Dehumidification by sorption methods

To achieve very low dew points it is necessary to dehumidify by a process of absorption or adsorption, the former being a chemical change and the latter a physical one. In either case, the process of dehumidification is termed desiccation. Refrigeration plant is often still needed.

Dew points as low as -70°C can be achieved by sorption methods.

72 The psychrometry of air conditioning processes

(a) Absorption

Most substances take up moisture from the atmosphere to some extent but some materials have the ability to absorb very large amounts of moisture from the ambient air. An example of this is lithium chloride, which can remove from the atmosphere up to 100 times its own dry weight, under suitable circumstances. The amount of moisture absorbed from air by an aqueous solution of lithium chloride depends on the strength of the concentration, the temperature of the solution, the vapour pressure of the ambient air and the extent to which the air comes in contact with the absorbent solution. The stronger the solution of lithium chloride in water, the more is its ability to absorb moisture. Conversely, moisture can be driven off from the solution and the concentration of lithium chloride increased, by raising its temperature until the solution vapour pressure exceeds that of the ambient air. If the strong solution is then cooled it may be used again to absorb moisture from the air. This provides the opportunity for a regeneration cycle.

Two methods are in common use to effect desiccation by absorption:

(i) Figure 3.16 illustrates the first of these methods. A strong solution of lithium chloride is pumped from a sump and is sprayed over the outside of the tubes of a cooler coil, fed with chilled water. Moisture is absorbed from the airstream and a weak solution drains back to the sump. The concentration in the sump is kept at an adequately high value by pumping a proportion of it (10 per cent to 20 per cent) to a re-activation unit, where it is sprayed over the outside of the tubes of a heater coil, fed with steam at a gauge pressure of from 0.2 to 1.7 bar, or an equivalent heating medium.

Strong solution

 

Regenerator

Coil

 

Cooler

Coil

 

Vti

L_l_ I

‘ I ‘ I

Hir.

Dry

Air

 

Humid

Air

 

Scavenging air

 

Discharge

Air

 

Dehumidification by sorption methods

O

 

Strong

Solution

 

Weak solution

 

Dehumidification by sorption methods

Fig. 3.16 Absorption process of drying using aqueous lithium chloride solution sprayed

Over a cooler coil.

The moisture content of the air supplied to the treated space is controlled by regulating the flow rate of chilled water through the cooling coil.

The heat removed by the cooling coil comprises the latent heat of condensation, plus the heat of solution, plus the sensible heat removed from the air, plus the heat added to the solution by the regeneration process. A sensible cooler coil may be needed after the desiccating unit.

(ii) The second method of desiccation by absorption uses a continuously rotating drum
containing a material that has been impregnated with lithium chloride. Figure 3.17 illustrates this. A sector of the drum is used for continuous re-activation by passing high temperature air through it. This air is usually heated by means of steam or electricity. LTHW is inadequate.

Dehumidification by sorption methods

Steam

Chilled coil water Schematic

Fig. 3.17 Continuously re-activating, rotating absorption dehumidifier.

Continuous

Re-activating

Sector

Sector

Section

подпись: continuous
re-activating
sector
 
sector
section
Control over the moisture content of the air supplied to the treated space is effected by regulating the energy input to the re-activation heater battery.

The heat removed by the desiccating sector of the drum consists of the latent heat of condensation plus the sensible heat conducted from the re-activation sector. A sensible cooler coil is usually required, after the rotary absorption drum, in the supply air to the treated space.

(b)Adsorption

This involves no chemical changes. The ability of solid surfaces to retain gaseous molecules in contact, is exploited by providing materials that have been manufactured specifically to have an enormously large internal surface area: one gram of some substances can have an internal surface area of more than 4000 m2. Adsorbent materials are re-activated by heating them to a high temperature in order to drive off the moisture from the internal surfaces.

Typical substances are silica gel and activated alumina for drying, and activated carbon for the removal of other gases. Adsorbent materials can be tailor-made to take up specific gases, selectively.

EXAMPLE 3.17

1 m3 s-1 of air at 30°C dry-bulb, 20°C wet-bulb (sling) is to be cooled and dehumidified to a state of 10°C dry-bulb and 4.70 g kg-1 by means of a continuously rotating, absorption dehumidifier. Making use of Figure 3.18 determine the sensible cooling load after the dehumidifier.

Answer

From psychrometric tables or a chart, the entering moisture content is found to be 10.38 g kg-1, and the specific volume to be 0.8728 m3 kg-1. This value of moisture content is identified on the abscissa of Figure 3.18 and read upwards to the first curve, for an entering

Dehumidification by sorption methods

Entering moisture content, g/kg dry air Fig. 3.18 Typical performance of a rotating absorption drier.

