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re: When they start to close the valves...
Posted on 7/14/10 at 10:14 am to Tiger-Striped-Bass
Posted on 7/14/10 at 10:14 am to Tiger-Striped-Bass
quote:
The goal is to make P(hydro) greater than P(formation) so that the flow stops. (That's a bit oversimplified, but close enough.)
this will put the system in balance...
0
Posted on 7/14/10 at 10:33 am to Tiger-Striped-Bass
quote:
Here's what I don't. It's formation pressure that is felt throughout the system and is what we're fighting, right? Any pressure equalization, regardless of where flow is stopped, will be whatever the formation pressure is, right? If you stop the flow on top with the cap, pressure throughout the system equalizes to formation pressure. If you stop flow on the bottom, from that point down, pressure equalizes to formation pressure. So what difference does it make of where you stop flow in regards to creating other leaks in the sea floor/subsurface if the integrity of these areas is suspect?
I think this is the source of your confusion. The pressure is not the same everywhere. Pressure at any point in the system is the reservoir pressure minus the hydrostatic head between those two points. Go back to the equation I posted. So the higher up the well, the lower the pressure. The reason is that the weight of the oil column "pushes down" on the formation and reduces the pressure at that point. The more oil in between, the more the pressure reduction and the lower the pressure at that point. In the case of oil, it's not heavy enough to completely overcome the reservoir pressure. This is why they want to see 8000-9000 psi at the surface when they do the integrity test. Kill mud is heavier and it will overcome that pressure.
Posted on 7/14/10 at 10:37 am to MountainTiger
Here's a post I made in another thread that should help explain what's going on.
quote:
Imagine your 50' garden hose, one end hooked up to a faucet and the other end raised straight up into the air. You turn on the faucet and if your water pressure is high enough, water will flow out of the of the hose 50' up. Now replace that hose with a 100' hose. Now maybe your water pressure isn't high enough to overcome the weight of the water in the hose. No water flows.
OK, now go back to the 50' hose but this time there is a T a few feet above the faucet and on the tee there is a valve that connects to another hose. At first the valve is closed. You turn on the faucet and water flows out the top, just like before. But now you open the valve and inject mud that weighs twice as much as water into the hose. At first water continues to flow but as the hose fills up with heavy mud, the water pressure can't overcome the weight anymore. It's as if the hose was 100' long instead of 50'.
That's an oversimplified idea of what the relief well will do. They will come in from the side and penetrate the existing well (open the valve). Then they will push heavy mud into the well until the pressure of the formation can no longer push against the weight of the mud. At that point the flow stops.
Posted on 7/14/10 at 11:23 am to MountainTiger
Plus having control at the top will make the killing process much better. More control and faster kill.
Posted on 7/14/10 at 11:30 am to SD 71
that is the way I see it too. the valve stack will be a big help in the kill.
Posted on 7/14/10 at 12:25 pm to MountainTiger
Mountain,
I understand all of that and I guess I haven't been clear. Please bear with me because I really would like to know the explanation.
To adapt the garden hose analogy, you left out the key to my question - the rubber gasket at the faucet upstream from the tee, which we can use to represent the sea floor.
If the rubber gasket is damaged and will leak if we shut off the flow at the top of the hose, how will stopping the flow within the hose from the tee on the bottom, keep the pressure from the source from leaking out of the damaged gasket? Won't the same source pressure that makes the water overcome all of that head pressure now be exerted on the gasket and thus leak in either scenario? If you have 50psi of water at the faucet, no matter how you stop it, with a cap, with extra hose to increase head capacity or by pumping heavy mud in from near the bottom, there will still be 50psi on the damaged gasket from the source.
What isolates that formation pressure from the sea floor ? If there is 20,000 psi in the formation, there will still be 20,000 psi exerted on the suspect sea floor after bottom kill. Or does it put mud/cement in the formation to fill any cracks around the well too?
If they're afraid to pinch off the top because th eseafloor can't handle a couple thousand extra psi of head pressure next to the 20,000 from the source, I just don't see the seafloor holding back the source pressure w/o head either, once flow is stopped.
I understand all of that and I guess I haven't been clear. Please bear with me because I really would like to know the explanation.
To adapt the garden hose analogy, you left out the key to my question - the rubber gasket at the faucet upstream from the tee, which we can use to represent the sea floor.
