Problem 2.3 on the final

In part (a), the excess entropy should be a function of T and v (or rho), since these are the constant parameters of the canonical ensemble.

In part (b), the term "integral equation" simply means that your final equation that determines rho(x) will contain an integral involving rho(x). "Self-consistent" suggests that one side of this equation will be rho(x) = ....

MSS

PS9, problem 5c

The problem should indicate that the system is held at T0, not both T0 and P0.
MSS

PS9 correction, again

I meant to say that on problem 4 it should refer to problem 2, not problems 3 or 5.
MSS

TA office hour

I am available at the same time on Thursday.

PS9

Class,

PS9 is online in case you did not get a paper copy in class. Also, there is a correction in problem 4. Where it refers to "problem 5" it should read instead "problem 3".

MSS

Office hours tomorrow

Dear class,

I will hold office hours tomorrow immediately after our class, at 1:30pm.

MSS

TA Office hour

Due to the change of PS due date, I think I need to change TA office hour to Monday after class, 3:00-4:00.

Is it good for you? 

Hint for 2.2 on midterm

You are looking for the variation of T with V in a certain kind of process. Thus, you might want to consider a derivative of the form (dT/dV).
MSS

Extra office hours

If there are further questions about PS6, I will be available in my office from 3-4pm today (11/6) to answer questions.
MSS

Correction to today's lecture

I inadvertently omitted a part of an equation derived today in class.

Recall that we had for the pure-component chemical potential:

d(mu/T)/dT = -h / T^2

If we assume h is constant with temperature, we can integrate from Tm to T, where Tm is the pure melting temperature, to get:

mu(T)/T - mu(Tm)/Tm = h * (1/T - 1/Tm)

Or, rearranging

mu(T) = h * (1 - T/Tm) + mu(Tm) * (T/Tm)

In class, I left off the rightmost term. When we apply this equation to both the pure liquid and crystal and take the difference, we get:

mu_L(Tm) - mu_X(Tm) = (h_L - h_X) * (1 - T/Tm) + [mu_L(Tm) - mu_X(Tm)] * (T/Tm)

However, mu_L(Tm) = mu_X(Tm) by the conditions for phase equilibrium at Tm, so the rightmost term vanishes.

MSS

Clarification on osmotic pressure

In today's lecture, we came used the following equation:

Pi = kB T x_solute / v

Then, we substituted in to get

Pi = kB T x_solute * N_solvent / V_solvent

Here, N_solvent and V_solvent apply to the right hand side of the osmotic pressure experiment; that is, the solution (non pure) phase. Since v is for the pure phase, here we are making the approximation that the molar volume of the solvent is the same on the left and right hand side of the experiment. This is almost certainly the case for small, dilute amounts of solvent.

MSS

Saturation pressure

In case the term "saturation pressure" is causing you confusion, here is something to think about. The saturation pressure at a given temperature is the pressure at which a gas will turn into a liquid upon compression; alternatively, it is the pressure at which a liquid will vaporize upon expansion. The "saturation temperature" is similar: at a given pressure, it is the temperature at which a gas will condense upon cooling, or a liquid will boil upon heating. Thus, there is nothing more complex about a saturation pressure than simply the pressure of liquid-vapor phase equilibrium.

MSS

PS4 5b

In the PS4 5b, showing dCp/dV=0, some used

d/dV [ (dS/dT)_P ]_T = d/dT[ (dS/dV)_T]_P,

However, this converstion does not give 0.

You can't switch the order of the derivatives because they are not compatible (you have different independent variables). d/dV [ (dS/dT)_P ]_T not equal to d/dT[ (dS/dV)_T]_P

To do the switch, you have to have the same set of independent variables. You could do this, for example:

d/dV [(dS/dT)_P]_T
= (dP/dV)_T * d/dP[(dS/dT)_P]_T
= (dP/dV)_T * d/dT[(dS/dP)_T]_P

And then you will find that the right hand term is zero.

Hope that helps,
Scott

Test yourself

Just with a quick look, without consulting any textbooks or resources, see if you are able to fill in the following in your head:

dS = ? ; S = ? ; maximized at constant ?,?,?
dE = ? ; E = ? ; minimized at constant ?,?,?
dA = ? ; A = ? ; minimized at constant ?,?,?
dH = ? ; H = ? ; minimized at constant ?,?,?
dG = ? ; G = ? ; minimized at constant ?,?,?

If you weren't able to rattle off the correct answers without hesitation, I would strongly suggest you invest a short amount of time committing these equations to memory. The reason that I want you to have these equations so down pat is that having them at the forefront of your memory will enable you to analyze and tackle problems so much more efficiently: for example, you will be able to quickly access Maxwell relations, integrate out thermodynamic properties, and connect derivatives to measurables, without having to flip through notes and handouts. The ability to call upon such basic analyses quickly and intuitively is not only important to this course, but a key part of your success as a careful researcher, no matter what field.

