22-Mec-B2 Environmental Control in Buildings · May 2016

Question 5 of 8: Induced-draft counterflow cooling tower

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, May 2016, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. Psychrometric charts and an R-134a p-h diagram are appended to the paper. All eight problems are solved here, because the set is intended as a study resource rather than an examination script.

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. 5–6 (heating and 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. 6 (heating loads and infiltration), Ch. 8 (energy estimating and the degree-day method), Ch. 12–13 (fluid flow, fans and duct design).
ASHRAE Handbook — Fundamentals (2021) — Ch. 1 (psychrometrics), Ch. 16 (ventilation and infiltration), Ch. 18 (residential and non-residential load calculations), Ch. 21 (duct design), Ch. 30 (thermophysical properties of refrigerants).
Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics, 9th ed., Wiley — Ch. 10 (vapour-compression refrigeration, multistage systems and heat pumps).
ANSI/ASHRAE Standard 62.1, Ventilation for Acceptable Indoor Air Quality; ANSI/ASHRAE Standard 55, Thermal Environmental Conditions for Human Occupancy.
Canadian frame: National Building Code of Canada 2020 and its Appendix C design temperatures; National Energy Code of Canada for Buildings 2020; Environment and Climate Change Canada heating degree-day normals; CSA B52 Mechanical Refrigeration Code; Canadian federal halocarbon regulations (SOR/2003-289) for the refrigerant discussion in Problem 7.

Check: assumptions carried through this paper. Cover-page instruction 1 invites a clear statement of any assumption. Standard barometric pressure of 101.325 kPa is used throughout; moist-air properties follow the ASHRAE Handbook — Fundamentals Ch. 1 formulation (Hyland–Wexler saturation pressure, so results agree with the appended chart to chart-reading accuracy rather than being read off it); R-134a properties are on the IIR datum and agree with the appended p-h diagram. Problem-specific assumptions — climate data, fuel prices, emission factors, air-change rates, occupant density, duct roughness and the coil bypass factor — are stated where they are first used.

Question 5: Induced-draft counterflow cooling tower (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-draft counterflow tower cooling 5.5 L/s of water entering at 44 °C, with a fan drawing 9 m³/s of air and absorbing 4.75 kW; air enters at 18 °C and 60 % relative humidity and leaves saturated at 26 °C, with the pressure constant at 1.013 bar throughout.

Given data
Quantity	Symbol	Value
Water volume flow, entering	$\dot V_w$	5.5 L/s
Water inlet temperature	$t_{w1}$	44 $^\circ$C
Air volume flow induced	$\dot V_a$	9 m³/s
Fan power absorbed	$\dot W_{fan}$	4.75 kW
Air entering	$t_{a1},\ \phi_1$	18 $^\circ$C, 60 %
Air leaving	$t_{a2},\ \phi_2$	26 $^\circ$C, saturated
Barometric pressure	$p$	1.013 bar

Find. The temperature of the water leaving the tower, and the make-up water rate needed to replace what evaporates.

Induced-draft counterflow tower serving a refrigeration plant condenser. Water falls through the fill against the rising air; a temperature controller on the leaving water modulates fan speed, which is how the tower regulates condensing pressure.

Approach. Treat the tower as one control volume. A moisture balance on the air stream gives the evaporation directly, which is also the make-up; an energy balance on the same control volume, with the water stream reduced by that evaporation and the fan work booked into the air, gives the leaving water temperature.

Regulating the refrigeration plant

The tower rejects the condenser heat of the refrigeration plant, and the temperature of the water it returns sets the condensing pressure and hence the compressor's pressure ratio and power. Control is therefore exercised on the leaving water temperature, and there are three usual means, shown on the sketch in order of preference. Fan speed is the primary control: a variable-frequency drive modulates the air flow so that the leaving water is held at set point, and because fan power varies roughly with the cube of speed, the energy saving in mild weather is large. Fan cycling, or staging multiple cells, is the cruder version of the same idea and is used where a drive cannot be justified. A bypass valve around the tower returns a fraction of the warm water straight to the condenser, and is the control of last resort in freezing weather, where it also keeps water moving through the basin. The set point itself should float downwards with the outdoor wet bulb rather than being fixed: a lower condensing temperature reduces compressor power by roughly 2 –3 % per kelvin, so letting the tower produce the coldest water it can, down to the chiller's minimum permitted condensing pressure, is usually worth far more than the fan energy it costs.

