Induced-draught counter-flow cooling tower: 22-Mec-B2 Environmental Control in Buildings, May 2017, Q3

Question 3: Induced-draught counter-flow 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-draught counter-flow tower rejecting the condenser heat of a refrigeration plant, worked in inch-pound units as the question is set.

Design data, Problem 3
Quantity	Symbol	Value
Entering air, dry bulb / wet bulb	$t_{1}$ / $t_{1}^{*}$	70 / 55 $^{\circ}\text{F}$
Leaving air, dry bulb / relative humidity	$t_{2}$ / $\phi_{2}$	90 $^{\circ}\text{F}$ / 95 %
Water flow	$\dot m_{w}$	3 200 lb/min
Water on / off (range 30 $^{\circ}\text{F}$)	$t_{w1}$ / $t_{w2}$	105 / 75 $^{\circ}\text{F}$
Drift loss	—	0.3 % of the circulating water

Find. The tower sketch and its control scheme; then the entering-air enthalpy, specific volume and relative humidity, the inlet volumetric flow, the evaporative loss as a percentage of the circulating water, and the total make-up rate.

Induced-draught counter-flow tower serving a refrigeration condenser. Air is drawn in through louvres at the base, passes upwards through the fill against the falling water and is discharged by the fan at the top through drift eliminators. Control: a temperature transmitter (TT) in the cold-water line to the condenser signals the controller (TC), which stages or modulates the fan — two-speed motor, or a variable-frequency drive — to hold the leaving-water temperature at set point. A basin by-pass valve maintains condensing head in cold weather when the fan alone cannot be throttled far enough.

Approach. Fix both air states psychrometrically; write the steady-flow energy balance across the tower, remembering that the evaporated water leaves in the air stream and must be replaced by make-up entering at the cold-water temperature; solve for the air mass flow, then convert to volume at the inlet state.

Part (a) — the entering air state. At 70 / 55 $^{\circ}\text{F}$ the inch-pound adiabatic-saturation relation gives $$W_{1}=\frac{(1093-0.556\,t^{*})W_{s}^{*}-0.240\,(t-t^{*})}{1093+0.444\,t-t^{*}} =0.005770\ \text{lb/lb}$$ from which $$h_{1}=0.240(70)+0.005770\,[1061+0.444(70)]=\boxed{23.10\ \text{Btu/lb}}$$ $$v_{1}=\frac{0.370486\,(70+459.67)(1+1.6079\,W_{1})}{14.696}=\boxed{13.48\ \text{ft}^{3}/\text{lb}}$$ $$\phi_{1}=\frac{p\,W_{1}}{(0.621945+W_{1})\,p_{ws}(70^{\circ}\text{F})}=\boxed{37.2\ \%}$$
The leaving air state. At 90 $^{\circ}\text{F}$ and 95 % relative humidity, $W_{2}=0.029428$ lb/lb and $h_{2}=0.240(90)+0.029428[1061+0.444(90)]=54.00$ Btu/lb. The air therefore gains 30.90 Btu per pound of dry air and picks up 0.023658 lb of water per pound of dry air.
Part (b) — energy balance and air flow. The water side gives up its sensible heat; the make-up entering at the cold-water temperature carries back part of what the evaporation removed, so $$\dot m_{a}\,(h_{2}-h_{1})=\dot m_{w}\,c_{pw}\,(t_{w1}-t_{w2}) +\dot m_{a}\,(W_{2}-W_{1})\,c_{pw}(t_{w2}-32)$$ Substituting, $$\dot m_{a}\,(30.90)=3200(1.0)(30)+\dot m_{a}(0.023658)(43)$$ $$\dot m_{a}=\frac{96\,000}{30.90-1.017}=\boxed{3\,213\ \text{lb/min of dry air}}$$ and at the inlet specific volume this is $$\dot V_{1}=\dot m_{a}\,v_{1}=3213(13.48)=\boxed{43\,300\ \text{ft}^{3}/\text{min}}$$ The liquid-to-gas ratio is $L/G = 3200/3213 = 0.996$, squarely in the 0.8 to 1.5 band that counter-flow towers are built for, and the approach to the entering wet bulb is $75-55=20\ ^{\circ}\text{F}$ — loose, but ordinary for a condenser-water tower.
Part (c) — evaporative loss. The moisture the air carries away is $$\dot m_{ev}=\dot m_{a}(W_{2}-W_{1})=3213(0.023658)=76.0\ \text{lb/min}$$ $$\text{evaporative loss}=\frac{76.0}{3200}\times100=\boxed{2.38\ \%}$$ The familiar rule of thumb — about 1 % of the circulating flow for each $10\ ^{\circ}\text{F}$ of range — would predict 3 %, so the computed figure is the right order and slightly lower, as expected when the entering air is already at 37 % relative humidity rather than bone dry.
Part (d) — total make-up. Drift is water carried out as entrained droplets, not vapour, and is quoted as a fraction of the circulating flow: $$\dot m_{drift}=0.003(3200)=9.6\ \text{lb/min}$$ $$\dot m_{makeup}=\dot m_{ev}+\dot m_{drift}=76.0+9.6=\boxed{85.6\ \text{lb/min}}$$ which is 10.3 US gallons per minute, or 2.68 % of the circulating water.
Check the balance and comment on blowdown. The heat rejected is $3200(1.0)(30)=96\,000$ Btu/min = 1 688 kW, and the air stream carries away $3213(30.90)-76.0(43)=96\,000$ Btu/min — identical, as it must be. In service a third loss, blowdown, is added to control the concentration of dissolved solids: at $N$ cycles of concentration the blowdown is $\dot m_{ev}/(N-1)-\dot m_{drift}$, so at four cycles it would be $76.0/3-9.6=15.7$ lb/min and the make-up would rise to about 101 lb/min. The question asks only for evaporation and drift, so 85.6 lb/min is the answer, but the omission is worth naming.