As we progress into new material in the next few weeks, it's critical that you don't get lost in the new concepts as a result of difficulties with these fundamental equations. Just a little advice,

Cheers,
MSS

Notes on Maxwell relations

If you're confused about Maxwell relations, here is something of an additional explanation. Consider the derivative:

(dN/dT)_(V,mu)

The first thing you want to do is identify from which thermodynamic potential this derivative stems. To do that, we look at the independent variables. These are the derivative variable (T) as well as the constant condition variables (V,mu). Thus, we want a function whose natural variables are T,V,mu.

We don't actually know offhand such a function, but we can construct one. Look at the Helmholtz free energy,

dA = -S dT -P dV + mu dN

Here we have independent variables T,V, and N. We need to swap mu for N. To do this, we just perform a Legendre transform to get some new potential Y:

Y = A - mu N
dY = -S dT - P dV - N dmu

Now we can use Y to develop our Maxwell relation. We see that our original function is just

-d2Y/dT dmu = -d(dY/dmu)/dT = dN/dT

If we switch the order of the derivatives, we get

-d2Y/dmu dT = -d(dY/dT)/dmu = dS/dmu

Therefore, we find that:

(dN/dT)_(V,mu) = (dS/dmu)_(T,V)

So the general rule of thumb is:

1. Determine which thermodynamic potential underlies the derivative of interest.
2. Construct that potential via Legendre transforms, if necessary.
3. Find the second derivative giving rise to your derivative of interest, and evaluate it in two different cases where the order of the derivatives is switched.

There is one point on which you must be careful. Not every derivative of a thermodynamic property can be related by a Maxwell relation. For example, consider:

(dT/dS)_(V,N)

This derivative comes from the second derivative of the energy:

d2E/dS2

Note that this is not a cross-derivative, between two independent variables. Therefore (dT/dS)_(V,N) cannot be related to another derivative.

MSS

Solutions to practice problems today

Hi class-

You can find solutions to the practice problems you worked today in teams at:

www.engr.ucsb.edu/~shell/che210a/practice4.pdf

Cheers,
MSS

PS3, problem 5b/c

Here's a slightly clarified explanation of this part of the problem. Here, by constant pressure process, we mean that the entire system is at the same pressure throughout the process. That is, even though it may not be at equilibrium, the pressure that it exerts on the container walls remains constant. This should be all that you need to simplify the first law (differential form) to show that constant pressure processes are also constant enthalpy processes. Note that we don't need to assume quasi-static or reversible behavior; we only need to know that the pressure is constant.
MSS

PS3, problem 4c

In this problem, you are considering the case of a quasi-static, adiabatic expansion. A good place to start is by writing the differential first law for this problem, and then simplifying the remaining terms using measured relations for an ideal gas. You might have to integrate at some point. Hint, hint.
-MSS

PS3, problem 3

For part (b) in this problem, express your density of states and entropy functions in terms of x rather than L. This will simplify your life. Then, when you want to take an L-derivative of the entropy, you can use the chain rule.
-MSS

PS3

Hi all-

On problem 3, what is meant by "end-to-end length" is the distance between the first anchor of the first segment and last anchor of the last segment, regardless of what is happening in between. In other words, L is only determined by the end points, and does include parts of the chain that move to regions in the x direction beyond the end point.

Scott

Class on Friday

Dear ChE210A class members-

I am going to cancel class this Friday so that we all can attend the Chemical Engineering Graduate Student Symposium. I fully expect to see each of you there to support this important event.

You may place your problem sets in Dong Hun's mailbox on Friday. I will distribute PS3 by email.

Cheers,
MSS

Updated office hours

Dear class,

Dong Hun needed to make a slight update to his office hours, due to a course conflict. The revised schedule is below:

me - Wednesdays 10:30am-12pm, 3321 Engineering II
Dong Hun - Thursdays 12-1:30pm, 3315 Engineering II

MSS

Solutions to practice problems today

Hi class-

If you were are still feeling a bit unsure about the practice problems today, here is a written explanation.

Problem 1:

In this problem there are M things from which you are choosing: the sites to which you will assign particles. Assume you take the following procedure. You start with particle 1 and assign it to site 1 that you choose, then particle 2 to site 2, and so on and so forth.

(a) Since the particles are all identical, the order in which we choose the sites does not matter, since they will all eventually be assigned to a blue particle. Therefore, we use the combination formula, MCN = M!/N!(M-N)!.

