22-Mec-B2 Environmental Control in Buildings · May 2013

Question 1 of 8: Mixed-air plant with reheat — cycle, loads and plant energy (20 marks)

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

Notes on this paper

Paper format. Professional Engineers of Ontario / EGBC annual examination, 07-Mec-B2 (now 22-Mec-B2) Environmental Control in Buildings, May 2013 sitting. Three hours, open book. Eight problems of 20 points each; the candidate is instructed to solve five and to nominate which five are to be graded. Psychrometric charts and a pressure–enthalpy diagram for ammonia (R-717) are appended to the paper, and candidates are expected to bring an environmental-control text and steam tables. Instruction 1 invites the candidate to state any interpretation assumptions with the answer — that latitude is used explicitly below where the printed data are redundant.

All eight problems are worked here. Every psychrometric state has been recomputed from the ASHRAE formulation for saturation vapour pressure rather than scaled off a chart, so the numbers are tighter than a graphical solution would be; chart-quality agreement (about ±0.2 K and ±0.0002 kg/kg) is all that an examiner expects.

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), Ch. 6 (cooling loads), Ch. 10 (cooling towers), Ch. 15 (duct design).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 8 (energy estimating and degree-day methods), Ch. 12–13 (fluid flow and duct design).
ASHRAE Handbook – Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 21 (duct design), Ch. 25–27 (heat, air and moisture transfer in the envelope).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression and multistage refrigeration).
ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy, and ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality.
Canadian frame: National Building Code of Canada 2020, National Energy Code of Canada for Buildings 2020, and Environment and Climate Change Canada Canadian Climate Normals for degree-day data.

Psychrometric relations used throughout. At barometric pressure $p$, with saturation vapour pressure $p_{ws}(t)$ from the ASHRAE correlation,

$$W = 0.6220\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}, \qquad h = 1.006\,t + W\,(2501 + 1.86\,t)$$

in SI (kJ per kg of dry air), and in the inch-pound system $h = 0.240\,t + W\,(1061 + 0.444\,t)$ Btu per lb of dry air. The thermodynamic wet-bulb temperature is obtained from the adiabatic-saturation equation, which is what a chart's constant-wet-bulb lines represent.

Question 1: Mixed-air plant with reheat — cycle, loads and plant energy (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. A single-zone plant handling a mixture of return and outdoor air through a chilled-water cooling coil and a reheat coil, at sea level ($p = 101.325$ kPa).

Given data
Quantity	Symbol	Value
Zone sensible load	$\dot{Q}_s$	20.5 kW
Zone latent load	$\dot{Q}_l$	8.8 kW
Zone (room) state	R	25 °C, 50% RH
Supply air state and flow	S	14 °C, 60% RH, 1.8 kg/s
Outdoor design state	O	27 °C, 70% RH
Re-circulated : fresh air (by mass)	—	3 : 1
Cooling-coil apparatus dew point	ADP	5 °C
Refrigeration plant coefficient of performance	COP	2

Find. (a) the plant schematic, (b) the cycle plotted on the psychrometric chart, (c) a table of dry- and wet-bulb temperatures at every significant point, (d) the total air-conditioning load imposed by the room, (e) the total energy input to the plant, and (f) the energy input when the reheat coil is served by condenser heat recovery.

(a) Plant schematic. Outdoor air O mixes with re-circulated room air R in the ratio 1 : 3 to give state M; the chilled-water coil cools and dehumidifies M to W; the heating coil reheats W to the supply state S; the fan delivers S to the zone, which returns air at state R.

Approach. Fix R, S and O from the ASHRAE moisture-content relation, mix R and O by mass fraction to get M, locate the coil off-state W on the straight line from M to the apparatus dew point at the supply moisture content (the reheat process is sensible, so the coil must do all of the dehumidification), then close energy balances on the coil, the reheater and the refrigeration plant.

