22-Mec-B2 Environmental Control in Buildings · December 2017

Question 6 of 8: Overall U factor of a curtain wall with 40% glass

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Engineers Canada national examination 16-Mec-B2 Environmental Control in Buildings, December 2017, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. ASHRAE Psychrometric Chart No. 1 (SI and inch-pound) and a pressure–enthalpy diagram for R-717 are appended to the paper as pages 6–8.

All eight problems are worked here. The paper mixes SI and inch-pound units deliberately: Problems 1, 3 and 4 are SI, Problems 2, 6 and 7 are inch-pound, and Problem 8 is SI with a Canadian climate. Each solution is worked in the units the question uses, as the cover-page instructions require.

Reference texts for this subject.

W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann — the standard reference for this examination code; Ch. 2–3 (psychrometry and the psychrometric chart), Ch. 6 (air-conditioning plant cycles), Ch. 7 (the cooling coil, apparatus dew point and by-pass factor), Ch. 9 (cooling towers), Ch. 15 (fans).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 5 (heat transmission in building structures), Ch. 8 (energy estimating and the degree-day method), Ch. 12 (fans and duct design).
ASHRAE Handbook — Fundamentals — Ch. 1 (psychrometrics), Ch. 14 (climatic design information), Ch. 21 (duct design), Ch. 25–27 (thermal and moisture performance of the building envelope), Ch. 30 (fenestration).
Stoecker & Jones, Refrigeration and Air Conditioning, 2nd ed., McGraw-Hill — Ch. 10–12 (vapour-compression cycle, compressors, condensers and evaporators); ASHRAE Handbook — Refrigeration for ammonia plant practice.
National Building Code of Canada and the National Energy Code of Canada for Buildings (NRC), Appendix C climatic data; CSA and Canada Green Building Council material for Problem 5.

Property basis used throughout. Moist-air properties are computed from the ASHRAE Fundamentals ideal-moist-air relations, so every state quoted here can be read back off the psychrometric chart supplied with the paper:

$$W=\frac{0.621945\,p_w}{p-p_w},\qquad h_{\text{SI}}=1.006\,t+W\,(2501+1.86\,t),\qquad h_{\text{IP}}=0.240\,t+W\,(1061+0.444\,t)$$

with $h$ in $\text{kJ/kg}$ of dry air for $t$ in $\,{}^{\circ}\text{C}$ and in $\text{Btu/lb}$ of dry air for $t$ in $\,{}^{\circ}\text{F}$. Ammonia properties are quoted on the same datum as the attached ASHRAE p–h diagram ($h_f=200\ \text{kJ/kg}$ and $s_f=1.0\ \text{kJ/(kg}\cdot\text{K)}$ for saturated liquid at $0\,{}^{\circ}\text{C}$); only differences enter the answers, so any consistent chart or table gives the same duties.

Question 6: Overall U factor of a curtain wall with 40% glass (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.

Layer (outside to inside)	Thickness	Resistance $R$ (h·ft²·°F/Btu)
Outside air film, $15$ mph wind (winter)	—	$0.17$
Limestone	$3\ \text{in}$	$0.24$
Air space, vertical, non-reflective	$6\ \text{in}$	$1.01$
Lightweight-aggregate concrete block	$8\ \text{in}$	$2.00$
Air space (furring), non-reflective	$3/4\ \text{in}$	$1.01$
Gypsum plasterboard	$1/2\ \text{in}$	$0.45$
Inside air film, still air, horizontal flow	—	$0.68$
Glazing: $1/4$ in plate glass, single	—	$U_g=1.04\ \text{Btu/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)}$
Glass area fraction	—	$40\%$
Design temperature difference	$\Delta t$	$75-10=65\,{}^{\circ}\text{F}$

Find. The overall U factor of the wall assembly including its $40\%$ glazed area, and the resulting design heat loss per unit area.

The wall section as drawn, with its series-resistance network below. The glazed area, shown to the same scale, is a single 1/4 in plate-glass light occupying 40 % of the elevation.