(b) Here the order that we pick sites *does* matter because the particles are all colored. Therefore, we use the permutation formula, NPM = M!/(M-N)!.

(c) Here, we are picking sites with replacement; that is, it doesn't matter whether or not we have chosen a site before or not. Thus, for distinguishable, we have that there are M options for us to place each one. In total, there are M^N options.

(d) Now we have to consider that this is the same case as (c), but if we swap the locations of two particles, then we don't have a new configuration. In the case where M >> N, for the vast majority of configurations, we never have more than one molecule on a site. In this case, we have over-counted configurations using the M^N rule by a factor that depends on the number of ways there are to swap around particle identities, which is N!. Thus we have that the number of conformations is M^N/N!. On the other hand if M is of the same order as N, then we will have lots of configurations in which there is more than one particle in each lattice spot. In such cases, we have *not* overcounted by N! but by some number that is less that that factor. For example, consider the case when all particles are on the same lattice site. In the M^N counting of all configurations, this one is included only once. Therefore, we have not overcounted it at all. This is a subtlety to this problem that we did not discuss in class.

Problem 2:

Essentially, to count classical microstates (systems with continuous degrees of freedom), we have to introduce some discretization into space, which is what we did in this problem. Ultimately we will let this discretization be extremely small. Here, we want to count the number of microstates (combinations of x,y values) that lie witin a certain energy range U to U+dU. The potential energy function is u=a r^2.

To start, we realize that the number of combinations of x and y in the perscribed energy range is related to the area enclosed by two energy contours (U and U+dU) in the x,y plane: it is given by the number of grid points of our discretized x and y axes that lie in this area. For sufficiently small discretization dx, the number of points is just the area A divided by the size of a grid square, dx^2. Now the problem of finding the number of microstates is the same thing as finding the area between these energy contours.

Each contour is a circle, since u=a(x^2+y^2). Therefore, we can find the squared radii of the two circles as, r^2 = U/a and (U+dU)/a, respectively. The area of each circle is pi times these values, and the intermittant area is the difference between the two:

A = pi*((U+dU)/a - U/a)

The number of microstates is A/dx^2, such that our final answer simplifies to

Omega = pi * dU/ (a * dx^2).

We could have also determined this answer using a different approach: first finding the total number of microstates with energy less than or equal to U. This is simply the area of a circle of radius r^2 = U/a, divided by dx^2:

Xi = pi*U/(a* dx^2)

The number of microstates that lie within an energy range dU centered around an energy U is given by the derivative of this function

dXi/dU = # of microstates per energy U

(dXi/dU) * dU = # of microstates in a window of width dU centered around U

Therefore,

Omega = (dXi/dU)*dU = pi * dU / (a * dx^2)

And this is the same answer as before.

-MSS

Office hours

Dear 210A,

The office hours for our course are as follows:

me - Wednesdays 10:30am-12pm, 3321 Engineering II
Dong Hun - Thursdays 2-3:30pm, 3315 Engineering II

And of course, feel free to contact either of us if you would like to meet individually at another time, or stop by, if my office door is open.

Cheers,
MSS

Final exams

Class,

If you did not get back your final exams, you may stop by my office this week to get them.

In addition, if you have subscribed to this blog by email, you may want to remove yourself at this point, as it will be used for future courses that I teach.

Good luck in your future studies,

MSS

Final exam and course grades

Dear class,

Summaries of grade distributions for the final exam and the course grades can be found at the course website.

Also, I will be handing out the graded exams the first day of the 110B course next week. We will also discuss the solutions then.

MSS

Leftover HW9 sets

If you didn't get your graded HW9 back, you can pick it up at my office tomorrow morning.

MSS

HW9

HW9 solutions have been posted and the graded problem sets will be distributed at the review session.

MSS

Review Session

Badri and I will be holding a Review Session on Thursday Evening at 7 pm in Engr. II 3361. We will mainly be answering questions so bring any questions that you have.

I am not planning on holding any official office hours today or tomorrow, so if you have questions you want answered before the review session (or if you can't make the review session), feel free to email me (mblack@engr.ucsb.edu) with questions or to set up a meeting.

-Matt

How to study for the final

I've had a couple of folks inquire as to the best way to study for the final. Here are my thoughts on that.

It is very easy to feel like there are a tremendous number of equations, processes, and special cases to deal with, particularly in the last few chapters on engines that we covered this week and last. You should not try to memorize each and every case. Instead, you should try to understand what are the systematic approaches you take to solve these problems so that you can tackle any problem!

So what should you know? How should you approach a problem?

