I have a 6 ft diameter pipe that is sealed at the top (effectively air tight) that drains into a empty pond. The elevation change is 41 ft. The pipe is at a 45 degree angle and is about 58 ft long, the volume is 1639 cubic ft. When the pond fills with water, to a height of 58 ft (~25 psig) above the bottom of the pipe, air is trapped in the pipe. Do you know what the air pressure will be and how high will the water go in the pipe? Thanks
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Air Pressure in vertical pipe when water enters pipe and traps air inside of it 1
- Thread starter Butchsc
- Start date
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1
- #2
P1 * V1 = P2 * V2
P1 is 14.7 psia
P2 is depth to water surface inside pipe x 62.4/144
V2 is depth to water surface inside pipe x 1639/58
D = Depth to water surface inside pipe
P2 = D * 62.4/144 : D = 144 * P2 / 62.4
V2 = D * 1639/58 : D = 58 * V2 / 1639
144 * P2 / 62.4 = 58 * V2 / 1639
P2 = 58 * V2 / 1639 * 62.4 / 144
P1 * V1 = P2 * V2
P1 * V1 = 58 * V2 / 1639 * 62.4 / 144 * V2
14.7 psia * 1639 ft3 = 58 / 1639 * 62.4 / 144 * V2^2
V2^2 = 14.7 psia * 1639 ft3 / 58 * 1639 / 62.4 * 144
Can you handle it from here????
Independent events are seldomly independent.
P1 is 14.7 psia
P2 is depth to water surface inside pipe x 62.4/144
V2 is depth to water surface inside pipe x 1639/58
D = Depth to water surface inside pipe
P2 = D * 62.4/144 : D = 144 * P2 / 62.4
V2 = D * 1639/58 : D = 58 * V2 / 1639
144 * P2 / 62.4 = 58 * V2 / 1639
P2 = 58 * V2 / 1639 * 62.4 / 144
P1 * V1 = P2 * V2
P1 * V1 = 58 * V2 / 1639 * 62.4 / 144 * V2
14.7 psia * 1639 ft3 = 58 / 1639 * 62.4 / 144 * V2^2
V2^2 = 14.7 psia * 1639 ft3 / 58 * 1639 / 62.4 * 144
Can you handle it from here????
Independent events are seldomly independent.
The air-pressure can be calculated using the ideal gas law pV=nRT. With nRT being constant, pv=const and p1V1 = p2V2. With p1=1,013 bar, V1=1639 tf3 = 46,4m3.
The Volume V2 = V1 - h*Apipe
e.g. the original volume minus the waterfilled part of the pipe, h being the height over the inlet on the inside of the pipe.
The hydrostatic pressure can be calculated as
p2 = p,hydr = ro*g*(H-h) = 1000 kg/m3 * 9,81 m/s * (H-h)
where H is the height of the water pillar on the outside of the pipe.
Now you can combine and solve h for a given H
The Volume V2 = V1 - h*Apipe
e.g. the original volume minus the waterfilled part of the pipe, h being the height over the inlet on the inside of the pipe.
The hydrostatic pressure can be calculated as
p2 = p,hydr = ro*g*(H-h) = 1000 kg/m3 * 9,81 m/s * (H-h)
where H is the height of the water pillar on the outside of the pipe.
Now you can combine and solve h for a given H
LittleInch
Petroleum
The way I see it is this.
Forget about the actual volume of the pipe, assume it is all the same ID and its the realtive changes in volume that matter Forget about the angle until the end - it's the vertical height that matters.
The only thing that is equal is the pressure in the pipe caused by the water compressing the air balancing the water column on the outside.
For the air part, P2 = P1 x (V1/V2) = 1 x (HT/(HT-H1)), where H1 is height of water in the pipe from the bottom up to the equilibrium point and HT is the total vertical height of the pipe (not the actual length as it is at 45 degrees)
This pressure is matched by the hydrostatic head caused by the difference in head between the total depth of the water measured from the bottom of the tube (HW) minus H1, i.e. P2 = ((HW-H1) x SG x G ) / 100 (gives units in bar)
If P1 = 1bara and SG = 1, Then HT/(HT-H1) = ((HW-H1) x 1 x G)/ 100.
