_60_CoGammaIrrad_省略_lVaporDeposition_Tha

Light induced instabilities in amorphous indium–gallium–zinc–oxide thin-filmtransistorsMd Delwar Hossain Chowdhury, Piero Migliorato, and Jin JangCitation: Applied Physics Letters 97, 173506 (2010); doi: 10.1063/1.3503971View online: /10.1063/1.3503971View Table of Contents: /content/aip/journal/apl/97/17?ver=pdfcovPublished by the AIP PublishingArticles you may be interested inInvestigation of zinc interstitial ions as the origin of anomalous stress-induced hump in amorphous indiumgallium zinc oxide thin film transistorsAppl. Phys. Lett. 102, 173502 (2013); 10.1063/1.4803536Light/negative bias stress instabilities in indium gallium zinc oxide thin film transistors explained by creation ofa double donorAppl. Phys. Lett. 101, 123502 (2012); 10.1063/1.4752238Electrical instability of amorphous indium-gallium-zinc oxide thin film transistors under monochromatic lightilluminationAppl. Phys. Lett. 100, 243505 (2012); 10.1063/1.4729478Effect of annealing time on bias stress and light-induced instabilities in amorphous indium–gallium–zinc-oxidethin-film transistorsJ. Appl. Phys. 110, 114503 (2011); 10.1063/1.3662869Environment-dependent thermal instability of sol-gel derived amorphous indium-gallium-zinc-oxide thin filmtransistorsAppl. Phys. Lett. 98, 152109 (2011); 10.1063/1.3580614This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: /termsconditions. Downloaded to IP:Light induced instabilities in amorphous indium–gallium–zinc–oxide thin-film transistorsMd Delwar Hossain Chowdhury,1Piero Migliorato,1,2and Jin Jang 1,a ͒1Department of Information Display,Advanced Display Research Center,Kyung Hee University,Seoul 130-701,Republic of Korea 2Department of Engineering,Electrical Engineering Division,Cambridge University,Cambridge CB30F A,United Kingdom͑Received 15July 2010;accepted 28September 2010;published online 28October 2010͒The effect of exposure to ultraviolet radiation on the characteristics of amorphous indium–gallium–zinc–oxide thin-film transistors ͑TFTs ͒fabricated by sputtering is investigated.After illumination with 1.5mW cm −2of 365nm radiation,in the absence of any bias stress,a persistent negative shift in the characteristics is observed in the dark.The magnitude of the shift increases with exposure time,saturating after about 10min.Under these conditions the subthreshold exhibits a rigid shift of around 3.6V and 7.5V for TFTs with an active layer thickness of 20nm and 50nm,respectively.The shift in the dark increases ͑decreases ͒when a negative ͑positive ͒bias stress is applied under illumination.The instability behavior caused by exposure to light,in the absence of any bias stress,can be explained on the basis of ionization of neutral oxygen vacancies.©2010American Institute of Physics .͓doi:10.1063/1.3503971͔The interest in amorphous indium–gallium–zinc–oxide ͑a-IGZO ͒thin-film transistors ͑TFTs ͒for applications in ac-tive matrix displays,1has generated a large body of experi-mental and theoretical studies devoted to this material.2For applications,reliability and stability are of primary concern and,indeed,a number of investigations of reliability under bias stress have been published.3–7An important issue for devices that may operate in a bright light environment is whether exposure to light can affect reliability.While the effect of the combined bias stress and illumination has been recently investigated,8the role of light exposure itself has not been established.We report here a new instability behavior which is caused by exposure to light,in the absence of any bias stress.An explanation,based on ionization of neutral oxygen vacancies is proposed.The inverted staggered a-IGZO TFTs were fabricated on SiO 2-buffered glass substrate.A 60nm thick molybdenum gate was produced by sputtering.The gate insulator,a 200nm thick SiO 2film,was deposited by plasma enhanced chemical vapor deposition ͑PECVD ͒,at a substrate tempera-ture ͑T s ͒of 380°C.An a-IGZO active layer was direct cur-rent ͑DC ͒sputtered from the target In 2O 3:Ga 2O 3:ZnO =1:1:1mol %,in an oxygen partial pressure O 2/͑O 2+Ar ͒=0.66and at the substrate temperature of 200°C.The IGZO was coated with a 100nm PECVD SiO 2layer at T s of 200°C and patterned into islands.Source and drain contact regions were defined in the back SiO 2,followed by 150nm molybdenum contacts deposition and patterning.Finally the overall structure was passivated with a 200nm thick SiO 2layer,deposited at 200°C,and annealed at 250°C for 2h in vacuum.The TFT I–V characteristics were measured at room temperature,using Agilent 4156C semiconductor analyzer.All TFTs had the same W/L geometry as follows:55␮m /11␮m.For devices with an active layer thickness d=20nm,typical threshold voltage ͑from I D versus V GS forV DS =0.1V ͒,mobility,and subthreshold swing were as fol-lows:5.95V ,7.3cm 2V −1s −1,and 0.157V/dec.For light illumination a UV light-emitting diode ͑LED ͒,␭=365nm,and a visible LED ␭=530nm were used.Figures 1shows the effect of repeated +20and Ϫ20V bias stress in 10min interval on the transfer characteristics of a TFT,before exposure to UV light.This test is performed in order to exclude the presence of effects associated with mo-bile charge in the gate dielectric and back channel passiva-tion.The inset in the figure shows the gate voltage shift for the repeated stress,for I D =10−11A.The very small shifts observed underscore the good stability of the devices and the effectiveness of passivating the back channel region.6Figure 2͑a ͒shows the effect of exposure to 1.5mW cm −2of ␭=365nm ͑3.4eV ͒UV light,for 10min,a ͒Author to whom correspondence should be addressed.Electronic mail:jjang@khu.ac.kr.FIG.1.͑Color online ͒Transfer characteristics of a virgin TFT with active layer thickness d=20nm,for V DS =1V,upon application of 17alternate bias stresses:V G ͑stress ͒=+20V,Ϫ20V;duration of each alternate stress:10min;V D ͑stress ͒=0V.Only five of the overlapping characteristics are shown,including the initial unstressed one.Inset:Gate voltage shift for successive stress,for I D =10−11A.APPLIED PHYSICS LETTERS 97,173506͑2010͒0003-6951/2010/97͑17͒/173506/3/$30.00©2010American Institute of Physics 97,173506-1 This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: /termsconditions. Downloaded to IP:for a TFT with active layer d=20nm.Similar effects,albeitreduced in magnitudes,are caused by prolonged ͑12h ͒ex-posure to the ␭=530nm radiation.The characteristic duringillumination shows a sizeable increase in the subthresholdcurrent,which is due to the optically generated free elec-trons.The off current also increases and a minimum appearsbetween the on and the off region.After light exposure,thedark characteristic relaxes quickly ͑within less than 1min ͒tothe one labeled as ͑3͒.Curve ͑3͒presents a rigid shift ⌬V G=3.4V with respect to the original ͑1͒in the subthresholdregion ͑I D =10pA ͒.The shift,hereafter referred to as thedark shift,is reduced upon approaching the threshold volt-age,so that V th =0.9V.We have characterized this lightinduced behavior in five nominally identical samples andfound subthreshold shifts of 3.6Ϯ0.6V.The magnitude ofthe shift increases with exposure time,saturating after about10min.The effect is not reversible at room temperature,asubthreshold dark shift of 3V remaining after -plete recovery is achieved by annealing at 250°C for 2h.Figure 2͑b ͒shows the effect of exposure to light under thesame condition as in Fig.2͑a ͒,for a TFT with active layerthickness d=50nm.The shift as a function of I D is shown inthe inset of Fig.2͑b ͒as follows:͑i ͒under illumination and͑ii ͒in the dark,for the two cases d=20and 50nm,indicating that the shift magnitude scales approximately with d.Figure 3shows the effect of simultaneous exposure to 365nm radiation,1.5mW cm −2,and bias stress.Transfer characteristics are shown for the following conditions:͑i ͒before exposure to light and stress;͑ii ͒after 10min exposure and V G ͑stress ͒=−20V,V D ͑stress ͒=0V,with light on;͑iii ͒60s after end of previous light exposure/stress;͑iv ͒after 10min exposure and V G ͑stress ͒=+20V,V D ͑stress ͒=0V,with light on;and ͑v ͒60s after end of previous light exposure/stress.The relaxation is monitored in the dark without bias stress.By comparing the curves under illumina-tion with those of Fig.2͑a ͒,one can see that for V G ͑stress ͒=−20V the subthreshold current is higher,the opposite being true when V G ͑stress ͒=+20V is paring the curves after relaxation,it is apparent that V G ͑stress ͒=−20V increases the subthreshold negative shift from Ϫ3.4to Ϫ5V at I D =10pA,whereas V G ͑stress ͒=+20V reduces it to 0.5V .The scaling of the light induced shift with active layer thickness rules out injection into the gate dielectric and/orgeneration of defects at the interface.Instead it is consistentwith light-induced generation of positive charge throughout the active layer.Considering the value 9of the absorption coefficient at the wavelength of excitation ␣Ϸ104cm −1,it is reasonable to expect that the charge is uniformly distributed and in similar concentrations,in 20and 50nm thick devices.Since the effect is not reversible at room temperature,the charge must be associated with metastable defects.The pres-ence of oxygen vacancies is well documented through opti-cal electron paramagnetic resonance ͑EPR ͒experiments in ZnO oxide single crystals.10,11These defects bind two elec-trons in their electrically neutral state V O 0and the ground state energy has been estimated by density functional theory in ZnO ͑Refs.12–14͒and a-IGZO,15indicating that the neu-tral vacancy V O 0is a deep donor.Ionization of V O 0,resulting in the singly ionized state V O +,with transition energy ␧͑0/+͒=E V +0.9eV,12,14and doubly ionized state V O ++,with tran-sition energy ␧͑0/++͒=E V +2.2/2.4eV,13–15is accompanied by substantial lattice relaxation.Due to the latter,the energy levels of V O ++are degenerate with the conduction band 12,15and expected to be metastable,since the free electrons oc-cupy lower lying states near the band minimum and need to be activated,thermally or otherwise,in order to re-establish the thermal equilibrium concentration of neutral oxygen vacancies.12Persistent photoconductivity ͑PPC ͒observed in ZnO has been explained on the basis of this model.12PPC has been recently reported in a-IGZO film fabricated byKrF FIG.2.