Please post any questions you have here. I will be responsive on Sunday, but not much on Saturday.
It has been a pleasure working with all of you. I hope you do well on the final. Remember you can use 3 8.5" x 11" cheat sheets (with writing on both sides) during the final.
We will take the class photo before the final on Monday morning. Sorry I forgot about it today. It is optional, you can choose not to be in the photo if you wish.
Best of luck in your career and in life.
Adios,
Dr. B
48 comments:
Dr. B,
Is that Three 8.5x11 sheets, or Four? I thought it was four.
If you have multiple heaters (Qh1, and Qh2) and multiple condensors (Qc1 and Qc2), and you wish to apply the 1st law to the system that contains both, is Wcycle=Qh1+Qh2-Qc1+Qc2, and then Qh=Qh1+Qh2, Qc=Qc1+Qc2, when you wish to determine COPs?
does dS=0 immediately imply delta=gamma? or are there certain stipulations surrounding this claim?
Can you still download homework assignment files (w/out solutions) after they've been taken down?
Professor- in the 2001 final, I tried using dS=S2-S1-Rln(P2/P1) and pressure ratios to solve for the temperature at the exit of the turbine, and got the same answer: 585.51 K. This does not agree with the posted answer of 608 K. Could you verify that is the correct answer?
nevermind, i figured it out
what is the formula for regeneration effectivness. Its not on the summary pages and I've seen it several different ways in the book (i think there are some typos). Starting with #1 at the boiler feed what is the formula?
I'm a little confused about when you can assume constant heat capacities for gases. Is it solely based on whether you consider the gas ideal?
Parwiz:
You are right, the syllabus says FOUR cheat sheets !
Yes, in principle, but you had a typo.
Wcycle=(Qh1+Qh2)-(Qc1+Qc2).
Scattante:
If S = a constant AND you are dealing with an ideal gas AND the heat capacities of the IG are constant, then delta = gamma in the polytropic path equation.
Maria:
Yes. I just got one.
Scattante 1:33
Which problem ? I assume P2.
P2 uses STEAM. Steam is not generally considered to be an ideal gas. This eqn: dS=S2-S1-(R/MW)ln(P2/P1) is Gibbs 2nd Eqn for ideal gases with variable heat capacities. Data was GIVEN in the problem statement. This problem could be solved using entropy. If you wish to do so, use the full steam tables in your book or the NIST website. In 2001, I wanted them to solve this problem WITHOUT using entropy. To see how to do this, review HW6, P8.
What ? You doing 1B ? Doh !
I used the IG propery table for air from the textbook I used in 2001, so the H & S values do not match. But I think T2 is correct. Did you remember to use R/MW to make the units work ? Give me some more info to work with here and maybe I can be more helpful.
Scattante:
Good. Explain it to us if possible.
Magicmarker:
If you spot typos, PLEASE let me know.
The effectiveness of a regenerator is the ratio of the actual change in enthalpy of the stream to be preheated to the maximum possible change in enthalpy that could be achieved. So, in terms of Example 9F-1, actual change in enthalpy = H6 -H1 and maximum possible change in enthalpy = H3 - H1, so effectiveness =(H6-H1)/(H3-H1).
I hope this helps and I hope you will let me know if you see something in the book or program that contradicts this.
Lance Armstrong:
It is not up to you to decide whether to assume heat capacities are constant. The problem statement must TELL you if they are constant. If you are given values for Cp and Cv or just for gamma, then you should assume they are constants.
Good question.
Can we automatically assume that the stream leaving a condenser is saturated liquid?
For hw 9 problem 3, in the diagram, shouldn't state 3 be between the point of saturated liquid and saturated vapor because its saturated? and not in the superheated vapor region as it shows on the solutions? If its in superheated region as shown, you wont have quality.
and does the word "condensate" imply that the stream is only saturated liquid, or a mix?
In problem 4 on the HW, going from state 4 to 7 through closed feed water heater, there is Q rigt? and would that be Q(H) or Q(c) or is ther e no Q? and why?
So i guess my question is how do you know when there is a heat transfer? I know there will always be heat transfer through condenser, boiler and evaporator and HEX and combustor correct?
Does a gas do flow work?
I ask because in the test question #3, you did not use Q-W=dH, which is the form of the 1st law for flow work, but you used Q-W=dU.
Giro:
In general, this is not a safe assumption. But, when we defined the Rankine Cycle and the Rankine VCR Cycle, we said the condenser effluent was a saturated liquid. If there is any confusion on the final, plase ASK ME !
