1. 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

2. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization)  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. Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

3. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization)  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. Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering *

4. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) 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 Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

5. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

6. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering Example: Oxalic acid H 2 C 2 O 4 is burned at atmospheric pressure with 248% excess air, so that 65% of the carbon burns to CO 2. Calculate the dew point of the product gas. Solution: Basis: 1 mol H 2 C 2 O 4 Chemical reaction equations: H 2 C 2 O 4 + 0.5O 2 → 2 CO 2 + H 2 O H 2 C 2 O 4 → 2CO + H 2 0 + 0.5O 2 O 2 required= 0.5 mol Mol O 2 entering: (1 + 2.48)(0.5 mol O 2 ) = 1.74 mol O 2

7. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering Component Mol in Mol out With n co 2 = (0.65)(2) = 1.30, Component Mol nH2OnH2O n co 2 n co nc2nc2 nN2nN2 1.0 1.3 0.7 1.59 6.55 11.4 Total

8. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering y H 2 O = 1 mol H 2 O/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 = y H2O (p Total ) = 0.0898(101.3 kPa) = 9.09 kPa (1.319 psia) From steam tables, the dew point temperature: T=316.5 K

9. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) 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). Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

10. Example: If a pound of saturated air at 75 o F and 1 atm (the vapor pressure of water is 0.43 psia at 75 o F) 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. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

11. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization)  Effect of an increase of pressure on saturated air, removal of condensed water, and a return to the initial pressure at constant temperature. Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

12. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering 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 75 o F 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.

13. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering 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 p H 2 O = 0.105 psia p air = 14.6 p total = 14.7 psia The pressure of the water vapor represents a dew point of about 36°F.

14. 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? Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

15. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering The vapor pressure of benzene at 26 o C p*=99.7 mm Hg Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

16. 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 p total = 99.7 / 0.916 x 10 -3 = 109 x 10 3 mm Hg Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

17. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) 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. Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

18. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Evaporation of water at constant pressure and a temperature of 65°C. Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering

19. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering 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 o C 100 kPa Air 100 kPa saturated Air-alcohol mixture Basis: 6.0 kg of alcohol

20. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering The minimum volume of air means that the resulting mixture is saturated; any condition less than saturated would require more air. Once you calculate the number of moles of air, you can apply the ideal gas law. Since p Alcohol = 5.93 kPa p Air = p total - p Alcohol = (100 - 5.93)kPa = 94.07 kPa

21. Two-Phase Gas-Liquid Systems (Saturation, Condensation, Vaporization) Dr Saad Al-ShahraniChE 201: Introduction to Chemical Engineering