Re: hydrogen for how much?
- From: "bill" <ford_prefect42@xxxxxxxxxxx>
- Date: 8 Feb 2007 14:13:54 -0800
On Feb 8, 3:46 pm, "RichD" <r_delaney2...@xxxxxxxxx> wrote:
On Feb 4, The Ghost In The Machine <e...@xxxxxxxxxxxxxxxxxxxxxxx>
wrote:
www.nationalvapor.com
Electrolysis = electricity
Electricity =/= free
The amount of hydrogen liberated during electrolysis represents
less energy that it took to make it.
https://www.cia.gov/cia/publications/factbook/geos/xx.html
At present we're consuming about 82.59 million bbl of oil per day,
worldwide.
http://www.gravmag.com/oil.html
stipulates that 1 42-gallon barrel of oil will yield 19.5
gallons of gasoline, though if one assumes a barrel of
oil is mostly carbon (even were it full of solid methane
it would be 75% carbon by weight; if it were pure x-chain
aliphatics -- (CH2)x -- it would be 86%, so it needs to be
somewhere between the two), one can multiply by 44/12 (since
CO2 is 44 g/mol) and get the amount of CO2 dumped from consuming
that barrel.
http://www.allmeasures.com/Formulae/static/materials/66/density.htm
suggests 820 kg/m^3 for oil density, and that a barrel of oil will
therefore weigh 130.37 kg. (The actual density will probably depend
on the field.)
So...130.37 kg of carbon times 0.86 divided by 0.012 kg/mol
gives me 9343.18 moles of C, yields 9343.18 moles CO2, or
411.1 kg CO2 from that barrel, or 33.952 million metric tonnes
of CO2 thrown into the atmosphere per day.
C-C: +347 kJ/mol
O=O: +498 kJ/mol
C=O: +805 kJ/molhttp://www.webchem.net/notes/how_far/enthalpy/bondenthalpy.htm
therefore C + O2 = CO2 + (2*805-498-1.5*347) = CO2 + 591.5 kJ/mol
http://hypertextbook.com/facts/2002/KarolShepelsky.shtml
suggests 45-46 MJ/kg of crude. So OK; we'll work with that;
the general idea is that we must replace 45-46 MJ/kg of crude,
times 130.37 kg, times 82.59 million, or almost half an exajoule
(4.95 * 10^17 J) a day.
Hydrogen is a lot simpler, from an enthalpy standpoint.http://www.webelements.com/webelements/elements/text/H/enth.html
2H2 + O2 = 2H2O + 4*427.6 - 2*435.99 - 498.36 = 2H2O + 340.06 kJ.
Since 2H2 is two moles divide in 2, or 170.03 kJ/mol H2,
or 85 MJ/kg H2.
In order to generate 4.95 * 10^17 J/day from hydrogen we will need
to generate 5.82 * 10^9 kg per day. If we assume 8 hours/day
electrolysis, 90% efficiency, and 1250 W/m^2 on our PV panels,
that gives us an area of 1.528 * 10^10 m^2 or 1.528 * 10^6 hectare
or 15280 km^2 or 5900 mi^2.
This is bigger than Connecticut.
The going cost of a PV panel is currently about $6.5/peak
watt of generated power. This cost may go down as
production ramps up; one might assume it goes down
to $2.5/peak watt (this is very ad hoc, admittedly).
Given that assumption, one gets $38.2B in construction
costs, though it could easily balloon to almost $100B
if one uses the original $6.5/W figure.
I attended a lecture by Nathan Lewis, Caltech chemist,
a few years ago. He's one of the leading researchers
into the carbon reduction problem. A very impressive talk,
with all the numbers.
He emphasizes solar as the solution, with nukes, wind, etc.
as helpful but minor. He displayed a map, with Arkansas
covered by photovoltaics, and claims that would do it, one
of those on every continent. So your CT estimate looks
right. Expensive, but if there's no alternative...
He assumes a 50 year time frame, which makes everything
seem eminently feasible.
He also made the interesting point that electric or
hydrogen cars are inefficient, because you must carry the
complete product chain; whereas, with petroleum, the waste
vents to the atmosphere, lightening the load. Electricity is
'highly organized' energy, which should be used to power
your stereo, not transport or heating.
while covering arkansas with pv cells sounds feasible stated like
that, have you looked at how much the wretched things cost?? and how
about storage for that much energy for the 18 hours a day when the sun
isn't shining.
.
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