元素分析与热值计算
The direct pyrolysis and catalytic pyrolysis of Nannochloropsis sp.residue for renewable bio-oils
Pan Pan a ,Changwei Hu a,*,Wenyan Yang a ,Yuesong Li a ,Linlin Dong a ,Liangfang Zhu a ,Dongmei Tong a ,Renwei Qing b ,Yong Fan b
a Key Laboratory of Green Chemistry and Technology,Ministry of Education,College of Chemistry,Sichuan University,Chengdu,Sichuan 610064,China b
School of Life Science,Sichuan University,Chengdu,Sichuan 610064,China
a r t i c l e i n f o Article history:
Received 14September 2009
Received in revised form 17January 2010Accepted 19January 2010
Available online 12February 2010Keywords:Pyrolysis Bio-oils Microalgae
Nannochloropsis sp.HZSM-5catalyst
a b s t r a c t
Nannochloropsis sp.(a kind of green microalga)residue was pyrolyzed without catalyst or with different amount of HZSM-5catalyst in a ?xed bed reactor in nitrogen ?ow.The effects of pyrolysis parameters such as temperature and catalyst-to-material ratio on product yields were studied.The bio-oils obtained were analyzed by elemental,GC–MS and FTIR analysis.The results indicated that the bio-oils from cat-alytic pyrolysis of Nannochloropsis sp.residue (BOCP)had lower oxygen content (19.5wt.%)and higher heating-value (32.7MJ kg à1)than those obtained from direct pyrolysis (BODP)which had an oxygen con-tent of 30.1wt.%and heating-value of 24.6MJ kg à1.The BODP mainly consisted of long carbon chain compounds with various terminal groups (LCTG),while the BOCP mainly consisted of aromatic hydrocar-bons.These properties of bio-oils demonstrated that the Nannochloropsis sp.residue can be used as a renewable energy resource and chemical feedstock.
ó2010Elsevier Ltd.All rights reserved.
1.Introduction
Presently,the widely used energy sources are still limited to the conventional fossils such as coal,petroleum and natural gas,which are depleting,and their usage causes serious environmental prob-lems.The exploitation of renewable and environmentally friendly energy resources is urgent and signi?cant.Biomass,mainly includ-ing crops,forestry products and marine products,is widespread on the earth.With the exacerbation of energy crisis and environmen-tal deterioration,biomass as an environmentally friendly and renewable energy resource has attracted more and more interests (Cercel,2002;Goyal et al.,2008).
Biomass can be converted into fuel products by biological (fermentation and anaerobic digestion)or thermochemical (gasi?-cation,liquefaction)route.Among these conversion processes,pyrolysis is considered to be an effective technology,by which bio-mass can be converted to valuable bio-oils,char and gaseous prod-ucts (Mohan et al.,2006).Especially,pyrolysis bio-oils are very attractive because of their high energy density and convenience in usage,storage and transport (Islam et al.,2004).Though numer-ous progresses on producing pyrolysis bio-oils have been reported with lignocellulosic biomass as raw materials (Karaosmanoglu et al.,1999;Mohan et al.,2006;Muller-Hagedorn and Bockhorn,2007;Putun et al.,2004),the bio-oils obtained can not be used
directly as fuel due to their high oxygen content,high viscosity,high corrosiveness and relative instability,and need to be up-graded by complicated process (Czernik and Bridgwater,2004;Zhang et al.,2007).The poor quality of bio-oils is mainly attributed to the chemical components of lignocellulosic biomass (cellulose,hemicellulose and lignin),thus exploiting more appropriate bio-mass is necessary and valuable.
Microalga is a widely distributed low-grade water https://www.wendangku.net/doc/c718270259.html,-pared with lignocellulosic biomass,microalgae as energy resource have the following advantages:(1)microalgae have higher photo-synthetic ef?ciency and higher biomass production (Peng et al.,2001;Schenk et al.,2008);(2)microalgae can be cultivated in an aquatic medium,and do not occupy arable land (Rodol?et al.,2009);(3)the chemical composition of microalgae can be modu-lated easily by varying cultivation conditions,and high lipid con-tent can be obtained (Rodol?et al.,2009);(4)microalgae can utilize the salt and organic matter derived from waste water as fer-tilizers (Schenk et al.,2008).In addition,microalgae can effectively reduce greenhouse gas concentration in the atmosphere by their high capability in ?xing carbon dioxide (Chiu et al.,2009;Kishimoto et al.,1994).So the exploitation and utilization of mic-roalgae for fuel production can gain both economic and environ-mental bene?ts.However,for the large-scale usage of microalgae,there are still some fundamental challenges to be dealt with,such as the high cost of microalgae cultivation and its collec-tion,as well as the ef?ciency in microalgae utilization (Li et al.,2007;Liliana et al.,2009).
0960-8524/$-see front matter ó2010Elsevier Ltd.All rights reserved.doi:10.1016/j.biortech.2010.01.070
*Corresponding author.Tel.:+862888835525;fax:+862885411105.E-mail addresses:chwehu@https://www.wendangku.net/doc/c718270259.html, ,gchem@https://www.wendangku.net/doc/c718270259.html, (C.Hu).Bioresource Technology 101(2010)
4593–4599
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Generally,microalgae contain varying amounts of lipids,sugars, proteins and pigments etc.Presently,the attentions about the con-version of microalgae to fuel products mostly focus on the lipid which can be used to produce high-quality bio-diesel by the con-ventional esteri?cation and transesteri?cation(Chisti,2007;Li et al.,2007;Miao and Wu,2006).However,after lipid is extracted from microalgae cells(Li et al.,2007),numerous residues of micro-algae which mainly contain almost all soluble polysaccharide,pro-tein and some residual lipid are thrown away or sometimes used as animal feed.(Liliana et al.,2009)With the development of technol-ogy in producing low-cost fuel products derived from microalgae, the utilization of algal residues will become signi?cant.
Nannochloropsis sp.has a high lipid content(about46%,deter-mined with Bligh–Dyer method,using methyl alcohol–chloroform mixture(2/1,v/v)as the extraction solvent),and is considered to be a promising green microalgae for fuel products(Gouveia and Oliveira,2009;Rebolloso-Fuentes et al.,2001).In the present work, the direct pyrolysis of Nannochloropsis sp.residue was studied.
In addition,based on numerous researches on catalytic pyroly-sis of lignocellulosic biomass,catalytic pyrolysis by using molecu-lar sieves as catalysts can not only effectively upgrade the quality of bio-oils but also adjust the components of bio-oils to meet dif-ferent demands(Adjaye and Bakhshi,1995;Demiral and S?ens?z, 2008;Vitolo et al.,1999;Williams and Horne,1995a;Williams and Nugranad,2000).Molecular sieves are well-known heteroge-neous catalysts used in petroleum industry,and have been suc-cessfully applied in the interconversion of hydrocarbons,such as alkylation,isomerization,aromatization and thermal cracking reactions.Among various molecular sieve catalysts,HZSM-5has gained great favors.ZSM-5is an aluminosilicate zeolite with a high silica-to-alumina ratio and strong acidity.Its structure is based on channels with insecting tunnels.The substitution of H+for Na+ made the acidity of zeolite stronger,which facilitates the conver-sion of hydrocarbons,such as thermal cracking(Pujadóa et al., 1992).At present,most studies are aiming at the pyrolysis or cat-alyzed pyrolysis of lignocellulosic biomass to obtained bio-oil(De-miral and S?ens?z,2008).And there is no literature available concerning the direct pyrolysis or catalyzed pyrolysis of algal res-idue to produce bio-oils.This paper will give some primary results.
2.Experimental
2.1.Material and catalyst
Nannochloropsis sp.was cultivated in laboratory.The culture medium was arti?cial seawater concocted by ourselves(using f/2 medium)(Guillard,1975).Irradiance provided by?uorescent lamps was constant at70l mol mà2sà1and the ambiance temper-ature was maintained at23±0.2°C.Cells of Nannochloropsis sp. were collected by centrifugation and washed with distilled water, and then dried in a vacuum desiccator at65°C for24h.The consti-tution of dry Nannochloropsis sp.was listed in Table1.
Lipid in cells of Nannochloropsis sp.was extracted by the mixed solvent of ethylether and petroleum ether(1/2,v/v),and then used for the production of bio-diesel by esteri?cation and transesteri?-cation.After residual solvent in algal residue was removed by vac-uum distillation,the algal residue which occupied approximately 70wt.%in cells dry weight was used as pyrolysis material.The proximate and ultimate analysis of Nannochloropsis sp.residue was shown in Table1.Since metal salts contained in biomass ex-erted a signi?cant impact on the pyrolysis process(Ross et al., 2008;Williams and Horne,1994a),the contents of metals in the sample were determined by Inductively Coupled Plasma Spec-trometry(ICP).The result was also listed in Table1.
The HZSM-5used in the catalytic pyrolysis of Nannochloropsis sp.residue was purchased from Nankai University.It had elliptical pore of0.56nm diameter,with a silica–alumina ratio of about25 and pore volume of0.19cm3gà1.The surface area was 347.9m2gà1.Before use,HZSM-5was activated in a muf?e furnace at400°C under air atmosphere for4h,with a heating rate of 10°C minà1.
2.2.Experimental apparatus
A?xed bed reactor designed by ourselves was used as experi-mental apparatus.The schematic diagram of the reactor was shown in Fig.1.The outer tube was35mm diameter?600mm height.Nannochloropsis sp.residue(1g)and catalyst(different amount)were mixed and packed in the inner vessel(25mm diam-eter?120mm height),and were heated by an electric furnace. Temperature was controlled by a temperature controller(SKW-100).Nitrogen as a carrier gas was ventilated to the reactor from the inlet located in the top of the reactor and its?ow rate was con-trolled by a?ow meter.A condenser(ice trap)connected at the exit of the reactor was used to collect the liquid products.
2.3.Experimental procedure
The direct pyrolysis of Nannochloropsis sp.residue was per-formed in the?xed bed reactor mentioned above.The material was heated by an electric furnace with a heating rate of 10°C minà1from room temperature to the?nal temperature,and then kept for2h at the?nal temperature.The volatiles produced in the pyrolysis process were swept out by carrier gas with a?ow rate of30ml minà1.The condensable components in volatiles formed liquid products which were collected in the condenser, and the incondensable components in volatiles formed gaseous products which were collected in a gas bag.
