09 Chapter 6-C PDF

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Summary

From the Last Class Vapor Pressure Equations  Cox chart & Duhring plots: log p* vs. T  Antoine equation: (Table B.4) ☺☺ B log10 P*  A  TC  Wagner equation  “Properties of Gases and Liquids” ln P*  (A  B1.5  C3  D6 ) / Tr   1  Tr Note : the vapor pressure equations should be use...

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From the Last Class

Vapor Pressure Equations  Cox chart & Duhring plots: log p* vs. T  Antoine equation: (Table B.4) ☺☺ B log10 P*  A  TC

 Wagner equation  “Properties of Gases and Liquids” ln P*  (A  B1.5  C3  D6 ) / Tr   1  Tr

Note : the vapor pressure equations should be used within the specified temperature range !

6.2 Gibbs Phase Rule ☺☺☺  Types of Process Variables  Extensive Variables  depend on the size of the system (mass and volume)

 Intensive Variables  do not depend on the size of the system (T,P, density, and mass fraction)

 Gibbs Phase Rule  Degree of freedom (F) is the number of intensive variables that can be specified independently for a system at equilibrium. No reaction

F  2m The number of phases The number of chemical species The number of degree of freedom

Raoult’s Law for GLE: Single Condensable Component ☺☺☺  Water evaporation into dry air  Saturation (GLE)  Gibbs Phase Rule

A+B B GLE

F  2m  222  2 두 개의 변수가 결정되면 다른 값들은 결정될 수 있다.

 Raoult’s Law for GLE ☺☺☺ Partial pressure of vapor in the gas = Pure-component vapor pressure at the system temperature

pi  yi P  pi* (T )

P&Ty P&yP T&yP

Example 6.3.-2 Material Balance Around a Condenser ☺☺☺

 A stream of air at 100oC contains 5260 mmHg contains 10% water by volume. (a) Calculate the dew point and degrees of superheat of the air. * pH 2O  yH 2O P  pH 2O (Tdp )

pH 2O  0.1 5260  526 mmHg 증기압이 526mmHg인 온도를 찾음. Antoine equation

Tdp 

log10 p*  A 

B T C

Table B-4

B 1668.21  C   228  90o C * A  log10 p 7.96681  log10 526

degree of superheat = 100 -90 = 10 oC

Example 6.3.-2: A stream of air at 100oC contains 5260 mmHg contains 10% water by volume. ☺☺☺ (b) Calculate the percent of vapor that condenses and the final composition of the gas phase if the gas is cooled to 80oC at constant pressure. Basis : 100 mol feed gas

100 mol feed

Q2 mol

y H2O 1-y BDA

0.1 H2O 0.9 BDA

Q1 mol H2O Number of component = 2 Number of unknown = (Q1, Q2, y) = 3

BDA : Bone-Dry Air

D.O.F. = 1: cannot be solved Should we stop?

Basis : 100 mol feed gas

100 mol feed

Q2 mol y H2 O 1-y BDA

0.1 H2O 0.9 BDA T = 100oC, P = 5260 mmHg

T = 80oC, P = 5260 mmHg Q1 mol H2O

Use thermodynamic information: A gas in equilibrium with liquid must be saturated with the liquid . * p  y P  p (T ) o i i i thus, y is a saturated condition at 80 C, 5260 mmHg

yH 2O  pH* 2O (80o C ) / P  355 / 5260  0.0675 yBDA  1  yH 2O  0.9325 Number of component = 2 Number of unknown = (Q1, Q2) = 2

D.O.F. = 0

Material Balance BDA Balance

100×0.9 = Q2×0.9325

Q2=96.5 mol

Total Balance

100=Q1+Q2

Q1=3.5 mol

% Condensed

3.5/(100×0.1) ×(100) = 35 %

Example 6.3.-2: A stream of air at 100oC contains 5260 mmHg contains 10% water by volume. ☺☺☺ (c) Calculate the percentage condensation and the final gas-phase composition if the gas is compressed isothermally to 8500 mmHg. Basis : 100 mol feed gas

100 mol feed

Q2 mol

y H2O 1-y BDA

0.1 H2O 0.9 BDA

Q1 mol H2O Number of component = 2 Number of unknown = (Q1, Q2, y) = 3

BDA : Bone Dry Air

Cannot be solved !

Basis : 100 mol feed gas

100 mol feed

Q2 mol y H2 O 1-y BDA

0.1 H2O 0.9 BDA T = 100oC, P = 5260 mmHg

T = 100 oC, P = 8500 mmHg Q1 mol H2O

Use thermodynamic information : A gas in equilibrium with liquid must be saturated with the liquid . * p  y P  p (T ) o i i i thus, y is a saturated condition at 100 C, 8500 mmHg

yH 2O  pH* 2O (100o C ) / P  760 / 8500  0.0894 1  yH 2O  0.9106

Number of component = 2 Number of unknown = (Q1, Q2) = 2

D.O.F. = 0

Material Balance BDA Balance

100×0.9=Q2×0.9106

Total Balance

100=Q1+Q2

% Condensed

1.2/(100×0.1) ×(100) = 12 %

Q2=98.8 mol Q1=1.2 mol

Let’s Start!

