SOLUTION AND EQUILIBRIUM THERMODYNAMICS Prof. Angel Darío González-Delgado
OBJETIVOS •
•
•
•
•
Conocer y utilizar las aplicaciones aplicaciones de la termodinámica termodinámica en condiciones de equilibrios físicos y químicos en sistemas de composición variable, conocer los fundamentos fundamentos básicos de la Cinética Química. Entender las relaciones relaciones de propiedades fundamentales fundamentales para las soluciones homogéneas de composición variable. Aprender a calcular calcular las propiedades en equilibrios termodinámicos. Realizar análisis termodinámicos termodinámicos de procesos. procesos. Comprender el impacto de una solución de ingeniería utilizando la termodinámica de equilibrio y soluciones en un contexto mundial, económico, ambiental y social
REFERENCIAS BIBLIOGRAFICAS •
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•
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•
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Smith, J. M., Van Ness, H. C. & Abbott, M.M. (2005) Introduction Introduction to Chemical Engineering Thermodynamics. Ed. McGraw –Hill. 7a. Edición. New York. Cengel, Y & Boles, M. (2002) Termodinámica. Termodinámica. Editorial McGraw Hill. 4ª edición. México Van Wylen, Wylen, G.J. & Sonntag, R.E. (1993), Fundamentals of classical classi cal thermodynamics, SI version, 5th edition, 744 pp., (John Wiley & Sons, New York). Wark & Richards. Termodinámica, Termodinámica, Sexta Edición. México. McGraw-Hil McGraw-Hill. l. 2006. 1164 pp. ISBN884812829X. Jones, J.B.& Dugan,R.E. (1997): Ingeniería Termodinámica. Prentice Hall. pp. 328. Sherwin,K. (1995): Introducción Introducción a la Termodinámica. Addison Wesley Iberoamericana. pp.148. Prausnitz J, Lichtenthaler RN, Gomes de Azevedo E. (2000): Termodinámica Termodinámica molecular de los equilibrios de fases 3a. ed. Prentice Hall. Dincer, Dincer, I. & Rosen, M.A. Exergy —Energy, Environment and Sustainable Development, 2nd ed. Elsevier: Oxford, UK, 2012.
Semana No. 1
2
3
SABER
HACER
Relación de propiedades fundamentales. Relaciones de Maxwell. Ecuaciones Tds, Potencial Químico como criterio para el equilibrio de fases. Desarrollo de ejercicios de verificación de relaciones de maxwell y cambios de temperatura, presión y entropia.
Comprender y analizar la relación entre propiedades termodinámicas fundamentales y el potencial químico en el equilibrio de fases. .
Propiedades molares parciales. Mezcla de Conocer y aplicar las propiedades gases ideales. parciales a situaciones reales. Desarrollo de ejemplos y actividades para el cálculo de propiedades molares parciales y a dilución infinita. Revisión de problemas resueltos (Hora tutoría) Fugacidad y Coeficiente de fugacidad para Calcular los coeficientes de una especie pura. Fugacidad y Coeficiente de fugacidad para especies puras y en fugacidad para una especie en solución. solución. Desarrollo de ejercicios de cálculo de fugacidades de sustancias puras y mezclas multicomponente utilizando excel. Revisión de problemas resueltos (Hora tutoría)
Semana No. 4
5
SABER
HACER
Correlaciones generalizadas para el Conocer la importancia de la coeficiente de fugacidad. La solución ideal. solución ideal y las propiedades en Propiedades en exceso. Propiedades de las exceso. fases liquidas a partir de datos EVL. Desarrollo de ejercicios de cálculo de propiedades de exceso. Revisión de problemas resueltos (Hora tutoría). Modelos de composición global y local para la Conocer y aplicar los distintos energía de Gibbs en exceso. modelos de solución para calcular Taller: Casos de estudio utilizando Excel, la energía de Gibbs en exceso. Matlab, Unisim u otros. Desarrollo de ejercicios propuestos. Revisión de problemas resueltos (Hora tutoría)
6
Primer examen. Cambio de propiedades en el mezclado. Entender el cambio de las Bases moleculares para el comportamiento de propiedades fundamentales en los mezclas. procesos de mezclado. Análisis de artículos científicos Desarrollo de ejercicios de cálculo de efectos
Semana No. 7
8
9
SABER
HACER
Naturaleza del equilibrio. Regla de las fases. Conocer, entender y saber aplicar la Teorema de Duhem. EVL. Comportamiento regla de las fases a problemas cualitativo. específicos de ingeniería.
