[Stoves] Calculation help

Fri Jan 6 17:33:51 CST 2012

Another question..

It takes j energy to go from 20c to 100c and from 100c liquid to 100c vapor.
But doesn’t it also take in more energy when  water vapor is heated from 100
to 450 (?) and that is removed before it is used for work?

You did not include that in your calculations. It seems no one does. 

Thanks

Again.

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: Wednesday, December 28, 2011 11:01 PM
To: 'Discussion of biomass cooking stoves'
Subject: Re: [Stoves] Calculation help

Dear Frank

In order to do this sensibly (I won't say 'properly' because one could
always do it without mistakes) please restate the analysis of the sample on
a dry basis.

The reason is that it is virtually impossible to make a quick comparison
between fuel samples (by mental calculation) without it being on a common
basis. As each fuel sample will vary slightly in moisture, it is necessary
for practical purposes to subtract the moisture first, then express the
contents on a % basis for the mass that remains.

Carbon

Nitrogen

Hydrogen

Ash

You did not show a value for Oxygen so if you have to guess, use 92% of the
carbon value and wee what it looks like.

As Received (AR) MASHCON

Moisture

Ash

Sulphur

Hydrogen

Carbon

Oxygen

Nitrogen

That should add up to 100%

Dry it and weigh again (dry at 5 Deg C above ambient boiling point for 24
hours).

The Moisture content is normally expressed on a Wet Weight Basis (WWB) which
means the moisture is expressed as a % of the total AR mass.

[Beware: some labs do not do this, particularly with coal. Never assume
anything. Always check to see that they did.]

After Drying (AD)

Ash

Sulphur

Hydrogen

Carbon

Oxygen

Nitrogen

That should add up to 100% with larger values that before because the
moisture is no longer diluting the samples.

Moisture compensation:

Please note that there are two moistures to be considered: the moisture in
the fuel (water) which is evaporated when drying. This moisture reduces the
heat available for cooking because it is difficult to condense the moisture
on the pot yielding its heat. With alcohol stoves there is some condensation
which messes up the heat transfer calculations because that is working at
the HHV not the LHV (more another time).

The LHV is not the correction for water in the fuel. It is the correction
for not condensing water vapour manufactured in the fire by burning Hydrogen
and Oxygen together to make Water (H2O). The amount of this water vapour
depends on the mass of Hydrogen in the fuel. As you have measured Hydrogen
directly as a % of the total mass, you know how much there is. A mole of H2
weighs 2 g so you can quickly calculate the mass of water that will be
produced H2O weighs 18 g per mole so for every g of H you will get 9 g of
Water. If one kg of dry fuel has 6% Hydrogen in it, it will make 6% * 1000 g
* 9 = 540 g of water vapour. The heat to be deducted from the HHV to give
the LHV is the heat not recovered by condensing this water vapour. It is 540
g x 2257 J = 1.22 MJ per kg of that fuel burned.

As the Hydrogen content of different fuels varies, you need a new
calculation for each fuel. There is a rule of thumb you can use if you have
an HHV value and do not have the ultimate analysis of that fuel. It is 1.32
MJ/kg (dry). That corresponds to a Hydrogen content of 6.5%.

The LHV now calculated, the heat content of the fuel is then further reduced
by the moisture required to be removed. Andrew mentioned 25 to 100 degrees
of heating but this is not normal, 20° is normal temperature (68° F) and in
some cases 0°C but that would be a cold lab. ‘Normal’ temperature is not 77°
F. 

There are four commonly used standard temperatures:

0° C for chemical and physical processes. Standard Temperature and Pressure
(STP)

15° C for aircraft (ISA and ICAO)

20° C for Normal testing of lots of ‘ordinary’ things for example, fans and
things that work at room temperature like stoves intaking air and pots of
water lying around. Normal Temperature and Pressure (NTP)

25° C is used for chemistry as well. Standard Ambient Temperature and
Pressure (SATP). Must be warmer in the chemistry lab or the reactions are
heating the beaker.

http://www.engineeringtoolbox.com/stp-standard-ntp-normal-air-d_772.html

Pick one. Use it. Report it.  Whatever you pick, someone will object. Water
out of a tap or a bucket is not 25 degrees, if you consider the whole world.
In Waterloo it is often 3°.

The calculation of the heat available from fuel with moisture in it has been
covered many times before here. For a kg of moist fuel here is a practical
example:

Fuel mass 1000 g

HHV 19.25 MJ/kg

LHV 18 MJ/kg

Moisture content 20%

Dry fuel mass in the 1 kg = 1000 - (0.2 * 1000) = 800 g

Heat in the dry mass = 0.8 kg x 18 MJ/kg = 14.4 MJ (total Joules, not per kg
because it is for 800 g)

Water mass in the fuel = 0.2 * 1000 = 200 g

Heat to boiling point Normal Temp to Normal Boiling Point (20°-100° C) = 80
* 4.186 * 200 = 66,976 J

