Vapor-Liquid Equilibria
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1 31 Introduction to Chemical Engineering Calculations Lecture 7. Vapor-Liquid Equilibria
2 Vapor and Gas Vapor A substance that is below its critical temperature. Gas A substance that is above its critical temperature (but below the critical pressure). For the same pressure, a vapor is more easily condensed while a gas is normally non-condensable. 2
3 The PT Diagram 3
4 Definition of Terms Vapor Pressure Pressure of the vapor when the vapor and liquid of a pure component are in equilibrium. Also referred to as saturated pressure. Saturated Temperature Temperature at which the vapor and liquid will co-exist at a given pressure. Commonly referred to as the boiling or condensation point. Dew Point The temperature when the vapor starts to condense. Bubble Point The temperature when the liquid starts to vaporize. 4
5 Definition of Terms Saturated Vapor and Saturated Liquid Vapor and liquid at the saturated P and saturated T. Subcooled Liquid A liquid that is below its saturated T at a given pressure. Compressed Liquid A liquid that is above its saturated P at a given temperature. Superheated Vapor A vapor that is above its saturated T at a given pressure. Quality For a saturated mixture, quality refers to the mass fraction of the vapor in the mixture. 5
6 The PT Diagram 6
7 The PT Diagram 7
8 Saturated and Superheated Steam Tables 8
9 Properties of Saturated Mixture For a saturated mixture, V ˆ (1 x) V ˆ (x) Vˆ l g mixture H ˆ (1 x) H ˆ (x) Hˆ l g mixture U ˆ (1 x) U ˆ (x) Uˆ l g mixture where x = mass fraction of vapor in the mixture. 9
10 Example 7/8-1. Vapor-Liquid Properties of Water For each of the conditions of temperature and pressure listed below for water, state whether the water is a solid, liquid, saturated mixture, or superheated vapor. If it is a saturated mixture, calculate the quality. State P (kpa) T (K) V (m 3 /kg)
11 Example 7-1. Vapor-Liquid Properties of Water Using the saturated steam table, State P (kpa) T (K) Phase Liquid Vapor Vapor Saturated 11
12 Example 7-1. Vapor-Liquid Properties of Water From the saturated steam pressure table, By 2-point linear interpolation, properties at P = can be estimated: V ˆ m / kg and V ˆ m / kg l 3 3 g 12
13 Example 7-1. Vapor-Liquid Properties of Water If x is the quality of the saturated mixture, then m m m 1 x x kg kg kg Solving for x, x =
14 Change of Vapor Pressure with Temperature Many functional forms have been proposed to predict vapor pressure (p*) from temperature. One of these functions is the Antoine equation: B ln (p*) A C T Where p* = vapor pressure of the substance, mmhg T = temperature, K A, B, C = constants for each substance 14
15 Change of Vapor Pressure with Temperature The Clapeyron equation is another function relating vapor pressure p* and temperature T: dp* ĤV dt T(Vˆ V ˆ ) g l Where p* = vapor pressure of the substance T = temperature H V = latent heat of vaporization V g, V l = specific molar volumes of vapor and liquid, respectively 15
16 Change of Vapor Pressure with Temperature Simplifications of the Clapeyron equation: 1. Specific molar volume of liquid is very small compared to specific molar volume of the vapor, such that (Vˆ V ˆ ) Vˆ g l g 2. The vapor is assumed to behave ideally. ˆV g RT P * 16
17 Change of Vapor Pressure with Temperature The Clapeyron equation then becomes, dp* dt P* Hˆ RT 2 V Separating the variables and integrating to yield the Clausius-Clapeyron equation, dp* ĤV dt 2 P* R T Ĥ ln(p*) RT V B 17
