22-Mec-B2 Environmental Control in Buildings · December 2017

Question 3 of 8: Induced-draft counter-flow cooling tower

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Engineers Canada national examination 16-Mec-B2 Environmental Control in Buildings, December 2017, three hours, open book. Eight problems of 20 points each; candidates are required to solve five, and all questions carry the same value. ASHRAE Psychrometric Chart No. 1 (SI and inch-pound) and a pressure–enthalpy diagram for R-717 are appended to the paper as pages 6–8.

All eight problems are worked here. The paper mixes SI and inch-pound units deliberately: Problems 1, 3 and 4 are SI, Problems 2, 6 and 7 are inch-pound, and Problem 8 is SI with a Canadian climate. Each solution is worked in the units the question uses, as the cover-page instructions require.

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 and the psychrometric chart), Ch. 6 (air-conditioning plant cycles), Ch. 7 (the cooling coil, apparatus dew point and by-pass factor), Ch. 9 (cooling towers), Ch. 15 (fans).
McQuiston, Parker & Spitler, Heating, Ventilating and Air Conditioning: Analysis and Design, 6th ed., Wiley — Ch. 3 (moist air), Ch. 5 (heat transmission in building structures), Ch. 8 (energy estimating and the degree-day method), Ch. 12 (fans and duct design).
ASHRAE Handbook — Fundamentals — Ch. 1 (psychrometrics), Ch. 14 (climatic design information), Ch. 21 (duct design), Ch. 25–27 (thermal and moisture performance of the building envelope), Ch. 30 (fenestration).
Stoecker & Jones, Refrigeration and Air Conditioning, 2nd ed., McGraw-Hill — Ch. 10–12 (vapour-compression cycle, compressors, condensers and evaporators); ASHRAE Handbook — Refrigeration for ammonia plant practice.
National Building Code of Canada and the National Energy Code of Canada for Buildings (NRC), Appendix C climatic data; CSA and Canada Green Building Council material for Problem 5.

Property basis used throughout. Moist-air properties are computed from the ASHRAE Fundamentals ideal-moist-air relations, so every state quoted here can be read back off the psychrometric chart supplied with the paper:

$$W=\frac{0.621945\,p_w}{p-p_w},\qquad h_{\text{SI}}=1.006\,t+W\,(2501+1.86\,t),\qquad h_{\text{IP}}=0.240\,t+W\,(1061+0.444\,t)$$

with $h$ in $\text{kJ/kg}$ of dry air for $t$ in $\,{}^{\circ}\text{C}$ and in $\text{Btu/lb}$ of dry air for $t$ in $\,{}^{\circ}\text{F}$. Ammonia properties are quoted on the same datum as the attached ASHRAE p–h diagram ($h_f=200\ \text{kJ/kg}$ and $s_f=1.0\ \text{kJ/(kg}\cdot\text{K)}$ for saturated liquid at $0\,{}^{\circ}\text{C}$); only differences enter the answers, so any consistent chart or table gives the same duties.

Question 3: Induced-draft counter-flow cooling tower (20 points)

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.

Quantity	Symbol	Value
Entering air	$t_1,\ t_{wb1}$	$23\,{}^{\circ}\text{C}$ db, $17\,{}^{\circ}\text{C}$ wb
Leaving air	$t_2$	$25\,{}^{\circ}\text{C}$, saturated
Water circulated	$\dot m_w$	$15\ \text{kg/s}$
Water temperatures	$t_{w,in},\ t_{w,out}$	$27\,{}^{\circ}\text{C}$ to $21\,{}^{\circ}\text{C}$
Fan air power	$\dot W_{fan}$	$5\ \text{kW}$
Water specific heat	$c_{p,w}$	$4.186\ \text{kJ/(kg}\cdot\text{K)}$

Find. A sketch of the tower and its control connection to the refrigeration plant, the induced air mass flow, the evaporative loss as a percentage of the circulating water, and the make-up water rate.

Induced-draft counter-flow tower. Air is drawn in through the louvres at the base, upward through the fill against the falling water, past the drift eliminators and out through the fan. The cold-water temperature controller modulates fan speed — this is the link that regulates the refrigeration plant.