Solve for H1 as all other inputs are known.
This allows for the fact that HW is more than HT, i.e.t he pipe is totally submerged
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
Forget about the actual volume of the pipe, assume it is all the same ID and its the realtive changes in volume that matter Forget about the angle until the end - it's the vertical height that matters.
The only thing that is equal is the pressure in the pipe caused by the water compressing the air balancing the water column on the outside.
For the air part, P2 = P1 x (V1/V2) = 1 x (HT/(HT-H1)), where H1 is height of water in the pipe from the bottom up to the equilibrium point and HT is the total vertical height of the pipe (not the actual length as it is at 45 degrees)
This pressure is matched by the hydrostatic head caused by the difference in head between the total depth of the water measured from the bottom of the tube (HW) minus H1, i.e. P2 = ((HW-H1) x SG x G ) / 100 (gives units in bar)
If P1 = 1bara and SG = 1, Then HT/(HT-H1) = ((HW-H1) x 1 x G)/ 100.
Solve for H1 as all other inputs are known.
This allows for the fact that HW is more than HT, i.e.t he pipe is totally submerged
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
- Thread starter
- #7
I think you guys are close
BIG INCH I am not sure your equation is completely correct if D in your equation is the Depth of the water in the pipe, then the air deptch would be 41-D, Vw = D*1639/41 and Va = (41-D)* 1639/41
LITTLE INCH I am not sure this is a valid equation
If P1 = 1bara and SG = 1, Then HT/(HT-H1) = ((HW-H1) x 1 x G)/ 100.
because you said HT = the total height (41), HW = height of water and H1 is also the height of the water ?
If there is not water (pond empty) then there is no water pressure
let me know what you think
BIG INCH I am not sure your equation is completely correct if D in your equation is the Depth of the water in the pipe, then the air deptch would be 41-D, Vw = D*1639/41 and Va = (41-D)* 1639/41
LITTLE INCH I am not sure this is a valid equation
If P1 = 1bara and SG = 1, Then HT/(HT-H1) = ((HW-H1) x 1 x G)/ 100.
because you said HT = the total height (41), HW = height of water and H1 is also the height of the water ?
If there is not water (pond empty) then there is no water pressure
let me know what you think
LittleInch
Petroleum
From the data given by the OP - a sketch would resolve all this...., my HT is the vertical height of the pipe. HW is the height of the water from the bottom edge of the pipe to the water level - 58 feet. H1 is the vertical height of the water inside the tube. In this case HW > HT and the pipe becomes fully submerged, not that it makes ny difference. My guess is that as HW increases, H1 is not proportional, but is some kind of inverse power relationship as the gas becomes more compressed.
I'll post a sketch later if I have the time and inclination....
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
I'll post a sketch later if I have the time and inclination....
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
Butchs, a sketch from you would make things clearer.
As I envisioned this, D_ft is the vertical depth from surface of the pond to the water surface inside the pipe.
Who cares what 41-D is? 41-D, the vertical distance from water surface inside the pipe to the bottom of the pipe, the outlet, has nothing to do with air pressure inside the pipe.
P2 = D_ft * 62.4 lbs/ft3 * ft2/144in2 = 0.433 psi/ft_D of fresh water.
P to V is a linear relationship, as long as compressibility and temperature can be ignored. Compressibility can usually can be at these low pressures, but there are thermoclines that typically occur at 20 ft and another at 40 ft. If surface water temperature is 80F, the 20 ft thermocline will be around 68-72 degF and the 40 ft thermocline will be near 55-60F, in a pond in a summer, temperate climate zone.
Independent events are seldomly independent.
As I envisioned this, D_ft is the vertical depth from surface of the pond to the water surface inside the pipe.
Who cares what 41-D is? 41-D, the vertical distance from water surface inside the pipe to the bottom of the pipe, the outlet, has nothing to do with air pressure inside the pipe.
