22-Mec-B2 Environmental Control in Buildings · May 2015

Question 1 of 8: Mixed-air plant with reheat — cycle, plant loads and energy input

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 2015 sitting. Three hours, open book. Eight problems of 20 points each; the candidate is instructed to solve five and to nominate on the cover of the first workbook which five are to be graded. Psychrometric charts and a pressure–enthalpy diagram for the refrigerant are appended to the paper, and candidates are expected to bring an environmental-control text and steam tables. Instruction 1 invites the candidate to submit a clear statement of any interpretation assumptions with the answer — that latitude is used explicitly below wherever the printed data are redundant or incomplete.

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 below are tighter than a graphical solution would be; chart-quality agreement (about $\pm 0.2$ K in temperature and $\pm 0.0002$ kg/kg in humidity ratio) is all that an examiner expects, and a candidate reading the appended charts should reproduce every answer to within that band.

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 (fans and 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, fans and duct design).
ASHRAE Handbook – Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 21 (fans and duct design), Ch. 25–27 (heat, air and moisture transfer in the envelope), Ch. 30 (refrigerant properties).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression refrigeration and heat pumps).
ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality, and ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy.
Canadian frame: National Building Code of Canada 2020 (Part 6, and Appendix C for design temperatures and degree-days), National Energy Code of Canada for Buildings 2020, Health Canada Residential Indoor Air Quality Guidelines, and Environment and Climate Change Canada Canadian Climate Normals.

Psychrometric relations used throughout. At barometric pressure $p$ with saturation vapour pressure $p_{ws}(t)$ from the ASHRAE correlation, the humidity ratio, specific enthalpy and humid volume of moist air are

$$W = 0.621945\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}, \qquad h = 1.006\,t + W\,(2501 + 1.86\,t), \qquad v = \frac{0.287042\,(t + 273.15)\,(1 + 1.6078\,W)}{p}$$

with $h$ in kJ per kg of dry air, $t$ in $\,{}^{\circ}$C and $p$ in kPa. In inch-pound units the enthalpy becomes $h = 0.240\,t + W\,(1061 + 0.444\,t)$ Btu per lb of dry air with $t$ in $\,{}^{\circ}$F. Wet-bulb temperatures are obtained by solving the adiabatic-saturation equation, not by eye. Mixing two air streams is exact in moisture and in enthalpy, so $W$ and $h$ of the mixture are the mass-weighted averages and the mixed dry bulb follows from them; weighting the dry bulb directly is the usual shortcut and differs here by about $0.01$ K.

Question 1: Mixed-air plant with reheat — cycle, plant loads and energy input (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 drawing outdoor air into a mixing chamber with return air, cooling and dehumidifying the mixture on a chilled-water coil of known apparatus dew point, reheating to the specified supply state, and delivering it by fan to the zone. Sea level, so $p = 101.325$ kPa throughout.

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	24 °C, 50% RH
Supply state and mass flow	S	14 °C, 60% RH, 1.8 kg/s
Outdoor design state	O	28 °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 operating cycle on the psychrometric chart; (c) dry- and wet-bulb temperatures at every significant point; (d) the total air-conditioning load on the room; (e) the total energy input to the plant; and (f) the energy input when the reheat is taken from the condenser cooling water instead of from a separate source.

(a) Plant schematic. Outdoor air O mixes with return air R in the ratio 1 : 3 to give the mixed state M; the chilled-water coil cools and dehumidifies M to the off-coil state C on the condition line towards the 5 °C apparatus dew point; the heating coil reheats C to the supply state S, which the fan delivers to the zone.

Approach. Fix the three stated states from the ASHRAE property relations, mix O and R by mass to locate M, run the coil condition line from M towards the apparatus dew point until it reaches the supply humidity ratio (reheat is sensible, so the coil must already deliver $W_S$), and then take each coil duty as a mass flow times an enthalpy difference.