The sketch and its control (first part of the question). Warm condenser water enters the spray header at the top and falls through a film fill against the rising air; the cooled water collects in the basin and is pumped back to the condenser. The fan sits above the drift eliminators and pulls air through, which is what makes the tower induced draft rather than forced draft: the fan handles saturated discharge air rather than fouling-prone inlet air, and the low inlet velocity avoids recirculation of the plume back into the louvres. Regulation of the refrigeration plant is achieved by controlling the cold-water temperature leaving the basin, because that temperature sets the condensing pressure and hence the compressor power. A temperature sensor in the basin outlet drives, in increasing order of refinement: a two-speed or variable-frequency drive on the fan motor, the staging of individual cells in a multi-cell tower, and a modulating bypass that diverts part of the flow around the fill in cold weather. The last is essential in the Canadian climate — without it the head pressure falls so far in winter that the expansion valve is starved, and the basin is at risk of freezing.

Approach. Write a steady-flow energy balance on the whole tower including the fan work and the enthalpy the make-up water brings in, solve it for the air mass flow, and then take the moisture balance for the evaporation.

Fix the two air states. The entering state comes from the dry- and wet-bulb pair through the adiabatic-saturation relation, and the leaving state is simply saturated air at $25\,{}^{\circ}\text{C}$:
Point Description $t$ (°C) $W$ (g/kg) $h$ (kJ/kg) $\phi$ (%) $v$ (m³/kg)
1 air entering the louvres 23.0 9.634 47.64 55.0 0.8520
2 air leaving the fan 25.0 20.086 76.32 100.0 0.8719
Heat rejected by the water. $$\dot Q_w=\dot m_w\,c_{p,w}\,(t_{w,in}-t_{w,out})=15\times4.186\times(27-21) =376.74\ \text{kW}$$ The tower is working over a range of $6\ \text{K}$ down to an approach of $21-17=4\ \text{K}$ — a close approach, so this is a generously sized tower.
Part (a) — energy balance for the air mass flow. Everything that enters the control volume must leave it. The air carries in $\dot m_a h_1$, the water gives up $\dot Q_w$, the fan motor puts in $\dot W_{fan}$, and the make-up water arrives at the basin temperature carrying $\dot m_a(W_2-W_1)\,h_f(21\,{}^{\circ}\text{C})$: $$\dot m_a h_1+\dot Q_w+\dot W_{fan}+\dot m_a (W_2-W_1)\,h_f =\dot m_a h_2$$ $$\dot m_a=\frac{\dot Q_w+\dot W_{fan}} {(h_2-h_1)-(W_2-W_1)\,h_f} =\frac{376.74+5}{(76.32-47.64)-0.010452\times88.14}$$ $$\dot m_a=\boxed{13.76\ \text{kg/s of dry air}}$$ which is $11.72\ \text{m}^3/\text{s}$ at the inlet and $12.00\ \text{m}^3/\text{s}$ at the fan. The liquid-to-gas ratio is $L/G=15/13.76=1.09$, squarely in the $0.8$–$1.5$ band that counter-flow film fill is designed for. Omitting the fan work would give $13.31\ \text{kg/s}$ and omitting the make-up enthalpy as well $13.14\ \text{kg/s}$; both are within a few per cent, but the full balance is the defensible answer.
Part (b) — evaporative loss. The moisture balance alone gives the water that actually leaves as vapour: $$\dot m_{ev}=\dot m_a\,(W_2-W_1)=13.76\times(0.020086-0.009634) =\boxed{0.1438\ \text{kg/s}}$$ $$\frac{\dot m_{ev}}{\dot m_w}\times100 =\frac{0.1438}{15}\times100=\boxed{0.958\%}$$ The familiar rule of thumb — about $1\%$ of the circulating water for every $5.5\ \text{K}$ of range — predicts $1.09\%$ here, so the answer is the right size. It comes out slightly below the rule because part of the tower duty is carried as sensible heating of the air (its dry bulb rises from $23$ to $25\,{}^{\circ}\text{C}$) rather than as evaporation.
Part (c) — make-up water. The instruction to take account of the moisture gained by the air is the whole point: the water lost from the circuit is not $\dot Q_w/h_{fg}$, which would credit the entire tower duty to evaporation, but exactly the moisture the air carries away. That gives $$\dot m_{mu}=\dot m_{ev}=0.1438\ \text{kg/s} =\boxed{517.6\ \text{kg/h}\ (517.6\ \text{L/h})}$$ For comparison, the naive latent-only estimate $\dot Q_w/h_{fg}=376.74/2454=0.1535\ \text{kg/s}$ overstates the requirement by about $7\%$. In service the make-up must also replace drift and blowdown; at four cycles of concentration the blowdown is $\dot m_{ev}/(4-1)=0.0479\ \text{kg/s}$, so the float valve should be sized for a total of about $0.192\ \text{kg/s}$, or $690\ \text{kg/h}$.