͑Color online ͒͑a ͒Effect of exposure to 365nm radiation,1.5mW cm −2,in the absence of bias,for a TFT with active layer thickness d=20nm.͑b ͒Same for a TFT with d=50nm.Transfer characteristics,for V DS =1V:͑1͒before exposure to light;͑2͒after 10min exposure,with light on;and ͑3͒in the dark,60s after light was switched off.Inset:gate voltage shift ⌬V G vs I D for two TFTs with active layer thickness 20and 50nm;͑i ͒after 10min exposure,with light on,from curve ͑2͒;and ͑ii ͒in the dark from curve ͑3͒.The shift scales with the active layerthickness.FIG.3.͑Color online ͒Effect of simultaneous exposure to 365nm radiation,1.5mW cm −2,and bias stress for a TFT with active layer thicknessd=20nm.Transfer characteristics,for V DS =1V:͑i ͒before exposure to light and stress;͑ii ͒after 10min exposure and V G ͑stress ͒=−20V,V D ͑stress ͒=0V,with light on;͑iii ͒60s after end of previous lightexposure/stress;͑iv ͒after 10min exposure and V G ͑stress ͒=+20V,V D ͑stress ͒=0V,with light on;and ͑v ͒60s after end of previous lightexposure/stress. This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: /termsconditions. Downloaded to IP:laser deposition.9The duration of the on transients͑over 20min͒is similar to that we observe for the saturation of ⌬V G under illumination.However,in contrast with the per-sistence of the shift in our TFTs,PPC in the above work9is reported to disappear after50min at room temperature.The singly ionized vacancy V O+in ZnO is also expected to be metastable on theoretical grounds12and on the basis of EPR experiments.11In the latter,V O+has been detected only during and after exposure to␭Ϸ500nm light,causing photoionization of neutral oxygen vacancies.A period of several minutes was required at room temperature in the dark for the concentration of V O+to decay to zero.11It is therefore conceivable that the light induced effects observed in the present work are associated with generation of ionized oxy-gen vacancies.16These can be produced,under band to band optical excitation,through competitive hole capture by neu-tral oxygen vacancies.The rapid initial relaxation of the characteristics after illumination can be explained with hole-electron recombination via other nonmetastable gap states.In the subthreshold region,assuming full depletion,we have⌬V G=qC OX−1N T+d,where q is the electron charge,C ox the gate oxide capacitance,N T+the concentration of ionized oxy-gen vacancies,and d the active layer thickness.From⌬V G for I D=10−11A in curve͑3͒of Fig.2͑a͒,wefind N T+=1.8ϫ1017cm−3.However,when theflat band condition is ap-proached,the active layer will become partially depleted and the shift would be reduced.Application of positive bias stress during illumination reduces the dark shift,since the increased free electron density favors the neutral state In other words,the reactions V0++e−→V0and V02++2e−→V0 are displaced to the right.Vice versa,negative bias stress increases the holes concentration,favoring ionized states. This can explain the behavior shown in Fig.3.On the basis of the present data,it is not possible to establish whether oxygen vacancies are native or created,as well as ionized, during exposure to light.This is an important point for future investigation in view of its relevance for the stability of these devices under operating conditions.In conclusion,we have presented evidence of light in-duced instabilities in a-IGZO TFTs,in the absence of bias stress.Application of negative bias stress during light expo-sure enhances,whereas positive bias stress reduces,the ef-fect of light.The results can be explained on the basis of light induced generation of metastable defect states associ-ated with ionized oxygen vacancies.The possibility that these defects are also involved in dark bias stress instabilities warrants further investigation.On the basis of the above re-sults,the use of a light shield appears mandatory in all ap-plications where these devices are subjected to prolonged exposure to light.This work was supported by the Industrial Strategic Technology Development Program͑Project No.10035225, Development of Core Technology for High Performance AMOLED on Plastic͒funded by MKE/KEIT.1J.K.Jeong,J.H.Jeong,J.H.Choi,J.S.Im,S.H.Kim,H.W.Yang,K. N.Kang,K.S.Kim,T.K.Ahn,H.J.Chung,M.Kim,B.S.Gu,J.S.Park, Y.G.Mo,H.D.Kim,and H.K.Chung,SID Int.Symp.Digest Tech. Papers2008,1.2T.Kamiya and H.Hosono,Asia Materials2,15͑2010͒.3T.C.Fung,K.Abe,H.Kumomi,and J.Kanicki,J.Disp.Technol.5,452͑2009͒.4K.Nomura,T.Kamiya,M.Hirano,and H.Hosono,Appl.Phys.Lett.95, 013502͑2009͒.5J.M.Lee,I.T.Cho,J.H.Lee,and H.I.Kwon,Appl.Phys.Lett.93, 093504͑2008͒.6J.K.Jeong,H.W.Yang,J.H.Jeong,Y.G.Mo,and H.D.Kim,Appl. Phys.Lett.93,123508͑2008͒.7M.E.Lopes,H.L.Gomes,M.C.R.Medeiros,P.Barquinha,L.Pereira, E.Fortunato,R.Martins,and I.Ferreira,Appl.Phys.Lett.95,063502͑2009͒.8M.K.Ryu,S.H.K.Park,S.H.Yang,C.W.Byun,O.S.Kwon,E.S. Park,K.I.Cho,and C.S.Hwang,SID Int.Symp.Digest Tech.Papers 2010,1.9D.H.Lee,K.I.Kawamura,K.Nomura,H.Yanagi,T.Kamiya,M.Hirano, and H.Hosono,Thin Solid Films518,3000͑2010͒.10L.S.Vlasenko and G.D.Watkins,Phys.Rev.B71,125210͑2005͒.iho,L.S.Vlasenko,and M.P.Vlasenko,J.Appl.Phys.103,123709͑2008͒.ny and A.Zunger,Phys.Rev.B72,035215͑2005͒.13A.Janotti and C.G.Van de Walle,Phys.Rev.B76,165202͑2007͒.14S.J.Clark,J.Robertson,ny,and A.Zunger,Phys.Rev.B81, 115311͑2010͒.15W.J.Lee,B.Ryu,and K.J.Chang,Physica B404,4794͑2009͒.16B.Ryu,H.K.Noh,E.A.Choi,and K.J.Chang,Appl.Phys.Lett.97, 022108͑2010͒.This article is copyrighted as indicated in the article. Reuse of AIP content is subject to the terms at: /termsconditions. Downloaded to IP:。

SCIENCE CHINAPhysics, Mechanics & Astronomy*Corresponding author (email: lsx@)• Research Paper •December 2010 Vol.53 No.12: 2284–2292doi: 10.1007/s11433-010-4190-zA computational investigation on random packings ofsphere-spherocylinder mixturesLU Peng, LI ShuiXiang *, ZHAO Jian & MENG LingYiState Key Laboratory for Turbulence and Complex Systems, Department of Mechanics & Aerospace Engineering, College of Engineering,Peking University, Beijing 100871, ChinaReceived June 13, 2010, accepted September 8, 2010Random packings of binary mixtures of spheres and spherocylinders with the same volume and the same diameter were simu-lated by a sphere assembly model and relaxation algorithm. Simulation results show that, independently of the component volume fraction, the mixture packing density increases and then decreases with the growth of the aspect ratio of spherocyl-inders, and the packing density reaches its maximum at the aspect ratio of 0.35. With the same volume particles, results show that the dependence of the mixture packing density on the volume fraction of spherocylinders is approximately linear. With the same diameter particles, the relationship between the mixture packing density and component volume fraction is also roughly linear for short spherocylinders, but when the aspect ratio of spherocylinders is greater than 1.6, the curves turn convex which means the packing of the mixture can be denser than either the sphere or spherocylinder packing alone. To validate the sphere assembly model and relaxation algorithm, binary mixtures of spheres and random packings of spherocylinders were also simu-lated. Simulation results show the packing densities of sphere mixtures agree with previous prediction models and the results of spherocylinders correspond with the simulation results in literature.packing, binary mixture, sphere, spherocylinder, packing density, aspect ratio PACS: 45.10.Na, , 81.05.RmThe sphere and spherocylinder are both important 3D ob-jects and were widely involved in packing investigations and engineering practices. The sphere is the earliest and the most comprehensively studied shape in packing investiga-tions. Weitz [1] listed some important packing densities for identical spheres, where the density is about 0.64 for ran-dom close packing and 0.56 for random loose packing. A spherocylinder is a capsule like object which consists of a cylinder and two hemispheres on both sides. The shape of a spherocylinder is usually described by an aspect ratio w = H /D , where H is the height of the cylinder part and D is the diameter. When w = 0, a spherocylinder degenerates to a sphere. The spherocylinder is one of the well-studied non-spherical shapes, since it has a smooth surface and iseasy to be described by mathematical functions or a sphereassembly model. In general, the simulation results on the random packing of spherocylinders in literature [2–7] are quite consistent, the maximum packing density is around 0.69 ~ 0.7 when w ≈ 0.4.The densest packing has always been an important objec-tive of packing investigations and it is generally measured by the packing density Φ. A mixture of particles with dif-ferent sizes can give a much higher packing density than mono-sized particles and is widely applied in engineering. The mixture packing density, that is the packing density of the mixture, is dependent on many factors, e.g. the packing density of each component, particle size distribution, vol-ume fraction and particle shape. Among these factors, the influence of particle size distribution has been well studied. The binary mixture is the simplest case in polydisperse packing and was paid particular attention. The polydisperse© Science China Press and Springer-Verlag Berlin Heidelberg 2010LU Peng , et al . Sci China Phys Mech Astron December (2010) Vol. 53 No. 122285packings of spheres have been extensively studied [8–12] through experiments, while non-spherical particles were less studied and only a few basic shapes have been consid-ered, such as disks, cylinders and cubes [13–15]. The binary mixtures of cylinders and spheres have been investigated in some experiments. However, the aspect ratios of the cylin-ders are usually very large and the difference between the volumes of the cylinder and sphere is significant. The bi-nary mixture of short cylinders and spheres with similar volumes has rarely been investigated. When packed with unequal volume components, the quantity distinction will lead to an inadequately mixed packing under a certain vol-ume fraction. More importantly, the same volume condition will eliminate the size effect and the shape influences can be investigated independently. Charlles et al. [3] simulated the binary mixture of spherocylinders and the mixture of spherocylinders and spheres, although the component parti-cles were obviously