Arithmetic help
Arithmetic help
(OP)
It has been too long.
A train leaves .... no that is not it.
I have 500' of 1/4" OD, 1/8" ID , 0.921 specific gravity, polyethylene tube going down a well. It will be used as a depth sensor by pressurizing with air to indicate water depth (around 220 psi max). The tubing weighs about 7.4 lb (probably includes the spool).
Roughly how many lbs of force will this exert on the weight pulling it down into the well. Positive (float) or negative (sink). I need to determine the proper weight of the device on the end of the tube and it is not handy to just add weight. Thre are some other issues with that.
A train leaves .... no that is not it.
I have 500' of 1/4" OD, 1/8" ID , 0.921 specific gravity, polyethylene tube going down a well. It will be used as a depth sensor by pressurizing with air to indicate water depth (around 220 psi max). The tubing weighs about 7.4 lb (probably includes the spool).
Roughly how many lbs of force will this exert on the weight pulling it down into the well. Positive (float) or negative (sink). I need to determine the proper weight of the device on the end of the tube and it is not handy to just add weight. Thre are some other issues with that.





RE: Arithmetic help
1. Determine the volume occupied by the tubing and the weight of in-situ fluid that is displaced.
2. Determine the weight of the tubing full of your expected fluid.
Most of the time I put a check valve on the end of a cap string to give me a positive (on/off) indication of bottom-hole pressure. If you do that then the calculation is pretty easy (i.e., the whole tubing string is full of air). With a check valve, you can pick a maximum-possible liquid level and compare the weight of the displaced fluid to your tubing weight and add enough weight to make your string 110% of the displaced fluid.
Without a check valve the answer is a lot tougher and I don't even know how I'd start to solve it.
David Simpson, PE
MuleShoe Engineering
www.muleshoe-eng.com
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
The harder I work, the luckier I seem
RE: Arithmetic help
Without splitting hairs:
The displacement is 0.049in2 x 500ft x 12in = 294.5 in3.
The bouyant force is 294.5in3 x 0.0361 lbs/in3 = 10.63 lbs.
The weight as you say is 7.4 lbs. Net up = 10.63-7.4 = 3.2
Assume air dens. is 0. So I would start with a 10 lb weight, at least, to be conservative.
I hope the weight sits on the bottom. Otherwise the tube may stretch over time and you will need to recal. We used to use these all the time, but not for 500 ft. depth and not flex tubing with weight on end. We always tacked metal tube to a clip at the bottom.
RE: Arithmetic help
RE: Arithmetic help
The weight is not on the bottom; that is another 105' down. But there will be a rope supporting the device on the end. I have put the low water sensor in some pieces of stainless steel pipe and a coupler. I drilled a hole through the pipe to make the sensor hang vertical so the little internal float won't bind. This is my weight.