Hello, Benhogan.
This was a fun question to answer! I'm sorry it took so long, but
I got started and had to go to bed...I find that math calculations
and sleep depravation usually result in the wrong answer. :) But
having had a good night's sleep, I have confidence in the veracity
of these figures. I am going to answer your question using the
second form in which you asked it--that is, by using the volumes
instead of BTU. I feel that the answer will be more accurate that
way. I assume that you already have data which states that 1
gallon of liquid propane equals approximately 100,000 BTU, and 1
cubic meter of natural gas equals the same. I didn't verify that
information since you already have it. It isn't too hard to
calculate the number of BTU produced by these reactions using the
standard enthalpies of formation of the products and reactants,
but I won't go through that unless you specifically request it.
Note that my calculations will probably look crummy unless you
view this page with a fixed width font in your browser! Viewed
with a fixed width font (for example, Courier), the calculations
should line up nicely.
Here are the facts we need to solve the problem:
Propane
-------
Density: 585.3 g/L (liquid)
Molar Mass: 44.094 g/mol
Chemical Formula: C3H8
Chemical reaction: C3H8 + 5O2 --> 3C02 + 4H20 + heat
One mole of Propane produces 4 moles of water.
Water
-----
Density: 1 g/cm^3
Molar Mass: 18.016 g/mol
Methane (natural gas)
---------------------
Chemical Formula: CH4
Chemical reaction: CH4 + 2O2 --> CO2 + 2H2O + heat
Molar Mass: 16.042 g/mol
One mole of Methane produces 2 moles of water.
Conversions and constants
-------------------------
1 gal = 3.7854 L
1 lb = 2205 g
0 deg C = 273.15 K
25 deg C = 298.15 K
R = 0.0821 L*atm/(mol*K)
Since the propane is in liquid state, the procedure we will follow
is to go from volume of propane --> mass of propane --> moles of
propane --> moles of water --> mass of water:
1 gal C3H8 3.785 L 585.3 g 1 mol
---------- x --------- x --------- x ---------- = 50.25 mol C3H8
1 1 gal 1 L 44.09 g
50.25 mol C3H8 4 mol H20
-------------- x ------------ = 201.0 mol H20
1 1 mol C3H8
201.0 mol H20 18.016 g 1 lb
------------- x ----------- x -------- = 1.642 lb H20
1 1 mol H20 2205 g
So one gallon of propane burns to produce 1.64 pounds of water
(three significant figures).
Natural gas (methane) is a little different because it is in a
gaseous state. The number of gas molecules present (and therefore
the amount of water which can be produced) depends on the
temperature and pressure. I'll assume standard temperature (25
deg C) and pressure (1 atm) in this case because those are the
most commonly used figures (e.g., if you look up how many BTU are
contained in 1 cubic meter of methane, it will probably be given
at 25 deg C and 1 atm--also referred to as STP). The procedure to
follow in this case is to go from volume of methane --> moles of
methane --> moles of water --> mass water. The formula we need to
use to convert from volume to moles is the ideal gas law, PV=nRT.
But first we need to convert 1 m^3 to liters.
1000000 cm^3 1 L
1 cubic meter x -------------- x ----------- = 1,000 L
1 m^3 1000 cm^3
PV 1 * 1000
n = -- = ----------------- = 40.85 moles
RT 0.0821 * 298.15
40.85 mol CH4 2 mol H20
------------- x ----------- = 81.71 mol H20
1 1 mol CH4
81.71 mol H20 18.016 g 1 lb
------------- x ----------- x -------- = 0.6676 lb H20
1 1 mol H20 2205 g
So when 1 cubic meter of methane gas at STP burns, it produces
0.668 pounds of water (three significant figures).
Additional Links:
These pages contain some information on the two compounds:
http://elifritz.members.atlantic.net/compounds/propane.htm
http://elifritz.members.atlantic.net/compounds/methane.htm
This page gives the chemical formulas for the combustion
reactions:
http://home.att.net/~lfretzin/notes08.html
Search Strategy:
Google Search:
chemical reaction propane combustion
://www.google.com/search?hl=en&lr=&ie=UTF-8&q=chemical+reaction+propane+combustion&spell=1
Google Search:
heat OR energy obtained OR produced combustion propane
://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&q=heat+OR+energy+obtained+OR+produced+combustion+propane
Google Search:
water produced combustion propane
://www.google.com/search?q=water+produced+combustion+propane
Used my college chemistry and physics books to find the necessary
constants, formulas, and conversion factors.
I hope this is a satisfactory answer. If you have any questions,
please feel free to ask for clarification.
Best regards,
Bikerman |
Clarification of Answer by
bikerman-ga
on
18 Oct 2003 06:47 PDT
Hi,
I just wanted to clarify that this is the amount of water produced
when burning 1 gallon and 1 cubic meter of propane and methane
respectively. Technically speaking, there is no water contained in
either propane or methane--the water is formed by the combination of
the hydrogen in the gases and oxygen in the air. I think that is what
you meant by "...how much water is in a cubic meter of natural gas or
a gallon of propane", but I just wanted to make it clear...in case my
chemistry professor ever looks at this. :-)
Thanks,
Bikerman
|
Clarification of Answer by
bikerman-ga
on
18 Oct 2003 09:07 PDT
Thank you racecar-ga! I am very sorry--I checked my work several
times and still missed that. Indeed it should read 1 kg = 2.205 lb
and therefore 1 lb = 453.5 g.
