Heat of Compression with CO2 System 3

[JavaMoose]

I've looked through here and out on the web and couldn't find the solution (maybe I was asking the wrong question).

We're setting up a CO2 system feeding some equipment. We have waste CO2 piped to us with the following specs: .35 - .48 bar; 150*F; 123SCFM; 0.034 molar H20; .95 molar CO2. Ambient air temperature will be between 75*F and 90*F - this system is skid mounted outdoors.

The system as it's designed so far is using an after-cooler to bring the temperature down to 100*F and lower the dew point, filter/water trap to recover the gross condensate, then into a single stage compressor (Sullair ES-6 10H-24KT Oil Flooded) and compressed to 6.89 bar. After leaving the compressor it goes to a coalescing particulate filter then on to a refrigerated air dryer.

My problem is that I am trying to determine what the outlet temperature of the CO2 will be leaving the compressor? The air dryer has a max input temperature of 140*F and I want to be sure I'm not exceeding that.

 

[zdas04]

The calculation is really ugly. I've got an example on my web page MuleShoe Engineering Sample on slide 237 (Acrobat page 238) that goes through the steps. [It is a 7.7 MB file so it doesn't open really quickly, sorry about that] You have to know quite a bit (most importantly you have to know the oil flow rate), but all the steps are there.

I just noticed that some of the "=" signs don't show up on the version on the web page. On the top line, you need to put in "=" signs after the first variable, before the "80", before the "770R", and before the "310F". The second line is Q=M*cp*[Δ]T and the answer is 1116000 BTU/hour

David

 

[JavaMoose]

You. Rock.

Also, you're right - that is pretty ugly. What if I don't know how much I am moving per day?

It's an on demand system, it's compressed and stored in an 80gal tank (after the refrigerated dryer) - the system should only kick on a few times a day.

 

[zdas04]

That is the worst of all worlds. The compressor oil is hydrophilic (it really wants the water vapor that is in the CO2), so if you just run for a few minutes a few times a day, your oil is going to get progressively more contaminated and in a few weeks (days?) it will look like milk, be very viscous, have the lubricity of sand paper, and the surface tension of an alligator.

I would put a centrifugal oil purifier in series with a heating element and circulate the oil to keep it hot enough to cook the water out. It is expensive, but a lot cheaper than wiping bearings once a week.

David

 

[JavaMoose]

David, any chance you could give me a little insight on that equation. I'm setting it up in EES and having some trouble using your example (form the PDF) and getting the results that you did.

 

[zdas04]

I'm not sure what EES is, but can you post a pdf of the output so I can look at it?

David

 

[JavaMoose]

Sorry, EES is a software package - Engineering Equation Solver (

http://www.fchart.com/ees/index.php).

I'll post a PDF later tonight showing what I have so far.

I guess the main issue I am having, with your example, is that I'm not sure where some of the variables came from and what some of the constants are. For example, where did the 460 come from on the T_in? Same with the 770? Is 'k' in your equation the thermal conductivity of the gas? Is 'R' the individual gas constant?

 

[zdas04]

I'm sorry, I keep forgetting how foreign the FPS unit system is to the SI community.

The 460 is the conversion from Fahrenheit to Rankine.

"k" is the ratio of specific heats (cp/cv), and is widely used in any adiabatic calculation. I see it as [γ] frequently in the compressible flow literature, but mostly it is "k" in the incompressible flow world.

770R is the result of the first equation, I subtract 460R in the first line to show that without the oil the gas would have gone from 80F to 310F.

I've checked the math a few times and I think it is right.

David

 

[JavaMoose]

Ok, so here is a screenshot of the first equation in EES. However, using the same variables as in your PDF example, I am getting a different result. See anything I'm missing?

 

[zdas04]

Well, it seems that there is a mistake in both my .pdf and in your file. Your mistake is easier to find, in the last step you should have parentheses around the Tin+Rankine, you were multiplying the compression ratio term times 460 instead of times 540.

My mistake looks to be harder to find. As I was developing the example, I started out with atmospheric pressure of 12.3 psia (atmospheric pressure where the compressor was located), developed the example then decided to move the problem to sea level and didn't pay enough attention to intermediate results.

