If you have any questions related to Baratuci's Final Exam, please post them here instead of using email.
I can answer Thermo questions for Castner's students, but I cannot tell you anything about your final exam. The answer is...I don't know.
It has been a pleasure working and blogging with you.
Best of luck on the final exam !
Adios,
Dr. B
29 comments:
I am working on practice exam for the spring 2001 called the "brayton refrigeration cycle: ideal and real" and for the life of me I cannot fine the efficiency for part a because either my picture is wrong and/or I just cannot find the temperature before the Qc and the temperature after the Qh. I have all the information for the input of the compressor and output, but everything else is caput. Please help with any equations I am missing.
Anon 11:47 AM
This problem is screwed up. Part (a) is not possible. You should feel good that you did not find a way to do this !
Part (b) is ok, so try that and skip part (a).
I will put a note about this problem on the course website.
why can a closed cycle not be a picton and cylinder device? (pg. 223)
If a Carnot vapor power-cycle is impractical, what is a practical Carnot cycle?
dumb question im sure but, when finding the wsh for a turbine for example and making the cancelations as usual (assuming of course what we usually assume) its h4-h3 for instance. But how come when i go to find Q and i do the first law its somehting like mdot(h4-h3). How come we can "ignore" the mass flow rate for the turbine for example. In many problems i assume this and it somes out correct, but in the first law its min times (energy stuff) - mout (energy stuff). Even when i cancel terms i dont get rid of the mdot. I thought it could be something about ss but arent both the turbine and the condenser for example ss?? confused
ah its the difference between finding qdot and just q isnt it? ie finding qdot you need to mulitply by the mass flow rate and just q you dont...i think
could you help me out and double check my reasoning for these two questions??
A cascade refrigeration cycle has a higher COP than a basic vapor compression refridegeration cycle that operates between the same two reservoir temperatures— TRUE because you have a greater Qc?
A multi stage refridegeration cycle has a higer COP than a basic vapor compression ref cycle that operates between the same two reservoir temperatures— TRUE
im trying to double check on thermo CD but i thought id post this asap so i can get your comments
Anon 3:16 PM
A closed cycle is a cycle in which the working fluid is NOT discarded (think of your refrigerator where R-134a is the working fluid).
An open cycle is a cycle in which fresh working fluid undergoes a cycle and is then discarded (think of an automobile engine where air is the working fluid).
On page 223 I was trying to emphasize that closed cycles are generally a series of open systems like pumps, turbines and heat exchangers. Just because the cycle is closed does NOT mean it must take place in a piston-and-cylinder device.
Good question.
Anon 3:24 PM
The Rankine Cycle is a practical vapor power cycle is. It is NOT a Carnot cycle and it is not reversible.
Chris 3:28 PM
1st Law, open system with Ekin and Epot negligible is:
Qdot - Wsh,dot = Mdot * deltaHhat
You cannot ignore the Mdot for a turbine.
But another way to write this eqn is by dividing by Mdot:
Qhat - Wsh,hat = deltaHhat
where Qhat = Qdot / Mdot and Wsh,hat = Wsh,dot / Mdot.
The issue is the difference between RATE terms like Qdot and Wsh,dot and specific quantities like Qhat and Wsh,hat.
Once you have the difference between terms like Q, Qdot and Qhat, everything will work well for you.
Chris 3:49
Yep, you've got it. Good work.
Anon 5:02 PM
1- True: Qc is bigger and Qh is smaller. COP = Qc/Wcycle = Qc/(Qh-Qc) or COP = 1/[(Qh/Qc)-1]
So, as Qh/Qc drops and approaches 1, COP approaches...infinity. So COP definitely increases.
Anon 5:02 PM
2- True for the same reasons as on the cascade cycle. See page 265 in the notebook.
im working on the 2004 final and im on question number 3. Im curious as to why i have to use the MEBE to determine the pump work by assuming incompressiable. If the pump is isentropic, cant i just fund entropy at 4 set that equal top entropy at 1 and solve for the enthalpy at one beacuse i know the pressure, which is 3.5MPA???
Anon 7:53 PM
Either method is fine. The results should be very similar (within 1% unless the change in P is very large).
The isentropic method is easy if you have NIST of the Excel plug-in available, but it is not so fun with tables.
So on the first quiz answer thingy it says that a temperature change of one degree celsius is equal to a temperature change of one degree farenheit. That's bad
i'm sure this is a stupid question but why is "The determination of the change in molar internal energy for an ideal gas by integrating the constant volume heat capacity from T1 to T2 is aplicable only to constant volume processes." false? btw 2 p's in applicable
How do we interpolate between the subcooled liquid tables and the saturated liquid values? Like if I have a pressure below 5 Mpa and an entropy value.
In Rankine cycle, is the inlet of the pump always sat.liq?
Ninja 11:54 AM
Doh ! No. A temperature change of 1 degree celsius is equal to a temperature change of 1 degree Kelvin.
Anon 12:19 PM
Thanks for spotting that typo in applicable.
The U of an ideal gas is a fxn of T only. So, deltaU = INT{Cv dT} for any forcess, not just constant volume processes.
Whats the difference between little q and big Q?!?
Anon 2:27 PM
How about an example. Calculate Hhat for subcooled liquid H2O at 100 C and 1 MPa.
At T = 100:
P H
101.42 419.17
1000 Hhat
5000 422.85
Hhat = 419.845 kJ/kg
Anon 2:54 PM
Yes, the fluid leaving the condenser is a saturated liquid unless the problem tells you otherwise.
help 8:26 PM
q = heat flux = Q/Area = W/m^2
That is a nice and useful blog there. Do visit mine at
http://steamcenter.blogspot.com
I have put a link to this blog fom mine at
http://steamcenter.blogspot.com
under the title thermodynamics
As regards the question about carnot cycle being impractical. it is so because it is difficult to pump and expand mixtures of liquid and vapor. Therefor it is modified so that pumping and expansion can be done with a single phase. It then becomes a rankine cycle. However now heating the pumped liquid involves temperature change and that cannot be done reversibaly so the cycle is a little less efficient. Using feed water heaters reduces this irreversibility but there is a limit to the number of feed water heaters one can employ.To make the cycle perfectly reversible one would need an infinite number. Modern power plants use far too many feed water heaters in any case. For more on that read my blog
Dr. B thanks for dropping into my blog at http;//steamcenter.blogspot.com/
I have a link to your blog from mine because it is very useful for those learning thermodynamics
Post a Comment