Yeah MIT for doing classroom repairs during lecture. (1 student) Yes. One might think that the first thing they would check is the class schedule... but it is nice to get the boards fixed and they did help us get rolling again with the duct tape, etc. So in the end it worked out.
The above, combined with my forgetting to bring the 3x5 index cards took some of the polish off the lecture, but I think the content was good. Judging from the muddiest part of the lecture cards, most people were comfortable with the First Law written as a rate equation (SFEE). Remember that there are three common areas of difficulty for the SFEE. The first is units and signs (as demonstrated in the PRS question responses). The other two we will talk about in T9 and T10.
Time to return with cards: 6min 23s.(1 student) Are you kidding? Our TA's are much better than that. For the record it was 1:45 ... and all for a coaster.
Responses to 'Muddiest Part of the Lecture Cards'
1) Can mass flow in and out do work or transport heat into and out of the system? (1 student) Great question. The mass flow in and out carries with it some amount of internal energy (not heat). As we have defined it heat is the transfer of energy at a system boundary by virtue of a temperature difference only. In this case energy is transferred by the flow into and out of the control volume. So the answer to the heat part of your question is "no". The answer to the work part of your question is "yes" and this is the topic of tomorrow's lecture (T9). In preparation for this, please read the section of the notes related to shaft work and flow work.
2) No mud, just unclear how to get u for a gas entering/exiting an engine. (1 student) and Without knowing q and w, are there any easier formulas for getting u, like with p, v, T ....? (1 student) You get u the same way we have for the other problems we have worked in the class. But first note that what you really need is Du. For an ideal gas Du is related to the change in temperature by cv. If you want to find the value of u at a particular condition (versus finding Du), you should first look up in the back of S, B, & VW what u is for the gas you are working with at a reference temperature (usually 296K). Then use Du = cvDT to get the value at the temperature you need. (The same holds true for enthalpy except you would use cp and start with the reference enthalpy at 296K.) I don't understand how you know the entering parameters for internal energy, velocity and height of the entering stream. (1 student) These are typically given, or solved for if enough other information is given. They would be specific to whatever device you were studying.
3) Physically, what is the difference between and eout-ein = q-w. and related questions on units (2 students) Physically there is no difference. The two equations are the same with the exception of the choice of variables used (intrinsic versus extrinsic). Sometimes one is more convenient than the other. The second equation is wholly expressed in terms of intrinsic variables (J/kg), so in order to find Watts out of a particular system, one would have to multiply by the mass flow (kg/s) for that system.
4) c=velocity, not just the speed of light? (1 student) In a never-ending search for a symbol we haven't used yet, I have chosen c for velocity of the flow (we already use v for specific volume, V for volume, and u for internal energy---which are the common ones used for velocity in the Fluids lectures).
5) Why isn't "gh" used in the SFEE in the reading notes? (1 student) Good observation. I left it off for the same reason we typically leave it off in other versions of the First Law we have worked with. For gases, changes in potential energy are typically small relative to changes in other forms of energy. You all showed this on T5 part a. You have to lift a gas an long way (14km) to change its potential energy an equivalent amount to the work done in a compression of 5x. In the example we did in class and in the derivation on the board I included it so you would feel comfortable using it for problems where the change in potential energy is important (like water flows).
6) Units in [u + (c^2)/2 + gh], the latter two were neglected and u taken to be J/kg anyway. Don't understand it. (1 student) All the units are consistent. c^2 and gh have units of m^2/s^2 which is the same as J/kg. (A Joule is a kg-m^2/s^2).
7) What is e? (1 student). The same e we defined in Chapter 4.
8) The derivation of the new version of the First Law (for mass flow) was a little rushed. (1 student) Please review the derivation in the notes, and read through the appropriate section of S, B & VW. If there are parts that are still unclear contact me or a TA to work through them.
9) When we say d/dt = 0, does that mean both dEc.v./dt = 0 and dm/dt = 0? (1 student) Yes.
10) Working with equation (1 student) This was the first time through. You will get much more practice. Just remember, be careful with signs and units.
11) I am still having difficulty trying to figure out the sign convention for heat added. (1 student) Heat added to the system is positive. Heat removed is negative. In the example given in class the heat was removed from the system so it was negative.
12) Why does steady flow imply no change in E as well as mdotout=mdotin? (1 student) If the mass within the control volume does not change over time, then what comes in must equal what comes out. The same is true for the energy. If the energy within the control volume does not change over time then all of the energy flows to and from the system (ein, eout, q and w) must all balance to zero. Muddy part: before you narrowed it to steady ---> the open system w/changing mdot? (1 student). For an unsteady problem the mass (and the energy) contained in the control volume can change as a function of time. For example if the mass flow in is greater than the mass flow out, the mass within the control volume will increase.
13) If the mass is constant in a closed system, why do we rarely find the mass? I mean, we always use specific volume. (1 student) Quite often it is more convenient to work in intrinsic variables rather than extrinsic variables. But the concepts behind the equations are the same either way.
14) How can the entering and exiting stream heights be so different? (-10m & 300m) or is it just the example? (1 student). Basically, it was just an example, but a difference in heights like this would be relevant for the flow over a hydroelectric dam (like the Hoover Dam).
15) How does [u + (c^2)/2 + gh] sum up to be e? ( 2 students). It expresses the total energy per kg as the sum of the internal energy, the kinetic energy and the potential energy (all per kg).
16) From the lecture notes, why is the flow work p2v2-p1v1? (1 student). I am pleased you read the material. This is the topic of our lecture tomorrow (T9).
17) In the calculation I get that the rate of work is approximately equal to 18kW instead of 20kW. Are the calculations right? (1 student). Yes they are. The correct answer is 20005 W ( I just checked it again to make sure).
18) No mud (34 students). All-time high for this semester.