22-Mec-B2 Environmental Control in Buildings · December 2017

Question 2 of 8: Winter heating and humidification plant — preheater rating and washer efficiency

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

Notes on this paper

Paper format. Engineers Canada national examination 16-Mec-B2 Environmental Control in Buildings, December 2017, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. ASHRAE Psychrometric Chart No. 1 (SI and inch-pound) and a pressure–enthalpy diagram for R-717 are appended to the paper as pages 6–8.

All eight problems are worked here. The paper mixes SI and inch-pound units deliberately: Problems 1, 3 and 4 are SI, Problems 2, 6 and 7 are inch-pound, and Problem 8 is SI with a Canadian climate. Each solution is worked in the units the question uses, as the cover-page instructions require.

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 and the psychrometric chart), Ch. 6 (air-conditioning plant cycles), Ch. 7 (the cooling coil, apparatus dew point and by-pass factor), Ch. 9 (cooling towers), Ch. 15 (fans).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 5 (heat transmission in building structures), Ch. 8 (energy estimating and the degree-day method), Ch. 12 (fans and duct design).
ASHRAE Handbook — Fundamentals — Ch. 1 (psychrometrics), Ch. 14 (climatic design information), Ch. 21 (duct design), Ch. 25–27 (thermal and moisture performance of the building envelope), Ch. 30 (fenestration).
Stoecker & Jones, Refrigeration and Air Conditioning, 2nd ed., McGraw-Hill — Ch. 10–12 (vapour-compression cycle, compressors, condensers and evaporators); ASHRAE Handbook — Refrigeration for ammonia plant practice.
National Building Code of Canada and the National Energy Code of Canada for Buildings (NRC), Appendix C climatic data; CSA and Canada Green Building Council material for Problem 5.

Property basis used throughout. Moist-air properties are computed from the ASHRAE Fundamentals ideal-moist-air relations, so every state quoted here can be read back off the psychrometric chart supplied with the paper:

$$W=\frac{0.621945\,p_w}{p-p_w},\qquad h_{\text{SI}}=1.006\,t+W\,(2501+1.86\,t),\qquad h_{\text{IP}}=0.240\,t+W\,(1061+0.444\,t)$$

with $h$ in $\text{kJ/kg}$ of dry air for $t$ in $\,{}^{\circ}\text{C}$ and in $\text{Btu/lb}$ of dry air for $t$ in $\,{}^{\circ}\text{F}$. Ammonia properties are quoted on the same datum as the attached ASHRAE p–h diagram ($h_f=200\ \text{kJ/kg}$ and $s_f=1.0\ \text{kJ/(kg}\cdot\text{K)}$ for saturated liquid at $0\,{}^{\circ}\text{C}$); only differences enter the answers, so any consistent chart or table gives the same duties.

Question 2: Winter heating and humidification plant — preheater rating and washer efficiency (20 points)

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.

Quantity	Symbol	Value
Space state	$t_R,\ t_{dp}$	$68\,{}^{\circ}\text{F}$ db, $48.5\,{}^{\circ}\text{F}$ dew point
Space heating load (all sensible)	$\dot Q_R$	$250{,}000\ \text{Btu/h}$
System air flow	$\dot V$	$7250\ \text{cfm}$
Outdoor air for ventilation	$\dot V_O$	$2400\ \text{cfm}$
Outdoor design state	$t_O,\ \phi_O$	$30\,{}^{\circ}\text{F}$ db, $60\%$ RH
Main heater rating	$\dot Q_{mh}$	$360{,}000\ \text{Btu/h}$
Plant	$-$	preheater → adiabatic spray cabinet → main heater

Find. The plant diagram with every state point characterised, the cycle on the chart, the Btu/h rating of the preheater, and the adiabatic saturation efficiency of the spray cabinet together with its make-up water rate.

Part (a) — the winter plant. Outdoor air mixes with return air at M, is preheated to 2, adiabatically humidified to 3 in the spray cabinet, and raised to the supply temperature S by the main heater.

Approach. Everything in this problem is determined — there is nothing to assume. The all-sensible room load fixes the supply state, the stated main-heater rating fixes the state leaving the washer, the washer is adiabatic so it fixes the state leaving the preheater, and the preheater duty is then simply what is left between the mixed state and that point.

