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

Question 6 of 8: Induced-draught cooling tower — leaving water temperature and make-up rate

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 6: Induced-draught cooling tower — leaving water temperature and make-up rate (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. An induced-draught tower in which warm water falls through a counter-current air stream, losing heat by both sensible transfer and evaporation. The air leaving is stated to be saturated, which closes the problem completely.

Given data
Quantity	Symbol	Value
Water flow rate	$\dot{V}_w$	6 L/s
Water inlet temperature	$t_{w1}$	45 °C
Air volume flow rate	$\dot{V}_a$	9 m$^3$/s
Fan power absorbed	$\dot{W}_f$	4.75 kW
Air entering	—	18 °C, 60% RH
Air leaving	—	26 °C, saturated
Barometric pressure	$p$	1.013 bar

Find. A diagram of the tower, the final (leaving) temperature of the cooling water, and the make-up water required per second.

The induced-draught tower. Water is sprayed over the fill and falls counter-current to air drawn upward by the fan; the air leaves saturated at 26 °C, having gained both sensible heat and evaporated moisture, and make-up replaces exactly what evaporates.

Approach. Fix the two air states, convert the air volume flow to a dry-air mass flow through the humid volume at inlet, take the make-up directly from the moisture balance, and then apply a steady-flow energy balance over the whole tower — including the fan work, which ends up in the air stream — to find the leaving water temperature.

Air state entering. At 18 °C and 60% RH with $p = 101.3$ kPa,$$W_1 = 0.007699 \text{ kg/kg}, \qquad h_1 = 37.620 \text{ kJ/kg}, \qquad v_1 = 0.83521 \text{ m}^3\text{/kg dry air}$$ and the entering wet-bulb temperature, which is the thermodynamic limit on how cold the water could ever be made, is $13.41\,{}^{\circ}$C.
Dry-air mass flow. The fan handles a volume, so the mass of dry air it carries follows from the humid volume at the inlet:$$\dot{m}_a = \frac{\dot{V}_a}{v_1} = \frac{9}{0.83521} = 10.7757 \text{ kg dry air/s}$$
Air state leaving. Saturated at 26 °C,$$W_2 = 0.021357 \text{ kg/kg}, \qquad h_2 = 80.604 \text{ kJ/kg}$$ so each kilogram of dry air leaves carrying $W_2 - W_1 = 0.013659$ kg more water and $h_2 - h_1 = 42.984$ kJ more energy than it entered with.
Make-up water from the moisture balance. Whatever moisture the air gains must have evaporated from the circulating water and must be replaced:$$\dot{m}_{up} = \dot{m}_a\,(W_2 - W_1) = 10.7757\,(0.021357 - 0.007699) = \boxed{0.1472 \text{ kg/s}}$$ that is $529.9$ kg/h, or about $2.48$ % of the circulating flow — the usual order for a tower with this range, before any blowdown for dissolved solids is added.
Water mass flows. At 45 °C the density of water is 990.2 kg/m$^3$, so$$\dot{m}_{w1} = 6 \times 10^{-3} \times 990.2 = 5.9412 \text{ kg/s}$$ and the stream leaving the basin is smaller by exactly what evaporated,$\dot{m}_{w2} = 5.9412 - 0.1472 = 5.7940$ kg/s. The liquid-to-gas ratio is $L/G = 0.5513$.
Energy balance over the tower. The fan work is dissipated into the air stream, so it appears on the input side. Taking liquid enthalpies as $c_w t$ with $c_w = 4.18$ kJ/kg·K,$$\dot{m}_{w1} c_w t_{w1} + \dot{W}_f + \dot{m}_a h_1 = \dot{m}_{w2} c_w t_{w2} + \dot{m}_a h_2$$ The air-side term is the dominant one:$$\dot{m}_a (h_2 - h_1) = 10.7757\,(80.604 - 37.620) = 463.19 \text{ kW}$$
Leaving water temperature. Rearranging for $t_{w2}$,$$t_{w2} = \frac{\dot{m}_{w1} c_w t_{w1} + \dot{W}_f - \dot{m}_a(h_2 - h_1)}{\dot{m}_{w2} c_w}$$$$t_{w2} = \frac{5.9412 \times 4.18 \times 45 + 4.75 - 463.19}{5.7940 \times 4.18} = \boxed{27.2\,{}^{\circ}\text{C}}$$ The tower therefore achieves a range of $45 - 27.2 = 17.8$ K, removing $\dot{m}_{w2} c_w (t_{w1} - t_{w2}) = 430.7$ kW of heat from the water.
Sanity-check the result against tower theory. The leaving water temperature must lie above the entering air wet bulb, and it does: the approach is $27.2 - 13.4 = 13.8$ K. An approach of this size is poor for a counter-flow tower, but it is what the stated exit air state forces — air leaving saturated at only 26 °C cannot carry away enough enthalpy to bring 5.94 kg/s of water much lower. Note also that evaporation does the great majority of the work: the latent term $\dot{m}_{up} \times 2450 \approx 361$ kW out of the $463$ kW total air-side gain, with the balance sensible.

