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

Question 2 of 8: Winter heating cycle with a steam humidifier

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 2: Winter heating cycle with a steam humidifier (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 winter plant mixing outdoor and return air, heating the mixture sensibly and then injecting dry saturated steam to make up the moisture the building loses. Worked in inch-pound units throughout, as printed, at standard barometric pressure.

Given data
Quantity	Symbol	Value
Inside design state	R	68 °F dry bulb, 50% RH
Outdoor design state	O	32 °F dry bulb, 10% RH
Building sensible heat loss	$\dot{Q}_s$	220 000 Btu/h
Building latent heat loss	$\dot{Q}_l$	45 000 Btu/h
Outdoor air : mixed air, by mass	$x$	0.45
Supply air dry-bulb temperature	$t_S$	100 °F
Humidifier steam	—	dry saturated at 20 psia

Find. (a) and (b) the characteristic points and the cycle on the chart; (c) the supply air state and quantity in lb/h; (d) the heater rating in Btu/h; and (e) the steam mass flow rate.

(a) Plant schematic for the winter cycle. Outdoor air O joins return air R in the mixing box (45%% outdoor by mass) to give M; the heater raises M sensibly to H; the steam humidifier adds moisture to reach the supply state S.

Approach. The supply air must carry both the sensible and the latent load, so the sensible loss and the stated supply temperature fix the mass flow, and the latent loss then fixes the supply humidity ratio. Working backwards through the humidifier (which adds moisture at the steam enthalpy) locates the state leaving the heater, and the heater duty is the sensible rise from the mixed state to it.

Fix the room and outdoor states. At 68 °F and 50% RH, $W_R = 0.007262$ lb/lb and $h_R = 0.240(68) + W_R(1061 + 0.444 \times 68) = 24.244$ Btu/lb. At 32 °F and 10% RH the outdoor air is very dry: $W_O = 0.000375$ lb/lb, $h_O = 8.084$ Btu/lb.
(c) Supply air quantity from the sensible load. The supply air offsets the sensible loss by arriving above room temperature. With the moist-air sensible specific heat $c_p = 0.240 + 0.444\,W_R = 0.2432$ Btu/lb·°F,$$\dot{m} = \frac{\dot{Q}_s}{c_p\,(t_S - t_R)} = \frac{220\,000}{0.2432\,(100 - 68)} = \boxed{28266 \text{ lb/h}}$$ At the supply state the humid volume is $v_S = 14.307$ ft$^3$/lb of dry air, so the volumetric supply rate is $\dot{m}\,v_S/60 = 6740$ cfm.
(c continued) Supply humidity ratio from the latent load. The same air must carry the moisture the building loses, at the latent heat of the water vapour it delivers:$$W_S = W_R + \frac{\dot{Q}_l}{\dot{m}\,(1061 + 0.444\,t_S)} = 0.007262 + \frac{45\,000}{28266 \times 1105.4} = 0.008702 \text{ lb/lb}$$ so the supply air is at $100\,{}^{\circ}$F and $21.3$ % RH, with $h_S = 33.619$ Btu/lb. As a check, $\dot{m}(h_S - h_R) = 265000$ Btu/h, which is exactly the 265 000 Btu/h total loss.
Locate the mixed state M. With 45% outdoor air by mass, the moisture and energy balances give $W_M = 0.45(0.000375) + 0.55(0.007262) = 0.004163$ lb/lb and $h_M = 0.45(8.084) + 0.55(24.244) = 16.972$ Btu/lb, whence $t_M = (h_M - 1061\,W_M)/(0.240 + 0.444\,W_M) = 51.91\,{}^{\circ}$F at $51.1$ % RH.
(e) Steam mass flow. The humidifier lifts the humidity ratio from the mixed value to the supply value, and every pound of that increase is a pound of steam:$$\dot{m}_{st} = \dot{m}\,(W_S - W_M) = 28266\,(0.008702 - 0.004163) = \boxed{128.3 \text{ lb/h}}$$ At 20 psia the steam is dry saturated at $227.9\,{}^{\circ}$F with $h_g = 1156.2$ Btu/lb.
(d) Heater rating. Steam injection is not adiabatic: the steam carries its own enthalpy into the air stream, so the heater has less to do than the full mixed-to-supply enthalpy rise. Taking an energy balance across the humidifier, the state H leaving the heater satisfies$$h_H = h_S - (W_S - W_M)\,h_g = 33.619 - (0.004539)(1156.2) = 28.371 \text{ Btu/lb}$$ At the unchanged humidity ratio $W_M$ that corresponds to $t_H = 99.05\,{}^{\circ}$F — the steam is worth only about $0.5$ K of the temperature rise. The heater duty is therefore$$\dot{Q}_{heater} = \dot{m}\,(h_H - h_M) = 28266\,(28.371 - 16.972) = \boxed{322214 \text{ Btu/h}}$$ equivalently $94.4$ kW.
Check the whole plant. The heater and the steam together must cover the building loss plus the energy needed to warm the ventilation air from outdoor to room conditions:$$\dot{Q}_{heater} + \dot{m}_{st} h_g \;\overset{?}{=}\; \dot{Q}_s + \dot{Q}_l + \dot{m}\,x\,(h_R - h_O)$$$$322214 + 148341 = 470555 \quad\text{versus}\quad 265\,000 + 205555 = 470555 \text{ Btu/h}$$ which closes to $0.00$ % — the residue is the small non-linear cross term in the enthalpy relation, not an error.

