Winter plant — preheat, mix, heating coil and steam humidifier: 22-Mec-B2 Environmental Control in Buildings, Undated paper, Q3

Problem 3: Winter plant — preheat, mix, heating coil and steam humidifier (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. A winter plant with an outdoor-air preheater, a mixing box, a main heating coil and a steam humidifier, all at 101.325 kPa.

Quantity	Symbol	Value
Total heating load of the building	$Q$	155 kW
Space design condition	$R$	21 °C, 30 % RH
Outdoor design condition	$O$	−15 °C, ~0 % RH
Preheat coil exit	state 1	16 °C
Supply condition	$S$	40 °C, 30 % RH
Humidifying steam	—	saturated vapour, 1.13 bar absolute
Ventilation air	—	1/3 of the supply air, by volume

Find. The supply air quantity, the temperature rise across the heating coil, the steam quantity, and the duties of the heating coil and preheater.

Part (a) — plant sketch. Outdoor air is preheated from −15 °C to 16 °C, mixed with return air, heated at constant humidity ratio to state 4, then humidified with saturated steam to the supply state S.

Approach. Fix R and S from the psychrometric relations, size the supply air on the total enthalpy rise the question specifies, take one third of that volume as ventilation air at the mixing-box entry, mix exactly, then work backwards from S through the steam-injection line to find the heating-coil exit and the two duties.

Fix the space and supply states. At 21 °C, $p_{ws} = 2.487$ kPa, so $p_w = 0.746$ kPa and $W_R = 0.004615\ \text{kg/kg}$, $h_R = 32.85\ \text{kJ/kg}_{da}$. At 40 °C, $p_{ws} = 7.384$ kPa, so $W_S = 0.01390\ \text{kg/kg}$, $h_S = 76.04\ \text{kJ/kg}_{da}$ and $v_S = 0.9069\ \text{m}^3/\text{kg}_{da}$.
Part (b) — size the supply air on the stated total load. The supply air must deliver the whole 155 kW between its own state and the space state: $$\dot m_{da}=\frac{Q}{h_S-h_R}=\frac{155}{76.04-32.85}=\boxed{3.59\ \text{kg}_{da}/\text{s}}$$ which is $\dot V_S = 3.59 \times 0.9069 = 3.26\ \text{m}^3/\text{s}$ (11,700 m³/h) at the supply condition. Of that, 70.4 kW arrives as sensible heat and 84.6 kW as the moisture the humidifier put in — a humidification-dominated load, which is what a space held at 30 % RH against −15 °C dry outdoor air actually looks like.
Ventilation air, by volume. One third of the supply volume is $\dot V_{OA} = 3.26/3 = 1.085\ \text{m}^3/\text{s}$. It enters the mixing box at the preheat exit, 16 °C and $W=0$, where $v = 0.8191\ \text{m}^3/\text{kg}$: $$\dot m_{OA}=\frac{1.085}{0.8191}=1.32\ \text{kg}_{da}/\text{s}\quad\Rightarrow\quad x=\frac{1.32}{3.59}=0.369$$ The mass fraction (36.9 %) exceeds the volumetric third because the ventilation air is cooler and drier, hence denser, than the supply air it is measured against.
Mix exactly to state M. With $h_1 = 1.006(16) = 16.10\ \text{kJ/kg}$ and $W_1 = 0$, $$W_M=0.369(0)+0.631(0.004615)=0.002912,\qquad h_M=0.369(16.10)+0.631(32.85)=26.67\ \text{kJ/kg}$$ $$t_M=\frac{h_M-2501W_M}{1.006+1.86W_M}=19.16\,{}^{\circ}\text{C}\ \ (21.3\ \%\ \text{RH})$$
Part (e) — the steam quantity. The heating coil cannot change $W$, so all the moisture is added by the humidifier between $W_M$ and $W_S$: $$\dot m_w=\dot m_{da}(W_S-W_M)=3.59(0.01390-0.002912)=0.0394\ \frac{\text{kg}}{\text{s}} =\boxed{142\ \text{kg/h}}$$
Part (c) — work back through the steam line to state 4. Saturated vapour at 1.13 bar absolute is at 103.1 °C with $h_g = 2{,}680\ \text{kJ/kg}$. Steam injection is not adiabatic — it carries its own enthalpy in — so $$h_4=h_S-(W_S-W_M)h_g = 76.04-0.010988(2680)=46.59\ \text{kJ/kg}$$ and at the unchanged $W_M$, $$t_4=\frac{46.59-2501(0.002912)}{1.006+1.86(0.002912)}=38.86\,{}^{\circ}\text{C}$$ so the heating coil raises the air from 19.16 °C to 38.86 °C, a rise of $$\boxed{\Delta t = 19.7\ \text{K}}$$ Note that the steam contributes only 1.1 K of dry-bulb rise on top of its 11 g/kg of moisture; treating the humidifier as a constant-temperature process would put state 4 at 40 °C and overstate the coil duty by about 4 kW.
Part (f) — heating-coil capacity. $$Q_{hc}=\dot m_{da}(h_4-h_M)=3.59(46.59-26.67)=\boxed{71.5\ \text{kW}}$$
Part (f) — preheater capacity. The preheater sees only the outdoor-air branch, from −15 °C ($h_O = -15.09\ \text{kJ/kg}$) to 16 °C: $$Q_{ph}=\dot m_{OA}(h_1-h_O)=1.32(16.10+15.09)=\boxed{41.3\ \text{kW}}$$ Its practical job is freeze protection as much as capacity — it is what keeps the mixing box and the downstream coil above freezing on a design night.
Close the whole plant. Everything entering must equal everything leaving: $$\dot m_{da}h_S \;\overset{?}{=}\; \dot m_{OA}h_O+(\dot m_{da}-\dot m_{OA})h_R+Q_{ph}+Q_{hc}+\dot m_w h_g$$ $$272.9 = -19.99+74.38+41.3+71.5+105.7 = 272.9\ \text{kW}\ \checkmark$$ The balance closes to nine figures, which validates the two duties, the steam rate and the mixed state together.

