Posted: 5/8/2015 2:12:29 PM EDT
| So I am sitting here bored as hell on the frac site. We are currently pumping 91 barrels a minute (1bbl=42 gallons) , pressure is sitting at 8410 psi. If the wellhead were to snap off at the base, how much thrust would it be pushing and how high do you think it would fly if it magically still had that rate and pressure flowing through it? (5" wellhead diameter) |
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Okay, it's been a few years since I took physics, but I'm taking a stab at it. Probably fucked up something along the lines, but I did my best.
Given a pressure of 8410 psi (57,984,908 pascals) on a .0635 square meter wellhead comes out to about 3,682,041 N of total force. This is roughly equal to 422 200lb men standing on the same wellhead. Now how high will it fly? I've got to make some assumptions here, but the concept should still apply. Given the equation F= m a, we can figure out how much acceleration the wellhead would have when it shears off. So using our calculated force above, and making the assumption that it weighs 100kg, we find that it would accelerate at 36,820 m/s squared. a = 3,682,041 / 100 = 36,820 m/s squared. This means that if the force acted on the wellhead for one whole second, it would be traveling 36,820 m/s. This is something like 5 times faster than a space shuttle travels when orbiting the earth. Or Mach 107. But since the farther it travels from the break point, the less force will be cating on it, I have to make an assumption about how long the force actually acts on the wellhead. For a bit of armchair math, I'm assuming it acts on the wellhead for .01s. So the formula from here is Vf = Vo + at. Final velocity = starting velocity + acceleration x time. Since we don't have a starting velocity, that'll be zero. But we do have time and acceleration values that allow us to calculate its final velocity. Vf = 3,682,041 x .01 = 368.2 m/s. So your wellhead after shearing off will be traveling at 368.2 m/s, slightly over the sound barrier. So you'll have a sonic boom as it hurtles upward at a terrifying velocity. But how high will it go? Now we've got to figure out how long it will be in the air. Ignoring air resistance and using the negative acceleration due to gravity. , we find that it will travel upwards for 37.57s. V = Vo + at 0 = 368.2 + 9.8 t t = 37.57s. So how far will it have traveled upwards during those 37.57s? X = Xo + VoT + 1/2 aT^2. X = 0 + (368.2)(37.57) + 1/2(9.8)(37.57)^2 X = 20,749 m. So your wellhead will travel 20,749 meters in the sky. This is nearly 13 miles, and is higher than most planes fly. So if you're unlucky, you'll take down an airliner and get fracking banned in the US. |
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Pretty good physics there. But the weld head weighs a lot more than 100 kg. Maybe 20,000 pounds, depending on a lot of factors.
But a simple estimate of how much force is there is the area of the bore, 5"/2 times 3.14 times the pressure, and I get 163,154 pounds of force acting on said well head. My son used to run frac jobs for Schlumberger, they finally forbid their hands from casually walking around the piping and head end of the pumps while pumping, for just this reason. At the pump head end, it is more than possible to have 15,000 psi present, any kind of a leak or failure, is doom for personnel in a bad spot, i.e., down range. A fifty pound chunk of pump is pretty effective shrapnel. |
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Pretty good physics there. But the weld head weighs a lot more than 100 kg. Maybe 20,000 pounds, depending on a lot of factors. But a simple estimate of how much force is there is the area of the bore, 5"/2 times 3.14 times the pressure, and I get 163,154 pounds of force acting on said well head. My son used to run frac jobs for Schlumberger, they finally forbid their hands from casually walking around the piping and head end of the pumps while pumping, for just this reason. At the pump head end, it is more than possible to have 15,000 psi present, any kind of a leak or failure, is doom for personnel in a bad spot, i.e., down range. A fifty pound chunk of pump is pretty effective shrapnel. Yeah, running it through quickly, I must have messed up the force calculations by fudging the conversions to SI. And 20,000 pounds? Man, I had an image of like a little 5" welded cap on there. And I was thinking 8500 psi was an incredible amount, 15000 is just unthinkable. |
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Yeah, running it through quickly, I must have messed up the force calculations by fudging the conversions to SI. And 20,000 pounds? Man, I had an image of like a little 5" welded cap on there. And I was thinking 8500 psi was an incredible amount, 15000 is just unthinkable. Quoted:
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Pretty good physics there. But the weld head weighs a lot more than 100 kg. Maybe 20,000 pounds, depending on a lot of factors. But a simple estimate of how much force is there is the area of the bore, 5"/2 times 3.14 times the pressure, and I get 163,154 pounds of force acting on said well head. My son used to run frac jobs for Schlumberger, they finally forbid their hands from casually walking around the piping and head end of the pumps while pumping, for just this reason. At the pump head end, it is more than possible to have 15,000 psi present, any kind of a leak or failure, is doom for personnel in a bad spot, i.e., down range. A fifty pound chunk of pump is pretty effective shrapnel. Yeah, running it through quickly, I must have messed up the force calculations by fudging the conversions to SI. And 20,000 pounds? Man, I had an image of like a little 5" welded cap on there. And I was thinking 8500 psi was an incredible amount, 15000 is just unthinkable. Wait, you were being serious? You really think that well head would rocket upwards at mach 107? 