P2 = D_ft * 62.4 lbs/ft3 * ft2/144in2 = 0.433 psi/ft_D of fresh water.
P to V is a linear relationship, as long as compressibility and temperature can be ignored. Compressibility can usually can be at these low pressures, but there are thermoclines that typically occur at 20 ft and another at 40 ft. If surface water temperature is 80F, the 20 ft thermocline will be around 68-72 degF and the 40 ft thermocline will be near 55-60F, in a pond in a summer, temperate climate zone.
Independent events are seldomly independent.
- Thread starter
- #10
A sketch would help but I do not have one I can post at this time...perhaps tonight
In my mind D is the height of the water inside the pipe, the pipe has 58 ft of water over it and the top of the pipe is 58-41 = 17 ft below the surface. Another way to look at this is that the pressure in the pipe, at the bottom of the pipe, must equal the static pressure of the 58 ft of water (or about ~25 psig). We can call this Pt for total pressure on the pipe system. To counteract this pressure we have the pressure caused by the column of water inside the pipe (in feet this is D), and the pressure of the compressed air, the volume here is (41-D)*1639/41.
The balance is Pt = Pw + Pa where Pw is the pressure caused by the water column, Pa is P2 in P1*V1 = P2*V2....
to my analysis this becomes a quadratic equation....no this is real life and not homework.thanks
In my mind D is the height of the water inside the pipe, the pipe has 58 ft of water over it and the top of the pipe is 58-41 = 17 ft below the surface. Another way to look at this is that the pressure in the pipe, at the bottom of the pipe, must equal the static pressure of the 58 ft of water (or about ~25 psig). We can call this Pt for total pressure on the pipe system. To counteract this pressure we have the pressure caused by the column of water inside the pipe (in feet this is D), and the pressure of the compressed air, the volume here is (41-D)*1639/41.
The balance is Pt = Pw + Pa where Pw is the pressure caused by the water column, Pa is P2 in P1*V1 = P2*V2....
to my analysis this becomes a quadratic equation....no this is real life and not homework.thanks
LittleInch
Petroleum
This has been annoyoing me, but then I realised that I needed to modify the second part (pressure created by the water column) by adding 1 to get both sides into bara.
Then I solved by iteration to arrive at H1 in attached diagram to be 6.7 m and a pressure of 2.1 bara. Both accurate to +/- 0.1.
Has the right feel to it as well, but feel free to comment on it.
Because the pipe is at 45 degrees, the length along the pipe you need to work out from the upper tangent line as this is when the air starts to be compressed, but that is simple trigonometry
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
Then I solved by iteration to arrive at H1 in attached diagram to be 6.7 m and a pressure of 2.1 bara. Both accurate to +/- 0.1.
Has the right feel to it as well, but feel free to comment on it.
Because the pipe is at 45 degrees, the length along the pipe you need to work out from the upper tangent line as this is when the air starts to be compressed, but that is simple trigonometry
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
- Thread starter
- #13
LittleInch
Petroleum
So pretty much the same as mine then?. The distance ~D and the vertical distance for calcualtion purpose must start at the top lip of the pipe as nothiong will be compressed until the woater overlaps the pipe. Normally not relevant, but for a 6 foot diameter pipe....
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
My motto: Learn something new every day
Also: There's usually a good reason why everyone does it that way
- Thread starter
- #15
There are two ways to solve this problem in my opinion.
