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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link=blue vlink=purple><div class=WordSection1><p class=MsoPlainText>Dear Frank<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Carbon<o:p></o:p></p><p class=MsoPlainText>Nitrogen<o:p></o:p></p><p class=MsoPlainText>Hydrogen<o:p></o:p></p><p class=MsoPlainText>Ash<o:p></o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>As Received (AR) MASHCON<o:p></o:p></p><p class=MsoPlainText>Moisture<o:p></o:p></p><p class=MsoPlainText>Ash<o:p></o:p></p><p class=MsoPlainText>Sulphur<o:p></o:p></p><p class=MsoPlainText>Hydrogen<o:p></o:p></p><p class=MsoPlainText>Carbon<o:p></o:p></p><p class=MsoPlainText>Oxygen<o:p></o:p></p><p class=MsoPlainText>Nitrogen<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>That should add up to 100%<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Dry it and weigh again (dry at 5 Deg C above ambient boiling point for 24 hours).<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>[Beware: some labs do not do this, particularly with coal. Never assume anything. Always check to see that they did.]<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>After Drying (AD)<o:p></o:p></p><p class=MsoPlainText>Ash<o:p></o:p></p><p class=MsoPlainText>Sulphur<o:p></o:p></p><p class=MsoPlainText>Hydrogen<o:p></o:p></p><p class=MsoPlainText>Carbon<o:p></o:p></p><p class=MsoPlainText>Oxygen<o:p></o:p></p><p class=MsoPlainText>Nitrogen<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>That should add up to 100% with larger values that before because the moisture is no longer diluting the samples.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Moisture compensation:<o:p></o:p></p><p class=MsoPlainText>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).<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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 (H<sub>2</sub>O). 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 H<sub>2</sub> weighs 2 g so you can quickly calculate the mass of water that will be produced H<sub>2</sub>O 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 <i>not</i> recovered by condensing this water vapour. It is 540 g x 2257 J = 1.22 MJ per kg of that fuel burned.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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%.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>There are four commonly used standard temperatures:<o:p></o:p></p><p class=MsoPlainText>0° C for chemical and physical processes. Standard Temperature and Pressure (STP)<o:p></o:p></p><p class=MsoPlainText>15° C for aircraft (ISA and ICAO)<o:p></o:p></p><p class=MsoPlainText>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)<o:p></o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><a href="http://www.engineeringtoolbox.com/stp-standard-ntp-normal-air-d_772.html">http://www.engineeringtoolbox.com/stp-standard-ntp-normal-air-d_772.html</a><o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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°.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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:<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Fuel mass 1000 g<o:p></o:p></p><p class=MsoPlainText>HHV 19.25 MJ/kg<o:p></o:p></p><p class=MsoPlainText>LHV 18 MJ/kg<o:p></o:p></p><p class=MsoPlainText>Moisture content 20%<o:p></o:p></p><p class=MsoPlainText>Dry fuel mass in the 1 kg = 1000 - (0.2 * 1000) = 800 g<o:p></o:p></p><p class=MsoPlainText>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)<o:p></o:p></p><p class=MsoPlainText>Water mass in the fuel = 0.2 * 1000 = 200 g<o:p></o:p></p><p class=MsoPlainText>Heat to boiling point Normal Temp to Normal Boiling Point (20°-100° C) = 80 * 4.186 * 200 = 66,976 J<o:p></o:p></p><p class=MsoPlainText>Heat to evaporate the 200 g of water = 200 x 2257 J = 451,400 J<o:p></o:p></p><p class=MsoPlainText>Total heat lost to uncondensed moisture = 451,400 + 66.976 = 518,376 J<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Net heat content of the fuel = 18 MJ - 0.518376 = <u>17.481 MJ/kg Net</u><o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Oh well, who cares…<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>To review: express the chemical contents as a % of the mass as received including moisture.<o:p></o:p></p><p class=MsoPlainText>Calculate the moisture in the fuel by drying it.<o:p></o:p></p><p class=MsoPlainText>Get the HHV from a bomb calorimeter<o:p></o:p></p><p class=MsoPlainText>Get the ultimate analysis from whatever you have available (XRD/XRF/GC etc)<o:p></o:p></p><p class=MsoPlainText>Calculate the ultimate analysis excluding moisture<o:p></o:p></p><p class=MsoPlainText>Calculate the LHV based on the HHV and the Hydrogen content<o:p></o:p></p><p class=MsoPlainText>Calculate the Net heat available by compensating for moisture.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Bob’s your Uncle.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>The fuel should have two values reported for general use: the net heat content (AR) and the LHV (AD).<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>May yours always be correct.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Regards<o:p></o:p></p><p class=MsoPlainText>Crispin<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>E&OE, as they say.