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914World.com _ 914World Garage _ OT. WAY OT
Posted by: jfort Aug 27 2008, 10:55 AM
My son and I have a real problem and we need some help. My son is a licensed blaster. He has been asked to stimulate oil wells with dynamite. I am trying to help him. In short, we will fabricate a "torpedo." Currently, we are thinking of using a 5' length of 1.5" PVC with the down end made into a point. The torpedo will be filled with 2 sticks of dynamite and filled with as dense a sand/gravel as we can find. It will be dropped down a 5" pipe filled with water. It is estimated that it is 640' to the bottom. Can you help me calculate how long it will take the torpedo to reach the bottom? (We need to make sure it gets to the bottom before the fuse ignites the dynamite!!) I am sure the efficiency through the water, drag, is an important and not known for sure variable, but perhaps there is a range that would get us close enough.
Any guidance or leads would be most appreciated.
Posted by: ericread Aug 27 2008, 11:21 AM
I believe you should find all of the necessary documentation in the following link:
http://looney.goldenagecartoons.com/tv/rrshow/
Eric Read
"If you don't ask, how are you ever going to learn?"
Posted by: Lavanaut Aug 27 2008, 12:25 PM
Ooooh fluid dynamics. Going to be very tough to estimate given the fact that you're delivering the dynamite in a homemade vehicle with unknown properties. I might think about triggering it via remote control after allowing "more than enough" time for it to reach the bottom. 
Radio Shack is your friend!
![popcorn[1].gif](style_emoticons/default/popcorn[1].gif)
Posted by: jfort Aug 27 2008, 12:41 PM
My son is a sophisticated blaster, having done many projects: quarries, bridges, sewers, etc., and he knows all about setting off blasting caps electrically or with det chord. Caps have depth limitations, apparently, so we've been looking into the old fashioned torpedo method that uses a fuse. Old timers say there is enough time, but I'd like to rough it out mathematically if I can. I found a kid's science fair project on line where he dropped different shapes down a water-filled pipe. A sphere had a falling rate of 1.6 ft/sec, which seems slow to me.
Posted by: pin31 Aug 27 2008, 12:45 PM
Check out this web site: http://www.newton.dep.anl.gov/askasci/phy99/phy99x88.htm
Discussion on objects falling through water.
Bottom line is there are too many variables to calculate a definite answer (from the info provided).
But assuming the medium is water, the torpedo is streamlined and very dense you can guesstimate a ball park answer.
I'm assuming that you don't need to know exactly when the torpedo reaches the bottom (as long as it does).
The fastest (best case scenario) would be dropping through air (32ft/sec2).
It would take 20 sec to fall 640ft (ignoring terminal velocity etc..)
Assuming the water provides a 50% resistance, you should be ok at 40 sec.
Good question.....hope this helped
Posted by: angerosa Aug 27 2008, 12:50 PM
Use a 640' fuse and don't worry about it. They have fuses that burn under water.
If you don't like that solution which is sure fire and guarantees you've reached the right depth, buy some wider PVC pipe and fill with water and do test runs to see how fast it traverses a fixed distance - then use that figure to calculate how long it will take to go 640'.
Some things to be careful of: If the actual 640' hole filled with water isn't exactly straight, even if it's off at an angle of .001 degrees it will kind of slide down the inside of the tube instead of flow through water and the sliding friction will slow down it's descent. Also if the water gets "murkier" as you go deeper. The increased density of the water will slow down the descent.
I would totally use a long string, maybe some fishing line to verify how deep you are. You could also tug the torpedo back and fourth to get it to go over a rough spot.
Posted by: drgchapman Aug 27 2008, 12:54 PM
I would be thinking a dry run or three to determine the unknown variables. Fishing line to retrieve and verify time to bottom. I don't think you can calculate as the variables are too many.
Posted by: Todd Enlund Aug 27 2008, 01:26 PM
The fastest (best case scenario) would be dropping through air (32ft/sec2).
It would take 20 sec to fall 640ft (ignoring terminal velocity etc..)
640=0(t)+.5(32t²)
640=16t²
40=t²
√40=6.32 seconds. You /2 instead of √.
Assuming the water provides a 50% resistance, you should be ok at 40 sec.
Good question.....hope this helped
On what did you base this assumption?
I agree with Gary... if you HAVE to do it this way, I'd be testing my ass off.
I like the remote trigger or the 640' fuse better.