In the catalytic pyrolysis experiment of Nannochloropsis sp.res-idue,the same?xed bed reactor was used.Catalyst and material were mixed directly at the ratio from0.2/1to1/1,and then placed in the inner vessel.Other experiment parameters were the same as those in the direct pyrolysis experiment.
Table1
The proximate,ultimate and metal analysis of Nannochloropsis sp.residue a.
Proximate analysis(wt.%)Ultimate analysis(wt.%)Metal analysis(ppm)
Moisture7.0C44.10Ca5503 Volatile63.5H7.09K1660 Fixed C19.6N 5.51Na893 Ash9.9O b33.40Mg7915
Heating-value c20.7MJ kgà1
HHV(MJ kgà1)=(3.55C2à232Cà2230H+51.2C?H+131N+20600)?10à3.
a Analyzed as received,after lipid in cells of Nannochloropsis sp.was extracted according to the depiction in the text.
b By difference,O(%)=100-C–H–N-Ash sed as received.
c Heating-value was calculate
d by th
e following formula(Friedl et al.,2005).
4594P.Pan et al./Bioresource Technology101(2010)4593–4599
The total yield of liquid products was calculated by the weight difference of the condenser before and after reaction.Oily phase (de?ned as bio-oils)and aqueous phase of liquid product were sep-arated and weighted respectively.Mass of char was determined by weight,and the weight of catalyst needed to be subtracted for the catalytic process.Coke on the catalyst was reckoned to the char be-cause of the dif?culty of separating coked catalyst powder from reaction residue in this study.And the gas yield was calculated by overall mass balance.The yields of various products were calcu-lated as follow:
Y L?eW L=W MT?100e%T;Y A?eW A=W MT?100e%T;
Y o?eW o=W MT?100e%T?Y LàY A;
Y C?eW C=W MT?100e%T;Y G?1àeY LtY CT
where Y L denoted the yield of liquid product;W L,the weight of li-quid product;W M,the weight of material;Y A,the yield of aqueous product;W A,the weight of aqueous product;Y o,the yield of bio-oil; W o,the weight of bio-oil;Y C,the yield of char;Y G,the yield of gas-eous product.
In the present work,experiments were divided into three series. The?rst series of experiments was performed to investigate the ef-fect of temperature on product yields in the direct pyrolysis pro-cess.The?nal pyrolysis temperature was maintained at300,350, 400,450and500°C,respectively.The second series of experiments was performed to investigate the effect of catalyst-to-material ra-tio(0/1,0.2/1,0.4/1,0.6/1,0.8/1and1/1,wt/wt)on product yields. In this series of experiments,the?nal temperature of400°C was chosen.The third series of experiments was carried out to deter-mine the effect of temperature(from300to500°C)on product yields in the catalytic pyrolysis process.The catalyst-to-material ratio was kept at1/1in all experiments in the third series.
2.4.Characterization
Gaseous product collected in the gas bag was analyzed by two different gas chromatography analyzers.CO,CO2and H2were determined by GC7890T using a5A molecular sieve column with a thermal conductivity detector(TCD)and nitrogen as the carrier gas(column temperature:80°C;inlet temperature:90°C;detector temperature:120°C).Gaseous hydrocarbons were analyzed by SP-6800A in temperature programming using a column packed with porapak Q with a thermal conductivity detector(TCD)and nitrogen as the carrier gas(column temperature:column was heated to 40°C for2min then to90°C with the heating rate of25°C minà1, and kept for15min at90°C;inlet temperature:70°C;detector temperature:140°C).An Elementar Vario El IV analyzer was used to determine the elemental composition of bio-oils.The infrared spectra analysis of bio-oils was conducted using a NEXUS670-FTIR infrared spectrometric analyzer.The compounds in bio-oils were determined by a gas chromatography–mass spectrometry(Agilent 5973GC–MS:column:DB-5;inlet temperature:250°C;detector temperature:230°C;helium?ow:16ml minà1;mode:split).
3.Results and discussion
3.1.The effect of temperature on the direct pyrolysis of Nannochloropsis sp.residue
The yields of various products obtained from the direct pyroly-sis of Nannochloropsis sp.residue as a function of?nal pyrolysis temperature were presented in Fig.2.As can be seen,the yield of liquid product was low at300°C,and then markedly increased with increasing temperature,while the highest yield of47.6wt.% was obtained at400°C.Above400°C,the yield of liquid product gradually decreased.The liquid product obtained in our experi-ments was a mixture of an oily phase(upper phase)and an aque-ous phase(lower phase),which separated automatically.The change of the bio-oil yield in upper phase had a similar trend to that of liquid product with increasing temperature.And the maxi-mum bio-oil yield of31.1wt.%was obtained at400°C.The yield of char markedly decreased from45.3to24.2wt.%,while the yields of aqueous and gaseous product gradually rose from13.8to20.0wt.% and from18.9to33.5wt.%,respectively,as the pyrolysis tempera-ture rose from300to500°C.Table2showed the yields of various gaseous products.It could be seen that the gaseous product con-sisted large amount of CO2.Among various gaseous hydrocarbons, the content of CH4was the highest.The total yield of gaseous hydrocarbons markedly increased from0.12to5.12wt.%as the pyrolysis temperature increased from300to500°C.
At lower temperature,the yield of oil/gas was low.This might be caused by the low conversion of raw material on the one hand, and additionally,various components in algal residue could not be vaporized and pyrolyzed adequately.Thus the yield of‘‘char”was relatively higher.With increasing temperature,the volatiles de-rived from the pyrolysis of materials gradually increased.These
P.Pan et al./Bioresource Technology101(2010)4593–45994595
volatiles could undergo various reactions like dehydration and thermal cracking at higher temperature,and generated polar and nonpolar organic compounds with low molecule weight such as water(Mohan et al.,2006;Putun et al.,2004).Consequently,the yields of aqueous product,gaseous product and bio-oil gradually increased as the temperature increased from300to400°C.How-ever,once the pyrolysis temperature exceeded400°C,a part of vol-atiles which should form bio-oils at lower pyrolysis temperature could be cracked further into more incondensable gaseous prod-ucts,and then caused the decrease of bio-oils yield.
These results indicated that the optimal temperature range un-der the present conditions for the direct pyrolysis of Nannochlorop-sis sp.residue was350–450°C,which was much lower than that for the direct pyrolysis of lignocellulosic biomass(commonly 450–650°C)(Karaosmanoglu et al.,1999;Qi et al.,2006).It could be mainly attributed to the following three factors.Firstly,Nanno-chloropsis sp.residue mainly contains lipid,soluble polysaccharide and protein,which can be pyrolyzed more easily than lignocellu-losic feedstocks(Mohan et al.,2006).Secondly,the cell structure of Nannochloropsis sp.is relatively incompact compared to ligno-cellulosic biomass,and may be destroyed further in the extraction process.Thirdly,Nannochloropsis sp.residue contains more inor-ganic ions than some typical lignocellulosic biomass such as wood, as demonstrated by the ICP analysis results shown in Table1.The higher metal salt content can also lower the pyrolysis temperature because of its notable catalytic effect(Ross et al.,2008;Williams and Horne,1994a).
3.2.The effect of the catalyst-to-material ratio on the catalytic pyrolysis of Nannochloropsis sp.residue
Based on the results of the?rst series of experiments,400°C was chosen as pyrolysis temperature in this series of experiments. The yields of various products at different catalyst-to-material ra-tio were given in Fig.3.As can be seen,the yield of liquid product had a notable decrease with a simultaneous increase of char from 28.01%to34.56%,as catalyst HZSM-5was added into reaction material.It can be explained as the coking trend over molecular sieves(Vitolo et al.,1999).When the volatile intermediates formed in the primary pyrolysis process passed over the catalyst surface or went deep into its micropores,the high acidity of catalyst might enhance various secondarily reactions of the intermediates such as repolymerization and aromatization.As a result,bio-oil reduced. Cokes were remained on the catalyst surface and in its micropores, with more gas and aqueous products produced.
When the catalyst-to-material ratio increased from0.2/1to1/1, the yield of liquid product had a slight change.The yield of bio-oil gradually decreased from25.1to20.7wt.%,while the yields of aqueous product and gaseous product gradually increased from 15.8to24.9wt.%and from24.5to29.8wt.%,respectively.The yields of gaseous products were given in Table3.It was worth not-ing that the total yield of gaseous hydrocarbons increased with increasing catalyst-to-material ratio.At400°C,the total yield of gaseous hydrocarbons,which was1.80wt.%in the direct pyrolysis process,increased to3.08wt.%when catalysts-to-material ratio was0.2/1,and further increased to 5.78wt.%when catalyst-to-material ratio was1/1.We gathered that the increase of catalyst-to-material ratio enhanced the direct contact between heterogenous catalyst and solid materials,and promoted the ther-mal cracking and secondarily reactions of reactants to produce more gas and aqueous products.
3.3.The effect of temperature on the catalytic pyrolysis of Nannochloropsis sp.residue
The catalyst-to-material ratio was chosen to be ratio1/1in all experiments of this series.The product yields obtained at different temperature were shown in Fig.4.It can be seen that the yield of liquid product sharply increased with increasing temperature when pyrolysis temperature was below400°C.Above400°C,the yield of liquid product had only a slight decrease.The yield of bio-oil rose from11.2to19.7wt.%as the temperature rose from 300to400°C,and then decreased to16.9wt.%at500°C.The yield of aqueous product continuously rose with increasing https://www.wendangku.net/doc/c718270259.html,pared with the?rst series of experiments,the yield of bio-oil was lower,whereas the yield of aqueous product was high-er in the presence of HZSM-5catalyst at all pyrolysis temperatures.
The yield of char markedly decreased as the pyrolysis tempera-ture https://www.wendangku.net/doc/c718270259.html,pared with the yield of char obtained in the di-rect pyrolysis process,the yield of char obtained in the catalytic pyrolysis process was higher when temperature was below 400°C,but was lower when temperature was above400°C.This phenomenon was not only resulted from more decomposition of char at higher temperature but also related to the decrease of coke over catalyst surface.It can be speculated that coke on catalyst sur-face could be easily formed at relatively low temperature,and partly decomposed at higher temperature over400°C when Nan-nochloropsis sp.residue is used as pyrolysis material.This tendency was similar to the results reported by Williams et al.that is,the coke formation on catalyst ZSM-5decreased with temperature increasing from400to600°C when rice husks was used for pyro-lysis(Williams and Nugranad,2000).