Calculo de punto de rocío y punto de burbuja. Calcular el punto de roció y de Calculo de evaporación instantánea. Sistema burbuja de una mezcla binaria. soluto-solvente. Realizar cálculos de evaporación Taller de cálculo de puntos de rocío y burbuja instantánea. utilizando Excel, Matlab, Unisim u otros. Desarrollo de ejercicios propuestos de cálculo de vaporización instantánea. Revisión de problemas resueltos (Hora tutoría). Propiedades de los fluidos a partir de las ecuaciones viriales de estado. Propiedades de los fluidos a partir de las ecuaciones cúbicas de estado. Discusión de problemas y casos (Hora tutoría).
Comprender, aplicar y calcular las propiedades viriales. Calcular las propiedades de los fluidos a través de las ecuaciones cúbicas.
Semana No. 10
11
12
SABER
HACER
Correlaciones tipo Pitzer. EVL a partir de Entender y aplicar los conceptos de ecuaciones cúbicas de estado. Equilibrio y equilibrio entre fases. estabilidad. Equilibrio liquido-liquido. Equilibrio vapor- Entender y aplicar los conceptos de liquido-liquido. Equilibrio vapor sólido-liquido. equilibrio entre diferentes fases. Equilibrio sólido-vapor. •
Desarrollo de ejercicios propuestos. Revisión de problemas resueltos (Hora tutoría) Coordenadas de reacción. Aplicación de los Calcular la energía estándar de Gibbs criterios de equilibrio a las reacciones y la constante de equilibrio para químicas. Cambio de la energía estándar de reacciones químicas Gibbs y la constante de equilibrio. Segundo examen.
13
Efecto de la temperatura sobre la constante de equilibrio. Evaluación de las constantes de equilibrios. Relación de las constantes de equilibrio con la composición. Conversión de equilibrio para las reacciones individuales.
Resolver problemas de equilibrio en reacciones químicas aplicados a situaciones reales
Evaluar la dependencia de la constante de equilibrio con la Desarrollo de ejercicios propuestos. Revisión temperatura y la composición de problemas resueltos (Hora tutoría).
Semana No. 14
15
16
SABER
HACER
Regla de las fases y teorema de Duhem para Conocer, entender y saber aplicar la los sistemas reactivos. Equilibrio en regla de las fases para sistemas reacciones múltiples. reactivos a problemas específicos de Desarrollo de ejercicios de aplicación (Hora ingeniería. tutoría) Análisis termodinámico de procesos. Calculo Realizar análisis termodinámico de del trabajo ideal y trabajo perdido. procesos Desarrollo de taller de cálculo de trabajo perdido en una etapa de proceso. Revisión de problemas resueltos (Hora tutoría) Análisis exergético de procesos de flujo en Realizar el análisis exergético estado uniforme. Caso de estudio utilizando completo de un proceso químico, Excel. calculando la eficiencia exergética y por etapa, las Proyecto de aula: diagnostico energético de un global proceso químico utilizando análisis exergético. irreversibilidades totales y la exergía de los residuos del proceso. Parcial Final
OUTLINE 0. Introduction 1.
Fundamental Property Relation
2.
Chemical Potential & Phase Equilibria
3.
Partial Properties
4.
The Ideal-Gas Mixture Model
5.
Fugacity & Fugacity Coefficient: Pure Species
6.
Fugacity & Fugacity Coefficient: Species in Solution
7.
