MG1602
Well-Known Member
Have a 10 gallon polarware kettle with false bottom. Past two brews, my false bottoms have collapsed while recirculating and I don't know why. Can anyone help me figure this out why this happened?
Crushed false bottom
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Ok, what exactly are your "false bottoms" made from??
I've tried flexing my FB, and it takes a lot of force to get my FB to bend, way more than the pressure that 20-odd pounds of grain in 60-odd pounds of water could ever provide...
Cheers!
I've tried flexing my FB, and it takes a lot of force to get my FB to bend, way more than the pressure that 20-odd pounds of grain in 60-odd pounds of water could ever provide...
Cheers!
Hang Glider
Beer Drinker
Switch to a metal false bottom
OP
OP
MG1602
Well-Known Member
They are metal.
two_hearted
Well-Known Member
Can you describe them in more detail? Or post a pic?
I have a 10 gallon Polarware with the matching full width false bottom. More than a few times I have had a stuck mash and pulled down on the FB with a pump at full throttle and the FB did not collapse or show any sign of stress. It's not only the weight of the grain and water. It's mostly the atmospheric pressure pushing down on the FB that can cause it to collapse when sucking on it with a pump. Atmospheric pressure is about 14.7 psi. Assuming the pump cannot be 100% efficient, the force would probably be more on the order of maybe 10 psi or so. The Polarware FB has a surface area of about 160 square inches (This is a rough estimate). The total force pushing down on the FB at the assumed 10 psi would be 1,600 lbs. Even if we assume that my calcs are off to the high side by 1/3rd, it would be a half ton of total force on the FB. That's a hell of a load no matter how you look at it. I'm actually quite surprised that I haven't collapsed my FB yet, but it's well built with very substantial supports.
What kind of FB are you using? That should reveal a lot.
What kind of FB are you using? That should reveal a lot.
Umm...1600 pounds of force?
I have to believe there's something fatally flawed with that calculation, and my first guess is you're multiplying the pump draw (which, fwiw, unless you're using a positive displacement pump, is probably closer to zero than 10 psi) by the area of the FB, when you probably should be dividing the draw by the area.
Or maybe I'm wrong...but I have an excuse, I lost count how many drafts I've drawn tonight, and the buzz is fairly intense...
Cheers!
I have to believe there's something fatally flawed with that calculation, and my first guess is you're multiplying the pump draw (which, fwiw, unless you're using a positive displacement pump, is probably closer to zero than 10 psi) by the area of the FB, when you probably should be dividing the draw by the area.
Or maybe I'm wrong...but I have an excuse, I lost count how many drafts I've drawn tonight, and the buzz is fairly intense...
Cheers!
OP
OP
MG1602
Well-Known Member
I have the same false bottom as you Cat. One of the three legs just basically gave out.
I can't load the pic right now. Is this a faulty FB?
I can't load the pic right now. Is this a faulty FB?
OP
OP
MG1602
Well-Known Member
OP
OP
MG1602
Well-Known Member
Sorry. Paste did not work.
Fail.
IMO, yes there was something defective with the FB. A leg could have been bent and later straightened which could very well compromise it's integrity. I'm guessing on that though, but mine has been through the wringer with no sign of strain.
OP
OP
MG1602
Well-Known Member
I tried to bend it back, but no luck. Pretty damn strong, which suprises me that both of false bottoms collapsed!
Umm...1600 pounds of force?
I have to believe there's something fatally flawed with that calculation, and my first guess is you're multiplying the pump draw (which, fwiw, unless you're using a positive displacement pump, is probably closer to zero than 10 psi) by the area of the FB, when you probably should be dividing the draw by the area.
Or maybe I'm wrong...but I have an excuse, I lost count how many drafts I've drawn tonight, and the buzz is fairly intense...
Cheers!
The max pump suction is far above zero. As a matter of fact, it is about equal to the max head the pump is capable of. My LG shutoff head is 16 ft. That translates to just shy of 7 psi of suction. So, my guesstimate was off by about my predicted 1/3rd high as expected, but that would still put the total force at about 1/2 ton. Anyone familiar with fluid mechanics would know that you do, in fact, multiply the pump suction by the surface area of the FB. Obviously, you would only approach that much force if the FB was totally plugged up as can happen with a stuck mash. As I said, I've been there and been there more than once. My FB has not yet collapsed, but I've seen it happen on one of my friends system. I use a vacuum gauge on my pump for precisely this reason. I want to know how much suction is being applied to the FB. The vacuum gauge gives me an indication of how much the FB is restricted and how much force the FB is supporting.