18 Example 7-2. Estimation of Normal BP Using the Antoine Equation Determine the % error if the Antoine equation is used to estimate the normal boiling point of benzene. From literature value, the normal boiling point of benzene is K. Solution: For benzene, values of the constants for the Antoine equation are: A = B = C =
19 Example 7-2. Estimation of Normal BP Using the Antoine Equation Using the Antoine equation, ln (760 mmhg) T Solving for temperature, T = K The percent error is, K 353.3K % error % K 19
20 Example 7-3. Clausius-Clapeyron Equation The vapor pressure of benzene is measured at two temperatures, with the following results: T 1 = C T 2 = C P* 1 = 40 mm Hg P* 2 = 60 mm Hg Calculate the latent heat of vaporization and the parameter B in the Clausius-Clapeyron equation and then estimate P* at C using this equation. 20
21 Example 7-3. Clausius-Clapeyron Equation The Clausius-Clapeyron is a linear equation with: Solving for slope, m: 1 x y ln(p*) T Ĥ R V m b B m ĤV y ln(p* 2) ln(p* 1) T1 T2 ln(p* 2 /P* 1) R x 1 1 T1 T2 T2 T 1 21
22 Example 7-3. Clausius-Clapeyron Equation Using absolute values for temperature: T 1 = K and T 2 = K The slope is computed as: m m 60 mmhg K K ln ˆ 40 mmhg R K K HV Ĥ R V 4213K 22
23 Example 7-3. Clausius-Clapeyron Equation And the latent heat of vaporization is: J ĤV 4213K R 4213K , 030 molk J mol The intercept b (b = B) can be determined as follows: Hˆ B ln(p* ) ln(p* ) Hˆ V V 1 2 RT1 RT B ln(40)
24 Example 7-3. Clausius-Clapeyron Equation The Clausius-Clapeyron equation for benzene can now be written as: 4213 ln(p*) T At T = C = K, the vapor pressure of benzene is 4213 ln(p*) P* exp mmhg 24
25 Vapor-Liquid Equilibrium for Multi-component Systems Consider a binary mixture with components A and B. Vapor Y A (P A ) Y B (P B ) Liquid X A0 X B0 Heating Liquid X A X B Note: X, Y = mole fractions of the component in the vapor and liquid phases, respectively. 25
26 Vapor-Liquid Equilibrium for Multi-component Systems For multi-component systems, P i = f(x i ) This functional relationship is given by 1. Raoult s Law generally used when x i is close to Henry s Law generally used when x i is close to 0. 26
27 Vapor-Liquid Equilibrium for Multi-component Systems Henry s Law: P i = x i H i where P i x i H i = partial pressure of component i in the vapor phase. = y i P T (if the vapor behaves ideally) = mole fraction of component i in the liquid phase. = Henry s law constant 27
28 Vapor-Liquid Equilibrium for Multi-component Systems Raoult s Law: P i = x i P* i where P i x i P T = partial pressure of component i in the vapor phase. = y i P T (if the vapor behaves ideally) = mole fraction of component i in the liquid phase. = total pressure 28
29 Vapor-Liquid Equilibrium for Multi-component Systems If the vapor behaves ideally, the Raoult s law becomes Rearranging the equation, y i P T = x i P* i y x i i P * P T i K i where K i is the V-L equilibrium constant. 29
30 Example 7-4. Vapor-Liquid Equilibrium Calculation Suppose that a liquid mixture of 4.0% n-hexane (A) in n-octane (B) is vaporized. What is the composition of the first vapor formed if the total pressure is 1.00 atm? Values of the Antoine constants for n-octane are: A = B = C =
31 Example 7-4. Vapor-Liquid Equilibrium Calculation Solution: Assuming the vapor behaves ideally, the composition of the vapor is determined using the Raoult s law: y y A B P * P T P * P T A B x x A B 31
32 Example 7-4. Vapor-Liquid Equilibrium Calculation Upon formation of the first vapor, the composition of the liquid is essentially the same as the initial composition. Hence, x A = and x B = 1 x A =0.960 The vapor pressures of A and B are calculated using the Antoine equation: B ln (P*) A C T 32