The yield of gaseous product sharply increased with increasing temperature.The detailed components and their yields in gaseous
Table2
The yields(wt.%)of various gaseous products at different temperature in the direct pyrolysis process.
Temperature H2CO2CO CH4C2H4C2H6C n a
300°C0.0514.49 4.280.12nd b nd0.12 350°C0.1518.56 3.800.93nd tr c0.93 400°C0.2318.73 3.680.97tr0.83 1.80 450°C0.4521.40 3.66 2.75tr 1.79 4.54 500°C0.5225.01 2.86 3.440.840.84 5.12
a The total yields of gaseous hydrocarbons.
b Not detected.
c Trace amount.
4596P.Pan et al./Bioresource Technology101(2010)4593–4599
product were given in Table4.As can be seen,the total yield of gas-eous hydrocarbons obviously increased with increasing tempera-ture.Moreover,the yield of gaseous hydrocarbons obtained in the catalytic pyrolysis process was higher than that obtained in the direct pyrolysis process at all pyrolysis temperatures.
3.4.Analysis of the bio-oils
The bio-oils obtained in the direct pyrolysis process at400°C (BODP)and in the catalytic pyrolysis process with the catalyst-to-material of1/1at400°C(BOCP)were analyzed comparatively by elemental,FTIR and GC–MS analysis.
Table5showed the characteristics of the BODP and BOCP in comparison with wood pyrolysis bio-oils and heavy fuel oil.It can be seen from Table5that the contents of carbon and hydrogen in the BODP were56.13%and7.13%,respectively,which were much higher than those in original materials(44.10%and7.09%, respectively).After HZSM-5catalyst was introduced to pyrolysis process,the contents of carbon and hydrogen in the BOCP in-creased to65.21%and9.83%,respectively.Meanwhile,the oxygen concentration decreased from33.4%in original materials,to 30.09%in the BODP,and further to19.53%in the BOCP.The results demonstrated that the catalytic pyrolysis using HZSM-5as catalyst could ef?ciently decrease the oxygen content of bio-oils,and in-crease the heating-value of bio-oils,though the yield of BOCP was lower than that of BODP at the same reaction conditions.
Nitrogen content of bio-oils could derive from the protein and chlorophyll of algal residues,and was generally higher in those bio-oils obtained from lignocellulosic biomass.In addition,the re-sults implied that catalytic pyrolysis by using HZSM-5as catalyst could not reduce the nitrogen content under present condition.
According to the elemental analysis,bio-oils obtained from the Nannochloropsis sp.residue contained more carbon,more hydro-gen,less oxygen,and possessed higher heating value,compared to the bio-oils obtained from lignocellulosic biomass.And the for-mer bio-oil would be more suitable to be used as engine fuels.
The compounds identi?ed in the bio-oils were divided into four kinds–chain hydrocarbons,aromatic compounds,long carbon chain compounds with various terminal functional groups(LCTG) and others.Table6showed the total relative contents of the four kinds of compounds.The chain hydrocarbons mainly contained dodecane and3,7,11,15-tetremethyl-2-hexadecene.The aromatic compounds mainly consisted of phenol,alkylated benzenes and naphthalene such as toluene,p-xylene and2-methyl-naphthalene. The LCTG largely consisted of fatty acids and their methyl ester, with long carbon chain amide and nitrile.Other products were mainly the oxygenous and nitrogenous compounds such as dian-hydro mannitol,indole and pyrrole.
The results of GC–MS analysis showed that there were great dif-ferences between the BODP and the BOCP in chemical component. The relative content of the aromatic compounds in the BOCP reached48.60%,and was much higher than that in the BODP (10.89%).A small quantity of aromatic compounds like phenol,4-methyl-phenol and toluene,were detected with low concentra-tions in the BODP,whereas numerous aromatic hydrocarbons, which were various alkylated benzene and alkylated naphthalene, were identi?ed with high concentrations in the BOCP.Among the compounds in the BOCP,the relative content of p-xylene was the highest and reached12.24%.However,the relative content of LCTG in the BOCP markedly decreased compared to the BODP.Most of LCTG,which were present in the BODP with high relative concen-trations,such as pentadecanenitrile and hexadecanenitrile,were not detected in the BOCP.
The FTIR analysis of the BODP and BOCP were also in consistent with the results mentioned above(see to the Supplementary mate-rial).It was proved that BOCP contained more alkyl functional groups(especially substituted aromatic compounds)and less oxy-genated functional groups than BODP.
Table3
The yields(wt.%)of various gaseous products at different catalyst-to-material ratios(wt/wt)at400°C.
Catalyst-to-material ratio H2CO2CO CH4C2H4C2H6C3H6C n a
0/10.2318.73 3.680.97tr b0.83nd c 1.80
0.2/10.2618.25 2.94 2.16tr0.92nd 3.08
0.4/10.1818.88 2.38 2.020.570.75tr 3.34
0.6/10.1419.97 1.39 1.95 1.05 1.050.23 4.28
0.8/10.1620.76 1.17 1.850.92 1.100.51 4.65
1/10.1420.400.72 2.38 1.42 1.230.75 5.78
a The total yields of gaseous hydrocarbons.
b Not detected.
c Trace amount.
Table4
The yields(wt.%)of various gaseous products at different temperature in the catalytic
pyrolysis process.
Temperature H2CO2CO CH4C2H4C2H6C3H6C n a
300°C0.059.55 2.580.13tr b tr tr0.13
350°C0.0814.78 4.370.820.200.21tr 1.23
400°C0.1420.400.72 2.38 1.42 1.230.75 5.78
450°C0.1822.70 1.56 2.67 1.250.93 1.04 5.89
500°C0.1727.470.63 3.50 1.99 1.210.447.14
a The total yields of gaseous hydrocarbons.
b Trace amount.
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3.5.Discussions
The bio-oils obtained from the direct pyrolysis of lignocellulosic biomass were multicomponent mixtures consisted of hundreds of compounds such as acids,aldehydes,ketones,alcohols,phenols and their derivatives and so on(Karaosmanoglu et al.,1999;Qi et al.,2006).Almost all these compounds were with low heating-value,high acidity and instability due to the oxygenous functional groups.The BODP from microalgae residue were absolutely differ-ent from the bio-oil from lignocellulosic biomass pyrolysis in chemical composition.The most attractive chain hydrocarbons, nearly absent in lignocellulosic biomass pyrolysis bio-oils,were detected with high concentration(12.25%)in the BODP.And alde-hydes,ketones and phenolic compounds which were present with high concentrations in lignocellulosic biomass pyrolysis bio-oils were hardly detected in the BODP.The main oxygenated com-pounds in the BODP were fatty acid and dehydrated sugars.This re-sult was attributed to the different composition of reaction materials.The main components in lignocellulosic biomass were cellulose,lignin and hemi-cellulose,while microalgae contained largely residual lipid(mainly long chain fatty acid ester and free fatty acid),which could be converted into various hydrocarbons through decarboxylation and thermaldegradation processes.
In previous researches,HZSM-5catalyst was used to upgrade vapors derived from the pyrolysis of lignocellulosic biomass in or-der to improve the quality of bio-oils,and the increase in the con-centration of aromatic compounds in the bio-oils was observed (Williams and Horne,1995a;Williams and Nugranad,2000).In our experiments,the same increase in the concentration of aro-matic compounds was found in the BOCP.Nevertheless,bio-oils obtained from the HZSM-5catalyzed pyrolysis of lignocellulosic biomass contained various types of aromatic compounds such as benzene,naphthalene,phenanthrene,pyrene,phenol and their derivatives,while the BOCP contained only several alkylated ben-
Table5
The characteristics of BODP and BOCP.
BODP BOCP Wood pyrolysis bio-oils b Heavy fuel oil b
Ultimate analysis(wt.%)
C56.1365.2154–5885 H7.639.83 5.5–7.011 N 5.34 5.430–0.20.3 O a30.0919.5335–40 1.0 Elemental ratio
H/C 1.63 1.89––O/C0.400.22––Empirical formula CH1.63O0.40N0.08CH1.89O0.22N0.06––Heating-value(MJ kgà1)24.4c32.2c16–1940
a By difference,O(%)=100-C–H–N.
b Derived from the reference Czernik and Bridgwater(2004).
c Heating-value was calculate
d using th
e following formula(Friedl et al.,2005)HHV(MJ kgà1)=(3.55C2à232Cà2230H+51.2C?H+131N+20600)?10à3.
Table6
The molecular formula and relative content of typical compounds in the bio-oils.
Categories Components Formula Relative content(%)a
BODP BOCP
Chain hydrocarbons12.258.91
Dodecane C12H26 6.32 3.22
3,7,11,15-Tetremethyl-2-hexadecene C20H40 4.59 1.28 Aromatic compounds10.8948.6
Benzene C6H6nd b0.67
Toluene C7H80.53 3.24
p-Xylene C8H100.2212.24
1-Ethyl-2-methyl-benzene C9H12nd 3.41
1,2,3-Trimethyl-benzene C9H12nd 2.54
2-Methyl-naphthalene C11H10nd 6.54
1,4,6-Methyl-naphalene C13H14nd 1.14
Phenol C6H6O 1.43nd
4-Methyl-phenol C7H8O 1.84nd LCTG62.2424.85
Tetradecanoic acid C13H27COOH 3.62 1.96
Methyl palmitate C16H34O2 1.17nd
(Z)-11-Hexadecenoic acid C15H29COOH13.0410.06
n-Hexadecanoic acid C15H31COOH10.038.27
Oleic acid C17H33COOH 1.940.98
(Z)-9-Hexadecenoic-acid methyl ester C17H34O2 1.06nd
Stearic acid C17H35COOH 4.40nd
Tetradecanamide C13H27CONH2 1.45nd
Hexadecanamide C15H31CONH2 4.06 2.14
Pentadecanenitrile C14H31CN 1.94nd
Hexadecanenitrile C16H33CN 1.64nd Others14.6217.64
Dianhydro mannitol C6H10O48.53 4.25
Indole C8H6N 3.97nd
a The amount of the compounds was evaluated through the peek area;no response factors were introduced.
b Not detected.