(b) The winter cycle on the psychrometric chart. O–M and R–M are the mixing line, M–H the sensible heating, H–S the steam humidification (very nearly vertical, since dry saturated steam adds little sensible heat), and S–R the room ratio line along which the air gives up its heat and picks up its moisture deficit.

Final results
Quantity	Symbol	Result
Mixed state M	$t_M$, RH	51.91 °F, 51.1% RH
State leaving the heater H	$t_H$, RH	99.05 °F, 10.6% RH
(c) Supply air state	$t_S$, RH	100 °F, 21.3% RH ($W_S = 0.008702$ lb/lb)
(c) Supply air quantity	$\dot{m}$	28266 lb/h (6740 cfm at the supply state)
(d) Heater rating	$\dot{Q}_{heater}$	322214 Btu/h (94.4 kW)
(e) Steam mass flow	$\dot{m}_{st}$	128.3 lb/h at 20 psia, 227.9 °F

Topic. Winter sensible-plus-latent conditioning by heating and steam injection.

Key relations. Sensible: $\dot{Q}_s = \dot{m}\,c_p\,(t_S - t_R)$ with $c_p = 0.240 + 0.444\,W$. Latent: $\dot{Q}_l = \dot{m}\,(W_S - W_R)(1061 + 0.444\,t_S)$. Humidifier energy balance: $h_S = h_H + (W_S - W_M)\,h_g$. Heater: $\dot{Q} = \dot{m}(h_H - h_M)$.

Why this works. In winter the building loses both heat and moisture, and one air stream has to replace both. The temperature the air is delivered at settles the mass flow through the sensible balance; once that flow is known, the moisture balance has only one unknown left, the supply humidity ratio, so the supply state is fully determined without any assumption. The humidifier is then the only component that changes $W$, and because dry saturated steam enters at about 1156 Btu/lb it brings in real energy — treating the injection as adiabatic would leave the heater to supply that energy instead and overstate its rating.

Common pitfall. Sizing the heater from the full mixed-to-supply enthalpy rise, which double-counts the enthalpy the steam brings with it and inflates the duty to 470555 Btu/h instead of 322214. A second trap is using 0.240 rather than $0.240 + 0.444W$ for the moist-air specific heat; here it shifts the supply flow by about 380 lb/h.

Source. McQuiston, Parker & Spitler, HVAC: Analysis and Design, 6th ed., Ch. 3 (moist air processes); ASHRAE Handbook – Fundamentals (2021) Ch. 1; steam properties from the ASME steam tables at 20 psia.

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