Posted by: pin31 Aug 27 2008, 01:44 PM
Thanks Todd, you're correct...I used the wrong formula.
t=(2d/g)^1/2
t=(2*640/32)^1/2 = 6.323 sec.
The 50% was a swag (i.e. wild ass guess)!!
That's why I've been moved up to engineering management....too many dead brain cells !!!
Posted by: ClayPerrine Aug 27 2008, 02:40 PM
Thanks Todd, you're correct...I used the wrong formula.
t=(2d/g)^1/2
t=(2*640/32)^1/2 = 6.323 sec.
The 50% was a swag (i.e. wild ass guess)!!
That's why I've been moved up to engineering management....too many dead brain cells !!!
Those who can't do, Teach.
Those that can't Teach, Manage.
Maybe you need to contact these guys....
They can certainly help you blow stuff up.
Posted by: carr914 Aug 27 2008, 02:56 PM
Isn't the PVC going to have some bouyancy?
Posted by: jfort Aug 27 2008, 02:57 PM
this is what i've come up with thus far, though my units seem to be 1 cm short:
terminal velocity is proportional to the ratio of the density
of the object to the density of the medium, the square of the size of
the object, and inversely to the viscosity of the object:
v_terminal ~ (p_object/p_fluid) R^2
----------------------
n_fluid
density of broken granite is 1.65 g/cm3
density of water is roughly 1 g/cm3
1.5” PVC
cross area = πr squared
= 3.1416 x 1.905 cm2
=3.14 x 3.629025 cm2
= 11.40094 cm2
water has a nominal viscosity of 1.0 x 10-3 Pa s
the pascal-second (Pa·s), which is identical to kg·m-1·s-1
therefore, terminal velocity =
1.65 g/cm3 x 11.4 cm2 over
1 kg/m s (or 1000 g/100 cm s) (or 10 g/cm s)
= 18.81 g/cm times 10 cm s/g
= 188.1 cm/s
=6.17 feet/sec
for 640 feet = 103 seconds
Posted by: jfort Aug 27 2008, 03:07 PM
right. PVC will have some bouyancy, but i discounted it. we'll pack the PVC with gravel (I used the density for crushed granite). i wish one of you young students would check my math and units. a decimal point could have big consequences if i am off. but not bad for an old attorney, i think
Posted by: Lavanaut Aug 27 2008, 03:22 PM
No offense, but there are way too many unknowns here to even bother trying to calculate it to the second. Like...
Not off to a good start, and that's the least difficult variable to determine.
Go with something that doesn't require all the variables. Connect the dynamite to an anchor. Drop it in. Wait 5 (or 25!) minutes. Detonate by remote. To me this doesn't sound like the line of work where you want to be guessing or worrying about decimal places unless absolutely necessary.
Posted by: jfort Aug 27 2008, 03:25 PM
i forgot to square the PVC cross section area: so, is this correct? i am losing it:
thus terminal velocity is proportional to the ratio of the density
of the object to the density of the medium, the square of the size of
the object, and inversely to the viscosity of the object:
v_terminal ~ (p_object/p_fluid) R^2
----------------------
n_fluid
density of broken granite is 1.65 g/cm3
density of water is roughly 1 g/cm3
1.5” PVC
cross area = πr squared
= 3.1416 x 1.905 cm2
=3.14 x 3.629025 cm2
= 11.40094 cm2
squared = 129 cm2 cm2
water has a nominal viscosity of 1.0 x 10-3 Pa_s
the pascal-second (Pa·s), which is identical to kg·m_1·s_1
therefore, terminal velocity =
1.65 g/cm3 x 129 cm2 cm2 over
1 kg/m s (or 1000 g/100 cm s) (or 10 g/cm s)
= 214.4 g cm over 10 g/cm s
= 214.4 g cm times 0.1 cm s/g
= 21.44 cm/s
=0.7 feet/sec
for 640 feet = 914 seconds or 15.2 minutes
Posted by: carr914 Aug 27 2008, 03:29 PM
No offense, but there are way too many unknowns here to even bother trying to calculate it to the second. Like...
To me this doesn't sound like the line of work where you want to be guessing or worrying about decimal places unless absolutely necessary.
To Me this doesn't sound like the line of work where you want imput from us idiots
T.C.
