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Posted: 3/30/2006 10:10:02 AM EDT
say you have a 4’ tall 50 gal tank full of water. it' can be sealed and can be pressurized.
it also has a drain line on the bottom and the drain runs back up alongside the tank to the top. So the drain vent is 4’ tall.
You vent it to atmos and fill the tank with water and the drain line fills. (simple) The pressure at the base of the tank is about 2 PSI and 50 gal also holds 6.6 standard cubic feet.

Let’s say you attach an air cylinder to the vent on the top of the tank and begin applying pressure. The drain is open. As you apply more and more pressure water begins flowing out the drain. First of all how many standard cubic feet of air must you have in the cylinder and what pressure must it be at to completely evacuate the tank?



Thanks.

Link Posted: 3/30/2006 10:14:26 AM EDT
Trying to make a flame thrower huh....
Link Posted: 3/30/2006 10:15:42 AM EDT

Originally Posted By hk940:
say you have a 4’ tall 50 gal tank full of water. it' can be sealed and can be pressurized.
it also has a drain line on the bottom and the drain runs back up alongside the tank to the top. So the drain vent is 4’ tall.
You vent it to atmos and fill the tank with water and the drain line fills. (simple) The pressure at the base of the tank is about 2 PSI and 50 gal also holds 6.6 standard cubic feet.

Let’s say you attach an air cylinder to the vent on the top of the tank and begin applying pressure. The drain is open. As you apply more and more pressure water begins flowing out the drain. First of all how many standard cubic feet of air must you have in the cylinder and what pressure must it be at to completely evacuate the tank?



Well, I'm not going to crunch the numbers, but you would need enough pressure to lift the entire 50 gallons 4 feet off the ground (in order to get it out through the vent). So, the weight of the water divided by the surface area in the tank will give you the pressure.

If you know the pressure in the tank once it is completely evacuated, then you can calculate the amount of gas you need.

The problem is that as the tank empties, you'll need less and less pressure (less water to lift). That throws the calculations off if you want to remain economical. You'd need to do some calculus.

That's what occurs to me off-hand. Hope it helps.
Link Posted: 3/30/2006 10:19:09 AM EDT

Originally Posted By Zaphod:

Originally Posted By hk940:
say you have a 4’ tall 50 gal tank full of water. it' can be sealed and can be pressurized.
it also has a drain line on the bottom and the drain runs back up alongside the tank to the top. So the drain vent is 4’ tall.
You vent it to atmos and fill the tank with water and the drain line fills. (simple) The pressure at the base of the tank is about 2 PSI and 50 gal also holds 6.6 standard cubic feet.

Let’s say you attach an air cylinder to the vent on the top of the tank and begin applying pressure. The drain is open. As you apply more and more pressure water begins flowing out the drain. First of all how many standard cubic feet of air must you have in the cylinder and what pressure must it be at to completely evacuate the tank?



Well, I'm not going to crunch the numbers, but you would need enough pressure to lift the entire 50 gallons 4 feet off the ground (in order to get it out through the vent). So, the weight of the water divided by the surface area in the tank will give you the pressure.

If you know the pressure in the tank once it is completely evacuated, then you can calculate the amount of gas you need.

The problem is that as the tank empties, you'll need less and less pressure (less water to lift). That throws the calculations off if you want to remain economical. You'd need to do some calculus.

That's what occurs to me off-hand. Hope it helps.



BESIDES THAT

It's going to fail when the air gets to the bottom of the tank and bubbles rush to the top of the outlet tube.

Link Posted: 3/30/2006 10:29:00 AM EDT

Originally Posted By jchewie:

Originally Posted By Zaphod:

Originally Posted By hk940:
say you have a 4’ tall 50 gal tank full of water. it' can be sealed and can be pressurized.
it also has a drain line on the bottom and the drain runs back up alongside the tank to the top. So the drain vent is 4’ tall.
You vent it to atmos and fill the tank with water and the drain line fills. (simple) The pressure at the base of the tank is about 2 PSI and 50 gal also holds 6.6 standard cubic feet.

Let’s say you attach an air cylinder to the vent on the top of the tank and begin applying pressure. The drain is open. As you apply more and more pressure water begins flowing out the drain. First of all how many standard cubic feet of air must you have in the cylinder and what pressure must it be at to completely evacuate the tank?



Well, I'm not going to crunch the numbers, but you would need enough pressure to lift the entire 50 gallons 4 feet off the ground (in order to get it out through the vent). So, the weight of the water divided by the surface area in the tank will give you the pressure.

If you know the pressure in the tank once it is completely evacuated, then you can calculate the amount of gas you need.

The problem is that as the tank empties, you'll need less and less pressure (less water to lift). That throws the calculations off if you want to remain economical. You'd need to do some calculus.

That's what occurs to me off-hand. Hope it helps.



BESIDES THAT

It's going to fail when the air gets to the bottom of the tank and bubbles rush to the top of the outlet tube.




I know. I was trying to make the problem as simple as I could.
What I am doing is building a sealed water cistern for my cabin and I need a quick and dirty way to get enough pressure to overcome the height of the tank and get it to flow into the bathtub. Sure I can put in an electric pump but I wanted to see if I could pressurize the cistern and run it on air. I can pressurize a 5 gal propane cylinder to about 100 psig and I have a regulator that will knock the pressure down to about 5-40 psi.
Link Posted: 3/30/2006 10:31:52 AM EDT
[Last Edit: 3/30/2006 10:47:17 AM EDT by danno-in-michigan]

Originally Posted By Zaphod:

Originally Posted By hk940:
say you have a 4’ tall 50 gal tank full of water. it' can be sealed and can be pressurized.
it also has a drain line on the bottom and the drain runs back up alongside the tank to the top. So the drain vent is 4’ tall.
You vent it to atmos and fill the tank with water and the drain line fills. (simple) The pressure at the base of the tank is about 2 PSI and 50 gal also holds 6.6 standard cubic feet.

