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AR15.COM
8/30/2005 3:24:36 PM EDT
"Carbon Dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 2.2 kg/s and exits at 600 kPa and 450 K.  Neglecting the kinetic energy changes, determine the isentropic efficiency of the compressor."


I can't figure out the h terms becuase my book's table doesn't go down to 200 kPa.

Any help is appreciated.




Thanks
8/30/2005 3:32:18 PM EDT
[#1]
I'd go for a reciprocating compressor myself....

Taffy
8/30/2005 3:36:51 PM EDT
[#2]
just add everything together and multiply it by the page number.


... i mean, free bump.
8/30/2005 3:38:36 PM EDT
[#3]
Ask your TA or somthing.

shouldn't you know how to do this
8/30/2005 3:43:20 PM EDT
[#4]
Is that a soda machine or a beer keg?
8/30/2005 3:45:25 PM EDT
[#5]
Sorry, my Thermo book is at work.  
8/30/2005 3:50:18 PM EDT
[#6]

Quoted:
"Carbon Dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 2.2 kg/s and exits at 600 kPa and 450 K.  Neglecting the kinetic energy changes, determine the isentropic efficiency of the compressor."

Any help is appreciated.




Thanks




  Look in the tables for S and H or something like that.  Continue to multiply numbers until you get something between 0 and 1.  Isn't adiabatic constant enthalpy?  It has been a long time..........
8/30/2005 3:51:00 PM EDT
[#7]
42. the answer is always 42.
8/30/2005 3:56:21 PM EDT
[#8]

Quoted:
42. the answer is always 42.

...I thought the answer was  "B Shizznit " or "E all of the above"
8/30/2005 3:58:41 PM EDT
[#9]

Quoted:
Sorry, my Thermo book is at work.  



that sucks
8/30/2005 4:00:09 PM EDT
[#10]

Quoted:
Sorry, my Thermo book is at work.  



that sucks.

I would help but i still have a year or two before I have to take thermo.
8/30/2005 4:03:19 PM EDT
[#11]
read the chapter and do your own homework
no wonder this country is losing tech and engineering jobs to India
8/30/2005 4:12:14 PM EDT
[#12]
Here's the formula www.taftan.com/thermodynamics/ISENEFF.HTM
use the one for compression from P3 to P4.  

the mass rate m is 2.2 kg/s

You have to look up the values for the initial enthalpy (h3) and the final enthalpy (h4)

Now figure out WTF the term h3s means in that formula and you've got it made.

This explains more:  www.massengineers.com/Documents/isentropic_efficiency.htm

...well...I got farther than anybody else.  
8/30/2005 4:18:03 PM EDT
[#13]

Quoted:
read the chapter and do your own homework
no wonder this country is losing tech and engineering jobs to India



Ah, give him a break.  He's training for a management position

+1 for doing your own work.  How did I ever get my degrees before blogs existed?

8/30/2005 4:19:36 PM EDT
[#14]
I have all those formulas.  I can't figure out the h terms becuase my book's table doesn't go down to 200 kPa for carbon dioxide.  If it was water, I could do it.
8/30/2005 4:29:03 PM EDT
[#15]
Extrapolate  =  SWAG  =  guess
8/30/2005 4:38:03 PM EDT
[#16]
Thermodynamics is the devil.  I have to take it for a 3rd time next semester.  good luck my friend
8/30/2005 5:19:05 PM EDT
[#18]

Quoted:
Thermodynamics is the devil.  I have to take it for a 3rd time next semester.  good luck my friend



No, heat and mass transfer is the devil. Thermo is more like a rabid wolverine.
8/30/2005 6:39:55 PM EDT
[#19]
Remember you are dealing with an Idealized gas ( I assume)  I always screwed up and used tables to try and find gas state values for s and h.

I would look in my book, but I gotta go ta bed. Gotta be up early tomorrow.

*EDIT*

I remember that PV=mRT, and IIRC, P1V1/T1=P2V2/T2.  Thus, in adiabadic compression, P1V1=P2V2. Find the volume of the gas from the ideal gas law and use that to find the work of compression. via the Pressurte/Volume relation above.

the efficnecy is the work, not nessicarily the enthalpy. Isentropic Compressor Efficency is the

Work(Isentropic)/Work(Actual)

Since you are compressing, it will be more work to compress it for real than Ideally. IIRC, temprature is not a factor in the efficency since the process is isentropic.

So find the works of compression. That will lead you to the answer.


I dont rtemeber using tables for CO2, only for air in Thermo II.

