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# Vitodens Condensate - for Paul Pollets

Member Posts: 223
only on the wall!!!!!!

Leo G

• Member Posts: 6,928

Paul. Did some more thinking and added some math this time...

The math in your original post appears to be correct. "Drop" actually seems to be a standard unit of measure equivalent to 0.0648524 milliliter. 1/20th of a milliliter is 0.05 milliliter. Won't matter as I'll use 1/20th to keep things in the same scale, but be aware if you use a conversion program that has "drop" as a unit of measure.

Something very important is missing however. Namely, how may drops of condensate do you get by burning a therm of gas.

---------------------------------------

This is probably a weird way to go about it, but I'll try to find the weight of water produced by burning a therm of natural gas.

Assumption: The "natural gas" is in fact pure methane and we've achieved perfect combustion: CH4 + 2xO2 = CO2 + 2xH2O.

Molecular weight of the gasses involved:

CH4 (methane) = 16.043

O2 (oxygen) = 32

CO2 (carbon dioxide) = 44.01

H20 (water vapor) = 18.016

When you add the weights together on both sides of the equation you get a molecular weight of 80.04 on both sides of the equation.

On the fuel side, the methane made up 20% of the weight, the oxygen 80%. On the combustion side the water vapor made up 45% and the carbon dioxide 55%.

At normal temperature and pressure (NTP) of 68F and 30" of mercury one cubic meter of methane has a density of 0.0417 pounds. So one therm (100 cubic feet) of methane will weigh 4.17 pounds. (NTP of natural gas seems to be 60F and 30" but I won't even attempt the conversion--surely it can't be that much difference...)

Using the proportions from the molecular weight equation, it thus takes 4.17 / 0.2 = 20.85 pounds of total weight on the fuel side.

Since the combustion product side must also weight 20.85 pounds the weight of water produced is 20.85 * .45 = 9.3825 pounds of water vapor produced by burning 100 cubic feet of methane.

At atmospheric pressure (pray I can use that here) the water vapor has 9.3825 * 971.7 = 9,117 btus of latent heat.

Methane has a high heat value of about 1,000 btu/cu.ft. (Not sure why I keep finding a range--perhaps because they're thinking about natural gas, not pure methane) but anyway...

100 * 1,000 = 100,000 btu high heat value of the methane. 9,117 btus of vaporization energy so vaporization accounts for 9.12% of the total energy in that 100 cubic feet of methane. (Seems quite close to the 10% generally stated for natural gas, so surely I'm not too far off. Perhaps that difference in temperature between the "standards" accounts for more than I thought...)

Now, FINALLY (and I'm going to use therm and natural gas here):

Burning one therm of natural gas made 9.3825 pounds of water vapor. This is about 1.13 gallons.

1.13 gallons = 4,277 milliliters.

4,277 / 0.05 = 84,540 "drops" as measured here (1/20th milliliter)

At a cost per therm of \$1.25 and 84,540 theoretical drops per therm, each drop has a value of \$1.25 / 84,540 = \$0.0000147

The original calculation of the value of a "drop" was \$0.00000134

Unless one of us has misplaced a decimal point, it looks like the true value of a "drop" is about 10.1 times greater with a month worth of drops thus worth about \$35.15 and a years worth about \$421.78

IMPORTANT: This last paragraph is assuming that EVERY drop of condensation is actually recovered. IMPOSSIBLE! If you're willing to take Viessmann's word, a Vitodens in a "typical" installation will yield about .81 gallons of condensate per therm. The theoretical production of condensate (as presented here) is 1.13 gallons or about 80% "typical" compared to "theoretical".

I did not however even attempt to add in the "extra" value of the drop of condensate that comes its specific heat below 212F nor the overall reduction in temperature of all of the flue gasses. While I doubt that such could amount to the "missing" 20%, I suspect it could be awfully close under optimal conditions.
• Member Posts: 1,935
NTP = STP

should read standard temperature and pressure, FYI.

never seen it as NTP?
• Member Posts: 6,928

Nope. Normal Temperature Pressure (NTP) and Standard Temperature Pressure (STP) values are BOTH given for mass of methane as well as many other gasses. See this link. The NTP value was MUCH closer than used for "standard" natural gas.
• Member Posts: 1,935
hmmm

guess I better re-read my chemistry book.

how many gallons theoretically

how many gallons observed, via drips?

how were the drip determine? drips /min or drips/ 10 miutes?