[Δ]T(gas)=746.5R=286.8F
Q(gas)=1.175X10^6 BTU/hr
[Δ]T(oil)=14.07 R
T(oilOut)=194.4F

I'm not going to apologize to you for this, I will apologize to the students who paid for the course.

David

 

[dcasto]

Just download the software from sullair or coolware for a frick screw compressor, all screws are really close in the design and operation.

 

[JavaMoose]

No need to apologize to me either way, you've helped me a lot. Thanks for pointing out my error. I wasn't paying attention either, it seems. Looks like I have it from here out, I'll post my final results in a bit.

 

[zdas04]

"Just" download Frick software. I had to send them my resume and three references that were current Coolware users (and the @6$%&^%(* checked the references), then they took 3 weeks to approve my request. With Sulgas I just pretend I work for a (now defunct)fabricator (I'm running a 1997 version that doesn't have any model produced since 1996).

The only one that is really free is Gardner Denver's Rotosize, and in my opinion it is the best (as in most straight forward and matches field conditions closest).

Even with a decent program, I think it is useful to understand where the heat goes. The link I started with is from a 2-day general class I teach that is usually attended by non-facilities (and mostly non-)engineers. I still go through the math even though none of them will ever use it.

David

 

[Montemayor]

David:

I believe this is a very important subject related to a practical application of a screw compressor and I think you have given it the importance that it deserves.

I also think you are correct in your last remark as to knowing where the heat goes and I would offer the following algorithm as a possible estimated solution to the discharge temperature in the form of a mass + heat balance. I would solve for the screw compressor’s discharge temperature in the following manner:

The compressor process is adiabatic and reversible, so I have assumed it as isentropic and therefore can employ the adiabatic discharge temperature equation – as you have done:

T2 = (T1) (Rc)^((k-1)/k)

Where
T2 = adiabatic discharge temperature, oR
T1 = compressor inlet temperature, oR
Rc = Compression Ratio
k = Cp/Cv

Knowing T2, we can solve for the heat of compression:

Q = (Wgas) (Cp gas) (T2 –T1)

Where
Q = Heat of Compression, Btu/hr
Wgas = mass flow rate of the gas, lb/hr
Cpgas = Heat Capacity of the gas, Btu/lb-oF
T2 = theoretical compressor discharge temperature of gas, oF
T1 = compressor inlet temperature, oF

Assume an adiabatic system (no heat loss or gain – just the heat generated by the compression) and a heat balance around the compressor assuming steady state:

(Wgas) (Cpgas) (Tout –T1) + (Woil) (Cpoil) (Tout – t1) = Heat of compression, Btu/hr

Where,
Tout = the final temperature of the gas+oil mixture exiting the compressor, oF
Woil = mass flow rate of the oil, lb/hr
Cpoil = Heat Capacity of the oil, Btu/lb-oF
t1 = oil inlet temperature, oF

I believe we know all the variables except Tout, the temperature value you are seeking.

I am not going to be naïve and expect that this will be the exact, correct exit temperature, but I think this estimate is more correct than the one that you show in your calculations. However, I may be wrong, and if so, would like to see someone else’s resolution to the estimated temperature.

I hope this helps.

 

[zdas04]

Art,
Accounting for the different variables (I used m(dot) for mass flow rate instead of W), that is exactly what I showed in my class handout (well, it is what I intended to show, the conversion from MathType to .pdf was a bit shaky) up until the last step.

My last step was [Δ]T(oil)=Q(gas)/(m*c(oil)) which I'm questioning now. The two inlet temperatures are very different, so your equation works out to:

T(Out)=[Q(gas)+T(inGas)*cp(gas)W(gas)+T(inOil)*cp(oil)*W(oil)]/[cp(gas)*W(gas)+cp(oil)*W(oil)]

In my example changing the final equation to this results in the temp out being 7.4[°]F cooler than my final equation. I like yours better.

Thanks

David

David

 

[zdas04]

If anyone wants to see my example with most of the nonsense gone, I've uploaded it.

David