Problem 3 — results
Quantity	Value
(a) entering air enthalpy $h_{1}$	23.10 Btu/lb dry air
(a) entering air specific volume $v_{1}$	13.48 $\text{ft}^{3}/\text{lb}$
(a) entering air relative humidity $\phi_{1}$	37.2 % ($W_{1}=0.00577$ lb/lb)
(b) air mass flow / inlet volumetric flow	3 213 lb/min / 43 300 cfm
(c) evaporative loss	76.0 lb/min = 2.38 % of the water flow
(d) drift / total make-up	9.6 lb/min / 85.6 lb/min (10.3 USgpm)
Heat rejected; L/G; approach	96 000 Btu/min (1 688 kW); 0.996; 20 $^{\circ}\text{F}$

Topic. Counter-flow cooling-tower heat and mass balance, and head-pressure control of the refrigeration plant it serves.

Key relations. Energy: $\dot m_{a}(h_{2}-h_{1})=\dot m_{w}c_{pw}(t_{w1}-t_{w2})+\dot m_{ev}c_{pw}(t_{w2}-32)$. Mass: $\dot m_{ev}=\dot m_{a}(W_{2}-W_{1})$. Make-up = evaporation + drift + blowdown. Range $=t_{w1}-t_{w2}$; approach $=t_{w2}-t_{1}^{*}$; the wet-bulb temperature, not the dry bulb, is the thermodynamic limit.

Why this works. A cooling tower is a direct-contact device, so the air leaves close to saturation and its enthalpy rise — not its temperature rise — measures the heat it has absorbed. That is why the Merkel formulation works in enthalpy and why the entering wet bulb is the performance datum: two atmospheres with the same wet bulb cool identically however different their dry bulbs. The make-up correction term matters more than it looks: the water evaporated leaves the system at the leaving-water temperature and is replaced by mains water, so the tower is not a closed circuit and a naive balance omitting it over-predicts the air flow by about 3 %. On the control side, the leaving-water temperature is what the chiller cares about: too high and the condensing pressure and compressor power rise; too low and the expansion device loses the pressure difference it needs, which is why fan staging is supplemented by a basin by-pass in winter.

Common pitfall. Working the balance with the dry-bulb temperature difference of the air instead of the enthalpy difference, which ignores the latent transfer that does most of the cooling. A second, subtler error is quoting a make-up rate that omits drift or blowdown; drift is a droplet loss and is fixed by the eliminators, blowdown by water chemistry, and neither scales with the tower duty the way evaporation does.

Source. Jones, Air Conditioning Engineering, 5th ed., Ch. 9 (cooling towers); ASHRAE Handbook — Systems and Equipment, Ch. 40; CTI ATC-105 for the range/approach/drift definitions.

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

Question 3: Induced-draught counter-flow cooling tower (20 marks)