Anon 11:35,
Yes, I made the TS diagram before I completed the calculations and discovered that state 3 lies within the 2-phase envelope. I generally consider a SKETCH of a phase diagram like TS to be a tool to help me solve the problem. In that case, it is not critical that you know whether state 3 lies within the 2-phase region or out in the superheated vapor region. But it can also be a way to present results. Then, it deos matter where you position state 3...and 3S for that matter !
Giro:
The word condensate implies a liquid. It can be saturated or subcooled but it is not a 2-phase mixture.
But it is possible to have a 2-phase mixture leave a condenser. In that case, only the liquid would be called condensate.
Don't focus too much on this at this point. Just ask if you have any doubts on this issue during the final.
In HW number 9, problem 5, how were u able to use the polytropic relationship there to find T1, and where did the relationshp gamma = Cp/Cv come from, can you remind me so i can go look into it?
Anon 12:30,
All the heat that stream 4 gives up goes into stream 11 to pre-heat it. If you apply the 1st law to streams 4/7 then Q_4,7 < 0. If you apply the 1st law to streams 11/1 then Q_11,1 > 0. Q_4,7 = - Q_11,1.
If you instead choose to apply the steady-state, MIMO form of the 1st law to the FWH, then Q = 0 because no heat is exchanged with the surroundings...the FWH is adiabatic in this case.
This last approach shows why the heat transferred int eh FWH is NOT part of QC and it is NOT part of QH either. The heat transferred in the FWH is NOT exchanged with a thermal reservoir. It is an internal heat transfer and therefore does NOT show up (directly) in the efficiency or COP calculations.
When 2 process streams exchange heat with each other, we generally assume the HEX is adiabatic (no exchange with the surroundings) unless we are told otherwise.
If you get confused during the final, please ASK me !
Flower Girl:
Yes. Any flowing fluid does flow work.
I am not sure which test you are talking about. Give me a bit more info and I will be more helpful !
In the form of the 1st law that you gave, Q-W=dU, W = total work. This would include flow work. Does that help ?
Anon 1:54,
Gamma is defined as the heat capacity ratio = Cp/Cv.
I used the PVT relationship for a polytropic process and replaced delta with gamma. I can do this because:
1- the gas is ideal
2- the heat capacities are constant
3- the process is isentropic.
Because the heat capacities are constant I could have used:
S1s - S4 = Cp * Ln{T1S/T4} - (R/MW) * Ln{P1/P4} = 0
for the isentropic turbine and then solved for T1S.
If the heat capacities were not constant, I would have used
S1s - S4 = So1S - So4 - (R/MW) * Ln{P1/P4} = 0
and the IG Property Tables.
Are we required to know how all the variations of Brayton/Rankine ref/power cycles work? Or will we be given these diagrams on the test?
It has been an overwhelming amount of information we have covered.
Yeah, will we be given the schematics? I think going from a schematic to a TS diagram is easier than going from a problem statement to a schematic
What is the key mechanism of feed water heater(FWH) that we need to know? Also, can anyone explain the difference between the open and closed FWH.
I don't understand why sometimes the entropy is different even it is isentropic...
For example, in the book, example( 9C-1), S1=S6...
However, in example 9E-1,
Why isn't S1s=S4??
Then I thought because for eg.9E-1,
the isentropic efficiency is not 100%...
But I found that the question 3 in final exam of 2006 spring, even the compressor is isentropic, for example,S1 is not equal to S2...
They are sometimes the same and sometimes not...I am totally lost..
Please let me know why it is like that!!or let me know which one i should follow...thank you!
Pwiz:
You need to know how all of them work. You may or may not be given the flow diagrams on the test.
Yes, Ch 9 & 10 have a lot of information in them, but, fortunately, almost no new concepts.
Magicmarker.
I agree. But you may or may not be given the schematic on the test. So, you need to know how to both steps.
Anon 3:38,
A feed water heater is a heat exchanger.
A closed feedwater heater transfers heat from a hot stream into the cold boiler feedwater stream WITHOUT mixing with the boiler feed. This is a regular old process heat exchanger like we have used since Ch 5.
An open feedwater heater preheats the boiler feed by MIXING a hot stream (possibly VLE or even superheated) with the boiler feedwater. When you analyze an open feedwater heater, use the steady-state, MIMO form of the 1st Law.
Good question.
Thermo 4:41 PM,
This is a VERY important point !
The FUNDAMENTAL difference between example 9C-1 and 9E-1 is that the fluid in 9C-1 is WATER and you use the steam tables and the fluid in 9E-1 is AIR treated as an IDEAL GAS with variable heat capacities.
