24-Bld-A4 Building Environmental Control Systems · May 2017

Question 1 of 14: Air-Handling Unit with Cooling-Coil / Desiccant-Bypass Dehumidification

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

Notes on this paper

National Exams May 2017 — 07-Bld-A4, Building Environmental Control Systems (Building Engineering). 3-hour closed-book exam: Section 1 (four 20-mark essay/calculation questions) + Section 2 (ten 2-mark multiple-choice questions).

Reference texts: ASHRAE Handbook — Fundamentals (2021 ed.); ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality (2022 ed.); McQuiston, Parker & Spitler, Heating, Ventilating, and Air Conditioning: Analysis and Design (6th ed.); National Building Code of Canada 2020 (NBCC); National Energy Code of Canada for Buildings 2020 (NECB).

Question 1.1: Air-Handling Unit with Cooling-Coil / Desiccant-Bypass Dehumidification (20 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. Room (IA): 26°C, 60% RH, total load ΔH = 40 kW, moisture load ΔW = 10 g/s. Outdoor air (OA): 34°C, 80% RH, with ṁ_OA = 0.25 ṁ_SA. Supply air (SA) leaves at 20°C. Point A (leaving cooling coil C1): 17°C. Point B (leaving desiccant dehumidifier D): 38°C. The C1 outlet stream splits 50%/50% into a bypass leg (unchanged, state A) and a desiccant leg (state B), which remix adiabatically to state C before coil C2. Atmospheric pressure 101.325 kPa.

Given data — key points
Point	Description	Known value
IA	Room (return) air	26°C, 60% RH
OA	Outdoor air	34°C, 80% RH, 0.25 ṁ_SA
A	Leaving cooling coil C1	17°C
B	Leaving desiccant dehumidifier D	38°C
SA	Supply air to room	20°C
ΔH, ΔW	Room total / moisture load	40 kW, 10 g/s

Find. The dry-bulb temperature and relative humidity of every key point on the psychrometric chart; the supply-air mass flow rate ṁ_SA; and the heat-removal duties Q_C1 and Q_C2 of the two cooling coils.

Fig. 1.1a — Air-handling process flow: RA and OA mix to MA, cool through C1 to state A, split 50/50 between a direct bypass and desiccant dehumidifier D (state B), remix to state C, and cool through C2 to SA.

Fig. 1.1b — Psychrometric-chart plot of the process (part 1). OA–IA is the mixing line (MA lies on it); MA→A is the C1 cooling/dehumidifying process; A→B (dashed) is the desiccant dehumidifier's constant-enthalpy line; A and B remix along the straight line to C; C→SA is the C2 cooling/dehumidifying process.

Approach. Apply steady-flow mass and energy balances at each device using standard moist-air psychrometric relations (saturation pressure, humidity ratio W, enthalpy h); locate each state, then use the room's total-to-moisture load ratio to pin down the supply state and mass flow, and coil enthalpy differences to get the coil duties.