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If the immense pressure acted on it for a total of 1 second, yes. But these sorts of things happen on an incredibly short timescale. I just made the assumption that it'd fly off and the liquid below would stop acting on it within .01s. With this in mind, it'd be flying off a around 360 m/s, or a little over mach 1. But again, that's with the assumption that it weighs 100 kg. If it weighs 1,000 kg, it'd be flying at about 36 m/s, or 80 mph.
Basically, there are too many variables here to make an accurate determination. There's no way to know how long the pressure will act on the wellhead, and I don't know how heavy it is, I don't know how much drag it would have, etc. I took a page out of xkcd's "what if" page and tried to make it fun with the numbers I had. |
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So I am sitting here bored as hell on the frac site. We are currently pumping 91 barrels a minute (1bbl=42 gallons) , pressure is sitting at 8410 psi. If the wellhead were to snap off at the base, how much thrust would it be pushing and how high do you think it would fly if it magically still had that rate and pressure flowing through it? (5" wellhead diameter) Need to know the mass of the wellhead. ETA: And the frontal area and drag coefficient. We can assume it "flies" without tumbling. |
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Okay, it's been a few years since I took physics, but I'm taking a stab at it. Probably fucked up something along the lines, but I did my best. Given a pressure of 8410 psi (57,984,908 pascals) on a .0635 square meter wellhead comes out to about 3,682,041 N of total force. This is roughly equal to 422 200lb men standing on the same wellhead. Now how high will it fly? I've got to make some assumptions here, but the concept should still apply. Given the equation F= m a, we can figure out how much acceleration the wellhead would have when it shears off. So using our calculated force above, and making the assumption that it weighs 100kg, we find that it would accelerate at 36,820 m/s squared. a = 3,682,041 / 100 = 36,820 m/s squared. This means that if the force acted on the wellhead for one whole second, it would be traveling 36,820 m/s. This is something like 5 times faster than a space shuttle travels when orbiting the earth. Or Mach 107. 
91 bbl/min x 42 gal/bbl = 3,822 gal/min, or 14,468 liters per minute. This is being pushed through a 5" diameter hole. How fast is the oil going? A 5" circle has an area of Pi * diameter. Let's stick with metric for the moment, 'cause it's easier. A 5" diameter pipe has an opening of right at 40cm². The well is pushing 14,468 liters, 144,680 cm³ through that hole, at a velocity of 3,617cm/sec or 36.17 m/sec. Converting back to SAE, 119 feet per second. The casing isn't going to leave the wellhead much faster than that, and once the pressure is relieved, the stream of oil won't go much faster either. |
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Which should have told you right off the bat that your model was in error.
