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If you start with ice you could define the enthalpy being 0 at 0 C.
You always work with enthalpy differences so you should choose a
convenient zero condition.<br>
<br>
Peter<br>
<br>
On 19/01/2012 04:32, Frank Shields wrote:
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<p class="MsoNormal"><span style="color:#1F497D">Dear Crispin,<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">Not that this
has anything to do with stoves, but if water has 0 enthalpy
at 0 deg C and increases 4.186 j/c/g what would be the
enthalpy of ice? And ice at -10 deg C.? And water being the
densest at 4 deg C does this 4.186 j/c/g constant through
the 4 deg C down to 0 deg C? Just wondering.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">And is the LHV
ever calculated using the temperature of the stack gas? It
seems it would not be because some water may have condensed
on the cold surface meaning the total water would not all be
in the gas phase.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">Thanks for your
patience.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">Frank<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">Frank Shields<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">42 Hangar Way<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">Watsonville,
CA 95076<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">(831) 724-5244
tel<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D">(831) 724-3188
fax<o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><a class="moz-txt-link-abbreviated" href="mailto:frank@bioCharlab.com">frank@bioCharlab.com</a><o:p></o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
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<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="color:#1F497D"><o:p> </o:p></span></p>
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<p class="MsoNormal"><b><span
style="font-size:10.0pt;font-family:"Tahoma","sans-serif"">From:</span></b><span
style="font-size:10.0pt;font-family:"Tahoma","sans-serif"">
<a class="moz-txt-link-abbreviated" href="mailto:stoves-bounces@lists.bioenergylists.org">stoves-bounces@lists.bioenergylists.org</a>
[<a class="moz-txt-link-freetext" href="mailto:stoves-bounces@lists.bioenergylists.org">mailto:stoves-bounces@lists.bioenergylists.org</a>] <b>On
Behalf Of </b>Crispin Pemberton-Pigott<br>
<b>Sent:</b> Tuesday, January 10, 2012 2:20 PM<br>
<b>To:</b> Stoves<br>
<b>Subject:</b> Re: [Stoves] Calculation help<o:p></o:p></span></p>
</div>
</div>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">Dear
Frank<o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">There
is no need to over-complicate the issues for HHV.<o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">There
are three stages of calculation involved:<o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">From
liquid to boiling,<o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">From
boiling to evaporated water at boiling<o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">From
boiled vapour to hotter vapour<o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA">>I realize for
biomass the difference between J used in evaporation (2256<o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA">j/g) and energy in
water at 450c (2854 j/g) is well within the 'noise' of
biomass fuel so doesn't matter which one we use.<o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">The
energy 'at 450' includes parts of all three segments of the
calculation. The number 2854 is the sum of three things. I
was showing how to calculate them separately.<o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA">> And if we all
are using the 2256 value that is what I will do. <o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">2257
is probably the most accurate. Often people use 2260 for
convenience being a rounder number.<o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA">>But it seems
since the hydrogen in biomass never is in a liquid state or
never even water vapor before going from solid state to
water vapor at ~450 that we should be using 2854 ((2256<o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA">+(1.72 X 350)). <o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">Hydrogen
is in the solid state in the fuel (not the H2 in the
moisture, the H2 in the biomass). It is liberated as a gas
and most of it burns to water, but not all. There can be
quite a lot of H2 in the exhaust. See this graph. The lower
line is free H2(EF) which means it is the measured
concentration multiplied by (Excess Air +100%). You can see
there is nearly a fixed ratio between the CO and the H2.
This is common though not universal. Even after 3 hours,
there is still lots of free floating H2.<o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><img
id="Picture_x0020_1"
src="cid:part1.09000402.07000206@bigpond.com" height="174"
width="378"><o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA">>For the LHV
calculation of methane; I see wiki says the HHV is product
of water in liquid form <span style="color:#1F497D">[and
cooled to 0 degrees C] </span>and LLV is product of water
in vapor form - same way you calculate biomass. <span
style="color:#1F497D">[It has nothing to do with the fuel
type, it is how to calculate the heat content of
combustion products] </span>But, of course water is never
in the liquid form (until it completely condenses at below
100c.). Hydrogen is held by carbon or then water vapor at
the combustion temp. so I am surprised they calculate
(estimate) it the way they do. <o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:black" lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">In
fact it is pretty reasonable. Lower heating value is what
heating can be done by the fire. That is a pretty real world
attitude. Yes, it has not been done exactly. But it is easy
to calculate.<o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA">>So I will not
bother anyone with more questions, will calculate as
everyone does (subtract energy from water liquid to water
vapor) <o:p></o:p></span></p>
<p class="MsoPlainText"><span lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">Yes.
The HHV from a bomb calorimeter minus the latent heat of
evaporation of the moisture from combustion of Hydrogen
gives the useable dry fuel heat called LHV. So far no one
has come up with a difference between the latent heat of
evaporation and latent heat of condensation. If you can
create one, you will have a perpetual motion heat engine!<o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA"><o:p> </o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">Regards<o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA">Crispin<o:p></o:p></span></p>
<p class="MsoPlainText"><span style="color:#1F497D" lang="EN-CA"><o:p> </o:p></span></p>
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