State the room air (IA) and outdoor air (OA). Using the Buck saturation-pressure correlation at atmospheric pressure 101.325 kPa, $W = 0.622\,p_w/(P-p_w)$ with $p_w = \text{RH}\cdot p_{ws}(T)$, and $h = 1.006T + W(2501+1.86T)$ kJ/kg (T in °C, W in kg/kg): $$W_{IA}=12.64\ \text{g/kg}, \qquad h_{IA}=58.37\ \text{kJ/kg}$$ $$W_{OA}=27.29\ \text{g/kg}, \qquad h_{OA}=104.17\ \text{kJ/kg}$$
State point A (leaving C1). A cooling/dehumidifying coil driven well below the incoming dew point leaves air very close to saturation; taking A at 100% RH, 17°C gives $W_A = 12.13$ g/kg, $h_A = 47.81$ kJ/kg. Check: state A assumed saturated (100% RH) at the given 17°C leaving temperature — the standard simplification for a coil with a low bypass factor, since the exam gives no coil bypass factor or apparatus dew point.
State point B (leaving desiccant dehumidifier D). Solid-desiccant adsorption is essentially adiabatic: it trades latent heat for sensible heat along a constant-enthalpy line, so $h_B = h_A = 47.81$ kJ/kg. At the given $T_B = 38\,{}^{\circ}\text{C}$, solving $h_B = 1.006T_B + W_B(2501+1.86T_B)$ for $W_B$ gives $W_B = 3.73$ g/kg (9.1% RH — very dry, as expected for desiccant air).
State point C (bypass + B remixed, 50/50). Since the split is 50/50 and $h_B=h_A$ exactly, the mixture enthalpy is $$h_C = 0.5h_A+0.5h_B=h_A=47.81\ \text{kJ/kg}$$ regardless of the split fraction — a useful check. The dry-bulb and humidity ratio are the straight mass-weighted average of A and B: $T_C = 0.5(17)+0.5(38)=27.5\,{}^{\circ}\text{C}$, $W_C = 0.5(12.13)+0.5(3.73)=7.93$ g/kg (34.7% RH).
Locate SA and the mass flow rate via the room condition line. The room process line SA→IA has slope $\Delta H/\Delta W = 40/0.010 = 4000$ kJ/kg (per kg of water), i.e. $ (h_{IA}-h_{SA})/(W_{IA}-W_{SA}) = 4000$. Combined with $h_{SA}=1.006(20)+W_{SA}(2501+1.86\times20)$ at the given $T_{SA}=20\,{}^{\circ}\text{C}$, solving the two equations simultaneously gives $$W_{SA}=8.41\ \text{g/kg},\qquad h_{SA}=41.47\ \text{kJ/kg}\ (57.8\%\ \text{RH})$$ $$\boxed{\dot m_{SA} = \dfrac{\Delta H}{h_{IA}-h_{SA}} = \dfrac{40}{58.37-41.47} = 2.37\ \text{kg/s}}$$
State the mixed air MA and find Q_C1. With $\dot m_{OA}=0.25\dot m_{SA}$ and $\dot m_{RA}=0.75\dot m_{SA}$ (return air = supply less the outdoor-air fraction, balanced by an equal exhaust EA), the mixed state is the mass-weighted average $W_{MA}=0.75W_{IA}+0.25W_{OA}=16.30$ g/kg, $h_{MA}=0.75h_{IA}+0.25h_{OA}=69.82$ kJ/kg ($T_{MA}\approx28.0\,{}^{\circ}\text{C}$, 68.3% RH). Coil C1 removes: $$\boxed{Q_{C1} = \dot m_{SA}(h_{MA}-h_A) = 2.37(69.82-47.81) = 52.1\ \text{kW}}$$
Find Q_C2. Coil C2 cools state C down to SA: $$\boxed{Q_{C2} = \dot m_{SA}(h_C-h_{SA}) = 2.37(47.81-41.47) = 15.0\ \text{kW}}$$

Final Results — Question 1.1
Point	T (°C)	RH (%)
IA (room)	26.0	60.0
OA	34.0	80.0
MA	28.0	68.3
A (leaving C1)	17.0	100.0
B (leaving D)	38.0	9.1
C (remixed)	27.5	34.7
SA (supply)	20.0	57.8
Supply air mass flow, ṁ_SA = 2.37 kg/s
Cooling coil C1 output, Q_C1 = 52.1 kW
Cooling coil C2 output, Q_C2 = 15.0 kW

Topic: Desiccant/bypass dehumidification cycle on the psychrometric chart.

Key relations: $W=0.622p_w/(P-p_w)$; $h=1.006T+W(2501+1.86T)$; adiabatic mixing is mass-weighted in $W$ and $h$; solid-desiccant adsorption is (approximately) a constant-enthalpy process.

Why this works: A conventional cooling coil cannot dehumidify below its own leaving dew point without massive reheat, so many systems bypass part of the air around a desiccant wheel/dehumidifier that strips moisture at (nearly) constant enthalpy, trading sensible heat for latent heat. Remixing with the untouched bypass air brings the temperature back down before the final trim coil (C2) does the last sensible cooling to the supply condition. Because the bypass leg is literally state A, the mixture's enthalpy is a weighted average of two streams that are ALREADY at the same enthalpy (h_A=h_B), so h_C=h_A independent of the split ratio — a property worth recognizing to avoid needless algebra.

Common pitfall: Forgetting that return air is NOT the full supply flow (ṁ_RA=ṁ_SA−ṁ_OA, balanced by exhaust), and using the room's sensible or latent load alone rather than the combined slope ΔH/ΔW to locate the true supply-air state.

Source: ASHRAE Handbook — Fundamentals, Ch. 1 (Psychrometrics) & Ch. 24 (Air-Handling); McQuiston, Parker & Spitler, Ch. 3 & Ch. 10 (dehumidification systems).

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