Processes such as air conditioning, water cooling, humidification, drying operations, etc., where the chemical engineer has to deal with vapours, the vapour will be present in association with a
non-condensable gas or a mixture of non-non-condensable gases, the most familiar example being water vapour present in air. Air conditioning is the process of maintaining the water content in the air at some
desired level. Maintaining the humidity and temperature of air at some specified levels is important not only for houses and buildings, but also for certain plant areas in process industries. This is
achieved by either condensation of water vapour from air or by evaporation of liquid water into an unsaturated air stream. Cooling of process water is achieved in cooling towers by evaporation of water into air stream. Air conditioning and water cooling operations fall under the general category of
operations designated in chemical engineering parlance as humidification-dehumidification
operations. Another important unit operation which also makes use of an air stream or a stream of hot flue gases picking up moisture by evaporation is the drying operation. In drying, a wet material will be exposed to a relatively dry hot air or flue gas. The material gets dried by giving off its moisture to the unsaturated gas stream with which it is contacted. The air-water vapour system is not the only system of importance involving mixtures of condensable vapours and non-condensable gases. Frequently, vapours of organic compounds such as benzene, carbon tetrachloride, acetone, etc. present in air or process gases are to be recovered as part of a solvent recovery operation or are to be removed from the gas stream as part of a purification step. A clear understanding of the concepts of humidity and
saturation, degree of saturation and the conditions under which the gas stream remains saturated are therefore very essential to a chemical engineer.
8.1 SATURATION
Consider a liquid A (say, water) in contact with a non-condensable gas B (say, air) in a closed container at constant temperature and pressure. Initially the air is dry so that the partial pressure of water vapour in the air is zero. When the dry air comes in contact with liquid water, it picks
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up moisture from the liquid and as a result the partial pressure of water in the gas increases. Finally when equilibrium between the liquid and the gas is established, the partial pressure of water vapour in the gas stream remains at a constant value. Even if we provide an infinitely long period for contact between the liquid and the gas, the equilibrium partial pressure exerted by the vapours of the liquid in the gas will not change so long as the temperature is not altered. This is the state at which we say that the gas is saturated with the vapours of the liquid. The partial pressure exerted by the vapour under saturation is equal to the vapour pressure of the liquid at the given temperature. Since the total
pressure P is constant, we have under saturation, the partial pressure of vapour, pA = PAS, where PAS is the vapour pressure of component A, and the partial pressure of the gas, pB = P – PAS. If the gas is unsaturated, the partial pressure of the vapour is less than the vapour pressure of the liquid at the temperature of the gas.
EXAMPLE 8.1 The vapour pressure of acetone at 295 K is 13.25 kPa. For a mixture of nitrogen gas saturated with the vapours of acetone at 295 K and 105 kPa, calculate the following: (a) The mole percent of acetone in the mixture
(b) The percent composition by weight
(c) The amount of vapour in kilograms per m3 of the mixture.
Solution Since the nitrogen gas is saturated with acetone vapours, the partial pressure of acetone in the gas at 295 K, pA = the vapour pressure of acetone at 295 K, PAS = 13.25 kPa.
(a) The mole fraction of acetone is partial pressure
pPS 13.25
AA 0.1262total pressure == = = PP 105
The mole percent = mole fraction ¥ 100 = 12.62%
(b) The molecular weight of acetone = 58.048 and the molecular weight of nitrogen = 28.0.
One mole of the gas mixture contains 0.1262 mole of acetone and 0.8738 mole of nitrogen.
Or one mole of the gas mixture contains 0.1262 ¥ 58.048 = 7.3257 g of acetone and 0.8738 ¥ 28 = 24.4664 g of nitrogen. Therefore, the composition in weight percent is:
acetone =7.3257 100 23.04%(7.3257+24.4664)×=
nitrogen = 24.4664 100 76.96%(7.3257 +24.4664)×=
(c) Volume of one kmol gas at 295 K and 105 kPa is P0 T 101.3 2953
V PT
××= × × =23.35 m
0 105 273.15
The mass of vapour present in one kmole of gas is 7.3257 kg.