Part (d) — the process on the psychrometric chart. O→1 preheat at constant $W$; 1 mixes with R to give M; M→4 sensible heating; 4→S steam injection along a near-vertical line of slope $h_g$; S→R is the space process.

Result	Value
(b) Supply air, dry-air mass flow	3.59 kgₕₐ/s
(b) Supply air, volume flow at 40 °C / 30 % RH	3.26 m³/s (11,700 m³/h)
Ventilation air	1.085 m³/s = 1.32 kgₕₐ/s (36.9 % by mass)
Mixed state M	19.16 °C, $W$ = 0.00291 kg/kg, 21.3 % RH
(c) Heating-coil temperature rise (19.16 → 38.86 °C)	19.7 K
(e) Water vapour (steam) required	0.0394 kg/s = 142 kg/h
(f) Heating-coil capacity	71.5 kW
(f) Preheater capacity	41.3 kW

Check: the paper is over-specified, and cover-page instruction 1 requires the choice to be declared. It states the space state, the supply state and a 155 kW “total heating load”. Read as a total-enthalpy duty, the supply air is 3.59 kg/s (used above). Read as a sensible duty it would be $155/[(1.006+1.86W_S)(40-21)] = 7.91$ kg/s, more than twice as much, and the humidifier would then have to inject 313 kg/h. The word “total” is taken at face value here, and it is also the only reading under which both stated states are used; the sensible reading is quoted so a marker can see the alternative. Two further assumptions: the ventilation third is measured at the mixing-box entry (16 °C) — measuring it instead at the raw outdoor state would give 1.42 kg/s and move the coil duty by about 3 kW — and the outdoor air is taken as exactly dry, as the paper's “essentially 0 % relative humidity” states.

Topic. Winter air-conditioning plant with preheat, adiabatic mixing, sensible heating and steam humidification.

Key relations. $\dot m = Q/(h_S-h_R)$ on a total-enthalpy basis; adiabatic mixing in $W$ and $h$; sensible heating at constant $W$; steam injection $h_2 = h_1 + (W_2-W_1)h_g$ with $W$ increasing; $\dot m_w = \dot m_{da}\Delta W$; preheat duty on the outdoor-air branch alone.

Why this works. Each item of plant changes exactly one thing on the chart, and that is what lets the states be chased in either direction. A heating coil moves horizontally (constant $W$); a steam humidifier moves almost vertically, its slope set by the enthalpy of the injected vapour; a mixing box lands on the straight line between its two inlets at the mass-fraction point. Because the supply state and the mixed state are both fixed by the data, the intermediate state 4 is over-determined in principle and is found by walking backwards down the steam line — which is also why the steam enthalpy matters and cannot be ignored. Steam injection is nearly isothermal only because $h_g$ happens to be close to the enthalpy of saturated vapour at the air temperature; at 1.13 bar it still adds about a kelvin.

Common pitfall. Treating the humidifier as exactly isothermal, which puts state 4 at the supply temperature and inflates the coil duty; and applying the preheater duty to the whole supply stream instead of the outdoor-air branch, which triples it. A third is converting the one-third ventilation ratio as though it were a mass ratio — here that would give 1.20 kg/s instead of 1.32 and shift the mixed state by nearly a kelvin.

Source. Jones, Air Conditioning Engineering, 5th ed., Ch. 3 and Ch. 8; ASHRAE Handbook — Fundamentals, Ch. 1 and Ch. 22 (Humidifiers); McQuiston et al., §3.6–3.9.

Question 3 of 8: Winter plant — preheat, mix, heating coil and steam humidifier

Problem 3: Winter plant — preheat, mix, heating coil and steam humidifier (20 points)