[Stoves] Calculation help

[Frank Shields]

Dear Crispin,

 

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.

 

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.

 

Thanks for your patience.

 

Frank

 

Frank Shields

42 Hangar Way

Watsonville,  CA  95076

(831) 724-5244 tel

(831) 724-3188 fax

frank at bioCharlab.com

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

From: stoves-bounces at lists.bioenergylists.org
[mailto:stoves-bounces at lists.bioenergylists.org] On Behalf Of Crispin
Pemberton-Pigott
Sent: Tuesday, January 10, 2012 2:20 PM
To: Stoves
Subject: Re: [Stoves] Calculation help

 

Dear Frank

 

There is no need to over-complicate the issues for HHV.

 

There are three stages of calculation involved:

 

>From liquid to boiling,

>From boiling to evaporated water at boiling

>From boiled vapour to hotter vapour

 

>I realize for biomass the difference between J used in evaporation (2256

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.

 

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.

 

> And if we all are using the 2256 value that is what I will do. 

 

2257 is probably the most accurate. Often people use 2260 for convenience
being a rounder number.

 

>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

+(1.72 X 350)). 

 

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.

 



 

>For the LHV calculation of methane;  I see wiki says the HHV is product of
water in liquid form [and cooled to 0 degrees C] and LLV is product of water
in vapor form - same way you calculate biomass. [It has nothing to do with
the fuel type, it is how to calculate the heat content of combustion
products] 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. 

 

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.

 

>So I will not bother anyone with more questions, will calculate as everyone
does (subtract energy from water liquid to water vapor) 

 

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!

 

Regards

Crispin

 

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