Pipeline Cooldown 3

[limeister]

How does one go about doing this?

If I have a 1" pipe that is 50 meters long made out of stainless steel.

Its at ambient temp. Then you flow liquid O2 at -186 C through it. If it takes about 30 minutes to bring the pipe from ambient to operation temp of -186C. How do I calculate the amount of flash?

Yes I am still learning the stuff.

What do I need to know and how do I do it?

Thank you

 

[quark]

Make a heat balance of steel and oxygen.

Latent heat of oxygen is 50.94 kCal/Kg
Density of Steel is 7843 Kg/M³

Specific heat of stainless steel is 0.11 kCal/Kg deg.C
for 1" pipe assuming sch. 40 pipe ID is 0.027M (1.049&quot
and OD is 0.033M (1.315&quot

So mass of 50 M pipe is
3.142(OD²-ID²)xL*7843/4 = 110.89 Kgs

Heat to be removed from pipe to cool it to -186⁰C from 30⁰C (assuming ambient) is therefore

Q = mCp(T1-T2) = 110.89 x 0.11 x [30-(-186)] = 2634.7 kCal
Amount of oxygen to be evaporated comes out to be
2634.7/50.94 = 51 kgs. (Heat to be removed/Latent Heat)

I didn't consider sensible heat rise for a temperature difference of 3 ⁰C. (boiling point of oxygen is
-183 ⁰C)

If you want to maintain the pipe temperature within 30 minutes you have to evaporate oxygen at the rate of 1.7kg/min.

Note: This is the minimum quantity of oxygen because in practical case you have to consider heat flow into the system from ambience.

 

[limeister]

Hello, Quark. Who ever you are. Thanks alot.

I have just recently started my job as a chemical engineer. thing is I don't have any background. I majored in chemistry in university. I guess this was a an easy problem to solve.

This forum will be a big help to me.

I learn new things about the chemical engineering field everyday.

 

[25362]

Is the pipe insulated ? In what surroundings, humidity of air, etc. ?

 

[kenvlach]

Quark,
it doesn't make sense to ignore sensible heat.
Isn't the vaporizing O2(g) exiting at the far end of the pipe? And absorbing heat as it goes? And from thermodynamics, gases have higher Cp's per mole than condensed phases & they usually weigh less per mole.

I made a rough estimate that it only takes 40 Kg of O2, with an av. O2(g) exit T of -117.4 C (a rise of 68.83 C).
From a thermodynamics book:
Cp(O2,av.123 K) = 6.96 cal/oC/mole = 0.2175 Kcal/oC/Kg
and used your value for SS.
Then Latent heat + sensible heat for oxygen
= 40 Kg x 50.94 Kcal/Kg + 0.2175 Kcal/oC/Kg x 68.63oC x 40 Kg
= 2037.6 + 597.1 Kcal
= 2634.7 = Heat removed from SS

Does it make sense for modelling that the boiling of the O2 begins at the front end and progresses 5/3 meter/min?
Initially, the O2(g) flows the 50 meters & exits at -49 C, from the halfway point it exits at -114 C, and at the finale, O2(liq) exits at -186 C.
Of course, it would have helped to find a a heat transfer coefficient, and then do a few iterations to get a fairly precise value.
And, it could be that 51 Kg is the right amount since non-adiabatic conditions, but presume they use good insulation.

limeister, what kind of insulation's on that pipe?
And, are you using Sch. 40 pipe? I used Sch. 10S on my steam system.
Ken

 

[quark]

Ken!

I fully appreciate your efforts for perfecting the solution and you are right that we have to consider sensible heat rise too.

You are also right that the sensile heat gain starts approximately from 5/3meters for 40kg/hr flow rate.

How could you arrive at the final temperature of gaseous oxygen? I just wanted to know the basis of your calculation. Assuming some convective coefficients I am getting it much higher.

However I have the following concerns.
1. It cannot be assumed that for a full pipe cross section oxygen is being evaporated. This may affect superheating process.
2. I don't know whether there is sufficient pressure drop in the system to evacuate gaseous oxygen at the rate of 35m/s. If not so error in my calculation increases , because heat from pipeline is utilized to superheat gaseous oxygen rather.

Once again thanks.

 

[CHD01]

While you may be able to evaluate the boundary conditions and limits for operation; I don't believe one can solve this problem without the mass flowrate and initial operating pressure for liquid O2 given? Not only is it important for heat balance calculations to determine time for cooldown; but it determines if flashing occurs and how much as I see it?

This is a very complex problem, we do need to know the insulation thickness as well, plus whether solar heating affects are to be included; and if so what the orientation of the pipe is to the sun?

The initial pipe wall temperature will have to be determined based on a heat balance (again ignore solar affects) between atmosphere and OD of insulation due to convection, conductive heat through insulation, conductive heat through pipe wall, and convective heat between inside pipe wall and flowing fluid. We will assume scaling to be ignored also.

