Pressure drop in downflow vessel, force applied to vessel support 1

[RLarsenST]

I searched around for about a half an hour and couldn't find a definitive answer to this question, so I thought I'd make the leap into asking my first question on these forums.

I'm posting this in the mechanical engineering section even though I am a ChE because I believe a mechanical engineer might be better suited to put this to rest.

Here's the question:

I have a 120" ID x 30' S/S vessel loaded with ~26' of granulated media. Assume the mass of the vessel and media is ~400,000lbs. I am flowing a volume of gas (downflow) through the vessel and media and out the side of the vessel near the bottom. The pressure drop across the media is known to be 25psi.

What is the impact on the loading of the vessel support and concrete support pad from the pressure drop? Is it nearly equivalent to:

((60in)^2 * pi) * 25 lbs/in^2 = ~283,000 lbs

So is the pad actually seeing an additional 283,000lbs. of load while the gas is flowing through the vessel, on top of the 400,000lbs due to the mass of vessel and media?

My opinion is that it does, or very nearly does (of similar magnitude), however I am having a hard time convincing my colleagues since I don't know for sure myself.

Any help would be greatly appreciated!

 

[jte]

RLarsen-

Probably a question more suited for forum794 but that’s not a big deal.

Simply put, no, the support does not know whether or not you are flowing any fluid through the media.

This may be easiest to explain if you picture the vessel as two separate chambers, with the top one being at say 100 psi and the lower one being at 75 psi. The 25 psi pressure drop you are taking through the media is countered not by the support, but by the extra 25 psi that the top head sees and carries through longitudinal pressure. Basically, the top head is pushing up as hard as the pressure drop is pushing down as it passes through the media.

jt

 

[RLarsenST]

Alright, setting pressure drop aside for a second, you're saying there will be no effect even from drag of the gas through the bed on the product? That there is no net "force vector" for lack of a better term in the downward direction adding to the apparent mass of the vessel and product to the support?

In other words, a scale mounted under the bed will show an identical reading whether the gas is flowing or not?

For instance, if the vessel were mounted sideways on a frictionless surface or something, and the gas flowing through horizontally, since the pressure is higher on one side of the media, would it not push the vessel in the direction of flow?

Sorry if there's something blatantly obvious that I'm missing here. I do understand what you're saying about the pressure of the gas pushing in all directions against the vessel at the same time effectively cancelling the force of the gas pressing on the product, I'm just having trouble wrapping my head around the entire concept.

Appreciate the help.

 

[jte]

I think you understand… you just don’t believe. How ‘bout we look at it a bit differently. Take a car with a fuel tank which has an internal fuel pump which pushes gasoline to the fuel injectors. That gas goes through a fuel filter on its way to the engine. As it goes through the filter, there is a dP across it. Does the fuel line see a change in loading as the car uses more gas and the gas flow rate and dP go up?

Maybe you have some reasonably small basket strainers on your piping. Clean out the basket, and set the strainer back with a bathroom scale under it. Let it run until it clogs. Note any change in weight that is not attributable to the crud clogging the filter.

jt

 

[BigInch]

All internal pressure stresses are resisted by the vessel wall. Pressure acting laterally on the vessel wall is resisted by hoop tension. Pressure acting upwards on the top eliptical head and pressure acting downwards on the bottom eliptical head are resisted by tension in the vessel wall that holds the heads together.

Lets say you have 50 psi in the upper half of a vessel and you have 25 psi in the lower half. There is a close packed filter (weighing nothing) at the middle. The vessel has 1000 in2 of crossectional area. Flow across the filter has a pressure drop of 25 psi, so 50 psi - 25 psi = 25 psi. Then you have 50,000 lbs pushing up on the top eliptical head, 50,000 lbs pushing down on the top of the filter. In the lower half, you have 25,000 lbs pushing down on the lower eliptical head and 25,000 pushing up on the bottom of the filter. The wall of the vessel holds the heads together, so the wall in the top half of the vessel has a tension stress = 50,000 lbs divided by the crossectional area of steel of the vessel wall. The bottom half of the vessel has 25,000 lbs / crossectional area of steel. So, you notice there is a difference in wall stress at the middle of the vessel (where the filter attaches to the vessel wall). The change in vessel wall stress of 50,000 - 25,000 = 25,000 lbs is caused by the forces transferring from the filter to the vessel wall at that point, which should be 25,000 lbs coming from the filter support. Let's check. 50,000 lbs down on the filter, minus 25,000 lbs up on the bottom of the filter = net 25,000 lbs down. That divided by the vessel crossectional area is the same as the change in vessel wall stress. All of those forces are resisted by the vessel wall tension stresses. None of which thereby need to be resisted either by the foundation or the columns holding the vessel up.

Any clearer now???

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[TGS4]

Like jte says - you understand but don't believe. Try an exercise that I use when confronted by problems like this.

Start by drawing a free-body diagram. Your FBD will include the entire vessel, but will not intersect the vessel. Draw the external forces and the weight. What's the reaction at the bottom? Now, modify your FBD so that the vessel is now intersected at the mid-way point. Include all forces (and pressures, including longitudinal stress pressures from the vessel wall) at the FBD intersections ONLY. include the weight of the vessel inside the FBD. What's the reaction at the bottom? Is it different? It shouldn't be.

