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CHAPTER 7
DRYING
(cont'd)

PSYCHROMETRY


Wet-bulb Temperatures
Psychrometric Charts
Measurement of Humidity


The capacity of air for moisture removal depends on its humidity and its temperature. The study of relationships between air and its associated water is called psychrometry.

Humidity (Y) is the measure of the water content of the air. The absolute humidity, sometimes called the humidity ratio, is the mass of water vapour per unit mass of dry air and the units are therefore kg kg-1, and this will be subsequently termed just the humidity.

Air is said to be saturated with water vapour at a given temperature and pressure if its humidity is a maximum under these conditions. If further water is added to saturated air, it must appear as liquid water in the form of a mist or droplets. Under conditions of saturation, the partial pressure of the water vapour in the air is equal to the saturation vapour pressure of water at that temperature.


The total pressure of a gaseous mixture, such as air and water vapour, is made up from the sum of the pressures of its constituents, which are called the partial pressures. Each partial pressure arises from the molecular concentration of the constituent and the pressure exerted is that which corresponds to the number of moles present and the total volume of the system. The partial pressures are added to obtain the total pressure.


EXAMPLE 7.5. Partial pressure of water vapour
If the total pressure of moist air is 100 kPa (approximately atmospheric) and the humidity is measured as 0.03 kg kg-1, calculate the partial pressure of the water vapour.

The molecular weight of air is 29, and of water 18
So the mole fraction of water = (0.03/18)/(1.00/29 + 0.03/18)
                                          = 0.0017/(0.034 + 0.0017)
                                          = 0.048
                                                                                   

Therefore the water vapour pressure
                                          = 0.048 x 100 kPa
                                          = 4.8 kPa.

The relative humidity (RH) is defined as the ratio of the partial pressure of the water vapour in the air (p) to the partial pressure of saturated water vapour at the same temperature (ps). Therefore:

                                    RH = p/ps

and is often expressed as a percentage = 100 p/ps


EXAMPLE 7.6. Relative humidity
If the air in Example 7.5 is at 60°C, calculate the relative humidity.

From steam tables, the saturation pressure of water vapour at 60°C is 19.9 kPa.
Therefore the relative humidity = p/ps
                                            = 4.8/19.9
                                            = 0.24
                                             or 24%.

If such air were cooled, then when the percentage relative humidity reached 100% the air would be saturated and this would occur at that temperature at which p = ps = 4.8 kPa.

Interpolating from the steam tables, or reading from the water vapour pressure/temperature graph, this occurs at a temperature of 32°C and this temperature is called the dew point of the air at this particular moisture content. If cooled below the dew point, the air can no longer retain this quantity of water as vapour and so water must condense out as droplets or a fog, and the water remaining as vapour in the air will be that corresponding to saturation at the temperature reached.

The humidity Y can therefore be related to the partial pressure pw of the water vapour in air by the equation:

Y = 18 pw /[29(Ppw)]                                                                                            (7.4)

where P is the total pressure. In circumstances where pw is small compared with P, and this is approximately the case in air/water systems at room temperatures, Y 18pw/29P.

Corresponding to the specific heat, cp, of gases, is the humid heat, cs of moist air. It is used in the same way as a specific heat, the enthalpy change being the mass of dry air multiplied by the temperature difference and by the humid heat. The units are J kg-1 °C-1 and the numerical values can be read off a psychrometric chart. It differs from specific heat at constant pressure in that it is based only on the mass of the dry air. The specific heat of the water it contains is effectively incorporated into the humid heat which therefore is numerically a little larger than the specific heat to allow for this.


Wet-bulb Temperatures

A useful concept in psychrometry is the wet-bulb temperature, as compared with the ordinary temperature, which is called the dry-bulb temperature. The wet-bulb temperature is the temperature reached by a water surface, such as that registered by a thermometer bulb surrounded by a wet wick, when exposed to air passing over it. The wick and therefore the thermometer bulb decreases in temperature below the dry-bulb temperature until the rate of heat transfer from the warmer air to the wick is just equal to the rate of heat transfer needed to provide for the evaporation of water from the wick into the air stream.

Equating these two rates of heat transfer gives

hcA(Ta - Ts) = lk'gA(YsYa)
hc(Ta - Ts) =
lk'g(YsYa)

where a and s denote actual and saturation temperatures and humidities; hc is the heat-transfer coefficient and k'g the mass transfer coefficient from the air to the wick surface; l is the latent heat of evaporation of water.

