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01.5 Real Fluids and Tabulated Properties Click here. 100 2

Double interpolation (uakron, 8min) is exactly what it sounds like: to find a steam property when neither the pressure nor temperature are among the tabulated values, you need to interpolate twice. We interpolate first on pressure, then on temperature. It is a bit tedious, but straightforward.

Comprehension Questions:
1. Describe how you would use double interpolation to obtain H if given T=275 C and P=0.45MPa.
2. Describe how you would use double interpolation to obtain H if given T=275 C and V=0.555m3/kg.

09.05 - Fugacity and Fugacity Coefficient Click here. 100 1

What is fugacity? (10min) ( Defines fugacity in terms of Gibbs Energy and describes the need for defining this new property as a generalization of how pressure affects ideal gases.
Comprehension Questions
1. The phases in this video start with concentrations 0.0007kg/L and 1.0 kg/L, when not at equilibrium. What are the equilibrium concentrations?
2. Why is concentration an unreliable indicator for the direction of mass transfer?
3. Name two indicators for the direction of mass transfer that are superior to concentration.  

17.05 - Effect of Pressure, Inerts, Feed Ratios Click here. 100 1

Partial pressures and reactor sizing are among the keys to chemical engineering calculations (, 7 min, review from Section 1.6). Partial pressures (, 7 min) also play an essential role in reaction equilibrium calculations. Partial pressure calculations basically involve straightforward mass balances, but specific vocabulary and a need for systematic precision can cause difficulty. The calculations involve six elements that must be carefully computed:

(1) Stoichiometry - the reaction equation must be stoichiometrically balanced such that the number of atoms of each element are the same on both sides of the equation. This balance is achieved by adjusting the stoichiometric coefficients. The change in the number of moles of each component must be in correct stoichiometric proportions relative to the "key component." Inert compounds (see below) are NOT included in the stoichiometric equation. For the example in this presentation, the objective of the reactor is to oxidize carbon monoxide (CO) in a catalytic converter by reacting it with oxygen (O2). So, CO + 0.5 O2 = CO2.
(2) Limiting reactant (aka. "key component") - It is common to feed an excess of one of the components in order to promote complete conversion of the other components. The limiting reactant is the component that is NOT in excess. For this example, O2 is fed in excess so that CO conversion can be promoted. CO becomes the limiting reactant in that case and conversion must be computed relative to CO, NOT O2. If you think about it, expressing the conversion with respect to the excess component would mean that 100% conversion could result in a negative mole number for the limiting reactant. Such an implication is obviously physically impossible (and potentially embarrassing if you appear not to know that).
(3) %Excess - The number of moles of an excess component in the feed is (1+Xs) times the stoichiometric amount relative to the key component, where the stoichiometric amount is the number of moles necessary to perfectly balance the key component, and Xs is the fractional form of the %excess. For this example,  the stoichiometric ratio of CO:O2 would be 1:0.5 and for 50% excess, Xs = 0.50, and the actual ratio would be 1:0.75.
(4) %Conversion - the %conversion is the fraction of the entering amount of the limiting reactant that is transformed into product(s). Note that this might be different from the "extent of reaction," ξ. For example, if 50 moles/h of CO enter the reactor and the conversion is 90%, then 5 moles of CO exit the reactor. If you express the number of moles of CO as 50-ξ, you might conclude that the moles of CO exiting the reactor is 49.1. Take a minute to think about what the words mean before you start to calculate, then make a mental estimate of what the results should be, then get out your calculator. Another common mistake is to apply the % conversion to all the components, wrongly including the excess component. For example, if 45 moles of CO react, then 22.5 moles of O2 react. With 50% excess O2 in the feed, the O2 exiting should be 37.5-22.5=15, NOT 3.75. This is what it means to be careful and systematic. You must compute the conversion of limiting reactant first, then compute the conversion of other components relative to the limiting reactant.
(5) Inerts - These are components that may enter the reactor by coincidence or convenience but do not participate in the reaction. Therefore, their number of moles exiting the reactor is simply equal to their number of moles entering the reactor. A common mistake is to apply the %conversion to all components entering the reactor, including the inerts. In this example, the source of O2 is air, with roughly 4:1 ratio of nitrogen (N2) to O2. The N2 is inert.
(6) Total Pressure - Once the mole numbers and mole fractions have been computed, don't forget to multiply the mole fractions by the total pressure to get the partial pressure. The partial pressure is equal to the mole fraction only in the case that the reactor operates at 1.00 bar.

Comprehension Questions:

1. What is the value of the total pressure (bar) applied in the presentation of this example?
2. What equation is used to compute the mole number of O2? What is the final overall equation used to compute PO2?
3. Suppose 100 moles/h of ammonia (NH3) at 100bars is to be produced from N2 and hydrogen (H2) with 10% excess N2. Methane (CH4) is included with the N2+H2 as a result of the synthesis process with a ratio of 1:10 CH4:H2. (a) Write a stoichiometrically balanced equation (b) Identify the limiting reactant (c) Calculate the number of moles and partial pressures of each component entering the reactor. (d) Calculate the number of moles and partial pressures of each component exiting the reactor assuming 25% conversion.

10.07 - Nonideal Systems Click here. 100 1

Nonideal Mixtures (4:58) (

Raoult's law is an easy way to calculate VLE, but it is inaccurate for most detailed VLE calculations. This screencast provides an overview of the problems, and introduces the concept of an azeotrope. The VLE K-ratio is shown to be less than one or greater than one dependenting on the overall system concentration relative to the azeotrope composition where K=1. The concept of positive and negative deviations is introduced.

11.02 - Calculations with Activity Coefficients Click here. 100 2

This example shows how to incorporate activity calculations into Excel for solutions that follow the Margules 1-parameter (M1) model.(9min,

You should be able to adapt this procedure along with the procedure for the multicomponent ideal solutions to create a multicomponent M1 model. If you are having trouble, the video for the multicomponent SSCED model illustrates a very similar procedure. You can check your answers by putting in the same component twice. For example, instead of an equimolar binary mixture, input a quaternary mixture with 0.25 moles of methanol, 0.25 methanol (ie. type it as if it was another component), 0.25 of benzene and 0.25 of benzene. If you don't get the same results as for the binary equimolar system, check your calculations.Note: This is a companion file in a series. You may wish to choose your own order for viewing them. For example, you should implement the first three videos before implementing this one. Also, you might like to see how to quickly visualize the Txy analog of the Pxy phase diagram. If you see a phase diagram like the ones in section 11.8, you might want to learn about LLE phase diagrams. The links on the software tutorial present a summary of the techniques to be implemented throughout Unit3 in a quick access format that is more compact than what is presented elsewhere. Some students may find it helpful to refer to this compact list when they find themselves "not being able to find the forest because of all the trees."

Comprehension Questions: Assume the SCVP model (Eq. 2.47).
1. Develop a Pxy diagram for the IPA+water system like Figure 10.8c, guessing values of A12 until you match the maximum pressure (azeotrope). What value of A12 did you find? (Hint: A12 is not the same as A12*RT.)
2. Develop a Pxy diagram for the acetone+chloroform system like Figure 10.9c, guessing values of A12 until you match the minimum pressure (azeotrope). What value of A12 did you find? (Hint: A12 is not the same as A12*RT.)
3. Develop a Pxy diagram for the acetone+acetic acid system like Figure 10.9a, guessing values of A12 until you match the pressure at x1=0.5 (305mmHg). What value of A12 did you find? (Hint: A12 is not the same as A12*RT.)

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.

08.07 - Implementation of Departure Functions Click here. 100 1

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.

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?

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.

14.09 - 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.