Point	Description	$t$ (°C)	$W$ (g/kg)	$h$ (kJ/kg)	$\phi$ (%)	$v$ (m³/kg)
1	air entering the louvres	23.0	9.634	47.64	55.0	0.8520
2	air leaving the fan	25.0	20.086	76.32	100.0	0.8719

The air state change plotted on the SI chart: the air is driven from 1 up to saturation at 25 °C, and the enthalpy rise carries the 376.7 kW of water heat plus the 5 kW of fan work.

Final results.

Quantity	Result
Heat rejected by the water	$376.74\ \text{kW}$
(a) Induced air mass flow	$\mathbf{13.76\ \text{kg/s}}$ dry air ($11.72\ \text{m}^3/\text{s}$ at inlet)
Liquid-to-gas ratio	$L/G=1.09$
(b) Evaporation	$0.1438\ \text{kg/s}$, i.e. $\mathbf{0.958\%}$ of the circulating water
(c) Make-up (evaporation only)	$\mathbf{517.6\ \text{L/h}}$
(c) Make-up including blowdown at 4 cycles	$690\ \text{kg/h}$
Range / approach	$6\ \text{K}$ / $4\ \text{K}$

Check: assumptions declared under cover-page instruction 1. (i) The $5\ \text{kW}$ of fan air power is added to the air stream, which is correct for an induced-draft machine because the fan sits in the leaving air; on a forced-draft tower it would be added at inlet, with the same effect on the balance. (ii) The make-up water is taken to enter the basin at the cold-water temperature of $21\,{}^{\circ}\text{C}$. (iii) Drift and blowdown are excluded from the answer to part (c), which asks only for the loss associated with the moisture gained by the air; the four-cycle figure is quoted separately as design guidance. (iv) The air is assumed to leave genuinely saturated as stated, with the drift eliminators removing entrained droplets so no liquid water leaves with the plume.

Topic. Cooling-tower heat and mass balance, range and approach.

Key relations. Energy: $\dot m_a\big[(h_2-h_1)-(W_2-W_1)h_f\big]=\dot m_w c_{p,w}\Delta t +\dot W_{fan}$. Moisture: $\dot m_{ev}=\dot m_a(W_2-W_1)$. Blowdown at $n$ cycles of concentration: $\dot m_{bd}=\dot m_{ev}/(n-1)$. Range $=t_{w,in}-t_{w,out}$; approach $=t_{w,out}-t_{wb,in}$; $L/G=\dot m_w/\dot m_a$.

Why this works. A cooling tower is an enthalpy exchanger, not a temperature exchanger. The driving potential is the difference between the enthalpy of saturated air at the water temperature and the enthalpy of the bulk air, which is why the entering wet-bulb temperature — not the dry bulb — sets the lowest water temperature the tower can ever reach. The approach can be made small only by making the tower large, and it can never be zero. Two-thirds or more of the duty leaves as latent heat in the evaporated water, which is what makes evaporative cooling so effective and also what makes the make-up water bill a real operating cost. Because the condensing pressure follows the cold-water temperature, the tower is not a passive accessory: controlling it is controlling the refrigeration plant, and that is the connection the question is asking about.

Common pitfall. Computing the make-up as $\dot Q/h_{fg}$, which assumes the whole tower duty is evaporative and overstates the answer by roughly the fraction of the duty carried as sensible heat — here about $7\%$. The other frequent slip is leaving the fan work out of the energy balance when the question has deliberately supplied it; on this tower it is $1.3\%$ of the duty, small but not zero, and an examiner who gives you a number expects to see it used.

Source. Jones, Air Conditioning Engineering, 5th ed., Ch. 9 (cooling towers); ASHRAE Handbook — Systems and Equipment, Ch. 40; McQuiston, Parker & Spitler, 6th ed., §13-6.

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