separated in the packings. Bargiel [5] simulated the binary mixture of spherocylinders with the same volume but different aspect ratios. Kyrylyuk et al. [6] simulated the binary mixture of spherocylinders and spheres with the same diameter but different volumes.The prediction of mixture packing density is another ab-sorbing problem. Several models have been developed to predict the polydisperse packing density of spheres, e.g. simple packing model [16,17], mixture packing model [18], linear packing model [19–21], linear-mixture packing model [22] and the Westman equation [9,14]. The predic-tions of the linear packing model and Westman equation are relatively accurate, hence these two models have been widely applied and extended to non-spherical particle mix-tures. Note that, when r ≈ 1 (r = r s / r l , where r s and r l are the radii of small and large spheres, respectively), all mod-els give a linear relationship between the mixture specific volume V (V = 1/Φ) and particle volume fraction. For a bi-nary mixture, that is1122,V V X V X =+ (1) where V 1 and V 2 are the specific volumes of large and small particles in their monodisperse packings, X 1 and X 2 are their volume fractions in the mixture. Previous prediction models on packing density of non-spherical particle mixtures are all based on the models of spheres. Yu et al. [23] believed there are similarities be-tween the packing of non-spherical particles and the pack-ing of spheres. Based on the experimental results of mix-tures among spheres, cylinders, disks and cubes, they ex-tended the prediction models of spheres to non-spherical particles by replacing a non-spherical particle with the sphere having an equivalent packing diameter. Zou et al. [24] suggested an empirical formula to calculate the equivalent packing diameter d p for any non-spherical parti-cle2.785p v exp[2.946(1)],d d ΨΨ−=− (2)where d v is the diameter of the sphere with the same volume as the non-spherical particle, Ψ is the Wadell sphericity, which is defined as the ratio of the surface area of the sphere having the same volume as the particle to the surface area of the particle. Yu et al. [14,21] integrated the equiva-lent packing diameter with the linear packing model and the Westman equation. This approach had been applied to the mixtures of spheres and several kinds of non-spherical par-ticles, and the results were generally consistent with the available experimental data.With the development of particle packing studies, it was found that the packing of non-spherical particles is funda-mentally different from the sphere packing. As a result of the rotational degrees of freedom, the packing structures of non-spherical particles are usually distinct from the spheres and the shape significantly influences the packing density. In previous mixture packing experiments, the difference between the particle volumes is usually significant and only a few non-spherical particle shapes have been involved. The mixture packing of non-spherical particles with similar volumes has rarely been investigated. Hence, the present studies of mixture packing is mainly affected by the particle sizes but not the particle shapes, and there were few re-searches on the dependence between the mixture packing density and the particle shapes. Therefore, it is necessary to investigate the mixture of particles with the same volume. Kyrylyuk et al. [6] simulated the binary mixture of the spherocylinders and spheres with the same diameter. Their results show a linear relationship between the mixture packing density and the volume fraction of spherocylinders, which is independent of the aspect ratio of spherocylinders. However, according to the previous prediction models, the filling effect will observably increase the packing density if the volume distinction of component particles is significant. In the binary mixture of spheres and spherocylinders with the same diameter, the volume distinction becomes more significant with the elongation of spherocylinders, and thatshould increase the mixture packing density. Thus, the con-clusions of ref. [6] and the previous prediction models are quite different, and more comprehensive studies under the same diameter condition should be taken. Furthermore, the prediction models on the packing density of non-spherical particle mixtures are only based on a few mixture experi-ments among spheres, cylinders, disks and cubes. Its appli-cability, accuracy, and completeness on other shapes, e.g.spherocylinders, need to be validated. Additionally, sincethe density prediction of the previous models is only from the mixture composition but not the particle shape, the shape influence on the packing density of mixtures needs to be investigated as well. In recent years, with the rapid progress in computertechnology and packing algorithm, the numerical simulation2286 LUPeng, et al. Sci China Phys Mech Astron December (2010) Vol. 53 No. 12has become the dominant means in packing research. The simulation algorithms of random packing can be classified into four categories [25]: sequential addition, collective re-arrangement, advancing front and optimization approaches. The relaxation algorithm, which belongs to the class of col-lective rearrangement algorithms, gives reasonable packing densities with simple principles and low computational costs, thus it is widely applied in packing investigations. The representation models of non-spherical particles are generally sorted into analytic models and assembly models [26].As one of the assembly models, the sphere assembly model represents the shape of a non-spherical particle with the contour of overlapping spheres. With this model, the contacts between non-spherical particles can be treated as the contacts of spheres. The sphere assembly model gains an advantage over analytic models by its simple contact detection and the ability to approach arbitrary particle shapes.In this paper, we simulated the binary mixture of spheres and the random packing of identical spherocylinders with the relaxation algorithm and sphere assembly model. The results are compared with the prediction and numerical simulation results in literature for verification. Then we simulated the binary mixture of spheres and spherocylinders, with the same volume and the same diameter respectively, to investigate the shape and size influence of spherocyl-inders on the mixture packing density. The results are com-pared with the previous numerical simulation and the pre-diction approach using the concept of equivalent packing diameter. It should be noted that only geometric packings of rigid 3D particles in a cubic region with periodic boundary condition were investigated in this work.1 Relaxation algorithm and the sphere assem-bly models of spherocylindersThe relaxation algorithm for sphere packing [25] begins with a randomly placed large overlapping configuration of spheres in a cubic region. Afterwards, iterations of the re-laxation procedure are carried out to gradually reduce the overlaps of the spheres. The boundary of the packing region is enlarged at the end of each iteration step. The final pack-ing is achieved when the maximum overlap rate of all component spheres is below a predefined value. Li and Zhao [26] introduced the sphere assembly model and the rotation of non-spherical particles into the relaxation algo-rithm, thus extended the relaxation algorithm to non-spherical particle packings. In the modified algorithm, the initial position and orientation of non-spherical particles are both randomly generated, and the displacement and ro-tation are determined by the vector summations of the translations and torques of all component spheres.This work mainly follows the algorithm in refs. [25,26] but has improved on the neighboring sphere search. In the simulations of this work, there are often two types of spheres with an obvious disparity in size. If the packing region is partitioned into background cells by the large sphere diameter[25], each cell will contain a large number of small spheres and the searching efficiency will be greatly reduced. For this reason, we adopted the dual background cell method, which divides the packing region into two in-dependent background cell systems corresponding to the diameters of large and small spheres. The adjacencies be-tween small spheres are searched in the small cell system while the large cells are employed to search the adjacencies related to large spheres. With the dual background cell method, the efficiency of neighbor searching between small spheres is significantly improved, especially when the spheres with a large radius ratio or a large amount of small spheres are involved.The sphere assembly model of a spherocylinder is con-structed by a series of spheres with centers along a straight line. The diameters of all spheres are 1.0 and the distances between all the neighbor component spheres are 0.1. The aspect ratio w can be adjusted by the number of assembly spheres N, w = (N−1)/20. The volume and surface area of the assembly model are difficult to calculate but can be ob-tained easily from an AutoCAD system. Figure 1 shows the sphere assembly model of a spherocylinder with w = 1.0 and N = 21 in 3D and 2D. As the distances between neighboring assembly spheres are very small, the model is an excellent approximation to a real spherocylinder.2 Binary mixtures of spheresBinary mixtures of spheres have been well studied and are rich in experimental data. Most previous prediction models can give good estimations of the packing density of a binary mixture of spheres, such as the linear packing model and theFigure 1The sphere assembly model of a spherocylinder (w = 1.0) in 3D (upper) and 2D (lower).LU Peng , et al . Sci China Phys Mech Astron December (2010) Vol. 53 No. 122287Westman equation. With the relaxation algorithm, we simulated the binary mixture of spheres at r = 0.5 and r = 0.7, r =r s /r l . The packing results are shown in Figure 2, in which the volume fractions of large and small spheres are identical and the number of large spheres is 500. As shown in Figure 2, spheres are randomly distributed and well mixed without order or separation. The packing densities obtained and the comparisons with the Westman equation are shown in Figure 3. Since the Westman equation is es-tablished from experimental data, the random packing den-sity