(1) Sketch out complete diagrams. Make flow diagrams, labeling streams in complete detail, if it is an open system. Make PV diagrams, labeling all points and processes, including work and heat flows and their signs. Making complete diagrams should tell you immediately what is known and what is unknown. Moreover, this process helps you translate the textual description of the problem into a clearer more organized format that will be easier to work from.

(2) Apply the fundamental equations. We wrote these out for you on a handout so you could see what these are. Each and every problem you have tackled can essentially be worked out starting from these equations. The book spends a lot of time deriving special case formulas based on these, but it is much more productive for you to understand how to get from the fundamental equations to the final answer without having to flip through the book looking for the specific case formula (which can often cause you to make errors during exams if you are time crunched and don't read the text surrounding the equations). On the exam, our main intent for having open book is so that you have access to reference data like heat capacities, equation of state parameters, etc. You should not have to flip through the book to find equations for each problem.

(3) Use variables as much as possible, and plug in numbers at the very end. This helps prevent numerical and units mistakes.

(4) Do consistency checks. Do the units work out? Are the numbers physically reasonable?

In studying for the final, I would recommend you go back to all of the homework problem statements. Look at the problems, without looking at the solutions, and ask yourself: "Can I solve this right now from the fundamental equations, without looking up a bunch of formulas in the book or reading the text?" If you can do that, you are good to go. If you can't, I would try to sketch an outline for the solution to that problem again starting from fundamentals.

A good way to test yourself would be to pick one of the application-oriented problems from the book in the later chapters we've been studying, and see how far you can get through it without having to rely on text in the book.

Prof. Chmelka and I will both be around next week if you would like to see us. You can come in if our office doors are open; otherwise, feel free to shoot us an email to find a good time to meet.

MSS

Some information on summer jobs

Dear class-

I happen to come across a couple of links regarding summer internship and research opportunities, and I'm passing these on to you in case you are interested. Now actually is the time to get going on these, as many deadlines for internships and research occur in March.

The NSF runs a program called Research Experience for Undergraduates (REU). Many departments in different scientific and engineering disciplines around the country run this program. I believe Professor Chmelka mentioned one such program at U. Delaware earlier in the course. The following is a student-centric website provided by NSF that has a search function for all of the participating departments, as well as links to the individual programs:

http://www.nsf.gov/crssprgm/reu/

If you are interested in an internship, the AIChE website has a list of links to companies that commonly offer such opportunities to chemical engineering students. For many of these, you can click through to get to the company website, and then follow the link to "careers" to get to an internship-specific information section. The list is at:

http://www.aiche.org/Students/Careers/internships.aspx

MSS

TA Office Hours Switch

TA Office hours will be held next week on Wednesday at 4 and Thursday at 9 in the TA office (there will be no office hours on Monday or Tuesday). I will also be available immediately after class on Wed. If you have other questions, you can email me (mblack@engr.ucsb.edu).

Thanks,

Matt

Calculating residual properties using equations of state

Yesterday in class I worked an example for how to calculate residual properties using the van der Waals equation of state. The first step of performing such calculations was to convert the integrals for the residual properties to a form that allowed easy substitution of an equation of state in the form P(T,rho). This meant we had to convert the integrals in P to integrals in rho.

I know we went through this very quickly, due to time constraints, but just keep in mind that this is just a mathematical manipulation of the equations to swap rho for P. It is just like any other variable substitution you have done in integral calculus. The book has a detailed derivation on pages 216 and 217 that shows exactly how to do this manipulation.

The final equations for the residual properties are given in 6.58, 6.59, and 6.60. These equations are fundamentally no different than those we discussed earlier (shown in the book in 6.46, 6.48, and 6.49); they simply involve integrals using rho rather than P.

This should be helpful in your working with problem 44 on the homework.

MSS

A question about ideal gas heat capacities

I was asked why the heat capacity in problem 38 is Cp = 5/2R, rather than 7/2R. Here is a short explanation of heat capacities for ideal gases:

Generally, an ideal gas is something that obeys PV=RT and that has U(T) only (i.e., U is not a function of pressure or volume). Saying that something is an ideal gas does not immediately specify the heat capacities, although it does say that they are independent of P or V since U(T) and H(T) only.

In some specific instances, however, we can use some approximations to the heat capacities for the ideal gas if we know about the molecular structure:

Cp=5/2R for a monoatomic system
Cp=7/2R for a diatomic system

Thus it depends on the kind of ideal gas that you have (monoatomic or diatomic). So in problem 38, it is likely that the problem refers to an ideal gas with a monoatomic molecular structure.