Approach. Sum the series resistances of the opaque path to get its U value, take the glazing U value from the fenestration tables, and combine the two paths in parallel by area weighting — the two paths run side by side between the same two air temperatures, so their conductances add.

Read the section and assign resistances. The values above are the standard ASHRAE Fundamentals figures for winter conditions. Two deserve comment. The $6\ \text{in}$ cavity is given $R=1.01$ rather than something proportional to its width, because the resistance of a plane air space is almost independent of thickness beyond about $3/4\ \text{in}$: once the gap is wide enough for a convection cell to form, widening it further increases convection as fast as it increases the conducting path, and radiation across the gap does not depend on the gap at all. The $8\ \text{in}$ lightweight-aggregate block is given $R=2.00$, roughly three times the value of a normal-weight block of the same size.
Sum the series resistances of the opaque wall. The layers lie one behind another, all carrying the same heat flux, so resistances add: $$\sum R = 0.17+0.24+1.01+2.00+1.01+0.45+0.68 =\boxed{5.56\ \text{h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F/Btu}}$$ $$U_{wall}=\frac{1}{\sum R}=\frac{1}{5.56} =\boxed{0.180\ \text{Btu/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)}}$$ The $8\ \text{in}$ block contributes $36\%$ of the total resistance and the two air spaces together another $36\%$ — in an uninsulated masonry wall of this kind the cavities are doing as much work as the masonry.
The glazing path. A single light of $1/4\ \text{in}$ float glass has almost no resistance of its own: the conductivity of glass is about $5.5\ \text{Btu}\cdot\text{in/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)}$, so the glass itself is worth only $R\approx0.05$, and the U value is set almost entirely by the two surface films: $$U_g\approx\frac{1}{0.17+0.05+0.68}=1.11 \;\Rightarrow\;\text{take the ASHRAE winter design value } U_g=1.04\ \text{Btu/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)}$$ Single glazing is therefore about six times as conductive as the wall beside it.
Combine the two paths in parallel. The opaque wall and the glass span the same two air temperatures, so their conductances add in proportion to area: $$U_{overall}=(1-A_g)\,U_{wall}+A_g\,U_g =0.60\times0.180+0.40\times1.04$$ $$U_{overall}=0.108+0.416 =\boxed{0.524\ \text{Btu/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)}}$$ which is $2.98\ \text{W/(m}^2\cdot\text{K)}$ in SI. The arithmetic makes the point of the question: the glass occupies $40\%$ of the elevation but carries $79\%$ of the heat loss.
Design heat loss. With the stated $\Delta t=75-10=65\,{}^{\circ}\text{F}$, $$q=U_{overall}\,\Delta t=0.524\times65 =\boxed{34.1\ \text{Btu/(h}\cdot\text{ft}^2)}$$ or $107\ \text{W/m}^2$ — several times what the National Energy Code of Canada for Buildings would permit today, which is exactly why this construction is no longer built.
The concrete column as a thermal bridge. The $14\times14\ \text{in}$ column shown in the spandrel is a parallel path of much lower resistance than the wall around it. Normal-weight concrete has $k\approx12\ \text{Btu}\cdot\text{in/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)}$, so $14\ \text{in}$ of it is worth only $R=1.17$, and with the two films $$U_{col}=\frac{1}{0.17+1.17+0.68}=0.496\ \text{Btu/(h}\cdot\text{ft}^2 \cdot{}^{\circ}\text{F)}$$ — nearly three times the wall value and close to the glass. The question does not give a column spacing, so the column area fraction cannot be computed; but as a sensitivity, if the columns occupied $5\%$ of the elevation at the expense of the opaque wall the overall U factor would rise from $0.524$ to $0.540$ (up $3.0\%$), and at $10\%$ to $0.556$ (up $6.0\%$). The effect is real but secondary to the glazing.

Final results.