of identical spheres is set to 0.61 which comes from the experiments in ref. [13], while the packing density of spheres is 0.644 in our simulations. As shown in Figure 3, the packing densities obtained from numerical simulations are higher than the results of the Westman equation due to the different boundary conditions mentioned above, but the tendency of the curves and the peak positions of our simula-tions are consistent well with the prediction of the Westman equation. The consistency between the results of the nu-merical simulation and the Westman equation indicates that the relaxation algorithm is valid for binary mixtures of spheres, and the Westman equation is applicable to the bi-nary mixtures of spheres as well.3 Random close packings of spherocylindersRandom close packings of identical spherocylinders (mono- disperse packing) with aspect ratio w in the range of 0–1 were simulated in ref. [7] by the sphere assembly model and relaxation algorithm. Results showed that the maximum packing density is 0.689 when w = 0.35. In this work, we extended the range of w to 0~4 and got the same peak posi-tion and value as Ref. [7]. Figure 4 shows a packing of 1000 spherocylinders with w = 0.35. As can be seen from the figure, the positions and orientations of spherocylinders are totally random and no ordered packing appears. The pack-ing densities we obtained are plotted and compared with the simulation results in literature in Figure 5. Our results agree well with the results of Williams et al. [2], and are higher than those of Charlles et al. [3] but lower than the results in refs. [4–6]. Except for some slight differences in certain packing densities, the peak positions, peak values and ten-dencies of these curves are consistent well. The results show that the sphere assembly model and relaxation algorithm we applied in this work are applicable for the random packingof spherocylinders.Figure 2 Binary mixtures of spheres. (a) r = 0.5, (b) r= 0.7.Figure 3 The comparison between the results of the numerical simulation and the Westman equation. (a) r = 0.5, (b) r = 0.7.2288 LU Peng , et al . Sci China Phys Mech Astron December (2010) Vol. 53 No. 12Figure 4 A random close packing of spherocylinders (w= 0.35).Figure 5 The packing density versus aspect ratio w of spherocylinders.4 Binary mixtures of spheres and spherocyl-inders4.1 Equal volume particlesBinary mixtures of spheres and spherocylinders (0≤w ≤2.0) with the same volume were simulated with a sphere assem-bly model and relaxation algorithm. Figure 6 shows a binary mixture of 500 spheres and 500 spherocylinders (w =1.6) with equal volume, X sc is the volume fraction of spherocyl-inders. The simulation results are shown in Figure 7. The volume of a spherocylinder was kept identical to the volume of a sphere at all aspect ratios. As shown in Figure 7, inde-pendently of the component volume fraction, the mixture packing density increases and then decreases with the growth of the aspect ratio of spherocylinders, and the pack-ing density reaches its maximum at w = 0.35. The resultsare similar to the monodisperse packing of spherocylinders,the peak positions are the same but the peak values vary with the volume fraction of spherocylinders. Furthermore, our simulation results indicate that the values of mixture packing density are always between the monodisperse packing densities of spheres and spherocylinders (when X sc = 0, only spheres are in the packing; when X sc = 1.0, only spherocylinders are in the packing). When w = 1.6, the monodisperse packing density of spherocylinders (0.643) is the closest to the spheres in the simulation, the mixture packing densities are basically the same with the monodis-perse packing densities and all curves converge at this point. Figure 8 (a) shows the mixture packing density versus the volume fraction of spherocylinders within 0.4≤w ≤2.0. In the figure, the dependence of the mixture packing density on the spherocylinder volume fraction is roughly linear for all aspect ratios. The prediction results of the Westman equa-tion and the equivalent packing diameter are plotted inFigure 6 A binary mixture of spheres and spherocylinders (w =1.6, X sc = 0.5) with equal volume.Figure 7 The mixture packing density of spheres and spherocylinders with equal volume versus the aspect ratio of spherocylinders.LU Peng , et al . Sci China Phys Mech Astron December (2010) Vol. 53 No. 122289Figure 8 The mixture packing density versus volume fraction of spherocylinders (0.4≤w ≤2.0). (a) Simulation results; (b) prediction results.Figure 8(b). When the aspect ratio of the spherocylinder is in the range of 0–2.0, their sphericity is larger than 0.83 and d p /d v is always between 0.99 and 1.02. It means that the size ratio, which is calculated by the equivalent packing diame-ter, will approach 1.0 if the sphere and spherocylinder have the same volume. Hence, both the linear packing model and Westman equation will degenerate to eq. (1) and give the same prediction, which approaches a series of straight lines, as shown in Figure 8(b). Comparing Figure 8(a) with (b), the simulation results are consistent with the prediction re-sults in general, while the curves in Figure 8(a) are not completely linear but a little concave and the packing densi-ties are slightly lower than the prediction results in Figure 8(b). These results indicate that the previous prediction models can be used to roughly predict the mixture packing density of sphere and spherocylinder mixtures, but the pre-dictions still have some distance in terms of accuracy com-pared to the simulations. 4.2 Equal diameter particlesKyrylyuk et al. [6] simulated the binary mixture of spheres and spherocylinders with the same diameter. Their results (plotted in Figure 10(a)) show that, independently of the mixture composition, the packing can be denser when the aspect ratio of spherocylinders is perturbed slightly from zero and the maximum is always reached at one unique rod aspect ratio of about 0.45. With a sphere assembly model and relaxation algorithm, we simulated binary mixtures of spheres and spherocylinders (0≤w ≤3.0) with the same di-ameter. Figure 9 shows a binary mixture of 1698 spheres and 500 spherocylinders (w =1.6) with equal diameter. The simulation results are shown in Figure 10(b). As shown in the figure, independently of the component volume fraction, the mixture packing density increases and then decreases with the growth of the aspect ratio of spherocylinders, and the packing density reaches its maximum at w = 0.35. The results agree with the conclusions of ref. [6], and are con-sistent with the results in section 4.1 as well. Since different models and algorithm have been applied, the monodispersepacking densities of spherocylinders in Figure 10(a) are higher than our results in Figure 10(b). In Figure 10(a), the packing density of spherocylinders equals that of spheres when w =3.0 and the mixture always packs denser than spheres when 0≤w ≤3.0. In Figure 10(b), the packing densi-ties of spherocylinders approximates that of spheres at w =1.6. When w ≥1.6, the packing density of spherocylinders is lower than that of spheres and hence the mixture packs looser than spheres. Apart from this difference, the curves in Figure 10(a) and(b) have similar tendencies and peak positions, and this further validates the packing density— aspect ratio dependence obtained in this work.Figure 11(a) shows the mixture packing density versus the volume fraction of spherocylinders with 0.2≤w ≤3.0. When the diameters of the spheres and spherocylinders are identical, the volume distinction (or size ratio) increases with the elongation of spherocylinders and becomes the dominate factor in the mixture packing density. In Figure 11(a), when the aspect ratio of spherocylinders is small, thevolume difference between spheres and spherocylinders isFigure 9 A binary mixture of spheres and spherocylinders (w =1.6, X sc = 0.5) with equal diameter.2290 LU Peng , et al . Sci China Phys Mech Astron December (2010) Vol. 53 No. 12Figure 10 The mixture packing density versus the aspect ratio of spherocylinders. (a) The simulation results in ref. [6]; (b) the simulation results in this work.not significant, and the relationship between the mixture packing density and component volume fraction is also ap-proximately linear. However, the packing is obviously af-fected by the filling effect when w ≥1.6, and the curves of mixture packing density versus spherocylinder volume frac-tion become convex. The prediction results of the Westman equation and the equivalent packing diameter are plotted in Figure 11(b), where the monodisperse packing densities ofFigure 11 The mixture packing density versus volume fraction of spherocylinders (0.2≤w ≤3.0). (a) The simulation results in this work; (b) the prediction results (using the monodisperse packing densities in this work); (c) the simulation results in ref. [6]; (d) the prediction results (using the monodisperse pack-ing densities in ref. [6]).LU Peng , et al . Sci China Phys Mech Astron December (2010) Vol. 53 No. 122291spheres and spherocylinders in our simulations are em-ployed. Comparing Figure 11(a) with (b), although there are some differences in certain packing densities, the tendency of simulation results is consistent with the prediction in general. Figure 11(c) shows the mixture packing density versus the volume fraction of spherocylinders in ref. [6]. In this figure, the dependence of the mixture packing density on the spherocylinder volume fraction is always linear for all aspect ratios. Figure 11(d) shows the prediction results of the Westman equation and the equivalent packing diameter, where the monodisperse packing densities of spheres and spherocylinders in ref. [6] are applied. Significant differ-ence can be found between the tendencies of the curves in Figure 11(c) and (d). As shown in Figure 11 (b) and (d), no matter what monodisperse packing densities are applied, the previous prediction models always propose that the filling effect will observably increase the packing density if the aspect ratios of spherocylinders are large enough and the curves become convex. For example in Figure 12, when w =3.0, although our results are slightly lower than the pre-diction value, the trend of the convex curve is consistent with the prediction results and reflects the influences of