Air dry-bulb temperature of 30°C. Reading across to the scale on the left-hand side gives a leaving moisture content of 4.7 g kg-1 dry air. (If a different leaving moisture content were wanted then reference would have to be made to another curve, not shown in the figure, for different operating conditions. Possibly a drier of different size would have to be considered.) Reference to the upper curve, for the same entering moisture content, gives a leaving dry-bulb temperature of 56°C.

Figure 3.19 illustrates the related psychrometry, the process being shown by the full lines; air enters the drier at state O and leaves it at state B. A sensible cooler coil is then used to cool the air from state B at 56° to state C at 10°. From tables or a chart the enthalpy at state B is determined as 68.60 kJ kg-1. Similarly, at state C, the enthalpy is found to be

21.67 kJ kg’1.

Hence the sensible cooling duty is

0^28 X (68’60 “ 2L67) = 53J7 kW An alternative approach would be first to cool and dehumidify the air by conventional means and then to pass it through the rotary drier. Sensible cooling is still required and the overall energy removal is the same because the enthalpies on and off the plant are unchanged but the economics might be different. This alternative process is shown by the broken lines in Figure 3.19: air is first cooled and dehumidified from O to W, then dried from W to E and finally sensibly cooled from E to C.

Dehumidification by sorption methods

Dew-point

10.38 g kg

4.70 g kg

Fig. 3.19 Psychrometry for Example 3.17.

If an air washer is fitted in the airstream extracted from the conditioned room the air may be adiabatically cooled to a lower dry-bulb temperature. The air can then be passed through a rotary sensible heat exchanger in order to reduce the temperature of the hot, fresh airstream leaving the absorption dehumidifier without the use of mechanical refrigeration (see Figures 3.20(a) and (b)). Further mechanical refrigeration is certainly needed but it will be less than if the adiabatic cooling and sensible heat exchange described had not been adopted.

The air extracted from the conditioned space is hot after passing through the rotary sensible cooler and may be discharged to waste. However, the thermal efficiency of the process may be improved if it is fed through the heater battery used to regenerate the absorption dehumidifier.

Figure 3.20(a) shows the plant arrangement for this. Supply and extract fans are not shown in the figure, as a simplification. When such necessary fans are included, the air handled by them suffers a temperature rise of 1 K per kPa of fan total pressure if the driving motor is outside the airstream, or a rise of 1.2 K per kPa of fan total pressure if the driving motor is within the airstream. See section 6.5.

EXAMPLE 3.18

See Figures 3.20(a) and (b). It is assumed that the room treated by the plant considered in Example 3.17 has sensible and latent heat gains that give a state in the room (the point R in Figure 3.20(b)) of 20.5°C dry-bulb, 34 per cent saturation, 5.177 g kg-1 dry air and 12.0°C wet-bulb (sling) when 1 m3 s“1 air is supplied at 10°C dry-bulb and 4.70 g kg-1 (the point S in Figure 3.20(b)). Air is extracted from the room, passed through an air washer having a humidifying efficiency of 90 per cent and undergoes adiabatic humidification from state R to state W. The air then flows through one side of a rotary sensible heat exchanger having an efficiency of 80 per cent. Air from the absorption dehumidifier, at 56°C dry-bulb and 4.70 g kg-1 dry air (state B), flows through the other side of the rotary sensible heat exchanger, leaving at state C. The air then passes through a sensible cooler coil and is supplied to the room at 10°C dry-bulb, 4.70 g kg-1 dry air (state S). The air

Rotary absorption Steam

Dehumidification by sorption methods

Coil

(a)

Dehumidification by sorption methods

Fig. 3.20 (a) Plant arrangement for Example 3.18. Supply and extract fans are omitted as a simplification, (b) Psychrometry for Example 3.18. Air temperature rises due to supply and extract fans are omitted as a simplification.

Extracted from the room and leaving the washer at state W flows through the extract side of the rotary sensible heat exchanger, emerging at state D, with a high temperature. This air is fed through the steam heater battery used to regenerate the rotary absorption dehumidifier to improve the overall thermal efficiency of the operation.

Determine:

(a) The state of the air leaving the washer (W).

(b) The state of the dehumidified air leaving the supply side of the rotary sensible heat exchanger (C).

(c) The state of the humidified extract air leaving the extract side of the rotary sensible cooler (D).

(d) The load on the necessary sensible cooler coil.

Answer

(a) The adiabatic change of state through the air washer is along a wet-bulb line from

20.5°C dry-bulb, 12°C wet-bulb, and 5.177 g kg-1 (state R). From tables or a chart the

Moisture content at 12°C saturated is 8.763 g kg-1 dry air and its enthalpy is 34.18 kJ kg-1 dry air. Since the humidifying efficiency is 90 per cent the state leaving the washer (W) is determined as:

T = 20.5 — 0.9(20.5 — 12) = 12.85°C dry-bulb g = 5.177 + 0.9(8.763 — 5.177) = 8.404 g kg“1 dry air

Ft = 34.18 kJ kg“1 dry air

(because the points A and W are so close together their enthalpies are virtually the same).

