How do they get this Exergy Value?
How do they get this Exergy Value?
(OP)
Hi , i hope for all thermodynamics experts (especially in the fields of exergy) out there, please help me. I am curious about a certain value, i am not an experienced exergy finder for a mixture (in this case ammonia-water) , here is the link to the paper :
http:/ /ecow.engr .wisc.edu/ cgi-bin/ge t/me/761/k lein/lectu re/kalinac ycl/kalina simulation -vidal.pdf
What got me confused in the paper was in page 8 about the table. I know that X is mole fraction, so it is given, alongside m, h, and s. But i really don't know how they got the exergy value. I tried using equation (2), but didn't get the result, and also if i use equation (1), do i have to use h0 and s0 of the ammonia-water mixture? If so, where does X (mole fraction) fit in?
Can anyone give me a link on how to calculate the entropy and enthalpy of ammonia-water mixture?
Thanks
http:/
What got me confused in the paper was in page 8 about the table. I know that X is mole fraction, so it is given, alongside m, h, and s. But i really don't know how they got the exergy value. I tried using equation (2), but didn't get the result, and also if i use equation (1), do i have to use h0 and s0 of the ammonia-water mixture? If so, where does X (mole fraction) fit in?
Can anyone give me a link on how to calculate the entropy and enthalpy of ammonia-water mixture?
Thanks





RE: How do they get this Exergy Value?
RE: How do they get this Exergy Value?
As far as I can see, it is a good paper with lots of information.
I can only provide you with a few hints that might help you with the subject matter covered by the paper.
When dealing with thermodynamics, one needs to know that physical properties are well defined and numerical values known -- for some reason, basic thermodynamics classes never get a chance to cover physical properties. For example, the reference temperature is 25 [C] and/or 298 [K], while pressure is 1 [atm]. At such conditions, enthalpy of carbon, hydrogen, sulfur, oxygen and nitrogen is equal to zero. Based upon experimental measurements, all other enthalpy values are obtained. Furthermore, enthalpy, entropy and specific heat values are known. As a result, all other physical properties can be calculated.
Here is a URL that can provide some additional explanation to you when it comes to physical properties:
http://www.engineering-4e.com/properties.htm
At this point, my suggestion to you is to go back to the paper and read it a few more times.
The paper provides one with a phone number, authors' names and school URL. If you need additional explanation, my suggestion is to go directly to the authors -- in my opinion, they will be more than happy to address your specific questions ...
http://www.engineering-4e.com
RE: How do they get this Exergy Value?
http://www.engineering-4e.com
RE: How do they get this Exergy Value?
http://www.engineering-4e.com/calc1.htm
For exergy, exergy = h - ho - T(s-so)
http://www.engineering-4e.com
RE: How do they get this Exergy Value?
RE: How do they get this Exergy Value?
http://www.engineering-4e.com