Then you will have to set up an excekk sheet to evaluate exit pipe wall temperature based on a known flow. YOu have one check for your calculations, which is that it takes about 30 minutes to cool down the pipe.

If I was an instructor, this might be the perfect problem for a final (this might be solved in 2 hours but I doubt it).

Like I said, this is a tough one.



The more you learn, the less you are certain of.

 

[kenvlach]

Quark,
It’s a very interesting problem, and I’m sure you now realize its complexity.
I maybe had an advantage in that I have calculated both cooling coil (Freon system) and heating coil (low pressure steam) lengths for cooling & heating process tanks, respectively. In both cases, rate determining step is on the exterior surface of coil. None of my coils are anywhere near 50 meters, nor is the temperature difference quite so great (especially not for my cooling coils), so I take sensible heat into consideration. Also, calculated steam condensed during initial heatup of my steam header in order to size the dripline trap.
Now, for limeister’s pipe, I figured the vaporizing O2 cannot escape fast enough to avoid some back pressure, so there is some mixed phase O2(g,l) range. On a chiller system, the compressor creates suction to pull the warmed gas out, whereas for the O2 I presume it vents to 1 atm at the end of pipe. Hence, I assumed that there is a linear pressure profile extending from the end of the single phase O2(l) to the end of the pipe, and corresponding O2(g) temperature & velocity profiles. I didn’t know the pressure, but I knew that it increased residence time & hence heat transfer. It seemed reasonable to take the length of the O2(g) path as 50 – (5/3)*t in meters for t in minutes. And, like I said, I remembered from thermodynamics that gases usually have much higher Cp’s than do solids.

However, I saw right away from the moving boundary conditions that an analytical solution is not possible; even if I had the necessary heat transfer coefficients (which vary as functions of varying dT & dP), and could assume perfect insulation. It would take a reasonable model and a few computer iterations to get a good number.

So, in true engineering fashion, I made a wild a__ guess about the what fraction of sensible heat would be transferred, assumed it would be linear with the length that the O2(g) traveled and from that an average dT for an av. path of 25 meters. Used your figure for the SS sensible heat, solved for an O2 mass using both the latent & sensible heats, then I rounded off the mass to 40 Kg and redid the dT. Although the 40 Kg figure is a only a rough estimate, I knew it was better than the 51 Kg.

Re your concerns: Yes, there is a significant back pressure and a certain amount of mixed phase flow. If the 40 Kg/30 minutes was correct, it would generate a significant gas volume, 41.67 moles/min = 933.9 L/minute if at STP (0 C, 1 atm). Since the pipe ID gives a flow area of 22.9 cm2 = 0.0229 L/cm, the gas volume is 407.8meters/min which is pretty fast, huh?
The true instantaneous gas volume is less & viscosity & density higher since lower actual T. By calculating convection or using information from a refrigeration book, I’m sure that you can get a more accurate value for the O2.

Please post your final results.
Ken

 

[kenvlach]

Oops, forgot to divide by 4 when calculating the pipe inside xs area. area is 5.7255 cm2 = 0.0057255 L/cm = 0.57255 L/m).
The flow rate should be (933.9 L/min)/(0.57255 L/m) = 1631 meters/min (for gas idealized to STP) or about 60 miles per hour.

 

[quark]

Ken!

Greatest disadvantage for us is, we lack the flexibility of having a physical look at the system and without that I personally feel anything I do is a guess work.

What I intended to do in my first post was to give an idea about the calculation method. If I have to actually calculate the process, I first doubt the cooling time of 30minutes.Secondly I consider this process as transient conduction process and I have no references where I can get some correlations with inner flow of fluid.Finally I couldn't still understand the actual significance (is it only to calculate wastage of liquid oxygen or something related to process)

But you really did a good work.I feel I benefit more from you in future.

Cheers

 

[CHD01]

I agree the 30 minute cooldown sounds wrong. But again, you can do preliminary calcs as you are doing them to establish boundary operation; but how can you conclude anything without the intial and final pressure? The cool-down time will be determined by the imposed upstream and downstream pressure which determines the mass flowrate and the overall heat balance as mentioned and whether any flashing is going to occur - or am I missing something? Attached are physical properties for O2 at various temperatures:

Deg C Vapor Pres (Psia) Liq Den lbs/ft3 Liq Vis cP
-118 and hotter O2 cannot exist as a liquid
-125 569 44.4 0.07
-150 178 59.4 0.10
-186 10.6 72.2 0.21
-200 1.6 76.3 0.32


For an example of the problems we get into (please excuse the english units - but my spreadsheets were built this way prior), with 200 psig @ -186 Deg C initial conditions, exhausting to atmosphere; flashing and 2-phase flow will certainly occur. If the downstream pressure was held at 150 psig, the flow would reduce to about 19M pph for the 50 psi drop with a 12 fps velocity for the 1-in sch 40 pipe. At 19M pph doing the heat balance without any insulation installed, the pipe wall (very little resistance)has to be cooled down in about 14 seconds (50 meters of pipe at a fluid velocity of 12 fps) - so we know the flow is a good bit less for a 30 minute cooldown (if true as stated). For a 30 minute cooldown, the flow required is about 147 pph (about 0.25 gpm equivalent) with a velocity of 0.09 fps. The problem is that when we go back and run a heat balance at this reduced flowrate the ending liquid O2 temperature is the atmospheric temperature which we know will not be the case since O2 cannot exist as a liquid above -118 Deg C. So flashing and 2-phase flow is indicated again, and my assumption that O2 remains a liquid at this pressure is wrong. Put another way as flow decreases the environmental heat gain becomes significant! So now a difficult problem has just increased in complexity.

