"BOILING" vs LIQUID LEVEL dP INSTRUMENT Indication

[CHD01]

Given a vaporizer with an external sight glass providing indication of liquid level in vaporizer, it can be observed that on sudden loss of vaporizer pressure, that the liquid in the sightglass boils with liquid swell occuring which lifts the liquid in the sightglass. The appearance of the liquid is that it is foamy - much like opening a soda bottle. The question is what is the net affect of this boiling relative to what the liquid dP instrument will indicate as a liquid level. The arguments are that: 1) the boiling reduces the density (SG) of the liquid resulting in an indicated liquid level in the sightglass lower than it actually is in the vaporiser, 2) that the pressure of the vapor displacing the liquid results in a greater level in the sightglass than actually exists in the vaporizer or 3) that both are occuring. If both occurs, which is more pre-dominate?

Any ideas or comments will be appreciated The more you learn, the less you are certain of.

 

[CHD01]

Sorry, the attached revision will correct an error in the aurguments of the original memo.

Given a vaporizer with an external sight glass providing indication of liquid level in vaporizer, it can be observed that on sudden loss of vaporizer pressure, that the liquid in the sightglass boils with liquid swell occuring which lifts the liquid in the sightglass. The appearance of the liquid is that it is foamy - much like opening a soda bottle. The question is what is the net affect of this boiling relative to what the liquid dP instrument will indicate as a liquid level. The arguments are that: 1) the boiling reduces the density (SG) of the liquid resulting in an indicated instrument liquid level lower than it actually is in the vaporiser, 2) that the pressure of the vapor displacing the liquid results in a greater instrument indicated liquid level than actually exists in the vaporizer or 3) that both are occuring. If both occurs, which is more pre-dominate? Is there a good way to calculate this?

Any ideas or comments will be appreciated The more you learn, the less you are certain of.

 

[joerd]

It seems to me that there are two effects at the same time, as you indicated under 3. The initial dP = SG*g*h where h = liquid level and g = gravitational constant. SG*h can also be viewed as m/A where m = mass of liquid in the cell and A = cross sectional area of the cell.
If the pressure is suddenly lost, the liquid will swell and there will actually be a two-phase mixture in the cell with some mixture density in between the vapor and liquid density. Initially, however, the total mass will remain the same. Therefore, the cell's indication will not change and will still represent the liquid level in the vaporizer.
Provided that the vaporization rate will be more or less the same in the vaporizer and in the dp cell, the indication should be close to the actual level in the vaporizer. The pressure drop of the vapor flowing from the cell to the vaporizer will of course result in a slightly lower indication of the level than it should be.

I am curious why you would still be interested in seeing the level as the pressure suddenly drops, since I would assume that an operator will have other things to worry about in such an event? Regards,

Joerd

 

[CHD01]

It is during startup of the vaporizer and the rest of the process that we are intereted in seeing the level as the pressure drops. We must initially bring the vaporizer up in pressure, then crack open the vapor supply valve to the rest of the process; no matter how carefully done this results in a lowering of vaporizer pressure and the observed boil-off.

Rather than infect your opinion with what I think, let me give you a few more facts:

1. The dP instrument is tied into the bottom and top connection of the sightglass for the vaporizer. Most piping is 2-in, there is one 1-in tie-in valve top and bottom of SG. The SG itself is connected with 2-in piping to the vaporizer.
2. The SG is as tall as the vaporizer, so mass could flow out of the top of the SG into the vaporizer.
3. We think the boiling in the SG inhibits the flow from the vaporizer into the SG and causes a difference between the indicated liquid level and the actual vaporizer level; but we are not sure.
4. The fluid involved is propylene, it has 7% oil which can build up on the bottom of the vaporizer during a shutdown and would therefore preferentially enter the bottom of the SG until equilibrium is reached during normal operation. We have no physical properties on this oil except we know its heavier than propylene.
5. After equilibrium is obtained the indicated instrument liquid level and the actual vaporiser liquid level match IF we calibrate the instrument for a SG of 0.41 versus what should be a SG for propylene of 0.53.
6. The SG was un-insulated at the time, while the vaporizer was insulated; this was due to process piping changes.
The more you learn, the less you are certain of.