear inner light. The tabulated shading coefficient for that combination in an insulating unit is $SC = 0.55$, so $$q_{\text{solar}}=A\times SC\times \text{SHGF}=2(0.55)(398)=\boxed{438\ \text{W}}$$ Roughly speaking, the heat-absorbing outer light intercepts the energy in the glass itself and then re-radiates and convects most of it back outside, which is why $SC$ is barely half.
6. Conduction: build the resistance network. The question gives $h_i$, which is an instruction to assemble $U$ from resistances rather than to look one up: $$R_o=\frac{1}{22.7}=0.044,\quad R_{\text{glass}}=\frac{0.006}{0.9}=0.0067\ \text{(each light)},$$ $$R_{\text{air space}}=0.17\ \ (16\ \text{mm vertical, ordinary glass surfaces}),\quad R_i=\frac{1}{8}=0.125$$ $$R_{\text{tot}}=0.044+0.0067+0.17+0.0067+0.125=0.352\ \frac{\text{m}^2\text{K}}{\text{W}} \Rightarrow U=\frac{1}{0.352}=2.84\ \frac{\text{W}}{\text{m}^2\text{K}}$$
7. Conduction gain. $$q_{\text{cond}}=UA(t_o-t_i)=2.84(2)(35-26)=\boxed{51\ \text{W}}$$
8. Total instantaneous heat gain. $$q=q_{\text{solar}}+q_{\text{cond}}=438+51=\boxed{489\ \text{W}}$$ Solar accounts for 90 % of it. That ratio is the engineering message of the question: on a west facade in late afternoon the conduction term is almost irrelevant, and money spent on shading or on a lower shading coefficient buys nine times what the same money spent on a better $U$-value does. Note also that this is the instantaneous heat gain, not the cooling load — converting it requires a solar cooling load factor to account for the mass of the room storing and releasing part of it.