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Thermodynamics question (smart people chime in)

Posted: Thu Jul 07, 2011 2:21 pm
by geardog32
Im taking a Thermodynamics course and skipped the intro course, I am fine on the concepts and problems with the exception of knowing how to read the damn property tables for instance: I have superheated refrigerant 134a at 800kpa at 35 degrees Celsius but the problem is that my tables temperatures go by tens so I have it at 800kpa 30 degrees and 800kpa 40 degrees is there a simple way of getting the 35 degrees C enthalpy value accurately at 800 kpa? This is something explained in the into course that I did not take and I obviously need to know how to get these values to do well in the course.

here is an example of such a table http://users.abo.fi/rzevenho/Thermodyna ... R-134a.pdf

If its something stupid simple just excuse my momentary ignorance as being tired and trying to finish my homework.

And don't recommend any online calculators that do it for you because I need to know how to do it the old fashioned way for exams.

Posted: Thu Jul 07, 2011 6:48 pm
by Fnord
Having no college physics knowledge I am going to speak from my posterior and ask you, would a simple PV=nrt calculation not work in this scenario? You really need to ask DYI or SB15 or maybe Ragnarok... they seem to have this stuff with tea and a bagel every morning.

Edit: After two mintues skimming research, You may need to do a specific heat calculation also, though I have no idea what the SH of 134a is. Good luck :P

Posted: Thu Jul 07, 2011 7:20 pm
by geardog32
I'm trying to find the enthalpy (H value) for r134a and all I have is that it is 35 degrees Celsius at 800 kpa and I have a plethora of tables with this data. The problem is that the tables do not have data points at every single temperature and pressure because that would be (theoretically) an infinite amount of tables. So I need to find an easy way to sort of extrapolate the data in between the given data in the tables to get the data I need for the problem.

Posted: Thu Jul 07, 2011 9:23 pm
by saefroch
Plug a bunch of points into your calculator and find a curve that fits, then use the curve to interpolate the missing value. No physics knowledge required :P

Posted: Thu Jul 07, 2011 9:28 pm
by geardog32
Thought about it, but so far in my college career I have not been allowed to use my graphing calculator in 90% of my classes. I even had a calculus professor that wouldn't let us use any type of calculator on exams and quizzes.

Posted: Thu Jul 07, 2011 9:30 pm
by saefroch
That. Is. CRAZY!

So far as I know, there is no other way to find out what it is...

Posted: Thu Jul 07, 2011 11:40 pm
by Technician1002
Graph paper and a French Curve is an alternative to a graphing calculator. This is the way we did it before graphing calculators existed. Align the curve to 3 or more points crossing the desired area and add the missing point. It should be on the curve.

Posted: Fri Jul 08, 2011 3:09 am
by al-xg
And that can also be done analytically, linear interpolation is the simplest form.

As I'm guessing you have more then just the two data points you can use more accurate models. Quadratic interpolation, general form of Newton's interpolating polynomials, Lagrange's interpolating polynomials...
Not really that complicated to do, especially for someone studying thermodynamics.

Posted: Fri Jul 08, 2011 10:41 am
by geardog32
Well I talked to my professor today and to my disappointment he told me its a best guess scenario using linear interpolation. I thought that i had to be more accurate than that but I guess my final answers just need to be + or - 5%

Thanks to everyone for your creative answers.

Posted: Sat Jul 09, 2011 11:15 am
by jimmy101
Linear interpolation is what I would have guessed, either that or you have to find the analytical equation that the graph was created with. (Which may or may not related to the underlying thermodynamics.)

Often in thermo problem 5% is more than accurate enough since there are almost always assumptions made that aren't exactly true. Isothermal, isentropic, ideal behavior etc. are often assumed and rarely true. In addition, several of the measured values tend to be of limited accuracy (and precision). Pressure and temperature in particular are tricky to measure to high precision and accuracy.