Heat to evaporate the 200 g of water = 200 x 2257 J = 451,400 J

Total heat lost to uncondensed moisture = 451,400 + 66.976 = 518,376 J

Net heat content of the fuel = 18 MJ - 0.518376 = 17.481 MJ/kg Net

Sam Baldwin and Piet Visser and the American heating and refrigeration
engineers and I agree that the LHV deduction as performed above is not
realistic but it is widely used for convenience. We stovers think the value
should be about 1.38 MJ not 1.32. The difference of 0.6 MJ is the
consideration of the heat that would be released when cooling the condensed
moisture from 100° down to 20°. As we are not in a position to inflict real
engineering into the stove calculations (save for our internal use) it
remains as it is. ASHRAE uses the temperature of 150° C as the lowest useful
heat available from a gas stream and regards all heat contained in that to
be a loss. Thus their LHV is lower than ‘ours’ by 50 degrees worth of
heating the gases and water vapour. They do that because if gives accurate
mathematical modelling of heat exchangers which as you probably can tell by
now, is not delivered by commonly used stove formulas.

Oh well, who cares

To review: express the chemical contents as a % of the mass as received
including moisture.

Calculate the moisture in the fuel by drying it.

Get the HHV from a bomb calorimeter

Get the ultimate analysis from whatever you have available (XRD/XRF/GC etc)

Calculate the ultimate analysis excluding moisture

Calculate the LHV based on the HHV and the Hydrogen content

Calculate the Net heat available by compensating for moisture.

Bob’s your Uncle.

The fuel should have two values reported for general use: the net heat
content (AR) and the LHV (AD).

Please note that coal people do not ‘dry’ their samples, they ‘air dry’ them
to an unknown moisture level (but above 0%) and call the remaining moisture
and combustion moisture ‘inherent moisture’. This IM varies from sample to
sample even within the same small pile because the air drying effectiveness
depends on the relative humidity in the lab. As a result the published LHV’s
for coal are usually wrong.

May yours always be correct.

Regards

Crispin

E&OE, as they say.

-----Original Message-----
From: stoves-bounces at lists.bioenergylists.org
[mailto:stoves-bounces at lists.bioenergylists.org] On Behalf Of Frank Shields
Sent: Wednesday, December 28, 2011 7:48 PM
To: 'Discussion of biomass cooking stoves'
Subject: Re: [Stoves] Calculation help

Thanks for adding this to the server as I had planned. Not sure what
happened but I see I did use the word 'compost' a few times when I meant
'biomass'. Guess I should have had that second cup of java. 

-----Original Message-----

From: stoves-bounces at lists.bioenergylists.org
[mailto:stoves-bounces at lists.bioenergylists.org] On Behalf Of
ajheggie at gmail.com

Sent: Wednesday, December 28, 2011 2:54 PM

To: Discussion of biomass cooking stoves

Subject: [Stoves] Calculation help

Hi all please find below a post from Frank Shields that didn't propagate to
the list for some reason. My uneducated way of addressing this would be to
consider what would the outcome be with a fossil fuel with negligible water.
So I would calculated on the oven dry mass and then subtract the latent heat
of the water from the hydrogen in the fuel at the temperature it was
discarded as vapour. This would give the LHV of the oven dry sample. I tend
to look on un chemically bound water a bit like salts of hydration and would
then deduct their latent heat from the LHV to give me the actual available
heat.

Mind I'm not looking at losses to the 4th decimal place so I think in
approximations and genrally use a figure of -2.7MJ for water discarded up
the flue. 

I'm open to argument for a more precise approach.

Also there is a potential ambiguity if we don't specify the basis on which
our moisture content is calculated, for this purpose I feel it should be on
the gross wet weight of the sample.

AJH

********************

Dear Stovers,

A few questions and checks on some calculations if you don't mind.

1)      Should the HHV value be reported as if the biomass is 100 grams of
oven dried material or reported on the dry fraction of a 100 g sample
received? 

2)      Having an analysis: Dry wt.

Percent dry wt.

N = 1.5

C = 42.0

H = 6.2

S = 0.03

Ash = 1.6

Water in receiving sample

Water = 17.6%

The HHV = 20.37 based on the dry sample and 16.79 based on the weight of an
as-Received sample and the formula I am using.

Now questions about determining the LHV:

17.6 g water per 100 grams wet compost or 0.176 g water per gram wet
compost.

It takes 0.0552 kj to take the temperature from 25c to 100c

It takes 0.3964 kj to 100c water to 100c vapor

It takes 0.0907 kj to 100c vapor to 400c vapor

Total = 0.5423 kj lost from the water moisture

Then we have the water produced from the hydrogen in the biomass. 

If we start with 6.2 % hydrogen and find the char has retained 2.2 so 4.0%
hydrogen has converted to water vapor.

That is 0.04 g hydrogen per gram biomass

That is 0.18 g water vapor starting at 450c per gram biomass 

Total 0.5901 kj energy lost in the vapor at 450c

The HHV = 20.37 kj/g on a dry sample

The HHV = 16.79 kj/g on the dry fraction of an as-received biomass sample

The LLV = 16.79-(0.5423 + 0.5901) =15.66 kj/g as-Received biomass

Thanks 

Frank

Frank Shields

42 Hangar Way

Watsonville,  CA  95076

(831) 724-5244 tel

(831) 724-3188 fax

frank at bioCharlab.com

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