22-Mec-B2 Environmental Control in Buildings · May 2018
Question 2 of 8: Condensation on the walls of an enclosed balcony (10 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Eight problems, three hours, open book.
Problem 1 carries 30 marks, Problem 2 carries 10 marks and Problems 3 to 8
carry 20 marks each; candidates answer any five, and indicate their choice on
the cover of the first workbook. Psychrometric charts (SI and inch-pound) and
an R-134a pressure–enthalpy diagram are appended to the paper.
All eight problems are solved below, because the set as a whole
is the study resource.
Reference texts.
ASHRAE, Handbook — Fundamentals (2021): Ch.1 psychrometrics,
Ch.16 ventilation and infiltration, Ch.18 heating and cooling load
calculations, Ch.21 duct design, Ch.26 heat, air and moisture transmission.
McQuiston, Parker and Spitler, Heating, Ventilating and Air
Conditioning: Analysis and Design, 6th ed. — loads, psychrometric
processes, duct and air distribution design.
W. P. Jones, Air Conditioning Engineering, 5th ed. — the
percentage-saturation convention, apparatus dew point and coil by-pass factor,
face-and-by-pass plant.
Shan K. Wang, Handbook of Air Conditioning and Refrigeration, 2nd
ed. — single-duct reheat and VAV systems, discriminator (zone-demand)
control.
ASHRAE Standard 55-2023, Thermal Environmental Conditions for Human
Occupancy; ASHRAE Standard 62.1-2022, Ventilation for Acceptable
Indoor Air Quality.
National Building Code of Canada 2020 and NRCan / Environment and Climate
Change Canada climatic design data for the Canadian design conditions used in
Problems 5 and 7.
Check: two readings taken from the printed
paper.
(1) The length label on the Problem 6 duct sketch is printed as
“L1 = =100 ft == 6ft”. The equation editor
has dropped the symbols after each equals sign; the pattern
“something = something = 100 ft” and
“something = something = 6 ft” means four named
lengths in two equal pairs, and the sketch shows exactly four duct runs. The
solution therefore takes L1 = L2 = 100 ft
(plenum to tee, and tee to elbow) and L3 = L4 =
6 ft (the two drops to the ceiling diffusers), and states the reading
as an assumption under cover-page instruction 1. Only the pressure totals in
parts (c) and (d) depend on it; the duct diameters in part (a) do not.
(2) Problem 1 gives the outdoor air as “percentage saturation
50 %” but the room as “RH 50 %”. These are different
quantities and the difference is deliberate: percentage saturation is
$\mu = W/W_{s}$, relative humidity is $\phi = p_{w}/p_{ws}$. At 26 °C
the 50 % saturation state is 50.8 % RH, so treating them as
interchangeable shifts the outdoor humidity ratio by about
0.2 g/kg.
Question 2: Condensation on the walls of an enclosed balcony
(10 marks)
Given. Indoor air at 72 °F and 50 %
relative humidity, at a barometric pressure of 14.5 psia; the enclosed
balcony is an unconditioned extension whose walls are cooled by the outdoor
air.
Find. (a) the surface temperature at which condensation
begins, and (b) the construction detailing that prevents it.
Approach. Condensation begins when a surface falls to the
dew point of the air touching it, so part (a) is the dew-point temperature of
the stated indoor state; part (b) is then a question of keeping every interior
surface above that temperature and keeping vapour out of the assembly.
Part (a) — get the vapour pressure. Relative humidity
is defined on partial pressures, so at 72 °F (22.22 °C) the
ASHRAE saturation pressure is 2.681 kPa and
$$p_{w} = \phi\,p_{ws}(t) = 0.50 \times 2.681 = 1.340\ \text{kPa}
\;(0.194\ \text{psia})$$
Note that the barometric pressure does not enter here: relative humidity fixes
$p_{w}$ on its own. The 14.5 psia matters only if the humidity ratio is
wanted, which at this state is $W = 0.00845\ \text{lb/lb}$ — about
1.4 % higher than the sea-level value, and worth quoting because it is the
number a load calculation would use.
Part (a) — invert the saturation curve. The dew point
is the saturation temperature at that vapour pressure, i.e. the root of
$p_{ws}(t_{dp}) = p_{w}$:
$$p_{ws}(t_{dp}) = 1.340\ \text{kPa} \quad\Longrightarrow\quad
t_{dp} = 11.31\ ^\circ\text{C} = \boxed{52.4\ ^\circ\text{F}}$$
Any surface in contact with this air that sits at or below 52.4 °F will
collect liquid water; below 32 °F it will collect frost instead.
Diagnose the complaint before prescribing. The balcony has
been enclosed but not conditioned, so its walls and glazing run close to
outdoor temperature while the indoor air — and its 52.4 °F dew
point — migrates freely into the new space through the old exterior door
and window openings. On a Canadian winter design day the balcony surfaces are
30 to 50 °F below the dew point, so condensation is not a
defect of workmanship but the inevitable consequence of putting warm, moist air
against a cold, uninsulated enclosure.
Part (b) — the construction technique. Two things
have to be true at once: the interior surface temperature must stay above the
dew point, and moisture must not be able to reach a cold surface inside the
assembly. In order of effect:
Insulate on the outside of the mass and make the insulation
continuous. Exterior (or split) insulation keeps the whole structural
wall, including slab edges and balcony cantilevers, on the warm side of the
thermal break, which is what raises the interior surface temperature.
Break the thermal bridge at the balcony slab. An
uninterrupted concrete slab running from inside to outside is the single
largest cause of localized condensation in this detail; a structural thermal
break at the slab penetration is the standard remedy.
Put a continuous air and vapour barrier on the warm (interior)
side of the insulation, sealed at every penetration. Air leakage
carries one to two orders of magnitude more moisture into an assembly than
vapour diffusion does, so the air seal matters more than the vapour
permeance.
Upgrade the glazing to sealed double or triple units with
a low-conductivity spacer and a thermally broken frame; single glazing cannot
be kept above a 52 °F dew point at Canadian design temperatures.
Decide what the balcony is. Either bring it fully inside
the thermal envelope — insulate, air-seal and condition it, so its
surfaces sit at room temperature — or keep it outside, seal the original
exterior wall so that indoor air cannot reach it, and ventilate the balcony to
outdoors. The failure here is the halfway state.
Manage the source. Exhaust ventilation at the kitchen and
bathrooms, and a winter setpoint nearer 30 to 35 % RH, drop the
dew point to roughly 39 to 43 °F and give the assembly a
useful margin.
Close the loop with a design criterion. Expressed as a
temperature index, the requirement is
$$I = \frac{t_{\text{surface}} - t_{o}}{t_{i} - t_{o}} \ge
\frac{t_{dp} - t_{o}}{t_{i} - t_{o}}$$
At the Toronto heating design temperature of 0 °F this needs
$I \ge 0.73$, which a single-glazed, uninsulated enclosure cannot reach and a
thermally broken, insulated one comfortably can.
Problem 2 — results
Part
Quantity
Result
(a)
Vapour pressure of the room air
1.340 kPa (0.194 psia)
(a)
Onset of condensation
52.4 °F (11.3 °C) surface temperature
(a)
Humidity ratio at 14.5 psia
0.00845 lb/lb dry air
(b)
Remedy
continuous exterior insulation, structural thermal break at the slab, continuous warm-side air and vapour barrier, insulating glazing with a broken frame, and a clear decision to place the balcony either inside or outside the envelope