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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.

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)

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. 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.

  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. 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
PartQuantityResult
(a)Vapour pressure of the room air1.340 kPa (0.194 psia)
(a)Onset of condensation52.4 °F (11.3 °C) surface temperature
(a)Humidity ratio at 14.5 psia0.00845 lb/lb dry air
(b)Remedycontinuous 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
(b)Required temperature index at 0 °F outdoorsI ≥ 0.73