# Top-rated ScreenCasts

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When expressing the derivative of the total Gibbs energy by chain rule, there is one particular partial derivative that relates to each component in the mixture: the "chemical potential." By adapting the derivation from Chapter 9 of the equilibrium constraint for pure fluids, we can show that the equilibrium constraint for mixtures is that the chemical potential of each component in each phase must be equal. That is fine mathematically but it is not very intuitive. By translating the chemical potential into a rigorous definition of fugacity of a component in a mixture, we recognize that an equivalent equilibrium constraint is that the fugacity of each component in each phase must be equal. (8min, Live, uakron.edu) This offers the intuitive perspective of, say, molecules from the liquid escaping to the vapor and molecules from the vapor escaping to the liquid; when the "escaping tendencies" are equal, the phases experience no net change and we call that equilibrium.

Solid-liquid Equilibria using Excel (7:38min, msu)

The strategy for solving SLE is discussed and an example generating a couple points from Figure 14.12 of the text are performed. Most of the concepts are not unique to UNIFAC or Excel. This screeencast shows how to use the solver tool to find solubility at at given temperature.

VLE Routines - General Strategies (4:49) (msu.edu)

Deciding which routine to use is more challenging than it appears. Also understanding the strategy used to solve the problems is extremely helpful in being able to develop the equations to solve instead of trying to memorize them.

Concepts for General Phase Equilibria (12:33) (msu.edu)

The calculus used in Chapter 6 needs to be generalized to add composition dependence. Also, we introduce partial molar properties and composition derivatives that are not partial molar properties. We introduce chemical potential These concepts are used to show that the chemical potentials and component fugacities are used as criteria for phase equilibria.

Derive the internal energy departure function (uakron.edu, 20min) for the following EOS:
P = (RT(1+V1.5)/V1.5)*(1+sqrt(V)) - a/(V^2T^1.3)/(1+sqrt(V)) This sample derivation is more complicated than average, but the usual procedure still works. We begin by rearranging to obtain an expression for Z and finding the Helmholtz departure, then differentiating to get the internal energy.

Comprehension: Given (A-Aig)TV/RT = -2ln(1-ηP) - 16.49ηPβε/[1-βε(1-2ηP)/(1+2ηP)^2 ]

1. Derive the internal energy departure function.

2. Derive the expression for the compressibility factor.

3. Solve the EOS for Zc.

The Flory and Flory-Huggins Models (7:05) (msu.edu)

Flory recognized the importance of molecular size on entropy, and the Flory equation is an important building block for many equations in Chapter 13. Flory introduced the importance of free volume. The Flory-Huggins model combines the Flory equation with the Scatchard-Hildebrand model using the degree of polymerization and the parameter χ. The Flory-Huggins model is used widely in the polymer industry.

Comprehension Questions:

Assume δP=δS for polystyrene, where δS is the solubility parameter for styrene. Also, polystyrene typically has a molecular weight of about 15,000. Room temperature is 25°C.

1. Estimate the infinite dilution activity coefficient of styrene in polystyrene.
2. Estimate the infinite dilution activity coefficient of toluene in polystyrene.
3. Estimate the infinite dilution activity coefficient of acetone in polystyrene.
4. Which of the above would be the "best" solvent for polystyrene? Explain quantitatively.

Entropy Balances: Solving for Turbine Efficiency Sample Calculation. (uakron.edu, 10min) Steam turbines are very common in power generation cycles. Knowing how to compute the actual work, reversible work, and compare them is an elementary part of any engineering thermodynamics course.

Comprehension Questions:

1. An adiabatic turbine is supplied with steam at 2.0 MPa and 600°C and it exhausts at 98% quality and 24°C. (a) Compute the work output per kg of steam.(b) Compute the efficiency of the turbine.

2. A Rankine cycle operates on steam exiting the boiler at 7 MPa and 550°C and expanding to 60°C and 98% quality. Compute the efficiency of the turbine.

14.09 - Numerical procedures for binary, ternary LLE Click here. 100 1

LLE flash using Matlab/Chap14/LLEflash.m (5:54) (msu.edu)

An overview of the LLE flash routine in Matlab, including an overview of the program logic and then an example of how to run the program.

See also - Supplement on Iteration of LLE with Excel and Matlab.

Example 17.4 and 17.5 solved using Kcalc.xlsx (6:01) (msu.edu)

The full form of the temperature dependence of Ka is implemented in Kcalc.xlsx and Kcalc.m. This screecast covers the use of Kcalc.xlsx for Example 17.4 and Example 17.5 of the textbook.

Comprehension Questions:

1. CO and H2 are fed in a 2:1 ratio to a reactor at 500K and 20 bars with a catalyst that favors only CH3OH as its product. Calculate ΔGRº and ΔHRº.
2. CO and H2 are fed in a 1:1 ratio to a reactor at 500K and 20 bars with a catalyst that favors only CH3OH as its product. Calculate ΔGRº and ΔHRº.
3. CO and H2 are fed in a 1:1 ratio to a reactor at 600K and 20 bars with a catalyst that favors only CH3OH as its product. Calculate ΔGRº and ΔHRº.
4. CO and H2 are fed in a 1:1 ratio to a reactor at 500K and 20 bars with a catalyst that favors only CH3OH as its product. Calculate ΔGTº and ΔHTº. Check your answer for ΔGTº using the value given for Ka in Example 17.1.
5. CO and H2 are fed in a 1:1 ratio to a reactor at 600K and 10 bars with a catalyst that favors only CH3OH as its product. Calculate Ka, ΔGTº and ΔHTº.
6. CH3OH is fed to a reactor at 200ºC and 1 bar with a catalyst that produces CO and H2. Calculate Ka, ΔGTº and ΔHTº for this reaction and compare to the literature values given in Example 17.6 of Section 17.10.
7. CH3OH is fed to a reactor at 300ºC and 1 bar with a catalyst that produces CO and H2. Calculate Ka for this reaction and compare to the value given in Example 17.6 of Section 17.10. Give two reasons why the two estimates are not identical.