OT. WAY OT, Engineering Problem. Need some help |
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OT. WAY OT, Engineering Problem. Need some help |
jfort |
Aug 27 2008, 10:55 AM
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#1
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Senior Member Group: Members Posts: 1,137 Joined: 5-May 03 From: Findlay, OH Member No.: 652 Region Association: Upper MidWest |
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. |
jfort |
Aug 27 2008, 02:57 PM
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#2
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Senior Member Group: Members Posts: 1,137 Joined: 5-May 03 From: Findlay, OH Member No.: 652 Region Association: Upper MidWest |
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 |
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