[fl7782]

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Hello, new member here.  I am a physicist and have a need for some chemistry calculations.  I would appreciate your review of the following calculations and your commentary on any of it.  Thank you.

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This is a calculation of the pressure increase in a pressure vessel which is heated externally.  There is some paper-like material inside the vessel which will combust until the oxygen in the vessel is depleted.  My intention is to estimate the increase in pressure, including the effects of heat and increase in gas molecules.

A couple of comments: (1) This is a calculation only, we will not be doing this experimentally... no worries.  (2) An estimate is acceptable; an answer within 30% or 40% would be fine.

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Initial air inside vessel, mole fraction:

Nitrogen 0.78
Oxygen 0.21
Argon etc 0.01

Ideal Gas Law PV = nRT

n/V [=] moles/L (has units of)

n/V = P/RT

Example, for Nitrogen:

(.78 atm)/[(.08206 L.atm/K.mol)(0+273)K] = .035 moles/L N2

where 0 C has been chosen as standard temperature.

Result:

.035 moles/L N2
.0094 moles/L O2
.00045 moles/L Argon etc

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The volume of the vessel is V = 3.5 L

So initially,

.123 moles N2
.033 moles O2
.0016 moles Argon etc

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Treat the paper-like material as cellulose: C6H10O5

Molecular weight: 162.14 g/mole

Assume there is about 1/4 lb = 113 g of paper-like material in the vessel:

113 g / (162 g/mole) = 0.70 mole cellulose

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Assume complete combustion:

C6H10O5 + 6O2 = 6CO2 + 5H2O
.0055 moles, .033 moles, .033 moles, .028 moles

The moles listed above were determined per:
There is only .033 moles of O2 in the closed vessel; once that is depleted, combustion must cease.
Cellulose will lose 1/6 as many moles as O2 = (1/6)(.033) = .0055
CO2 moles created match O2 moles used.
H2O moles = (5/6)(.033) = .028

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The ideal gas law PV=nRT may be written

P = nRT/V = (n.total)RT/V = (na+nb+...)RT/V = (na)RT/V + (nb)RT/V + ... = pa + pb +...

So the pressure is the sum of the partial pressures of the individual gases.

For the initial gas mix: n1 = n.total.initial = .123 moles N2 + .033 moles O2 + .0016 moles Argon etc = .158 moles

For the final gas mix: n2 = n.total.final = .123 moles N2 + .033 moles CO2 + .028 moles H2O + .0016 moles Argon etc = .186 moles

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PV = nRT

We have P/(nT) = R/V = constant

P1/(n1T1) = P2/(n2T2)

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Assume the vessel is heated externaly to 1250 C:

P2 = (n2/n1)(T2/T1)P1 = (.186/.158) [(1250+273)/(0+273)] P1 = 6.6 P1

Assume initial pressure was 29.92 inHg = 14.7 psi (with apologies for the mixed units, but I've been careful within the formulas--it's the same ratio of temperatures using British units of Rankine rather than Kelvin)

P2 = 6.6 P1 = 97 psi

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So... heating the vessel expands the gases inside, increasing the internal pressure.  Less than 10% of the paper-like material combusts, using up all of the oxygen in the vessel, increasing the internal pressure by less than 20%.  The initial 14.7 psi internal pressure reaches a maximum of 97 psi.  This assumes that the vessel and contents are actually heated to 1250 C and that the combustion that occurs is complete combustion.

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I will probably re-work this with a standard temperature of 70 F.

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How does this look--do you see any bad assumptions or incorrect applications of the theory?  Thank you in advance, your input is very much appreciated.



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[fl7782]

Me again... a couple of things I should've included:

- I have assumed the amount of heat released by the combustion of the small amount of paper-like material is insignificant compared to the external heating up to 1250 C (partially because I believe that's true, and partially because at this point, I haven't estimated the heat released).

- Does the generation of water vapor / steam have to be handled in any special way?  I just recall that steam is a special case.

Thanks.

[Borek]

Get rid of that "moles per liter" concept (go directly for pV=nRT with V=3.5 L). It makes your calculations quite difficult to follow. To be honest, that's what threw me off around the cellulose combustion part, up to this moment everything was mostly correct. Chances are in the later calculations you tricked yourself into using these moles/L instead of real moles.

If you heat the vessel up to 1250°C combustion heat doesn't matter - it will just let you use less external heat when heating. That's Physics 101, conservation of energy, isn't it?

[fl7782]

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Thanks for the reply.

I see your point on the moles/L approach.  I did that because I don't have the exact volume figure yet.  But I can simplify the calculation once I have it.