scottland
Well-Known Member
Your math is fatally flawed, sorry. The suction of the pump applied to 160sq inches nets close to zero. You're thinking of fluid dynamics as if the suction was still applied to a 1/2" tube. A march pump might have one inch of head when applied to a 12" tube(i.e. A keg) Put you thumb over your garden hose if you want a visual of the same concept(obviously the pressure is reversed, but pressure is pressure.) to say it another way: I can stick a straw on your neck, suck on the other end, and you'll get a hickie. If we put a 12" diameter pipe on your back and was somehow able to suck on it (jokes aside) it would do a damn thing. Get the idea?
So now that we can forget about the suction from the pump, we're talking about about 10-15lbs of grain and roughly 40-50lbs of water. Thats 50-65lbs any way you slice it. Get the idea out of your head that your pump is applying a 1/2 ton of force to the false bottom, it's counter-productive.
It sounds to me like a leg on your false bottom is collapsing. Obviously a picture of your setup would help. Also a picture of the collapsed false bottom.
So now that we can forget about the suction from the pump, we're talking about about 10-15lbs of grain and roughly 40-50lbs of water. Thats 50-65lbs any way you slice it. Get the idea out of your head that your pump is applying a 1/2 ton of force to the false bottom, it's counter-productive.
It sounds to me like a leg on your false bottom is collapsing. Obviously a picture of your setup would help. Also a picture of the collapsed false bottom.
brewingmeister
Well-Known Member
Weight of more than half a car is exerted on a fb? I guess i don't know about fluid dynamics. I'm not familiar with that LG pump but my magnetic drive pump could barely pull a fart out of a squirrel. It pumps what is fed to it. I use a grant so i guess my fb will never see that much pressure. Let's see this collapse.
-Cheers
-Cheers
Sorry, but you don't know what you are talking about. Ever seen that video of the rail tank car imploding? It was crushed by atmospheric pressure only.
The weight of the grain and water on a FB is trivial relative to the force that can be applied by pump suction if the FB is plugged up.
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The common mag drive pumps are not self priming and I think that's pretty much general knowledge, but once primed, they do indeed pull a suction. They would not be able to move even an ounce of fluid if they did not do so.
I also question 1/2 ton. Long before the false bottom would collapse any tubing connected between the pump and the tank would collapse, flow would stop and the pump would lose all suction. Your math may be correct but the strength of the other equipment negates the theory.
JuanMoore
Getting the banned back together
I don't understand why people think that the weight of the grains and water are pressing down on the FB during the mash. My old cooler MLT had a flimsy SS braid manifold that would flatten under the slightest pressure, but never collapsed even with 24# of grain and 7 gal of water sitting on top of it. The grains are pretty much bouyant in solution, and the water is only putting pressure on the bottom of the container, not the things inside of the container (false bottom, manifold, etc). Think of it this way, if you put a cheap disposable pie tin upside down at the bottom of a lake, does it get crushed from the millions of pounds of water sitting on top of it?
Any downward pressure on the false bottom is caused by pressure differentials created through pump suction. I'm a big guy, and I've seen false bottoms that could hold my weight completely collapse from pump suction. I understand that many of the mag-drive pumps we use for brewing don't seem very powerful, but given the right circumstances it doesn't take much force to create huge amounts of hydraulic pressure. Proof of this can be found in the brake system of most modern automobiles.
To the OP, at what point is the FB collapsing, right away or as the fluid level is dropping when the vessel is being emptied? How fast are you trying to recirculate? Are you slowly increasing flow until you get to the desired level, or trying to start out at full flow? Is your FB perforated stainless, or the kind with the circular indents? Is the FB flat or domed?
Any downward pressure on the false bottom is caused by pressure differentials created through pump suction. I'm a big guy, and I've seen false bottoms that could hold my weight completely collapse from pump suction. I understand that many of the mag-drive pumps we use for brewing don't seem very powerful, but given the right circumstances it doesn't take much force to create huge amounts of hydraulic pressure. Proof of this can be found in the brake system of most modern automobiles.
To the OP, at what point is the FB collapsing, right away or as the fluid level is dropping when the vessel is being emptied? How fast are you trying to recirculate? Are you slowly increasing flow until you get to the desired level, or trying to start out at full flow? Is your FB perforated stainless, or the kind with the circular indents? Is the FB flat or domed?
