22-Mec-B2 Environmental Control in Buildings · December 2013
Question 2 of 8: Summer air conditioner — apparatus dew point and coil capacity
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.
Question 2: Summer air conditioner — apparatus dew point and coil capacity (20 points)
90% saturation (degree of saturation $\mu = 0.90$)
Room sensible heat factor
$\text{RSHF}$
0.68
Barometric pressure
$p$
14.696 psia (sea level)
Find. The apparatus dew point, the dew-point temperature of the air leaving the coil and
the off-coil dry-bulb temperature; the total coil capacity in Btu/hr; and the split of that capacity into
sensible and latent components — with the system diagram and the cycle drawn on the I-P chart.
Figure 2.1 — System schematic. 2000 cfm outdoor air and 6000 cfm return air mix at M, pass over the cooling coil to S, and are delivered to the space which returns at R.
Figure 2.2 — Operating cycle. O–R is the mixing line carrying M; S–R is the room line of slope RSHF = 0.68, extended to the saturation curve at the apparatus dew point ADP; M–S is the coil process, which passes through ADP.
Approach. Convert both entering streams to mass flows through their specific volumes and
mix them to fix state M. Draw the room sensible-heat-factor line through the room state R; where it meets
the saturation curve is the apparatus dew point, and where it meets the 90%-saturation curve is the
off-coil state S. The coil load is then the enthalpy drop from M to S on the total mass flow.
Fix the two entering states. From the 90 °F DB / 73 °F WB pair the
adiabatic-saturation relation gives $W_o = 0.01352$ lb/lb da, $h_o = 36.48$ Btu/lb da and
$v_o = 14.16\ \text{ft}^3/\text{lb}$. For the return air at 75 °F, 50% RH,
$W_r = 0.00924$ lb/lb da, $h_r = 28.11$ Btu/lb da, $v_r = 13.68\ \text{ft}^3/\text{lb}$
(room dew point 55.1 °F).
Convert to mass flows and mix. Working in mass rather than volume avoids the small
error the “cfm-weighted” shortcut introduces when the two streams have different densities:
$$\dot m_o = \frac{2000 \times 60}{14.16} = 8476\ \text{lb/h},\qquad
\dot m_r = \frac{6000 \times 60}{13.68} = 26\,317\ \text{lb/h}$$
$$\dot m_t = 34\,793\ \text{lb/h},\qquad
W_m = \frac{8476(0.01352) + 26\,317(0.00924)}{34\,793} = 0.01028\ \text{lb/lb da}$$
$$h_m = \frac{8476(36.48) + 26\,317(28.11)}{34\,793} = 30.15\ \text{Btu/lb da}
\;\Rightarrow\; t_m = 78.7\ \text{°F DB},\ 65.3\ \text{°F WB}$$
Construct the room sensible-heat-factor line. Along a line of constant SHF through
the room state, sensible and latent enthalpy changes keep the ratio 0.68, so
$$\frac{\Delta W}{\Delta t} = \frac{c_{pa}\,(1 - \text{SHF})}{\text{SHF}\;h_g}
= \frac{0.240\,(1 - 0.68)}{0.68 \times 1061} = 1.065 \times 10^{-4}\ \text{lb/lb per °F}$$
Every point on the supply line satisfies $W = W_r - 1.065\times10^{-4}(75 - t)$. (This is the line the
protractor on the chart draws for SHF = 0.68.)
Apparatus dew point (part a). The ADP is where that line, extended, meets the
saturation curve, i.e. where $W_{\text{sat}}(t) = W_r - 1.065\times10^{-4}(75 - t)$. Solving,
$$t_{\text{ADP}} = \boxed{42.9\ \text{°F}}$$
This is the effective surface temperature the coil must reach for the room to be held at 75 °F, 50% RH
with a sensible heat factor of 0.68.
Off-coil state (part a). Air leaving at 90% saturation lies on the same room line
where $W = 0.90\,W_{\text{sat}}(t)$:
$$t_s = \boxed{48.0\ \text{°F DB}},\qquad W_s = 0.00636\ \text{lb/lb da},\qquad
h_s = 18.39\ \text{Btu/lb da}$$
Its dew point — the “air dew point” asked for — follows from $W_s$:
$$t_{dp,s} = \boxed{45.2\ \text{°F}}$$
(the wet bulb is 46.5 °F). As expected the three temperatures nest,
$t_{\text{ADP}} < t_{dp,s} < t_s$. The corresponding coil bypass factor is
$$\text{BF} = \frac{t_s - t_{\text{ADP}}}{t_m - t_{\text{ADP}}}
= \frac{48.0 - 42.9}{78.7 - 42.9} = 0.14$$
a realistic value for a four- to six-row coil.
Cooling coil capacity (part b). The coil must take the whole mixture from M to S:
$$q_t = \dot m_t (h_m - h_s) = 34\,793\,(30.15 - 18.39)
= \boxed{409\,000\ \text{Btu/hr}} \;(34.1\ \text{tons})$$
Checking by parts: the room load is $\dot m_t(h_r - h_s) = 338\,000$ Btu/hr and the outdoor-air load is
$\dot m_o(h_o - h_r) = 71\,000$ Btu/hr, summing to 409 000 Btu/hr.
Sensible and latent split (part c). Splitting the same enthalpy drop into its
dry-air and moisture parts,
$$q_s = \dot m_t\,(c_{pa} + c_{pv}W_s)(t_m - t_s) = 34\,793\,(0.2428)(78.7 - 48.0)
= \boxed{259\,000\ \text{Btu/hr}}$$
$$q_l = \dot m_t\,(W_m - W_s)(h_g) = 34\,793\,(0.01028 - 0.00636)(1096)
= \boxed{150\,000\ \text{Btu/hr}}$$
so the coil is 63.4% sensible and 36.6% latent. Its own sensible heat factor, 0.634, is lower than the
room value of 0.68 because the 2000 cfm of humid outdoor air adds mostly latent load.
Check — reading of part (a). The phrase “the apparatus dew point (ADP) the air dew point and the air of coil dry bulb temperature” is read as three quantities: the ADP, the dew point of the air leaving the coil, and the off-coil dry-bulb temperature. The room dew point (55.1 °F) is also reported in case the second item was intended to mean the room air.