4598P.Pan et al./Bioresource Technology101(2010)4593–4599
zene and alkylated naphthalene(Williams and Horne,1995a;Wil-liams and Nugranad,2000).
The increase in the concentration of aromatic hydrocarbons in the BOCP is due to the catalytic effect of HZSM-5catalyst.Accord-ing to Williams and Horne(1994b,1995a,b),the catalyst plays a role of reducing the oxygen content and increasing aromatic hydrocarbon content of bio-oils through the unique pore structure and high activity of the catalysts.The increased aromatic hydrocar-bons in the BOCP may mainly derive from the residual lipid and polysaccharide in algal residue.The residual lipid is vaporized and cracked to numerous volatiles.Some higher molecular weight compounds in volatiles,which should be condensed to LCTG in the direct pyrolysis process,react on the surface of HZSM-5catalyst, and generate many small molecular weight hydrocarbons such as C2H4and C3H6.And these small molecular weight hydrocarbons subsequently enter to three-dimensional pore system of HZSM-5 catalyst,and undergo further polymerization and aromatization to from aromatic hydrocarbons.In addition,the pyrolysis of poly-saccharides in algal residue may produce some low molecular weight species which can be converted to aromatic hydrocarbons under the effect of HZSM-5catalyst.The pore size of HZSM-5cat-alyst used is0.56nm which is equivalent to molecular size of naphthalene,so large molecules could not pass into and out of HZSM-5catalyst.Consequently,it is reasonable that the aromatic hydrocarbons in the BOCP are mainly alkylated benzene and alkyl-ated naphthalene in the present work.However,the enhanced acidity and micropore structure over catalyst surface also aggra-vates coking through secondarily reactions between those inter-mediates.The phenomenon is surely harmful to the usage of catalyst,and should be further studied in future researches.
4.Conclusion
The pyrolysis of Nannochloropsis sp.residue could be performed at mild temperature of about400°C,which was lower than that of lignocellulosic material.And the reaction temperature had great effect on product yields.The yield of the BOCP from Nannochlorop-sis sp.residue was lower than that of the BODP at each tempera-ture.And it gradually decreased as the catalyst-to-material ratio further https://www.wendangku.net/doc/c718270259.html,pared with the BODP,the BOCP had lower oxygen content,markedly different chemical composition,and so higher heating-value.The BODP mainly consisted of LCTG,while the BOCP mainly consisted of aromatic hydrocarbons.
Acknowledgements
This work is?nancially supported by the National Basic Re-search Program of China(973program,No.2007CB210203)and the Special Research Fund for the Doctoral Program of Higher Edu-cation of China(No.20050610013).The characterization of the cat-alyst and bio-oils from Analytical and Testing Center of Sichuan University are greatly appreciated.The supervision of academician Qingshi Zhu is highly acknowledged.
Appendix A.Supplementary data
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燃烧值与计算题
燃烧值 发布日期：2010年8月04日浏览次数：42 来源：网络物理编辑 【燃烧值】1千克某种燃料完全燃烧放出的热量，叫做这种燃料的燃烧值。燃烧值的公式是：q=Q/m，其中q为燃料的燃烧值，m为这种燃料的质量，Q为这种燃料完全燃烧时放出的热量。当热量的单位用焦耳，质量的单位用千克时，燃烧值的单位为“焦/千克”。常用的几种燃料的燃烧值可从下表中查到。 实际上，燃料很难做到完全燃烧，因此燃料燃烧时放出的热量往往少于根据燃烧值计算出来的热量。 通过燃料的燃烧值可计算一定质量的某种燃料完全燃烧时放出的热量。 例题1 质量为0.5千克的汽油完全燃烧时，能放出多少热量？ 根据燃烧值公式q=Q/m，得Q=mq。从燃烧值表中查出汽油的燃烧值为4.6×10^7焦/千克。将已知条件代入公式即可得热量Q=2.3×10^7焦。 根据实际需要的热量，可通过燃烧值公式计算出燃料的质量. 题2 把2千克水从20℃加热到沸腾，需要完全燃烧多少克酒精（不考虑热量损失）？ 题目中明确的告诉我们两个条件，其一：酒精是完全燃烧的；其二：没有热量损失，即酒精放出的热量全部被水吸收， 本文来自马博士教育网，转载请标明出处：https://www.wendangku.net/doc/c718270259.html,/wl/wlsc/cz/wlcd/rx/12197.html
燃烧，燃烧产物冷却到燃烧前的温度（一般为环境温度）时所释放出来的热量。 固体或液体发热量的单位是千卡/千克（kcal/kg）、千焦耳/千克（KJ/kg）或兆卡/千克（Mcal/kg）、兆焦尔/千克（MJ/kg）；气体燃料的发热量单位是千卡/标准立方米（kcal/Nm3）、千焦耳／标准立方米（KJ/Nm3）或兆卡/标准立方米（Mcal/Nm3）、兆焦尔/标准立方米（MJ/Nm3）。燃料热值有高位热值与低位热值两种。 高位热值是指燃料在完全燃烧时释放出来的全部热量，即在燃烧生成物中的水蒸汽凝结成水时的发热量，也称毛热。 低位热值是指燃料完全燃烧，其燃烧产物中的水蒸汽以气态存在时的发热量，也称净热。 高位热值与低位热值的区别，在于燃料燃烧产物中的水呈液态还是气态，水呈液态是高位热值，水呈气态是低位热值。低位热值等于从高位热值中扣除水蒸汽的凝结热（在1个标准大气压和100℃情况下，水的汽化热为2253.02焦耳／克，在常温常压下为2441.12焦耳／克；水汽凝结成液态水时放出相同的热量）。 燃料大都用于燃烧，各种炉窑的排烟温度均超过水蒸汽的凝结温度，不可能使水蒸气的凝结热释放出来，所以在能源利用中一般都以燃料的应用的低位发热量作为计算基础。各国的选择不同，日本、北美各国均习惯用高位热值，而我国、前苏联、德国和经济合作与发展组织是按低位热值换算的，有的国家两种热值都采用。 煤和石油的高低位热值相差约5％，天然气和煤气为10％左右。 各种燃料燃烧值的资料2008-12-13 10:38:28| 煤=3×107J/kg 煤气的燃烧值是4.2×107J/Kg,1焦=0.024卡路里 标准煤：7000大卡/kg＝7000*4.18＝29260kJ/kg＝29.26MJ/kg 焦炉煤气：4000大卡/m3左右，煤气密度0.54kg/标准m3 所以，4000大卡/标准m3/(0.54kg/标准m3)≈7400大卡/kg 显然，煤气的热值较高。 各种燃料热值 燃料名称热值MJ/kg 折算率 固体燃料 焦炭25.12-29.308 0.857-1.000 无烟煤25.12-32.65 0.857-1.114 烟煤20.93-33.50 0.714-1.143 褐煤8.38-16.76 0.286-0.572