Not true. You are confusing the total load with the force applied per square inch. Suction of as little as 1 psi applied to the entire surface area of the FB, 160 sq inches for example, would be 160 lbs. Ten psi would be 1,600 lbs. I've collapsed a FB connected with tubing. The tubing did not collapse as you suggest it would have and neither did the pump lose all suction prior. Negate the theory my ass. It ain't a theory kid. It's a reality.
You've got it basically right Juan. There is no weight on a FB until a pressure differential is created below it with a pump or simply opening a drain port. Yes, it's a fundamental hydraulic principle. Small force applied over large surface area, essentially a piston and cylinder effect. I've heard that it only takes an air pressure differential of a few psi to lift an entire airliner. Can this be true? Sounds preposterous to me.
scottland
Well-Known Member
Catt22: Holding other variables constant: Pressure drops when volume increases. The 10psi of suction the pump is applying is within the 1/2" diameter hose. As the volume increases to a 12" diameter pot, the pressure(suction) drops significantly. I don't really feel like arguing about this because it's turning into a waste of time, but I promise you that you're over-thinking this.
Well, I promise you that you are under-thinking this. We are talking about a stuck mash situation here. The pressure won't be relieved so long as the FB is plugged up. The 10 psi would be applied across the entire surface area of the FB. This stuff is fundamental mechanics. It's very straight forward. Take another look at what atmospheric pressure did to that tank car. You could apply suction to evacuate that tank car through a very tiny pipe and the effect would be the same. Might take considerable time to pump the air out of the tank, but eventually it would still collapse. The diameter of the suction hose is essentially irrelevant.
scottland
Well-Known Member
Dude, your tank car is a horrible example. There was a big temperate drop within that tanker for that to happen. Either:
A. The tanker was filled with a pressurized gas which was rapidly discharged. That would cause a rapid drop in temperature, and then a rapid drop in pressure. The difference between the extremely low pressure inside and the atmosphere crushed the tanker.
B. The tanker was filled with steam or something of the sort and allowed to cool. Again, the drop in temperature causes a rapid drop in pressure, and just like in example A, the drastic difference between the near vacuum inside the tank and the 1bar of atmosphere would crush it.
Neither of those scenarios have anything to do with what we are talking about.
A. The tanker was filled with a pressurized gas which was rapidly discharged. That would cause a rapid drop in temperature, and then a rapid drop in pressure. The difference between the extremely low pressure inside and the atmosphere crushed the tanker.
B. The tanker was filled with steam or something of the sort and allowed to cool. Again, the drop in temperature causes a rapid drop in pressure, and just like in example A, the drastic difference between the near vacuum inside the tank and the 1bar of atmosphere would crush it.
Neither of those scenarios have anything to do with what we are talking about.
A. No dude, it is a perfect example of what atmospheric pressure is capable of. It matters not how you reduce the internal pressure of the tank. The effect would be the same. I fully understand the principle. You don't seem to.
B. Yes, that is correct. Again, it doesn't matter what means is used to reduce the internal tank pressure. The effect would be the same. The pressure drop need not be rapid. Slowly reducing the internal pressure would eventually have the same result.
Shockerengr
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Guys, Catt22 is absolutely correct on the forces possible.
Force equals pressure times area. period. For a false bottom, as long as it's not clogged, wort can flow through, and the pressure differential is very low/close to zero. As it clogs, the pressure differential can rise, and it's simply the math of surface area times pressure differential. A couple PSI over a large area is a lot of force.
A couple examples, I routinely lift trucks (around 10000lbs once you add the lift bits in) with an old centerpost lift using about 45-60 PSI of air pressure alone. it's roughly a 16" post, and it's the pressure against that post that develops all that force.
For aircraft, the pressure delta is only a couple PSI spread over the entire wing. The doors on aircraft have to be designed to handle many tens of tons of force because of a 6-10 PSI cabin pressure.
For the tanker, how the pressure change happened is irrelevant, the fact is even with a perfect vacuum inside, the pressure delta could have been at most 14.7 PSI (or atmospheric pressure)
Put yet another way, with the three PSI that a march pump can produce, I can lift a 5,000 pound car with 5ft x 3ft plate, and lift it as high as I like.
Force equals pressure times area. period. For a false bottom, as long as it's not clogged, wort can flow through, and the pressure differential is very low/close to zero. As it clogs, the pressure differential can rise, and it's simply the math of surface area times pressure differential. A couple PSI over a large area is a lot of force.