Check: two stated-assumption points. First, the printed sentence “Calculate the final temperature of the amount of cooling water make-up required per second” is garbled in the original; it is read here as asking for both the final water temperature and the make-up rate, and both are given. Second, 6 L/s has been converted at the density of water at the inlet temperature (990.2 kg/m$^3$ at 45 °C) rather than at 1000 kg/m$^3$; taking 6.00 kg/s instead would give a leaving temperature of 27.4 °C, a difference of only 0.2 K, and would not change the make-up rate at all since that depends solely on the air side.

Final results
Quantity	Symbol	Result
Entering air humidity ratio	$W_1$	0.007699 kg/kg ($h_1 = 37.62$ kJ/kg)
Leaving air humidity ratio	$W_2$	0.021357 kg/kg ($h_2 = 80.60$ kJ/kg)
Dry-air mass flow	$\dot{m}_a$	10.776 kg/s
Air-side heat gain	$\dot{Q}$	463.2 kW
Make-up water required	$\dot{m}_{up}$	0.1472 kg/s (530 kg/h, 2.48% of circulation)
Final (leaving) water temperature	$t_{w2}$	27.2 °C
Cooling range	—	17.8 K
Approach to the entering wet bulb (13.4 °C)	—	13.8 K

Topic. Mass and energy balances on an evaporative cooling tower.

Key relations. Moisture: $\dot{m}_{up} = \dot{m}_a (W_2 - W_1)$. Dry air from the humid volume: $\dot{m}_a = \dot{V}_a / v_1$. Energy: $\dot{m}_{w1} c_w t_{w1} + \dot{W}_f + \dot{m}_a h_1 = \dot{m}_{w2} c_w t_{w2} + \dot{m}_a h_2$ with $\dot{m}_{w2} = \dot{m}_{w1} - \dot{m}_{up}$. Range $= t_{w1} - t_{w2}$; approach $= t_{w2} - t_{wb,1}$.

Why this works. A cooling tower works because the air leaves at a higher enthalpy than it entered, and enthalpy is the only air-side property that matters — which is why the whole of Merkel's tower theory is written in terms of enthalpy driving force rather than temperature. Most of that enthalpy gain is latent: evaporating roughly 2.5% of the circulating water carries away the great majority of the heat, which is why a tower can cool water below the air's dry-bulb temperature. The thermodynamic floor is the entering wet-bulb temperature, since air can never be cooled below its own wet bulb by contact with water; how close a real tower comes to that floor is its approach, and approach is what tower size and fill depth buy. The fan work appears on the input side because it is dissipated into the air stream, adding to the load the tower must reject.

Common pitfall. Forgetting that the water leaving is smaller than the water entering by the evaporated mass — using $\dot{m}_{w1}$ on both sides here would give 26.54 °C instead of 27.2 °C. Also common is computing the dry-air mass flow from the standard density 1.2 kg/m$^3$ rather than from the humid volume, and omitting the fan power from the balance.

Source. Jones, Air Conditioning Engineering, 5th ed., Ch. 10 (cooling towers and evaporative cooling); ASHRAE Handbook – Systems and Equipment, Ch. 40; McQuiston, Parker & Spitler, 6th ed., Ch. 14.

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