Why should diatomic gases have a higher heat capacity? Recall that the heat capacity measures how much the internal energy or enthalpy changes with temperature. In a diatomic ideal gas, energy can be stored in the bond that connects the two atoms, in addition to the kinetic energy of the molecule. By stretching and compressing like a spring, the bond can contain potential energy. Therefore, the heat capacity is higher than for a monatomic gas because there is this additional place to store energy as the temperature increases.

In other more complicated systems, the heat capacity for an ideal gas can be a function of temperature, which is usually well-fitted by the form that we've been using in class. Note, however, that still for an ideal gas, these forms do not have a P or V dependence. This is why they are still called "ideal gas heat capacities". The way in which we figure out what the properties of real gases are, for example the enthalpy, is to first figure out what the ideal gas enthalpy would be at the given temperature and pressure using these heat capacity expressions, and then to add back in the part that might need to account for deviations from ideal gas behavior:

Delta H = Delta Hig + Delta Hresidual

where the Delta Hig comes from the ideal gas heat capacity integral and Delta Hresidual comes form the expressions developed recently in class that use PVT data.

MSS

Common Mistakes on HW7

On problem 32, some people didn't look up the correct Tn to plug into the equations 4.12 and 4.13.
Overall, make sure that you are computing the integrals for Cp correctly. It seemed there a lot of people that had the right set up, but did not get the correct answer.
On 33, some people ignored the fact that there were 3 different deltaH's that needed to be calculated (heating the liquid to 368, delta H of vaporization, and heating the gas from 368 to 500).

On 35, make sure all your signs are correct - the heat is being removed from the system (so Q is negative).

On 36, you only needed to concern yourself with the amount that was reacted. While it was technically correct to cool the steam and non-reacting species to 298 and then heat them up again, you should realize that if they do not react, these actions will cancel each other out (since it was isothermal). So you only need to compute all the deltaH's for the .33 moles of butene that reacts (for a 1 mole basis feed).

-Matt

HW8, problem 43

The efficiency is defined differently in this problem than in the case for a heat engine. The reason for this difference is that what we are putting in is a work (the work required to refrigerate) rather than a heat (the boiler in the Carnot cycle, for example). Here, the efficiency is closer to the "isentropic efficiency" we discussed in class for the turbine. For cases in which we expect the nonideal work to be less than the ideal work (as in a turbine), we have:

eta = (actual work required) / (work required in the ideal reversible case)

On the other hand, for cases in which we expect the nonideal work to be greater (as in refrigeration), we have:

eta = (work required in the ideal reversible case) / (actual work required)

Regardless, eta is always defined so that it is less than one.

In general, efficiencies tend to measure works relative to an ideal case. In the special case of a heat engine (like a Carnot engine), the efficiency we discussed is relative to a sort-of "super ideal" case in which all of the boiler heat is converted to work.

So the general way in which one treats problems like this is almost always the same:
(1) Solve the system as if it were ideal (reversible processes) -- in this case, it is a straightforward application of the open energy balance and the entropy generation equation.
(2) Use the efficiency to relate the actual work to the ideal work.

MSS

Hint on HW8, problem 41

There might be some confusion in this problem about whether or not it is an open system. Even though the problem states that it is a "steady flow" process, none of the values or conditions given involve rates, and in fact, the amount of substance given is a constant (not a rate). Therefore, you might consider whether or not you really think this is an open system.

MSS

Two quick notes from class today

(1) I found a small typo in the heat capacity calculations handout. In the last equation on the first page, there is a T1^2 that should be a T1^3. A corrected version is posted on the website. Also, if you find any more typos, please let me know so that I can correct the document.

(2) In discussing the microscopic derivation of the entropy, I neglected to mention the following: since S = k ln Omega, and since Omega counts the number of configurations of molecules, the second law really says that molecules like to maximize their number of configurations. This underlies the statement that the entropy always increases; molecules never spontaneously decide to choose a smaller subset of configurations than the total number that are available to them.

Earlier we had said that the second law is not about increasing disorder. Instead, it is about increasing numbers of configurations. The two are not the same. For example, if one very quickly cools a liquid below its freezing temperature and then suddenly places it in an isolated container, as the liquid starts to freeze, the entropy increases. This actually happens because there are more configurations when some of the liquid becomes frozen. This seems counterintuitive, since one intuitively wants to think of a liquid as having more configurations than a crystal. But for very low temperature liquids, there are simply less configurations for the molecules than there would be if some of those molecules could vibrate around lattice sites in a crystal. One can even calculate this and show it to be true. Therefore, our intuitive notion of "order" and "disorder" is not really quantitative, and not always related to the number of configurations, which is what is really important.

Question box: more interactive problem sessions?

We received a question box comment asking if we could have more group-worked problems similar to those done during the interactive problem session, presumably during recitation hours.

