Page 1 Chemical Engineering Thermodynamics A Brief Review SHR Chapter 2 1 Thermo.key - February 7, 2014 Page 2 Chemical Engineering Thermodynamics A Brief Review SHR Chapter 2 1 Thermo.key - February 7, 2014 “Rate” vs. “State” Thermodynamics tells us how things “end up” (state) but not how (or how fast) they “get there” (rate) Limitations on getting to equilibrium: • Transport (mixing): diffusion, convection ? example: why doesn’t more of the ocean evaporate to make higher humidity in the deserts? (heat transfer alters dew point locally, atmospheric mixing limitations, etc.) • Kinetics (reaction): the rate at which reactions occur are too slow to get to equilibrium ? example: chemical equilibrium says trees (and our bodies) should become CO 2 and H 2 O (react with air) ? example: chemical equilibrium says diamonds should be graphite (or CO 2 if in air) Sometimes, mixing & reaction are suf?ciently fast that equilibrium assumptions are “good enough.” • if we assume that we can use thermodynamics to obtain the state of the system, we avoid a lot of complexity. ? do it if it is “good enough!” • Most unit operations (e.g. distillation) assume phase equilibrium, ignoring transport limitations. ? good enough if things are mixed well and have reasonable residence/contact times. 2 Thermo.key - February 7, 2014 Page 3 Chemical Engineering Thermodynamics A Brief Review SHR Chapter 2 1 Thermo.key - February 7, 2014 “Rate” vs. “State” Thermodynamics tells us how things “end up” (state) but not how (or how fast) they “get there” (rate) Limitations on getting to equilibrium: • Transport (mixing): diffusion, convection ? example: why doesn’t more of the ocean evaporate to make higher humidity in the deserts? (heat transfer alters dew point locally, atmospheric mixing limitations, etc.) • Kinetics (reaction): the rate at which reactions occur are too slow to get to equilibrium ? example: chemical equilibrium says trees (and our bodies) should become CO 2 and H 2 O (react with air) ? example: chemical equilibrium says diamonds should be graphite (or CO 2 if in air) Sometimes, mixing & reaction are suf?ciently fast that equilibrium assumptions are “good enough.” • if we assume that we can use thermodynamics to obtain the state of the system, we avoid a lot of complexity. ? do it if it is “good enough!” • Most unit operations (e.g. distillation) assume phase equilibrium, ignoring transport limitations. ? good enough if things are mixed well and have reasonable residence/contact times. 2 Thermo.key - February 7, 2014 Some T erminology... M= C X i=1 y i ¯ M i Gibbs-Duhem Equation: ? @M @ P ? T,y dP + ? @M @ T ? P,y C X i=1 y i d ¯ M i =0 “Partial Molar” property M - arbitrary thermodynamic property (per mole) N i - moles of species i y i - mole fraction of species i (liquid or gas) Note at constant T and P, C X i=1 y i d ¯ M i =0 “Excess” propertyM E ? M M ideal how much M deviates from the ideal solution behavior. “Residual” property M ideal = C X i=1 y i M i how much M deviates from the ideal gas ( M ig ) behavior. M ig = C X i=1 y i M ig i M R ? M M ig change in M due to adding a differential amount of species i. For ideal mixtures, ¯ M i =M i ¯ M i ? ? @ NM @ N i ? T,P,N j 3 Thermo.key - February 7, 2014 Page 4 Chemical Engineering Thermodynamics A Brief Review SHR Chapter 2 1 Thermo.key - February 7, 2014 “Rate” vs. “State” Thermodynamics tells us how things “end up” (state) but not how (or how fast) they “get there” (rate) Limitations on getting to equilibrium: • Transport (mixing): diffusion, convection ? example: why doesn’t