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Okay, it's been a few years since I took physics, but I'm taking a stab at it. Probably fucked up something along the lines, but I did my best. Given a pressure of 8410 psi (57,984,908 pascals) on a .0635 square meter wellhead comes out to about 3,682,041 N of total force. This is roughly equal to 422 200lb men standing on the same wellhead. Now how high will it fly? I've got to make some assumptions here, but the concept should still apply. Given the equation F= m a, we can figure out how much acceleration the wellhead would have when it shears off. So using our calculated force above, and making the assumption that it weighs 100kg, we find that it would accelerate at 36,820 m/s squared. a = 3,682,041 / 100 = 36,820 m/s squared. This means that if the force acted on the wellhead for one whole second, it would be traveling 36,820 m/s. This is something like 5 times faster than a space shuttle travels when orbiting the earth. Or Mach 107.
I was just having some fun with it. Obviously it's not going to have that force acting on it for a full second. Keep reading, I make the assumption that it stops accelerating .01 seconds after the failure. |
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Okay, it's been a few years since I took physics, but I'm taking a stab at it. Probably fucked up something along the lines, but I did my best. Given a pressure of 8410 psi (57,984,908 pascals) on a .0635 square meter wellhead comes out to about 3,682,041 N of total force. This is roughly equal to 422 200lb men standing on the same wellhead. Now how high will it fly? I've got to make some assumptions here, but the concept should still apply. Given the equation F= m a, we can figure out how much acceleration the wellhead would have when it shears off. So using our calculated force above, and making the assumption that it weighs 100kg, we find that it would accelerate at 36,820 m/s squared. a = 3,682,041 / 100 = 36,820 m/s squared. This means that if the force acted on the wellhead for one whole second, it would be traveling 36,820 m/s. This is something like 5 times faster than a space shuttle travels when orbiting the earth. Or Mach 107. But since the farther it travels from the break point, the less force will be cating on it, I have to make an assumption about how long the force actually acts on the wellhead. For a bit of armchair math, I'm assuming it acts on the wellhead for .01s. So the formula from here is Vf = Vo + at. Final velocity = starting velocity + acceleration x time. Since we don't have a starting velocity, that'll be zero. But we do have time and acceleration values that allow us to calculate its final velocity. Vf = 3,682,041 x .01 = 368.2 m/s. So your wellhead after shearing off will be traveling at 368.2 m/s, slightly over the sound barrier. So you'll have a sonic boom as it hurtles upward at a terrifying velocity. But how high will it go? Now we've got to figure out how long it will be in the air. Ignoring air resistance and using the negative acceleration due to gravity. , we find that it will travel upwards for 37.57s. V = Vo + at 0 = 368.2 + 9.8 t t = 37.57s. So how far will it have traveled upwards during those 37.57s? X = Xo + VoT + 1/2 aT^2. X = 0 + (368.2)(37.57) + 1/2(9.8)(37.57)^2 X = 20,749 m. So your wellhead will travel 20,749 meters in the sky. This is nearly 13 miles, and is higher than most planes fly. So if you're unlucky, you'll take down an airliner and get fracking banned in the US. You missed that the acceleration due to gravity is NEGative 9.81. so that should be +1/2(-9.81) that's also the range in the x direction...or in other words, your X's should be Y's Y=v*t*sin(theta)-1/2*g*t^2 Assuming straight vertical, sin(90)=1.. Y=v*t-1/2*g*t^2 Pressure is Area is pi*r^2 or pi*2.5^2 = 19.63 sq.in P=f/a or force=p*a force = 8410psi* 19.63 sq.in) = 165,129.96 lbs |
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If the immense pressure acted on it for a total of 1 second, yes. But these sorts of things happen on an incredibly short timescale. I just made the assumption that it'd fly off and the liquid below would stop acting on it within .01s. With this in mind, it'd be flying off a around 360 m/s, or a little over mach 1. But again, that's with the assumption that it weighs 100 kg. If it weighs 1,000 kg, it'd be flying at about 36 m/s, or 80 mph. Basically, there are too many variables here to make an accurate determination. There's no way to know how long the pressure will act on the wellhead, and I don't know how heavy it is, I don't know how much drag it would have, etc. I took a page out of xkcd's "what if" page and tried to make it fun with the numbers I had. You should work for NASA |
