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Advanced Thermodynamics Note 11 Solution Thermodynamics: Applications

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Presentation on theme: "Advanced Thermodynamics Note 11 Solution Thermodynamics: Applications"— Presentation transcript:

1 Advanced Thermodynamics Note 11 Solution Thermodynamics: Applications
Lecturer: 郭修伯

2 Liquid-phase properties from VLE data
Fugacity For species i in the vapor mixture: Vapor/liquid equilibrium: The vapor phase is assumed an ideal gas: Therefore: The fugacity of species i (in both the liquid and vapor phases) is equal to the partial pressure of species i in the vapor phase. Its value increases from zero to Pisat for pure species i

3 Fig. 12.1 Table 12.1 The first three columns are P-x1-y1 data. Columns 4 and 5 are: Column 6 is:

4 Fig 12.3 Fig 12.2 Henry’s constant, the limiting slope of the curve at xi = 0. Henry’s law expresses: , it is approximate valid for small values of xi

5 Henry’s law Lewis/Randall rule Gibbs/Duhem equation x1 → 0 x2 → 1
Gibbs/Duhem equation for binary mixture at const. T and P: The Lewis/Randall rule, Division by dx1 when x1 = 1, limit

6 Excess Gibbs energy Table 12.2 Column 6:

7 Fig. 12.5 Positive deviation from Raoult’s law behavior: The dimensionless excess Gibbs energy: The value of GE/RT is zero at both x1= 0 and x1 =1

8 From Fig 12.5(b), linear relation:
Similarly, The Margules equations Limiting conditions:

9 VLE data for diethyl ketone (1) / n-hexane (2) at 65°C are given in the first three columns of Table Reduce the data. Table 12.4 Fig 12.7 The solid lines. Not consistency! Omit Barker’s method Fig 12.7(b) for (GE/x1x2RT) fitting:

10 Models for the excess Gibbs energy
weak GE/RT = f (T, P, composition) At constant T: Data fitting, convenient, but only for binary system The Redlich/Kister expansion The Margules equation The van Laar equation

11 Local composition models
Can be applied to multi-component systems The Wilson equation: The NRTL(Non-Random-Two-Liquid) equation: The UNIQUAC equation and the UNIFAC method: App. H.

12 Property changes of mixing
Excess properties The M change of mixing Because of their direct measurability, V and H are the property changes of mixing of major interest.

13 Fig 12.10

14 The excess enthalpy (heat of mixing) for liquid mixture of species 1 and 2 at fixed T and P is represented by the equation: Determine expressions for and as functions of xi. The partial properties:

15 Fig 12.13 1. Each ΔM is zero for a pure species. 2. The Gibbs energy change of mixing ΔG is always positive. 3. The entropy change of mixing ΔS is positive.

16 Heat effects of mixing processes
Heat of mixing: For binary systems: When a mixture is formed, a similar energy change occurs because interactions between the force fields of like and unlike molecules are different. Heat of solution based on 1 mol of solute dissolve in liquids:

17 Calculate the heat of formation of LiCl in 12 mol of H2O at 25°C.
Fig 12.14

18 A single-effect evaporator operating at atmospheric pressure concentrates a 15% (by weight) LiCl solution to 40%. The feed enters the evaporator at the rate of 2 kg/s at 25°C. The normal boiling point of a 40% LiCl solution is about 132°C, and its specific heat is estimated as 2.72 kJ/kg °C. What is the heat transfer rate in the evaporator? Feed at 25°C 2 kg/s 15% LiCl 1.25 kg superheated steam at 132°C and 1 atm 0.75 kg 40% LiCl at 132°C Q The energy balance: the total enthalpy of the product streams minus the total enthalpy of the feed stream liquid water is vaporized and heated to 132°C 0.75 kg of 40% LiCl solution is heated to 132°C mixing of 0.45 kg of water with 0.30 kg of LiCl(s) to form a 40% solution at 25°C separation of 2 kg of a 15% LiCl solution into its pure constituents at 25°C

19 Enthalpy/concentration diagrams
The enthalpy/concentration (Hx) diagram Fig 12.17

20 Solid NaOH at 70°F is mixed with H2O at 70°F to produce a solution containing 45% NaOH at 70°F. How much heat must be transferred per pound mass of solution formed? Energy balance: 45% NaOH: on the basis of 1 (lbm): 0.45(lbm) of solid NaOH dissolved in 0.55 (lbm) of H2O. Fig 12.19, x1 = 0: Fig 12.19, x1 = 45%:

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