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Re: small tube filling pressures
In a message dated 96-04-10 14:02:09 EDT, you write:
> >>Morgan wrote:
> >>Life (hours)= (Axp^4xV)/ (kxi^2)
>>>where A= inside surface area of each electrode
> >> p= filling pressure of neon
> >> V= tube volume
> >> i= tube current
> >> k= emperical factor based on the coating material
>
> Interesting formula, Jeff. If I may apply it to a unit of glass,
>say 8'- 6" (2 meters) long of 15mm clear glass with 2 Eurobrite
>electrodes and 10mm fill, on a 30ma trans.
>
> The electrode shell is approx. 30mm long by 10mm in diameter
> Cir.= Pi x D
> Cir.= 3.14159 x 10mm = 31.42mm
> Area of 1 elec.= 31.42mm x 30mm =942.6 sq. mm
>
> The tube volume is (Pi x r^2) x 2m or 2000mm
> Inside dia. of 15mm tube is 12mm/2 =6mm
> 3.14159 x 6^2 x 2000 = 226,194 cu.mm
>
> So:
> Where A=942.6sq. mm and; (is it one elec. or both?)
> p=10mm and;
> V=226,194 cu. mm and;
> i=10ma and;
> k=???? say 100?
>
> (942.6 x 10^4 x 226,194) / (100 x 30^2) or
> (942.6 x 10,000 x 226,194) / (100 x 900) = 23,690,050 hrs.
> Hmmm! Either I should get back to the drawing board or seriously
>consider extending my warranty period.
>
2,704 years? Hope you have a lot of great great great....great grandkids
interested in keeping the family business alive. Your mistake is in
arbitrarily choosing k value, see below.
> Any idea what units of measure are to be used for the different
>variables ? (millimeters, centimeters etc.) and what would be the
>range of the k factor.
>
That's where you seem to get lost. If I did it right, [k] must have units of
cm^5 torr^4 / { ma ^2 hrs}. There's no suggestion of what k is numerically,
so your calculation is highly speculative from there on. k is an empirical
constant which is not defined here.
> Also would appear that identical tubes on 30ma and 60ma
>transformers, the 30ma tube would last 4 times longer than the one
>running at 60ma.
Yes, this is the only meaningful use you can get from this formula (assuming
it's correct?), which is to make comparative ratios between different
situations.
For example, if correct, since V=La, where L is the length of the tube and a
is the internal cross-sectional area, you can get the rule:
p2/p1 = {V1/V2}^-4 = (L1/L2)^-4
which says that for constant lifetime you should raise the pressure of a
shorter tube to about the fourth root (square root of square root) of the
ratio of the length of a longer tube of the same diameter (more standard
tabular value) to the length of the shorter tube. This is a very, very slow
function of length.
The same rule applies to diameters. That is why I question this rule. A
fourth root is too slow to model the tables we all use, assuming we have the
goal of comparable lifetimes.
Somewhat apocryphal, perhaps?
>
> Would be interested in a copy of those tables by Frank Montroy.
Yes, me too. Anyone have a copy?
Jeff Golin
>
>