This HW covers CB chapter 7 or LT chapter 7.

The homework consists of 13 problems for a total of 59 pts.

Please begin your question with the problem number you are asking about.

Cengel & Boles: Ch 7: 14(2pts), 15(2pts), 18(2pts), 28E(4pts), 29(6pts), 66(6pts)

WB-1(4pts) , WB-2(6pts), WB-3(4pts) , WB-4(4pts), WB-5(8pts) , WB-6(5pts), WB-1(6pts)

## 39 comments:

On 7-28E why is the heat transfer in part a positive?

7.28E , Anon 3:06pm

I constructed a tie-fighter diagram for this problem and all Qs are positive in this setting. The heat is rejected by the HEX to the cold reservoir.

For WB-1 it states that there is heat flowing also where the system is at a temperature of 1000K. Do we need to figure out which direction the heat is going in order to properly place it into the first law? If so, how do we do that if we do not know the temperature of the surroundings?

On WB-3 part b when T1 = -40 degC and H1 = 211.44; I tried ramping the pressure up to the max on NIST, but the enthalpy still was not high enough should I just use the plugin? or am I doing something wrong? Is anybody else having trouble here?

WB-1 , Jayson , 2:43pm

Yes. Assume one direction for this 3rd heat flow and if you get a negative value, then you guessed the wrong direction. Tsurr is not relevant to either the 1st Law or the 2nd Law. In the 2nd Law, you are only interested in what is foing on inside the system and whether it is INTERNALLY reversible. So, when you evaluate INT{dQ/T} the is the T inside the system at which the heat transfer occurs.

I hope this helps.

WB-3 , Jayson , 5:08pm

Make sure your reference state is the same as mine !

I am on NIST now and Hsatvap at -40 degC is 374 kJ/kg which is far greater than 211.44 kJ/kg.

Consider the possibility that the answer lies inside the 2-phase envelope.

WB-3 got that on lock but WB-1 we shall see... I may need to shift gears to math now though I am not doing all that I should for that class hard to balance some times but yes overall I believe I will be needing to meet with you sometime after class to get help with the problems you write. Cause it is hard to put it into words on a blog. Not that I do not appreciate the blog cause I certainly do.

Thanks and sorry for rambling.

Jayson,

No worries. It will all get done by Th AM. See you tomorrow.

on WB-5 I found Qc from the area under the T-s diagram. I now need either Qh or W. Which one should I find and how would I go about doing that?

WB-4...Is the first law involved? If so is Qsurr + QH = QC then does Qsurr need to be replaced by the Integration of TdS? I know none of the ideal gas can be used because it is not an ideal gas...I think. That eliminates a bunch of eqns including Gibbs, plus it is not adiabatic. I believe you are supposed to assume it is internally reversible, but that is about it. I was thinking maybe go back to using the Carnot efficiency eqns but I know entropy and entropy gen are supposed to be involved.

Thanks for helping!

Also I believe WB-7 does not involve a Turbine, even though that is in the name? I assume that is a copying error.

In WB_5 I have no problem getting the values for S1 S2 S3, the question is which reference state did you use did you use the book, the plug in, or NIST? I am guessing the plug in but the number for Snot is way different in the table for ideal properties of air, compared to the number I got using the plug in...?

WB-5 Anon 2:16

I do not think you need Q_H. This is because it is internally reversible. So, you get to use the equation including T_C and T_H like a Carnot cycle. The only question is if it is a power cycle or a refrigeration cycle and I got that the process is going counter-clockwise, which indicates a refrigeration cycle. BUT that still leaves two options with two different equations: A heat pump or a refrigerator... unfortunately I am not as positive on whether it is a HP or a refrigerator, but based on temperature going up in the second step I would guess that it is a Heat pump but I am still analyzing.

Hope it helps (Also hope I am right)

WB-2

Ok, so I know that Sgen = 0 for reversible processes and Sgen > 0 for irreversible processes and that ΔS = ∫ dQ/T + Sgen...but where do I go from there?

WB-4 , Jayson , 6:08pm

I did not use the 1st law on this one. But yes, your statement of the 1st Law is correct. You need to show that if it is int rev and Sgen=0, then Qc is maximized. So, use the latest and greatest form of the 2nd Law that includes Sgen !

I hope this helps!

WB-7 , Jayson , 6:11pm

LOL. Sorry, bud. There is no turbine in this problem. I made a classic cut-and-paste error in the title.

I think WB-6 has some problems... We are supposed to assume adiabatic, and reversible which leads to assuming isentropic. However after calculation on NIST, using the tables, and the plug in, there is no way that it is isentropic because the data does not match up or else there is some CRAZY interpolation required. Plus I would not know how to approach the problem if it were not isentropic.

Frazzle Snazzle Puddin Pop.

I have been working on hw too much...

Somehow though I got the right answer on WB-6. Some luck there cause I was mixing and matching like crazy.