This is a very useful comment to us, and we're particularly glad to hear that the interactive problem session was well appreciated. In terms of incorporating more such activities during class time, it is a bit challenging for us to do that at this point because it would require a restructuring of the remaining classes (both lectures and recitation), which are tightly balanced to cover the remaining material. This is, however, a helpful note for us in planning out course material for future years.

On the other hand, the problem sets are an excellent source of examples that you can utilize in a manner similar to the interactive problem session. You should of course attempt these problems on your own at first, but working with other students in small groups after this first attempt can help you resolve difficulty and enhance your understanding through discussions. Furthermore, Prof. Chmelka, Matt, and I hold weekly open office hours which you can use by yourself or with a group to work interactively through problem spots in the homework sets. These times have been dedicated explicitly to helping you, and we also enjoy chatting with and getting to know you better.

Cheers,
MSS

Statistics for the midterm

Here is a summary of the midterm results:

Average: 56
Standard deviation: 13.5
High: 78

If you are trying to get a sense of how you did, here are some breakdowns of different point ranges:

• Greater than 70 (9 count): excellent job on the exam; you're showing that you are familiar with and can apply the material in class.
• 60-70 (11 count): good job on the exam; you have most concepts down, but there are just a few stumbling blocks where you need improvement.
• 50-60 (23 count): your performance was average; you are grasping some concepts, but you should spend some time making sure you have a firmer understanding of the material.
• Less than 50 (17 count): your performance was below average; you need to identify areas that are problematic for you and work on them by either reading the book, working through examples, or seeing the professors or TAs.

Some words of advice might be helpful:

First, try not to get too attached to the absolute value of the numbers. We take into account the performance of the entire class on the exam, and the statements above should let you know how you're doing. What's most important is that you focus on the mistakes you made, and not on the specific numbers, so that you can get a sense of what you need to improve.

Second, if you didn't do as well as you had hoped on the exam, don't let that unmotivate you. There is plenty of room to make progress on the final exam, and in the homework sets from this point on. Just make sure you identify ways to help you improve your work and studying, and you utilize all of the resources available to you (including the office hours).

MSS

HW 5

Here are some of the common mistakes on homework 5:

On number 20, many people did not account for the units - you need to change Cp into m^2/(s^2*C). (Cp for water = 4.184X10^3 m^2/(s^2*C)). Or you could have changed the kinetic energy into kJ/kg. Make sure you pay attention to the signs (temperature increased in this problem).

On 21, the problem was a liquid. Therefore you cannot use the ideal gas equations for work. Also, although there was a small change in volume, there was a very large pressure change so you cannot say that there was no work done. Also, be careful you copy the equations done correctly. The second term in kappa was multiplied by P, so kappa depended on P. Therefore when you take the integral, you cannot pull kappa outside of the integral.

On 22, many people struggled with taking the derivatives. The first step was to solve the equation for V in part a and P in part b. Then differentiate. The chain rule and multiplication (or quotient) rules were necessary on part b. Many people first solved for beta and used other properties to get dP/dT. While this was correct, I believe it was a little more difficult than solving for (dP/dT)v directly. Also, the problem specified that V should not be in the answer, so you should have substituted in what V equaled in the end.

On 23, make sure you say specifically how you solved for the answer. It was OK to use your calculator or computer, but make sure you say how you got the answer. Also, make sure your answers make sense - all three methods should have given similar results. If you got something way different, make sure to check it.

On 24 and 25, you can not assume ideal behavior or constant density for the liquid. The point of these problems was to use the more detailed equations in the text. There were mutliple ways of solving these problems (RK, Lee-Kesler, generalized correlations if appropriate), but ideal gas was not a valid assumption.

Resources for class

Dear class-

Since we are now in the middle of the quarter, it is a good idea for you to step back and take a broad assessment of your understanding of the course material. If you feel like you are missing key concepts or are having trouble solving problems in the homework sets and exam, you should take action to get up to speed. Here are some things you can do:

• If you are not already reading the book, this needs to become a priority. In some courses, you may have been able to get by without reading the book. This is not one of those courses.
• Prepare a summary sheet of the key equations of the course. Make sure to distinguish between which are fundamental and which are specific to a particular substance (e.g., ideal gas).
  
• Take advantage of either the TA's, Professor Chmelka's, or my office hours to help you resolve lingering confusion about the material. If you are unable to make these, schedule an appointment instead.
• Tau Beta Pi also offers tutoring specifically for this course, if you are looking for a student perspective. Tutoring takes place on a regular basis in the Undergraduate Conference Room in Bldg 698. I can give you more details if you are interested, or consult the webpage http://www.engineering.ucsb.edu/~tbp/.
  