Generalized Correlations for Fugacity Coefficient
0. Introduction
0. Introduction
Composition (α)=f(P, T, μ, xi)
0. Introduction 1. Capacity: 80000 ton/year
% w/w
Compound
2. Thermodynamics package
11.3
PLO
12.7
PLP
6.6
OOO
26.2
POO
33.0
POP
7.2
PPP
3.0
Palmitic acid
- Chao Seader - NRTL - Peng-Robinson - BWRS 3. Crude palm oil composition (CPO) 4. Group contribution method
0. Introduction Esterification
Transesterification Washing
Methanol Glycerol separation recovery 40-80% of plant FCI corresponds to separation equipment
0. Introduction The phase equilibria problem:
Source: www.cchem.berkeley.edu
1. Fundamental Property Relation Remember from calculus course: if Z f ( x, y )
dz dz d ( z) dx dy dx y dy x d ( z ) Mdx Ndy
dN dM ? dx dy y x
1. Fundamental Property Relation dN dM dx y dy x Remember from calculus course (the cyclic relation): if
f ( x, y , z ) 0 x x( y , z ) y y ( x, z ) z z ( x, y )
dx dy dz 1 dy z dz x dx y Which properties of a system do you remember?
1. Fundamental Property Relation Let’s replace x, y and z for P, V and T (direct measurable basic properties of a system): f ( P, V , T ) 0 P P(V , T ) V V ( P, T ) T T ( P, V )
dP d V dT 1 dV T d T P dP V
1. Fundamental Property Relation Example 1.1. Consider air at 300 K and 0.86 m 3/kg. The state of air changes to 302 K and 0.87 m3/kg as a result of some disturbance. Find the change in pressure of air using:
dz dz d ( z) dx dy dx y dy x
Geometric representation of the disturbance discussed in the above example
1. Fundamental Property Relation Example 1.1. Consider air at 300 K and 0.86 m3/kg. The state of air changes to 302 K and 0.87 m3/kg as a result of some disturbance. Find the change in pressure of air using:
dz dz d ( z) dx dy dx y dy x 1. 2. 3. 4.
Choose an EoS Find dT and dV Derive EoS Replace terms
1. Fundamental Property Relation Solution: The temperature and specific volume of air changes slightly during a process. The resulting change in pressure is to be determined. Assumptions: Air is an ideal gas Analysis: Due to changes are small, the changes in T and v can be expressed as:
≅ ∆ = 302 − 300 = 2 ≅ ∆ = 0.87 − 0.863 / = 0.013 /
1. Fundamental Property Relation And ideal gas obeys the relation PV=RT. Solving for P yields: =
Note that R is a constant and P=P(T, v). Therefore:
=
+
=
−
2
1. Fundamental Property Relation =
−
−
2
=
2
− − − 3
= 0.287
301
2
3
3
0.865
= 0.664 =
0.865
1.155
0.491
3
0.01
2
1. Fundamental Property Relation Therefore, Therefore, the pressure pressure will decrease decrease by 0.491 kPa as kPa as a result of this disturbance. Notice that if the temperature temperature had remained constant (dT=0), the pressure the pressure would decrease by 1.155 kPa as kPa as a result of the 0.01 m3/kg increase in specific volume.
− =
=
1.155
1. Fundamental Property Relation However, if the specific the specific volume had remained constant constant (dv=0), (dv=0), the pressure the pressure would increase by 0.664 kPa as kPa as a result of the 2 K rise in temperature. temperature.
=
= 0.664
And,
+
=
= 0.664 − 1.155
=
−0.491
1. Fundamental Property Relation Example 1.2. Using the ideal-gas equation of state, verify a) the cyclic relation and b) the reciprocity relation at constant P. Analysis: The The idea ideall-g gas equa equati tion on of stat state e Pv=R Pv=RT T invo involv lves es the three variables P, v and T. Any two of these can be taken as the inde indepe pend nden entt varia riables bles,, with ith the the remain mainin ing g one one bein being g the the dependent variable.
