22-Mec-B2 Environmental Control in Buildings · May 2013

Question 6 of 8: Mechanical-draught cooling tower — leaving water temperature and make-up (20 marks)

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 2013 sitting. Three hours, open book. Eight problems of 20 points each; the candidate is instructed to solve five and to nominate which five are to be graded. Psychrometric charts and a pressure–enthalpy diagram for ammonia (R-717) are appended to the paper, and candidates are expected to bring an environmental-control text and steam tables. Instruction 1 invites the candidate to state any interpretation assumptions with the answer — that latitude is used explicitly below where the printed data are redundant.

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 are tighter than a graphical solution would be; chart-quality agreement (about ±0.2 K and ±0.0002 kg/kg) is all that an examiner expects.

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 (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 and duct design).
ASHRAE Handbook – Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 21 (duct design), Ch. 25–27 (heat, air and moisture transfer in the envelope).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression and multistage refrigeration).
ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy, and ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality.
Canadian frame: National Building Code of Canada 2020, National Energy Code of Canada for Buildings 2020, and Environment and Climate Change Canada Canadian Climate Normals for degree-day data.

Psychrometric relations used throughout. At barometric pressure $p$, with saturation vapour pressure $p_{ws}(t)$ from the ASHRAE correlation,

$$W = 0.6220\,\frac{\phi\,p_{ws}(t)}{p - \phi\,p_{ws}(t)}, \qquad h = 1.006\,t + W\,(2501 + 1.86\,t)$$

in SI (kJ per kg of dry air), and in the inch-pound system $h = 0.240\,t + W\,(1061 + 0.444\,t)$ Btu per lb of dry air. The thermodynamic wet-bulb temperature is obtained from the adiabatic-saturation equation, which is what a chart's constant-wet-bulb lines represent.

Question 6: Mechanical-draught cooling tower — leaving water temperature and make-up (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 small mechanical-draught cooling tower operating at 1.013 bar throughout.

Given data
Quantity	Value
Water entering	5.5 L/s at 44 °C
Air volume flow drawn by the fan	9 m³/s at inlet conditions
Fan power absorbed	4.75 kW
Air entering	18 °C, 60% RH
Air leaving	saturated at 26 °C
Barometric pressure	1.013 bar

Find. the tower schematic, the temperature of the water leaving the tower, and the make-up water flow required per second.

The tower. Water is sprayed over the fill and falls counter-current to the air; the fan draws 9 m³/s through the pack and its 4.75 kW of shaft power ends up in the air stream. Make-up replaces exactly the water evaporated.

Approach. Fix both air states, convert the volumetric air flow to a dry-air mass flow through the humid specific volume at inlet, get the make-up from the water-vapour mass balance, then apply the steady-flow energy equation to the whole tower (including the fan work) to find the leaving water temperature.