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span lang=EN-US>-----Original Message-----<br>From: stoves-bounces@lists.bioenergylists.org [mailto:stoves-bounces@lists.bioenergylists.org] On Behalf Of Frank Shields<br>Sent: Wednesday, December 28, 2011 7:48 PM<br>To: 'Discussion of biomass cooking stoves'<br>Subject: Re: [Stoves] Calculation help</span><o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>-----Original Message-----<o:p></o:p></p><p class=MsoPlainText>From: stoves-bounces@lists.bioenergylists.org [mailto:stoves-bounces@lists.bioenergylists.org] On Behalf Of ajheggie@gmail.com<o:p></o:p></p><p class=MsoPlainText>Sent: Wednesday, December 28, 2011 2:54 PM<o:p></o:p></p><p class=MsoPlainText>To: Discussion of biomass cooking stoves<o:p></o:p></p><p class=MsoPlainText>Subject: [Stoves] Calculation help<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>I'm open to argument for a more precise approach.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>AJH<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>********************<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Dear Stovers,<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>A few questions and checks on some calculations if you don't mind.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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? <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>2) Having an analysis: Dry wt.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Percent dry wt.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>N = 1.5<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>C = 42.0<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>H = 6.2<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>S = 0.03<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Ash = 1.6<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Water in receiving sample<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Water = 17.6%<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>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.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Now questions about determining the LHV:<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>17.6 g water per 100 grams wet compost or 0.176 g water per gram wet compost.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>It takes 0.0552 kj to take the temperature from 25c to 100c<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>It takes 0.3964 kj to 100c water to 100c vapor<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>It takes 0.0907 kj to 100c vapor to 400c vapor<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Total = 0.5423 kj lost from the water moisture<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Then we have the water produced from the hydrogen in the biomass. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>If we start with 6.2 % hydrogen and find the char has retained 2.2 so 4.0% hydrogen has converted to water vapor.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>That is 0.04 g hydrogen per gram biomass<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>That is 0.18 g water vapor starting at 450c per gram biomass <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Total 0.5901 kj energy lost in the vapor at 450c<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>The HHV = 20.37 kj/g on a dry sample<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>The HHV = 16.79 kj/g on the dry fraction of an as-received biomass sample<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>The LLV = 16.79-(0.5423 + 0.5901) =15.66 kj/g as-Received biomass<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Thanks <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Frank<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Frank Shields<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>42 Hangar Way<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Watsonville, CA 95076<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>(831) 724-5244 tel<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>(831) 724-3188 fax<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>frank@bioCharlab.com<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>_______________________________________________<o:p></o:p></p><p class=MsoPlainText>Stoves mailing list<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>to Send a Message to the list, use the email address stoves@lists.bioenergylists.org<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>to UNSUBSCRIBE or Change your List Settings use the web page http://lists.bioenergylists.org/mailman/listinfo/stoves_lists.bioenergylists.org<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>for more Biomass Cooking Stoves, News and Information see our web site:<o:p></o:p></p><p class=MsoPlainText>http://www.bioenergylists.org/<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>_______________________________________________<o:p></o:p></p><p class=MsoPlainText>Stoves mailing list<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>to Send a Message to the list, use the email address<o:p></o:p></p><p class=MsoPlainText>stoves@lists.bioenergylists.org<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>to UNSUBSCRIBE or Change your List Settings use the web page<o:p></o:p></p><p 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