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You missed that the acceleration due to gravity is NEGative 9.81. so that should be +1/2(-9.81) that's also the range in the x direction...or in other words, your X's should be Y's Y=v*t*sin(theta)-1/2*g*t^2 Assuming straight vertical, sin(90)=1.. Y=v*t-1/2*g*t^2 Pressure is Area is pi*r^2 or pi*2.5^2 = 19.63 sq.in P=f/a or force=p*a force = 8410psi* 19.63 sq.in) = 165,129.96 lbs Quoted:
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Okay, it's been a few years since I took physics, but I'm taking a stab at it. Probably fucked up something along the lines, but I did my best. Given a pressure of 8410 psi (57,984,908 pascals) on a .0635 square meter wellhead comes out to about 3,682,041 N of total force. This is roughly equal to 422 200lb men standing on the same wellhead. Now how high will it fly? I've got to make some assumptions here, but the concept should still apply. Given the equation F= m a, we can figure out how much acceleration the wellhead would have when it shears off. So using our calculated force above, and making the assumption that it weighs 100kg, we find that it would accelerate at 36,820 m/s squared. a = 3,682,041 / 100 = 36,820 m/s squared. This means that if the force acted on the wellhead for one whole second, it would be traveling 36,820 m/s. This is something like 5 times faster than a space shuttle travels when orbiting the earth. Or Mach 107. But since the farther it travels from the break point, the less force will be cating on it, I have to make an assumption about how long the force actually acts on the wellhead. For a bit of armchair math, I'm assuming it acts on the wellhead for .01s. So the formula from here is Vf = Vo + at. Final velocity = starting velocity + acceleration x time. Since we don't have a starting velocity, that'll be zero. But we do have time and acceleration values that allow us to calculate its final velocity. Vf = 3,682,041 x .01 = 368.2 m/s. So your wellhead after shearing off will be traveling at 368.2 m/s, slightly over the sound barrier. So you'll have a sonic boom as it hurtles upward at a terrifying velocity. But how high will it go? Now we've got to figure out how long it will be in the air. Ignoring air resistance and using the negative acceleration due to gravity. , we find that it will travel upwards for 37.57s. V = Vo + at 0 = 368.2 + 9.8 t t = 37.57s. So how far will it have traveled upwards during those 37.57s? X = Xo + VoT + 1/2 aT^2. X = 0 + (368.2)(37.57) + 1/2(9.8)(37.57)^2 X = 20,749 m. So your wellhead will travel 20,749 meters in the sky. This is nearly 13 miles, and is higher than most planes fly. So if you're unlucky, you'll take down an airliner and get fracking banned in the US. You missed that the acceleration due to gravity is NEGative 9.81. so that should be +1/2(-9.81) that's also the range in the x direction...or in other words, your X's should be Y's Y=v*t*sin(theta)-1/2*g*t^2 Assuming straight vertical, sin(90)=1.. Y=v*t-1/2*g*t^2 Pressure is Area is pi*r^2 or pi*2.5^2 = 19.63 sq.in P=f/a or force=p*a force = 8410psi* 19.63 sq.in) = 165,129.96 lbs Ehh, it was armchair math done quickly. Fun question though. |
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You should work for NASA Quoted:
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If the immense pressure acted on it for a total of 1 second, yes. But these sorts of things happen on an incredibly short timescale. I just made the assumption that it'd fly off and the liquid below would stop acting on it within .01s. With this in mind, it'd be flying off a around 360 m/s, or a little over mach 1. But again, that's with the assumption that it weighs 100 kg. If it weighs 1,000 kg, it'd be flying at about 36 m/s, or 80 mph. Basically, there are too many variables here to make an accurate determination. There's no way to know how long the pressure will act on the wellhead, and I don't know how heavy it is, I don't know how much drag it would have, etc. I took a page out of xkcd's "what if" page and tried to make it fun with the numbers I had. You should work for NASA At least I know how to get a spaceship moving at Mach 107 now. We'll be on Mars in no time. |
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Pi. The cause of and solution to all of life's problems. Quoted:
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3. 1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 Pi. The cause of and solution to all of life's problems. No, that's alcohol according to Homer J Simpson |