发热量计算
（2）煤的各种发热量名称的含义 a.煤的弹筒发热量（Qb） 煤的弹筒发热量，是单位质量的煤样在热量计的弹筒内，在过量高压氧（25～35个大气压左右）中燃烧后产生的热量（燃烧产物的最终温度规定为25C）。 由于煤样是在高压氧气的弹筒里燃烧的，因此发生了煤在空气中燃烧时不能进行的热化学反应。如： 煤中氮以及充氧气前弹筒内空气中的氮，在空气中燃烧时，一般呈气态氮逸出，而在弹筒中燃烧时却生成N2O5或NO2等氮氧化合物。这些氮氧化合物溶于弹筒税种生成硝酸，这一化学反应是放热反应。另外，煤中可燃硫在空气中燃烧时生成SO2气体逸出，而在弹筒中燃烧时却氧化成SO3，SO3溶于弹筒水中生成硫酸。SO 2、SO3,以及H2SO4溶于水生成硫酸水化物都是放热反应。所以，煤的弹筒发热量要高于煤在空气中、工业锅炉中燃烧是实际产生的热量。为此，实际中要把弹筒发热量折算成符合煤在空气中燃烧的发热量。 b.煤的高位发热量（Qgr） 煤的高位发热量，即煤在空气中大气压条件下燃烧后所产生的热量。实际上是由实验室中测得的煤的弹筒发热量减去硫酸和硝酸生成热后得到的热量。 应该指出的是，煤的弹筒发热量是在恒容（弹筒内煤样燃烧室容积不变）条件下测得的，所以又叫恒容弹筒发热量。由恒容弹筒发热量折算出来的高位发热量又称为恒容高位发热量。而煤在空气中大气压下燃烧的条件湿恒压的（大气压不变），其高位发热量湿恒压高位发热量。恒容高位发热量和恒压高位发热量两者之间是有差别的。一般恒容高位发热量比恒压高位发热量低8.4～20.9J/g,实际中当要求精度不高时，一般不予校正。 煤的低位发热量，是指煤在空气中大气压条件下燃烧后产生的热量，扣除煤中水分（煤中有机质中的氢燃烧后生成的氧化水，以及煤中的游离水和化合水）的汽化热（蒸发热），剩下的实际可以使用的热量。
热值计算练习
热值计算练习 1、酒精的热值为×107J/kg,30g酒精完全燃烧用来加热3kg5℃的水，不计热损失，能使水温升高到多少摄氏度 2、不计热损失，燃烧20g的焦炭能使5kg水温度升高到80℃，求水原来的温度（焦炭的热值为×107J/kg） 3、质量为0.4kg的汽油完全燃烧所放出的热量有50%被25kg10℃的水吸收，那么水温升高多少摄氏度（汽油的热值为×107J/kg） 4、利用煤气灶加热5kg20℃的水至沸腾，若加热过程中煤气完全燃烧所放出的热量由80%被水吸收，则能消耗多少煤气（煤气的热值为×107J/m3） 5、质量为500g的铁锅中盛有3L水，把它们从15℃加热到90℃；现在利用天然气加热这锅水，已知在加热过程中由40%的热损失，则需消耗多少热值为×107J/m3的天然气 6、在1标准大气压下完全燃烧100g的干木柴放出的热量只有40%被1kg6℃的水吸收，则水温可以升高到多少摄氏度（干木柴的热值为×107J/kg） 7、酒精的热值为×107J/kg，40g酒精完全燃烧用来加热3kg20℃的水，已知此酒精炉子的效率为90%，则在1标准大气压下能使水温升高多少摄氏度 8、一只铜球的体积为250 cm3，当温度由20℃升高到60℃时吸收的热量为3900J，问此球是否实心若为空心的，则空心体积为多大[铜的密度为×103kg/ m3，铜的比热容为×103J/(kg·℃)] 9、一台汽油机效率为30%，现需要做6×107J的有用功，需要消耗多少千克的汽油（汽油的热值为×107J/kg） 10、一辆汽车发动机的功率为23kw，热机效率为25%，如果车速为36km/h，行驶100km，至少需要消耗多少汽油（汽油的热值为×107J/kg） 11、质量为1kg的水从400m高空中落下，若重力对水做的功有25%用来使水的内能增加，可使这些水的温度升高多少 12、一个雨滴从100m的空中落下，若重力做的功全部转变成热并被水
实验三燃料热值测定
实验三 燃料热值的测定 一、 实验的理论基础 燃料的燃烧热（或热值）是指单位质量（g 或gmol ）的燃料在标准状态下与氧完全燃烧时释放的热量。完全燃烧是指燃料（常指碳氢燃料）中的C 完全转变为二氧化碳，氢转变为水，硫转变为二氧化硫。如果燃烧发生于定压过程，这是的燃烧热为定压燃烧热，又称燃烧焓，如果燃烧过程保持容积不变，这是的燃烧为定容燃烧热。 假定有N 中组分参与反应的方程式为： [][] γγ11 11 '"M M I n I n ==∑ ∑ → 式中[]M 代表组分分子式，γ1为分子前指数，“'”，“"”分别为反应物和产物，则 ()() ()()Q E T Q H T C i i i I o o P i i i i o o = -= -∑∑γ γγ γ ' " '" ()E T i o o ，()H T i o o 分别为标准定容生成热或生成焓（kcal/gmal ，kcal/kg ）。上标“o ”代表标准 H E R T N i o i o o o =+? R o 为通用气体常数，?N=0， 一般情况下，由于E R T N i o o o ???，常常可以用生成焓代替生成热，即 H E i o i o ≈ 根据反应产物中水的状态不同，热值又有低热值和高热值之分。如产物水为蒸汽，这 是的热值为低热值，如产物为液态水，热值为高热值，两者的差值为水的蒸发潜热（Qr=10.52kcal/gmal ）。 工业上常用燃料的元素分析法确定高低热值的关系。若用符号Q gw y 和Q dw y 表示应用基 () Q Q H W dw y gw y y y =-+69（kcal / kg ） H W y y ,分别为应用基氢百分含量和全水份含量。 本实验测定的是分析基弹筒热值，用Q Dr f 表示。它与高位热值间的关系为 () Q Q W W Q S a Q W W gw y gw f y f DT f DT f DT f y f =--=-+?? --100100225100100. 式中：S DT f 为分析基硫百分含量；W f 为分析基水份含量；a 为修正系数；无烟煤和贫煤取 0.001，其它煤种取0.0015。 二、 实验原理
12.2 热量与热值 练习题精选 附答案
《12.2热量与热值》同步练习 一、填空题： 1、将热量、温度、内能三个物理量分别填入下面的空格中： （1）今天天气很热，这里的“热”是指； （2）物体吸热升温，这里的“热”是指； （3）摩擦生热，这里的“热”是指。 2、在物理学中，把物体在热传递过程中内能改变的多少叫，用符号表示，它的单位是。 3、1kg某种燃料放出的，叫做这种燃料的热值；酒精的热值为 3.0×107J/kg，它表示的物理意义是。 4、有一箱木炭，其质量是10kg，木炭热值为3.4×107J/kg。若用掉一半，剩下的木炭的热值为 ，燃烧剩下的木炭最多可放出热量。 5、实验表明，对同种物质组成的物体，其吸收或放出的热量与物体的和 成正比。 6、煤油的热值为4.6×107J/kg，完全燃烧200g煤油，可放出热量为，若要获得2.3×108J 的热量，则需完全燃烧kg的煤油。 二、选择题： 7、关于热量，下列说法正确的是（） A．热水比冷水含有的热量多B．一大桶水比一小桶水含有的热量多 C．一个物体内能越多，含有的热量越多D．热量是热传递过程中内能的改变量 8、当一个物体温度升高时，下列说法正确的是（） A．它的热量一定增加B．它的内能一定增加 C．它的状态一定改变D．它的体积一定增大 9、一个物体温度降低了，则（） A．它含有的热量减少B．它一定对外放出了热量 C．外界一定对物体做了功D．它的内能减少 10、下列关于热值的说法中，正确的是（） A．任何物体都有热值B．燃料燃烧时才有热值 C．燃料的热值与燃料是否完全燃烧无关D．燃烧时放出热量多的燃料热值大 11、现代火箭用液态氢做燃料，是因为它具有（） A．较小的密度B．较低的沸点 C．较大的体积D．较高的热值 12、酒精的热值比干木材大，下列说法中正确的是（） A．酒精含有的内能比干木材高 B．燃烧酒精比燃烧干木材放出的热量多 C．燃烧酒精比燃烧干木材产生的温度高 D．要使它们完全燃烧放出同样多的热量，所需酒精比干木材质量小 13、关于温度、热量和内能的说法中正确的是（） A．0℃的冰块没有内能 B．温度高的物体，内能一定大 C．物体吸收热量，温度一定升高 D．物体吸收热量，内能一定增大
热量与热值教案(九年级物理)
14.2热量与热值 一、情景引入 小明一家四口一个月如果用干木柴作燃料需要90kg，若用煤炭需要60kg，若用煤气需要30kg。为什么不同的燃料所需的质量不一样呢？ 二、教材研究 问题1——什么是热量？ 1、使温度不同的物体相互接触，低温物体温度，高温物体温度，这个过程叫做热传递。 2、在热传递过程中，叫做热量。它的单位是。物体吸收热量，内能；物体放出热量，内能。 问题2——如何理解热值的概念及物理意义？ 1、在物理学中，把叫做这种燃料的热值。它的单位是。 2、干木柴的热值是 1.2×107J／kg，它的物理意义是。完全燃烧500g的干木柴所放出的热量是。 三、典例分析 平时我们所说的“热”，在不同情况下的含义不同，下述各句中“热”的含义是： （1）“今天天气很热”中的“热”是指：。 （2）“摩擦生热”中的“热”是指：。 （3）“这是一块散热片”中的“热”是指：。 （4）“小明感冒发热”中的“热”是指：。 解析“热”在不同情况下的含义不同，天气很热是指温度高；摩擦生热是指内能增加； 吸收(或放出)热是指热量；感冒发热是指体温高。 答案（1）温度高（2）内能（3）热量（4）温度高 四、达标训练 目标知识1——理解热量的概念 1、关于热量，下列说法正确的是（） A、热水比冷水含的热量多 B、一大桶水比一小桶水含的热量多 C、物体内能越大，含的热值越多 D、热量是热传递过程中内能的改变量，没有热传递过程就没有热量 目标知识2——热值的概念及物理意义 2、燃料燃烧时放出热量的多少（） A、只与燃料的种类有关 B、只与燃料的质量有关