1. Fundamental Property Relation Replacing x, y and z in the following equation by P, v and T, respectively, we can express the cyclic relation for and ideal gas as:
− −
=
1
=
1
1. Fundamental Property Relation
= , = → = − Where, = , = → = = , = → = Substituting yields, − − − − 2
=
2
=
1
=
1
1. Fundamental Property Relation U: Internal energy (molecular energy) H: Enthalpy (energy in a thermodynamic system, includes internal energy and the amount of energy required to make room for it by displacing its environment). = +
A: Helmholtz energy (the useful work obtainable from a closed thermodynamic system at a constant temperature).
=
−
1. Fundamental Property Relation G: Gibbs energy (maximum or reversible work that may be performed by a thermodynamic system at a constant temperature and pressure). G W
H
Re v
TS
G2
G1
(closed system, constant T and P)
1. Fundamental Property Relation d (u )
TdS
PdV
d (H )
TdS
PdV ( PdV
d ( H )
TdS
VdP
d ( A)
d ( A)
PdV
d (G )
d (G )
VdP
TdS
TdS
PdV (TdS
VdP
SdT )
SdT
VdP
SdT
)
(TdS SdT )
1. Fundamental Property Relation U, H, A and G (state functions) can be used for tell us if a chemical reaction is spontaneous or not, subject to constraints.
Parameter
Chemical reaction is spontaneous if
U, Internal Energy
dU (Constant V, S) < 0
H, Enthalpy
dH (Constant P, S) < 0
A, Helmholtz energy
dA (Constant V, T) < 0
G, Gibbs energy
dG (Constant P, T) < 0
1. Fundamental Property Relation d (u )
d ( H )
TdS
TdS
PdV
Vd P
d ( A)
PdV
d (G )
VdP
dN dM dx y dy x
SdT
S dT
dT dP dV dS s v
dT dV dP s dS P
dP dS dT dV V T
dV dS dT P dP T
1. Fundamental Property Relation Maxwell relations are extremely valuable in thermodynamics because they provide a means of determining the change in entropy, which cannot be measured directly, by simply measuring the changes in properties P, v, and T
1. Fundamental Property Relation Example 1.3. Verify the validity of the following Maxwell relation for steam at 250 °C and 300 kPa.
− ?
=
Analysis. The Maxwell relation states that for a simple compressible substance, the change in entropy with pressure at constant temperature is equal to the negative of the change in specific volume with temperature at constant pressure.
1. Fundamental Property Relation Analysis. If we had explicit analytical relations for the entropy and specific volume of steam in terms of other properties, we could easily verify this by performing the indicated derivations. However, all we have for steam are tables of properties listed at certain intervals. Therefore, the only course we can take to solve this problem is to replace the differential quantities with corresponding finite quantities, using property values from the tables or about the specified state.
1. Fundamental Property Relation
= − ∆ ∆ ≅ − ?
?
=250°
400− − 200 400
200
=300
≅ − 300 −−200 ° ?
=250°
300°
200 °
=300
7.3804 − 7.7100 (0.87535 − 0.71643) ≅ − 400 − 200 300 − 200 °
3
?
−0.00165 ≅ −0.00159 3
3
1. Fundamental Property Relation d (u )
d ( H )
TdS
TdS
PdV
VdP
d ( A)
PdV
d (G )
VdP
T
du du ; P dS v dV s
T
dH dH ; V dS P dP s
SdT
S dT
dA dA P ; S dV T dT V
V
dG dG ; S dP T dT P
1. Fundamental Property Relation T-dS Equations:
S f (T , V )
dz dz d ( z) dx dy dx y dy x dS dS d (S ) dT dV dT V dV T dS dS T d ( S ) T dT T dV dT V d V T
1. Fundamental Property Relation T-dS Equations:
In a constant volume process: Td ( S )
In a constant pressure process: Td ( S ) Then:
dP (1)Td ( S ) CV dT T dV dT V
CV dT
C P dT
1. Fundamental Property Relation T-dS Equations:
S
S (T , P)
dS dS d (S ) dT dP dT P dP T dS dS T d ( S ) T dT T dP dT P dP T dV (2)Td ( S ) C P d T T d P d T P
1. Fundamental Property Relation Combining two T-dS Equations:
dV dP (CP CV )dT T dP T dV 0 dT P dT V dV dP (3)(CP CV ) T dT P dT V dP dV dT Remember that: 1 dV T dT P dP V
1. Fundamental Property Relation Some algebra:
dP dV dP dV dT dT T P V 2
Replacing in (3):
dV dP (CP CV ) T (4) dT P dV T For all substances (dP/dV) is always negative, which means that Cp-Cv is always positive, or Cp>Cv
1. Fundamental Property Relation G
H TS
d G S dT P As S is always a positive quantity, G decreases when temperature increases (despite is an energy!) Re v
WS
G2
G1
If T increases, G decreases and reversible work decreases.