33 Example 7-4. Vapor-Liquid Equilibrium Calculation For n-hexane: P* A exp T For n-octane: P* B exp T Since vaporization temperature is not given, the next step in the calculation is to determine its value. 33
34 Example 7-4. Vapor-Liquid Equilibrium Calculation For the vapor mixture, Using Raoult s Law, P T = P A + P B P T = P* A x A + P* B x B = 1.00 atm = 760 mmhg Using the expressions for the vapor pressures as defined by the Antoine equation T 63.63T e e
35 Example 7-4. Vapor-Liquid Equilibrium Calculation The last equation is a non-linear equation. To find the value of T, Newton s method is used by defining f(t) as: T 63.63T f(t) = 0.040e 0.960e Similar to example 6-4, the temperature value that will satisfy this equation is determined using the following iteration formula: T k1 Tk f (T k ) f '(T ) k 35
36 Example 7-4. Vapor-Liquid Equilibrium Calculation Differentiating f(t): f '(T) T 63.63T e e = T T Any value for the temperature may be used as initial guess. For this example, the following initial guess is used: T 0 = 1000 K 36
37 Example 7-4. Vapor-Liquid Equilibrium Calculation Step T K (K) f(t) f'(t) T K+1 (K) E
38 Example 7-4. Vapor-Liquid Equilibrium Calculation For T = K, the vapor pressures of n-hexane and n-octane are: P* A = mm Hg P* B = mm Hg And the composition of the first vapor formed is: y y A B
39 Vapor-Liquid Equilibrium for Multi-component Systems Consider a mixture of a vapor (A) and a non-condensable gas (B). Gas Mixture Y A (P A ) Y B (P B ) Cooling Vapor Y A (P A ) Y B (P B ) Liquid X A 39
40 Vapor-Liquid Equilibrium for Multi-component Systems (Terms and symbols in parenthesis refer specifically to air-water system.) Relative Saturation (Relative Humidity) PV sr hr 100 P * Absolute Saturation (Absolute Humidity) V s a h a mass of vapor PVMWV massof vapor-free gas P P MW T V G Molal Saturation (Molal Humidity) s m h m moles of vapor PV molesof vapor-free gas P P T V 40
41 Example 7-5. Humidity of Air Humid air at 75 0 C, 1.1 bar, and 30% relative humidity is fed into a process unit at a rate of 1000 m 3 /h. Determine: a. the molar flow rates of water, dry air, and oxygen entering the process unit. b. the molal humidity and absolute humidity. c. the dew point. Assume ideal gas behavior. 41
42 Example 7-5. Humidity of Air Solution for (a): Determine the mole fraction of water: y P 2 H2O H O PT The partial pressure of water is calculated from relative humidity: P P* H2O h 100 r H2O 42
43 Example 7-5. Humidity of Air At T = 75 0 C, P* H2O = 289 mm Hg And the partial pressure of water is, P H2O = (289 mm Hg)(0.3) = 86.7 mm Hg For a total pressure of 1.1 bar = 825 mm Hg, the mole fraction of water is, 86.7 mm Hg mol H2O yh2o mm Hg mol HA 43
44 Example 7-5. Humidity of Air Determine the molar flow rate of humid air: 3 1.1bar 1000 m / h P V kmol na RT m bar h K kmol K The molar flow rate of water can now be obtained: kmol HA 0.105kmolH2O kmol H2O n H2O h kmol HA h 44
45 Example 7-5. Humidity of Air The molar flow rate of dry air: kmol kmol DA kmol DA n DA h kmol HA h The molar flow rate of O 2 : kmol DA 0.21kmol O kmol O n O h kmol DA h
46 Example 7-5. Humidity of Air Solution for (b): The absolute humidity (h a ): T H2O DA PH2O MWH2O kg H2O ha P P MW kg DA The molal humidity (h m ): PH2O 86.7 mol H2O hm P P mol DA T H2O 46
47 Example 7-5. Humidity of Air Solution for (c): At the dew point, P H2O = P* H2O = 86.7 mmhg From vapor pressure data for water, this vapor pressure of water occurs at: T = C This is the dew point since at this temperature, water will start to condense. 47
vap H = RT 1T 2 = 30.850 kj mol 1 100 kpa = 341 K
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