C、只与燃料是否完全燃烧有关 D、跟上述三个因素有关 3、一堆木炭烧了一半，则剩下的木炭（） A、热值变为原来的一半 B、密度变为原来的一半 C、质量变为原来的一半 D、热量变为原来的一半 五、中考试题 （2007年茂名市）用无烟煤来做饭时，若无烟煤的热值是3.4×107J／kg，完全燃烧0.5kg无烟煤能放出J热量；饭菜做好时，厨房弥漫着饭菜的香味，说明。 解析：本题要点是对燃料热值物理意义和扩散现象的理解。无烟煤的热值是 3.4×107J／kg,，的物理意义是质量为1kg的无烟煤完全燃烧时放出的热量是 3.4×107J，即燃料燃烧放出的热量=热值×质量 答案： 1.7×107J 分子在不停地做无规则运动。 六、课堂训练 1、燃料在燃烧过程中，燃料的能转化为能，质量是kg 的某种燃料完全燃烧放出的热量，叫做这种燃料的，其单位是。 2、酒精的热值是 3.0×107J/kg，它表示的物理意义是，完全燃烧kg酒精，可以放出1.5×108J的热量。 3、在燃料燃烧过程中放出热量的计算为：Q=mq，其中Q指：，m 指：，q指：。 4、燃料的燃烧是一种变化，在燃烧过程中燃料的能转化为能。 5、汽油的热值是4.6×107J/kg，读作：，它表示的意思是1kg的燃烧放出的热量是。 6、燃料在燃烧时，实际放出的热量按计算得出的热量，有效利用的热量实际放出的热量。(填:“大于”“小于”或“等于”) 7、将一杯石油倒出一半后，剩余石油 A、密度不变，比热容和热值减半 B、质量、比热容和热值都变为原来一半 C、比热容不变，热值和密度减半 D、比热容和密度不变，质量为原来的一半 8、两物体相互接触放在一起没有发生热传递，这是因为它们有相同的 （） A、温度 B、质量 C、内能 D、热量 9、两物体发生热传递时，下面说法正确的是（） A、质量大的传给质量小的 B、体积大的传给体积小的 C、温度高的传给温度低的 D、内能大的传给内能小的 10、一个物体温度升高了，这是因为
发热量计算公式
发热量计算公式 以煤工业分析结果，创立计算煤炭低位发热量新公式的原理与方法，不再详述。仅就实际应用的计算公式介绍如下： 1.计算烟煤低位发热量新公式 以焦耳表示的计算方式： Qnet.ad＝35859.9-73.7Vad-395.7Aad-702.0Mad+173.6CRC 焦/克 或用卡制表示的计算式： Qnet.ad＝8575.63-17.63Vad-94.64Aad-167.89Mad+41.52CRC卡/克Qnet.ad——分析基低位发热量； Vad——分析基挥发分（%）； Aad——分析基灰分（%）； Mad——分析基水分（%）； CRC——焦渣特征。 2.计算无烟煤低位发热量新公式 以焦耳表示的计算方式： Qnet.ad=34813.7-24.7Vad-382.2Aad-563.0Mad焦/克 或者以卡制表示的计算式： Qnet.ad=8325.46-5.92Vad-91.41Aad-134.63Mad卡/克
如果有条件能测定H值，或者从固定用煤矿区取得矿区以往H值的 平均值，用下式计算的无烟煤低位发热量结果精度更高。 以焦耳表示的计算式： Qnet.ad=32346.8-161.5Vad-345.8Aad-360.3Mad+1042.3Had 焦/克 或者用卡制表示的计算式： Qnet.ad=7735.52-38.63Vad-82.70Aad-86.16Mad+249.27Had 卡/克 3.计算褐煤低位发热量新公式 以焦耳表示的计算式： Qnet.ad=31732.9-70.5Vad-321.6Aad-388.4Mad焦/克 或者用卡制表示的计算式： Qnet.ad=7588.69-16.85Vad-76.91Aad-92.88Mad卡/克 4.在水泥生产使用中，计算标准煤耗时，按上述公式计算的分析基低 位发热量（Qnet.ad）用下式换算成应用煤低位发热量（Qnet.ar）后，再 计算标准煤耗。 应用煤低位发热量计算公式 100-Mad100-Mar Qnet.ar=Qnet.ad×──────-23（Mar－Mad×─────） 焦/克 100-Mad100-Mad 煤经挥发分测定后遗留在坩埚内固体残渣的特征。 焦渣特征（CRC）煤炭热分解以后剩余物质的形状。根据不同形状分为8
(2020精选)中考物理试题分类汇编_专题24_热量与热值(含解析)
专题24 热量与热值 一．填空题（共12小题） 1．（2018?泰安）标准大气压下，质量为0.5kg、温度为70℃的水放出4.2×104J的热量，水的温度降低了20 ℃[c水=4.2×103J/（kg?℃）]。 【分析】知道水的质量、放出的热量、比热容，利用Q放=cm△t求水温度降低值。 【解答】解： 由Q放=c水m△t可得，水降低的温度： △t===20℃。 故答案为：20。 2．（2018?福建）合理分类和利用垃圾可以变凌为宝。在一定条件下，1吨垃圾能”榨”出170kg燃料油，若燃料油的热值为4.0×107J/kg，则这些燃料油完全燃烧释放出的热量为 6.8×109J，燃烧时燃料油的化学能转化为内能。 【分析】（1）利用Q放=mq求燃料油完全燃烧释放出的热量； （2）燃料在燃烧时，发生化学变化，把燃料的化学能转化为内能。 【解答】解： （1）燃料油完全燃烧释放出的热量： Q放=mq=170kg×4.0×107J/kg=6.8×109J； （2）燃料油燃烧过程中，将燃料油的化学能转化为内能。 故答案为：6.8×109；内。 3．（2018?哈尔滨）如图是“研究水沸腾规律”的实验，试管中水的质量为0.2kg，这些水由20℃升高到70℃吸收的热量是 4.2×104J；水沸腾过程中，温度计示数将不变。【c水=4.2×103J/（kg?℃）】 【分析】（1）知道水的质量、水的比热容、水的初温和末温，利用吸热公式Q吸=cm△t求水吸收的热量；（2）水沸腾时，吸收热量、温度不变。 【解答】解：
（1）水吸收的热量： Q吸=c水m△t=4.2×103J/（kg?℃）×0.2kg×（70℃﹣20℃）=4.2×104J； （2）水沸腾过程中，吸收热量、温度不变，温度计示数将不变。 故答案为：4.2×104；不变。 4．（2018?怀化）汽车用水作为冷却剂，是因为水的比热容较大。汽车在行驶过程中消耗了2kg汽油，若汽油完全燃烧则放出的热量为9.2×107J（汽油的热值为4.6×107J/kg）。 【分析】（1）相同质量的水和其它物质比较，吸收或放出相同的热量，水的温度升高或降低的少；升高或降低相同的温度，水吸收或放出的热量多。 （2）知道汽油的质量和热值，利用Q放=mq求汽油完全燃烧产生的热量。 【解答】解： （1）因为水的比热容较大，相同质量的水和其它物质比较，升高相同的温度，水吸收的热量多，水能带走较多的热量，冷却效果好，所以用水做汽车发动机的冷却剂。 （2）2kg汽油完全燃烧产生的热量： Q放=mq=2kg×4.6×l07J/kg=9.2×l07J。 故答案为：比热容；9.2×l07。 5．（2018?连云港）天然气热水器将40kg的水由22℃加热到42℃，水的内能增加了 3.36×106J，水的内能增加是通过热传递方式实现的，若天然气燃烧放出的热量有84%被水吸收，需要消耗0.1 kg的天然气。已知c水=4.2×103J/（kg?℃），q天然气=4×107J/kg。 【分析】（1知道水的质量、水的比热容、水的初温和末温（在一个标准大气压下，水的沸点为100℃），利用吸热公式Q吸=cm△t求水吸收的热量； （2）改变物体内能的方法：做功和热传递； （3）利用η=求天然气完全燃烧放出的热量，再利用Q放=mq求需要天然气的质量。 【解答】解：（1）水吸收的热量： Q吸=cm△t=4.2×103J/（kg?℃）×40kg×（42℃﹣22℃）=3.36×106J； （2）水从火吸收热量，内能增加、温度升高，是利用热传递改变水的内能； （3）由η=得，天然气完全燃烧放出的热量： Q放===4×106J， 由Q放=mq得，需要天然气的质量： m===0.1kg。
燃料燃烧及热平衡计算参考
燃料燃烧及热平衡计算参考 3.1 城市煤气的燃料计算 3.1.1 燃料成分 表2.2 城市煤气成分（%）[2] 成分 CO 2 CO CH 4 C 2H 6 H 2 O 2 N 2 合计 含量 10 5 22 5 46 2 10 100 3.1.2 城市煤气燃烧的计算 1、助燃空气消耗量[2] （1）理论空气需要量 Lo=21O O 0.5H H 3.5C CH 20.5CO 2 2624-++?+ Nm 3/Nm 3 (3.1) (3.1)式中：CO 、CH 4 、 C 2H 6 、 H 2 、 O 2——每100Nm 3湿气体燃料中各成分的体积含量（Nm 3）。则 Lo=21 2465.055.322255.0-?+?+?+? = 4.143 Nm 3/Nm 3 （2）实际空气需要量 L n =nL 0, Nm 3/Nm 3 (3.2) (1.2)式中：n ——空气消耗系数，气体燃料通常n=1.05 1.1 现在n 取1.05，则 L n =1.05×4.143=4.35 Nm 3/Nm 3 （3）实际湿空气需要量 L n 湿 =（1+0.00124 2H O g 干） L n ， Nm 3/Nm 3 （3.3） 则 L n 湿=（1+0.00124×18.9）×4.35=4.452 Nm 3/Nm 3 2、天然气燃烧产物生成量 （1）燃烧产物中单一成分生成量 CO)H 2C CH (CO 0.01 V 6242CO 2+++?=’
(3.4) 2 O V 0.21(=?′0n-1)L (3.5) 2 2n N V (N 79L )0.01=+?′ (3.6) )L 0.124g H H 3C (2CH 0.01V n 干 O H 2624O H 22+++?= (3.7) 式中CO 、CH 4 、 C 2H 6 、 H 2 ——每100Nm 3湿气体燃料中各成分的体积含量。 则 0.475)5222(100.01V 2CO =+?++?= Nm 3/Nm 3 4.4131)(1.050.21V 2O ?-?==0.046 Nm 3/Nm 3 01.0)35.47910(V 2N ??+==3.54 Nm 3/Nm 3 4.35)18.90.124465322(20.01V O H 2??++?+??==1.152 Nm 3/Nm 3 （2）燃烧产物总生成量 实际燃烧产物量 V n = V CO2+V O2+V N2+V H2O Nm 3/Nm 3 (3.8) 则 V n =0.47+0.046+3.54+1.152=5.208 Nm 3/Nm 3 理论燃烧产物量 V 0=V n -（n -1）L O (3.9) V 0=5.208-（1.05-1）×4.143=5.0 Nm 3/Nm 3 (3) 燃料燃烧产物成分[2] %100V V CO n CO 22?= (3.10) %100V V O n O 22?= (3.11) %100V V N n N 22?= (3.12) 100%V V O H n O H 22?= (3.13)