Cheers,
MSS

Correction: sign convention for work done

Hi all,

There is another correction. There is an error in the earlier post about the sign convention used for calculating work done.

The sign convention for work done, as adopted in class and in the book (pg. 9) is expressed as:
dW = - P dV

For example, for an ideal gas, going from V1 to V2 in a mechanically reversible isothermal process,
W = -RT ln(V2/V1) = RT ln(V1/V2)
W = -RT ln(P1/P2) = RT ln(P2/P1)

Hence a compression process results in positive work, and expansion process results in negative work.

Sorry for the error in the earlier post. The solutions to HW 4 as posted use the correct sign convention, so you can refer to those and you should be OK.

thanks
badri

Graded HW4 set

Hi class-
Your graded homework 4 sets will be outside of my office until 6:30pm today. I don't want to leave them there overnight, but you will be able to pick them up tomorrow morning 8:30am and on as well.
MSS

HW 4 solution error

There is an error in the solution set posted for HW 4. The solution to Problem 17 is NOT 119. 15 F, it is 109.24 F. Sorry!

Badri

HW 3 and HW 4 common mistakes, notes

Hi everyone,

A few common problems people faced in HW 3:

1. Integrating dW = PdV correctly to find the work done:

For constant volume processes, W = 0. (If the process is also adiabatic, Q = 0 and therefore delta(U) = 0 by First Law).

When there is work done, we use an equation of state (EOS) or similar relation to find P as a function of V so that we can integrate the equation, e.g. for an ideal gas we use P = RT/V and obtain the equation W = R.T. ln(V2/V1). If you have a different EOS as in Problem 12, you have to use the appropriate relation for P as a function of V.

2. Sign convention for work/heat:

Be very careful about the sign! I noticed a lot of errors (especially when you add up work done or heat associated with multi-step processes) due to incorrect sign. Remember, work done by a system (i.e. expansion work) is positive and work done on a system (i.e. compression work) is negative.

now, on to HW 4:

Open vs. closed systems: with regard to Problem 17:

You cannot apply the relations derived for adiabatic closed systems to relate T2 to T1 in this case! There is continuous mass flow in and out of the system, and that is an immediate indication that you have to perform mass and energy balances for the open system. This will allow you to calculate the change in enthalpy as a function of change in flows, work done, heat inputs, etc. The enthalpy, of course, depends on the state variables (P and T) and this will let you find their values.

Energy balance units:

You have to convert all the terms in the equation to the same system of units. Note that the delta(u^2) terms has units of m^2/s^2, which is equivalent to J/kg. If your enthalpy units are kJ/kg, you must convert before plugging into the energy balance equation or your answers will be way off.

Finally, I was a bit disappointed to see answers on Problem 18 with exit diameters greater than 6 cm. The problem statement itself specifies that the system has a converging nozzle, i.e. D2 MUST be less than D1! If your calculations say otherwise, they are obviously wrong and you should try to find the error. Common sense is your biggest ally while performing engineering calculations.

Good luck on the midterm.
Badri

Review session

The review session for the midterm will be at 7pm on Thursday Feb 7. The room is tentatively set to 3361.

HW3 common mistakes

Here were some of the common issues on HW 3:

1. Some had trouble integrating dW = PdV to find the work done under various conditions. Be sure to know when pressure is constant and when it varies (and how). On Problem 11, many people failed to realize that the system is adiabatic and hence there is no change in temperature/internal energy since it is an ideal gas. On Problem 12, many people tried to find W as a function of P1 and P2, i.e. they didn't know which variable to substitute between P and V (several people also tried to substitute V as f(P) after integration).

2. There were many 'intuitive' answers for Problem 9 based on Raoult's law etc. Credit was given if the reasoning was accurate, but the phase rule is a more general and powerful explanation since it can be applied to ternary and higher systems etc.

3. On Problem 13, there was some confusion about the sign for W.

Web resource for thermophysical properties

In addition to the book, there are a number of websites that maintain property databases. The National Institute of Stantards and Technology (NIST) maintains an extensive listing of properties for many substances that are easily user-customized in terms of ranges of temperature or pressure, choice of units, etc. That website can be located here:

http://webbook.nist.gov/chemistry/

MSS

HW5, problem 21

There may be a little bit of confusion about how to proceed on this problem. You will have to make an approximation at some point in order to evaluate the solution. You should first try to carry your equations as far as possible towards the end goal, and then make your approximation at the end. (Hint: think about what aspects of the process will change by a large and small amounts).
-MSS

Office Hours

In my office hours on Monday, I wrote that (TP)^((1-y)/y) = constant, for an adiabatic process. It should be T*(P)^((1-y)/y) = const, where y = cp/cv. Sorry for any confusion.