(b) The dry-bulb temperature on the extract side of the rotary sensible heat exchanger receiving adiabatically cooled air from the washer at state W is 12.85°C. On the supply side of the rotary sensible heat exchanger air enters from the rotary absorption dehumidifier at 56°C dry-bulb and 4.70 g kg-1 dry air (state B). Since the rotary sensible heat exchanger is 80 per cent efficient the dehumidified air is cooled to:

56 — 0.80(56 — 12.85) = 21.48°C

But its moisture content remains at 4.70 g kg-1 dry air, giving state C. From tables or a chart the corresponding enthalpy is 33.55 kJ kg-1 dry air.

(c) On the opposite side of the rotary sensible heat exchanger the air has been warmed and its temperature is calculated as 12.85 + 0.8(56 — 12.85) = 47.37°C dry-bulb (state D). The moisture content remains at 8.404 g kg-1 dry air and the air is discharged into the steam heater battery to assist in the regeneration of the rotary absorption dehumidifier.

(d) Since 1 m3 s"1 of air is supplied to the conditioned room at 10°C dry-bulb, 4.70 g kg-1 the corresponding enthalpy (from tables or a chart) is 21.67 kJ kg_I dry air. The residual refrigeration load is:

[1/0.8728] x (33.55 — 21.67) = 13.61 kW

This is about 25 per cent of the sensible cooling load found in the answer to Example 3.17.

Exercises

1. Dry saturated steam at 1.2 bar and 104.8°C is injected at a rate of 0.01 kg s’1 into an airstream having a mass flow rate of dry air of 1 kg s-1 and an initial state of 28°C dry-bulb,

11. 9°C wet-bulb (sling). Calculate the leaving state of the airstream if all the injected steam is accepted by it. Use equation (2.24) as necessary.

Answers

29.3°C, 11.937 g kg-1.

2. (a) Define the term ‘humidifying efficiency’, as applied to an air washer. Explain why

This is only approximately true if expressed in terms of dry-bulb temperature.

(b) Write down an equation that defines an adiabatic saturation process in an air washer, identifying all symbols used.

(c) 2 m3 s_1 of air at 23.5°C dry-bulb, 11°C wet-bulb (sling) and 2.94 g kg-1 dry air enters an air washer that has a humidifying efficiency of 85 per cent. If the airstream undergoes a process of adiabatic saturation use your definition from part (b) to determine the dry-bulb temperature and moisture content of the air leaving the washer. It is given that the specific heat capacity of dry air is 1.012 kJ kg-1 K-1, the latent heat of evaporation of water is 2450 kJ kg-1 and the specific heat of water vapour is 1.89 kJ kg-1 K"1. Do not use psychrometric tables or a chart.

Answers

12.9°C, 7.36 g kg-1.

3. (a) An operating theatre is maintained at an inside temperature of 26°C dry-bulb when

The outside air is at 45°C dry-bulb, 32°C wet-bulb (sling) and the sensible and latent heat gains are 9 kW and 3 kW, respectively. Determine the cooling load if 100 per cent fresh air is handled, the air temperature leaving the cooler coil being 14°C and the apparatus dew point 12°C. Assume a rise of 1.5°C across the supply fan (which is located after the cooler coil) and a further rise of 2°C because of heat gains to the supply duct.

(b) Determine the percentage saturation maintained in the theatre under the conditions of part (a).

Answers

72.67kW, 50.9 per cent.

Unit

J kg-1 K1 kJ kg’1 KT1

подпись: unit
j kg-1 k1 kj kg'1 kt1
Notation

Symbol Description

Ca specific heat of dry air

E

8

подпись: e
8

Kg per kg dry air or

подпись: kg per kg dry air orEffectiveness of an air washer or spray chamber moisture content

H

K

Hw

Ma

Ms

подпись: h
k
hw
ma
ms

Enthalpy of moist air enthalpy of steam enthalpy of feed-water mass or mass flow of dry air mass flow of steam injected g per kg dry air kJ per kg dry air

KJ kg 1 kJ kg’1

Kg or kg s 1 kg s’1

Fflw

Mass flow of feed water evaporated

Kg s"1

T

Dry-bulb temperature

°C

F

Wet-bulb temperature

°C

K

Dew point temperature

(or the dry-bulb temperature at a state D)

°c

Tw

Feed-water temperature

°c

Я

Contact factor

N

Humidifying efficiency

Percentage saturation

%

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