Fix the three stated states. With $W = 0.621945\,\phi\,p_{ws}/(p - \phi\,p_{ws})$ and $h = 1.006\,t + W(2501 + 1.86\,t)$, the room, outdoor and supply states are
$W_R = 0.009299$, $h_R = 47.815$ kJ/kg; $W_O = 0.016687$, $h_O = 70.771$ kJ/kg; $W_S = 0.005944$, $h_S = 29.105$ kJ/kg. Solving the adiabatic-saturation equation at each state gives wet bulbs of $17.07$, $23.69$ and $9.95\,{}^{\circ}$C respectively.
Locate the mixed state M. A 3 : 1 recirculation ratio makes the outdoor fraction $x = 1/4 = 0.25$ by mass. Moisture and energy balances on the mixing chamber give$$W_M = x\,W_O + (1-x)\,W_R = 0.25(0.016687) + 0.75(0.009299) = 0.011146$$$$h_M = x\,h_O + (1-x)\,h_R = 0.25(70.771) + 0.75(47.815) = 53.554 \text{ kJ/kg}$$ Inverting the enthalpy relation for the dry bulb, $t_M = (h_M - 2501\,W_M)/(1.006 + 1.86\,W_M) = 25.01\,{}^{\circ}$C, whose wet bulb is $18.90\,{}^{\circ}$C.
Run the coil condition line to the off-coil state C. The reheat coil that follows adds no moisture, so the cooling coil must already have brought the air to the supply humidity ratio: $W_C = W_S = 0.005944$. The coil condition line is the straight line from M to the apparatus dew point (saturated air at $5\,{}^{\circ}$C, $W_{ADP} = 0.005402$), so the contact factor is$$\text{CF} = \frac{W_M - W_C}{W_M - W_{ADP}} = \frac{0.011146 - 0.005944}{0.011146 - 0.005402} = 0.9056$$ a bypass factor of $0.0944$, which is a realistic four-to-six-row chilled-water coil. The off-coil dry bulb follows on the same line, $t_C = t_M + \text{CF}\,(t_{ADP} - t_M) = 6.89\,{}^{\circ}$C ($96.5$ % RH, wet bulb $6.62\,{}^{\circ}$C), and $h_C = 21.872$ kJ/kg.
(c) Tabulate the significant points. The six points that must be identified on both the schematic and the chart are O, R, M, C, S and the apparatus dew point; their dry- and wet-bulb temperatures are collected in the table below and plotted on the psychrometric chart that follows.
(d) Total air-conditioning load on the room. The room load is what the question states it to be — the sensible and latent gains the supply air must absorb:$$\dot{Q}_{room} = \dot{Q}_s + \dot{Q}_l = 20.5 + 8.8 = \boxed{29.3 \text{ kW}}$$ at a sensible heat ratio of $\text{SHR} = 20.5/29.3 = 0.700$.
(e) Total energy input. The cooling coil must remove the enthalpy difference between the mixed and off-coil states, and the reheat coil must restore the air from C to S:$$\dot{Q}_c = \dot{m}\,(h_M - h_C) = 1.8\,(53.554 - 21.872) = 57.03 \text{ kW}$$$$\dot{Q}_h = \dot{m}\,(h_S - h_C) = 1.8\,(29.105 - 21.872) = 13.02 \text{ kW}$$ The refrigeration plant delivers the cooling at a COP of 2, so it absorbs $\dot{W} = \dot{Q}_c/\text{COP} = 57.03/2 = 28.51$ kW; adding the reheat, which in this arrangement is bought separately,$$\dot{E} = \frac{\dot{Q}_c}{\text{COP}} + \dot{Q}_h = 28.51 + 13.02 = \boxed{41.53 \text{ kW}}$$
(f) Reheat recovered from the condenser cooling water. The condenser must reject everything the evaporator absorbs plus the compressor work,$$\dot{Q}_{cond} = \dot{Q}_c + \dot{W} = 57.03 + 28.51 = 85.54 \text{ kW}$$ which is more than six times the $13.02$ kW the reheat coil needs. The reheat is therefore available at no additional energy cost — it is heat that would otherwise go to the cooling tower — and the plant input falls to the compressor duty alone,$$\dot{E}_{(f)} = \frac{\dot{Q}_c}{\text{COP}} = \boxed{28.51 \text{ kW}}$$ a saving of $13.02$ kW, or $31.3$ % of the answer to part (e). The only practical caveat is that the condenser water leaves at roughly $30$–$35\,{}^{\circ}$C, which is warm enough to lift air from $6.9\,{}^{\circ}$C to $14\,{}^{\circ}$C, so the recovery is thermodynamically as well as arithmetically available.