Convert the air quantities to dry-air mass flow. Using standard air ($13.33\ \text{ft}^3/\text{lb}$, i.e. $4.5$ lb of dry air per hour per cfm), which is also the basis of the familiar $1.10\,\dot V\,\Delta t$ sensible relation: $$\dot m_{da}=4.5\times7250=32{,}625\ \text{lb/h},\qquad x_O=\frac{2400}{7250}=0.3310$$ so $10{,}800\ \text{lb/h}$ of outdoor air mixes with $21{,}825\ \text{lb/h}$ of return air.
Part (a) — fix the two given states. A $48.5\,{}^{\circ}\text{F}$ dew point means the space humidity ratio is the saturation value at that temperature:
Point Description $t$ (°F db) $W$ (gr/lb) $h$ (Btu/lb) $\phi$ (%) $t_{wb}$ (°F)
O outdoor air 30.00 14.58 9.44 60.0 26.08
R space / return 68.00 50.48 24.19 49.6 56.72
Mix the two streams at M. Mass-weighting $W$ and $h$ and deriving the dry bulb: $$W_M=0.3310(14.58)+0.6690(50.48)=38.59\ \text{gr/lb}=0.005513\ \text{lb/lb}$$ $$h_M=0.3310(9.44)+0.6690(24.19)=19.31\ \text{Btu/lb}$$ $$t_M=\frac{h_M-1061\,W_M}{0.240+0.444\,W_M}=55.50\,{}^{\circ}\text{F} \quad(59.2\%\ \text{RH})$$
Find the supply state S from the room balance. The room load is entirely sensible, so the air adds no moisture to the space and leaves it with the moisture it arrived with: $W_S=W_R=50.48\ \text{gr/lb}$. The energy balance on the room gives $$h_S=h_R+\frac{\dot Q_R}{\dot m_{da}}=24.19+\frac{250{,}000}{32{,}625} =31.85\ \text{Btu/lb} \;\Rightarrow\; t_S=99.51\,{}^{\circ}\text{F}\ (18.0\%\ \text{RH})$$ As a check against the familiar shortcut, $\Delta t=250{,}000/(1.10\times7250)=31.3\ \text{F}^{\circ}$, giving $99.3\,{}^{\circ}\text{F}$ — the same answer to within the rounding built into the $1.10$ coefficient.
Work backwards through the main heater to state 3. The main heater is sensible, so it too leaves $W$ unchanged, and its stated rating fixes how far below the supply temperature the washer must deliver: $$h_3=h_S-\frac{\dot Q_{mh}}{\dot m_{da}}=31.85-\frac{360{,}000}{32{,}625} =20.82\ \text{Btu/lb} \;\Rightarrow\; t_3=54.14\,{}^{\circ}\text{F}\ (81.2\%\ \text{RH})$$ This is the step that makes the problem determinate. Without the stated rating the split of duty between the preheater and the main heater would be a free choice.
Work backwards through the adiabatic washer to state 2. The spray cabinet exchanges no heat with its surroundings, so the only energy crossing its boundary is the enthalpy of the make-up water that replaces what evaporates: $$h_2=h_3-(W_3-W_2)\,h_f(t_3) =20.82-(0.007211-0.005513)(22.3)=20.78\ \text{Btu/lb}$$ The washer adds moisture but no dry air, so $W_2=W_M=38.59\ \text{gr/lb}$ and $$t_2=\frac{h_2-1061\,W_2}{0.240+0.444\,W_2}=61.58\,{}^{\circ}\text{F} \quad(47.6\%\ \text{RH})$$
Part (c) — the preheater rating. The preheater takes the air from the mixed state M to state 2: $$\dot Q_{pre}=\dot m_{da}\,(h_2-h_M)=32{,}625\,(20.78-19.31) =\boxed{48{,}100\ \text{Btu/h}}$$ This is the same duty whether the preheater sits in the outdoor-air branch or in the mixed stream, because it is the same air raised through the same total enthalpy rise. Placed in the outdoor-air branch alone — the usual arrangement, since that is where freeze protection is wanted — it would raise the $2400$ cfm of outdoor air by about $18\ \text{F}^{\circ}$, from $30$ to roughly $48\,{}^{\circ}\text{F}$.
Part (d) — adiabatic saturation efficiency. An adiabatic washer moves the air along its own thermodynamic wet-bulb line, and the wettest state it could possibly deliver is saturation at the adiabatic saturation temperature of the entering air. For state 2 that is $t^{*}=51.00\,{}^{\circ}\text{F}$, at which $W^{*}=55.52\ \text{gr/lb}$. The efficiency is the fraction of that potential actually realised, and it may be written either on humidity ratio or on temperature: $$\eta_{\text{sat}}=\frac{W_3-W_2}{W^{*}-W_2} =\frac{50.48-38.59}{55.52-38.59}=0.704 \qquad \eta_{\text{sat}}=\frac{t_2-t_3}{t_2-t^{*}} =\frac{61.58-54.14}{61.58-51.00}=0.703$$ $$\eta_{\text{sat}}=\boxed{70.4\%}$$ The two forms agreeing to a tenth of a per cent is the check that states 2 and 3 really do lie on one wet-bulb line — a typical figure for a single bank of sprays.
Part (d) — make-up water. The water the cabinet must replace is exactly the moisture the air leaves with: $$\dot m_w=\dot m_{da}\,(W_3-W_2)=32{,}625\,(0.007211-0.005513) =\boxed{55.4\ \text{lb/h}}$$ which is $6.6$ US gal/h, or about $25\ \text{kg/h}$. Confirm the whole plant: the two heaters supply $48{,}100+360{,}000=408{,}100\ \text{Btu/h}$, and the air gains $\dot m_{da}(h_S-h_M)=408{,}900\ \text{Btu/h}$, of which $800\ \text{Btu/h}$ arrived as the enthalpy of the spray water. The balance closes.