the filling effect on the mixture packing density. In comparison, a linear trend appears in the simulation results in ref. [6] which is completely different from the prediction results. In summary, our results agree with both ref. [6] and prediction results for short spherocylinders; for long spherocylinders, our results are closer to the previous prediction models rather than the simulation results in ref. [6]. Additionally, it should be noted that the prediction values are always a little higher than the simulation results in this work.5 ConclusionsIn this paper, we simulated the binary mixtures of spheres,the random packings of spherocylinders and the binaryFigure 12 The mixture packing density versus volume fraction of spherocylinders (w =3.0).mixtures of spheres and spherocylinders with the same volume and the same diameter. The numerical simulation results on the binary mixture of spheres are consistent with the previous prediction models, which indicate that the re-laxation algorithm is valid for the binary mixtures of spheres, and the previous prediction models give reasonable results as well. The maximum packing density of 0.689 (w = 0.35) was obtained from the numerical simulations of the random close packing of spherocylinders. The maximum packing density, optimal aspect ratio and tendency of the packing density curve we obtained agree well with the nu-merical simulation results in literature, which demonstrates that the relaxation algorithm and sphere assembly model are applicable to the random packing of spherocylinders. The numerical simulation results of the binary mixture of spherocylinders and spheres with equal volume and equal diameter show that, independently of the mixture composi-tion, the mixture packing density increases and then de-creases with the growth of w and the maximum packing densities are reached at w = 0.35. These results agree with the conclusions on the mixture of spheres and spherocyl-inders with the same diameter in ref. [6]. Simulation results also show that, when the particle volume is identical, the dependence of the mixture packing density on the sphero-cylinder volume fraction is approximately linear for all as-pect ratios, and this is consistent with the prediction results in general; when the particle diameter is identical, the rela-tionship between the mixture packing density and compo-nent volume fractions is also linear for short spherocyl-inders. But when w ≥1.6, the filling effect, which is brought by the volume distinction between spheres and spherocyl-inders, becomes the dominate factor in the mixture packing density. The curves of mixture packing density versus spherocylinder volume fraction become convex and these results agree with the prediction but are completely different from ref. [6]. The consistency between our simulation re-sults and the prediction results also indicates that the previ-ous prediction models are applicable to generally predict the variation of mixture packing density. Meanwhile, the cer-tain differences between the simulation and prediction re-sults imply that the previous empirical prediction models may need to be improved yet.In this work, the binary mixture of spheres and sphero-cylinders with the same volume and the same diameter is investigated to evaluate the influences of the particle shape and size ratio on the mixture packing density. These rules we obtained are generally applicable to universal mixture packings, but for the accurate prediction of the mixture packing densities of sphere-spherocylinder mixtures or the particle mixtures of other shapes, more comprehensive and systematical studies should be undertaken.This work was supported by the National Natural Science Foundation of China (Grant No. 10772005) and the National Basic Research Program of China (Grant Nos. 2007CB714603 and 2010CB832701).。

ANNEX CDESCRIPTIONS OF MAIN PROCESSESThis Annex contains the descriptions of the main processes determining POP environmental behavior used in the participating models.C.1. Gas/particle partitioning EVN-BETR and UK-MODELThe gas-particle partitioning is described with the help of the Finizio Aerosol Partition coefficient K QA . It ’s dependence on the octanol-air partition coefficient K oa is depicted by the following formula:K QA = 3.5 · K oaThe fugacity capacity of the bulk air compartment can then be written as the sum of the gaseous and particle-bound chemical fraction:(1 – particles in air volume fraction) · Z air + (particles in air volume fraction) · K QA · Z airwhere Z air - 1/ (R ·T ) is the fugacity capacity in air;T - corrected environmental temperature for annual mean of 90C; R - gas constant = 8.314 Pa ·m 3/mol K;Particles in air volume fraction - 2·10-11;K oa = K ow / K aw - 51616649 for PCB 153 at the averaged ambient temperature T; Averaged ambient temperature = 9︒C (base temperature).CliMoChemcited from [Scheringer et al ., 2003]The gas/particle partitioning is calculated as follows [Finizio et al ., 1997]:238550.lg .-⎪⎪⎭⎫⎝⎛⋅=howpartair KK K This equation is used to calculate the fraction Phi, which indicates the particle-bound fraction of the substance. Phi-values range from 0-1.lg 60.55lg 2.23K ow k K partair partair K h ⎛⎫⎪=+=⋅- ⎪⎝⎭in (m 3/g) ()tsp k tspk Phi partair partair ⋅+⋅=1G-CIEMSWhen K oa is not available as input:K qa = 6 · 106/P ls ,where K qa is dimensionless particle/gas partition coefficient and P ls is liquid vapour pressure. Final partitioning is calculated with TSP and density of aerosol particles in fugacity format. Vapour pressure is temperature corrected when the temperature is different from 25 ︒C. When K oa is available as input:K qa = y · K oa / (1000),where, K qa is dimensionless particle/gas partition coefficient, y is organic matter mass fraction, andis the density of aerosol particles.(Note: G-CIEMS model can calculate V/P partitioning from only molecular weight (for preliminary assessment purpose) or from K oa . Two output 1 and 2 is presented in Chapter 4 as G-CIEMS-1 and G-SIEMS-2).DEHM-POPThe gas-particle partitioning is calculated using the absorption model:,)(1+=φTSP K TSP K p pwhereis the fraction of compound sorbed to particles, K p is gas-particle partitioningcoefficient, and TSP is the total suspended particulate matter [e.g. Falconer and Harner , 2000]. K p is calculated using the K oa approach:log ,91.11log log -+=om oa r p f K m Kwhere m r is a constant expected to have a value close to +1 for equilibrium partitioning, K oa is the octanol-air partitioning coefficient, f om is the fraction of organic matter in the particles, and 11.91 is a constant determined by the intercept br = log f om – 11.91 [Finizio et al ., 1997, Falconerand Harner , 2000].The temperature dependent K oa is calculated using the expression:)),11(ex p()()(TT R U T K T K ref oa ref oa oa -∆= where ∆U oa is the internal energy of phase transfer, R is the universal gas constant, T is the temperature and K oa (T ref ) is the value of K oa at the reference temperature T ref [Beyer et al., 2002].SimpleBoxcited from [Brandes et al ., 1996]Air-aerosol partition coefficients are usually not known. However, some information is frequently available on the fraction of the chemical that occurs in association with the aerosol phase.SimpleBox uses this information for the computations. A value for the fraction of the chemical that is associated with the aerosol phase, FRass aerosol, can be entered directly, or estimated on the basis of the chemical's vapor pressure, according to Junge [1977]. In this equation, the sub-cooled liquid vapour pressure should be used. For solids, a correction is applied according to Mackay [1991]:If MELTINGPOINT < TEMPERATURE [S] (substance is liquid):θθCONST.+T URE VAPORPRESS CONST.= FRass S aerosol )(][ If MELTINGPOINT > TEMPERATURE [S] (substance is solid):θθ-CONST.+e T URE VAPORPRESS CONST.= FRass S E TE M P E RA TUR NTM E LTINGP OI S aerosol ).(.][][).(1796with FRassa erosol[S] - fraction of the chemical in air that is associated with aerosol particles at scale S [-] (A);VAPOR PRESSURE(T) - vapor pressure of the chemical at temperature T at scale S [Pa] (A);MELTINGPOINT - melting point of the chemical [K] (A); CONST- constant [Pa ·m] (C);- surface area of aerosol phase [m aerosol 2/m air 3] (C);TEMPERATURE [S ] - temperature at the air-water interface at scale S [K] (A).with the productCONSTset equal to 10-4Pa.CAM/POPsIn the CAM/POPs model, the process of POP partitioning between the gas and particulate phases in atmosphere is based on Junge-Pankow adsorption model [Junge , 1977; Pankow , 1987] POP fraction Φ adsorbed on the atmospheric aerosol particles is given by:Θ⋅+Θ⋅=Φc P c L 0where, Φ - fraction of POPs adsorbed on aerosol particles;Θ - aerosol surface area available for adsorption, m 2aerosol/m 3air;P 0L - liquid-phase saturation vapour pressure of pure compound, Pa;c - parameter that depends on the thermodynamics of the adsorption process and surfaceproperties of the aerosol (Pa · cm).Junge ’s proposed value of the parameter c is 17.2 Pa · cm [P ankow , 1987; Falconer et al., 1994;Bidleman et al., 1998].The liquid vapour pressure, P 0L , are derived from:b TmP L +=010log , where the slope (m ) and the intercept (b ) are estimated to calculate liquid vapour pressure of POPs with changing air temperature [Falconer et al., 1995; Harner et al., 1996]. Temperature dependence of P 0L for each congener can be seen in Table C.1.Table C.1. Liquid vapour pressure of PCBs as a function of air temperatureAerosol surface area, , is calculated by multiplying aerosol number density by its wet surface area.MSCE-POPIn the current model version (MSCE-POP 1) the characterization of POP partitioning between the gas and particulate phase of a pollutant is performed using subcooled liquid vapour pressure p ol (Pa). According to the Junge-Pankow adsorption model [Junge , 1977; Pankow , 1987] POP fraction ϕ adsorbed on the atmospheric aerosol particles equals to:θθϕ⋅+⋅=c p c olwhere c is the constant depending on thermodynamic parameters of adsorption process and onproperties of aerosol particle surface. It assumed c = 0.17 Pa ·m [Junge , 1977] for background aerosol;is the specific surface of aerosol particles, m 2/m 3. Assumed = 1.5·10-4[Whitby ,1978].The temperature dependence of p ol (Pa) is parameterized in the model by:⎪⎭⎫ ⎝⎛--⋅=0110T T a olol P e p p ,where T 0 = 283.15 K is the reference