6.4 Multicomponent Vapor-Liquid Equilibria  Gas-Liquid Processes    

Chemical reactions Distillation (증류) Gas  Liquid : Absorption (흡수) Liquid Gas : Stripping (탈기)

 VLE information  From literature, databases  Raoult’s Law & Henry’s Law ☺☺☺  Rigorous calculation using model equations

 Distribution of components between vapor and liquid phases  Phase-Equilibrium Thermodynamics

Raoult’s Law and Henry’s Law ☺☺☺ Vapor (P and yi)

 Raoult’s Law pi  yi P  xi pi* (T )

Liquid (xi)

 Ideal gas  Ideal Solution

pi* = vapor pressure

 Valid for almost pure liquid (xi 1)  Valid for mixture of similar substances (over entire range of compositions: 0 < xi < 1)

 Henry’s Law yi P  xi H i (T )

Hi = Henry’s law constant

 Valid for dilute solution (xi0)

Example 6.4-2 ☺☺ Use either Raoult’s law or Henry’s law to solve the following problems. 1. A gas containing 1 mole% ethane is in contact with water at 20oC and 20 atm. Estimate the mole fraction of dissolved ethane.

N2, O2, CO2, …. CH4, C2H6, ….

From Perry’s Handbook

Dilute solution

Apply Henry’s Law

H C 2 H 6 (20o C )  2.63 104 atm/mole fraction yi P  xi H i (T )

xi  yi P / H i (T ) 

0.01 20 6  7 . 6  10 2.63 104

Example 6.4-2 ☺☺ Use either Raoult’s law or Henry’s law to solve the following problems. 2. An equimolar liquid mixture of benzene (B) and toluene (T) is in equilibrium with its vapor at 30oC. What is the system pressure and the composition of the vapor? Benzene + Toluene

pi  yi P  xi pi* (T )

P   pi

Similar Substances

Apply Raoult’s Law

Table B.4

pB* (30o C )  119 mmHg

Antoine Eq’n

pT* (30o C )  36.7 mmHg

pB  xB pB*  0.5 119  59.5 mmHg pT  xT pT*  0.5  36.7  18.35 mmHg yB  xB pB* / P  59.5 / 77.9  0.764 yT  xT pT* / P  18.35 / 77.9  0.236

P  59.5 mmHg  18.35 mmHg  77.9 mmHg

Phase diagrams for binary VLE ☺☺  Txy diagram (at a fixed P) T

 Pxy diagram (at a fixed T) P

vapor

liquid

Bubble P

Dew T

V+L

Bubble T

V+L

Dew P

liquid vapor

x1

x or y

y1

x1

y1

x or y

Bubble and Dew Points    

Bubble Point Temperature Bubble Point Pressure Dew Point Temperature Dew Point Pressure

Bubble point

: Constant P, T : Constant T, P : Constant P, T : Constant T, P

Dew point

Bubble P Pressure Liquid

P

Vapor

yi

xi Given T,x  Calculate P,y

Composition

Dew P Pressure Liquid

P

Vapor

yi

xi Given T,y  Calculate P,x

Composition

Bubble T Temperature

Vapor

T

Liquid

xi

yi Given P,x  Calculate T,y

Composition

Dew T Temperature Vapor

T

Liquid

xi

yi Given T,x  Calculate P,y

Pressure

VLE Calculations  Bubble Point Temperature Calculation  Given P, x  Calculate T,y yi 

xi pi* (Tbp )

P P  xa pa* (Tbp )  xb pb* (Tbp )  ...

 Dew Point Temperature Calculation  Given P,y  Calculate T,x xi 

yi P pi* (Tdp )

 xi   i

i

yi P 1 pi* (Tdp )

VLE Calculation  Iterative calculation required  Not explicit form

 Iterative calculation techniques  Trial and error method  Newton-Raphson Method  Secant Method OBJ ( X )  0

find X that satisfies given relation

Algorithm for Bubble/Dew Point Calculations Example : Bubble T Calculation Objective Function Start

x

i

1 Given P,x

i

y

i

1

i

OBJ   xi  yi  0 i

Assume T

yi  xi pi* (T ) / P

i

Calculate OBJ

i

Phase Equilibrium Calculate new T,y

yi P  xi pi* (T ) Ki  yi / xi  pi* (T ) / P

OBJ   xi  yi

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