I have no problem with the calcs done by all - its just that any of our calcs (including mind) are kind of worthless without more information as I see it. Again, have I missed something?


The more you learn, the less you are certain of.

 

[limeister]

My goodness. The response on this question has been overwhelming. I must apologise for the lack of information I have given to you all.

This was a problem my boss wanted me to do in a very crude manner. We were just wanting to know the many different scenarios that could result in flash/loss of O2.

Sorry to say but all these calculations have totally confused me. I have been at my job for about half a year and I am still trying to learn about the plant.

There are some things that you may want to know:

1) The 30 minuted cooldown was an approximation. I simply asked the operators how long it usually took. Since factories aren't meant to be cooled down/warmed up very often I could only estimate what the value is.

2) Obviously there is insulation around the pipe. My boss just told me to ignore it.

3) The pipe is actually less than 50 m but then again my boss just said "use 50 meters." The pipe is not straight but actually has several bends in it.

4) I really appreciate you time and effort into this matter but the depth of all your knowledge is certainly beyond me.

 

[kenvlach]

Comment for limeister & Quark & other new engineers:

Engineers are pretty much only needed to design systems (and to analyze failures afterwards, if applicable), not for operation. If it wasn't for solving transient condition problems, there would be much less need for engineers. For example, a factory owner will want new equipment designed for a specific heat-up or cool-down period for production requirements. You don't want to pay workers to wait while a process solution reaches temperature. And, that process is connected to others, so a ripple effect. Hence, you must learn to solve transient problems, either numerically or empirically using nomographs such as from HXer mfrs. like Tranter or in Electroplating Engineering Handbook, etc.

 

[quark]

Ken!

I think this is morethan what I can take subjective to the actual post. Now I will tell you frankly why I was so careless in my first reply.

I exactly figured out why the original post was asked (perhaps from my experience)

When your boss asks questions of this sort, there are some standard assumptions.

1. First of all he doesn't know how to do otherwise he could not have given you such a difficult problem.
2. He somehow wants to give you some work so that you don't roam around the plant.
3. He wants to check whether you can take your future responsibilities in a proper way.

The general practice what I did was (this is a suggestion to Limeister)to trivialize the solution, showing it to boss, asking for his estimation and correcting the solution as per his estimation. So everybody is happy at the end of the day.

Designing without operational knowledge is like a blind man trying to hit bulls eye.

By the way, which plant owner asks you to desing after commissioning of the plant?

Note: Neither I am a new engineer nor a student.

 

[kenvlach]

Quark,
maybe I was a bit preachy. I have some 'boss experiences' to share. My experience is rather different, the boss always assumes a quick and easy solution to a difficult problem.
1) "What's the matter with our crane? It won't lift this part."
--- turn out boss had estimated weigh by rule of thumb for aluminum, but the part was SS. 1-ton crane couldn't lift 2+ tons.
2) "Why are we spending so much money on electroless nickel (EN) chemicals?"
--- I did some bookkeeping & found a) we were spending less than the industry norm for chemical costs as % of sales, and b) the EN plater was due a bonus based upon 3 months steady increase in sales. Big chill, then CFO shifted some figures to cheat the plater out of his bonus.
3) "You were wrong, we didn't have to do your [fill in the blank for various preventative maintenance] and everything is fine."
---[months later] "It's costing me money while [whatever piece of equipment] is down]. Fix it now."
4) "What do you mean we can't do this electropolish job because we don't have enough electricity? Are you so dumb you don't know we can rent a couple diesel [powered] generators and bring the power in through the window?"
--- none of the electricians we brought in would risk their licenses to make the connections he wanted, anyway.

P.S. My experience is that lots of modification goes on as owners always want to squeeze in more process lines & equipment, handle bigger parts, more colors, run longer hours, & skimp on maintenance so lots of opportunity for repairs under pressure (both figuratively & literally --
"we can't afford to let the boiler cool down&quot. This probably varies by industry, though.
Guess I'm both work hardened and fatigued.

 

[quark]

Ken!

More or less it is same all over.These things won't change untill and unless we make our bosses fold their sleeves and dig their hands into work. But coping up with these things is what an engineer should get from experience. Technical acumen is rather easier to get.

That is why my famous line is "I want to be TC the Top Cat" and sometimes I take inspiration from that cartoon hero.

Regards,