• I think there is an error in your calculation. It seems that you are assuming that the total value of the therm of gas is incorporated into the value of the condensate. In reality, the value of the condensate is only represented by the fraction of heat which is recovered by condensing. You are ignoring the value of the 90% which is obtained by combustion without condensing. Thats why your value is about 10 times what was obtained in the previous calculation, which only included the value of the heat gained by condensing the water vapor.
• Member Posts: 1,935
typo?

funny units?

you have 1 cubic meter weighing .0417lbs, but 100 cubic feet weghing 100 times the cubic meters weigh?
• Member Posts: 6,928

No need to re-read your chemistry book. If you're looking at the mass of a gas it's all about temperature and pressure and there are different "normal", "standard" and "utilized" temperature/pressure combinations. One of the websites I was looking through today made a BIG point of this saying that it VERY frequently causes SEVERE problems. Don't forget that NASA supposed messed up a Mars probe by using 0F instead of 0C (or is that the other way around...)

Am pretty sure that I came close to the actual value of a "drop". The OVERALL value (say monthly/yearly) is however based on recovering EVERY possible "drop". As I said, IMPOSSIBLE.

1.13 gallons per 100 cubic feet of perfectly combusted methane is what I came up with for the "theory" as presented here. There are CERTAINLY some assumptions and short-cuts and I did my best to point them out.

I gave NO estimate of observed "drops".

Other than the very end where I applied the difference over time to the original post, time has nothing to do with anything here--I tried for the value of an actual "drop". I kind of assumed that the original poster had a time value in mind.
• Clarification

Mike, I think I found where the error is:

"At a cost per therm of \$1.25 and 84,540 theoretical drops per therm, each drop has a value of \$1.25 / 84,540 = \$0.0000147" This is the total heating Dollar value per drop, sensible + latent heat.

But:

"100 * 1,000 = 100,000 btu high heat value of the methane. 9,117 btus of vaporization energy so vaporization accounts for 9.12% of the total energy in that 100 cubic feet of methane."

The value of \$0.0000147 is based on the full heating value of the gas.

The value of the latent heat of condensation is only 9.12% of the full heating value, so:

\$0.0000147 * .0912 = \$0.00000134 which is exactly the same answer which was obtained by the first method!
• Member Posts: 6,928

Don't see that I mentioned cubic meters anywhere--if so it's a typo as it would be cubic feet. I was careful to find reference sources in feet, pounds and inches.
• Member Posts: 6,928

Thank you Mike! You CERTAINLY found an error but something tells me it might have occurred at the very end when I re-applied to the original way that the value of a drop was computed--I may have reintroduced the initial error of therms/drop vs. drops/therm at the very end.
• Member Posts: 6,928

This might end up with the same value, but let's see.

Let's find the value (based on the original) of the condensate produced by burning a therm of gas compared to the cost of a therm and then convert to drops...

Burning 1 therm of methane produced 9.3825 pounds of water with 971.7 = 9,117 btus of latent heat.

With a high heat value of 100,000 for that therm those 9,117 btus made up 9.12% of the available energy.

A therm cost \$1.25 so the value of the condensate is \$1.25 * .0912 = \$0.114

We had 84,540 "drops" in this condensate so the value of a single drop is \$0.114 / 84,540 = 0.00000134847 the same as the original calculation!!!

Did I waste my time? NO! At least I hope not--all I did was "prove" something that everyone knows to begin with even if I did put it in a way you WON'T find on the web and might be understandable and despite assumptions winds up EXACTLY the same. Does make me wonder how many assumptions are made in supposed "rules"...

HERE'S THE PROBLEM WITH THE ORIGINAL POST

IT ASSUMMES FAR TOO FEW DROPS

86,400 assumed drops per day with one therm of gas providing 84,540 drops per day. In other words it's assuming consumption of just a touch over one therm per day or about 155 therms per 5-month heating season!

If your house can achieve this then you should be using solar with wood or electric backup!!! ANY available boiler would be WILDLY oversized.

I've reduced my absolutely worst-case one month consumption in a big, old house with a Vitodens (6-24) to less than 155 therms but frankly I STILL consider the boiler a bit oversized...

[EDIT] What was I thinking? My absolutely worst-case consumption in a month will more than double 155 therms.
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not that it matters