Compressor Efficiency

Organized by textbook: learncheme.com/
Calculate the efficiency of a compressor that uses an ideal gas. Made by faculty at the University of Colorado Boulder, Department of Chemical and Biological Engineering. Reviewed by faculty from other academic institutions. Check out our Thermodynamics playlists: www.youtube.co...
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[@LearnChemE]

@harish eswar To calculate T_2rev, you start with the equation at time 2:28. Since you know all other values and that it equals zero, you can solve the equation. Move the second term to the other side of the equation: 7/2R(ln(T_2rev/T_1)=Rln(P_2/P_1). The R's cancel and P_2/P_1 = 8, so 7/2(ln(T_2rev/T1)=ln(8)=2.079. Multiply both sides by 2/7, so ln(T_2rev/T_1)=0.5941. Take the exponential of both sides: T_2rev/T_1=e^0.5941=1.811. T_1=373K, so multiply both sides by that and you have the answer: T_2rev = 675.7 or 676 K. Thank you for your question.

[@thetmonaye]

i m not clear about 1:30.. how do u get the yellow bit?

[@LearnChemE]

This screencast has been reviewed by faculty from other academic institutions.

[@justinemangachena7341]

Great presentation. It's straight to the point

[@deadawake123]

How do you use Cp when the pressure is NOT constant through the process?

[@nikan4now]

Thanks for the video. Perhaps it must be noted that reversible doesn't mean dS=0. Internally reversible and adiabatic mean dS=0

[@LearnChemE]

Yes, compressors are typically assumed to be adiabatic.

[@leofranko2657]

yes...as in reversible process only we dont know whether there is heat transfer or not...thus we can only say that S(gen) generation of entropy is zero..but not the entire entropy change of the system

[@dhrumilvaland3807]

can I use this for screw compressor calculation?

[@26jasti]

Why did you use Cp, when the pressure is not constant through the process. Should not you use Cv

[@johnspacetime3264]

Thank you so much kind sir

[@ankurjahagirdar]

This is true for which type of compressor?

[@harisheswar6319]

how do u get the T2 in rev case can u plz solve elaborate

[@dume85]

lnx-lny = lnx/y
5lnx=lnx^5
e^lnx=e^0 =) x=1
with that you should be able to rearrange the equation to (T2/373)^3.5=8
(8*373^3.5)^(1/3.5)=676
oh also T2^3.5/373^3.5 =(T2/373)^3.5

[@petersnowden1332]

How do you get T2rev = 676K? By my calculation it's 3971K every time, what am I doing wrong?

[@LearnChemE]

T2=T1[P2/P1]^(2/7)= 373(0.8/0.1)^0.286=675.7, where 2/7 is R/Cp

[@petersnowden1332]

Thanks! Got it now, I was just applying my ln rules incorrectly!

[@danjumajoseph2890]

Pls how can I get the molar flow rate of the compressor

[@edwardgustaf7399]

Danjuma Joseph it must be a known data.
Practically, in a plant you'll see normal volumetric flow (Nm^3/h) that can be converted into molar flow by using equations of state. Usually, Normal condition refers to 273.12K and 1 bar. So that you can calculate the molar flow.
Hope this answer your question.

[@Luzt.]

@@edwardgustaf7399 Normal conditions refer to 20degC, standard - 0degC.

[@edwardgustaf7399]

Luzt it depends what standard you are using. In term of Nm3/h the standard conditions are as I mentioned before.

[@Luzt.]

@@edwardgustaf7399 I did a little research and:
- "normal" generally refers to normal in a sense of laboratory ambient conditions, i.e. 20degC,
- "standard" generally refers to 0degC,
- both Sm3/h and Nm3/h are in use,
- it makes sense to use S for Standard and N for Normal.
I actually don't care who is right/wrong. I just wanted to clarify it for myself. Have a nice day.