Wednesday, May 10, 2006

HW #8, P#2 - Performance of a Reversible Compressor Along Different Process Paths - 8 pts

Nitrogen gas is compressed from 80 kPa and 27°C to 480 kPa by a 10-kW compressor. Determine the mass flow rate of nitrogen through the compressor, assuming the compression process to be ...
a.) isentropic
b.) polytropic with n = 1.3
c.) isothermal
d.) ideal two-stage polytropic with n = 1.3

14 comments:

Dr. B said...

Graham 6:17 PM
You can use P1/P2=Pr(T1)/Pr(T2) for part (a) because the process is isentropic. You cannot use the relative pressure to solve the other parts of this problem.

For the non-isentropic processes, I suggest you use a polytropic PVT relationship from page 186 in the TCD Notebook.

You may be missing something else. I am not sure. But this should get you moving along the right track.

Anonymous said...

For part a) we are given y but my friend told me we have to use a R specific to Nitrogen, how do we find this, I don't see it in the tables. There book has tables for the R values.

Anonymous said...

I tried to find T2 using the polytropic P & T relationship. My friend's book has an equation specific to this which is: W comp = m(dot) * (y*R*T1)/(y-1)*[(P2/P1)^((y-1)/y)-1]

I was not getting the right answer using the polytropic relationship and then our isentropic work equation but I got the right answer using the formula from there book.

How do we get this formula? Should we not be using this formula? I'm kind of confused...

Anonymous said...

Ok, those R tables they have are just R/MW, so nevermind about that, but I am still confused about using the equation I made in my second post.

Anonymous said...

what is n???

Dr. B said...

Anon 11:33 AM
In the book they use in section A of this course, they use a different R value for each chemical. These R values are just the universal gas constant, 8.314 J/mol-K, divided by the molecular weight of the chemical.

You can see this (R/MW) term in the equations for Wsh for IG's in the table on page 201 of the TCD Notebook.

Dr. B said...

Tyler 11:39 AM
The equation you wrote is a combination of the equation for shaft work for an IG undergoing an isentropic process on page 201 (derived in Thermo-CD Lesson of the TCD Notebook and the PVT relationship for IG's undergoing isentropic processes presented on page 186. I derived this equation in Lesson 8C on page 17 and presented in in the TCD Notebook on page 208.

You can use this equation because we do indeed have an IG undergoing an isentropic expansion in part (a).

The only explanation I can think of for for what you have told me is that you did not correctly use the polytropic relationships to determine T2. You should have used T2 = T1 {P1/P2}^{(1-gamma)/gamma}.

Dr. B said...

Anon 6:20 PM
Sorry bud, I took that problem from another book. In TCD, I use delta. So, delta = 1.3 in part (b). Sorry for the confusion. I will fix that on the HW page right now.

Anonymous said...

iam using the polytropic IG eqn. on page 201, i have T2 =771 but i keep getting .016 for the mdot. am i making a simple units error? anyone else get that?
thanks

Anonymous said...

for part b that is. part a worked out fine.

Anonymous said...

yeah this is really strange, for both b and d i am getting the answer several different ways, but it seems to be off by a factor of about two. R is 8.31J/molK isnt it? what else could i be doing wrong?
in d i found the Px by the geometric avg ~ 62kpa and then using the eqn on pg 208.
is there a trick???

Dr. B said...

AdamAnt 9:34, 9:36, 12:30

It is correct to use the Wsh equations on page 201, but your T2 value is not correct ! How did you get it ? You should have used the equations on page 186 with delta = 1.3 (part b). T2 is NOT the same in part (b) as in part (a).

Yes, R = 8.314 J/mole-K.

In part (b), I think you are using the wrong T2. In part (d) you probably off by a factor of 2 because you did not take into account the fact that you have two compressors in series ? Your Px is not correct. Px = sqrt(80*480) which is not 62 kPa. Perhaps you used the wrong P ? I cannot tell, but there is no trick !

Anonymous said...

Thanks Dr.B that helps a ton! - i hate transcription errors

Anonymous said...

If anyone's off by a factor of two, make sure to remember it's N2 gas, so the molar mass is twice what you might think.