7.3257 =0.3137 kg/m3Concentration of the vapour = 23.35
EXAMPLE 8.2 It is desired to prepare a 4% benzene vapour–air mixture by saturating dry air with benzene from a container of the liquid at 101.3 kPa. What temperature should be used to achieve the desired composition? The vapour pressure of benzene is given by the Antoine equation
ln PS 13.8858 2788.51=− −52.36
Solution The partial pressure of benzene in a 10% benzene vapour–air mixture at 101.3 kPa is 101.3 ¥ 0.1 = 10.13 kPa. For saturated air, the partial pressure is equal to the vapour pressure. Using the
Antoine equation, the temperature of benzene at which the vapour pressure is 10.13 kPa is determined.
ln 10.1313.8858 2788.51 ⇒T 293.4 K=−T −52.36 =
EXAMPLE 8.3 Dry air is blown through acetone at 300 K and a constant pressure of 101.3 kPa. If it is desired that 5 kg of acetone be evaporated what is the minimum amount of dry air required in
kilograms? The vapour pressure of acetone at 300 K is 16.82 kPa.
Solution The minimum quantity of air that is required to contain 5 kg of acetone corresponds to the
condition when air is saturated with 5 kg of acetone vapours. Under the saturated condition, the ratio of moles of acetone to moles of dry air is
P
S
A 16.82 0.1991PPS==
− A 101.3−16.82
The ratio of mass of acetone to mass of dry air is 0.199158.048 0.398529×=
where 58.048 is the molecular weight of acetone and 29 is the molecular weight of dry air. One
kilogram of dry air under saturated state contains 0.3985 kg of acetone vapour. The amount of dry air required to evaporate 5 kg of acetone is
5 =12.55 kg0.3985
8.2 HUMIDITY, PERCENT HUMIDITY AND DEW POINT 8.2.1 Humidity
The term humidity is used to indicate the concentration of the vapour in a gas–vapour mixture. For the air-water vapour system it is defined as the amount of water vapour present per unit quantity of dry air and can be expressed on a mole basis or a mass basis. The molal humidity is defined as the number of moles of water vapour present per one mole of dry air. Let there be nA moles of water vapour and nB moles of dry air in a given sample of air–water vapour system. Then,
molal humidity, Y =
moles of water vapour =nA (8.1)moles of dry air nB
Assuming that the air–water vapour mixture behaves as an ideal gas, the following equations can be used to determine the number of moles of the constituents.
pVnA =A
RT (8.2) n
B = pV
B
RT
where V is the volume of the mixture and T is the temperature. Combining Eqs. (8.1) and (8.2), we get np p
molal humidity, Y
=
AA A
== − (8.3)np Pp
BB A
Equation (8.3) means that the molal humidity is equal to the ratio of the partial pressure of water vapour to the partial pressure of dry air. Though the definitions given here and in the following
sections refer to the air–water vapour system, they are equally applicable for all vapour–gas mixtures.
If the gas were saturated with the vapours, the molal humidity would correspond to molal humidity at saturation denoted as Ys. It is equal to the moles of water vapour per mole of dry air under saturation.
Since at saturation, the partial pressure of the vapour is equal to the vapour pressure, the molal saturation humidity can be written as
PS
Y PPA (8.4)S
= −A
where PS is the vapour pressure of water.A
The absolute humidity is defined as the mass of vapour present per unit mass of vapourfree gas. For the air–water vapour system, it is the kilograms of water present per one kg of dry air. If mA and mB are the weights of the vapour and vapour-free gas, then
absolute humidity, YmA (8.5)′=mB
Absolute humidity is related to the molal humidity as:
mn M M
AA A A
YY
′= = × = × (8.6) mn M M
BB B B
where MA and MB are the molecular weights of A and B respectively. Combining with Eq. (8.3) and noting that for air the average molecular weight is approximately 29 and the molecular weight of water is 18, the absolute humidity of the air–water vapour system can be related to the partial pressure as
pA × 18 (8.7)Y Pp− A 29
The absolute saturation humidity (Ys¢) denotes the kilograms of water vapour that one kg of dry air can hold at saturated conditions. Analogous to Eq. (8.4), we can write
Ys′= PMSAA(8.8)PPS × M
−BA
At the boiling point of the liquid the vapour pressure becomes equal to the total pressure so that the denominator in Eq. (8.8) becomes equal to zero. It means that as the boiling point of the liquid is approached, the saturation humidity tends to infinity. We can see from Eqs. (8.4) and (8.8) that at a given temperature, the molal humidity at saturation is independent of the nature of the gas and
depends only on the pressure whereas the absolute saturation humidity depends on the characteristics of the gas as well.
