Tank Implosion due to insufficient vent diameter 4

[tmengineer]

Hello,

A vessel has imploded due to having an insufficient vent diameter and I have been tasked with calculating the new appropriate diameter.

The original vent was 4", 20m long, with 3 elbows, had a fan which wasn't used in the pipe and increased in diameter to 5" after 5m.

The tank volume is 13.0 m^3 and was emptied with a flow rate of 900 l/min.

I started off by calculating the pressure drop in the pipe by using Bernoulli's equation and accounting for major and minor losses but the pressure drop I calculated was negative and insignificant (-39 Pa).

Either I have made a mistake in my calculations or my approach has been incorrect. I used incompressible flow models as the flow velocity was small (1.91 m/s) and so the Mach number < 0.1.

Is it a calculation error or is the method wrong?

Any feedback would be greatly appreciated!

 

[zdas04]

Why does everyone blame every fluid flow regime on poor old Bernoulli? This is such a simple problem.

You know that the ideal gas law says P*V=n*R*T (and it is safe to use ideal gas assumptions at these pressures). So:
[ul]
[li]At constant temperature, as you increase the gas volume (by lowering the liquid level), at the same time you are increasing the amount of gas ("n") with inflow through the vent. If the inflow volume equals the outflow volume then pressure will be constant.[/li]
[li]But pressure has to drop for there to be flow in the vent pipe. You should be able to calculate how much pressure can drop before you crush your tank (probably something like -15 kPag, but I would add a safety factor and call my target pressure -5 kPag)[/li]
[li]Use the Spitzglass formula to calculate volume flow rate at actual conditions (using the -5 kPag as the downstream pressure). If the answer is less than 0.9 m^3/min then you have "n" decreasing slower than "V" and the tank will crush.[/li]
[/ul]
Your vent pipe is really long and kind of small. If some of those elbows you talk about can create a water trap then you can be certain of crushing a tank the first time you try to pump from a full tank on a cold day.

I use Sptzglass for this because the rate of change of gas density below atmospheric pressure is very rapid and all of the incompressible flow equations are invalid at rapidly changing density. That goes double for Bernoulli. Spitzglass tries to represent compressible flow and does an OK job of it for reasonably small flow rates (it doesn't do great with accounting for friction drops).

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[bimr]

You have a volume of 900 l/min leaving the tank. That means you need the equivalent amount of air to enter the tank. To calculate the minimum size vent, assume that the air flow velocity is less than sonic velocity and determine the friction loss through the vent pipe.

It appears that the 4-Inch vent has adequate size as there should only be 2-3" wg loss across the vent pipe.

Is there a chemical reaction in the tank? Are you condensing steam in the tank?

What is the design pressure of the tank?

Was the vent blocked?

 

[tmengineer]

Thanks for the replies!

Zdas04, I've used two forms of the Spitzlgass equation and got flow rates of 19.66 m^3/ min and 43.74 m^3/min. Both seem quite high and I did them in a quickly so I'll double check the units to see if I've made an error there, but at the minute because the gas flow rate is bigger than 0.9 m^/min the vent should suffice.

Type 1 Spitzglass: Q = 0.09 (( dh * D^5)/(SL(1+ 3.6/D + 0.03D))^.5 where dh is the pressure drop in inches of water, D is the pipe internal diameter (inches), S is the specific gravity of the gas, L is the pipe length (ft), and Q is the volumetric flow rate in MM scf/D.

Type 2 Spitzlgass: Q = 3550 k (dh / L*SG)^0.5 where Q is the volumetric flow rate in cfh, dh is the pressure drop in inches of water, L is the pipe length in ft, S is the specific gravity of the gas and
k = (D^5 /(1+ 3.6/D + 0.03D))^0.5 where D is the pipe diameter (in).


I usually use PV elite to determine the maximum vacuum that the tanks can take but this tank is a wooden fermenter used for beer production and PV elite doesn't deal with wooden vessels. I might resort to Roarks for a flat beam - the wooden slat that broke is along the top where the vent pipe goes into the tank.


bimr, the tank is used in beer production and is cleaning with hot water at 80 degrees C then allowed to cool naturally. I have done similar vent calculations on stainless tanks where the tank is rinsed with 5 degree water immediately after the 80 degree rinse but this is not the case this time. Yes the vent had a fan in it which was not used and probably acted as an obstruction.

 

[1gibson]

How do you "know" that the failure was due to insufficient diameter?

A fermenter should have some sort of bubbler or water trap on the overflow, is there not also a trap on the vent (3 elbows?) I was under the impression that it shouldn't be exposed to open air. Even if it wasn't designed that way, maybe an employee thought the same and dumped water in the U-trap (3 elbows?) of the vent.