The 1250 C value is sort of an upper bound.  The vessel is exposed to a large gasoline fire, and the reported values for gasoline flame temperatures that I've found have been 900 to 950 C, with transient readings up to 1200 to 1250 C.  I have all sorts of room given the low pressure value that comes out of the calculation.  There is no heat control of any sort, and so any heat released by combustion would just be in addition to the external heat... but my point was combusting about 1 g of cellulose releases an insignificant amount of heat.

I had said, "Less than 10% of the paper-like material combusts...".  In fact, it's less than 1%:

(.0055 moles)/(.7 moles) = less than 1%, amounting to 0.9 g

So I must ask, is it reasonable that 3.5 L of air only contains enough oxygen to combust 1 g of cellulose?


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[fl7782]

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I think I have been able to answer my last question above, that yes, about 3.5 L of air is required to combust 0.89 g of cellulose.

I approached it as follows:

Molecular weights: C 12.0 g/mol, H 1.0 g/mol, O 16.0 g/mol

Hypothesis: Combusting 0.89 g of C6H10O5 uses the oxygen in about 3.5 L of air (std conditions)

1 lb of C is found in 162/72 lb = 2.25 lb cellulose = 1021 g cellulose

(1021 g cellulose) [ (3.5 L air) / (0.89 g cellulose) ] = 4015 L air = 142 ft^3 air

So my calculations estimate that 142 cubic feet of air is needed to combust cellulose containing 1 lb of carbon.

Page 155, "Analysis of Flue Gases" in Steam: Its Generation and Use, estimates that 153 cubic feet of air is required for the complete combustion of 1 lb of carbon.

So I believe my estimate is reasonable.

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[Borek]

0.89 g looks reasonable for 3.5 L of air at 0°C, if the air has a higher initial temperature, mass of cellulose would be lower.

[Enthalpy]

1 mol taking 22.4L at 0°C, I get 0.16 mol, of which 0.033 mol are O₂.

0.89g of C₆H₁₀O₅ (162g) contain 0.033mol of C.

This means that little CO will be produced, together with unused O₂ left. The moderate temperature favours a complete reaction.

Though, the cellulose fire is hotter than 1250K. I've let Propep run on 0.89g cellulose and 4.6g air at initially 298K, and the computed temperature is 2200K. Heating isn't the proper method to reach 1250K.

[fl7782]

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Thank you both for your replies.

Enthalpy, the cellulose is inside a steel pressure vessel with a volume of about 3.5 L.  About 0.89 g of the cellulose is assumed to combust due to the applied heat.  This small combustion, even at 2200 K, would contribute little to additionally heating the 3.5 L of gases and the steel vessel itself, correct?

Just fyi, the heating of the vessel isn't by choice--the vessel is assumed to be exposed to a large gasoline fire.

I'm curious if you or Borek have any comments on handling the pressure due to steam.  One case is the small amount of steam produced by the combustion of the 0.89 g of paper-like material.  Another case might be that some water had somehow leaked into the vessel prior to the heating event.

Thanks.

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[Borek]

Handling pressure - regardless of its source - is an engineering problem, isn't it?

If memory serves me well steam requires special treatment at low temperatures, when it can be wet. Taking into account how approximate your other estimates are, you won't introduce larger errors treating steam as an ideal gas when you are dealing with temperatures higher by over 1000°C than the boiling point.

[fl7782]

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Thank you for your reply.


> "Handling pressure - regardless of its source - is an engineering problem, isn't it?"

Ah, I must apologize for my poor choice of words.  I did not mean "handle" in the sense of the vessel design, I meant how to treat steam in the calculations.


> "If memory serves me well steam requires special treatment at low temperatures, when it can be wet. Taking into account how approximate your other estimates are, you won't introduce larger errors treating steam as an ideal gas when you are dealing with temperatures higher by over 1000°C than the boiling point."

Thank you, that is what I was looking for.  It seems the key in this situation is that the temperature is much higher than the boiling point, and so treating the steam as an ideal gas is acceptable.

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[Enthalpy]

The cellulose is inside a steel pressure vessel with a volume of about 3.5 L.  About 0.89 g of the cellulose is assumed to combust due to the applied heat.  This small combustion, even at 2200 K, would contribute little to additionally heating the 3.5 L of gases and the steel vessel itself, correct?

0.89g of cellulose is about 1/6 of a paper sheet, more than enough to heat 3.5L air, which weighs only 5 times more than the cellulose. Or put the other way, if the fire consumes all oxygen, the result is necessarily hot.

This will happen far before the vessel absorbs any heat.

Following that, over a long time, the vessel will absorb heat, and the combustion products may cool down from 2000 degrees. Over this time, between 2000 and 1250 degrees, the reactions will remain at equilibrium, so that you get the proportions resulting from the final temperature.