 

[BigInch]

jte, yes the fuel line sees different loads.. axial loads in the fuel line change. Example, take a partially open valve in a pipeline. There is a differential pressure drop across the valve, that differential pressure times the x-sect area of flow equals an axial force which must be resisted by the rest of the valve body, which in turn is transferred into the pipe wall where it manifests itself as a change of axial stress in the pipe wall... in fact, that's one of the reasons it needs to be bolted or welded in place. When the valve is being closed, the pressure drop increases until it reaches the magnitude of the "end capping force", full line pressure * cross sectional area of flow.


**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[jte]

BigInch-

In your example of a valve, please go through a free body diagram, let's say 2 meters upstream of my control valve. Given a fixed upstream pressure, the longitudinal force in the pipe remains constant, regardless of valve position: Fully open, it is pressure times flow area of the pipe. Halfway closed, it is... pressure times flow are of the pipe. Fully closed it is... pressure times flow area of the pipe.

But let's head back to my basket strainer. Ever seen a basket strainer lift off its support because of the pressure drop across it?

jt

 

[RLarsenST]

Thanks guys, with a clearer head it all seems pretty obvious now with regard to the pressure issue, some time with Roark's probably would've made it even more obvious.

I understood it clearly already with regards to a vessel without gas actually moving through it, but for some reason couldn't accept that the gas moving through the tortuous path of the granulated material wouldn't exert some force in the direction of flow due to shear and friction. Does this hold that there will be no effect from the gas (fluid) moving through the vessel, regardless of the fact that the pressure drop does NOT apparently have an effect?

In fact, it's funny that discussion of a FBD came up, that was exactly what I was telling my colleague we needed to do but was unsure exactly how to do it since I haven't had to physically draw one out since college physics.

So for my own clarification, with regard to the horizontal vessel on a frictionless surface theoretical example, the reason it wouldn't be shoved in the direction of flow is that the forces exerted by the pressure on both sides of the "drop" balance out in all directions?

I knew I should've taken a statics or dynamics course in college, seemed to be a fairly large gap even though it was a ChE curriculum. Guess it's never too late to start boning up on it though!

 

[BigInch]

jte,

Error #1 is the longitudinal force does NOT remain constant when passing anything giving a differential pressure. If your valve was fully open (no pressure drop), your statement is true. But in the upstream segment from your control valve, axial force = P * Ax-section. Let's say the control valve differential pressure is 10 % of P, therefore downstream of the valve the axial closed end force, using a free body diagram can only be 0.9 P * Ax-section.

Assuming constant upstream and constant downstream pressures makes that problem very easy. I want you to consider a 1000 m long pipeline each end with end caps. How do we get flow with end caps you say? We're injecting flow through a small tee near the inlet end cap and removing flow from a small tee near the outlet end cap. Now this pipeline has a high inlet pressure of 10,000 kPa, driving a high flowrate such that there is a considerable friction loss of 10 kPa / meter. To make this easy, let's say there are are NO valves giving any differential pressure loss. Pressure loss is only due to the constant frictional head loss. What are the axial stresses in the pipe near the inlet end cap, what are the axial stress of the pipe at 500 m, and what are the axil strsses near the outlet end cap?

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[BigInch]

Inside Diameter is 1 meter and X-sectional area of pipe steel is 100 cm2.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[JohnGP]

LOL ........but putting the practicalities of the scenario aside, it does vividly support the point that you are making.

 

[BigInch]

Here's a complete FBD for a vessel containing layers of close-packed spheres and a fluid flowing from top to bottom.

The vessel top section has a pressure of 100 psi. A 25 psi pressure drop through the spheres is transferred to the vessel floor via fluid drag forces acting on the spheres, resisted by contact with the next sphere layer and so on. The sum of the drag losses is 25 psi pressure differential across the layer. That is transmitted to the vessel floor by the sum of the contact stresses of the last layer of spheres as they are supported by the vessel floor.

Result on the floor is 25 psi worth of contact forces and the remaining 75 psi of fluid pressure. The load on the vessel floor is transferred to the vessel wall.

Weight of the vessel material, the spheres and the fluid was not considered, but since that cannot be resisted by the pressure at the top of the vessel, the only place those weights can be resisted is by the vessel supports holding up the whole thing (not shown), located at the "corners" of the vessel where the wall is connected to the floor.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[BigInch]

Answer to pipeline problem I suggested above,


**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[RLarsenST]

BigInch, thanks very much for taking the time to put that FBD together, all is clear now. Sadly, I seem to have forgotten the adage "for every action, there's an equal and opposite reaction."

Thanks again.

 

[BigInch]

Maybe this will help you remember,

http://www.chugulugames.com/consult-propose-jeux-gratuits-chugulu.php?jeu=9 Ball Billard&vote=5

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/

 

[RLarsenST]

Ha. Thanks. I suppose pool could be considered a fairly relaxing physics refresher.

 

[BigInch]

Depends on how much you're playin' for.

**********************
"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

http://virtualpipeline.spaces.live.com/