Posted by: angerosa Aug 27 2008, 03:47 PM
i forgot to square the PVC cross section area: so, is this correct? i am losing it:
thus terminal velocity is proportional to the ratio of the density
of the object to the density of the medium, the square of the size of
the object, and inversely to the viscosity of the object:
v_terminal ~ (p_object/p_fluid) R^2
----------------------
n_fluid
density of broken granite is 1.65 g/cm3
density of water is roughly 1 g/cm3
1.5” PVC
cross area = πr squared
= 3.1416 x 1.905 cm2
=3.14 x 3.629025 cm2
= 11.40094 cm2
squared = 129 cm2 cm2
water has a nominal viscosity of 1.0 x 10-3 Pa_s
the pascal-second (Pa·s), which is identical to kg·m_1·s_1
therefore, terminal velocity =
1.65 g/cm3 x 129 cm2 cm2 over
1 kg/m s (or 1000 g/100 cm s) (or 10 g/cm s)
= 214.4 g cm over 10 g/cm s
= 214.4 g cm times 0.1 cm s/g
= 21.44 cm/s
=0.7 feet/sec
for 640 feet = 914 seconds or 15.2 minutes
I guess it's embedded in your terminal velocity formula but I don't see where gravity is factored in only the viscosity and friction. If your torpedo scrapes against the side of the tube 50 times or 500 times will severly impact the speed at which it reaches the bottom.
If the cross section of your torpedo is too close to the cross section of the tube it's going through, you'll have a suction effect slowing down the torpedo as well because there won't be enough area for water to rush by the sides of the tube.
Posted by: r_towle Aug 27 2008, 03:58 PM
Have you considered using a large caliber gun instead?
If he is licensed and this is all on the up and up, you can get access to the correct caliber gun (would probably come with operator) and fire an exploding projectile into the hole....
It may be worth investigating this option because the service would work for lots of depths and differing variables....might be a good side business for your son.
We had a fairly large gun to shoot the headwall of a ski area every morning to start the avalances...so I know its possible to acquire these types of machines provided you are licensed.
Rich
Posted by: Jeffs9146 Aug 27 2008, 04:10 PM
Why use free fall numbers?
Well maintenance companies use pumps to force chlorine into the well. Why don’t you make the PVC close to the size of the torpedo and pump water into the well forcing the torpedo down to the bottom?
You could moniter the flow rate of the water to get your depth.
Just a thought! 
Posted by: Jeffs9146 Aug 27 2008, 04:20 PM
You could also use a depth guage to set your trigger too explode.
Depths (Atmospheres) are fairly consistant and accurate!
Posted by: drgchapman Aug 27 2008, 04:34 PM
OK, after all this mental masterbation, it seems unanimous that it is impossible to determine the time to the bottom. I especially like the reference to the water flow past the torpedo as it goes down creating a drag effect.
TEST
TEST
TEST AGAIN
Place dynamite, RUN LIKE HELL!
Posted by: orange914 Aug 27 2008, 05:10 PM
this looks like a job for mythbusters 
Posted by: davep Aug 27 2008, 08:58 PM
I believe that terminal velocity will be reached quickly enough that once that is calculated it should be used for the entire fall. The pipe diameter is sufficiently large that the "torpedo" should not experience undue drag (some yes). I'd add vanes to try and give the "torpedo" some spin.
650 feet of magnesium wire might make a good fuse.
You could do a test prior to the actual blast. Attach 100 feet of wire, and let it drop (with a dummy warhead of the same mass/density). Time the 100 foot fall for each ten feet. If the rate of decent becomes constant, then use that rate. Retrieve the dummy.
Posted by: Todd Enlund Aug 27 2008, 09:26 PM
The pipe diameter is sufficiently large that the "torpedo" should not experience undue drag (some yes).
I agree, but the pipe diameter is sufficiently small that there will be friction every time that the "torpedo" hits the tube. In 640 feet of production tubing, there will be 19 joints that the "torpedo" could hang up on. Not likely, but possible.
I'd add vanes to try and give the "torpedo" some spin.
I'm not sure what you'd gain. I don't think it would add sufficient stability to prevent the "torpedo" from hitting the sides, but it would provide additional points to foul the "torpedo".
The consequence of premature detonation is probably ruptured production tubing, and would pose a problem...
I still agree with Gary. I'd be testing my ass off.
Posted by: davep Aug 28 2008, 02:41 PM
My thought was that spin would help it get past some obstacles.