 

[zdas04]

tmengineer,
With 5 kPa I get 20 SCM/min. Since the tank is wooden, I would probably move the limit down to 0.5 kPa which works out to 7 SCM/min (I've seen that equation off by 20% but never by an order of magnitude). That makes me think that the pipe is big enough and I would start looking for a physical restriction.

The P-Trap from 3 elbows is a real possibility. So is restriction through the fan (you are only dealing with 500 Pa after all). I think I'd start the fan and measure the dP from the tank to the fan suction to see if there was a hidden gremlin in the 4-inch.

David Simpson, PE
MuleShoe Engineering

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist

 

[tmengineer]

Hi 1gibson,

I'm pretty sure that the vent is the problem. There is 8 fermenters, 4 with 4" -> 5" vents, and 4 with 6" vents (they were installed in 2 phases) and the 4 with 6" vents do not show as much sign of damage as the 4" -> 5" lines do.

The fermenters have "switchers" in the gas space which stop the foam from going up into the vent and overflow but I don't think that there is a U bend / bubble trap in the vent line and I think that it is open to the atmosphere.

 

[Compositepro]

Given your operating conditions, I would think that simply adding cold water to a warm tank would be a far worse case than pumping out at 900 l/min.

It is also unsanitary to be sucking air into a fermentation tank through a long vent pipe that cannot be cleaned regularly. The fan was almost certainly on your pipe to draw-off the CO2 from fermentation, but with an air gap between the tank and the pipe. The air gap prevents sucking air through a dirty pipe and removes the restriction of the pipe when the tank needs to vent in or out quickly. But the fan should pick-up normal CO2 vent rates. Although it should be a blower and not a fan. A fan would create almost no flow through 20 meters of 4" pipe.

 

[tmengineer]

Compositepro,

I think I mislead you slightly - the tank cools naturally without cold water addition, it is in a similar tank that cold water is added, and yes this causes a greater pressure drop than the outflow of 900 l/min.

I have made a spreadsheet for the rapid cooling scenario which I am quite happy with but my hand calcs I was using to model the vacuum inside the fermenter (which is not subject to rapid cooling) were not giving a large enough vacuum to cause the damage which occurred.

That is a good point about the cleanliness - the tank is at a customer site and I work for a company tasked with the repairs so I'm not sure what the cleaning regime is. I have only been tasked with finding out what size the vent needs to be.

Yes you are right - the fan is for CO2 extraction and not for venting and I suspect that it is the main obstruction that has lead to the vacuum. When the new vent pipe is added there will not be a fan in the line. I had been going down the route of calculating the minor loss caused by the fan as an obstruction (from Bernoulli's equation with frictional corrections) but I was guessing at the loss coefficient as I couldn't find a tabulated value for an unused fan.

Thanks again for your replies!

 

[bimr]

Is the 900 l/m pumped or is it draining by gravity?

 

[tmengineer]

It's pumped away. Why do you ask?

 

[katmar]

The problem is not in the information you have given us. The pressure drop at 900 l/min through the pipe you have described would be close to the 39 Pa that you calculated. You can use the incompressible Darcy-Weisbach formula very safely when the pressure drop is so small. I find it easier to put these small pressure drops into perspective if they are expressed in mm of water gauge - around 4 mm in this case. bimr has calculate 2-3" of WG, but I disagree with that.

The problem is much more likely something along the lines suggested by bimr and 1gibson. If there is a low point in the vent piping it WILL fill up with water if you are working with foaming beer and cleaning with 80 deg C water. If the fan is not working take it out. It will only be a further restriction or will catch plastic bags, rags and pieces of paper and cause a blockage.

I have seen tanks implode too often when hot cleaning is followed by a deliberate or accidental infill of cold water. Don't expect the operating staff to own up to that. If you don't have a pressure or temperature recorder you will never know what happened.

If this were a design problem you would be imploding tanks on a regular basis. You have had a vent blockage or an inadvertent vacuum generation through rapid temperature change.

Katmar Software - AioFlo Pipe Hydraulics

http://katmarsoftware.com

"An undefined problem has an infinite number of solutions"

 

[georgeverghese]

Looks like this is a multipurpose vent line ( normal operations outbreathing, inbreathing plus emergency underpressure / overpressure break line).

Others have identified several weaknesses in the design of the vent line you have now.

Would this implosion on this wooden beer fermenter have occured if you had kept this line dedicated to normal operating venting operations, and installed a backup PVSV for emergencies on the tank?

