Widening the comfort band from a fixed set point saves energy because
Answer: A
Every degree of set point relaxation in summer cuts the cooling hours and the load appreciably, and in a naturally ventilated building it may remove the plant's operation altogether for weeks. It costs nothing but the acceptance.
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A wall of total thermal resistance 0.5 m\( ^2 \)K/W has a U-value of
Answer: B
Correct answer: 2.0 W/m2K
\[ U = \frac{1}{R} = \frac{1}{0.5} = 2.0\ \mathrm{W/m^2K}. \] The U-value and the total resistance are simple reciprocals, so improving a wall means adding resistance and watching the U-value fall - with diminishing returns, since each further increment of resistance buys less.
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Adding 50 mm of insulation of conductivity 0.04 W/mK to that wall raises its resistance to
Answer: C
Correct answer: 1.75 m2K/W
The insulation contributes \[ R = \frac{t}{k} = \frac{0.05}{0.04} = 1.25\ \mathrm{m^2K/W}, \] and resistances in series add: \( 0.5 + 1.25 = 1.75 \). The U-value falls from 2.0 to 0.57, so 50 mm of insulation has removed more than seventy per cent of the heat loss through that wall.
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A wall of 100 m\( ^2 \) and U-value 2.0 W/m\( ^2 \)K with 15 \( ^\circ \)C across it conducts
Answer: D
Correct answer: 3000 W
\[ Q = UA\Delta T = 2.0 \times 100 \times 15 = 3000\ \mathrm{W}. \] The same wall insulated to the 1.75 resistance of the previous item would conduct 857 W instead, which is the whole case for insulation expressed in one comparison.
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Glazing of 20 m\( ^2 \) with a solar heat gain coefficient of 0.4 under 500 W/m\( ^2 \) of incident radiation admits
Answer: A
Correct answer: 4000 W
\[ Q = A \times \mathrm{SHGC} \times I = 20 \times 0.4 \times 500 = 4000\ \mathrm{W}. \] Note how this compares with the 3000 W conducted through 100 square metres of uninsulated wall - a fifth of the area admits more heat, which is why glazing rather than opaque fabric usually governs a cooling load.
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Replacing that glazing with a coefficient of 0.25 reduces the admitted gain to
Answer: B
The gain falls in proportion to the coefficient, so 500 times 20 times 0.25 is 2.5 kilowatts, a reduction of 1.5. External shading achieving the same reduction would usually cost less and also improve the daylight quality.
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A facade of 200 m\( ^2 \) gross wall area carrying 70 m\( ^2 \) of glazing has a window to wall ratio of
Answer: C
Correct answer: 35 per cent
\[ \frac{70}{200} = 0.35. \] Thirty-five per cent is a moderate figure for a commercial facade. It is taken against the gross wall area including the glazing, so dividing by the opaque area alone would give 54 per cent - a different quantity and not the one the codes define.
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An indoor illuminance of 300 lux under an outdoor illuminance of 15 000 lux gives a daylight factor of
Answer: D
Three hundred divided by fifteen thousand is 0.02, that is two per cent, which is a reasonable figure for a side-lit room away from the window. Five per cent or more would be a well daylit space needing no electric light in the day.
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A room with a window head 2.5 m above floor level can be daylit usefully to a depth of about
Answer: A
Correct answer: 5 m
The usual rule is that daylight penetrates usefully to about twice the height of the window head above the floor, so \[ 2 \times 2.5 = 5\ \mathrm{m}. \] It is the head height rather than the window area that governs, which is why a tall narrow window daylights a room more deeply than a wide low one of the same area.
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A roof of 500 m\( ^2 \) at U-value 1.2 W/m\( ^2 \)K under 20 \( ^\circ \)C of difference conducts
Answer: B
Correct answer: 12 kW
\[ Q = UA\Delta T = 1.2 \times 500 \times 20 = 12{,}000\ \mathrm{W}. \] A roof is usually the largest single element of the envelope and receives the strongest solar radiation, so it repays insulation more than any other surface - and is generally the easiest to insulate.
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