That does change the answers to
201.0 mol H20 18.016 g 1 lb
------------- x ----------- x -------- = 7.985 lb H20
1 1 mol H20 453.5 g
for the propane and
81.71 mol H20 18.016 g 1 lb
------------- x ----------- x -------- = 3.246 lb H20
1 1 mol H20 453.5 g
for natural gas.
I didn't, however, miss the step of converting from volume to BTU. As
I stated at the beginning of my answer, my assumption is that
benhogan-ga already has that data, and that is why he picked 1 gallon
and 1 cubic meter. However, if it turns out that I didn't understand
properly, I can calculate that figure from the standard enthalpies of
formation as stated above.
I apologize for the error, and thanks for pointing it out, racecar-ga.
Bikerman
|
Clarification of Answer by
bikerman-ga
on
18 Oct 2003 11:11 PDT
Hello,
I apologize once again for the above error, and I decided to go
ahead and calculate the BTU/volume for you in case you did in fact
need it. It isn't that difficult, so there's no point in not
going ahead and doing it.
The total heat of the reaction is the sum of the enthalpies of
formation of the products minus the sum of the enthalpies of the
reactants. Here is the relevant data:
Standard enthalpies of formation
--------------------------------
C3H8(g) : -103.8 kJ/mol
CH4(g) : -74.8 kJ/mol
O2(g) : 0.0 kJ/mol
CO2(g) : -393.5 kJ/mol
H20(g) : -241.8 kJ/mol
H20(l) : -285.8 kJ/mol
Conversions
-----------
1 BTU = 1.055 kJ
NB: when methane or propane burn, it is actually water vapor that
is produced in the exothermic reaction. However, under normal
circumstances the water condenses into it's liquid state, which
gives off even more heat. Because of this, I'm going to give two
figures for each gas: one is the BTU produced before condensation
occurs, and the other is the total ending BTU. Which you need
depends on exactly what you are doing.
For propane, burning one mole produces:
sum of products - sum of reactants =
3*(-393.5)+4*(-241.8)-[-103.8+5(0)] = -2043.9 kJ
If the water condenses, you have
3*(-393.5)+4*(-285.8)-[-103.8+5(0)] = -2219.9 kJ
(The negative sign indicates that this was an exothermic
reaction--heat was given off.)
1 BTU
-2043.9 kJ x ---------- = -1937.3 BTU
1.055 kJ
1 BTU
-2219.9 kJ x ---------- = -2104.2 BTU
1.055 kJ
Now we have already established that one gallon of propane is
equal to 50.25 moles. So, before the water vapour condenses, we
have (I'm dropping the minus signs):
1937.3 BTU 50.25 mol
---------- x ----------- = 97300 BTU/gal (3 significant figs)
1 mol 1 gal
After condensation:
2104.2 BTU 50.25 mol
---------- x ----------- = 106,000 BTU/gal (3 significant figs)
1 mol 1 gal
Burning one mole of methane produces:
sum of products - sum of reactants =
-393.5+2*(-241.8)-[-74.8+2(0)] = -802.3 kJ
If the water condenses, that becomes:
-393.5+2*(-285.8)-[-74.8+2(0)] = -890.3 kJ
Converting to BTU:
1 BTU
-802.3 kJ x ---------- = -760.5 BTU
1.055 kJ
1 BTU
-890.3 kJ x ---------- = -843.9 BTU
1.055 kJ
From my answer, we know that 1 cubic meter of methane contains
40.85 moles of the gas. So, before condensation,
760.5 BTU 40.85 mol
--------- x ----------- = 31,100 BTU/m^3 (3 significant figs)
1 mol 1 m^3
and after condensation occurs,
843.9 BTU 40.85 mol
--------- x ----------- = 34,500 BTU/m^3 (3 significant figs)
1 mol 1 m^3
Combining these figures with those from my answer, we get the
following:
Hydrocarbon Before Condensation After Condensation
------------------------------------------------------
propane 8.21E-5 lb/BTU 7.53E-5 lb/BTU
methane 1.04E-4 lb/BTU 9.41E-5 lb/BTU
So 100,000 BTU worth of propane with the BTU measurement being
taken before condensation produces 8.21 pounds of water. If the
water is allowed to condense before measuring the BTU, you only
get 7.53 pounds of water.
Likewise, 100,000 pre-condensation BTU of methane produces 10.4
pounds of water, while that many post-condensation BTU produces
9.41 pounds of water.
I hope this is to your satisfaction. If not, please ask for
clarification. I hope the pre/post condensation issue isn't too
confusing, but it has to be dealt with...the exact figures depend
on whether your water vapour becomes a part of the system you are
dealing with, or gets exhausted before it is allowed to condense.
This is easy to see if you consider that the steam produced
contains heat--if the steam is removed, so is the heat it
contains. If the steam is allowed to condense, it will give off
it's heat in turning back to liquid and this becomes part of your
system.
Best regards,
Bikerman
|
