Top-rated ScreenCasts

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01.6 Summary Click here. 100 1

Keys to the Kingdom of Chemical Engineering (, 11min) Sometimes it helps to reduce a subject to its simplest key elements in order to "see the forest instead of the trees." In this presentation, the entire subject of Chemical Engineering is reduced to three key elements: sizing a reactor (, 7min), sizing a distillation column (, 9min), and sizing a heat exchanger (, 9min). In principle, these elements involve the independent subjects of kinetics, thermodynamics, and transport phenomena. In reality, each element involves thermodynamics to some extent. Distillation involves thermodynamics in the most obvious way because relative volatility and activity coefficients are rarely discussed in a kinetics or transport course. In kinetics, however, the rate of reaction depends on the partial pressures of the reactants and their nearness to the equilibrium concentrations, which are thermodynamical quantities. In heat exchangers, the heat transfer coefficient is important, but we also need to know the temperatures for the source and sink of the heat transfer; these temperatures are often dictated by thermodynamical constraints like the boiling temperature or boiler temperature required to run a Rankine cycle (cf. Chapter 5). In case you are wondering about the subject of "mass and energy balances," the conservation of mass is much like the conservation of energy; therefore, we subsume this subject under the general umbrella of thermodynamics. Understanding the distinctions between thermodynamics and other subjects should help you to frame a place for this knowledge in your mind. Understanding the interconnection of thermodynamics with subjects to be covered later should help you to appreciate the necessity of filing this knowledge away for the long term, such that it can be retrieved at any time in the future.

If you would like a little more practice with reactor mass balances and partial pressure, more screencasts are available from, MichiganTech, and popular chemistry websites.

05.4 - Refrigeration Click here. 100 2

Refrigeration Cycle Introduction (, 3min) explains each step in an ordinary vapor compression (OVC) refrigeration cycle and the energy balance for the step. You might also enjoy the more classical introduction (USAF, 11min) representing your tax dollars at work. The musical introduction is quite impressive and several common misconceptions are addressed near the end of the video.
Comprehension Questions: Assume zero subcooling and superheating in the condenser and evaporator.
1. An OVC operates with 43 C in the condenser and -33 C in the evaporator. Why is the condenser temperature higher than than the evaporator temperature? Shouldn't it be the other way around? Explain.
2. An OVC operates with 43 C in the condenser and -33 C in the evaporator. The operating fluid is R134a. Estimate the pressures in the condenser and evaporator using the table in Appendix E-12.
3. An OVC operates with 43 C in the condenser and -33 C in the evaporator. The operating fluid is R134a. Estimate the pressures in the condenser and evaporator using the chart in Appendix E-12.
4. An OVC operates with 43 C in the condenser and -33 C in the evaporator. The operating fluid is R134a. Estimate the pressures in the condenser and evaporator using Eqn 2.47.
5. An OVC operates with 43 C in the condenser and -33 C in the evaporator. Assume the compressor of the OVC cycle is adiabatic and reversible. What two variables (P,V,T,U,H,S) determine the state at the outlet of the compressor?

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

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

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.

10.02 - Vapor-Liquid Equilibrium (VLE) Calculations Click here. 100 2

VLE Routines - General Strategies (4:49) (

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.

10.08 - Concepts for Generalized Phase Equilibria Click here. 100 1

Concepts for General Phase Equilibria (12:33) (

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.

08.07 - Implementation of Departure Functions Click here. 100 2

Derive the internal energy departure function (, 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.

12.04 - The Flory-Huggins Model Click here. 100 3

The Flory and Flory-Huggins Models (7:05) (

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.

05.2 - The Rankine cycle Click here. 100 1

Rankine Cycle Introduction (, 4min) The Carnot cycle becomes impractical for common large scale application, primarily because H2O is the most convenient working fluid for such a process. When working with H2O, an isentropic turbine could easily take you from a superheated region to a low quality steam condition, essentially forming large rain drops. To understand how this might be undesirable, imagine yourself riding through a heavy rain storm at 60 mph with your head outside the window. Now imagine doing it 24/7/365 for 10 years; that's how long a high-precision, maximally efficient turbine should operate to recover its price of investment. Next you might ask why not use a different working fluid that does not condense, like air or CO2. The main problem is that the heat transfer coefficients of gases like these are about 40 times smaller that those for boiling and condensing H2O. That means that the heat exchangers would need to be roughly 40 times larger. As it is now, the cooling tower of a nuclear power plant is the main thing that you see on the horizon when approaching from far away. If that heat exchanger was 40 times larger... that would be large. And then we would need a similar one for the nuclear core. Power cycles based on heating gases do exist, but they are for relatively small power generators.
     With this background, it may be helpful to review the relation between the Carnot and Rankine cycles. (, 6min) The Carnot cycle is an idealized conceptual process in the sense that it provides the maximum possible fractional conversion of heat into work (aka. thermal efficiency, ηθ).
Comprehension Questions:
1. Why is the Carnot cycle impractical when it comes to running steam through a turbine? How does the Rankine cycle solve this problem?
2. Why is the Carnot cycle impractical when it comes to running steam through a pump? How does the Rankine cycle solve this problem?
3. It is obvious which temperatures are the "high" and "low" temperatures in the Carnot cycle, but not so much in the Rankine cycle. The "boiler" in a Rankine cycle actually consists of "simple boiling" where the saturated liquid is converted to saturated vapor, and superheating where the saturated vapor is raised to the temperature entering the turbine. When comparing the thermal efficiency of a Rankine cycle to the Carnot efficiency, should we substitute the temperature during "simple" boiling, or the temperature entering the turbine into the formula for the Carnot efficiency? Explain.

14.10 Solid-liquid Equilibria Click here. 100 3

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.

07.06 Solving The Cubic EOS for Z Click here. 100 2

6. Solving for density (, 9min) An alternative to solving directly for Z is to solve for density then compute Z=P/(ρRT). This requires iterative solution and it is not very expedient for repetitive calculations, but it requires no rearrangement of the EOS and it is easy to visualize. This sample calculation is illustrated here for the vdW EOS, solving for the density of propane as: (a) liquid 25C,11bars (b) liquid 62C,35bars (c) vapor at 80C and 30bars.

Comprehension Questions:

1. Solve for the liquid density (mol/cm3) of n-pentane at 62C and 2.5 bars using the vdW EOS.
2. Solve for the Z-factor of liquid n-pentane at 62C and 2.5 bars using the vdW EOS.
3. What's the value of the Z-factor at 80C and 30 bars according to this presentation?