Moisture content of the three specified states. With $p_{ws}(25) = 3.170$ kPa, $p_{ws}(14) = 1.599$ kPa and $p_{ws}(27) = 3.568$ kPa, $$W = 0.6220\,\frac{\phi\,p_{ws}}{p - \phi\,p_{ws}}\;\Rightarrow\; W_R = 0.009881,\; W_S = 0.005944,\;W_O = 0.015715 \ \text{kg/kg}$$ The room dew point is therefore 13.9 °C and the supply dew point 6.4 °C, so the coil must leave the air no warmer than about 6.4 °C saturated — consistent with the 5 °C apparatus dew point specified.
Specific enthalpies of those states. $h = 1.006\,t + W(2501 + 1.86\,t)$ gives $h_R = 50.32$, $h_S = 29.10$ and $h_O = 67.25$ kJ/kg of dry air.
Mixed state M leaving the mixing chamber. A 3 : 1 re-circulation ratio means mass fractions of 0.75 room air and 0.25 outdoor air, and both temperature and moisture content mix linearly: $$t_M = 0.75(25) + 0.25(27) = 25.5^\circ\text{C},\qquad W_M = 0.75(0.009881) + 0.25(0.015715) = 0.011340\ \text{kg/kg}$$ Consistency check on enthalpy: $h_M = 0.75(50.32) + 0.25(67.25) = 54.55$ kJ/kg, the same value the property relation returns for 25.5 °C and 0.011340 kg/kg.
Coil off-state W. The heating coil is a sensible process, so the moisture content leaving the cooling coil already equals the supply value, $W_W = W_S = 0.005944$ kg/kg. The coil process is the straight line from M to the apparatus dew point (5 °C saturated, $W_{ADP} = 0.005402$ kg/kg), whose slope is $$\frac{W_M - W_{ADP}}{t_M - t_{ADP}} = \frac{0.011340 - 0.005402}{25.5 - 5} = 2.897\times10^{-4}\ \text{kg/kg per K}$$ so $t_W = 5 + (0.005944 - 0.005402)/2.897\times10^{-4} = \boxed{6.87^\circ\text{C}}$ and $h_W = 21.85$ kJ/kg. The implied coil contact factor is $(25.5 - 6.87)/(25.5 - 5) = 0.909$, a realistic four- to six-row chilled-water coil.
(d) Total air-conditioning load on the room. The room load is the sum of the two components the question supplies: $$\dot{Q}_{room} = \dot{Q}_s + \dot{Q}_l = 20.5 + 8.8 = \boxed{29.3\ \text{kW}}$$ with a room sensible heat ratio of $20.5/29.3 = 0.70$.
Cooling-coil duty. The coil sees the mixed air, not the room air, so it must also remove the fresh-air load: $$\dot{Q}_{cc} = \dot{m}_a\,(h_M - h_W) = 1.8\,(54.55 - 21.85) = \boxed{58.9\ \text{kW}}$$
Reheat duty. Raising the coil off-state to the supply condition at constant moisture content costs $$\dot{Q}_{rh} = \dot{m}_a\,(h_S - h_W) = 1.8\,(29.10 - 21.85) = \boxed{13.1\ \text{kW}}$$
(e) Total energy input. The refrigeration plant delivers the coil duty at an overall COP of 2, and the reheat energy is bought separately: $$\dot{W}_{ref} = \frac{58.85}{2} = 29.4\ \text{kW},\qquad \dot{E}_{total} = 29.4 + 13.1 = \boxed{42.5\ \text{kW}}$$
(f) Energy input with condenser heat recovery. The condenser rejects everything the plant absorbs plus the work put in, $$\dot{Q}_{cond} = \dot{Q}_{cc} + \dot{W}_{ref} = 58.85 + 29.43 = 88.3\ \text{kW}$$ which is 6.8 times the 13.1 kW the reheat coil needs. The condenser cooling water can therefore carry the whole reheat duty and the purchased input collapses to the compressor power alone: $$\dot{E}_{recovered} = \boxed{29.4\ \text{kW}}$$ a saving of 13.1 kW, or 30.7% of the original input.

(c) Significant points on the diagram and the chart
Point	Description	Dry bulb (°C)	Wet bulb (°C)	$W$ (kg/kg)	$h$ (kJ/kg)
O	Outdoor fresh air	27.0	22.8	0.015715	67.25
R	Room / return air	25.0	17.9	0.009881	50.32
M	Mixing-chamber outlet	25.5	19.2	0.011340	54.55
W	Cooling-coil outlet	6.87	6.6	0.005944	21.85
S	Supply air after reheat	14.0	10.0	0.005944	29.10
ADP	Coil apparatus dew point	5.0	5.0	0.005402	18.59

(b) The operating cycle. Grey dashed lines: adiabatic mixing of R and O to M. Blue: the cooling and dehumidifying process M→W, aimed at the 5 °C apparatus dew point. Red: sensible reheat W→S. Gold: the room process S→R along the room ratio line.

Check: the printed data are redundant and slightly inconsistent. Taking the room and supply states literally, 1.8 kg/s from S to R would offset $1.8(50.32 - 29.10) = 38.2$ kW at a sensible heat ratio of 0.53, not the 29.3 kW at 0.70 that the stated loads give. The two data sets cannot both be exact. Part (d) is answered from the loads the question states, because that is what “the total air conditioning load for the room” asks for, and parts (e) and (f) are answered from the specified states, mixing ratio and apparatus dew point, none of which depends on the load split. Per instruction 1 on the cover page, this reading is stated as an assumption; a candidate who instead scaled the supply flow to 1.38 kg/s to satisfy the stated loads would reach the same method with proportionally smaller duties.

Topic. Mixed-air plant with reheat: fixing the coil off-state from the apparatus dew point.

Key relations. Mixing: $t_M = x\,t_R + (1-x)\,t_O$ and the same weighting for $W$ and $h$. Coil: the process line runs from the entering state to the apparatus dew point on the saturation curve, and the contact factor is $\beta = (t_M - t_W)/(t_M - t_{ADP})$. Duties: $\dot{Q} = \dot{m}_a\,\Delta h$ for a wet coil, $\dot{Q} = \dot{m}_a\,c_{p,moist}\,\Delta t$ for a dry one. Plant: $\dot{W} = \dot{Q}_{evap}/\text{COP}$ and $\dot{Q}_{cond} = \dot{Q}_{evap} + \dot{W}$.

Why this works. A cooling coil cannot dehumidify below its apparatus dew point, and a real coil never reaches it, so the off-state lies on the chord between the entering state and the ADP. Because reheat changes only dry-bulb temperature, the moisture content leaving the coil is pinned to the supply requirement, which is what fixes the off-state uniquely and hence splits the total plant duty between the coil and the reheater. Heat recovery from the condenser is free in first-law terms because the condenser must reject that heat anyway; the only cost is the heat-exchanger and pumping penalty, which this question tells us to neglect.

Common pitfall. Sizing the cooling coil on the room load instead of the mixed-air load. Here that error would understate the coil by $58.9 - 29.3 = 29.6$ kW — a factor of two — because 25% outdoor air at 27 °C and 70% RH is a large latent burden and because the reheat energy must itself be removed by the coil.

Source. W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann, Ch. 3 (properties of moist air) and Ch. 6 §6.4 (the mixed-air plant with reheat); ASHRAE Handbook – Fundamentals (2021), Ch. 1, eqs. (20)–(39).

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