Question 5 of 8: West window solar heat gain and the infiltration load expression

Question 5: West window solar heat gain and the infiltration load expression (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.

Find. (a) The instantaneous heat gain through the glazing at that instant, comprising the transmitted-plus-absorbed solar component and the conduction component driven by the 6 °C air-to-air difference. (b) A general expression for the sensible and latent components of the infiltration load.

Approach. Get the solar altitude and wall-solar azimuth from the date, latitude and hour angle; use the ASHRAE clear-sky model to build the direct, diffuse and ground-reflected irradiation on the vertical west surface; multiply by the reference-glass solar heat gain factor and the shading coefficient of the actual glazing; then add the conduction term from a series resistance calculation of the double unit.

1. Solar geometry (part a). At 18:00 solar time the hour angle is $H = 15(18 - 12) = 90\text{°}$. The altitude follows from $$\sin\beta = \sin L \sin\delta + \cos L \cos\delta \cos H = \sin 45\sin 20.6 + \cos 45 \cos 20.6 \cos 90 = 0.2488 \;\Rightarrow\; \beta = 14.4\text{°}$$ and the solar azimuth from $$\cos\phi = \frac{\sin\beta \sin L - \sin\delta}{\cos\beta \cos L} = \frac{(0.2488)(0.7071) - 0.3518}{(0.9686)(0.7071)} = -0.2568 \;\Rightarrow\; \phi = 104.9\text{° west of south}$$ For a west-facing wall ($\psi = 90\text{°}$) the wall-solar azimuth is $\gamma = |\phi - \psi| = 14.9\text{°}$, so $$\cos\theta = \cos\beta \cos\gamma = (0.9686)(0.9664) = 0.936 \;\Rightarrow\; \theta = 20.6\text{°}$$
2. Clear-sky irradiation on the window. The ASHRAE clear-sky model gives the direct normal irradiation, and the vertical surface sees half the sky and half the ground: $$G_{ND} = \frac{A}{e^{B/\sin\beta}} = \frac{1085}{e^{0.207/0.2488}} = 472\ \text{W/m}^2$$ $$G_D = G_{ND}\cos\theta = 442\ \text{W/m}^2,\qquad G_d = C\,G_{ND}\,\frac{1+\cos\Sigma}{2} = 0.136(472)(0.5) = 32\ \text{W/m}^2$$ $$G_R = G_{ND}(C + \sin\beta)\,\rho_g\,\frac{1-\cos\Sigma}{2} = 472(0.136+0.2488)(0.2)(0.5) = 18\ \text{W/m}^2$$ $$G_t = 442 + 32 + 18 = \boxed{492\ \text{W/m}^2}$$
3. Solar heat gain factor and the shading coefficient. The solar heat gain factor is the gain through the ASHRAE reference (double-strength clear sheet) glass: its transmittance plus the inward-flowing fraction of what it absorbs, about 0.87 of the incident total at this near-normal incidence: $$\text{SHGF} = 0.87\,G_t = 428\ \text{W/m}^2$$ For the actual unit — gray heat-absorbing outer sheet with a clear inner sheet — the ASHRAE fenestration table gives a shading coefficient $\text{SC} = 0.55$. Hence $$q_{\text{solar}} = A\,(\text{SC})(\text{SHGF}) = 2(0.55)(428) = \boxed{471\ \text{W}}$$ The heat-absorbing outer sheet is doing real work here: clear double glass ($\text{SC} \approx 0.87$) would pass about 745 W.
4. Conduction through the glazing. Adding the resistances in series — outdoor film (summer, 3.4 m/s, $h_o = 22.7$ W/(m²·K)), two 6 mm glass sheets ($k = 1.05$ W/(m·K)), the 1.7 cm vertical air space (ASHRAE, non-reflective surfaces, $R = 0.170$ m²·K/W) and the given indoor film: $$R_{\text{tot}} = \frac{1}{22.7} + 2\frac{0.006}{1.05} + 0.170 + \frac{1}{8} = 0.044 + 0.011 + 0.170 + 0.125 = 0.350\ \text{m}^2\!\cdot\!\text{K/W}$$ $$U = \frac{1}{0.350} = 2.85\ \text{W/(m}^2\!\cdot\!\text{K)},\qquad q_{\text{cond}} = UA(t_o - t_i) = 2.85(2)(33-27) = \boxed{34\ \text{W}}$$
5. Total instantaneous heat gain (part a). $$q = q_{\text{solar}} + q_{\text{cond}} = 471 + 34 = \boxed{505\ \text{W}}$$ That is 253 W/m² of glass, of which 93% is solar. Note the distinction the question is testing: this is the instantaneous heat gain, not the cooling load — the radiant part of it is first absorbed by floor and furnishings and reappears as cooling load over the following hours, which is what the cooling-load-factor method accounts for.
6. Sensible and latent infiltration load (part b). Infiltrating air must be brought from outdoor to indoor conditions. Splitting the enthalpy change into a dry-bulb term and a moisture term, for a volume flow $\dot V$ of outdoor air of density $\rho$: $$q_{\text{sensible}} = \dot m\,c_p (t_o - t_i) = \rho\,\dot V\,c_p\,(t_o - t_i)$$ $$q_{\text{latent}} = \dot m\,h_{fg}(W_o - W_i) = \rho\,\dot V\,h_{fg}\,(W_o - W_i)$$ $$q_{\text{total}} = \rho\,\dot V\,(h_o - h_i)$$ Evaluated at standard air these collapse to the working forms used throughout HVAC design — in SI with $\dot V$ in L/s, $q_s = 1.23\,\dot V\,\Delta t$ and $q_l = 3010\,\dot V\,\Delta W$ (both in W), $q_t = 1.20\,\dot V\,\Delta h$; in I-P with $\dot V$ in cfm, $q_s = 1.10\,\dot V\,\Delta t$ and $q_l = 4840\,\dot V\,\Delta W$ (both Btu/h), $q_t = 4.5\,\dot V\,\Delta h$. The same expressions serve for mechanical ventilation air; only the means of establishing $\dot V$ differs.