A couple examples, I routinely lift trucks (around 10000lbs once you add the lift bits in) with an old centerpost lift using about 45-60 PSI of air pressure alone. it's roughly a 16" post, and it's the pressure against that post that develops all that force.
For aircraft, the pressure delta is only a couple PSI spread over the entire wing. The doors on aircraft have to be designed to handle many tens of tons of force because of a 6-10 PSI cabin pressure.
For the tanker, how the pressure change happened is irrelevant, the fact is even with a perfect vacuum inside, the pressure delta could have been at most 14.7 PSI (or atmospheric pressure)
Put yet another way, with the three PSI that a march pump can produce, I can lift a 5,000 pound car with 5ft x 3ft plate, and lift it as high as I like.
scottland
Well-Known Member
And here lies the issue. The March pump can't produce 3PSI of discharge against a 5ftx3ft plate. I'd love to see you lift a 5k lb car with a march pump.
The march pump can't produce 10psi of suction against a 12" diameter pot. That assumption is where the problem is. A march pump can apply 10psi of suction within a 1/2" tube.
My little air compressor can produce 120psi or more. If I release that pressure into a 10,000 gallon tank, the pressure inside the tank will hardly change. Pressure drops as volume increases. I know this example is positive pressure, but it works the same in reverse.
Yes, an actual measured 10psi of suction when applied to the bottom of a false bottom will crush it. But a march pump won't apply 10psi of suction to a false bottom.
Shockerengr
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Yes, it can. it's pumping a fluid. If I pump a fluid into a container , pressure will be equal throughout the container for each height in the fluid (due to gravity. 3 psi is 3 psi whether it's in a 2" pop can or a 40ft tank. It's a basic theory of hydraulics. It's how a bottle jack can lift a car
What is affected is that rate. Because the flow is low in a march pump, I wouldn't be able to lift it very fast, as I have to fill the volume as the cylinder expands. but it would lift it none the less.
I do have to make a small correction, because of the weight of the water, I wouldn't be able to lift it more than a couple feet above the height of the pump.
Here's another fun one, I could burst a keg using nothing more than water, gravity, and a tube with the same diameter as a soda straw. All i'd need is to use the tube connected to the keg to raise the water column height to 200-500 ft. I'd end up with between 86psi and 216psi
scottland
Well-Known Member
I'm still having an issue applying any of this practically. I've had sparges stick with a march pump pulling against a cpvc manifold. I'm pretty sure that cvpc would crush with roughly 1/2 ton of force.
JuanMoore
Getting the banned back together
Yes, pressure drops as volume increases, but the volume isn't changing in this situation. The only relevant volume is the volume enclosed between the pump and the (clogged) false bottom, which remains constant. The factor that we need to look at isn't the volume (since it stays constant), but rather the surface area that the pressure is being applied to. I think you may be confusing pressure and force. If we had a constant force and surface area, then you're correct in thinking that the pressure would decrease as the surface area increases, since the equation for pressure is P=F/A. In this case it's the pressure and area that are constant, and what we're trying to calculate is the force being applied. If you solve the same equation for force, you get F=P*A, so with a constant pressure, as the area increases the force also increases substantially. Rather than derailing this thread by arguing about a subject you're not very familiar with, how about we try to help the OP solve his problem?
At no point is the pressure (or vacuum) being released into a larger space, the volume is staying constant. Let's think about your compressor as an example. When you have it pumped up to 120psi, where is the pressure greater? In your compressor tank which has a large volume and diameter, or at the end of your air hose which is only 1/4" diameter? The answer is that the pressure is the same throughout. Think of the space under the false bottom as your compressor tank, and the transfer hose from the pump as the air hose attached to your compressor. The pressure (or vacuum) is the same in both places.
JuanMoore
Getting the banned back together
I've driven over pieces of CPVC in my truck without damaging them, and that's applying force from only one direction. The forces created by a stuck sparge would be spread evenly around all sides of the pipes, and the round shape is ideal for countering this type of force.
scottland
Well-Known Member
Let me run through my understanding, stop me where i'm incorrect.
A pumps total head is the difference of the discharge head, subtracted from the suction (inlet head)
If there is a large amount of static weight applied to the discharge, that will decrease the discharge head of the pump correct?
How much static weight needs to be applied before the pumps head reaches zero? I would think from practical knowledge that a 5,000lb car would provide enough static weight to reduce a march pumps head to zero. I would think that the 5,000lb car (static weight) would overpower the motor in the march pump that is powering the impeller. There has to be a certain amount of weight where this happens.