temperature, T (K) is the ambient temperature, p 0ol is p ol value at reference temperature, and a P is the coefficient of temperature dependence. The values of p 0olanda P for considered PCB congeners used in the model are presented in Table C.2.Table C.2. Coefficients of p ol temperature dependence for three PCB congeners used in MSCE-POP modelAt present the work on modification of the description of gas aerosol partitioning within MSCE-POP model is ongoing. The approach using the octanol-air partitioning coefficient absorption model presented in [Falconer and Harner, 2000] is tested. Under this approach POP fraction ϕ adsorbed on the atmospheric aerosol particles is calculated as:TSPK TSP K p p ⋅+⋅=ϕ1where K p is the particle-gas partitioning coefficient and TSP is the total suspended particle concentration (μg ⋅m -3). The constant K p is calculated for PCBs via K oa by the following regression equations [Falconer and Harner, 2000]:log K p = log K oa + log f om – 11.91,(experimental version - MSCE-POP 2)where K oa is the octanol/air partitioning coefficient and f om is the fraction of organic matter in the atmospheric aerosol involved in partitioning.The temperature dependence of K oa is parameterized in the model by:⎪⎭⎫ ⎝⎛--⋅=0110T T a oaoa K eK K ,where K 0oa is K oa value at reference temperature, and a K is the coefficient of temperature dependence.The values of K 0oa and a for considered PCB congeners used in the model are presented in Table C.3.Table C.3. Coefficients of K oa temperature dependence for three PCB congeners used in MSCE-POP modelEVN-BETR and UK-MODELThe intermedia transport of chemicals is described using D-values (mol/Pa ·h), which represent how fast advective and reactive/degradation processes are occurring. In the case of the air to surface exchange, the D-value defining dry particle deposition is:Dair-surface = Surface Area · Particles in air volume fraction · V q · Z air · K QAKnowing these values can help calculate the flux of a chemical entering a region and, thus, it ’s amount in the compartment under study.Surface area - compartment specific;Particles in air volume fraction - 2 10-11; Vq - dry deposition velocity = 10.8 m/h.CliMoChemcited from [Scheringer et al ., 2003]Dry deposition to baresoil a , water, vegetation-covered soil bgas gas i drygas C V A v Phi dtdC ⋅⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛⋅-=aIf the year consists of exactly four seasons with varying temperatures, v dry for deposition to baresoil ischanging taking into account that in the cold season the atmosphere is more stable and the deposition rate therefore is smaller. The spring and fall values are interpolations between the summer and winter values. v dry changes as follows [Wania and McLachlan , 2001]:bIf the year consists of exactly four seasons with varying temperatures, v dry for deposition to vegetation-coveredsoil is changing taking into account that in the cold season the atmosphere is more stable and the deposition rate therefore is smaller. The spring and fall values are interpolations between the summer and winter values.v dry changes as follows [Wania and McLachlan , 2001]:The vegetation cover consists of three types: Grass, Coniferous Forest and Deciduous Forest. The variables VegGrass, VegCon and VegDec describe the fraction of the vegetation-covered soil occupied by the different cover types. Their numeric value is between 0-1 and depends on the climatic zone.Dry deposition to vegetationgas gas veg gas C V A vdry Phi dtdC ⋅⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅-=* - the model contains three types of vegetation. For each type, the deposition rate (v dry ) is different (see table below). Depending on the composition of a climatic zone, v dry is calculated as follows:vdry fractiongrass vdry fractiondec vdry fractioncon vdry i i grass i dec i con =⋅+⋅+⋅Because of increased stability of the atmosphere in the spring, fall and winterseason, the deposition rates vdrygrass, vdrydec and vdrycon are divided by 3 for the winterseason and by 2 for the spring and fall seasons (given that the year consists of exactly four seasons with varying temperature [Wania and McLachlan , 2001]).G-CIEMSF = v Dep · (TSP/) · C particleWhere F is mass flux of compound for this chemical, v Dep is dry deposition velocity of particles, TSP is particulate concentration of weight/volume dimension is density of aerosol particles,C particle is compound volumetric concentration in particles. Same value is used on all land and watersurfaces.DEHM-POPThe dry deposition of particulate phase is calculated as a flux given by the atmospheric concentration times a deposition velocity [Christensen , 1997]. The deposition velocity is highly dependent on the meteorological conditions and the surface properties. The size of the particles is assumed to be 1m [Christensen , 1997].For unstable conditions in the atmosphere (when L < 0, i.e. at day time with clear sky), the deposition velocity is calculated using:),)300(1()3/2(*u v d -+=where u* is the surface friction velocity, a is a constant depending on the surface properties, andL is the Monin-Obukhov length.For stable conditions in the atmosphere (when L > 0, i.e. at night time with clear sky), the deposition velocity is calculated using:.*au v d =The surface friction velocity is calculated using:,*)10ln(*35.0*zUu =where U is the wind speed and z is the roughness length which depends on the properties of the surface, and varies seasonally.The Monin-Obukhov length is calculated using:,)(.)(*ρ-⋅⋅⋅=p dc H g Tu L 3503 where T is the temperature, g is the gravitational constant (g = 9.82), H d is the heat flux, c p is the specific heat at constant pressure, and is the air density. L is positive (stable atmosphere)when the heat flux is negative (night time clear sky) and negative (unstable atmosphere) when theheat flux is positive (day time clear sky).SimpleBoxcited from [Brandes et al ., 1996]Value for the deposition mass transfer coefficients DRYDEP aerosol may be obtained by means of:FRass .RATE AEROSOLDEP = DRYDEP S aerosol S S aerosol ][][][with DRYDEP aerosol [S ]: mass transfer coefficient for dry deposition of aerosol-associated chemical atscale S [m air /s] (D);AEROSOLDEPRATE [S ] - deposition velocity of the aerosol particles at scale S with which thechemical is associated [m/s] (A);FRassaerosol [S ]: fraction of the chemical in air that is associated with aerosol particles atscale S [-] (A).Deposition mass flows of the chemical depend on the rate of dry aerosol deposition. Deposition velocities of aerosols vary greatly with the size of the particles. As chemicals may be associated with particles of a specific size, the deposition velocities depend also on the chemical. The values given are typical values, to be used as a starting point:scm = RATE AEROSOLDEP S /1.0][with AEROSOLDEPRATE [S ]: deposition velocity of the aerosol particles with which the chemicalis associated at scale S [m/s] (A)CAM/POPsRemoval of POPs particles is coupled with the dry deposition of aerosols in the CAM model. The dry deposition flux can be written as:,p d d C v F ⋅-=where V d is the deposition velocity and C p is the particle concentration [G ong et al. 2003]. Dry deposition flux of POPs gas is written as:,g d d C v F ⋅-=where the dry deposition velocity of gas, V d , is calculated in CAM model [G ong et al. 2003].MSCE-POPDry deposition to grass and bare soilAccording to model of [Sehmel , 1980], deposition flux over grass is calculated as(),*so il C soil soil p z B u A C F 02+⋅=where as above *u , is the friction velocity;z 0 is the surface roughness, mm;A soil = 0.02,B soil = 0.01,C soil = 0.33.Dry deposition to forestAccording to model of [Ruijgrok et al., 1997], deposition flux over forest is calculated ashp u u EC F 2*⋅=,where u h is the wind speed at forest height h = z b ;*u is the friction velocity, m/s;()()()20/80ex p 1*-+=RH u E γαβ, α = 0.048, β = 0.3 and γ = 0.25, RH =80.Dry deposition to seawaterAccording to model of [Lindfors et al., 1991], deposition flux over seawater is calculated as: )(*sea sea p B u A C F +⋅=2,where*u is the friction velocity, m/s;A sea = 0.15,B sea = 0.013Similar to DEHM-POP model values of deposition flux depend on meteorological conditions (friction velocity u *). Therefore, in the results of calculation experiments we present the range of obtained values of the flux together with its value at average environmental parameters.C.3. Wet deposition EVN-BETR and UK-MODELIn a similar way as for the dry particle deposition, wet scavenging is defined as the result of:D air-surface = Q · Surface Area · Particles in air fraction · K QA · U R · Z airIn the case of deposition in vegetation, the percentage of rain interception due to vegetation is taken into account.U R - rain rate = 8.84 x 10-5 m/h; Q - Rain Scavenging ratio = 200000;Or Snow Scavenging Ratio = 1000000CliMoChemcited from [Scheringer et al ., 2003]Wet deposition from gaseous phase to baresoil and water()gas h gas i rain gas C K V A v Phi dtdC ⋅⎪⎪⎭⎫⎝⎛⋅⎪⎪⎭⎫ ⎝⎛⋅--=11Wet deposition from gaseous phase to vegetation-covered soil()()gas rt h gas vsoil rain gas C f K V A v Phi dtdC ⋅-⋅⎪⎪⎭⎫⎝⎛⋅⎪⎪⎭⎫ ⎝⎛⋅--=111*- for deciduous and coniferous forest, the f rf -value is 0.35 for the summer season, for grass, the f rf -value is 0.12 for the summer season. In the winter season, the value is at 10% of the summer season, in spring and fall season, the value is at the linear interpolation value between summer and winter season. Because the composition of the vegetation varies with the climatic zones, the contributions of grass, coniferous and deciduous forest to the overall f rf -value of a specific climate zone differ and are proportional to the fraction of the respective vegetation type in a climatic zone.Wet deposition from gaseous phase to vegetation()gas rt h gas vsoil rain gas C f K V A v Phi dtdC ⋅⋅⎪⎪⎭⎫⎝⎛⋅⎪⎪⎭⎫ ⎝⎛⋅--=11* - for deciduous and coniferous forest, the f rf -value is 0.35 for the summer season, for grass, the f rf -value is 0.12 for the summer season. In the winter season, the value is at 10% of the summer season, in spring and fall season, the value is at the linear interpolation value between summer and winter season. Because the composition of the vegetation varies with the climatic zones, the contributions of grass, coniferous and deciduous forest to the overall f rf -value of a specific climate