8/30/2005 6:48:11 PM EDT
[#20]
Man, this sounds too much like Chemical Engineering.  It has been over 20 years since I had to do that.   I'm glad I started working as a Civil Eng right out of school.

Steve
8/30/2005 7:16:24 PM EDT
[#21]

Quoted:
Remember you are dealing with an Idealized gas ( I assume)  I always screwed up and used tables to try and find gas state values for s and h.

I would look in my book, but I gotta go ta bed. Gotta be up early tomorrow.

*EDIT*

I remember that PV=mRT, and IIRC, P1V1/T1=P2V2/T2.  Thus, in adiabadic compression, P1V1=P2V2. Find the volume of the gas from the ideal gas law and use that to find the work of compression. via the Pressurte/Volume relation above.

the efficnecy is the work, not nessicarily the enthalpy. Isentropic Compressor Efficency is the

Work(Isentropic)/Work(Actual)

Since you are compressing, it will be more work to compress it for real than Ideally. IIRC, temprature is not a factor in the efficency since the process is isentropic.

So find the works of compression. That will lead you to the answer.


I dont rtemeber using tables for CO2, only for air in Thermo II.




How can I find a volume when there is flow?
8/30/2005 7:45:59 PM EDT
[#22]

Quoted:
"Carbon Dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 2.2 kg/s and exits at 600 kPa and 450 K.  Neglecting the kinetic energy changes, determine the isentropic efficiency of the compressor."


I can't figure out the h terms becuase my book's table doesn't go down to 200 kPa.

Any help is appreciated.




Thanks



Look at the stuff below and see whether any of it looks like stuff in your textbook and/or class notes.  There are a bunch of assumptions that have to be made in solving a problem like this, and your prof may have given you some clues.

adiabatic efficiency of compressor that operates between entrance (station 1) and exit (station 2) is given by the following formula:
Eff = (h2s-h1)/(h2-h1)  (where h2s is the enthalpy that would result from an isentropic process that had the same pressure ratio)

Assuming a perfect gas and constant specific heats, for an isentropic process: (p2/p1) = (T2/T1)^(gamma/gamma-1), where gamma is the ratio of specific heats

gamma for CO2 is 1.293759 (according to this website: LINK
(Does your textbook or class notes indicate the gamma for CO2?)

Using algebra, solve for T2, which would be the temperature at the exit in an isentropic process.  Coincidentally, I get 450 degrees, which means the effiicency of the compressor is 100% (because the isentropic temperature and actual temperature are the same, the isentropic enthalpy and the actual enthalpy are therefore the same).  

edited to fix the formula

8/30/2005 7:49:14 PM EDT
[#23]

Quoted:
42. the answer is always 42.



+1


en.wikipedia.org/wiki/The_Answer_to_Life%2C_the_Universe%2C_and_Everything
8/30/2005 7:51:18 PM EDT
[#24]


Look at the stuff below and see whether any of it looks like stuff in your textbook and/or class notes.  There are a bunch of assumptions that have to be made in solving a problem like this, and your prof may have given you some clues.

adiabatic efficiency of compressor that operates between entrance (station 1) and exit (station 2) is given by the following formula:
Eff = (h2s-h1)/(h2-h1)  (where h2s is the enthalpy that would result from an isentropic process that had the same pressure ratio)



This is familiar.  I just don't know how to get the h values


Assuming a perfect gas and constant specific heats, for an isentropic process: (p2/p1) = (T2/T1)^gamma, where gamma is the ratio of specific heats

gamma for CO2 is 1.293759 (according to this website: LINK
(Does your textbook or class notes indicate the gamma for CO2?)

Using algebra, solve for T2, which would be the temperature at the exit in an isentropic process.  Coincidentally, I get 450 degrees, which means the effiicency of the compressor is 100% (because the isentropic temperature and actual temperature are the same, the isentropic enthalpy and the actual enthalpy are therefore the same).  



Are you sure this is correct.  According to the 2nd law of Thermo, you cant have 100% efficiency.
8/30/2005 7:58:19 PM EDT
[#25]

Are you sure this is correct. According to the 2nd law of Thermo, you cant have 100% efficiency.


Am I sure this is correct?  No.  It's been more than a few years since I took this class.  
Did your prof give you the formula above that uses gamma?  Did he give you the gamma for CO2?  If not, he probably didn't intend for you to use this approach.

I agree that you can't have 100% efficiency in real life but thermo homework assignments are about as far removed from real life as you can get.  

Edited to add:  By the way, shoot me an IM or e-mail after your prof gives you the answer and let me know what it is.  I'm curious now.