Gas leak ?

Mar 07, 2008 24 Replies

Return to original calculation

for 0.25 mBar reduction in 20mB over 2 minutes

11m of 22mm pipe and 9m of 15mm pipe, no known dead runs.

volume of that pipe between meter and appliances 5,800,000 cu mm = 354 cu ins = 0.2 cu ft Volume of whole gas meter = 5x7x8 inches =.16 cu ft

Therefore maximum vol = .36 cu ft

By Boyle's law drop in pressure of 0.25/20 = 1/80 means an increase in volume (the leak) of 1/80 so 0.36/80 = 0.0045 cu ft in 2 minutes = 0.135 cu ft per hour well above 0.01 cu ft / hour let alone 0.003 cu ft per hr read on the watch dial pointer of the meter

Pilot light consumption about 0.5 cu ft / hour so 3 times the minimum resolvable amount in the soundness test. So manometer test determines a leak of gas of degree 1/3 of a pilot light or more. I'll settle for that unless anyone knows the consumer side volume of a gas meter .

So far so good.

No! No! No! it is 0.25mb change in an absolute pressure of atmospheric+20 say 1000+20 (varying on the weather by several percent +/-).

Boyles law applies to absolute pressures not gauge pressure.

Example a bicycle pump (in perfect working order) starts with the handle out and the pressure inside the pump at 0 bar gauge (as shown by the pressure gauge it is connected to), the absolute pressure is 1 bar. You move the handle in half way the pressure rises to 2 bar absolute (Boyles law) and 1 bar on the gauge.

The gas companies make a slight adjustment to the gas bill dependent on the average weather to allow for the fact that they supply slightly more gas in times of high atmospheric pressure or vice versa.

It is because the pressure decay method is showing the difference between the absolute pressures that small changes in the guage pressure correspond to very very small leaks.

The method is so sensitive that simply placing your hand on the gas pipe for 15 seconds can make the gauge move up a small amount (say 0.25 mbar). Try it! (or removing the meter box cover in direct sunlight).

By a similar calculation....

Temp rise of 200mm of pipe from 17 to 37C (290K to 310K). Fraction of total volume of warmed pipe 1/200 (say 20m of pipe + same volume in gas meter). Fractional pressure rise (Ideal gas law), for warmed section is about 20K/300K i.e. 1/15. The product or the fractions is the rise in absolute pressure of 1 part in 3000 (15*200). I.e. about 1/3 of a mbar.

I have given this my best shot of trying to convince you the pressure decay method is sensitive and effective. Watching the dial is NOT the way to test for leaks.

water, I would have thought .5mm or .05 mB was the resolvable movement?

Swapping manometer for Borden gauge as the water would boil.

What is your calculation for the following situations?

Volume of pipe V, at absolute pressure 1020mB leaking to air at 1000mB pressure and same pipe but at absolute pressure 20mB leaking to a vacuum ?

Same leak rate or different?

It's 2.5mm of /difference/ so the water level only moves half that on one side of the gauge.

Given the meniscus and parallax errors of observation a 1mm movement is a practical minimum detectable movement, although you could probably do a bit better.

The 0.25mbar allowance was introduced because the electronic manometers are so sensitive that the last digit (0.01 mbar) never settles.

The same leakage rate. But the there's 50x less gas in the pipe and the percentage drop in absolute pressure is 50x greater.

Taking 120mbar against 100mbar the same leakage rate but 10x less gas in pipe.

temp and pressure and there is a factor of 50 to take into account

Return to original calculation, hopefully last time

for 0.25 mBar reduction in 20mB over 2 minutes

11m of 22mm pipe and 9m of 15mm pipe, no known dead runs.

volume of that pipe between meter and appliances 5,800,000 cu mm = 354 cu ins = 0.2 cu ft Volume of whole gas meter = 5x7x8 inches =.16 cu ft

Therefore maximum vol = .36 cu ft

By Boyle's law drop in pressure of 0.25/1020 = 0.000245 means an increase in volume (the leak) of that factor so 0.36/(1/0.000245) = 0.000088 cu ft in 2 minutes = 0.0026 cu ft per hour

Pilot light consumption about 0.5 cu ft / hour so about 200 times the minimum resolvable amount in the soundness test.

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