settle out pressure/temp 1

[helixd]

thread1036-236204

Dear all,

I saw a previous post on calculating settle out pressures on compressor loops - The situation is such that we have a gas system consisting of a cooler, scrubber (remove free water after cooling) and a compressor unit. Now we assume that the ESDVs (located b4 cooler and after compressor) are activated. A recycle valve that fails open exists that connects the discharge of the compressor back to the suction (in fact the cooler first) thus resulting in a closed recycle loop. I need to determine the settle out pressure and temperature. We assume when the ESDV’s are activated, all the units stop working.

To simplify the whole question, I took in my calculations, 2 closed containers connected by a valve (to depressurise from 60 to 20barg). This can be used to generalize the above scenario. Container A is at 20 barg, 10m3, 45deg C while container B is at 60barg, 1m3, 120deg C. I also assumed z factor equal to 1.

I used the method given by the below link (similar to yours) but I’m not sure if this method is applicable to my situation. I’m not sure why some of the working done is based on standard temp and pressure (equation 4), i.e. 0 deg C and 1 atm. Also the energy balance (Q=n*cp*delta T) assumes constant specific heats, but how do the equations work out (equation 7)?

http://webwormcpt.blogspot.com/2008/11/simple-manual-method-for-settle-out.html

If I follow the procedure as per the simplified question given above, I obtain 60degC and 24barg as my settling Temp and Pressures. Kindly let me know if you think this method is applicable, or if there is any other method based on inputting P, V, T info in getting the results.

I have attached my excel working for your reference.

Appreciate your guidance to the question above. Thank you for your time.

Regards,
helix

 

[LittleInch]

Looks good to me. There are many simplifications made, e.g. red valves all close simultaneously, no metal temperatures or equipment latent heat etc etc, so your calculation can be pretty rough too. The volume calc simply gets all the different gas volumes at different pressures and temperatures all on the same basis before summing them.

The margins of error in this calc are quite large due to the simplifications involved so add a margin of 15% to the pressure and temp for any subsequent calculations such as Blowdown times etc

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[helixd]

Thank you LittleInch!

Regarding your input -

1.Is there another method to use without using this standard conditions approach? I tried doing it at room temp conditions, and got the same result.

2. For the heat balance, I understand that Cp would cancel out, but where would delta T be in the equation, as of now it Q =n*cp*T not Delta T?

3. And lastly, would you know how I can perform this simple simulation in hysys? Because its a closed system, Im not quite sure..:S

Thanks once again.

 

[LittleInch]

1) For this system I would imagine you could do it at any temp and pressure so long it was all the same one

2) Errrr, not sure what you mean, but you are just equalising the energy in the gas so need delta t somewhere

3) Dunno. I'm not a hysys jockey, I just ask the questions and understand the answers, not how it's actually done.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[helixd]

What I meant was for: Equation 7 in the link provided, how did they get that expression?

My understanding of energy/heat balance is Q=n*cp*delta T. What is provided is ns*cp*Ts=(n1*cp*T1)+(n2*cp*T2)+... where cp cancels out, and ns=n1+n2+n3+...

If I bring the Left hand side of the equation to the right hand side after cancelling cp, I obtain [n1*(T1-Ts)]+[n2*(T2-Ts)]+... =0

am i missing some physics behind this as it doesnt make sense to me.

 

[LittleInch]

What you describe with delta T is change of heat energy. What the calculation 7 is doing is calculating the overall sum of heat energy then dividing by the total of moles to obtain the average temperature

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way