First if we treat air as an ideal gas, then PV=nRT and if we assume that the temperature does not vary significantly (air is in pipe so it should remain close to the same temperature), then P1V1 = P2V2. The pipe, as shown is 58 ft long and 6 ft in diameter, ignoring the angle at the top the total volume is Length x Area = 58 x 32 x 3.14 = 1,639 ft3. The maximum height of the water is 58 ft or 58 ft x 62.4 lb/ft3/ 144 in2/ft2 = 25.1 or ~25 psig. Pascal’s Principle states pressures are transmitted evenly in a fluid at a given point. Thus at the bottom of the pipe, the effective pressure is ~25 psig. As the water fills the pond it will enter the pipe and trap the air and then begin to compress the air. The combined pressure of the air and water must equal the 58 ft head at the bottom of the pipe. The pressure balance is shown below
Pt = Pw + PA where
Pt = the total pressure at the bottom of the pipe (58 ft of head or ~ 25 psi)
Pw = the pressure of the water in this case D ft (or D ft x 62.4 lb/ft3/ 144 in2/ft2 in psi)
PA = Pressure of the air which is P2 or P1V1 / V2 and V2 can be expressed as 1639 x (1 -D/41), in absolute terms P1 = 14.7 psia or ~ 33.9 ftH2O, V1 = 1,639 ft3; P1V1 = 55726 ft4
Re-arranging these terms gives
Pt = D + P1V1 / 1639(1-D/41) (in feet) or
0 = D + P1V1 /1639(1-D/41) - Pt
0 = D + 33.9 ft x 1639 ft3/(1639 ft3 x (1-D/41)) – 58 (D in feet of H2O)
re-arranging and putting in numeric values to get to a quadratic equation we have
0 = -39.97 x D2 + 5304 x D – 95062 feet (may have a minor math error in the a, b and c terms)
Which has a solution of 21.27 and 111.44 ft for D, the 111.4 value is not applicable.
The second way to do this and perhaps somewhat easier is to set up a spread sheet and iterate on the answer until the pressure of the air and pressure of the water equals 58 ft of water.
58 ft = D + P1V1 /(1639 x (1-D/41) )
Then just build a spread sheet and keep increasing D until a value of 58 is obtained (see below). There is a slight difference in these two approaches, probably due to some minor math error in the quadratic equation algebra which with more time I could trouble shoot. In any case we see that the water column will be about 21 ft high and the air column about 20 ft.
D ft Pair ft Total Ft
1 0.85 1.85
2 1.74 3.74
3 2.68 5.68
10 10.94 20.94
18 26.53 44.53
19 29.28 48.28
20 32.29 52.29
21 35.59 56.59
21.3 36.65 57.95
Both approaches lead to a water height of about 21.3 ft, from this we can calculate the air pressure P2 from
P2 = P1V1 / V2 where V2 = 1639 x (1 -D/41) = 1639 x( 1 -.48) = 789 ft3 then
P2 = 33.9 x 1639/789 = 70.4 ftA or 36.5 ft G or 15.8 psig (A= absolute, G= gauge)
Finally there is another issue in peak pressure. The pressure calculated above, is about 16 psig. But this pressure could actually be substantially higher. Why because in addition to Pascal’s Principle, Bernoulli’s equation, the ideal gas law and the assumption of isothermal conditions, we have to deal with conservation of momentum. If the water enters the pipe too quickly (surge) we can get a much higher pressure spike. The water will enter the pipe until the internal air pressure and the weight of the water matches the driving force (external force of pressure (F= P X A), however when this force balance occurs, the water may or will still be in motion. To stop the water the momentum in the water (Mass x velocity) or kinetic energy (.5MVel2) must be absorbed, this shows up as a pressure spike in the air. The magnitude of the spike will depend in part, on how fast the water fills the pond and enters the pipe. Good papers on this can be found at:
This was for water surging into a horizontal pipe, but the same phenomena can occur for the angled pipe shown. Individuals familiar with fluid systems know this as water hammer or geysers in storm water systems. So the problem is not as simple as it would seem.
