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