22-Mec-B2 Environmental Control in Buildings · December 2013
Question 1 of 8: Natatorium — mixing, preheat and psychrometric cycle
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Professional Engineers of Ontario /
Engineers Canada annual examination 07-Mec-B2 Environmental Control in Buildings,
December 2013 — 3 hours, open book, non-communicating calculator permitted.
Eight problems of 20 points each; candidates answer any five.
ASHRAE psychrometric charts (SI and I-P) and an R-134a pressure–enthalpy diagram are
appended to the paper. All eight problems are solved below, because the set is
intended as a study resource.
Reference texts.
ASHRAE Handbook — Fundamentals (Ch. 1 Psychrometrics, Ch. 15 Fenestration,
Ch. 16 Ventilation and Infiltration, Ch. 18 Nonresidential Cooling and Heating Load Calculations,
Ch. 21 Duct Design);
McQuiston, Parker & Spitler, Heating, Ventilating, and Air Conditioning: Analysis and
Design, 6th ed.;
Çengel & Boles, Thermodynamics: An Engineering Approach, 9th ed. (R-134a
property tables);
ANSI/ASHRAE Standard 55 Thermal Environmental Conditions for Human Occupancy;
ANSI/ASHRAE Standard 62.1 Ventilation for Acceptable Indoor Air Quality.
Canadian practice: National Energy Code of Canada for Buildings (NECB) and CSA F280 where a
Canadian code reference is required.
Check — table-derived coefficients. This is an
open-book paper whose cover page instructs candidates to “submit a clear statement of the
assumption(s)” made. Where a solution needs a value that is read from a handbook table or
chart rather than given in the question — shading coefficients, air-space thermal
resistances, curtain-wall leakage rates, surface-averaged wind pressure coefficients, window and
door U-factors, CO2 generation rates — the value used is stated explicitly at the
point of use with its source. The method, not the table entry, is what the marks follow; a
different edition of the table shifts the arithmetic but not the answer structure.
Find. The supply air mass (and volume) flow at 35 °C, the outdoor-to-recirculated
air ratio, the preheat-coil duty and the heat exchanged in the mixing box — together with a
system schematic and the four state points plotted on the psychrometric chart with their dry- and
wet-bulb temperatures.
Figure 1.1 — System schematic. Outdoor air O mixes with recirculated room air R at M; the mixture is heated to the supply state S and delivered to the space, which returns to state R.
Figure 1.2 — Operating cycle on the psychrometric chart. O–R is the adiabatic mixing line (M lies on it); M–S is sensible heating at constant humidity ratio; S–R is the room process, which gains moisture and loses sensible heat.
Approach. Fix the room and outdoor states from the given dry-bulb/RH pairs, size the
supply air from the sensible balance at the stated 35 °C supply temperature, obtain the
supply humidity ratio from the latent balance, then work backwards: the heater does not change
humidity ratio, so the mixed-air humidity ratio equals the supply humidity ratio and fixes the
outdoor-air fraction by a moisture balance on the mixing box.
Fix the outdoor and room humidity ratios. With $p_{ws}$ from the ASHRAE
saturation-pressure correlation and $W = 0.622\,p_w/(p - p_w)$:
$$W_o = \frac{0.622\,(0.20)(0.7060)}{101.325 - 0.1412} = 0.00087\ \text{kg/kg da},\qquad
W_r = \frac{0.622\,(0.50)(2.985)}{101.325 - 1.4925} = 0.00930\ \text{kg/kg da}$$
The corresponding specific enthalpies are $h_o = 4.19$ and $h_r = 47.81$ kJ/kg da.
Size the supply air from the sensible balance. The room loses 130 kW of
sensible heat, so the supply air must deliver it while cooling from 35 °C to the room 24 °C:
$$\dot m_a = \frac{q_s}{c_p\,(t_s - t_r)} = \frac{130}{1.02\,(35 - 24)}
= \boxed{11.59\ \text{kg/s dry air}}$$
At the supply state this is $\dot V_s = \dot m_a v_s = 11.59 \times 0.878 = 10.2\ \text{m}^3/\text{s}$
(about 21\,600 cfm) — the answer to part (a).
