Air washers

A diagrammatic picture of an air washer is given in Figure 10.11. Essentially, a washer consists of a spray chamber in which a dense cloud of finely divided spray water is produced by pumping water through nozzles. On its path through the chamber the air first

Scrubber header

Perforated screen for smoothing airflow Airflow

Eliminator Plates

Overflow &

Emptying

Outlet

C. W.-

Strainer

Feed

Continuous bleed for Strainer

Water treatment

A

Automatic control valve

To chilled Water Store

From —► c chilled water store

Fig. 10.11 Air washer: single bank of nozzles blowing upstream. Note that it is highly desirable for hygienic reasons to empty the spray water tank completely once in each twenty-four hours and to refill with fresh water. The air washer is rarely used today, for hygienic reasons.

Stand pipes

Spray nozzles

Passes an array of deflector plates or a perforated metal screen, the purpose of which is to secure a uniform distribution of air flow over the cross-section of the washer and to prevent any moisture accidentally being blown back up the duct. Within the spray chamber itself, the nozzles which atomise the water are arranged in banks, usually one or two in number, but occasionally three. The efficiency of the washer depends on how many banks are used and which way they blow the spray water, upstream or downstream. The banks consist of stand pipes with nozzles mounted so as to give an adequate cover of the section of the chamber. The arrangement is usually staggered, as shown in Figure 10.12. Finally, the air leaves the washer by passing through a bank of eliminator plates, the purpose of which is primarily to prevent the carry-over of unevaporated moisture beyond the washer into the ducting system. A secondary purpose is to improve the cleaning ability of the washer by

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Fig. 10.12 Stand piping and nozzle arrangement for atomising water in an air washer.

Offering a large wetted surface on which the dirt can impinge and be wasthed away. In general, washers are poor filters; they should not be relied on alone to clean the air.

Provided the temperature of the water circulated is less than the dew point of the entering air, a washer can dehumidify, and its cooling and dehumidifying capacity can be modulated by varying the temperature of the water handled by the pump. The usual way of doing this is by means of a 3-port mixing valve, as shown in Figure 10.11. However, in general, air washers are less useful than cooler coils in dehumidifying and cooling air. The reason for this is that contra-flow does not occur. As a droplet of water leaves a spary nozzle it travels for a short distance either parallel or counter to the direction of airflow, after that, it starts falling to the collection tank, and the heat transfer is according to cross-flow conditions. Near the surface of the water in the tank, even at the downstream end of the washer the air encounters water which is at its warmest. The consequence of all this is that a leaving air temperature which is less than the leaving water temperature is unobtainable.

To secure effective atomisation, the pressure drop across the nozzles used in the spray­chamber type of washer must be fairly large. If an attempt is made to modulate the capacity of the washer by reducing the water flow through the nozzles, it will be unsuccessful, because proper atomisation ceases below a certain minimum pressure. Typical nozzle duties vary from 0.025 litres s-1 with a pressure drop of 70 kPa through a 2.5 mm orifice to 0.3 litres s“1 with a drop of 280 kPa through a 6.5 mm orifice.

Efficiencies vary with the arrangement of the banks, and typical percentage values are;

1 bank downstream	50
1 bank upstream	65-75
2 banks downstream	85-90
2 banks upstream	92-97
2 banks in opposition	90-95
3 banks upstream	>95

For face velocities of airflow over the section of 2.5 m s“1

These efficiencies are based on the assumption that the section of the spray chamber is adequately covered by nozzles and that these are producing a properly atomised spray. A typical nozzle arrangement is 45 per m2 yielding a water flow rate of about 2 litres s-1 over each m2 of section of each bank.

As with cooler coils, the velocity of airflow is also significant: a value of 2.5 m s-1 over the face area is usually chosen. Little variation from this is desirable, either way, if the eliminators are to be successful. A range of 1.75 m s-1 to 3.75 m s_1 has been suggested.

Good practice in the design and use of air washers for air conditioning requires a water quantity of 0.11 litres s_1 per nozzle and 22 nozzle per m2 of cross-sectional area in each bank, with a mean air velocity of 2.5 m s-1.

Two banks, blowing upstream, are adequate, so the total circulated water quantity should be 0.44 litres s“1 m-2 of cross-section.

Washers must be long enough to allow adequate contact between the spray water and the air and to avoid the penetration of spray through the downstream eliminators or the upstream smoothing screen. There is therefore commonly 0.8 to 1.5 m between the banks with spaces of about 0.3 and 0.5 m before and after the initial and final banks, respectively. The tank is usually 450 mm deep with about 100 mm of freeboard. Washers are generally designed for air velocities of 2.0 to 3.0 m s"1 over their cross-section. For higher velocities the eliminators may not be fully effective unless specially designed for the purpose.

Washers have fallen out of favour for commercial applications and are now very seldom used.

1. (a) Define the term ‘contact factor’ as applied to cooler coils and state the conditions under which it remains constant.

(b) Explain how the performance of a chilled water cooler coil varies under conditions of (i) varying water flow rate through the tubes, and (ii) varying water flow temperature.

(c) How is apparatus dew point, as applied to chilled water cooler coils, related to the flow and return temperatures of the chilled water passing along the tubes?

2. A cooler coil is chosen to operate under the following conditions:

Air on: 28°C dry-bulb, 20.6°C wet-bulb (sling), 28.3 m3 s_1.

Air off: 12°C dry-bulb, 8.062 g per kg dry air.

Chilled water on: 5.5°C.

Chilled water off: 12.0°C.

You are asked to calculate (a) the contact factor, (b) the design cooling load and (c) the water flow rate through the coil.

Answers

(a) 0.88; (b) 868 kW; (c) 31.8 kg s“1.

