Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Saturation When any noncondensable gas (or a gaseous mixture) comes in contact

Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization)

Saturation

When any noncondensable gas (or a gaseous mixture) comes in

contact with a liquid:

The partial pressure of the vapor in the gas will equal the vapor

pressure of the liquid at the temperature of the system.

The gas will acquire molecules from the liquid. If contact is

maintained for a sufficient period of time, vaporization continues

until equilibrium is attained.

At equilibrium, the rate of vaporization is equal to the rate of

condensation; therefore, the amount of liquid and the amount of

vapor remain constant.

Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering


After equilibrium is reached no more net liquid will vaporize into the

gas phase. The gas is then said to be saturated with the particular

vapor at the given temperature.

We also say that the gas mixture is at its dew point.

The dew point for the mixture of pure vapor and noncondensable gas

means the temperature at which the vapor just starts to condense. At

the dew point the partial pressure of the vapor is the vapor pressure.



When air is saturated with water, the ideal gas law can be applied to

both air and water vapor with excellent precision. Thus, we can say

that the following relations hold at saturation:

or

Because V and Tare the same for the air and water vapor.


*


Example: suppose you have a saturated gas, say water in air at 51°C, and the pressure on the system is 750 mm Hg absolute. What is the partial pressure of the air? If the air is saturated,

The partial pressure of the water vapor is p* at 51°C. You can look in a handbook, or use the steam tables, and find that p* = 98 mm Hg. Then

Solution




87.0750

652

total

airair p

py



Example: Oxalic acid H2C2O4 is burned at atmospheric pressure with 248% excess air, so that 65% of the carbon burns to CO2. Calculate the dew point of the product gas.

Solution:

Basis: 1 mol H2C2O4

Chemical reaction equations:

H2C2O4 + 0.5O2 → 2 CO2 + H2O

H2C2O4 → 2CO + H20 + 0.5O2

O2 required= 0.5 mol

Mol O2 entering: (1 + 2.48)(0.5 mol O2) = 1.74 mol O2



Component Mol in Mol out

With nco2= (0.65)(2) = 1.30,

Component Mol

nH2O

nco2

nco

nc2

nN2

1.0

1.3

0.7

1.59

6.55

11.4Total



yH2O = 1 mol H2O/11 .14 mol total = 0.0898

The partial pressure of the water in the product gas (at an assumed atmospheric pressure) determines the dew point of the stack gas:

p*H2O = yH2O (pTotal) = 0.0898(101.3 kPa) = 9.09 kPa (1.319 psia)

From steam tables, the dew point temperature: T=316.5 K


Condensation

Condensation is the change of vapor (in a noncondensable gas) to

liquid. Some typical ways of condensing a vapor in a gas are:

Cool it at constant system total pressure (the volume changes).

Cool it at constant total system volume (the pressure changes).

Compress it isothermally (the volume changes).


Example: If a pound of saturated air at 75oF and 1 atm (the vapor pressure

of water is 0.43 psia at 75oF) is compressed isothermally to 4 atm

(58.8 psia), almost three-fourths of the original content of water vapor

now will be in the form of liquid. and the air still has a dew point of

75°F. Remove the liquid water, expand the air isothermally back to 1

atm , and you will find that the dew point has been lowered to about

36°F. Here is how to make the calculations. Let 1 = state at 1 atm and

4 = state at 4 atm with z = 1.00 for both components.




Effect of an increase of pressure on saturated air, removal of condensed water, and a return to the initial pressure at constant temperature.




Pick as a basis 0.43 mol of H20. For saturated air at 75°F and 4 atm: I

For the same air saturated at 75oF and 1 atm:

Because the moles of air in state 1 and in state 4 are the same, the material balances simplify to

That is, 24.5% of the original water will remain as vapor after compression.



After the air-water vapor mixture is returned to a total pressure of 1 atm, to get the partial pressure of the water vapor the following two equations apply at 75°F:

From these two relations you can find that

pH2O = 0.105 psia

pair = 14.6

ptotal = 14.7 psia

The pressure of the water vapor represents a dew point of about 36°F.

00717.07.14

)43.0)(245.0(22 total

OH

total

OH

n

n

p

p

Example: Emission of volatile organic compounds from processes is closely

regulated. Both the Environmental Protection Agency (EPA) and the

Occupational Safety and Health Administration (OSHA) have established

regulations and standards covering emissions and frequency of exposure. This

problem concerns the first step of removal of benzene vapor from an exhaust

stream using the process shown in Figure El7.2a. The process has been

designed to recover 95% of the benzene from air by compression. What is the

exit pressure from the compressor?





The vapor pressure of benzene at 26 oC p*=99.7 mm Hg



Exiting components in the gas phase from the compressor:

mol of benzene = 0.018 (0.05) = 0.90 X 10-3 g mol

mol of air = 0.982 g mol

y Benzene exiting =0.9 x 10-3/0.983 = 0.916 X 10-3

total gas = 0.983 g mol

The partial pressure of the benzene is 99.7 mm Hg so that

ptotal = 99.7 / 0.916 x 10-3 = 109 x 103 mm Hg




Vaporization

Vaporization is the reverse of condensation, namely the

transformation of a liquid into vapor (in a noncondensable gas).

You can vaporize a liquid into a noncondensable gas, and raise the

partial pressure of the vapor in the gas until the saturation pressure

(vapor pressure) is reached at equilibrium.



Evaporation of water at constant pressure and a temperature of 65°C.




Example: What is the minimum number of cubic meters of dry air at 20°C and

100 kPa that are necessary to evaporate 6.0 kg of liquid ethyl alcohol if the

total pressure remains constant at 100 kPa and the temperature remains

20°C? Assume that the air is blown through the alcohol to evaporate it in

such a way that the exit pressure of the air-alcohol mixture is at 100 kPa.

Solution:

p*alcohol at 20°C = 5.93 kPa Mol. wt. ethyl alcohol = 46.07

Alcohol

6.0 kg

20 oC100 kPa

Air

100 kPa

saturated

Air-alcohol

mixture

Basis: 6.0 kg of alcohol



The minimum volume of air means that the resulting mixture is saturated; any condition less than saturated would require more air.

Air

Alcohol

Air

Alcohol

n

n

p

p

*

Once you calculate the number of moles of air, you can apply the ideal gas law. Since pAlcohol = 5.93 kPa

pAir = ptotal - pAlcohol = (100 - 5.93)kPa = 94.07 kPa

air mol kg 07.2alcohol mol kg 93.5

air mol kg 07.94

alcohol kg 07.46

alcohol kg 1alcohol kg 0.6



kPa 100 and C 20at m 4.50(kPa) 100

K)(293

mol)(K) kg(

)(kPa)(m 8.314air mol kg 07.2 o33

AirV

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Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Saturation When any noncondensable gas (or a gaseous mixture) comes in contact