1. Fundamental Property Relation
Remember that for a turbine: Re v
WS
H2
H 1
Source: www.turbinesinfo.com
So, if T2 increases and T1 remains the same, the maximum work obtainable increases, this contradicts what is written in the previous slide.
1. Fundamental Property Relation
Re v
WS
H2
H 1
Source: www.turbinesinfo.com
Answer: This equation only applies for Isentropic work, and Gibbs energy takes into account entropy term.
1. Fundamental Property Relation
Answer: This equation only applies for Isentropic work, and Gibbs energy takes into account entropy term.
1. Fundamental Property Relation If we consider the temperature dependence of G/T:
d dT
1
T 1
T
d dT
G T
1 dG d G P T dT P dT
1
T
P
dG G 1 dG G dT P T T dT P T 2
G 1 H T S S T T S T H T P T
G
2
1. Fundamental Property Relation d dT
G T
H
T P
2
Gibbs-Helmholtz Equation: Shows that we can obtain a measurement of H, by measuring the temperaturedependence of G.
H d G For any process with a defined deltaH: T dT T P 2
1. Fundamental Property Relation If we consider the Pressure dependence of G (constant temperature, dT=0) d (G ) VdP SdT d (G ) VdP Pf
G ( P final ) G ( Pinitial )
VdP Pi
For solids and liquids (incompressible fluids)
G ( P final ) G ( Pinitial )
V ( Pf
Pi )
Gibbs energies for solids and liquids are constant with pressure
1. Fundamental Property Relation For ideal gases: Pf
G ( P final ) G ( Pinitial )
Pi
P f dP RT ln P Pi
RT
Gibbs energies for ideal gases are function of temperature and pressure
1. Fundamental Property Relation
1. Fundamental Property Relation What does dG(P,T) mean:
1. Fundamental Property Relation
dG/dT (gases)>dG/dT (liquids)>dG/dT (solids)
1. Fundamental Property Relation Issues to consider: •
How do individual chemical species in a mixture, or reactants and products in a reaction system contribute to G?
•
Can we calculate G and dG from their concentrations?
•
How G is affected by fluxes of matter in/out of the system
1. Fundamental Property Relation Definition of Total Gibbs of any CLOSED system:
GH
TS
d (nG ) d (nH ) Td (nS ) (nS )dT from H
U
PV
obtain d(nH) d (nG ) (nV )dP (nS )dT
1. Fundamental Property Relation Total Gibbs energy change of any CLOSED system:
d nG
nV dP nS dT
*applied to a single phase fluid in a closed system wherein no chemical reactions occurs & the composition is constant.
Therefore,
nG P nV T , n
nG T
nS P n ,
No of moles of all chemical species are held constant.
1. Fundamental Property Relation For an open system: Consider two containers at constant T and P, one contains A and the other contains B, both connected, but initially the valve is closed.
Source: www.chegg.com
1. Fundamental Property Relation
Container I
Container II
Since G (Gibbs energy an extensive variable: Gtotal
GI
GII
G of all A and B in Container I nGtotal
G of all A and B in Container II nGI
nGII
1. Fundamental Property Relation Now we open the valve:
nG I
nGI ( n A , n B )
nG II
nGII (n A , nB )
After we open the valve, A and B will mix spontaneously, which means that total G will decrease.