Leaving water temperature and make-up

Entering air state. At 18 °C, $p_{ws}=2.064$ kPa, so at 60 % relative humidity $$W_1=0.621945\,\frac{0.60\times 2.064}{101.3-0.60\times 2.064} =0.007699\ \text{kg/kg}$$ $$h_1=1.006(18)+0.007699\,[2501+1.86(18)]=37.62\ \text{kJ/kg}$$ $$v_1=\frac{0.287(291.15)}{101.3}\,(1+1.6078\times 0.007699) =0.8352\ \text{m}^3/\text{kg}$$
Dry-air mass flow. The 9 m³/s is measured at the entering state, so $$\dot m_a=\frac{\dot V_a}{v_1}=\frac{9}{0.8352} =10.776\ \text{kg/s of dry air}$$
Leaving air state. Saturated at 26 °C, where $p_{ws}=3.363$ kPa, $$W_2=0.621945\,\frac{3.363}{101.3-3.363}=0.021357\ \text{kg/kg} \qquad h_2=80.60\ \text{kJ/kg}$$
Moisture balance gives the make-up. Every kilogram of water that leaves in the air must be replaced, $$\boxed{\;\dot m_{ev}=\dot m_a\,(W_2-W_1) =10.776\,(0.021357-0.007699) =0.1472\ \text{kg/s}\;(\approx 0.1476\ \text{L/s}, \ 529.9\ \text{kg/h})\;}$$ That is 2.70 % of the circulating water, which is the classic result: a cooling tower evaporates roughly 1 % of its throughput for every 5 –6 K of range.
Water mass flow. At 44 °C the density of water is 990.7 kg/m³, so $$\dot m_w=0.0055\times 990.7=5.449\ \text{kg/s}$$ giving a liquid-to-gas ratio of $L/G=0.506$, which is on the air-rich side for a small tower.
Energy balance on the tower. The air gains $$\dot m_a\,(h_2-h_1)=10.776\,(80.60-37.62) =463.19\ \text{kW}$$ but the induced-draft fan sits in that air stream and delivers its 4.75 kW into it, so the heat that actually came out of the water is $$q_w=463.19-4.75=458.44\ \text{kW}$$
Solve for the leaving water temperature. The water stream leaving is lighter than the stream entering by the evaporated mass, $$\dot m_w c_p t_{w1}-(\dot m_w-\dot m_{ev})\,c_p\,t_{w2}=q_w$$ $$5.449(4.18)(44)-(5.449-0.1472)(4.18)\,t_{w2}=458.44$$ $$\boxed{\;t_{w2}=24.53\ ^\circ\text{C}\;}$$
Sanity-check the result against the wet bulb. The entering air has a thermodynamic wet bulb of 13.41 °C, so the tower achieves a range of 19.47 K and an approach of 11.12 K. A large approach like this is what one expects when the leaving air is constrained to only 26 °C: the air leaves far from equilibrium with the entering water, so the fill is not being used to its thermodynamic limit. The driving potential remains positive at both ends — the saturation enthalpy at 24.5 °C is 74.5 kJ/kg against the entering air's 37.62, and at 44 °C it is 203 kJ/kg against the leaving air's 80.60 — so the counterflow arrangement is thermodynamically consistent.

Check: where the fan work is booked. The 26 °C saturated state is taken to be measured at the tower discharge, i.e. after the induced-draft fan, so the fan's 4.75 kW is already contained in $h_2$ and must be deducted before the remainder is charged to the water. If instead the leaving state were measured at the top of the fill, ahead of the fan, the water would have to supply the whole 463.19 kW and the leaving water would be 24.32 °C. The difference is 0.21 K and does not affect any conclusion, but the assumption should be stated.

Two practical riders belong with the make-up figure. The 0.1472 kg/s computed above is evaporation only; a real tower also loses drift, typically 0.005 –0.2 % of the circulating flow depending on the eliminators, and must be bled to control the dissolved solids that the evaporation concentrates. At the common design figure of three cycles of concentration the blowdown is half the evaporation again, so the total make-up to specify for the water service would be nearer 0.22 kg/s than 0.1472 kg/s. And because the tower is an open evaporative device operating at 24 –44 °C, it falls under provincial Legionella control requirements — in British Columbia and Ontario, a documented maintenance and water-treatment programme is mandatory, not optional.

Final results — Question 5
Quantity	Value
Entering air humidity ratio / enthalpy	0.007699 kg/kg, 37.62 kJ/kg
Leaving air humidity ratio / enthalpy	0.021357 kg/kg, 80.60 kJ/kg
Dry-air mass flow	10.776 kg/s
Water mass flow / $L/G$ ratio	5.449 kg/s, 0.506
Heat rejected by the water	458.44 kW
Final water temperature	24.53 $^\circ$C
Make-up (evaporation) required	0.1472 kg/s = 529.9 kg/h
Range / entering wet bulb / approach	19.47 K / 13.41 $^\circ$C / 11.12 K

Topic. Simultaneous heat and mass transfer in an evaporative cooling tower.

Key relations. Moisture balance $\dot m_{ev}=\dot m_a(W_2-W_1)$; energy balance $\dot m_wc_pt_{w1}-(\dot m_w-\dot m_{ev})c_pt_{w2} =\dot m_a(h_2-h_1)-\dot W_{fan}$; range $=t_{w1}-t_{w2}$; approach $=t_{w2}-t_{wb,in}$; liquid-to-gas ratio $L/G=\dot m_w/\dot m_a$; humid volume $v=R_aT(1+1.6078W)/p$.

Why this works. A cooling tower is not a heat exchanger in the ordinary sense; most of the cooling is achieved by evaporating a small fraction of the water, and the latent heat for that evaporation comes out of the water that remains. This is why the driving potential is the difference between the enthalpy of saturated air at the local water temperature and the enthalpy of the bulk air stream — Merkel's insight — rather than a temperature difference. It is also why the entering wet-bulb temperature, not the dry bulb, is the thermodynamic floor: the wet bulb is the temperature at which air and water are in equilibrium with no net transfer, so no amount of fill or air can take the water below it. The whole-tower balance used here needs neither Merkel's integral nor a transfer coefficient, because both end states of the air are given; the integral would be needed only to size the fill or to predict off-design behaviour.

Common pitfall. Forgetting that the leaving water stream is smaller than the entering one by the evaporation, which here would shift the answer by about 0.7 K. The second pitfall is the sign of the fan work: in an induced-draft tower the fan is in the leaving air stream, so its power appears as an enthalpy rise that did not come from the water; adding it to the water's duty instead of subtracting it makes a 9.5 kW error in a 460 kW balance. A third is evaluating the air's specific volume at 26 °C when the 9 m³/s is stated at the inlet.

Source. W. P. Jones, Air Conditioning Engineering, 5th ed., Ch. 10 (cooling towers and evaporative cooling); ASHRAE Handbook — Systems and Equipment, Ch. 40 (cooling towers); McQuiston, Parker & Spitler, 6th ed., §3.7 (adiabatic saturation and wet-bulb temperature).

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