S^o is NOT entropy it is the Ideal Gas Entropy Function. S^o is related to S by the 1st and 2nd Gibbs eqns. The 2nd Gibbs, for example, tells us that for an ideal gas with variable heat capacities, S_2 - S_1 = S^o_2 - S^o_1 - (R/MW) * Ln{P_2/P_1}
So even when S_2 = S_1,
S^o_2 is NOT= S^o_1
The issue here is not that the meaning of the word isentropic changes. The issue is that S^o is NOT entropy !
For test 2, number 3, part e and f, how do we know to assume that the evaporator to be internally reversible?Is this something we should always assume?
I misread a problem, but I was wondering, what if you knew the output of a turbine and were given the isentropic efficiency. Would there be a way to determine turbine inlet's enthalpy?
Flower Girl,
In this class, the answer is almost always yes. All heat exchangers are always considered to be isobaric in this class.
The special HEXs that exchange heat with thermal reservoirs are internally reversible as well.
A process HEX (like a FWH, is generally NOT internally reversible because the heat transfer that occurs inside almost always occurs through a finite temperature differenece.
Pwiz:
Yes. Just use S_S,in = S_out. Then, you know P_in and S_S,in and you can determine H_S,in from tables. Then use the isentropic efficiency to get H_in.
The underscores "_" mean subscript.
when dealing with a flash drum, esp problem 6 HW9, how do we know which streams are saturated liquid, and which are saturated vapor? For instance, in problem 6, can we assume that because the lower cycle stream was hotter when it arrived at the flash drum that it transfers all its heat to the upper cycle, leaving the exiting lower cycle stream as saturated liquid, and the exiting upper cycle stream as saturated vapor?
also regarding problem 6 of HW9, how does an evaporator reject heat? wouldn't it have to add heat to the system to evaporate a saturated mixture?
In previous exam of year 2003, problem number 1: It is assumed that first turbine expands to saturated vapor but second turbine is not. Why is that? when are we suppose to make assumptions on whether its saturated liquid or vapor or not?
So what im asking is how do you know what states they are in after compressor or turbine unless saturated?
In a HEX, I can find dS for both streams but how would I go about determining the entropy generation due to finite temp differnce inside? when can I assume the hex is int reversible?
Giro:
The one that comes out the top is sat'd vapor and the one that comes out the bottom is sat'd liquid. If that is not the case, the problem statement must tell you.
It is just a physical thing. The liquid is more dense, so it comes out the bottom.
Remember, tomorrow, when in doubt ASK.
Giro 8:48 PM,
Evaporators do not reject heat, they absorb heat in order to vaporize some of the liquid. Where did I say the evaporator rejected heat ? I need to fix that if I said it.
I will own...
There are SEVEN tests from 2003, so I am not sure which one you are looking at, but...
You cannot tell the phases present in a turbine effluent just by looking.
When I SKETCH at PV or TS diagram at the beginning of a problem, I do it to help guide me through the solution. It is VERY common when you make such a sketch that you are not certain whether some streams (especially compressor and turbine effluent streams) lie in the superheated or 2-phase region. That kind of detail is part of the nitty-gritty problem solution. You don't need to resolve that kind of detail before you can SKETCH the cycle.
You must determine the values of TWO intensive variables in order to know the phases present for ANY stream. For a compressor or turbine effluent, you usually know the pressure and you need to know one other intensive variable. Frequently this information is provided by giving you the isentropic efficiency of the compressor or turbine. With some effort, this allows you to determine H^_out for the compressor or turbine. Using this 2nd intensive property, you determine the phases present and the quality if the state lies in the 2-phase envelope.
Magicmarker:
Check out the "General" form of the S-balance eqn on page 173, bottom half of the page. For a HEX like an open feedwater heater, this S-balance eqn simplifies to:
dS/dt = SUM{mdot * S^)inlet streams - SUM{mdot * S^)outlet streams + SUM{Qdot/Tsys} + Sdot_gen. The HEX is adiabatic and we assume the device operates at steady-state, so:
Sdot_gen = SUM{mdot * S^)outlet streams - SUM{mdot * S^)inlet streams.
This Sgen is the INTERNAL Sgen. There is no external Sgen because the HEX is adiabatic. So, the total and internal Sgen are equal.
Good question !
That is it for me. I am going to sleep and I suggest you do the same. The law of diminishing returns says you get progressively less benefit from extra studying. So, if you have studied a lot this weekend, it is time to stop. Always sleep at least 6 hours before any test.
Best of luck to all of you tomorrow. I have Oreos in my bag ! Have a great summer !
Adios,
Dr. B
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