more of the ocean evaporate to make higher humidity in the deserts? (heat transfer alters dew point locally, atmospheric mixing limitations, etc.) • Kinetics (reaction): the rate at which reactions occur are too slow to get to equilibrium ? example: chemical equilibrium says trees (and our bodies) should become CO 2 and H 2 O (react with air) ? example: chemical equilibrium says diamonds should be graphite (or CO 2 if in air) Sometimes, mixing & reaction are suf?ciently fast that equilibrium assumptions are “good enough.” • if we assume that we can use thermodynamics to obtain the state of the system, we avoid a lot of complexity. ? do it if it is “good enough!” • Most unit operations (e.g. distillation) assume phase equilibrium, ignoring transport limitations. ? good enough if things are mixed well and have reasonable residence/contact times. 2 Thermo.key - February 7, 2014 Some T erminology... M= C X i=1 y i ¯ M i Gibbs-Duhem Equation: ? @M @ P ? T,y dP + ? @M @ T ? P,y C X i=1 y i d ¯ M i =0 “Partial Molar” property M - arbitrary thermodynamic property (per mole) N i - moles of species i y i - mole fraction of species i (liquid or gas) Note at constant T and P, C X i=1 y i d ¯ M i =0 “Excess” propertyM E ? M M ideal how much M deviates from the ideal solution behavior. “Residual” property M ideal = C X i=1 y i M i how much M deviates from the ideal gas ( M ig ) behavior. M ig = C X i=1 y i M ig i M R ? M M ig change in M due to adding a differential amount of species i. For ideal mixtures, ¯ M i =M i ¯ M i ? ? @ NM @ N i ? T,P,N j 3 Thermo.key - February 7, 2014 Phase Equilibrium G =G(T,P,N 1 ,N 2 ,...,N C ) Gibbs Energy (thermodynamic property) Phase Equilibrium is achieved when G is minimized. dG= SdT+VdP+ C X i=1 ? @ G @ N i ? P,T,N j dN i µ i ? ? @ G @ N i ? P,T,N j Chemical Potential (thermodynamic property, partial molar Gibbs energy) In phase equilibrium, T (1) = T (2) = ? = T (N) P (1) = P (2) = ? = P (N) For multiple phases in a closed system, G system = N X p=1 G (p) dG system = N X p=1 dG (p) For a nonreacting system, moles are conserved. dG system = N X p=2 " C X i=1 ? µ (p) i µ (1) i ? dN (p) i # T,P dG system = N X p=1 " C X i=1 µ (p) i dN (p) i # T,P T o minimize G, dG=0. But each term in the summation for dG is independent, so each must be zero. µ (p) i =µ (1) i =µ (2) i = ···=µ (N) i For phase equilibrium, the chemical potential of any species is equal in all phases. SHR §2.2 ¯ f i =Cexp ? µ i RT ? Solution: Partial Fugacity C is a temperature- dependent constant. ¯ f (p) i = ¯ f (1) i = ¯ f (2) i = ···= ¯ f (N) i For phase equilibrium, the fugacity of any species is equal in all phases. Problem: µ i !1 P! 0 as N X p=1 dN (p) i =0 dN (1) i = N X p=2 dN (p) i 4 Thermo.key - February 7, 2014 Page 5 Chemical Engineering Thermodynamics A Brief Review SHR Chapter 2 1 Thermo.key - February 7, 2014 “Rate” vs. “State” Thermodynamics tells us how things “end up” (state) but not how (or how fast) they “get there” (rate) Limitations on getting to equilibrium: • Transport (mixing): diffusion, convection ? example: why doesn’t more of the ocean evaporate to make higher humidity in the deserts? (heat transfer alters dew point locally, atmospheric mixing limitations, etc.) • Kinetics (reaction): the rate at which reactions occur are too slow to get to equilibrium ? example: chemical equilibrium says trees (and our bodies) should become CO 2 and H 2 O (react with air) ? example: chemical equilibrium says diamonds should be graphite (or CO 2 if in air) Sometimes, mixing & reaction are suf?ciently fast that equilibrium assumptions are “good enough.” • if we assume that we can use thermodynamics to obtain the state of the system, we avoid a lot of complexity. ? do it if it is “good enough!” • Most unit operations (e.g. distillation) assume phase equilibrium, ignoring transport limitations. ? good enough if things are mixed well and have reasonable residence/contact times. 