Point	Description	$t$ (°F db)	$W$ (gr/lb)	$h$ (Btu/lb)	$\phi$ (%)	$t_{wb}$ (°F)
O	outdoor air	30.00	14.58	9.44	60.0	26.08
R	space / return	68.00	50.48	24.19	49.6	56.72

Part (b) — the winter cycle on the inch-pound chart. M→2 is sensible preheat, 2→3 runs up the wet-bulb line through the adiabatic washer, 3→S is the main heater, and S→R is the all-sensible room load line.

Final results.

Quantity	Result
Dry-air mass flow	$32{,}625\ \text{lb/h}$ ($33.1\%$ outdoor air)
M — mixed air	$55.50\,{}^{\circ}\text{F}$ db, $59.2\%$ RH, $38.59\ \text{gr/lb}$
2 — leaving preheater	$61.58\,{}^{\circ}\text{F}$ db, $47.6\%$ RH, $38.59\ \text{gr/lb}$
3 — leaving spray cabinet	$54.14\,{}^{\circ}\text{F}$ db, $81.2\%$ RH, $50.48\ \text{gr/lb}$
S — supply to space	$99.51\,{}^{\circ}\text{F}$ db, $18.0\%$ RH, $50.48\ \text{gr/lb}$
(c) Preheater rating	$\mathbf{48{,}100\ \text{Btu/h}}$
(d) Adiabatic saturation efficiency	$\mathbf{70.4\%}$
(d) Make-up water	$\mathbf{55.4\ \text{lb/h}}$ ($6.6$ US gal/h)
Total heating input	$408{,}100\ \text{Btu/h}$

Check: assumptions declared under cover-page instruction 1. (i) The stated air quantities are taken as standard-air volumes, so $\dot m_{da}=4.5\dot V$ and the mixing ratio equals the volumetric ratio; this is the convention behind the $1.10$, $0.68$ and $4.5$ coefficients the question implies. (ii) The spray cabinet is adiabatic and the make-up water enters at the leaving air temperature, which is what recirculating spray water does; taking the make-up at mains temperature instead changes the preheater duty by well under $1\%$. (iii) Fan and duct gains are neglected, so the state leaving the main heater is the state entering the space. Note that the preheater duty is not a free choice here: the stated main-heater rating of $360{,}000$ Btu/h closes the one degree of freedom that otherwise exists between the two heaters.

Topic. The preheat / adiabatic-humidify / reheat winter cycle.

Key relations. Sensible processes: $\dot Q=\dot m_{da}\,\Delta h$ at constant $W$, or $\dot Q=1.10\,\dot V\,\Delta t$ for standard air. Adiabatic saturation: $h_2+(W_3-W_2)h_f(t_3)=h_3$, with $\eta_{\text{sat}}=(W_3-W_2)/(W^{*}-W_2) =(t_2-t_3)/(t_2-t^{*})$. Make-up water: $\dot m_w=\dot m_{da}(W_3-W_2)$. All-sensible room load: $W_S=W_R$ and $h_S=h_R+\dot Q_R/\dot m_{da}$.

Why this works. A winter plant has to do two things at once — raise the temperature and raise the moisture content — and the awkward part is that the only cheap humidifier available, a water spray, cools the air as it wets it. That is why the heat is split into two: the preheater puts in enough energy for the washer to give up as latent heat, and the main heater then lifts the now-humid air to the supply temperature. Because the washer runs along a line of constant wet-bulb temperature, its two end states are locked together, so fixing any one of the three duties fixes the other two. Here the examiner supplies the main-heater rating, which is why no assumption is needed; where nothing is supplied and the washer exit must be assumed (typically $95\%$ saturation for a single bank) before the problem will close.

Common pitfall. Assuming the washer saturates the air completely. It makes the arithmetic easy but forces the answer to part (d) to be $100\%$ by construction, which is plainly not what the question is asking for. The other frequent error is putting the room load into the moisture balance: the load is stated to be all sensible, so the supply humidity ratio equals the room humidity ratio, and any moisture the air is given in the plant is moisture it keeps.

Source. Jones, Air Conditioning Engineering, 5th ed., Ch. 3 (adiabatic saturation and the wet-bulb line) and Ch. 6 (winter plant cycles); McQuiston, Parker & Spitler, 6th ed., §3-4 and §3-7.

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