The vanes should not be designed to extend out past the body. I would have a bulbous nose for low drag, and a sharpened tail with a couple of tiny vanes to add clockwise spin (counter-clockwise below the equator).
Posted by: hcdmueller Aug 28 2008, 02:58 PM
Why not use commercially available shock tube as your line? The shock tube has a pretty high tensile strength. If you are nervous about that you could dual prime and just zip tie or tape the two strands shock tube together.(I recommend this for the redundancy) With a little bit of testing you can determine if the shock tube will support the weight or if you need a separate line to reinforce the setup. The reels of shock tube can be bought already setup with a cap.
There is no depth limitation on the cap it if is sealed in the torpedo device you are making. Just make sure that the the torpedo is sealed properly.
I am glad you decided to avoid the det cord as it can act funny if to much tensile stress is placed on it over a long distance possibly causing misfires.
I would actually recommend against using the time fuse as it has a tendency to burn differently in a confined space. The pressures that it will be exposed to at 640' can also have an effect on burn time.
Posted by: Travisfling Aug 28 2008, 03:07 PM
This just seems like a crazy question... I think the only thing that you'll KNOW to be true is going to be pressure. it doesn't matter how large the diameter of the tube is, just the depth. I would find the pressure for -640 of water, and then get a detonator based on that. Then you wouldn't have to worry about how long it took, or whatnot, it would just blow when it got to that pressure.
Posted by: Jeffs9146 Aug 28 2008, 03:37 PM
Thats what I was talking about earlier!!
Posted by: ArtechnikA Aug 28 2008, 04:01 PM
I think I'd put a camera and light source in the nose. CCD cameras are cheap and short-haul modem technology can handle a 1000' run. (I don't think I'd trust 232. 485 maybe...
That way you can *see* if it's safe to detonate.
(FWIW - i did some oilfield support engineering once upon a time in my storied career - drillrig simulation control systems and cementing (plug flow, laminar flow, density issues...) studies.)
If you don't want the 'unmanned vehicle' approach, i think I'd go with 650' of pipe and (effectively) positively drive it into place. Once it's bottomed, pull up the pipe, and hit it.
Posted by: YksKrad Aug 28 2008, 04:16 PM
I'm only an engineering student, and an Arch. E. major at that rate, but the pressure at depth would be dependent on material you're in which won't be pure water and likely won't be consistent over that range, so I would not trust any form of pressure trigger...
However time delay and fuses are a bad idea as well since you have no guarantee that the device will not get jammed upon the way down, or that it will travel in any predictable way (try dropping a nail onto a quarter at the bottom of a bucket of water... You'll find it's very difficult, likewise you're likely going to hit the walls of the well in no predictable manor...)
So based on those two assumptions/observations...
Manual detonation...physical depth meter....
On a side note... Don't they usually pump steam in several parallel side wells drilled in the vicinity nowadays, or is that only in certain situations(I could have sworn I had heard something about that)?
Posted by: ArtechnikA Aug 28 2008, 04:31 PM
On a side note... Don't they usually pump steam in several parallel side wells drilled in the vicinity nowadays, or is that only in certain situations(I could have sworn I had heard something about that)?
'Usually' is awfully general...
Steam injection is a common method of secondary recovery, but it's useful in only some applications. Steam is expensive to make - if it costs more to make the steam than the oil is worth, it's a bad tradeoff...
waterflood is a good method - you make some number of holes around the 'producing' well (a 5-spot waterflood is very common - 4 corners of a square for water, with the producer in the middle) and pump water down into the field. The water sinks below the oil and forces it up and into the producing well. Nice if you have a good supply of water, and it's pretty common to hit water before you hit oil, which is where the cementing comes into play...
but fracturing also works sometimes, where the oil is bound up in or behind strata. If you break up the big rock, you make channels for the oil to flow through. That sounds to me as if it's at least part of what's in play here.
Posted by: SirAndy Aug 28 2008, 04:51 PM
to add clockwise spin (counter-clockwise below the equator)
"Despite the large amount of misinformation, toilets — and even tornadoes — are too small to be affected by the Coriolis, whose force would only begin to directly influence a storm's swirling mass if it were approximately three times larger than the supercell storm systems that typically generate tornadoes."
"The Coriolis force only has a significant influence on the spin direction of Earth's largest atmospheric and oceanographic circulation systems, such as the Gulf Stream, jet stream and trade winds."