2 Thermo.key - February 7, 2014 Some T erminology... M= C X i=1 y i ¯ M i Gibbs-Duhem Equation: ? @M @ P ? T,y dP + ? @M @ T ? P,y C X i=1 y i d ¯ M i =0 “Partial Molar” property M - arbitrary thermodynamic property (per mole) N i - moles of species i y i - mole fraction of species i (liquid or gas) Note at constant T and P, C X i=1 y i d ¯ M i =0 “Excess” propertyM E ? M M ideal how much M deviates from the ideal solution behavior. “Residual” property M ideal = C X i=1 y i M i how much M deviates from the ideal gas ( M ig ) behavior. M ig = C X i=1 y i M ig i M R ? M M ig change in M due to adding a differential amount of species i. For ideal mixtures, ¯ M i =M i ¯ M i ? ? @ NM @ N i ? T,P,N j 3 Thermo.key - February 7, 2014 Phase Equilibrium G =G(T,P,N 1 ,N 2 ,...,N C ) Gibbs Energy (thermodynamic property) Phase Equilibrium is achieved when G is minimized. dG= SdT+VdP+ C X i=1 ? @ G @ N i ? P,T,N j dN i µ i ? ? @ G @ N i ? P,T,N j Chemical Potential (thermodynamic property, partial molar Gibbs energy) In phase equilibrium, T (1) = T (2) = ? = T (N) P (1) = P (2) = ? = P (N) For multiple phases in a closed system, G system = N X p=1 G (p) dG system = N X p=1 dG (p) For a nonreacting system, moles are conserved. dG system = N X p=2 " C X i=1 ? µ (p) i µ (1) i ? dN (p) i # T,P dG system = N X p=1 " C X i=1 µ (p) i dN (p) i # T,P T o minimize G, dG=0. But each term in the summation for dG is independent, so each must be zero. µ (p) i =µ (1) i =µ (2) i = ···=µ (N) i For phase equilibrium, the chemical potential of any species is equal in all phases. SHR §2.2 ¯ f i =Cexp ? µ i RT ? Solution: Partial Fugacity C is a temperature- dependent constant. ¯ f (p) i = ¯ f (1) i = ¯ f (2) i = ···= ¯ f (N) i For phase equilibrium, the fugacity of any species is equal in all phases. Problem: µ i !1 P! 0 as N X p=1 dN (p) i =0 dN (1) i = N X p=2 dN (p) i 4 Thermo.key - February 7, 2014 Chemical Potential, Fugacity, Activity SHR §2.2.1 & Table 2.2 P s i vapor pressure of species i. y i x i vapor mole fraction of species i. liquid mole fraction of species i. For phase equilibrium, the chemical potential, fugacity, activity (but not activity coef?cient or fugacity coef?cient) of any species is equal in all phases. ThermodynamicQuantity De?nition Description Limitingcaseofideal gasandidealsolution Chemicalpotential µ i ? ? ?G ?N i ? T,P,N j Partialmolarfreeenergy, g i µ i = g i Partialfugacity f i = Cexp( µ i/(RT)) Thermodynamicpressure f iV = y i P Fugacitycoef?cientofa purespecies f i ? f i/P Deviationoffugacitydueto pressure f iV = 1.0 f iL = P s i /P Fugacitycoef?cientofa speciesinamixture f iV ? f iV /(y i P) f iL ? f iL /(x i P) Deviationstofugacitydue topressureandcomposition f iV = 1.0 f iL = P s i /P Activity a i ? f i /f o i Relativethermodynamic pressure a iV = y i a iL = x i Activitycoef?cient g iV ? a iV/y i g iL ? a iL/x i Deviationoffugacitydueto composition. g iV = 1.0 g iL = 1.0 ¯ f iL = ¯ iL x i P = iL x i f o iL ¯ f iV = ¯ iV y i P = iV y i f o iV ¯ f iL = ¯ f iV ) y i x i = ¯ iL ¯ iV we will use this on the next slide... ¯ f i = f i for a pure species 5 Thermo.key - February 7, 2014Read More

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