Now i'm going to run through a scenario, correct my inaccuracies:
I can hook some hose up to a vacuum gauge and suck really hard, and probably pull around 5-6 psi. Now let's take that hose and attach it to a sealed corny keg. If I suck really hard, I won't be able pull 5-6psi. So even though my mouth is capable of 5-6psi of suction, that doesn't mean I can apply 5-6psi of suction to a corny keg. You guys keep saying that pressure is pressure, but unless I'm missing something, pressure is relative to volume.
And even though a march pump can probably pull 10psi of suction with a 1/2" hose that's plugged, as soon as you hook that hose up to a 12" diameter pot that psi is going to drop because it's pulling suction against a larger volume of fluid. Even if we're talking about the little bit of fluid sitting underneath the false bottom, that's still much more than is in the tube.
A pumps total head is the difference of the discharge head, subtracted from the suction (inlet head)
If there is a large amount of static weight applied to the discharge, that will decrease the discharge head of the pump correct?
How much static weight needs to be applied before the pumps head reaches zero? I would think from practical knowledge that a 5,000lb car would provide enough static weight to reduce a march pumps head to zero. I would think that the 5,000lb car (static weight) would overpower the motor in the march pump that is powering the impeller. There has to be a certain amount of weight where this happens.
Now i'm going to run through a scenario, correct my inaccuracies:
I can hook some hose up to a vacuum gauge and suck really hard, and probably pull around 5-6 psi. Now let's take that hose and attach it to a sealed corny keg. If I suck really hard, I won't be able pull 5-6psi. So even though my mouth is capable of 5-6psi of suction, that doesn't mean I can apply 5-6psi of suction to a corny keg. You guys keep saying that pressure is pressure, but unless I'm missing something, pressure is relative to volume.
And even though a march pump can probably pull 10psi of suction with a 1/2" hose that's plugged, as soon as you hook that hose up to a 12" diameter pot that psi is going to drop because it's pulling suction against a larger volume of fluid. Even if we're talking about the little bit of fluid sitting underneath the false bottom, that's still much more than is in the tube.
JuanMoore
Getting the banned back together
Yes pressure is relative to volume, but in the case we're discussing it's irrelevant because A) the volume is constant and B) as soon as the FB clogs it becomes a closed system.
In this scenario, whatever pressure the pump is applying stays constant, since the volume also stays constant. The pressure is distributed evenly throughout the fluid, so the pressure of the fluid inside the hose is the same as the pressure inside the fluid under the false bottom. Once again, you may be confusing pressure and force.
scottland
Well-Known Member
Correct, but I don't think anyone has measured the suction of the march pump applied to the false bottom. I'd bet it's nowhere near 10psi.
The 1/2 ton of force number came from multiplying the 10psi by the area of the false bottom. I don't believe the march pump is pulling anywhere near 10psi against a false bottom in a pot. At least any pot bigger than 1 inch in diameter.
Also, have any of your actually had a stuck sparge with a march pump in a home brew setting? All of the blackboard talk is nice, but it's not actually what happens.
When the grainbed compacts, and the sparge sticks, your tubing to the pump doesn't stay full of fluid. The sparge sticks, the pump sucks air, and the pump loses suction and you get cavitation. The pump doesn't continually pull suction against the compacted grain bed until the outlet stops.
Shockerengr
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Volume will affect the time it takes to hit those pressures.
If you hook your air compressor to a large tank, it will still b able to fill it to 120psi....it would just take a lot longer.
That's why pumps are rated in both pressure and flow rate. The fow rate will determin how fast it could fill to reach those rates.
If you hook your air compressor to a large tank, it will still b able to fill it to 120psi....it would just take a lot longer.
That's why pumps are rated in both pressure and flow rate. The fow rate will determin how fast it could fill to reach those rates.
I agree that you are still having issues. The surface area of your pvc manifold is nowhere near that of a full width false bottom. That fact coupled with the very strong circular cross section of the pipe would prevent it from collapsing when subjected to at most about 14.7 psi of atmospheric pressure. You still seem to be confusing the total load with unit force or pressure (as in psi).
My LG pump can generate 10 psi of suction and maybe more. I don't recall what the max has been as I try to avoid letting it get that high. I have a vacuum gauge on the suction side of my pump, so this is info is from actual experience, not simply conjecture. I also assisted one of my brewing buddies as we unintentionally collapsed his FB. His FB was poorly designed and lacked adequate support particularly around the perimeter. It flattened at much less than 10 psi suction.
A wise decision by any standard.
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