AndyPosted by: YksKrad Aug 28 2008, 05:12 PM
On a side note... Don't they usually pump steam in several parallel side wells drilled in the vicinity nowadays, or is that only in certain situations(I could have sworn I had heard something about that)?
'Usually' is awfully general...
Steam injection is a common method of secondary recovery, but it's useful in only some applications. Steam is expensive to make - if it costs more to make the steam than the oil is worth, it's a bad tradeoff...
waterflood is a good method - you make some number of holes around the 'producing' well (a 5-spot waterflood is very common - 4 corners of a square for water, with the producer in the middle) and pump water down into the field. The water sinks below the oil and forces it up and into the producing well. Nice if you have a good supply of water, and it's pretty common to hit water before you hit oil, which is where the cementing comes into play...
but fracturing also works sometimes, where the oil is bound up in or behind strata. If you break up the big rock, you make channels for the oil to flow through. That sounds to me as if it's at least part of what's in play here.
Thanks for the information!!!
Posted by: toon1 Aug 28 2008, 05:13 PM
Why not use commercially available shock tube as your line? The shock tube has a pretty high tensile strength. If you are nervous about that you could dual prime and just zip tie or tape the two strands shock tube together.(I recommend this for the redundancy) With a little bit of testing you can determine if the shock tube will support the weight or if you need a separate line to reinforce the setup. The reels of shock tube can be bought already setup with a cap.
There is no depth limitation on the cap it if is sealed in the torpedo device you are making. Just make sure that the the torpedo is sealed properly.
I am glad you decided to avoid the det cord as it can act funny if to much tensile stress is placed on it over a long distance possibly causing misfires.
I would actually recommend against using the time fuse as it has a tendency to burn differently in a confined space. The pressures that it will be exposed to at 640' can also have an effect on burn time.
Shock tube would be a great way to go and it's easy to use. You could also use a sperate line to support the the charge and run them down together. This would eliminate having to use sand or gravel in the PVC and not dampening the blast affect.
If he is trying to penetrate rock or other material Shape charges might be another option.
Explosives is how I make a living
Posted by: Lavanaut Aug 29 2008, 12:34 AM
to add clockwise spin (counter-clockwise below the equator)
"Despite the large amount of misinformation, toilets — and even tornadoes — are too small to be affected by the Coriolis, whose force would only begin to directly influence a storm's swirling mass if it were approximately three times larger than the supercell storm systems that typically generate tornadoes."
"The Coriolis force only has a significant influence on the spin direction of Earth's largest atmospheric and oceanographic circulation systems, such as the Gulf Stream, jet stream and trade winds."
Andy
Posted by: hcdmueller Aug 29 2008, 07:09 AM
Explosives is how I make a living
Me too!! (need a mad bomber smilie)
Probably a little different applications. toon1- what company do you work for and what application? I have a few years left but I am already trying to scout jobs on the "outside". I love the job. I am just getting tired of the Air Force corporate mentality.
BTW positive control of the shot is the only way to go.
Posted by: Phoenix-MN Aug 29 2008, 08:11 AM
In my previous life (job) I worked for a company that manufactured engineering borehole geophycical equipment and engineering seismic equipment. Most of the applications we worked with were 500' or less but we used standard blasting caps and HPD primers in water filled holes.
Getting a tool stuck in a hole always seemed to happen so I would not use a timed fuse. You could test very easily by making a dummy torpedo with only a cap in it and lower that to the depth needed and see if it detonates.
Good luck
(I miss blowing things up)
Posted by: toon1 Aug 29 2008, 09:29 AM
Explosives is how I make a living
Me too!! (need a mad bomber smilie)
Probably a little different applications. toon1- what company do you work for and what application? I have a few years left but I am already trying to scout jobs on the "outside". I love the job. I am just getting tired of the Air Force corporate mentality.
BTW positive control of the shot is the only way to go.
I work for LLNL S300 on the Hydro program. currenntly most of my work is doing coventional stuff as the hydro program is slowing down.
I had 10yrs experiance with shape charges working with the hellfire and TOW missles primaraly.
Things are'nt what they used to be and it's getting tough to stay with it due to the current gov't. budgets. It seems like they are slowly pulling the rug from under our feet.
The only thing keeping me there is the fact that I love what I do. Not many people can say that.
We have just gone through a transition and also a round of layoff's.we lost alot of good people and are still losing more. there is no hiring going on right now but if things tunaround there may be some openings
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