WB-5 , Jayson , 7:19pm

I hope you mean s~naught or S~o or S^o for states 1, 2 and 3. I used tables from another book. I will go back and repeat with the tables in your book. But the answer for T3 will not change more than 0.1 degC or so. You need to understand that S~o is NOT S~. S~o is just a function (the ideal gas entropy function) that is a convenience for ideal gases. It is NOT comparable to S from NIST or plug-ins. S~o = INT{Cp/T} dT from Tref to T. It is NOT entropy !

We probably need to discuss this face-to-face or in class.

WB-5 , Jayson , 7:29pm

The cycle must be completely reversible before you can use the reservoir temperatures alone to calculate the COP. So, yes, you DO need to determine Qh to answer this question.

Look at your PV or TS diagram to determine whether it is a power or refrigeration cycle. Clockwise or counter-clockwise ?

WB-6 , Jayson , 10:09pm

Yes, assume the turbine is adiabatic.

Yes, max work out comes from a reversible turbine. Adiabatic + reversible = isentropic.

You are on the right track. I do not understand what has you stuck. We can work on this in class and office hours tomorrow.

WB-6 , Jayson , 10:13pm

Congratulations. Tomorrow in class, we can make sure you understand why your answer is correct.

WB-2 , MUFfl3r , 8:49pm

Write our latest and greatest 2nd Law eqn (with Sgen). Write it for each cycle. Sgen for the reversible cycle is zero, right ? Sgen,irrev > 0. Evaluate the cyclic integral. It becomes Q/T for each interaction with a reservoir. Heat transferred into the system increases the entropy and is positive. Heat transferred out of the system decreases the entropy and is negative. Qh is the same for both cycles. A little algebra and you will have the answer to part (a) !

I hope this helps !

I don't get your saying in Problem WB-1 hint where you said that nth*Sgen=0. I got Sgen=0.35. Doesn't that mean nth=0?

So, how do I calculate the efficiency?

On 7-66 I set up a first law as Q-W = du and canceled out the Q and W leaving du = 0. After that I set up duwater +duiron = 0, which led to m*CpFe(T2-T1)+mCpWater(T2-T1) = 0. First off I'm wondering if im on the right track and secondly what is the cp of either substance? would I look it up at each of their individual starting temperatures to be used in this equation, or is there some other temperature or way that I should be looking up the cp value?

WB-1 , Anon , 11:20am

I learned yesterday that other students interpreted the problem statement the same way you did. There is a period in the problem statement, not a star indicating multiplication. Here is the text from the problem statement.

Use the 1st Law and the definition of ηth. Sgen = 0 within the system. ΔScycle = ?

Sorry for the confusion.

On W-B 4 is the answer simply an equation with Q_C isolated and all the other terms (Q_H, T_S, T_U, T_0 and Sgen on the other side?

7.66 , Camden, , 1:51pm

Yes, you are on the right track. I used CP,H2O = 4.18 kJ/kg-K and CP,Fe 0.45 = kJ/kg-K. You can look up values in the back of your textbook. Note T1Fe is not = T1H2O. I think you can safely assume Cp = constant for the solid Fe and the liquid H2O for that matter (at least over the modest range of temperatures in this problem.

I hope this helps.

I got -0.35 for the change in entropy whichever direction I assumed for the Q3. So, is that wrong?

Also, I recall that the efficiency equation=1-Tc/Th. Which one is Tc and which one is Th in this problem?

Anon , 3:12pm

I am not sure which problem you are working on. Just let me know.

oh, sorry, I meant problem WB-1

For WB-5 I am using the polynomial to find Qc and in the book it gives a Cp~ and I need integration for Cv_hat so to modify the Cp~ do you just subtract R from the a term, and divide the whole equation by the molecular weight or just divide R by molecular weight before integrating?

Also when integrating for Qc from T1 to T2 creates a problem because T2 and T1 are the same so T2-T1=0 and that ruins the integration.

On those last two comments I meant Q12, I think it represents Qc but I am not sure.

WB-1 , Anon , 3:19pm

Your Q3 is wrong. How did you get that ? Use the 2nd Law. Sgen,int = 0 b/c the cycle is internally reversible. Watch the signs. Q into a system increases the entropy. Use thermal efficiency= W/QH.

I hope this helps.

WB-5 , Jayson , 3:37 - 4:24

Yes, Qc = Q12.

Cp~ = Cv~ + R

Cp^ = Cp~/R

So, to answer your question, divide the whole eqn by R.

Check your integration using the ideal gas property table uo.

Use the TS Diagram for step 1-2 !! Q12 = T1 * (S2-S1)

Best of luck !

Oh, ok thanks! Also though since there is flow work shouldnt I use Cv instead of Cp? I thought that is what you said in class. Sorry this is probably redundant.

On WB-1 we don't know the third Q value or the work the cycle produces. How do we calculate these? The first law provides one equation, but what's the other equation?

WB-1 , Justin , 6:42 PM

The 2nd eqn is the 2nd Law !

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