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Ehh, it was armchair math done quickly. Fun question though. Quoted:
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Okay, it's been a few years since I took physics, but I'm taking a stab at it. Probably fucked up something along the lines, but I did my best. Given a pressure of 8410 psi (57,984,908 pascals) on a .0635 square meter wellhead comes out to about 3,682,041 N of total force. This is roughly equal to 422 200lb men standing on the same wellhead. Now how high will it fly? I've got to make some assumptions here, but the concept should still apply. Given the equation F= m a, we can figure out how much acceleration the wellhead would have when it shears off. So using our calculated force above, and making the assumption that it weighs 100kg, we find that it would accelerate at 36,820 m/s squared. a = 3,682,041 / 100 = 36,820 m/s squared. This means that if the force acted on the wellhead for one whole second, it would be traveling 36,820 m/s. This is something like 5 times faster than a space shuttle travels when orbiting the earth. Or Mach 107. But since the farther it travels from the break point, the less force will be cating on it, I have to make an assumption about how long the force actually acts on the wellhead. For a bit of armchair math, I'm assuming it acts on the wellhead for .01s. So the formula from here is Vf = Vo + at. Final velocity = starting velocity + acceleration x time. Since we don't have a starting velocity, that'll be zero. But we do have time and acceleration values that allow us to calculate its final velocity. Vf = 3,682,041 x .01 = 368.2 m/s. So your wellhead after shearing off will be traveling at 368.2 m/s, slightly over the sound barrier. So you'll have a sonic boom as it hurtles upward at a terrifying velocity. But how high will it go? Now we've got to figure out how long it will be in the air. Ignoring air resistance and using the negative acceleration due to gravity. , we find that it will travel upwards for 37.57s. V = Vo + at 0 = 368.2 + 9.8 t t = 37.57s. So how far will it have traveled upwards during those 37.57s? X = Xo + VoT + 1/2 aT^2. X = 0 + (368.2)(37.57) + 1/2(9.8)(37.57)^2 X = 20,749 m. So your wellhead will travel 20,749 meters in the sky. This is nearly 13 miles, and is higher than most planes fly. So if you're unlucky, you'll take down an airliner and get fracking banned in the US. You missed that the acceleration due to gravity is NEGative 9.81. so that should be +1/2(-9.81) that's also the range in the x direction...or in other words, your X's should be Y's Y=v*t*sin(theta)-1/2*g*t^2 Assuming straight vertical, sin(90)=1.. Y=v*t-1/2*g*t^2 Pressure is Area is pi*r^2 or pi*2.5^2 = 19.63 sq.in P=f/a or force=p*a force = 8410psi* 19.63 sq.in) = 165,129.96 lbs Ehh, it was armchair math done quickly. Fun question though. no problem. No one is faulting you. I miss my negatives sometimes too. ETA: I'm thinking you can probably get the initial velocity from the flow rate. Q=AV or m_dot = rho*A*V you have everything but Rho and V. solve for V, assume Tetradecane for heavy crude and you have your rho=0.7628 g/mL |
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Quoted: Pi. The cause of and solution to all of life's problems. Quoted: Quoted: 3. 1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 Pi. The cause of and solution to all of life's problems. |
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The big hurdle is figuring at how long it'll take to shear off the well cap and the resulting decay in the acting force as it escapes further from the end of the pipe.
Without knowing that, all quantitative calculations are just WAGs. You really have no choice but to figure out an empirical formula based on actual testing. |
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Quoted: No, that's alcohol according to Homer J Simpson Quoted: Quoted: Quoted: 3. 1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 Pi. The cause of and solution to all of life's problems. No, that's alcohol according to Homer J Simpson Shut the fuck up, Flanders. |
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Forgot to add in the weight. I have no idea what an entire frac stack weighs, but they are pretty damn heavy. I am thinking in the neighborhood of 20k pounds maybe. Our pipes are rated for 15k psi, we always test to 10k and we have burst discs on every pump that blow at 12,5 just to be safe. When that iron goes, nothing stops it. This is a frac stack that blew off a while back.  |