EXAMPLE 8.4 A mixture of acetone vapour and nitrogen gas at 101.3 kPa and 310 K contains acetone vapour to the extent that it exerts a partial pressure of 15 kPa. The vapour pressure of acetone is given by the Antoine equation
ln PS 14.5463 2940.46=− −49.19
where the pressure is in kPa and temperature is in K. Determine the following: (a) The mole fraction of acetone in the mixture
(b) The weight fraction of acetone in the mixture (c) The molal humidity
(d) The absolute humidity
(e) The molal saturation humidity (f) The absolute saturation humidity
(g) The mass of acetone in 100 m3 of the mixture.
Solution
(a) The partial pressure of acetone in the gas is 15 kPa.
partial pressure p 15 0.1481Mole fraction of acetone = total pressure P== = 101.3
(b) One mole of the mixture contains 0.1481 mole acetone and the rest (0.8519 mole) nitrogen. The molecular weight of acetone = 58.048 and the molecular weight of nitrogen = 28.
Therefore, the weight fraction of acetone in the mixture is
0.1481×58.048 =0.26490.1481 58.0480.8519 28×+ ×
(c) Molal humidity is the ratio of moles of acetone to moles of dry nitrogen, i.e.
Y0.1481 0.1738==
0.8519
(Note: Molal humidity may be calculated as the ratio of partial pressure of acetone to the partial pressure of nitrogen, i.e.
Y 15
==0.1738 moles of acetone)
101.3 −15 moles of nitrogen
(d) Absolute humidity is the ratio of kilograms of acetone to kilograms of nitrogen. It is obtained by multiplying the molal humidity by the ratio of the molecular weights 58.048=0.3603 kilograms of acetoneY′=0.1738× 28 kilograms of nitrogen
(e) The vapour pressure of acetone at 310 K, from the Antoine equation is 2940.46
ln
SS
PP = 26.36 kPa 310− 49.19
S
Saturation humidity, P
Y PPS = −
26.36
0.3517 moles of acetone==
101.3 moles of nitrogen−26.36 (f) Absolute saturation humidity, 58.048
Ys′= × =0.7292kilograms of acetone
28 kilograms of nitrogen (g) The number of moles of the mixture in 100 m3 is
VP T0 100 101.3 273.15 3.93 kmol22.414 0PT××= × × = 22.414 101.3 310
Since the mole fraction of acetone is 0.1481, the number of moles of acetone in the mixture is
0.1481 ¥ 3.93 = 0.5822 kmol The mass of acetone in 100 m3 of the total gas is 0.5822 ¥ 58.048 = 33.8 kg
8.2.2 Relative Humidity and Percent Humidity
The degree of unsaturation of a gas–vapour mixture can be expressed in two ways: the first is the percent relative saturation and the second is the percent saturation. The percent relative saturation or percent relative humidity is the ratio of the actual partial pressure exerted by the vapour to its vapour pressure at the same temperature, which is expressed as a percentage. Thus,
relative saturation (relative humidity) =
pA ×100% (8.9)PS
A
On the other hand, the percent saturation or percent humidity is the ratio of moles of vapour present in the gas per mole of dry gas to the moles of vapour present per mole of a vapour-free gas if it is
saturated at the temperature of the system, expressed as a percentage. It is same as the ratio of
kilograms of vapour present per kilogram of dry gas at the given temperature to kilograms of vapour that would be associated with one kilogram of dry gas if the gas mixture were saturated with the vapours at the same temperature.
Percent saturation (percent humidity) = (/ )
AB ×100% (8.10)(/ )
AB saturation
The numerator in Eq. (8.10) is the molal humidity and the denominator is the molal humidity at saturation. Therefore, the percent humidity can be defined as the ratio of the existing humidity of the mixture to the humidity if the gas mixture is saturated with the vapours.
YY′
Percent saturation = ×=×
ss
100% (8.11) YY′
A percent saturation of 50% means that the water content present in the given system per one kg of dry air is only half of the water that one kg dry air will contain if the air is saturated with water vapour at the same temperature. On the other hand, the relative saturation of 50 percent means that the partial