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THERMODYNAMICS II Thermodynamic Relations From: Koretsky, Milo. (2012). Engineering and Chemical Thermodynamics. Wiley. Notes. Dr. Diego Fernando Mendoza. Prausnitz J., Poling B., O’Connell J. (2001) The properties of Gases and Liquids, Fifth Edition. MgGraw- Hill. Moran, M. J., Shapiro, H. N., Boettner, D. D., & Bailey, M. B. (2010). Fundamentals of engineering thermodynamics. John Wiley & Sons

ADRIANA PATRICIA VILLEGAS Q. Ph. D. Chemical Engineering 1

Koretsky (2012)

Hypothetical Paths

The first step in manufacturing isobutene from isomerization of n-butane is to compress the feed stream of n-butane. It is fed into the compressor at 9.47 bar and 80ºC and optimally exits at 18.9 bar and 120ºC, so that it can be fed into the isomerization reactor. The work supplied to the compressor is 2100J/mol. Calculate the heat that needs to be supplied into the unit per mole of n-butane That passes through. Ws

Compressor

q

EoS:

Redlich–Kwong

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Hypothetical Paths Redlich–Kwong

Solving

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Koretsky (2012)

Koretsky (2012)

Hypothetical Paths

The first step in manufacturing isobutene from isomerization of n-butane is to compress the feed stream of n-butane. It is fed into the compressor at 9.47 bar and 80ºC and optimally exits at 18.9 bar and 120ºC, so that it can be fed into the isomerization reactor. The work supplied to the compressor is 2100J/mol. Calculate the heat that needs to be supplied into the unit per mole of n-butane That passes through. Ws

Compressor

q

4

Hypothetical Paths 1. Solution with T and v as independent properties (black)

From Fundamental properties

Maxwell relations

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Koretsky (2012)

Hypothetical Paths 1. Solution with T and v as independent properties (black)

Redlich–Kwong

Koretsky (2012)

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Hypothetical Paths

Koretsky (2012)

1. Solution with T and v as independent properties (black)

Redlich–Kwong

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Hypothetical Paths 1. Solution with T and v as independent properties (black)

Simplifying

Integrating

Koretsky (2012) 8

Hypothetical Paths 1. Solution with T and v as independent properties (black)

Koretsky (2012)

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Hypothetical Paths 2. Solution with T and P as independent properties (black)

Koretsky (2012)

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Hypothetical Paths

Koretsky (2012)

1. Solution with T and P as independent properties (black)

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Hypothetical Paths NOTE 1

Koretsky (2012)

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Hypothetical Paths NOTE 1

Koretsky (2012)

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Hypothetical Paths NOTE 1

vo

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Koretsky (2012)

Hypothetical Paths

Koretsky (2012)

NOTE 2 Other possibility to calculate the enthalpy change is using the internal energy change.

Redlich–Kwong

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Hypothetical Paths NOTE 3. Remember that you can use the general form of CEoS for solving problems associated to hypothetical paths.

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Hypothetical Paths NOTE 3. Remember that you can use the general form of CEoS for solving problems associated to hypothetical paths. In this order of ideas, the derivatives of interest are:

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Hypothetical Paths NOTE 3.

1. Solution with T and v as independent properties (Red)

It is recommendable to use PR to calculate the volume:

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Hypothetical Paths NOTE 3.

1. Solution with T and v as independent properties (Red)

Suggested Path for Enthalpy:

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Hypothetical Paths NOTE 3.

1. Solution with T and v as independent properties (Red)

Suggested Path for Entropy:

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Hypothetical Paths NOTE 3.

2. Solution with T and P as independent properties (Red)

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Hypothetical Paths NOTE 3.

2. Solution with T and P as independent properties (Red)

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Hypothetical Paths NOTE 3.

2. Solution with T and P as independent properties (Red)

Suggested Path for Enthalpy:

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Hypothetical Paths NOTE 3.

2. Solution with T and P as independent properties (Red)

Suggested Path for Entropy:

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Hypothetical Paths NOTE 4.

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Hypothetical Paths NOTE 4.

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Hypothetical Paths NOTE 4.

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Hypothetical Paths

Phase change: Δu, Δ h, Δ s α and β and can represent any phase: solid, liquid, or vapor. We may be interested in any of the following: vapor–liquid, liquid– liquid, liquid–solid, gas–solid,

Koretsky. (2012).

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Hypothetical Paths Phase change: Δu, Δ h, Δ s

Koretsky (2012)

Koretsky. (2012). The criterion for chemical equilibrium is when the Gibbs energy is at the minimum

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Hypothetical Paths Phase change: Δu, Δ h, Δ s

30 Koretsky. (2012).

Hypothetical Paths Phase change: (Wagner Correlation)

Reduced Vapor Pressure

Reduced Temperature

Prausnitz, 2001

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Hypothetical Paths

Prausnitz, 2001

Phase change: (Ambrose-walton)

This set of equations was fit to the vapor pressure behavior of the n-alkanes. The quantity, ƒ(2), is important only for fluids with large acentric factors and at low reduced temperatures 32

Hypothetical Paths

Phase change: Riedel Method

is the enthalpy of vaporization at the normal boiling point, and R Is 8.314J/molK and Pc is expressed in [bar].

Chen Method

Prausnitz, 2001 33

Hypothetical Paths

Prausnitz, 2001

(Wagner Correlation)

A common choice for n is 0.375 or 0.38

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Hypothetical Paths

Dahm, 2001

Examples: Five moles of gas are confined in a piston-cylinder device. At the beginning of the process, the gas has T = 300 K and V = 100 L. If the gas is compressed isothermally to a final volume of 15 L, how much work is required, and how much heat is added or removed? Assume the heat capacity is constant at Cv=30 J/mol K and that the gas is modeled by the van der Waals equation of state.

Energy Balance

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Hypothetical Paths

Dahm, 2001

Examples: Energy Balance

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Hypothetical Paths

Smith vannes, 2001

Examples: For an industrial process, is required 1-buteno at 293.15ºK. Assume a reference state of saturated liquid a 0ºC, in this point, h=0 and s=0. Using this information, indicate a isobaric thermodynamic path to reach the final state and determine the enthalpy change in each process. Tc= 420.0 K Pc= 40.43 bar ω=0.191 Tb=266.9 K (normal boiling point)

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Psat (293.15)

38 Psat (273.15)

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293.15K

Present the process In P-v and T-v coordinates

273.15K

1

2

Black vf

vg

v∞

Solution Slide 7-9 Redlich–Kwong or Solution Slides 18 and 19. PR (Please check the limits of integration)

Psat (293.15) Psat (273.15)

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293.15K

Red check Slide 14

273.15K

1

2 Check ideal gas

vf

vg

va

v∞

Solution Slide 18 and 19

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Psat (293.15)

40 Psat (273.15)

3

293.15K

273.15K

1

vf

2

vg

v∞

Explain the process in Production plants