9 replies to this topic

[junior1]

Hi there. I'm bit confused at the moment. Having read various different websites and documents I'm completely lost as to why for minimum blowdown temperature (on HYSYS) study you would consider LLLL (Low-Low Liquid Level) of the vessel being looked at. Can anyone help?

[djack77494]

junior,
I'm not totally sure we're "on the same page", but assuming you're trying to calculate the minimum temperature that will be reached in a vessel during a blowdown operation, my explanation would be as follows:

Minimum liquid level = Maximum amount of vapor phase
Cooling occurs mainly by Joule-Thompson effect in the vapor space
More vapor = more cooling
Less liquid = less thermal inertia in your system

Combined, these factors generate the controlling case at minimum liquid level. You didn't state whether or not you plan to do a more accurate dynamic simulation of your system. If you do, the effects of minimum liquid inventory should be even more apparent.

[Zauberberg]

Agree with Doug's post. In addition, you may need to consider HLL liquid inventory in the case you have vapor-liquid equilibrium inside the vessel. In such circumstances, the final temperature drop due to vaporization of hydrocarbons can result in much lower final temperature in the vessel and downstream of RO/blowdown valve. This will be the boiling/bubble point of hydrocarbon(s) at final pressure inside the drum.

[junior1]

Thank you very much djack77494 and Zauberberg for your explaination. Please do bare with me, I don't have much experience with this.

djack77494:

Yes I’m trying to calculate the minimum temperature that will be reached in a vessel during a blowdown operation.

“Cooling occurs mainly by Joule-Thompson effect in the vapor space”. – J-T effect does apply to both gases and fluids so the boil-off of liquid level also means cooling takes place. Or is this not true?

“Less liquid = less thermal inertia in your system” – Could you please explain this?

Zauberberg:

“In addition, you may need to consider HLL liquid inventory in the case you have vapor-liquid equilibrium inside the vessel” – Wouldn’t the vapour- liquid equilibrium exist in all blowdown cases? Or am I mission something.

Thanks
Junior1

[Zauberberg]

In addition, you may need to consider HLL liquid inventory in the case you have vapor-liquid equilibrium inside the vessel” – Wouldn’t the vapour- liquid equilibrium exist in all blowdown cases? Or am I mission something.

Theoretically - yes. But the effect of relatively small amount of liquid on heat transfer will be almost negligible, based on my experience. Furthermore, liquid will start to vaporize as soon as blowdown process is initiated so what will remain at the end - in most of the cases - is that the influence on temperature profile is very small (it is something similar to the term Doug has used, "thermal inertia"). As a rule of thumb, in systems where vapor presents more than 70% of vessel inventory the resulting temperature will be the one calculated by using Joule-Thomson effect; if system is liquid-dominated, the final temperature inside the vessel will reach bubble point of hydrocarbon mixture at the final blowdown pressure.

[djack77494]

“Cooling occurs mainly by Joule-Thompson effect in the vapor space”. – J-T effect does apply to both gases and fluids so the boil-off of liquid level also means cooling takes place. Or is this not true?

You may be correct if you are stating that something akin to the J-T effect applies to the liquid phase as well as the vapor phase. I'm not sure that the thermodynamic effect for a liquid would still be termed J-T effect. Pressure effects on the liquid phase however are small, but other things are taking place (see below).

“Less liquid = less thermal inertia in your system” – Could you please explain this?

The mass of the liquid phase is >>> the mass of the vapor phase.
The heat capacities of the two phases are not that dissimilar
Therefore (mCp)liquid >> (mCp)vapor
I think of the mCp term to be like a "thermal inertia".
Therefore, the thermal inertia is governed by the liquid phase.

Further Comments:
Perhaps wrongly, I was envisioning your system to be a liquid-vapor containing vessel. In most common situations, as you drop the pressure of a vapor, it expands and the temperature drops (the famous J-T effect). THe liquid will have a lower imposed pressure, but the effect of pressure on the thermodynamic properties of the liquid phase will be very small. The big contribution of the liquid phase is due to vaporization, which is NOT part of the J-T effect. Vaporization of a saturated liquid will occur as pressure drops, and its temperature will drop down to the point where it is at its equilibrium temperature at the new pressure. For liquified gases and light streams, that temperature (the boiling point for a pure component) could be quite low.