As the relative humidity of the air decreases, so the difference between the wet-bulb and dry-bulb temperatures, called the wet-bulb depression, increases and a line connecting wet-bulb temperature and relative humidity can be plotted on a suitable chart. When the air is saturated, the wet-bulb temperature and the dry-bulb temperature are identical.

Therefore if (TaTs) is plotted against (YsYa) remembering that the point (Ts, Ys) must correspond to a dew-point condition, we then have a wet-bulb straight line on a temperature/humidity chart sloping down from the point (Ts, Ys) with a slope of:

- (l k'g/hc)

A further important concept is that of the adiabatic saturation condition. This is the situation reached by a stream of water, in contact with the humid air, and where the temperature of the air and the humidity follow down a line called the adiabatic saturation line. Both ultimately reach a temperature at which the heat lost by the humid air on cooling is equal to the heat of evaporation of the water leaving the stream of water by evaporation.

Under this condition with no heat exchange to the surroundings, the total enthalpy change (kJ kg-1 dry air)

DH = cs(TaTs) + l(YsYa) = 0
 cs =   -  hc/k"g

where cs is the humid heat of the air.
Now it just so happens, for the water/air system at normal working temperatures and pressures that for practical purposes the numerical magnitude of the ratio:

cs = - hc/k"g (known as the Lewis number)                  1                        (7.5)                     

This has a useful practical consequence. The wet bulb line and the adiabatic saturation line coincide when the Lewis number = 1.

It is now time to examine the chart we have spoken about. It is called a psychrometric chart.


Psychrometric Charts

In the preceding discussion, we have been considering a chart of humidity against temperature, and such a chart is given in skeleton form on Fig. 7.3 and more fully in Appendix 9, (a) Normal Temperatures and (b) High Temperatures.


Figure 7.3 Psychrometric chart


The two main axes are temperature (dry bulb) and humidity (humidity ratio) . The saturation curve (Ts , Ys). is plotted on this dividing the whole area into an unsaturated and a two-phase region. Taking a point on the saturation curve (Ts, Ys) a line can be drawn from this with a slope:

- (lk'g/hc) = (l/cs)

running down into the unsaturated region of the chart (that “below” the saturation line). This is the wet bulb or adiabatic cooling line and a net of such lines is shown. Any constant temperature line running between the saturation curve and the zero humidity axis can be divided evenly into fractional humidities which will correspond to fractional relative humidities [for example, a 0.50 ratio of humidities will correspond to a 50% RH because of eqn. (7.4) if P » pw].

This discussion is somewhat over-simplified and close inspection of the chart shows that the axes are not exactly rectangular and that the lines of constant dry-bulb temperature are not exactly parallel. The reasons are beyond the scope of the present discussion but can be found in appropriate texts such as Keey (1978). The chart also contains other information whose use will emerge as familiarity grows.

This chart can be used as the basis of many calculations. It can be used to calculate relative humidities and other properties.


EXAMPLE 7.7. Relative humidity, enthalpy and specific volume of air

If the wet-bulb temperature in a particular room is measured and found to be 20°C in air whose dry-bulb temperature is 25°C (that is the wet-bulb depression is 5°C) estimate the relative humidity, the enthalpy and the specific volume of the air in the room.

On the humidity chart (Appendix 9a) follow down the wet-bulb line for a temperature of 20°C until it meets the dry-bulb temperature line for 25°C. Examining the location of this point of intersection with reference to the lines of constant relative humidity, it lies between 60% and 70% RH and about 4/10 of the way between them but nearer to the 60% line. Therefore the RH is estimated to be 64%. Similar examination of the enthalpy lines gives an estimated enthalpy of 57 kJ kg-1, and from the volume lines a specific volume of 0.862 m3 kg-1.

Once the properties of the air have been determined other calculations can easily be made.


EXAMPLE 7.8. Relative humidity of heated air
If the air in Example 7.7 is then to be heated to a dry-bulb temperature of 40°C, calculate the rate of heat supply needed for a flow of 1000 m3 h-1 of this hot air for a dryer, and the relative humidity of the heated air.