(b) The operating cycle on the psychrometric chart. O–M and R–M are the mixing line, M–C the cooling-coil condition line produced towards the 5 °C apparatus dew point, C–S the sensible reheat, and S–R the room ratio line.

Check: the printed data are deliberately over-specified and do not close. The question fixes the room loads (20.5 kW sensible, 8.8 kW latent, SHR 0.700) and the supply state and flow. Taken literally, air supplied at 14 °C / 60% RH and 1.8 kg/s would absorb $\dot{m}(h_R - h_S) = 33.68$ kW at a sensible heat ratio of 0.544, not the 29.3 kW at SHR 0.700 that the loads state. Neither figure is an extraction error — both are printed. No part of the question needs both, so nothing has to be reconciled: part (d) is answered from the stated loads, and parts (e) and (f) from the states, the mixing ratio and the apparatus dew point. Cover-page instruction 1 expressly invites this kind of stated assumption. A candidate who instead re-derived the supply flow from the room loads and the supply temperature would get $\dot{m} = 2.02$ kg/s and correspondingly smaller coil duties; that is a defensible alternative reading, but it discards the 1.8 kg/s the paper supplies.

Final results
Quantity	Symbol	Result
Mixed state M	$t_M$, $t_{wb,M}$	25.01 °C dry bulb, 18.90 °C wet bulb
Off-coil state C	$t_C$, $t_{wb,C}$	6.89 °C dry bulb, 6.62 °C wet bulb (96.5% RH)
Coil contact factor	CF	0.9056 (bypass 0.0944)
(d) Total room air-conditioning load	$\dot{Q}_{room}$	29.3 kW (SHR 0.700)
Cooling-coil duty	$\dot{Q}_c$	57.03 kW
Reheat-coil duty	$\dot{Q}_h$	13.02 kW
(e) Total energy input	$\dot{E}$	41.53 kW
Condenser heat rejection	$\dot{Q}_{cond}$	85.54 kW
(f) Energy input with recovered reheat	$\dot{E}_{(f)}$	28.51 kW

Topic. Mixing, coil condition line and apparatus dew point on a cool-and-reheat plant.

Key relations. Mixing: $W_M = xW_O + (1-x)W_R$ and $h_M = xh_O + (1-x)h_R$. Coil: contact factor $\text{CF} = (W_M - W_C)/(W_M - W_{ADP})$, equivalently $1 - \text{CF}$ is the bypass factor. Duties: $\dot{Q} = \dot{m}\,\Delta h$. Plant: $\dot{W} = \dot{Q}_c/\text{COP}$ and $\dot{Q}_{cond} = \dot{Q}_c + \dot{W}$.

Why this works. A cooling coil cannot deliver an arbitrary state. Its surface sits at one effective temperature — the apparatus dew point — and the air leaving is a mixture of air that touched the surface (and left saturated at the ADP) and air that bypassed it unchanged. That makes the process a straight line on the chart from the entering state towards the ADP, and the contact factor is simply how far along that line the air travelled. Because the reheat that follows is purely sensible, the humidity ratio leaving the cooling coil must already equal the supply humidity ratio, and that single condition fixes the off-coil point. The condenser-recovery question in part (f) then turns on the first law applied to the whole refrigeration machine: everything the evaporator absorbs plus everything the compressor puts in must leave through the condenser, so a plant rejecting 85.5 kW can certainly supply a 13.0 kW reheat coil.

Common pitfall. Reading the off-coil state as saturated air at the apparatus dew point. That is the limiting case of a perfect coil (CF = 1); a real coil stops short of it, and using the ADP state itself would overstate the cooling duty here by about 10%. A second trap is adding the reheat to the compressor input in part (f) after having already shown that the condenser can supply it free.

Source. W. P. Jones, Air Conditioning Engineering, 5th ed., Ch. 3 (psychrometry of mixing and coil processes) and Ch. 6; ASHRAE Handbook – Fundamentals (2021) Ch. 1.

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