3. An air-conditioning system comprises a chilled water cooler coil, reheater and supply air fan, together with a system of distribution ductwork. Winter humidification is achieved by steam injection directly into the conditioned space. It is considered economic, from an owning and operating cost point of view, to arrange that a constant minimum quantity of outside air is used throughout the year. The fixed quantity of outside air is 25 per cent by weight of the total amount of air supplied to the room. Using a psychrometric chart and the data listed below, calculate

(a) The contact factor of the cooler coil,

(b) The design load on the cooler coil in summer,

(c) The load on the cooler coil under winter conditions, and

(d) The winter reheat load.

Design data

25 kW 0.3 kW 4.4 kW 28°C dry-bulb, 19.5°C wet-bulb -1°C saturated 21°C dry-bulb, 50 per cent saturation 12.5°C dry-bulb 0.5°C

Sensible heat gain in summer Latent heat gain in summer Sensible heat loss in winter Outside state in summer Outside state in winter Inside state in summer and winter

Supply temperature in summer design conditions Temperature rise due to fan power and duct heat gains

Assume that the ratio of latent to total heat removal by the cooler coil is halved under winter design conditions. Take the specific heat of dry air as 1.012 kJ kg-1 K~’.

Answers

(a) 0.83; (b) 36.85 kW; (c) 20.05 kW; (d) 37.2 kW.

4. (a) Air at a constant rate and at a state of 20.5°C dry-bulb, 17.9°C wet-bulb (sling), flows on to a cooler coil and leaves it at 11°C dry-bulb, 10.7°C wet-bulb (sling), when the coil is supplied with a constant flow rate of chilled water at an adequately low temperature. Calculate the state of the air leaving the coil if the load is halved when the state of the air entering the coil changes to 20°C dry-bulb with 14.2°C wet-bulb (sling). Make use of a psychrometric chart.

(b) State briefly how the contact factor of a cooler alters when (i) the number of rows is reduced and (ii) the face velocity is reduced, assuming that the ratio of air mass flow to water mass flow is constant.

Answers

(a) 10.7°C dry-bulb, 10.2°C wet-bulb, 30.02 kJ kg-1, (b) (i) diminishes and (ii) increases.

5. Moist air at 28°C dry-bulb and 20.6°C wet-bulb flow over a 4-row cooler coil, leaving it at 9.5°C dry-bulb and 7.107 g per kg. Using a psychrometric chart, answer the following:

(a) What is the contact factor?

(b) What is the apparatus dew point?

(c) If the air is required to offset a sensible heat gain of 2.4 kW and a latent heat gain of

0.3 kW, in the space being conditioned, calculate the weight of dry air which must be

Supplied to the room in order to maintain 21°C dry-bulb therein.

(d) What is the percentage saturation in the room?

(e) If the number of rows of the coil is decreased to 2, the rate of waterflow, the rate of airflow and the apparatus dew point remaining constant, what temperature and humidity will be maintained in the room?

Ignore any temperature rise due to duct friction or heat gain.

Answers

(a) 0.925, (b) 8.1°C, (c) 0.204 kg s’1, (d) 49 per cent, (e) 26°C dry-bulb and 40 per cent saturation.

Notation

Unit M2 M2 M2 M2 M2 M2 KJ kg’

1 K-1

Description Face area Total surface area of the fins total internal surface area total surface area of the tubes total surface area per row total external surface area specific heat capacity of humid air
Symbol Af Ai Ar At C

Cw	Specific heat capacity of water	KJ kg“1 K“1
Di	Inside tube diameter	M
D0	Outside tube diameter	M
8	Moisture content of humid air	G kg-1 dry air
H	Enthalpy of humid air	KJ kg-1 dry air
K	Heat transfer coefficient for the surface film on the
	Air side of a cooler coil	W nT2 K~’
K	Constant	—
(lmtds	Logarithmic mean temperature difference between the airstream and the mean coil surface temperature	K
(LMTDW	Logarithmic mean temperature difference between the airstream and the water stream	K
Ma	Mass flow rate of dry air	Kg s’1
Mw	Mass flow rate of water	Kg s’1
Qs	Rate of sensible heat transfer	W or kW
Qt	Rate of total heat transfer	WorkW
R	Total thermal resistance	M2 K VT1
Ra	Thermal resistance of the air film	M2 K W“1
Rf	Thermal resistance of the fins	M2 K W“1
Rm	Thermal resistance of the metal of the tubes and fins	M2 K W“1
Rt	Thermal resistance of the tubes	M2 K W’1
R*	Thermal resistance of the water film	M2 K VT1
R	Number of rows of a cooler coil	—
S	Ratio of sensible to total heat transfer	—
T, h	Dry-bulb temperature of an airstream	°C
Td	Dew-point temperature of an airstream	°c
Hm	Mean coil surface temperature	°c
Fw	Chilled water or saturated air temperature	°c
Twa> Avb	Entering, leaving chilled water temperature	°c
*wm	Mean chilled water temperature	°c
Mvmin	Minimum chilled water temperature	°c
^wl> Av2	Mean coil surface temperature of the first, second, etc. rows of a coil	°c
T’	Wet-bulb temperature of an airstream	°c
U	Thermal transmittance coefficient	W nT2 K1
Ut	Biased [/-value for a cooler coil to account for total heat flow, air-to-air	W m“2 K-1
V	Mean chilled water velocity	M s"1
Vf	Face velocity over a cooler coil	M s-1
V	Rate of volumetric airflow over the face of a coil	M3 s“1
P	Contact factor	—
	Total surface effectiveness	_
K	Thermal conductivity of the metal of the tubes	W m“1 KT1
<t>	Fin efficiency	—
References

Posted in Engineering Fifth Edition