热值的综合计算
三．热量与热值综合计算 1．用天然气灶烧水，燃料0.5m3的天然气，使100kg的水从20℃升高到70℃。已知水的比热容c＝4.2×103J（kg·℃），天然气的热值q＝7.0×107J/m3。 求： （1）0.5m3的天然气完全燃料放出的热量Q 放 。 （2）水吸收的热量Q 吸 。 （3）燃气灶的效率η。 2．洒精是实验室里常用的燃料，现用酒精灯来加热水，若洒精完全燃烧产生的热量有50%被水吸收，现在把0.5kg 、20℃的水加热到100℃，需要燃烧多少克 酒精？［q 酒精=3×107J/（k g·℃），c 水 =4.2×103J/（k g·℃）] 3。太阳能热水器，水箱容积是200L．温度计测得自来水的温度为20℃，然后给热水器水箱送满水，中午时“温度传感器”显示水箱中的水温为45℃．请你求解下列问题： （1）水箱中水的质量； （2）水吸收的热量 （3）如果水吸收的这些热量，由燃烧煤气来提供，而煤气灶的效率为40％，求至少需要燃烧多少kg的煤气（煤气的热值为q=4.2×107J／kg） 4．某中学为学生供应开水，用锅炉将200kg的水从20℃加热到100℃，燃烧了4kg 的无烟煤．[水的比热容是4.2×103J/（kg?℃），无烟煤的热值是3.4×107J/kg]试求：（1）锅炉内200kg的水吸收的热量是多少焦耳？ （2）4kg无烟煤完全燃烧放出的热量是多少焦耳？ （3）此锅炉的效率是多少？ 拔高题 已知某型号的载重车在一段平直的高速公路上匀速行驶10.08km，所用时间是 8min，消耗燃油3L（假设燃油完全燃烧），汽车发动机在这段时间内的功率为63kW．若燃油的密度是0.8×103kg/m3，热值为3.15×107J/kg，求： （1）汽车行驶速度是多少？ （2）汽车行驶过程的牵引力是多大？ （3）汽车发动机的效率是多少？
沪科版九年级物理上册同步测试：专题训练(1) 热量、热值和热效率的综合计算
专题训练(一) 热量、热值和热效率的综合计算?类型一热量的计算 1．质量为500 g的金属块,温度从100 ℃降低到20 ℃共放出了3.52×104J的热量,金属块的比热容为( ) A．0.88×103 J/(kg·℃) B．0.44×103 J/(kg·℃) C．0.22×103 J/(kg·℃) D．0.11×103 J/(kg·℃) 2．甲、乙两金属块的质量之比是2∶1,比热容之比是3∶4,若它们吸收相同的热量,则它们升高的温度之比是( ) A．2∶3 B．3∶2 C．3∶8 D．8∶3 3．质量为1 kg、初温为20 ℃的水吸收4.2×105 J的热量后,它的温度在下列给出的四个温度中,最多有几个可能温度( ) ①80 ℃②100 ℃③120 ℃④130 ℃ A．1 B．2 C．3 D．4 4．向洗澡盆放水时,已知冷水为20 ℃,热水为80 ℃,想得到40 ℃的温水120 kg,应该分别放冷水和热水各多少千克？(不计热损失) ?类型二热值的计算 5．“可燃冰”作为新型能源,有着巨大的开发使用潜力.今年5月,我国已在世界上首次实现了可燃冰连续超过7天的稳定开采.若可燃冰的热值为 1.38×1010 J/m3,1 m3的可燃冰完全燃烧放出的热量相当于________kg汽油完全燃烧放出的热量.(q汽油＝4.6×107 J/kg) 6．质量为5 kg、初温为15 ℃的水,吸热后温度升高到95 ℃,则需吸收的热量为________J；如果这些热量由天然气燃烧来提供,则至少需要燃烧热值为4.0×107J/m3的天然气________m3.[水的比热容为4.2×103 J/(kg·℃)] 7．天然气具有热值高、投资成本低、价格实惠、污染少、安全等优点.钦州市今年天然气用户预计达到6万户,年用气量将达到1350万立方米以上.小李家5月份使用天然气10 m3,已知天然气价格为3.27元/m3,热值为8.4×107 J/m3.求：
热量热值分类计算题
有关热量热值的计算题 一．比热容基本计算 1．一个质量是200g的铁球，温度从150℃下降到100℃，求它放出的热量？ [铁的比热容等于0.46×103J/(kg·℃)] 2．一个电热水壶内装有2 kg的冷水，经过加热后水温从18℃升高到98℃， 求：电热水壶内的水吸收的热量是多少？ 3．烧杯内80℃的热水放出2.1×104J的热量后，水温降低到55℃，则烧杯内水的质量是多少？ 4．由实验测量．质量是100g、初温是24℃的实心金属球吸收2.3×103J的热量后，温度升高到74℃，则该小球的比热容是多少？查图表可知这种金属可能是？ 几种物质的比热[单位：J/(kg·℃)] 二．热值基本计算 1．天然气是一种清洁能源，2m3的天然气完全燃烧能放出多少热量？（天然气的热值取7.1×107J/m3） 2．木炭的热值是3.4x107J/kg ，完全然烧500g木炭，能放出多少的热量？ 3．酒精的热值是3×107J/kg，若要获得9×107J的热量，至少需要燃烧多少千克的酒精？ 三．热平衡计算 例：吃早饭的时候，妈妈用热水给小雪加热250g的袋装牛奶．为了使这袋牛奶的温度由12℃升高到42℃，妈妈用60℃的热水给牛奶加热． [水的比热容为4.2×103J/（kg?℃），该牛奶的比热容为2.5×103J/（kg?℃）]．问： （1）在加热过程中，牛奶吸收了多少热量？ （2）问妈妈至少要用多少千克热水给牛奶加热？ 四．热量与热值综合计算 1．用天然气灶烧水，燃料0.5m3的天然气，使100kg的水从20℃升高到70℃。 已知水的比热容c＝4.2×103J（kg·℃），天然气的热值q＝7.0×107J/m3。 求： （1）0.5m3的天然气完全燃料放出的热量Q 放 。 （2）水吸收的热量Q 吸 。 （3）燃气灶的效率η。 2．洒精是实验室里常用的燃料，现用酒精灯来加热水，若洒精完全燃烧产生的热量有50%被水吸收，现在把0.5kg 、20℃的水加热到100℃，需要燃烧多少克酒精？ ［q 酒精=3×107J/（k g·℃），c 水 =4.2×103J/（k g·℃）] 3．小星家的太阳能热水器，水箱容积是200L．小星进行了一次观察活动：某天早上，他用温度计测得自来水的温度为20℃，然后给热水器水箱送满水，中午时“温度传感器”显示水箱中的水温为45℃．请你求解下列问题： （1）水箱中水的质量； （2）水吸收的热量 （3）如果水吸收的这些热量，由燃烧煤气来提供，而煤气灶的效率为40％，求至少需要燃烧多少kg的煤气（煤气的热值为q=4.2×107J／kg） 4．某中学为学生供应开水，用锅炉将200kg的水从20℃加热到100℃，燃烧了4kg的无烟煤．[水的比热容是4.2×103J/（kg?℃），无烟煤的热值是3.4×107J/kg]试求： （1）锅炉内200kg的水吸收的热量是多少焦耳？ （2）4kg无烟煤完全燃烧放出的热量是多少焦耳？ （3）此锅炉的效率是多少？
专题训练(二) 热量、热值和热效率的综合计算
专题训练(二)热量、热值和热效率的 综合计算 ?类型一一般吸热、放热的计算 1．[2019·泸州] 2017年6月2日中央电视台报道，我国在南海海底实现“可燃冰”连续稳定产气22天，“可燃冰”燃烧时将化学能转化为________能，完全燃烧热值为5.6×109 J/m3的“可燃冰”2 m3，放出的热量是________J，相当于完全燃烧热值为2.8×107J/m3的天然气________m3放出的热量． 2．用两个相同的电加热器给质量同为2 kg的物质甲和水加热，它们的温度随时间的变化关系如图ZT－2－1所示，据此判断甲物质10 min吸收的热量为________J．[c水＝4.2×103 J/(kg·℃)] 图ZT－2－1 3．在标准大气压下，质量为10 kg的水，吸收1.26×106J的热量，温度能从72 ℃升高到________．[水的比热容为4.2×103J/(kg·℃)] 4．[2019·黔西南州] 某学校锅炉用天然气作燃料给水加热．已知天然气的热值为4.0×107J/m3.问：(不计热量损失) (1)完全燃烧4.2 m3的天然气可以获得多少热量？ (2)这些热量可以使1000 kg的水从30 ℃升高到多少摄氏度？[c水＝4.2×103J/(kg·℃)] 5．重庆温泉资源十分丰富，被誉为“温泉之都”，温泉水主要靠地热，温度大多在60 ℃以上．取质量为400 g、温度为60 ℃的温泉水加热，当吸收了5.04×104J的热量后，温度升高到多少摄氏度？此时这些水会不会沸腾？[假设当时的气压为标准大气压，水的比热容是4.2×103J/(kg·℃)] ?类型二热平衡方程的应用 6．甲、乙两杯冷水，质量相同，初温也相同．将质量和初温都相同，但温度比水高的铜球和铝球(c铜＜c铝)分别投入两杯冷水中，不计热量损失，水也不溢出，达到平衡时() A．盛铜球的那杯水末温高 B．盛铝球的那杯水末温高 C．两杯水温度一样高 D．以上情况均有可能 7．甲、乙两杯中分别盛有温度为60 ℃和20 ℃但质量相同的水，现将一个温度为20 ℃的铁球投入甲杯中足够长时间，取出后再投入乙杯中停留足够长时间．如果不计热量损失，比较甲、乙两杯水的温度变化，则() A．Δt甲＜Δt乙B．Δt甲＞Δt乙 C．Δt甲＝Δt乙D．无法判定 8．一根质量为2 g的烧红的铁钉，温度是600 ℃，若它的温度降低到100 ℃，释放的热量为________J．若这些热量全部用于加热100 g常温的水，则水温将升高________℃.[已知铁的比热容为0.46×103J/(kg·℃)，水的比热容为4.2×103J/(kg·℃)] 9．质量相等，初温均为10 ℃的甲、乙两物体，将它们分别投入温度均为60 ℃的等质量的两杯热水中，热平衡后甲所在杯中热水温度降低了5 ℃，乙所在杯中热水温度降低了20 ℃，不考虑热损失，则甲、乙两物体的比热容之比为________． 10．将质量为4 kg的冷水与质量为3 kg、温度为80 ℃的热水混合后，末温度为40 ℃，不计热量损失，则：[c水＝4.2×103J/(kg·℃)] (1)热水放出的热量是多少焦耳？
热值练习题