I apologize if this is not as clear as it could be. Please feel free to expand on any of these points and I'll try to provide a better explanation.

[junior1]

Thanks very much.

How does boil-off result in lower temperatures? (Just to ensure that I have correct understanding of this).

[djack77494]

It's not so much that boil-off drops the temperature. It's more a matter of reaching a new thermodynamic equilibrium point. With the lower pressure comes a lower temperature. As long as you have your 2 phases in equilibrium, the pressure will fix the temperature.

[Art Montemayor]

During a process vessel blowdown procedure, the sequence of event is exactly as Doug has explained. However the resulting lower temperatures realized are not only due to the principles that Doug has also detailed, but they affect different portions of the vessel and its connected components. The operation involves two basic unit operations:

1) The free expansion of a compressed gas; and,
2) The vaporization of any compressed liquid remaining in the vessel - leading up to a steadily changing phase equilibria.

The first effect is just as Doug has described it: it is a classic Joule-Thomson expansion resulting in a cool-down of the escaping, lower pressure vapors/gases. This cooling effect is downstream of the blowdown valve and differentially starts to cool down the blowdown piping, the blowdown valve and everything that can be found downstream of the blowdown. As long as the vessel is blowing down, there will be no cooling of the blowdown valve and the upstream, pressurized piping and vessel - at least not upstream of the blowdown valve's seat. The reason for this is that the constant flow of the relatively warmer vessel vapors that are flowing under pressure towards the valve's seat will pick up any coldness that the valve transmits upstream through metal conduction. As soon as the blowdown flow ceases, the cooling of the upstream piping will start (through conduction). This will be a relatively slow process and will not affect the contents that remain within the vessel.

The second effect is more pronounced in reducing the vessel's contents and walls temperature because it is a result of liquid vaporization due to a decrease in its pressure. This is the same effect that occurs in a normal mechanical refrigeration's vaporizer coil: the liquid acts as a boiling (actually vaporizing) refrigerant. As can be appreciated, this effect is much more pronouned and quicker.

Junior fails to identify what he calls the "minimum blowdown temperature" - is it the temperature of the vapors/gases exiting the blowdown valve? - or the temperature of the vessel itself? I believe he is referring to the former rather than the latter, in which case then he should concern himself with the Joule-Thomson effect - which is undergoing a differentially lower-and-lower temperature approach as the vessel contents themselves are feeding a progressively lower pressure as the vessel temperature decreases during the total blowdown procedure. This is a complex problem and involves differential equations to follow the complete progression of cold produced.

As Doug has inferred, the amount of original liquid (hydrocarbon, I must presume) has to be taken into account if one is to make a complete analysis and a heat and mass transfer of the operation. If the original liquid inventory is not sufficient to generate vapors during the blowdown time, then the Joule-Thomson effect will dominate at the end. However, if the amount of liquid inventory at the outset is sufficient to maintain a constant feed of steadily vaporized and progressively colder vapors to the blowdown valve, then the vaporization effect will dominate the operation in fixing the resultant low temperature at the end of the operation.

I hope I have succeeded in explaining what I believe is taking place during this type of blowdown.

[junior1]

Thank you very much guys, the explanations are fantastic!!

I have been asked to verify and highlight any problems with the flare design with the current design modifications. So I'm looking at all of the design constraints (i.e. mach no., back pressure and minimum temperature - both upstream and downstream of the blow down valve). So for back pressure investigation, for example, for a vessel depressuring I would take the HHLL (High-High liquid level) with fire scenario (the vessel is exposed to fire) as this will give me the highest depressuring rate and thus the highest back pressure. Now for the lowest possible temperature per depressuring of this vessel I would take the lowest ambient conditions possible and as I was lead to believe the lowest liquid level (LLLL) as this will give me the lowest temperature both upstream and downstream of the BD valve. Another colleague pointed out that in different company they used HHLL as this would give the lowest temperatures so I was confused as to which one of the approach is right and why.

My apologies for very little explanation to the problem I face, I should have explained the situation in detail to start of with.