On heating, the air condition moves, at constant absolute humidity as no water vapour is added or subtracted, to the condition at the higher (dry bulb) temperature of 40°C. At this condition, reading from the chart at 40°C and humidity 0.0125 kg kg-1, the enthalpy is 73 kJ kg-1, specific volume is 0.906 m3 kg-1 and RH 27%.

Mass of 1000 m3 is 1000/0.906 = 1104kg,
                                        
DH = (73 - 57) = 16 kJ kg-1.
So rate of heating required
                                              = 1104 x 16 kJ h -1
                                              = (1104 x 16)/3600 kJ s -1
                                              = 5 kW

If the air is used for drying, with the heat for evaporation being supplied by the hot air passing over a wet solid surface, the system behaves like the adiabatic saturation system. It is adiabatic because no heat is obtained from any source external to the air and the wet solid, and the latent heat of evaporation must be obtained by cooling the hot air. Looked at from the viewpoint of the solid, this is a drying process; from the viewpoint of the air it is humidification.


EXAMPLE 7.9. Water removed in air drying
Air at 60°C and 8% RH is blown through a continuous dryer from which it emerges at a temperature of 35°C. Estimate the quantity of water removed per kg of air passing, and the volume of drying air required to remove 20 kg water per hour.

Using the psychrometric chart (high-temperature version, Appendix 9(b) to take in the conditions), the inlet air condition shows the humidity of the drying air to be 0.01 kg kg-1 and its specific volume to be 0.96 m3 kg-1. Through the dryer, the condition of the air follows a constant wet-bulb line of about 27°C , so at 35°C its condition is a humidity of 0.0207kg kg-1.

Water removed = (0.0207 - 0.010)
                       = 0.0107 kg kg-1 of air.

So each kg, i.e. 0.96 m3, of air passing will remove 0.0107kg water,

Volume of air to remove 20 kg h-1
                       = (20/0.0107) x 0.96
                       = 1794 m3 h-1

If air is cooled, then initially its condition moves along a line of constant humidity, horizontally on a psychrometric chart, until it reaches the saturation curve at its dew point. Further cooling then proceeds down the saturation line to the final temperature, with water condensing to adjust the humidity as the saturation humidity cannot be exceeded.


EXAMPLE 7.10. Relative humidity of air leaving a dryer
The air emerging from a dryer, with an exit temperature of 45°C, passes over a surface which is gradually cooled. It is found that the first traces of moisture appear on this surface when it is at 40°C. Estimate the relative humidity of the air leaving the dryer.

On the psychrometric chart, the saturation temperature is 40°C and proceeding at constant humidity from this, the 45°C line is intersected at a point indicating:
        relative humidity = 76%

In dryers, it is sometimes useful to reheat the air so as to reduce its relative humidity and thus to give it an additional capacity to evaporate more water from the material being dried. This process can easily be followed on a psychrometric chart.


EXAMPLE 7.11. Reheating of air in a dryer
A flow of 1800 m3 h-1 of air initially at a temperature of 18°C and 50% RH is to be used in an air dryer. It is heated to 140°C and passed over a set of trays in a shelf dryer, which it leaves at 60 % RH. It is then reheated to 140°C and passed over another set of trays which it leaves at 60 % RH again. Estimate the energy necessary to heat the air and the quantity of water removed per hour.

From the psychrometric chart [normal temperatures, Appendix 9(a)] the humidity of the initial air is 0.0062 kg kg-1, specific volume is 0.834 m3 kg-1, and enthalpy 35 kJ kg-1. Proceeding at constant humidity to a temperature of 140°C, the enthalpy is found (high temperature chart) to be 160 kJ kg-1. Proceeding along a wet-bulb line to an RH of 60% gives the corresponding temperature as 48°C and humidity as 0.045 kg kg-1.

Reheating to 140°C keeps humidity constant and enthalpy goes to 268 kJ kg-1.

Thence along a wet-bulb line to 60 % RH gives humidity of 0.082 kg kg-1.

Total energy supplied   = DH in heating and reheating
                                 = 268 - 35
                                  = 233 kJ kg-1

Total water removed    = DY
                                 = 0.082 - 0.0062
                                  = 0.0758 kg kg-1

1800 m3 of air per hour = 1800/0.834
                                  = 2158 kg h-1
                                  = 0.6 kg s-1

Energy taken in by air = 233 x 0.6 kJ s-1
                                  = 140 kW

Water removed in dryer = 0.6 x 0.0758
                                  = 0.045 kg s-1
                                  = 163kgh -1

Exit temperature of air (from chart) = 60°C.