Obtain the supply humidity ratio from the latent balance. The pool evaporates
moisture into the space (a latent gain of 160 kW), so the supply air must arrive dry enough to
absorb it:
$$W_r - W_s = \frac{q_l}{\dot m_a\,h_{fg}} = \frac{160}{11.59 \times 2501} = 0.00552\ \text{kg/kg da}$$
$$W_s = 0.00930 - 0.00552 = 0.00378\ \text{kg/kg da}$$
State S is therefore 35 °C at $W = 0.00378$, i.e. only 10.9% RH — very dry air,
which is exactly what a natatorium needs.
Find the outdoor-air fraction from a moisture balance on the mixing box. Heating
is a constant-humidity-ratio process, so $W_m = W_s$. Writing the mixing balance with $x$ the outdoor
mass fraction,
$$W_m = x W_o + (1 - x) W_r \;\Rightarrow\;
x = \frac{W_r - W_m}{W_r - W_o} = \frac{0.00930 - 0.00378}{0.00930 - 0.00087} = 0.655$$
so 65.5% of the supply is outdoor air and 34.5% is recirculated:
$$\frac{\dot m_o}{\dot m_r} = \frac{0.655}{0.345} = \boxed{1.90 : 1}$$
In absolute terms $\dot m_o = 7.59$ kg/s and $\dot m_r = 4.00$ kg/s — the answer to part (b).
The high outdoor-air fraction is the direct consequence of the enormous latent gain: outdoor air at
2 °C is the cheapest available desiccant.
Locate the mixed state M. An adiabatic mixing box conserves enthalpy as well as
mass, so
$$h_m = x h_o + (1 - x) h_r = 0.655(4.19) + 0.345(47.81) = 19.24\ \text{kJ/kg da}$$
Inverting $h = 1.006\,t + W(2501 + 1.86\,t)$ at $W_m = 0.00378$ gives $t_m = 9.7\ \text{°C}$
(50.9% RH). Point M lies on the straight line O–R, at the fraction 0.345 of the way
from O to R — the graphical construction shown in Figure 1.2.
Preheat-coil duty (part c). The coil raises the mixture from M to S at constant $W$:
$$q_{\text{preheat}} = \dot m_a (h_s - h_m) = 11.59\,(44.90 - 19.24) = \boxed{297\ \text{kW}}$$
A useful cross-check: this must equal the space sensible loss plus the sensible energy needed to warm
the outdoor air from 2 °C to room temperature, $q = 130 + 7.59(1.02)(24 - 2) = 300$ kW, which agrees to
within the rounding of $c_p$.
Heat transfer rate for the mixing process (part d). The mixing box exchanges no
heat with its surroundings — it is adiabatic, so the external heat transfer rate is
$$q_{\text{mixing, external}} = \boxed{0\ \text{kW}}$$
What the process does transfer is energy between the two streams: the warm return air heats the
cold outdoor air. That internal exchange is
$$q_{o\to m} = \dot m_o (h_m - h_o) = 7.59\,(19.24 - 4.19) = 114\ \text{kW}
= \dot m_r (h_r - h_m) = 4.00\,(47.81 - 19.24)$$
Both forms agree, confirming the mixing balance. Quoting 114 kW as “the heat transfer rate for the
mixing process” is the intended reading; the point worth stating in the answer book is that this is
free recovered heat, not coil duty — without recirculation the preheat coil would have to supply
$q = 297 + 114 = 411$ kW.
The four state points, with the wet-bulb temperatures requested by the question, are collected below.
These are the values that would be annotated on the schematic and on the psychrometric chart.
State point
Dry bulb
Wet bulb
W (kg/kg da)
RH
O — outdoor air
2.0 °C
−3.1 °C
0.00087
20.0%
M — mixed air
9.7 °C
5.3 °C
0.00378
50.9%
S — supply air
35.0 °C
16.2 °C
0.00378
10.9%
R — room air
24.0 °C
17.1 °C
0.00930
50.0%
Check — assumptions stated per cover-page instruction 1. Sea-level barometric pressure 101.325 kPa; moist-air $c_p = 1.02$ kJ/(kg·K) and $h_{fg} = 2501$ kJ/kg at the 0 °C datum; fan, pump and duct heat gains neglected as the question directs; the pool-water make-up and deck losses are already embedded in the stated 130 kW / 160 kW; and the relief air leaving the space is at room state R, so the exhaust exactly balances the 65.5% outdoor-air intake.
