tag:blogger.com,1999:blog-20953269.post8793289317703674011..comments2023-04-22T16:35:23.188-07:00Comments on Thermodynamics: ENGR 224 - HW #5Unknownnoreply@blogger.comBlogger39125tag:blogger.com,1999:blog-20953269.post-58843826524205921542011-05-11T19:16:08.755-07:002011-05-11T19:16:08.755-07:00WB-1 , Justin , 6:42 PM
The 2nd eqn is the 2nd Law...WB-1 , Justin , 6:42 PM<br />The 2nd eqn is the 2nd Law !Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-12227995239632656332011-05-11T18:42:57.986-07:002011-05-11T18:42:57.986-07:00On WB-1 we don't know the third Q value or the...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?Justinhttps://www.blogger.com/profile/08562335712948005983noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-55724388972063805532011-05-11T17:00:11.847-07:002011-05-11T17:00:11.847-07:00Oh, ok thanks! Also though since there is flow wor...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.Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-13284919839240112622011-05-11T16:28:10.298-07:002011-05-11T16:28:10.298-07:00WB-5 , Jayson , 3:37 - 4:24
Yes, Qc = Q12.
Cp~ = C...WB-5 , Jayson , 3:37 - 4:24<br />Yes, Qc = Q12.<br />Cp~ = Cv~ + R<br />Cp^ = Cp~/R<br />So, to answer your question, divide the whole eqn by R.<br />Check your integration using the ideal gas property table uo.<br />Use the TS Diagram for step 1-2 !! Q12 = T1 * (S2-S1)<br /><br />Best of luck !Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-43636926607242463022011-05-11T16:24:26.762-07:002011-05-11T16:24:26.762-07:00WB-1 , Anon , 3:19pm
Your Q3 is wrong. How did yo...WB-1 , Anon , 3:19pm<br />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.<br /><br />I hope this helps.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-25712203193287161722011-05-11T16:10:20.780-07:002011-05-11T16:10:20.780-07:00On those last two comments I meant Q12, I think it...On those last two comments I meant Q12, I think it represents Qc but I am not sure.Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-82867819204505431752011-05-11T16:00:12.934-07:002011-05-11T16:00:12.934-07:00Also when integrating for Qc from T1 to T2 creates...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.Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-5562339218467638002011-05-11T15:37:11.500-07:002011-05-11T15:37:11.500-07:00For WB-5 I am using the polynomial to find Qc and ...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?Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-7000749675231613682011-05-11T15:19:50.807-07:002011-05-11T15:19:50.807-07:00oh, sorry, I meant problem WB-1oh, sorry, I meant problem WB-1Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20953269.post-84636901665394964062011-05-11T15:16:04.475-07:002011-05-11T15:16:04.475-07:00Anon , 3:12pm
I am not sure which problem you are ...Anon , 3:12pm<br />I am not sure which problem you are working on. Just let me know.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-28173232108145833702011-05-11T15:12:40.735-07:002011-05-11T15:12:40.735-07:00I got -0.35 for the change in entropy whichever di...I got -0.35 for the change in entropy whichever direction I assumed for the Q3. So, is that wrong?<br /><br />Also, I recall that the efficiency equation=1-Tc/Th. Which one is Tc and which one is Th in this problem?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20953269.post-66108505290019964972011-05-11T14:24:49.231-07:002011-05-11T14:24:49.231-07:007.66 , Camden, , 1:51pm
Yes, you are on the right ...7.66 , Camden, , 1:51pm<br />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.<br /><br />I hope this helps.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-5557358471676925632011-05-11T14:23:30.995-07:002011-05-11T14:23:30.995-07:00On W-B 4 is the answer simply an equation with Q_C...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?Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-49974044191707019052011-05-11T14:20:54.443-07:002011-05-11T14:20:54.443-07:00WB-1 , Anon , 11:20am
I learned yesterday that oth...WB-1 , Anon , 11:20am<br />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.<br /><br />Use the 1st Law and the definition of ηth. Sgen = 0 within the system. ΔScycle = ?<br /><br />Sorry for the confusion.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-36592007943335050882011-05-11T13:51:02.777-07:002011-05-11T13:51:02.777-07:00On 7-66 I set up a first law as Q-W = du and cance...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?Camdennoreply@blogger.comtag:blogger.com,1999:blog-20953269.post-23540804721923994682011-05-11T11:20:09.966-07:002011-05-11T11:20:09.966-07:00I don't get your saying in Problem WB-1 hint w...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?<br />So, how do I calculate the efficiency?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-20953269.post-11507484024642219962011-05-10T22:24:52.909-07:002011-05-10T22:24:52.909-07:00WB-2 , MUFfl3r , 8:49pm
Write our latest and great...WB-2 , MUFfl3r , 8:49pm<br />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) !<br /><br />I hope this helps !Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-21600894357521988572011-05-10T22:20:31.201-07:002011-05-10T22:20:31.201-07:00WB-6 , Jayson , 10:13pm
Congratulations. Tomorrow...WB-6 , Jayson , 10:13pm<br />Congratulations. Tomorrow in class, we can make sure you understand why your answer is correct.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-40622004958328483502011-05-10T22:19:40.223-07:002011-05-10T22:19:40.223-07:00WB-6 , Jayson , 10:09pm
Yes, assume the turbine is...WB-6 , Jayson , 10:09pm<br />Yes, assume the turbine is adiabatic.<br />Yes, max work out comes from a reversible turbine. Adiabatic + reversible = isentropic.<br />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.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-69143773189214884792011-05-10T22:16:47.691-07:002011-05-10T22:16:47.691-07:00WB-5 , Jayson , 7:29pm
The cycle must be completel...WB-5 , Jayson , 7:29pm<br />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.<br /><br />Look at your PV or TS diagram to determine whether it is a power or refrigeration cycle. Clockwise or counter-clockwise ?Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-59766093381352204662011-05-10T22:14:15.726-07:002011-05-10T22:14:15.726-07:00WB-5 , Jayson , 7:19pm
I hope you mean s~naught or...WB-5 , Jayson , 7:19pm<br />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 !<br /><br />We probably need to discuss this face-to-face or in class.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-10138007706661320472011-05-10T22:13:53.777-07:002011-05-10T22:13:53.777-07:00Somehow though I got the right answer on WB-6. Som...Somehow though I got the right answer on WB-6. Some luck there cause I was mixing and matching like crazy.Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-42906377531715447282011-05-10T22:09:24.789-07:002011-05-10T22:09:24.789-07:00I think WB-6 has some problems... We are supposed ...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.<br /><br />Frazzle Snazzle Puddin Pop.<br /><br />I have been working on hw too much...Jaysonhttps://www.blogger.com/profile/12297110225283255402noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-62906963750590643992011-05-10T22:08:56.115-07:002011-05-10T22:08:56.115-07:00WB-7 , Jayson , 6:11pm
LOL. Sorry, bud. There is...WB-7 , Jayson , 6:11pm<br />LOL. Sorry, bud. There is no turbine in this problem. I made a classic cut-and-paste error in the title.Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.comtag:blogger.com,1999:blog-20953269.post-69094311995177776902011-05-10T22:07:42.536-07:002011-05-10T22:07:42.536-07:00WB-4 , Jayson , 6:08pm
I did not use the 1st law o...WB-4 , Jayson , 6:08pm<br />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 !<br /><br />I hope this helps!Dr. Bhttps://www.blogger.com/profile/16971900402833290264noreply@blogger.com