OT. WAY OT, Engineering Problem. Need some help Options

[jfort]

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.

[ericread]

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?"

[Lavanaut]

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. (IMG:style_emoticons/default/idea.gif) Radio Shack is your friend!

(IMG:style_emoticons/default/popcorn[1].gif)

This post has been edited by Lavanaut: Aug 27 2008, 12:26 PM

[jfort]

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.

[pin31]

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

[angerosa]

Use a 640' fuse and don't worry about it. They have fuses that burn under water.

(IMG:style_emoticons/default/beer.gif)

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.

[drgchapman]

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.

[Todd Enlund]

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.

[pin31]

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 !!!

[ClayPerrine]

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.

(IMG:style_emoticons/default/poke.gif)



Maybe you need to contact these guys....

(IMG:http://dsc.discovery.com/fansites/mythbusters/about/gallery/mb_about.jpg)

They can certainly help you blow stuff up.

[carr914]

Isn't the PVC going to have some bouyancy?

[jfort]

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

[jfort]

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

[Lavanaut]

No offense, but there are way too many unknowns here to even bother trying to calculate it to the second. Like...

It is estimated that it is 640' to the bottom.

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. (IMG:style_emoticons/default/huh.gif)

[jfort]

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

[carr914]

No offense, but there are way too many unknowns here to even bother trying to calculate it to the second. Like...

It is estimated that it is 640' to the bottom.

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. (IMG:style_emoticons/default/huh.gif)

To Me this doesn't sound like the line of work where you want imput from us idiots (IMG:style_emoticons/default/biggrin.gif)

T.C.

[angerosa]

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.

[r_towle]

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

[Jeffs9146]

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! (IMG:style_emoticons/default/blink.gif)

[Jeffs9146]

You could also use a depth guage to set your trigger too explode.

Depths (Atmospheres) are fairly consistant and accurate!