1. Fundamental Property Relation
(nG I ) d ( nG I ) n A T , P , n
( nGI ) dnA n B T , P,n
B
(nG II ) d ( nG II ) n A T , P ,n
B
dnB
Container I
A
( nGII ) dnA n B T , P,n
dnB A
Container II
1. Fundamental Property Relation
If A flows to container II and B flows to container I (and mass is conserved) dn A (container.II ) dn B (container.I )
dnA (container.I )
dnB (container.II )
1. Fundamental Property Relation (nG I ) d ( nG I ) n T , P ,n A
( nGI ) dnA n T , P , n B
B
(nG II ) d ( nG II ) n A T , P ,n
B
dn A (container .II ) dn B (container .I )
dnB A
(nGII ) dnA nB T , P ,n
dnA (container .I )
dnB (container .II )
dnB A
1. Fundamental Property Relation Combining equations, for container I:
(nG ) nG ( ) I II dn A (container.I ) d ( nG ) n A T , P ,n nA T , P, n ( nG ) ( ) nG I II dn B (container.I ) n B T , P ,n nB T , P, n B
A
B
A
Mixing of A and B continues until dnG=0, this can happen if 0 or if the terms in blue are both zero. dn A dnB
1. Fundamental Property Relation •
So, when valve is open, mixing stops when:
(nG I ) (nGII ) n n A A T , P , n T , P,n B
B
(nG I ) ( nGII ) n n B B T , P , n T , P,n A
•
•
and
A
Derivatives respect to the number of moles of something are call partial molar quantities. The partial molar Gibbs free energy is so important it is given its own symbol. (nG ) Chemical potential of A n , , I
A
A
T P n B
1. Fundamental Property Relation (nG I ) ( nGII ) n n A A T , P , n T , P , n B
B
(nG I ) ( nGII ) n n B B T , P , n T , P , n A
•
•
and
A
The spontaneous mixing of A and B will continue until the chemical potential of a chemical specie in container I is equal to chemical potential of that chemical specie in container II. The uniform chemical potential of each component of a multicomponent system is a requirement for equilibrium.
1. Fundamental Property Relation We can give the chemical potential in terms of other variables: i
(nG ) (nH ) (nA) (nU ) ni T , P , n ni S , P, n ni V ,T , n ni S ,V , n j
j
j
So, our expression of dG must include these mass transfers, for an open system, the change of total Gibbs energy becomes: d ( nG ) ( nV ) dP ( nS ) dT
dn i
i
i
j
1. Fundamental Property Relation Chemical potential of component i : •
•
•
It correspond to the rate of change of G with ni when the component i is added to the system at fixed P,T and number of moles of all other species. The chemical potential is the partial molar Gibbs Free Energy (or U,H, A) of component i. Physically this corresponds to how the volume in the system changes upon addition of 1 mole of component ni at fixed P,T and mole numbers of other components.
1. Fundamental Property Relation Issues to consider: •
How do individual chemical species in a mixture, or reactants and products in a reaction system contribute to G? Can we calculate G and dG from their concentrations? d (nG )
dn i
i
i
•
How G is affected by fluxes of matter in/out of the system. d ( nG ) (nV )dP (nS )dT
dn i
i
i
1. Fundamental Property Relation
d nG nV dP nS dT
dn i
i
i
nG Since P
T n
nG nV , T
,
nS , P n ,
i
nG n i P T n ,
,
j
total change of Gibbs energy for OPEN system becomes:
nG nG nG d nG d ni dP dT ni P T n i P T n T P n ,
,
,
,
j
1. Fundamental Property Relation For special case of one mol of solution, n=1 & ni = x i ,
d nG nV dP nS dT
dn i
i
i
becomes d G VdP SdT i dxi i
G
G T , P, x1 , x2 ,.... xi ,..
At constant T & x:
G V P
G as a function of T, P and x i
T x ,
At constant P & x:
At constant P & x:
G S T P x ,
G H G T T P x ,
from H
G
TS
At T=253 K, we are 20 K below the melting point of ice. Water chooses to exist in the solid state. Reason:
<
<
At T=273.16 K, ice melts to form water and these two phases coexist. Reason:
=
<
At T=330 K, water is still in the liquid state. Reason:
<
<
At T=373.16 K, water boils and water and steam coexist. Reason:
=
<
