Potpourri starting May 17, 2010

[mickyd]

First thing to resolve tire filling exercise / nitrogen discussion is to use the actual elements that make up 'air'.

'AIR' is:
78% nitrogen
21% oxygen
1% argon
1% other 'schtuff'

A tire filled with 'air' contains 78% nitrogen already.

Clarification. You can not say that a nitrogen filled tire is 50% air. It would only be about 50% air if air was compressed and nitrogen was compressed and put in the tire together. To fill a tire they let all of the "compressed air" out and all that is left is residual air that is now under no more pressure than the air outside of the tire. As the "compressed" nitrogen is pushed into the tire that basically zero pressure air inside of the tire is also compressed as much as the final pressure of the tire and then amounts to a very tiny part of the compressed contents of the tire.........

Well, I am no engineer, and I may be wrong, but unless the tire is collapsed and has no air in it, I think that the volume of air that is not under pressure is more then the volume of nitrogen that is added under pressure. To me that means that more then 50% of the air in the tire is "regular" air. I am very willing to entertain other views if a qualified person can explain...(sorry farmer, but can you prove the part of your statement highlighted in red above?)

A 'normal' car tire holds about 4 cubic feet of gas at 32psig (gauge pressure) / 70 degrees F. When the tire is deflated (without sucking out the left over gas inside), it will probably contain about .25 cubic feet of gas left over. Since 78% (.78 proportion) of that gas left in the tire is nitrogen, we'll say that the tire contains .25 cu feet of x .78 proportion of it being nitrogen = .19 cubic feet of nitrogen and only about .06 cubic feet of oxygen. Now, if you fill the tire back up to pressure with 100% nitrogen, you can add 3.94 cu. ft. of nitrogen and the remainder will be the .06 cu. ft. left over oxygen OR in other words, 3.94 cu. feet of the 4.00 total (98.5%) will be nitrogen and .06 cu. ft. of the 4.00 total (1.5%) will be oxygen.

[JPG]

Since this IS potpouri, it is about time to alter the focus!

When I read the following, I had to think of Farmer's resin chair and electric fence thread(this thread is easier to find, and since the following has nothing to do with electric fences, I am placing it here)!

A Saskatchewan farm wife called the local phone company to report her telephone failed to ring when her friends called, and that on the few occasions, when it did ring, her dog always moaned right before the phone
rang.

The telephone repairman proceeded to the scene, curious to see this psychic dog or senile lady. He climbed the telephone pole nearest the house, hooked
in his test set and dialed the subscriber's house. The phone didn't ring right away, but then the dog moaned and the telephone began to ring.

Climbing down from the pole, the telephone repairman found:

1. The dog was tied to the telephone system's ground wire with a steel chain and collar;

2. The wire connection to the ground rod was loose;

3. The dog was receiving 90 volts of signaling current when the number was called;

4. After a couple of jolts, the dog would start moaning and then urinate;

5. The wet ground would complete the circuit, thus causing the phone to ring.
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All of which demonstrates that some problems CAN be fixed by pissing and moaning.

Just thought you'd like to know...

[JPG]

. . .





A 'normal' car tire holds about 4 cubic feet of gas at 32psig (gauge pressure) / 70 degrees F. When the tire is deflated (without sucking out the left over gas inside), it will probably contain about .25 cubic feet of gas left over.

. . .

Please explain the 16:1 inflated/deflated ft³ ratio!

[mickyd]

Please explain the 16:1 inflated/deflated ft³ ratio!

What aspect of it?? 4 cu. ft at 32 psig vs. .25 cu ft at 32 psig = 16:1

[reible]

OK rather then a tire I thought of this box that when measured is 4 cu ft.

The box is open so it contains 4 cu ft of air. Then given all other things stay the same (temperature etc.) I have a bases for do some simple math.

I close the box and seal it up. At this point the nitrogen content is 78% or by volume 3.12 cu ft. and from now on I will be adding 100% nitrogen only.

Now I want to end up with 32 psi and we will call this Pc, this will be standard air pressure of 14.7 (gauge reads 0) plus the 32 psi or 46.7.

Now to get from gauge reading of 0 to gauge reading of 32 psi we know we will need to be adding to the volume. Well this can be figured using Boyle's Law.

general form Pa * Va = Pc * Vc
Where Pa is at atmospheric pressure (14.7psi)
Va is volume of gas at atmospheric pressure
Pc is compressed pressure (plus 14.7psi)
and
Vc is volume of gas at compressed pressure

So

Va = ( Pc * Vc ) / Pa

Pc = 46.7
Vc = 4
Pa = 14.7

Va = (46.7 * 4 ) / 14.7
Va = 12.7075

The container has 4 cu ft in it so the most we can add is 12.7075 - 4 or 8.7075 cu ft of nitrogen.

Total nitrogen is then 3.12 + 8.7075 = 11.8275
This is then 93.07%

Now tires have a bunch of other factors happening including things like volume change, temperature changes, water vapor from a typical air compressor etc.

So that's my take on it.

Ed

[JPG]

Thank you Ed! I agree with both you and Boyle. 3:1(the previous 2:1 reference was a 'teaser').

Mickyd please explain 16:1(not merely point to it)!:D

[mickyd]

Thank you Ed! I agree with both you and Boyle. 3:1(the previous 2:1 reference was a 'teaser').] please explain 16:1[/color](not merely point to it)!:D

Do you mean how did I arrive at that ratio? Your question...it tain't clear. You need to add more verbiage to it.

Nice job Ed. I was working off my mental and physical model vs. Mr. Boyle's work.

[mickyd]

Got an email from PG-ZAC showing his new 20. He's got his own private suite vs. others bunking up 2 to 4 per 'room'. Here's some pics.
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[ATTACH]8986[/ATTACH]

[JPG]

Do you mean how did I arrive at that ratio? Your question...it tain't clear. You need to add more verbiage to it.

Nice job Ed. I was working off my mental and physical model vs. Mr. Boyle's work.

OK! verbiage How did your 'mental and physical model' end up with 4 ft³ and .25 ft³ as the volumes for the inflated and uninflated tire tire respectively?verbiage
verbiage
Methinks you are 'evading' the question!:rolleyes: verbiage verbiage verbiage verbiage verbiage verbiageverbiage verbiage verbiage verbiage verbiage...

verbage?

[mickyd]

[quote="JPG40504"]OK! verbiage How did your 'mental and physical model' end up with 4 ft&#179]
Now your question is clear. 4 cu ft came from an internet search. .25 cu ft. came from an estimate I made by taking my boats deflated inner tube, pressing all the air to one area, twisting the tube to block off the air and then continuing to twist, tightening it to a point were it was pressurized. Then, I squished it some more in a box with a board to further pressurize it to a point where I estimated it was at about 30psi by comparing the feel of then tube to the feel of my mountain bike tire which is at 30 psi. I then looked at this volume and determined that it was about 1/4 ft. cu. Crude method but any estimate is better than no estimate. My 98% vs. Ed Boyle's 93% wasn't too shabby 'ay?

Evading the question??? I wanted you to clarify your question since in the past, answering those vague questions leads to typing a response that was no where near what your were looking for in your head. Figured I'd save myself typing for nothing.