 

[bimr]

If the tank was draining from a valve into an open sump, you would get less of vacuum pulled.

The vacuum would be higher if you were pumping out of the tank, or if the discharge pipe was long enough to pull a siphon.

Katmar, sorry that you disagree with my headloss estimate. In cases like this, I usually have a more worst case estimate.

 

[tmengineer]

Zdas,

Sorry I didn't see your second comment until I was re-reading the thread this morning. That's good - if the pipe calculations suggest that an unrestricted pipe would provide an adequate flow rate then that's a useful result.

I'll focus on the possibility of an obstruction in the pipe, as Katmar, bimr and 1gibson suggested. The pipe diagram I have doesn't have a U bend or a P trap in it but I'm going to double check this with the production manager who is also working on the problem. I've had it before where a vent grill at the exit is clogged with junk, so there are a few possibilities as to how it could have been blocked.

I think I've got enough information to say that the pipe diameter and length is probably not the problem and that it is more likely to have been blocked - I'll continue on this approach. Thanks for all of your replies! There aren't any other chemical engineers working for my company and it's really useful for me to be able to discuss this problem with other engineers.

 

[LittleInch]

It would seem prudent to think about installing a vacuum relief valve right on the tank itself set at a few inches of water column. Venting I can see you don't want in an enclosed space as everyone would probably die, but inbreathing would seem to be OK direct.

Many tank companies supply these.

I was also interested in your comment that " the 4 with 6" vents do not show as much sign of damage "

What do you mean by this? It seems these vessels are incapable of any real negative pressure? Your requirement / ability to withstand negative pressure needs to be your first design consideration and everything else flows from there.

Good luck and let us know how it goes - pictures are always good!

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[tmengineer]

I'm sorry to change tack but the production manager told me that the tank failed due to the overpressure caused by the hot rinsing rather a vacuum from draining.

The vent was probably still blocked as discussed above but now I'm looking at it from a different approach. I want to know what the maximum overpressure for each vent size is. It'll either be 5", 6" or 8" and I'm told the new vent will not have a U bend or fan in it.

I have done an energy balance of the heat from the hot water coming in and heating the gas inside the tank to determine the temperature profile during heating. For this I've assumed that there is instantaneous heat transfer between the hot liquid and the gas and inspected 0.4s increments, and plotted the equilibrium temperature for each 0.4s increment.

I've used PV = nRT and assumed that the pressure is constant to determine the increase in volume for each 0.4s increment and turned that into a required volumetric flow rate through the vent pipe.

I've then used the "Adiabatic Frictionless nozzle flow" section of Perry's Chemical Engineers Handbook 8th Edition to determine the flow through the pipe caused by the overpressure and plotted the difference between relief required and relief available. Summing the difference for each time increment gives the accumulating volume, which I've turned into the over pressure using PV = nRT.

I have two concerns about this method: 1) It seems a little contrived and I'm not sure how to validate the assumptions / determine if they are even valid
2) The method from Perry's for the adiabatic frictionless nozzle flow relies on a graph which is difficult to interpolate accurately and small changes in values determined from interpolation result in large changes in the relief available, and so the end result is strongly vulnerable to interpolation errors.

Hmmm....

Is there a simpler method / correlation that relates increasing gas pressure to volumetric flow rate through a pipe? I'm quite stuck when I look at my two concerns above - I think they are significant problems.

Data for question: tank gas volume when empty = 87m^3, volumetric flow rate of hot liquid = 15 m^3/hr, Hot liquid temperature = 80 degrees C, vessel air temperature before cleaning = 25 degrees C.

 

[bimr]

Unless the hot rinsing is using steam in some manner, the relief calculation consists of a simple mass balance. Inflow - outflow equals relief volume required. Worst case scenario is maximum inlet with no outlet flow from tank.

 

[tmengineer]

Will the incoming hot water not heat up the air inside the tank causing it to expand and cause an overpressure?

 

[bimr]

The evaporation and condensation water in a closed container are proceeding at the same rate, so there is no net change, the system is in equilibrium. In the closed container, the pressure due to the water vapor reaches a maximum value (for a given temperature) called the vapor pressure.

The vapor pressure of water at 80 degrees C is 355.1 mm Hg. So that would be the maximum pressure on the tank if the water stayed at 80 degrees C and the tank was sealed off from the atmosphere.

That would be the worst case because there would be some cooling as the tank was filled.

If the tank was only partially filled with 80 degrees C water, the final equilibrium pressure would be less because the water would cool (with evaporation) somewhat prior to equilibrium being reached. In that case, the tank pressure would be the water vapor pressure at final water temperature.