Quantity	Result
Total resistance of the opaque wall	$\sum R=5.56\ \text{h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F/Btu}$
Opaque-wall U factor	$U_{wall}=0.180\ \text{Btu/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)}$
Glazing U factor (single $1/4$ in plate)	$U_g=1.04$
Overall U factor (60% wall + 40% glass)	$\mathbf{0.524\ \text{Btu/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)}}$ $=2.98\ \text{W/(m}^2\cdot\text{K)}$
Design heat loss at $\Delta t=65\,{}^{\circ}\text{F}$	$\mathbf{34.1\ \text{Btu/(h}\cdot\text{ft}^2)}$ $=107\ \text{W/m}^2$
Share of the loss through the glass	$79.4\%$ on $40\%$ of the area
Concrete-column path (thermal bridge)	$U_{col}=0.496$; $+3.0\%$ on the overall U if it is $5\%$ of the area

Check: reading of the drawing, declared under cover-page instruction 1. The section is a vertical cut showing both a spandrel and a window, and the seven labels have to be assigned between the two. The reading taken here is the one that makes a complete and conventional assembly: the opaque wall is $3''$ limestone / $6''$ cavity / $8''$ lightweight block / $3/4''$ furred cavity / $1/2''$ gypsum board, and the glazed area is a single $1/4''$ plate-glass light. The alternative reading — that the $3/4''$ air space belongs to the glazing, making it a double-glazed unit — would leave the gypsum board furred directly against the block with no cavity, and would give $U_{wall}=0.220$, $U_g\approx0.55$ and an overall $U=0.352$. That is a materially different answer, so the reading is stated rather than assumed silently. Two further points: (i) the $14\times14$ concrete column is quantified above as a sensitivity only, because no column spacing is given; (ii) the framing of the glazing is neglected, so $U_g$ is a centre-of-glass value — a real aluminium curtain-wall frame without a thermal break would make the overall figure worse, not better.

Topic. Series and parallel thermal resistance in building envelopes.

Key relations. Series (through the thickness): $\sum R=R_o+\sum(x/k)+\sum R_{air space}+R_i$ and $U=1/\sum R$. Parallel (side by side, same air temperatures): area-weighted conductances, $U_{overall}=\sum A_i U_i/\sum A_i$. Heat flux $q=U\,\Delta t$. Conversions: $1\ \text{Btu/(h}\cdot\text{ft}^2\cdot{}^{\circ}\text{F)} =5.678\ \text{W/(m}^2\cdot\text{K)}$.

Why this works. Resistances add in series and conductances add in parallel, for the same reason that they do in an electrical network: layers in series carry the same flux and their temperature drops add, whereas paths in parallel share the same driving temperature difference and their fluxes add. The practical consequence is that a wall is only as good as its worst path, and that an area-weighted U value is dominated by whichever element has the highest conductance rather than by whichever occupies the most area. Here the glass has six times the wall conductance, so at $40\%$ of the area it carries four-fifths of the loss; adding insulation to the opaque wall would barely move the overall figure, while double glazing would nearly halve it. That asymmetry is why modern energy codes regulate the window-to-wall ratio and the fenestration U value directly rather than trusting an area-weighted average, and why the assembly in this question — uninsulated masonry with single glazing — belongs to the era it was drawn in.

Common pitfall. Averaging the two U values arithmetically, or worse, averaging the resistances. Resistances may not be area-averaged: doing so here would give $\sum R=0.6(5.56)+0.4(0.96)=3.72$ and $U=0.269$, roughly half the true value, because it lets the good path shield the bad one. The other common slip is scaling the air space resistance with its width; a $6\ \text{in}$ cavity is worth essentially the same as a $3/4\ \text{in}$ one.

Source. ASHRAE Handbook — Fundamentals, Ch. 26 (heat, air and moisture transfer — assembly R-values and air-space resistances) and Ch. 15 (fenestration U factors); McQuiston, Parker & Spitler, 6th ed., Ch. 5.

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