Let’s say you attach an air cylinder to the vent on the top of the tank and begin applying pressure. The drain is open. As you apply more and more pressure water begins flowing out the drain. First of all how many standard cubic feet of air must you have in the cylinder and what pressure must it be at to completely evacuate the tank?



Well, I'm not going to crunch the numbers, but you would need enough pressure to lift the entire 50 gallons 4 feet off the ground (in order to get it out through the vent). So, the weight of the water divided by the surface area in the tank will give you the pressure.

If you know the pressure in the tank once it is completely evacuated, then you can calculate the amount of gas you need.

The problem is that as the tank empties, you'll need less and less pressure (less water to lift). That throws the calculations off if you want to remain economical. You'd need to do some calculus.

That's what occurs to me off-hand. Hope it helps.



No. The pressure required is only related to the height you're trying to lift the water (and its density). (excluding any "head loss" - and no, that's not a dirty term).

Edited to add: Quick and dirty calculation - one atmosphere (14.7 psi) is equivilent to a depth of 34 feet of fresh water (33 of sea water). Therefore, the minimum pressure needed to force water up 4 feet is approximately 14.7psi (4/34) = 1.72 psi. Now, since the top of the tank is sealed and the vent tube isn't, you also need to fight the atmospheric pressure (14.7 psi), so add that. Result - approximately 16.43 psi will lift a column of water 4 feet. As for the volume required, I have no idea. It'd obviously be greater than 6.6 cubic feet because the air is going to be slightly compressed by the pressure.

Edited again to add: just read about your intended application. I wouldn't go pressurizing things (like cisterns) unless you're darned sure they can take the pressure.
Link Posted: 3/30/2006 10:50:12 AM EDT

Originally Posted By danno-in-michigan:
I wouldn't go pressurizing things (like cisterns) unless you're darned sure they can take the pressure.



Is there a KABOOM thread in the future?
Link Posted: 3/30/2006 10:53:10 AM EDT

Originally Posted By Zaphod:

Originally Posted By danno-in-michigan:
I wouldn't go pressurizing things (like cisterns) unless you're darned sure they can take the pressure.



Is there a KABOOM thread in the future?



it's a never used pressure tank off of a sprayer.
at minimum it's good to 100 PSIG.
Link Posted: 3/30/2006 11:03:32 AM EDT
[Last Edit: 3/30/2006 11:04:48 AM EDT by Zaphod]

Originally Posted By danno-in-michigan:
No. The pressure required is only related to the height you're trying to lift the water (and its density). (excluding any "head loss" - and no, that's not a dirty term).




Uh, that's what I said.

He's trying to lift it the hight of the tank. Maybe he's only lifting the pipe volume (rather than the tank volume), but the weight is a function of the density. Divide the weight being lifted plus (14.7 psi ) by the surface area of the tank and bingo.
Link Posted: 3/30/2006 11:19:32 AM EDT
put the tank on top of the roof and use gravity f#$&^in designers making ingineers life miserable.....
Link Posted: 3/30/2006 11:31:59 AM EDT

Originally Posted By presto99:
put the tank on top of the roof and use gravity f#$&^in designers making ingineers life miserable.....




there again. i was going for simplicity in the calc so it’s not 50 gal. It’s 350 gal. so no worky in top of house or on a tower. Also with that design you have to use a pump to get the water in the tank. If I wanted a pump I would just put it at the base of the house and pump it where wanted it.
I could put it at the top of the hill and have 150’ of poly tube down to the house but trying to keep it from freezing in the winter would be a pain.
Link Posted: 3/30/2006 11:33:29 AM EDT
[Last Edit: 3/30/2006 11:37:23 AM EDT by hk940]

Originally Posted By hk940:

Originally Posted By presto99:
put the tank on top of the roof and use gravity f#$&^in designers making ingineers life miserable.....




there again. i was going for simplicity in the calc so it’s not 50 gal. It’s 350 gal. so no worky in top of house or on a tower. Also with that design you have to use a pump to get the water in the tank. If I wanted a pump I would just put it at the base of the house and pump it where wanted it.
I could put it at the top of the hill and have 150’ of poly tube down to the house but trying to keep it from freezing in the winter would be a pain.



150' of 1" ID pipe has how much volume?
bout 1.5 gal
Link Posted: 3/30/2006 11:53:35 AM EDT
You should only need to exceed the 2.0 psi to accomplish what you intend. The 4 feet of water in the drain pipe is the maximum pressure you will be fighting, therefore, if you exceed that pressure you will have water flowing out the drain pipe. However, the flow rate will be controlled by how much air you put into the tank.

I will say that it seems like you are getting more complicated than necessary. You would probably be better off just getting a small pump and be done with it.

Link Posted: 3/30/2006 12:14:10 PM EDT

Originally Posted By Zaphod:

Originally Posted By danno-in-michigan:
No. The pressure required is only related to the height you're trying to lift the water (and its density). (excluding any "head loss" - and no, that's not a dirty term).




Uh, that's what I said.

He's trying to lift it the hight of the tank. Maybe he's only lifting the pipe volume (rather than the tank volume), but the weight is a function of the density. Divide the weight being lifted plus (14.7 psi ) by the surface area of the tank and bingo.



Sorry, I misunderstood your original post.
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