Air entering, state 1. At 18 °C, $p_{ws} = 2.065$ kPa, so $p_w = 0.60(2.065) = 1.239$ kPa and $$W_1 = 0.6220\,\frac{1.239}{101.3 - 1.239} = 0.00770\ \text{kg/kg}, \qquad h_1 = 37.62\ \text{kJ/kg}$$ The inlet wet bulb is 13.4 °C, which is the thermodynamic floor for the leaving water temperature.
Air leaving, state 2. Saturated at 26 °C, $p_{ws} = 3.363$ kPa, so $$W_2 = 0.6220\,\frac{3.363}{101.3 - 3.363} = 0.02136\ \text{kg/kg}, \qquad h_2 = 80.60\ \text{kJ/kg}$$
Dry-air mass flow. The fan handles 9 m³/s of the entering mixture, whose specific volume per kilogram of dry air is $$v_1 = \frac{R_a T}{p - p_w} = \frac{0.287(291.15)}{101.3 - 1.239} = 0.8352\ \text{m}^{3}/\text{kg dry air}$$ so $$\dot{m}_a = \frac{9}{0.8352} = 10.78\ \text{kg/s}$$
Make-up water — the moisture balance. Everything the air picks up must be replaced: $$\dot{m}_{makeup} = \dot{m}_a\,(W_2 - W_1) = 10.776\,(0.02136 - 0.00770) = \boxed{0.147\ \text{kg/s}}$$ or 0.147 L/s — 2.7% of the circulating flow, which is the usual order for a tower with a 19 K range. The water leaving is therefore $5.5 - 0.147 = 5.353$ kg/s.
Energy balance on the whole tower. Taking liquid water enthalpy as $h_w = 4.187\,t$ from 0 °C, consistent with the datum built into the moist-air enthalpy, and counting the fan work as an input to the control volume, $$\dot{m}_{w1} h_{w1} + \dot{m}_a h_1 + \dot{W}_{fan} = \dot{m}_{w2} h_{w2} + \dot{m}_a h_2$$ The make-up is taken to enter at the leaving water temperature, so it does not appear separately. Substituting, $$5.5(4.187)(44) + 10.776(37.62) + 4.75 = \dot{m}_{w2} h_{w2} + 10.776(80.60)$$ $$1013.3 + 405.4 + 4.75 - 868.7 = 554.8\ \text{kW}$$
Leaving water temperature. $$t_{w2} = \frac{554.8}{5.353 \times 4.187} = \boxed{24.8^\circ\text{C}}$$ The range is $44 - 24.8 = 19.2$ K and the approach to the 13.4 °C entering wet bulb is 11.4 K — a wide approach, as expected of a small tower with a generous air-to-water mass ratio of $10.78/5.5 = 1.96$ and only a short fill depth.
Check the split of the heat rejected. The water gives up $1013.3 - 554.8 = 458.5$ kW while the air gains $10.776(80.60 - 37.62) = 463.2$ kW; the 4.75 kW difference is exactly the fan power, so the balance closes. Of the 463 kW carried away, the latent part $\dot{m}_a(W_2 - W_1)(2501) = 368$ kW is 79% — evaporation does nearly all of the work, which is why a tower can cool water below the ambient dry-bulb temperature.

Final results
Quantity	Result
Dry-air mass flow	10.78 kg/s
Air state leaving	26 °C saturated, $W = 0.02136$ kg/kg
Make-up (evaporation) rate	0.147 kg/s = 0.147 L/s
Water leaving the tower	5.353 kg/s at 24.8 °C
Cooling range	19.2 K
Approach to entering wet bulb (13.4 °C)	11.4 K
Heat rejected by the water	458 kW

Check: the litre-to-kilogram conversion is taken as 1 L = 1 kg. At 44 °C water is actually 990.6 kg/m³, so the true circulating mass flow is 5.45 kg/s; carrying that value through raises the leaving temperature by only 0.1 K, well inside chart-reading accuracy. The make-up is also assumed to enter at the basin temperature and blowdown is neglected, as the question gives no cycles of concentration; a real tower on Ottawa or Toronto make-up would bleed a further 0.5 to 1% to control dissolved solids.

Topic. Cooling-tower mass and energy balance.

Key relations. Moisture balance $\dot{m}_{makeup} = \dot{m}_a(W_2 - W_1)$; humid volume $v = R_a T/(p - p_w)$; overall energy balance $\dot{m}_{w1}h_{w1} + \dot{m}_a h_1 + \dot{W}_{fan} = \dot{m}_{w2}h_{w2} + \dot{m}_a h_2$; the entering air wet-bulb temperature is the theoretical limit for the leaving water.

Why this works. A cooling tower is an adiabatic direct-contact exchanger, so the only bookkeeping needed is that mass and energy both balance across the control volume. Because the air leaves saturated and much warmer, most of the heat leaves as latent heat in the added vapour rather than as a dry-bulb temperature rise; that is what lets the water be cooled towards the wet-bulb temperature rather than the dry-bulb. The fan power matters because it is dissipated inside the control volume and so must be rejected along with the process heat — ignoring it here would make the leaving water 0.2 K colder than it really is.

Common pitfall. Using the standard specific volume 0.83 m³/kg or dry-air density 1.2 kg/m³ instead of the humid volume at the actual inlet state, and forgetting that the fan handles the mixture, not the dry air alone. A larger error is to compute the make-up from a heat balance on the water alone: the make-up is a mass-balance result and depends only on the two moisture contents and the air flow.

Source. W. P. Jones, Air Conditioning Engineering, 5th ed., Butterworth-Heinemann, Ch. 10 §10.2–10.5 (cooling towers, with the same balance worked for a mechanical-draught tower); ASHRAE Handbook – Fundamentals (2021), Ch. 40 (cooling towers); CTI ATC-105 for performance terminology.

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