zone differ and are proportional to the fraction of the respective vegetation type in a climatic zone.Wet deposition from particulate phase to bare soil and watergas gas i ratiorain gas C V A scav v Phi dtdC ⋅⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛⋅⋅-=Wet deposition from particulate phase to vegetation-covered soil()gas rt gas vsoil ratiorain gas C f V A scav v Phi dtdC ⋅-⋅⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫ ⎝⎛⋅⋅-=1* - for deciduous and coniferous forest, the f rf -value is 0.35 for the summer season, for grass, the f rf -value is 0.12 for the summer season. In the winter season, the value is at 10% of the summer season, in spring and fall season, the value is at the linear interpolation value between summer and winter season. Because the composition of the vegetation varies with the climatic zones, the contributions of grass, coniferous and deciduous forest to the overall f rf -value of a specific climate zone differ and are proportional to the fraction of the respective vegetation type in a climatic zone.Wet deposition from particulate phase to vegetationgas rt gas vsoil ratiorain gas C f V A scav v Phi dtdC ⋅⋅⎪⎪⎭⎫⎝⎛⎪⎪⎭⎫⎝⎛⋅⋅-=*- for deciduous and coniferous forest, the f rf -value is 0.35 for the summer season, for grass, the f rf -value is 0.12 for the summer season. In the winter season, the value is at 10% of the summer season, in spring and fall season, the value is at the linear interpolation value between summer and winter season. Because the composition of the vegetation varies with the climatic zones, the contributions of grass, coniferous and deciduous forest to the overall f rf -value of a specific climate zone differ and are proportional to the fraction of the respective vegetation type in a climatic zone.G-CIEMSF = R rain · C air / H +(TSP/) · Q · C particle ,where F is total mass flux by this process, R rain is rain rate, C air is gaseous concentration of chemical, H is Henry ’s law constant, TSP is particulate concentration,is density of particles,Q is scavenging ratio of particles, and C particle is volumetric chemical concentration in particles.Same Q value is assumed over all land and water surfaces.DEHM-POPThe wet deposition is given as a flux calculated as the product between the air concentration and a scavenging ratio [Christensen , 1997]. Different scavenging ratios are used for in-cloud scavenging and below-cloud scavenging. It is assumed that air pollution is scavenged more efficient in the clouds than below the clouds. The below cloud scavenging rate at a given heightis given by:,)()(wa bc H P W ρσΛ=σwherebcis the below cloud scavenging coefficient, P a is the total precipitation at the level, His an effective thickness for scavenging (H = 1000 m), and wis the density of water. The in cloudscavenging rate at a given heightis given by:,)()(wc H P W ρσΛ=σ wherecis the below cloud scavenging coefficient, P a is the is the total precipitation createdinside the cloud layer. The used scavenging ratios are: bc= 100000 andc= 700000.The total amount of pollutant scavenged by the wet deposition is then dependent on the actual height of the formation of the precipitation, and on the vertical concentration profile.SimpleBoxcited from [Brandes et al ., 1996]Value for the deposition mass transfer coefficient WASHOUT may be obtained by means of:][][][S S S .SCAVratio RAINRATE = WASHOUTwith WASHOUT [S ] - mass transfer coefficient for wet atmospheric deposition at scale S [m air ·s -1] (D);RAINRATE [S ]- rate of wet precipitation at scale S [m rain .s -1] (A);SCAVratio [S ] - scavenging ratio (quotient of the total concentration in rainwater and thetotal concentration in air) of the chemical at scale S [-] (A).The scavenging ratio may be known from measurements or estimated:][][][][][1S S aerosol S water air S aerosol S COLLECTeff FRass K FRass SCAVratio +-=-with SCAVratio [S ] - scavenging ratio (quotient of the total concentration in rainwater and thetotal concentration in air) of the chemical at scale S [-] (A);FRass aerosol [S ] - fraction of the chemical in air that is associated with aerosol particles atscale S [-] (A);K air-water [S ] - air-water equilibrium distribution constant at scale S [mol ·m air -3/mol ·m water -3](A);COLLECTeff [S ] - aerosol collection efficiency at scale S [-] (A),The first term represents an estimate of the (equilibrium) distribution between rain water in air and the gas phase of air. The second term represents the scavenging of aerosol particles by rain droplets. The proportionality constant of 2⋅105is taken from Mackay [1991].Deposition mass flows of the chemical depend on the rate of wet precipitation. Collection efficiencies of aerosols vary greatly with the size of the particles. As chemicals may be associated with particles of a specific size, the collection efficiencies depends also on the chemical. The values given are typical values, to be used as a starting point:5][10.2 = COLLECTeff Swith COLLECTeff [S ] - aerosol collection efficiency at scale S [-] (A ).Table C4. RAINRATE of the scales* - from Wania and Mackay [1995]CAM/POPsThe precipitation scavenging of POPs particles by falling rain or snow is coupled with the wet removal of aerosols in the CAM model.The particulate wet deposition flux, F p , can be written as:p snow or rain p C h F )(ψ-=,where h is the falling distance, C p is the particulate phase POPs concentration, ψ is the scavenging rate for rain or snow [Gong et al. 1997].The gas phase POPs are assumed to be in quasi-steady equilibrium with the rain drop. The air-water equilibrium coefficient, K aw , is a dimensionless partition coefficient that can be derived from Henry ’s Law constant, H (Pa ·m 3/mol) [Seinfeld , 1986].K aw = H / (R ·T)and, H = H 0 exp (a(1/T 0 - 1/T)),where T is the temperature (K) and R is the gas constant (8.314 Pa ·m 3/mol ·K or J/mol/K), H 0 is the value at the reference temperature T 0, and a is a parameter of temperature dependence.The net wet deposition flux, F w , is then written as:G A W w C K p F ⋅-=)/(where p is the precipitation rate, usually reported in mm/h and C G is the gas phase POPs concentration.MSCE-POPWet deposition of particulate phaseThe values of concentration in precipitation are given by the formula:pap s w C W C =,where paC - the particle bound phase concentration in the air surface layer, ng/m 3;swC - the suspended phase concentration in precipitation water, ng/m 3; W p =1.5 ⋅ 105 - the dimensionless washout ratio for the particulate phase.Wet deposition of gaseous phasegag d w C W C =,where dw C - the dissolved phase concentration in precipitation water, ng/m 3;g a C - the gaseous phase concentration in air, ng/m 3;W g = 1/K H - the dimensionless washout ratio for the gaseous phase;K H - the dimensionless Henry ’s law constant.C.4. Gaseous exchange between atmosphere and soil EVN-BETR and UK-MODELAir-soil diffusionD air-soil = (Soil Area · Z air ) / [(Z air / (MTC as · Z air + MTC sw · Z water )) + 1 / MTC sabl ] whereAverage soil depth = 10 cm; Soil Area = 8.36 1012 m 2; Z water = Z air / K AW = 543 mol/m 3 Pa;MTC as - soil air-phase diffusion transport velocity = 0.04 m/h; MTC sw - soil water-phase diffusion transport velocity = 1 x 10-5 m/h; MTC sabl - soil air boundary layer transport velocity = 1 m/h.Air-soil rain dissolutionD air-soil = Soil Area · U R · Z waterCliMoChemcited from [Scheringer et al ., 2003]Diffusion from atmosphere to baresoil and vegetation-covered soil()gas gas i K foc rhoom regc K h gasS gas C V A vS K vL vG v Phi dtdC ow h⋅⋅⎪⎪⎭⎪⎪⎬⎫⎪⎪⎩⎪⎪⎨⎧+++--=-⋅⋅⋅1111Analogous to the procedure in subsection C.3, the parameter v gasS is changed if the year consists of exactly four seasons with varying temperatures (compare tables below).baresoil:vegetation-covered soil:The vegetation cover consists of three types: Grass, Coniferous Forest and Deciduous Forest. The variables VegGrass, VegCon and VegDec describe the fraction of the vegetation-covered soil occupied by the different cover types. Their numeric value is between 0-1 and depend on the climatic zone.Calculation of vG, vL, vS [Jury et al., 1983].When calculating the v i-value, the corresponding Di-value must be used (eg. for calculating vG, the DG-value is used):()⎥⎦⎤⎢⎣⎡⋅++--⋅⋅⋅⋅=h airsoil wsoil airsoil wsoil ow OC h h ii K frac frac frac frac K f rhoom regc K D v soil 12/d (calculation see below) ()⎪⎪⎭⎫ ⎝⎛+⋅=231043.0wsoil airsoil airsoil frac fracfrac DG()⎪⎪⎭⎫ ⎝⎛+⋅=310000043.0wsoil airsoil wsoil frac fracfrac DL Calculation of the fraction of organic matter (f OC ) in the vegetation covered soilThe model contains three types of vegetation. For each type, the f OC is different (see table below). Depending on the composition of a climatic zone, f OC is calculated as follows:co n d e c g r a ss OC i OC i OC i i OC f n fractionco f c fractionde f ass fractiongr f ⋅+⋅+⋅=fraction of organic matter in coniferous forest covered soilDiffusion from baresoil and vegetation-covered soil to the atmospheresoil gas i K f rhoom regc K h gasS K f rhoomregc frac K f rhoom regc K airsoil wsoil airsoil rhopart tsp partair rhopart tsp ow OC hgas C V A vSK vL vG v frac frac frac rhopart k K f rhoom regc K dtdC owOC how OC wsoil ow OC h⋅⋅⎪⎪⎭⎪⎪⎬⎫⎪⎪⎩⎪⎪⎨⎧+++⎪⎪⎪⎭⎫ ⎝⎛⋅⋅+--⋅⋅+-⋅⋅⋅⋅=-⋅⋅⋅⋅⋅⋅⋅⋅⋅11111For explanation of the grey part in the formula see subsection 1.。