First if we treat air as an ideal gas, then PV=nRT and if we assume that the temperature does not vary significantly (air is in pipe so it should remain close to the same temperature), then P1V1 = P2V2. The pipe, as shown is 58 ft long and 6 ft in diameter, ignoring the angle at the top the total volume is Length x Area = 58 x 32 x 3.14 = 1,639 ft3. The maximum height of the water is 58 ft or 58 ft x 62.4 lb/ft3/ 144 in2/ft2 = 25.1 or ~25 psig. Pascal’s Principle states pressures are transmitted evenly in a fluid at a given point. Thus at the bottom of the pipe, the effective pressure is ~25 psig. As the water fills the pond it will enter the pipe and trap the air and then begin to compress the air. The combined pressure of the air and water must equal the 58 ft head at the bottom of the pipe. The pressure balance is shown below
Pt = Pw + PA where
Pt = the total pressure at the bottom of the pipe (58 ft of head or ~ 25 psi)
Pw = the pressure of the water in this case D ft (or D ft x 62.4 lb/ft3/ 144 in2/ft2 in psi)
PA = Pressure of the air which is P2 or P1V1 / V2 and V2 can be expressed as 1639 x (1 -D/41), in absolute terms P1 = 14.7 psia or ~ 33.9 ftH2O, V1 = 1,639 ft3; P1V1 = 55726 ft4
Re-arranging these terms gives
Pt = D + P1V1 / 1639(1-D/41) (in feet) or
0 = D + P1V1 /1639(1-D/41) - Pt
0 = D + 33.9 ft x 1639 ft3/(1639 ft3 x (1-D/41)) – 58 (D in feet of H2O)
re-arranging and putting in numeric values to get to a quadratic equation we have
0 = -39.97 x D2 + 5304 x D – 95062 feet (may have a minor math error in the a, b and c terms)
Which has a solution of 21.27 and 111.44 ft for D, the 111.4 value is not applicable.
The second way to do this and perhaps somewhat easier is to set up a spread sheet and iterate on the answer until the pressure of the air and pressure of the water equals 58 ft of water.
58 ft = D + P1V1 /(1639 x (1-D/41) )
Then just build a spread sheet and keep increasing D until a value of 58 is obtained (see below). There is a slight difference in these two approaches, probably due to some minor math error in the quadratic equation algebra which with more time I could trouble shoot. In any case we see that the water column will be about 21 ft high and the air column about 20 ft.
D ft Pair ft Total Ft
1 0.85 1.85
2 1.74 3.74
3 2.68 5.68
10 10.94 20.94
18 26.53 44.53
19 29.28 48.28
20 32.29 52.29
21 35.59 56.59
21.3 36.65 57.95
Both approaches lead to a water height of about 21.3 ft, from this we can calculate the air pressure P2 from
P2 = P1V1 / V2 where V2 = 1639 x (1 -D/41) = 1639 x( 1 -.48) = 789 ft3 then
P2 = 33.9 x 1639/789 = 70.4 ftA or 36.5 ft G or 15.8 psig (A= absolute, G= gauge)
Finally there is another issue in peak pressure. The pressure calculated above, is about 16 psig. But this pressure could actually be substantially higher. Why because in addition to Pascal’s Principle, Bernoulli’s equation, the ideal gas law and the assumption of isothermal conditions, we have to deal with conservation of momentum. If the water enters the pipe too quickly (surge) we can get a much higher pressure spike. The water will enter the pipe until the internal air pressure and the weight of the water matches the driving force (external force of pressure (F= P X A), however when this force balance occurs, the water may or will still be in motion. To stop the water the momentum in the water (Mass x velocity) or kinetic energy (.5MVel2) must be absorbed, this shows up as a pressure spike in the air. The magnitude of the spike will depend in part, on how fast the water fills the pond and enters the pipe. Good papers on this can be found at:
This was for water surging into a horizontal pipe, but the same phenomena can occur for the angled pipe shown. Individuals familiar with fluid systems know this as water hammer or geysers in storm water systems. So the problem is not as simple as it would seem.
The pressure at the bottom of the pipe is 58'. What you need to determine is the volume change of air when compressed by 58' of water pressure. As you noted, the pressure at the bottom of the pipe is ~25 psig. P1*V1 = P2*V2 I think the trick is, after you've determined the volume change, you need to subtract out the difference in pressure from the height change of the water. So if the water rises 5 feet due to the compression of the air, the pressure of the air will be less than 25 psig since the pressure exterted by the water will be less than 58', it would be 53'. You need to iterate.
An interesting problem, to be sure. What's the application?
An interesting problem, to be sure. What's the application?
Oh, I meant to say look up information on a "diving bell". Might provide a clue. I didn't research this one much but it might be helpful:
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