HW 4, Problem 17 (SVA 2.27)

Just in case there is any confusion, the valve mentioned here is within the pipe itself, not a separate flow stream in or out of it.

-MSS

Common mistakes on HW2

Hi class-

Since you received HW2 back today, please take a few minutes of your time to go over the problems and pinpoint those places where you made mistakes. Just a small time investment reviewing each returned problem set will enhance your success in the course and ultimately save you many more hours later on when you are preparing for exams.

There were a couple of places where you seemed to have problems. In particular:

1) It's important to write down the equations first, before you fill in the values. This greatly reduces the chance for mistakes.

2) Some of you are having problems with the difference between lbf and lbm. Remember, lbf is a measurement of force. It has units of [mass*distance*time^(-2)]. On the other hand, lbm is a measurement of [mass]. You cannot say 1 lbf = 1 lbm, because there are differences of the basic units here. Instead, you can relate a lbf to a lbm times an acceleration, which has units of [distance*time(-2)]. However, you have to keep in mind that 1 lbf is not equal to 1 lbm * ft/s^2. Instead, 1 lbf = 32.17 lbm ft / s^2. You can just think of this as a unit conversion, just as if you would convert one kind of pressure to another.

3) There was some confusion about problem number 6 regarding cost scalings. Please pay particular attention to the solutions here.

4) For problem 7, remember that the universe as a whole is an isolated system (there is nothing else!). Therefore, according to the first law, the energy of the universe always remains constant.

-MSS

Problem 12 on HW3

Hi all-

On problem 12, there may be some confusion with the exponent n in the expression provided. This is not the number of moles. This is just a constant. In retrospect, we perhaps should not have picked a variable that had that dual meaning.

If it helps keep things clear, you can replace n in the exponent with c instead and work through the problem that way.

MSS

Enthalpy as a path-independent state function

Today in class we defined the enthalpy:

H = U + P V

We found that for an infinitely slow heating process at constant pressure, all of the heat that was added was accounted for a change in enthalpy:

Delta H^t = Q

I was asked how the enthalpy could be path-independent, when it is equal to Q, which is path-dependent. The reason is that the above equation only applies to the specific process of infinitely slow heating. If we do not heat in such a way that we are at equilibrium during the entire process, then Q is not necessarily accounted for by the change in enthalpy.

For a fast heating rate, that results in the same final state, the change in enthalpy will be the same (since it is a state function), but the value of Q can be different from it.

Correction to today's lecture

I think I made a mistake in today's lecture for the example we discussed involving sand being removed from the top of a piston. Here is a longer explanation that may be useful to you.

Recall that the two process we considered were:

(A) The sand is removed grain by grain. Each time a grain is removed, the piston inches up just a little bit higher as the gas expands a small amount.

(B) The sand is swept off instantly. The piston sharply rebounds and oscillates until it dampens out and comes to the same final state as for the process (A).

The work W is more negative for process A than for B. Recall that a negative work means the system does work on its environment. Thus, process A does more work on the environment than process B. The reason is because in A, the piston lifts much more sand against its weight--at each step, it's lifting an amount of sand that's slowly decreasing with time (as we remove grains). On the other hand, process B lifts only the residual sand that's left after we swept it off. Less sand means less gravitational force that is overcome, and hence less work.

Both the final states are the same for the two processes. Why? It is because the pressure of the gas is the same at the end of A and B (they have the same amount of sand on top of the piston) and because the temperature is the same (we're at constant temperature in this example.) That's 2 intensive variables for a single-component, single-phase system, which means the equilibrium state of the gas is completely specified by the Gibbs phase rule. That means that the change in internal energy for both processes--which is a state function--is the same, since the beginning and ending equilibrium states are the same.

The first law says that Delta U = Q + W. Since Delta U is the same for A and B and since W is less (more negative) for A, the heat transferred Q must be greater for A than B in order to achieve the same value of Delta U. (This was stated as the opposite in class).

You can rationalize this physically: when a gas expands, it tends to cool. Therefore, to bring it back to the same temperature, heat must be transferred to it. That heat comes from the environment. In the case of B, all of that lost work that we didn't use to raise the sand was used (by way of friction) to heat the gas back up to the same temperature. So case B needs less heat from the environment to counter the effect of cooling upon expansion.

MSS

CHE110A: Welcome!

Welcome students to CHE110A: Chemical Engineering Thermodynamics. The purpose of this blog is to archive and disseminate class announcements, post responses to questions about lecture material and homework problems, and provide suggestions and hints about class materials. You can subscribe to automatic updates through email using the box to the right.

-MSS