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

Question 2 of 8: All-outdoor-air winter plant — preheat, adiabatic saturation and reheat (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 2: All-outdoor-air winter plant — preheat, adiabatic saturation and reheat (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 once-through (100% outdoor air) winter make-up air unit: preheat coil, adiabatic saturator, reheat coil, then the space.

Given data
Quantity	Value
Space sensible heat loss	200,000 Btu/h
Space latent load	negligible
Space design state	75 °F, 50% RH
Outdoor air	saturated at 20 °F
Ventilation air (100% outdoor)	7000 scfm
Adiabatic-saturator leaving dry bulb	60 °F

Find. the dry-bulb temperature of the air entering the space, the preheat and reheat coil duties in Btu/h, and the humidification water in US gpm.

System schematic. State 1 is outdoor air, 2 the preheat-coil outlet, 3 the adiabatic-saturator outlet held at 60 °F dry bulb, and 4 the supply air after reheat. All the air is exhausted from the space, so there is no return duct.

Approach. Because the space has no latent load, the supply air must carry the space moisture content, which fixes state 3; the adiabatic saturator follows a constant thermodynamic-wet-bulb line, so state 2 is the intersection of that line with the outdoor moisture content; the two coil duties and the water make-up then follow from mass and energy balances.

Moisture content the supply air must carry. With no latent load in the space, $W_4 = W_3 = W_{space}$. At 75 °F and 50% RH, $p_{ws} = 0.4300$ psia, so $$W_{space} = 0.6220\,\frac{0.50(0.4300)}{14.696 - 0.50(0.4300)} = \boxed{0.00924\ \text{lb/lb}}$$ The corresponding space wet bulb is 62.6 °F.
Outdoor air state 1. Saturated at 20 °F (over ice, $p_{ws} = 0.0505$ psia): $W_1 = 0.002144$ lb/lb and $$h_1 = 0.240(20) + 0.002144\,[1061 + 0.444(20)] = 7.09\ \text{Btu/lb}$$
State 3 leaving the adiabatic saturator. It is at 60 °F dry bulb and $W_3 = 0.00924$ lb/lb, which is 84% RH — a realistic outlet for a spray washer of about 85% saturating efficiency. Its enthalpy is $h_3 = 24.45$ Btu/lb and the adiabatic-saturation equation returns a thermodynamic wet bulb of $t^{*} = 57.1^\circ\text{F}$. This wet-bulb line is the humidification path.
State 2 leaving the preheat coil. Preheating is sensible, so $W_2 = W_1 = 0.002144$ lb/lb; state 2 is where the $t^{*} = 57.1^\circ\text{F}$ line meets that moisture content. Solving the adiabatic-saturation relation for dry bulb, $$t_2 = \frac{(1093 - 0.556\,t^{*})W^{*}_s + 0.240\,t^{*} - W_1(1093 - t^{*})}{0.240 + 0.444\,W_1} = \boxed{91.3^\circ\text{F}}$$ with $h_2 = 24.27$ Btu/lb. The preheat coil must therefore lift the outdoor air more than 70 F° before the washer can bring it to the required humidity.
Dry-air mass flow. Standard air is taken at 0.075 lb per cubic foot, so $$\dot{m}_a = 7000 \times 0.075 \times 60 = 31{,}500\ \text{lb of dry air per hour}$$
(i) Temperature of the air entering the space. The supply air must offset the sensible loss, with humid specific heat $c_p = 0.240 + 0.444(0.00924) = 0.2441$ Btu per lb per F°: $$\Delta t = \frac{200{,}000}{31{,}500 \times 0.2441} = 26.0\ \text{F}^\circ \;\Rightarrow\; t_4 = 75 + 26.0 = \boxed{101.0^\circ\text{F}}$$
(ii) Preheat coil duty. $$\dot{Q}_{pre} = \dot{m}_a\,(h_2 - h_1) = 31{,}500\,(24.267 - 7.094) = \boxed{541{,}000\ \text{Btu/h}}$$ or 158.5 kW — nearly three times the space heat loss, which is the price of 100% outdoor air at 20 °F.
(iii) Reheat coil duty. From 60 °F to 101.0 °F at constant moisture content, $$\dot{Q}_{re} = \dot{m}_a\,c_p\,(t_4 - t_3) = 31{,}500 \times 0.2441 \times (101.0 - 60) = \boxed{315{,}000\ \text{Btu/h}}$$ or 92.4 kW.
(iv) Humidification water. The washer must supply the moisture picked up between states 2 and 3: $$\dot{m}_w = \dot{m}_a\,(W_3 - W_1) = 31{,}500\,(0.00924 - 0.002144) = 223\ \text{lb/h}$$ and at 8.34 lb per US gallon, $$\dot{V}_w = \frac{223.4}{8.337 \times 60} = \boxed{0.447\ \text{gpm}}$$ About half a gallon a minute is the entire humidification demand; the recirculating spray pump handles many times that flow, and only the evaporated fraction is made up.

Operating cycle. Red 1→2: sensible preheat at constant humidity ratio. Green 2→3: adiabatic humidification along the 57.1 °F thermodynamic wet-bulb line. Red 3→4: sensible reheat to the 101.0 °F supply condition. Dashed gold 4→space: the space process, horizontal because the latent load is negligible.

Final results
Quantity	Result
Preheat-coil leaving temperature, state 2	91.3 °F
Adiabatic-saturator leaving state 3	60 °F DB, 57.1 °F WB, $W = 0.00924$ lb/lb
(i) Temperature of air entering the space	101.0 °F
(ii) Preheat coil	541,000 Btu/h (158.5 kW)
(iii) Reheat coil	315,000 Btu/h (92.4 kW)
(iv) Humidification water	223 lb/h = 0.447 gpm

Topic. Preheat – adiabatic humidification – reheat for a 100% outdoor-air winter system.

Key relations. Sensible processes move horizontally on the chart ($W$ constant); adiabatic saturation moves along a constant thermodynamic-wet-bulb (very nearly constant-enthalpy) line; $\dot{Q} = \dot{m}_a\,\Delta h$ across a coil; $\dot{m}_w = \dot{m}_a\,\Delta W$ across a humidifier; $\dot{Q}_s = \dot{m}_a c_p \Delta t$ for the space balance.

Why this works. An adiabatic saturator adds water but no heat, so the air gives up sensible heat to evaporate it and the enthalpy stays almost constant (it rises only by the enthalpy of the injected liquid). That means the preheat coil, not the washer, does the work of raising the enthalpy: the sequence has to overshoot in dry-bulb temperature first so that the constant-wet-bulb path arrives at the required moisture content at the specified washer outlet temperature. The reheat coil then trims the dry bulb up to whatever the space sensible balance demands.

Common pitfall. Confusing a scfm figure with an actual cfm at 20 °F. The mass flow follows from standard density; using the cold-air density instead would inflate the mass flow by about 10% and every duty with it. A second trap is drawing the humidification along a constant-enthalpy line and then reading the preheat duty as $h_3 - h_1$: the difference is small here (0.18 Btu/lb) but it is conceptually the enthalpy of the make-up water.

Source. McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley, Ch. 3 §3.4 (adiabatic saturation) and Example 3.10; W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann, Ch. 3 §3.7 (air washers and their saturating efficiency).

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