If the rubber gasket is damaged and will leak if we shut off the flow at the top of the hose, how will stopping the flow within the hose from the tee on the bottom, keep the pressure from the source from leaking out of the damaged gasket? Won't the same source pressure that makes the water overcome all of that head pressure now be exerted on the gasket and thus leak in either scenario? If you have 50psi of water at the faucet, no matter how you stop it, with a cap, with extra hose to increase head capacity or by pumping heavy mud in from near the bottom, there will still be 50psi on the damaged gasket from the source.
What isolates that formation pressure from the sea floor ? If there is 20,000 psi in the formation, there will still be 20,000 psi exerted on the suspect sea floor after bottom kill. Or does it put mud/cement in the formation to fill any cracks around the well too?
If they're afraid to pinch off the top because th eseafloor can't handle a couple thousand extra psi of head pressure next to the 20,000 from the source, I just don't see the seafloor holding back the source pressure w/o head either, once flow is stopped.
This post was edited on 7/14/10 at 12:43 pm
Posted on 7/14/10 at 12:46 pm to Tiger-Striped-Bass
Ok, let's modify the garden hose analogy to include a cap on top. We screw a cap onto the top of the hose that can hold 5 psi. Anything over that will cause it to burst. Let's also assume that the pressure at your hose bib is 25 psi.
The pressure head of 50 feet of water is:
62.4 * 50 / 144 = 21.7 psi
Therefore the pressure at the cap is:
25 - 21.7 = 3.3 psi
Does the cap burst? No, it can handle 3.3 psi.
I think your problem is that you keep thinking that reservoir pressure will be felt by the BOP. It won't. The pressure there is the reservoir pressure minus the head.
The pressure head of 50 feet of water is:
62.4 * 50 / 144 = 21.7 psi
Therefore the pressure at the cap is:
25 - 21.7 = 3.3 psi
Does the cap burst? No, it can handle 3.3 psi.
I think your problem is that you keep thinking that reservoir pressure will be felt by the BOP. It won't. The pressure there is the reservoir pressure minus the head.
Posted on 7/14/10 at 1:09 pm to MountainTiger
My intitial post in this thread was in response to "why even try closing the top" and my comment was exclusively in regards to subsurface earth cracking, not flow control within piping.
Can the possibility of subsurface earth cracking be eliminated by bottom kill?
Can the possibility of subsurface earth cracking be eliminated by bottom kill?
Posted on 7/14/10 at 1:33 pm to Tiger-Striped-Bass
If the analogy is failing then let's just use the actual well. We'll have to makes some assumptions. The reservoir is about 13,500' from the wellhead. We don't know the density of the oil but let's say it weighs 40 lbs./ft.^3. That's a little low but there's a lot of gas in this flow that lowers it's specific weight. They've estimated the reservoir pressure at about 12,000 psi.
So the pressure at the wellhead is the reservoir pressure minus the weight of the oil. The oil head will be:
13,500 * 40 / 144 = 3750 psi.
Then the pressure at the wellhead will be 8250 psi. They are concerned that that pressure will be more than the BOP and/or the surface casing can handle.
Now you asked about the bottom kill and why wouldn't the wellhead feel the same pressure?
If the kill mud is 16 ppg, the specific weight is
16 / .1337 = 120 lbs./ft^3
So now the head is
120 * 13,500 / 144 = 11,200 psi and therefore the pressure on the surface pipe is only 800 psi, which it should be able to contain.
So the pressure at the wellhead is the reservoir pressure minus the weight of the oil. The oil head will be:
13,500 * 40 / 144 = 3750 psi.
Then the pressure at the wellhead will be 8250 psi. They are concerned that that pressure will be more than the BOP and/or the surface casing can handle.
Now you asked about the bottom kill and why wouldn't the wellhead feel the same pressure?
If the kill mud is 16 ppg, the specific weight is
16 / .1337 = 120 lbs./ft^3
So now the head is
120 * 13,500 / 144 = 11,200 psi and therefore the pressure on the surface pipe is only 800 psi, which it should be able to contain.
Posted on 7/19/10 at 11:32 pm to MountainTiger
If there's anybody that still doesn't understand how the relief well is going to work, watch this video: LINK
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