1、焦炭的热值是3×107J/kg，完全燃烧200g的焦炭放出的热量为 多少？ 2、已知天然气的热值是7.5×107J/m3，，若某家庭一个月用8m3的天然气，则这些天然气完全燃烧时放出的热量为多少J，若这些热量全部由热值是3.0×107J/kg的焦炭来提供，应完全燃烧多少kg的焦炭． 3、在一标准大气压下完全燃烧140克焦炭能放出多少热量？若这些热量的30%被水吸收，则能使10kg、20℃的水温度升高到多少度？（焦炭的热值为3.0×107J/kg） 4、质量为50千克的铜块，温度由30℃升高到100℃，需要吸收的热量是多少焦？若是燃烧焦炭给铜块加热，已知焦炭完全燃烧时放出的热量只有一半被铜块吸收，需要燃烧多少焦炭？[焦炭的燃烧值是3.0×107焦/千克，c铜=3.9×102焦/（千克?℃）]． 5、太阳能热水器内盛有50kg的水，在阳光照射下，水的温度升高了30℃，求：①水吸收了多少热量？ ②这些热量相当于完全燃烧多少质量的焦炭放出的热量？（q焦炭=3.0×107J/kg） 6、氢气是无污染的环保燃料，而焦炭燃烧时会产生有害气体．已知氢气的热值为1.4×108J/kg，焦炭的热值为3.0×107J/kg．完全燃烧3kg氢气放出的热量，跟完全燃烧多少kg焦炭放出的热量相等．1、焦炭的热值是3×107J/kg，完全燃烧200g的焦炭放出的热量为 多少？ 2、已知天然气的热值是7.5×107J/m3，，若某家庭一个月用8m3的天然气，则这些天然气完全燃烧时放出的热量为多少J，若这些热量全部由热值是3.0×107J/kg的焦炭来提供，应完全燃烧多少kg的焦炭． 2、在一标准大气压下完全燃烧140克焦炭能放出多少热量？若这些热量的30%被水吸收，则能使10kg、20℃的水温度升高到多少度？（焦炭的热值为3.0×107J/kg） 3、质量为50千克的铜块，温度由30℃升高到100℃，需要吸收的热量是多少焦？若是燃烧焦炭给铜块加热，已知焦炭完全燃烧时放出的热量只有一半被铜块吸收，需要燃烧多少焦炭？[焦炭的燃烧值是3.0×107焦/千克，c铜=3.9×102焦/（千克?℃）]． 5、太阳能热水器内盛有50kg的水，在阳光照射下，水的温度升高了30℃，求：①水吸收了多少热量？ ②这些热量相当于完全燃烧多少质量的焦炭放出的热量？（q焦炭=3.0×107J/kg） 6、氢气是无污染的环保燃料，而焦炭燃烧时会产生有害气体．已知氢气的热值为1.4×108J/kg，焦炭的热值为3.0×107J/kg．完全燃烧3kg氢气放出的热量，跟完全燃烧多少kg焦炭放出的热量相等．
热量热值知识点
热量与热值 1. 热量：在物理学中，把在热传递过程中物体内能改变的多少叫做热量。物体吸收热量，内能增加；放出热量，内能减少。 2. 热量用字母Q表示，单位是焦（J）。一根火柴完全燃烧放出的热量约为1000J。 3. 实验表明：对同种物质的物体，它吸收或放出的热量跟物体的质量大小、温度的变化多少成正比。 4. 热值：把1某种燃料在完全燃烧时所放出的热量叫做这种燃料的热值。 5. 热值是燃料的一种属性，与质量、是否完全燃烧等没有关系，只与燃料的种类有关，不同燃料的热值一般不同。 6. 燃料完全燃烧放出热量的计算公式：或 7. Q表示热量，单位是焦（J）, q表示热值，单位是焦/千克（）或焦/米3 （3） ; m表示质量，单位是千克（）;V表示体积，单位是米3 （m3） 8 .氢气的热值很大，为q氢=1.4 X 1083，表示的物理意义是：1m3的氢气在完全燃烧时所放出的热量为 1.4 X 108J。 9.提高炉子效率的方法：①改善燃烧条件，使燃料尽可能充分燃烧；②尽可能减少各种热量损失 巩固练习 1、热机是把____ 能转化为______ 能的机械，在热机里，转变为
的能量和燃料完全燃烧所释放的能量的比值称为热机效率。 2、如何提高热机效率，是减少能源消耗的重要问题，要提高热 机效率，其主要途径是减少热机工作中的各种___________ 损失，其次是保证良好的_________ ，减少机械损失。 3、为节约能源，需提高热机的效率，下列措施中不能提高效率的是 A、尽量使燃料充分燃烧 B、尽量增加热机的工作时间 C、尽量减少废气带走的热量 D、尽量减少热机部件间的摩擦 4、目前，社会上有一些旧的被淘汰的内燃机，从长远看，你认为有必要修一修再使用吗？说明原因？
热值计算问题
１．炉子的热量与的热量之比，叫做炉子的效率 2、现在火锅通常用一种被称为“固体酒精”的物质作为燃料，已知这种燃料的热值是 1×107J/kg，完全燃烧0.21kg“固体酒精”能放出多少热量？若放出的热量都被火锅中的水吸收，求能使多少kg的水温度升高50 ℃？ 3．一个家用节煤炉的效率为40%，完全燃烧一块煤球放出4x107J热量，那么有效利用的热 量是． 4．１kg的汽油完全燃烧放出的热量相当于kg干木柴完全燃烧放出的热量． 5.我国的地热资源相当丰富，已经发现的天然温泉就有2000处以上，温泉水的温度大多在 60 ℃以上，个别地方达到100—140℃。100kg的温泉水从60℃降低到20℃时所放出的热量 是多少J？这相当于多少m3煤气完全燃烧放出的热量？（煤气的热值为4.2×107J/m3）
6、某饰品加工厂有酒精喷灯加工饰品，张师傅在1分钟内用去酒精500g，被加工的一只质量为1kg的铜饰品温度从20℃升高到1020℃，求：?（1）铜饰品吸收的热量是多少？（2）酒精完全燃烧放出的热量是多少？（3）酒精喷灯的效率是多少？ [已知：℃铜 =0.39×103J/(kg·℃), 酒精热值为3.0×107J/kg， 例10：（2010兰州）一台单缸四冲程柴油机，飞轮转速为3600转/分，该柴油机活塞1s对外做功次。若其效率为40%，消耗5kg的柴油转化成的机械能是J。 (q柴油 =4.3×107J/kg) 例11（11·聊城）天然气灶烧水，燃烧0.5m3的天然气，使100kg的水从20℃升高到70℃。已知水的比热容为C=4 .2×103J/（kg?℃），，天然气的热值为q=7.0×107J/m3。求： （1）0.5m3天然气完全燃烧放出的热量Q放。 （2）水吸收的热量Q吸。（3）燃气灶的效率η。
燃料燃烧及热平衡计算参考
燃料燃烧及热平衡计算参考
L n 湿=（1+0.00124×18.9）×4.35=4.452 Nm 3/Nm 3 2、天然气燃烧产物生成量 （1）燃烧产物中单一成分生成量 CO) H 2C CH (CO 0.01V 6242CO 2+++?=’ (3.4) 2 O V 0.21(=?′0n-1)L (3.5) 2 2n N V (N 79L )0.01=+?′ (3.6) )L 0.124g H H 3C (2CH 0.01V n 干O H 2624O H 22+++?= (3.7) 式中CO 、CH 4 、 C 2H 6 、 H 2 ——每100Nm 3湿气体燃料中各成分的体积含量。 则 0.475)5222(100.01V 2CO =+?++?= Nm 3/Nm 3 4.4131)(1.050.21V 2O ?-?==0.046 Nm 3/Nm 3 01.0)3 5.47910(V 2N ??+==3.54 Nm 3/Nm 3 4.35)18.90.124465322(20.01V O H 2??++?+??==1.152 Nm 3/Nm 3 （2）燃烧产物总生成量 实际燃烧产物量 V n = V CO2+V O2+V N2+V H2O Nm 3/Nm 3 (3.8) 则 V n =0.47+0.046+3.54+1.152=5.208 Nm 3/Nm 3 理论燃烧产物量 V 0=V n -（n -1）L O (3.9) V 0=5.208-（1.05-1）×4.143=5.0 Nm 3/Nm 3 (3) 燃料燃烧产物成分[2] % 100V V CO n CO 22?= (3.10) % 100V V O n O 22?=
中考物理热量计算的压轴题(含答案)
中考物理热量计算的压轴题及答案 中考真题 人类的祖先钻木取火，为人类文明揭开了新的一页，钻木取火的一种方法如图所示，将削尖的木棒伸到木板的洞里，用力压住木棒来回拉动钻弓，木棒在木板的洞里转动时，板与棒互相摩擦，机械能转化为内能.而热集中在洞内，不易散发，提高了木棒尖端的温度，当达到约260℃时木棒便开始燃烧，因木头是热的不良导体，故受热厚度很薄，木棒受热部分的质量只有0.25g.已知：来回拉一次钻弓需1.0s，弓长为s=0.25m，人拉弓的力为16N，木头比热c=2×103 J/(kg·℃)，室温为20℃. 问： (1)人来回拉一次钻弓克服摩擦力所做的功为多少? (2)人克服摩擦力做功使机械能转化为内能，若其中有25%被木 棒尖端吸收，则1s内可使木棒尖端温度提高多少℃? (3)请你估算用多长时间才能使木棒燃烧起来? 【示范解析】 (1)人来回拉一次钻弓克服摩擦力所做的功W=2FS=2×16N×0.25m=8J;(2)木棒尖端吸收的热量Q=ηW=25%×8J=2J，木棒尖端升高的温度△t=Q/cm=2 J/[2×103J/(kg.℃)×0.25×10-3kg]=4℃;(3)使木棒燃烧起来的时间 t=(260℃-20℃)÷4℃/s=60s. 拓展延伸 1.阅读下面的短文，回答问题， 太阳能热水器 太阳能热水器是利用太阳能把水从低温加热到高温，以满足人们日常生活的需要.它具有安全、节能、环保等优点.
如图1所示，太阳能热水器主要由两部分构成：一部分是妾许多根玻璃吸热管组成的集热器，每根玻璃吸热管由双层玻璃管组成，双层玻璃管之间是真空.如图1所示是玻璃吸热管的工作原理图，它的结构与保温瓶的玻璃内胆相似，只是在玻璃吸热管的内表面涂了一层黑色的吸热材料;另一部分是保温储水箱，保温储水箱下部与玻璃吸热管相通. 玻璃吸热管工作原理如图2所示：当太阳光入射到黑色的吸热层上时，黑色吸热层能把太阳能更有效地转化为内能，将管内的水加热.向光一侧的水被加热后体积增大、密度变小而向上运动;背光一侧的水由于温度较低、密度较大而向下运动，形成冷热水循环，最终整箱水都升高至一定的温度.玻璃吸热管具有吸热保温作用 . 请回答以下问题： (1)玻璃吸热管内向光一侧的水吸热后体积增大、密度变小，所受到的浮力_______重力. (2)能否将黑色的吸热材料涂在玻璃吸热管的外表面上?简单说明理 由.___________ _____________________________________________________________________ _______. (3)将质量为100kg初温为20℃的水加热到60℃，求这些水需要吸收多少热量?[水的比热容是4.2×103J/(kg.℃)] (4)某品牌太阳能热水器集热器面积为S=2m2，热效率为η=40%(即热水器能将照射到玻璃吸热管上的太阳能的40%转化为水的热能)，该地点太阳能辐射到地面的平均功率为P=1.4×103W/m2(即平均每平方米每秒钟太阳辐射能为1.4×103J).在第(3)小题中水吸收的热量由该热水器提供，求需要加热多少小时?