Gas safe peeps

Ho ho ho!

tommyplumb! headman! and how many other aliases?

I think a lot of d1ck pulling going on here

I think you're right, it's not dick though. Just the same idiot whose been posting gas related queries under different usernames for the last few weeks.
You'd think admin would do something about him wouldn't you?

I have a feeling he is admin.

45yearsagasman said:

I hardly know you!

Click to expand...

Gasop said:

Ho ho ho!

tommyplumb! headman! and how many other aliases?

I think a lot of d1ck pulling going on here

Click to expand...

Not me mate, but it's still an interesting read. If you look back at all my aliases on here they all have the same theme....Sir Richard Puller/DeltaT/Richie Puller............etc etc etc

Granted I was unceremoniously booted out the combustion chamber pot, but that was my own fault............And I may return, one day.

No, this thread has all the markings of Joe 'Wellhard' 90.

Thanks for all the help,much appeciated.

How does the FGA calculate CO2% in POC?, is there a calculation for this?
How would one calculate the O2 in the POC?
How would one calculate the air% in POC?
I understand some saying the analyser calculates all these measurements so why bother with long hand, would it not be better to know how these measurments could be calculated without a FGA?
Is this not what you gas peeps should know?
Just like to further my knowledge.

Thanks again.

Please don't confuse me with other posters on here, thanks.

tommyplumb said:

.

How does the FGA calculate CO2% in POC?, is there a calculation for this?

Click to expand...

11.9 x (20.9 - O2)
---------------------------= CO2%
20.9

How would one calculate the O2 in the POC?

Click to expand...

20.9 x CO2/11.9 = ?....20.9-? = O2% (fuel lean)

How would one calculate the air% in POC?

Click to expand...

O2% x 100
-------------=XS air%
20.9 - O2

I understand some saying the analyser calculates all these measurements so why bother with long hand, would it not be better to know how these measurments could be calculated without a FGA?

Click to expand...

Yep.

Is this not what you gas peeps should know?

Click to expand...

Yep

Just like to further my knowledge.

Click to expand...

Thats good DeltaT.

Thanks Monoxide but this has just confused the issue even more so.
Maybe you have copied these calcs straight out of a book?
What does the 11.9 represent?,the same applies to the 20.9?
Please explain? as the calcs mean nothing without knowing what the numbers represent.
Anyone else have a take on this (monoxide excluded)

tommyplumb said:

Thanks Monoxide but this has just confused the issue even more so.
Maybe you have copied these calcs straight out of a book?
What does the 11.9 represent?,the same applies to the 20.9?
Please explain? as the calcs mean nothing without knowing what the numbers represent.
Anyone else have a take on this (monoxide excluded)

Click to expand...

.............no helping some folks , now **** off.

tommyplumb said:

What does the 11.9 represent?,the same applies to the 20.9?

Click to expand...

11.9 is the percentage of CO2 in flue gas with zero oxygen in it. You can deduce that from monoxide62's equation:

monoxide62 said:

11.9 x (20.9 - O2)
---------------------------= CO2%
20.9

Click to expand...

Just set O2 to zero.

Some GCE chemistry and physics might help. This is the reaction for the combustion of methane:

CH4 + 2O2 = 2CO2 + 2H2O

So half the O2 goes to make CO2 while the other half goes to make water. Now it's a property of gases that equal numbers of molecules occupy roughly equal volumes, regardless of what the gas is. So, from the equation above, we see that the volume of CO2 in the flue gas will be equal to half the volume of the O2 that was consumed. Air is 20.9% O2 (so now you know where 20.9 comes from. ). The rest is mostly nitrogen and other stuff that plays no part in combustion so 79.1% of the air going into a boiler comes straight out unchanged. Now for some maths:

For 100% of air going in, 79.1% emerges unchanged and half the O2, which is 10.45%, comes out as CO2. This would give us:

%CO2 = 100 x 10.45 / (79.1 + 10.45) = 11.7%

You'll notice two things: I ignored the H2O in the flue gas - because that would have given 9.5% CO2 - and 11.7% still isn't quite right. Leaving out the water would be correct for a condensing boiler and maybe gas analysers ignore it anyway. (Somebody out there will soon put me right if they don't!. ) The other thing I did was to start with pure methane, which natural gas isn't. It contains traces of other gases, notably ethane which gives more CO2 per unit of O2 than does methane. I'm not going into the fine details of that calculation; you can do it yourself. Let's just stick with the final figure of 11.9% CO2 that monoxide62 gave us.

Having measured the actual percentage CO2, you can derive the percentage O2 by rearranging the equation (GCE maths), which monoxide62 has done for you here:

monoxide62 said:

20.9 x CO2/11.9 = ?....20.9-? = O2%

Click to expand...

I'm afraid ! can't quite make sense of the last equation for percentage air. Shouldn't it be:

O2% x 100
------------- = XS air%
20.9

or am I missing something.

Dazzling stuff there Space cat...now forget this cr*p and get over to GD and see if my answer was right on that faster than light thread.

My brain hurts

Space cat said:

Having measured the actual percentage CO2, you can derive the percentage O2 by rearranging the equation (GCE maths), which monoxide62 has done for you here:

20.9 x CO2/11.9 = ?....20.9-? = O2%

Click to expand...

There is a slight floor in that equation , what if POC were fuel rich? hence the reason i added in brackets (fuel lean) , CO2% can be taken from either side of stoich bell chart , this mathmatical equation is only correct if CO2 is derived from right hand side of bell chart (fuel lean) , this is the reason CO2% alone should not be used when dealing with combustion analysis.

I'm afraid ! can't quite make sense of the last equation for percentage air. Shouldn't it be:

O2% x 100
------------- = XS air%
20.9

or am I missing something.

Click to expand...

Your missing something.

5% O2 x 100
-------------------------------= 31.4% XS air
20.9 - 5%

monoxide62 said:

There is a slight floor in that equation , what if POC were fuel rich? hence the reason i added in brackets 'fuel lean'

Click to expand...

Quite true. I was assuming complete combustion.

and also said:

Your missing something.

Click to expand...

Then I'd better go away and work it out.

Space cat said:

So half the O2 goes to make CO2 while the other half goes to make water. Now it's a property of gases that equal numbers of molecules occupy roughly equal volumes, regardless of what the gas is. So, from the equation above, we see that the volume of CO2 in the flue gas will be equal to half the volume of the O2 that was consumed. Air is 20.9% O2 (so now you know where 20.9 comes from. ). The rest is mostly nitrogen and other stuff that plays no part in combustion so 79.1% of the air going into a boiler comes straight out unchanged. Now for some maths:

Click to expand...

This may help.

Products of combustion leaving the appliance with sufficient O2 supplied (per 1 M3 nat gas.)

CO2 = 1.058
Water vapour = 2.019
Nitrogen in gas = 0.027
Nitrogen in air = 7.750
Total............= 10.84 m3

Remember , one cubic meter of methane requires 2 cubic metres of O2 (aprox 11 M3 of air).