Consideration of psychrometric charts, and what has been said about them, will show that they can be used for calculations focused on the air, for the purposes of air conditioning as well as for drying.


EXAMPLE 7.12. Air conditioning
In a tropical country, it is desired to provide processing air conditions of 15°C and 80% RH. The ambient air is at 31.5°C and 90% RH. If the chosen method is to cool the air to condense out enough water to reduce the water content of the air sufficiently, then to reheat if necessary, determine the temperature to which the air should be cooled, the quantity of water removed and the amount of reheating necessary. The processing room has a volume of 1650 m3 and it is estimated to require six air changes per hour.

Using the psychrometric chart (normal temperatures):
Initial humidity is 0.0266 kg kg-1.
Final humidity is 0.0085 kg kg-1.
Saturation temperature for this humidity is 13°C.
Therefore the air should be cooled to 13°C
At the saturation temperature of 13°C, the enthalpy is 33.5 kJ kg-1
At the final conditions, 15°C and 80 % RH, the enthalpy is 37 kJ kg-1
and the specific volume of air is 0.827 m3 kg-1.

Assuming that the air changes are calculated at the conditions in the working space.
       Mass of air to be conditioned = (1650 x 6)/0.827
                                                  = 11,970 kg h-1
       Water removed per kg of dry air
DY
                                                  = 0.0266 - 0.0085
                                                  = 0.018 kg kg-1
       Mass of water removed per hour
                                                  = 11,970 x 0.018
                                                  = 215 kg h-1

       Reheat required DH              = (37 - 33.5)
                                                  = 3.5 kJ kg-1
       Total reheat power required    = 11,970 x 3.5
                                                  = 41,895 kJ h-1
                                                  = 11.6 kJ s-1
                                                  = 11.6 kW.


Measurement of Humidity

Methods depend largely upon the concepts that have been presented in the preceding sections, but because they are often needed it seems useful to set them out specifically. Instruments for the measurement of humidity are called hygrometers.

Wet- and dry-bulb thermometers. The dry-bulb temperature is the normal air temperature and the only caution that is needed is that if the thermometer bulb, or element, is exposed to a surface at a substantially higher or lower temperature the possibility of radiation errors should be considered. A simple method to greatly reduce any such error is to interpose a radiation shield, e.g. a metal tube, which stands off from the thermometer bulb 1 cm or so and prevents direct exposure to the radiation source or sink. For the wet bulb thermometer, covering the bulb with a piece of wicking, such as a hollow cotton shoelace of the correct size, and dipping the other end of the wick into water so as to moisten the wet bulb by capillary water flow, is adequate. The necessary aspiration of air past this bulb can be effected by a small fan or by swinging bulb, wick, water bottle and all through the air, as in a sling psychrometer. The maximum difference between the two bulbs gives the wet-bulb depression and a psychrometric chart or appropriate tables will then give the relative humidity.

Dew-point meters. These measure the saturation or dew-point temperature by cooling a sample of air until condensation occurs. The psychrometric chart or a scale on the instrument is then used to give the humidity. For example, a sample of air at 20°C is found to produce the first signs of condensation on a mirror when the mirror is cooled to 14°C. The chart shows by moving horizontally across, from the saturation temperature of 14°C to the constant temperature line at 20°C, that the air must have a relative humidity of 69%.

The hair hygrometer. Hairs expand and contract in length according to the relative humidity. Instruments are made which give accurately the length of the hair and so they can be calibrated in humidities.

Electrical resistance hygrometers. Some materials vary in their surface electrical resistance according to the relative humidity of the surrounding air. Examples are aluminium oxide, phenol formaldehyde polymers, and styrene polymers. Calibration allows resistance measurements to be interpreted as humidity.

Lithium chloride hygrometers. In these a solution of lithium chloride is brought to a temperature such that its partial pressure equals the partial pressure of water vapour in the air. The known vapour pressure-temperature relationships for lithium chloride can then be used to determine the humidity of the air.

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Unit Operations in Food Processing. Copyright © 1983, R. L. Earle. :: Published by NZIFST (Inc.)
NZIFST - The New Zealand Institute of Food Science & Technology