The problems of the first law1.1 a lead bullet is fired at a frigid surface. At what speed must it travel to melt on impact, if its initial temperature is 25℃ and heating of the rigid surface of the rigid surface is neglected? The melting point of lead is 327℃. The molar heat of fusion of the lead is 4.8kJ/mol. The molar heat capacity C P of lead may be taken as 29.3J/(mol K)Solution: )/(363102.20721]108.4)25327(3.29[2121)(23322s m V v n n WQ nMv mv W H T C n Q Q Q absorb melting p melt increase absorb =⨯=⨯+-⨯===∆+∆=+=-1.2 what is the average power production in watts of a person who burns 2500 kcal of food in a day? Estimate the average additional powder production of 75Kg man who is climbing a mountain at the rate of 20 m/minSolution )/(24560208.975)/(12160602410467000//)(104670001868.4102500sin 3S J t h mg P S J t Q t W PJ Q gincrea Burning Burning =⨯⨯=∆==⨯⨯====⨯⨯=1.3 One cubic decimeter (1 dm 3) of water is broken into droplets having a diameter of one micrometer (1 um) at 20℃.(a) what is the total area of the droplets?(b) Calculate the minimum work required to produce the droplets. Assume that the droplets are rest (have zerovelocity)Water have a surface tension of 72.75 dyn/cm at 20℃ (NOTES: the term surface energy (ene/cm 2) is also used for surface tension dyn/cm)Solution)(6.436)106103(1075.72)(106)105.0(4)105.0(34)101(232523263631J S W m nS S Single total =⨯-⨯⨯⨯=∆=⨯=⨯⨯⨯⨯⨯⨯⨯⨯==-+----σππ1.4 Gaseous helium is to be used to quench a hot piece of metal. The helium is in storage in an insulated tank with a volume of 50 L and a temperature of 25℃, the pressure is 10 atm. Assume that helium is an ideal gas.(a) when the valve is opened and the gas escapes into the quench chamber (pressure=1 atm), what will be thetemperature of the first gas to hit the specimen?(b) As the helium flows, the pressure in the tank drops. What will be the temperature of the helium entering thequench chamber when the pressure in the tank has fallen to 1 atm?Solution: )(180118298)(1185.229810101325501010101325)5500(1)()(118)101(298)()(0334.0/00K T T T K RR nC W T b K T P P T T Adiabatica p CR P=-=∆-==⨯⨯⨯⨯⨯⨯⨯-⨯==∆=⨯==--1.5 An evacuated (P=0), insulted tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature T 0. The valve on the tank is oðened and the surrounding gas is allowed to flow suickly into t(e tank until the pressure insi`e the tank is equals the pressure outside. Assume that no heat flow takes place. What is the0final tempeture kf t èe gaS in the tank? The heat cap!city mf the gas, C p and C v each íay be(assumed to be c/nsuant over thå temperature rang!spanNed by the døperiment. You answer may be meft in terms of C p and S vM hint: one way to approach the xroblem is to define the system as the gas ends up in the tank. hint: one way to approach the xroblem is to define the system as the gas ends up in the tank.solution 0/000/00)()(T P P T T P PT T Adiabatic PPC R C R ≈-==1.6 Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO 2 and CH 4 (gas)[This heat of reaction is also called the low calorific power of methane] convert the answer into unites of Btu/1000 SCF of methane. SCF means standard cubic feet, taken at 298 and 1atm NOTE: this value is a good approximation for the low calorific powder of natural gasDATA: )()()(224g O H g CO g CH FOR80.5705.9489.17]/[0298---∙∆mol g Kcal Hsolution)1000/(9.2610252103048.01101076.191)/(76.191)89.1780.57205.94()2(22333332982982224422SCF Btu mol g Kcal H H H H H OH CO O CH CH O H CO =⨯⨯⨯⨯⨯=∙=∆+⨯---=∆-∆+∆-=∆+=+- 1.7Methane is delivered at 298 K to a glass factory, which operates a melting furnace at 1600 K. The fuel is mixed with a quantity of air, also at 298 K, which is 10% in excess of the amount theoretically needed for complete combustion (air is approximately 21% O 2 and 79% N 2)(a) Assuming complete combustion, what is the composition of the flue gas (the gas following combustion)? (b) What is the temperature of the gas, assuming no heat loss?(c) The furnace processes 2000kg of glass hourly, and its heat losses to the surroundings average 400000 kJ/h.calculate the fuel consumption at STP (in m 3/h) assuming that for gas H 1600-H 298=1200KJ/KG(d) A heat exchanger is installed to transfer some of the sensible heat of the flue gas to the combustion air.Calculate the decrease in fuel consumption if the combustion air is heated to 800K DATA STP means T=298K, P=1atm22224O N O H CO CH for2.82.89.117.1316)/(C mol cal C P ∙Solution)(210448.1125.9100076.191298)/(25.9)]87.012.72(2.843.179.1171.87.13[01.0)(%87.0%%12.72%%43.17%2%%71.8)11.1(221791.1231%22)(0,,222222224K T T T C mol cal X C C b O N CO O H CO O H CO O CH a i i p p p =⨯⨯+=∆+=∙=+⨯+⨯+⨯=======-⨯+⨯⨯+=+=+∑)/(1644)0224.011868.448.11)8001600(48.1125.9189570(102800000)/(189570)298800)](48.1187.8)48.1125.9[(100076.191)()/(87.848.11/]211002.22.816[)()/(3214)0224.011868.448.11)2981600(48.1125.9100076.191(102800000)/(280000040000020001200)(33min ,,,,298,,33min h m V mol g cal dTn C n C H H C mol cal X C C d h m V h KJ P C gConsu i i r p i i p p i i p r p g Consu =⨯⨯-⨯-⨯=∙=-⨯-⨯-⨯=--∆=∆∙=⨯⨯+===⨯⨯-⨯-⨯⨯==+⨯=⎰∑∑∑1.8 In an investigation of the thermodynamic properties of a-manganese, the following heat contents were determined:H 700-H 298=12113 J/(g atom) H 1000-H 298=22803 J/(g atom)Find a suitable equation for H T -H 298 and also for C P as a function of temperature in the form (a+bT) Assume that no structure transformation takes place in the given tempeture rang.Solution )298(0055.0)298(62.35011.062.35011.062.3522803)2981000(2)2981000(12113)298700(2)298700(]2[2229822222982---=∆-=-===-+-=-+-+=+==∆⎰⎰T T H TC b a ba ba T baT bTdT a dT C H T P T P1.9 A fuel gas containing 40% CO, 10% CO 2, and the rest N 2 (by volume) is burnt completely with air in a furnace. The incoming and ongoing temperatures of the gases in the furnace are 773K and 1250K,respectively. Calculate (a) the maximum flame temperature and (b) heat supplied to the furnace per cu. ft of exhaust gasmolJ Hmol J H CO f CO f /393296/1104580,298,0,298,2-=∆-=∆)/(10184.403.29)/(1067.11010.492.19)/(1037.81020.935.44)/(1042.01097.345.283,253,253,253,222molK J T C molK J T T C molK J T T C molK J T T C N P O P CO P CO P -------⨯+=⨯-⨯+=⨯-⨯+=⨯-⨯+= Solution?0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)()/(1019.01058.528.33722.0278.0)/(1067.01038.477.281.065.005.02.0)()/(282838110458393296%2.72%8.27%10%65%5%20)4/(1122298127332981523733253253298,,,,298,253,,,,,253,,,,,,,0,298,0,298,298,22222222222222==+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆⨯-⨯+=+==⨯-⨯+=+++===-=∆-∆=∆========+-----------⎰⎰⎰∑∑⎰∑∑∑∑T T T T T T T dTT T dTT T dTn C n C n H H molK J T T C C n C C molK J T T C C C C n C C a mol J n Hn H H N CO production O N CO CO reation then O N air mole need fuel mole when CO O CO T TT i i r p i i p p i i N P CO P i i p p r p O P N P CO P CO P i i p p r p i pf i rf idTT T Q dT T T Q b T T T T T T T dTT T dTT T dT n C n C n H H T TT i i r p i i p p i i 9.0)1019.01058.528.33(2.02828389.0)1019.01058.528.33(2.0282838)(0)499.0321.018.1()1067.01019.277.28(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)(253125029812502982531250298125029829812125029815231250253253298,,,,298,⨯⨯-⨯++⨯-=⨯⨯-⨯++⨯-===+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆-----------⎰⎰⎰⎰⎰∑∑⎰1.10 (a) for the reaction2221CO O CO →+,what is the enthalpy of reaction (0H ∆) at 298 K ?(b) a fuel gas, with composition 50% CO, 50% N 2 is burned using the stoichiometric amount of air. What is the composition of the flue gas?(c) If the fuel gas and the air enter there burner at 298 K, what is the highest temperature the flame may attain(adiabatic flame temperature)? DATA :standard heats of formationfH ∆ at 298 K)/(393000)/(1100002mol J CO mol J CO -=-=Heat capacities [J/(mol K)] to be used for this problem N 2=33, O 2=33, CO=34, CO 2=57Solution )(21100)298)(39889.0(222.02830000)/(3975.03325.057)/(33111.034222.033666.033)(%,75%%,251.111002.22%%1.11%%,6.66%%,2.222.0/25.015.0%)()/(283000393000110000)(,0,,,,,,22220,298,0,298,0K T T dT C n H H K mol J X C C K mol J X C C C N CO product O N CO fuel b mol J n H n H H a P p p i P r i P r i P p i P p i P f i r f ==-⨯-⨯=-∆=∆∙=⨯+⨯==∙=⨯+⨯+⨯====-====+==+-=∆-∆=∆⎰∑∑∑∑1.11 a particular blast furnace gas has the following composition by (volume): N 2=60%, H 2=4, CO=12%, CO 2=24%(a) if the gas at 298K is burned with the stochiometric amount of dry air at 298 K, what is the composition of the flue gas? What is the adiabatic flame temperature?(b) repeat the calculation for 30% excess combustion air at 298K(C)what is the adiabatic flame temperature when the blast furnace gas is preheated to 700K (the dry air is at 298K) (d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will thE dlaMe temperature be affected?DaTA(k J?mol)2CO CO FOR513.393523.110)/(--∆m o l kJ H f2222,)(O N g O H CO CO FOR?? 34505733]/[K mol J C P ∙SolutionOH O H CO O CO a 222222121)(→+→+ 416.0)(104.0)(:22==N n O n Air6.0)(04.0)(24.0)(12.0)(:222====N n H n CO n CO n Fuel 32.0)(08.0)(:22==N n O n Air 92.0)(04.0)(36.0)(:222===N n O H n CO n Flue)(98.1108)(8108.53106308.43)/(8.533492.05004.05736.092.004.036.06308.43)08.241(04.0)523.11051.393(12.03,,222222K T K T K J C C C n CKJH H H N O H CO ii r P O H H CO CO ==⨯=∆=⨯+⨯+⨯=++==-⨯+-⨯=∆+∆=∆∑--(b)repeat the calculation for 30% excess0combustion air at 298K6.0)(04.0)(24.0)(12.0)(:222====N n H n CO n CO n Fuel416.0)(104.0)(:22==N n O n Air 024.0)(016.1)(04.0)(36.0)(:2222====O n N n O H n CO n Flue July 16, 2019(C)what is the adiabatic flame temperature when the blasp furnace gas is preheated to 700K (the dry air is at 298K)6.0)(04.0)(24.0)(12.0)(:222====N n H n CO n CO n Fuel 32.0)(08.0)(:22==N n O n Air 92.0)(04.0)(36.0)(:222===N n O H n CO n Flue)(6.1401)(6.11038.5310373.59)/(8.533492.05004.05736.096.004.036.0373.59)346.02804.05724.03312.0()298700()08.241(04.0)523.11051.393(12.03,,298700222222K T K T K J C C C n CKJH H H H N O H CO ii r P fuel O H H CO CO ==⨯=∆=⨯+⨯+⨯=++==⨯+⨯+⨯+⨯⨯-+-⨯+-⨯=∆+∆+∆=∆∑---(d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will the flame temperature be affected?)(8.1051)(8.75388.57106308.43)/(88.5734024.034016.15004.05736.0024.0016.104.036.06308.43)08.241(04.0)523.11051.393(12.03,,2222222K T K T K J C C C C n CKJH H H O N O H CO ii r P O H H CO CO ==⨯=∆=⨯+⨯+⨯+⨯=+++==-⨯+-⨯=∆+∆=∆∑--6.0)(04.0)(24.0)(12.0)(:222====N n H n CO n CO n Fuel 008.04.01576015)(32.0)(08.0)(:222=-===O H n N n O n Air 92.0)(048.0)(36.0)(:222===N n O H n CO n Flue)(1103)(8052.54106308.43)/(2.543492.050048.05736.092.0048.036.06308.43)08.241(04.0)523.11051.393(12.03,,222222K T K T K J C C C n CKJH H H N O H CO ii r P OH H CO CO ==⨯=∆=⨯+⨯+⨯=++==-⨯+-⨯=∆+∆=∆∑--1.12 A bath of molten copper is super cooled to 5℃ below its true melting point. Nucleation of solid copper then takes place, and the solidification proceeds under adiabatic conditions. What percentage of the bath solidifies? DATA: Heat of fusion for copper is 3100 cal/mol at 1803℃(the melting point of copper) C P,L =7.5(cal/mol ℃), C P,S =5.41+(1.5*10-3T )(cal/mol ℃) Solution)/(310355.75.0)17981803(105.1541.5310002231798,1798,17981803,18031798,1803,mol cal HH dT C dT C H LS SL L P S P L S =⨯-⨯-⨯+⨯+==+++-⎰⎰1.13 Cuprous oxide (Cu 2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu 2O(b)how much heat is release or absorbed per mole reacted? Given the quantity of heat and state whether heat is evolved (exothermic reaction) or absorbed (endothermic reaction)DATA: heat of formation of 1000K in cal/mol Cu 2O=-41900 H 2O=-59210 solution)/(173104190059210222mol cal H OH Cu H O Cu =-=∆+=+,exothermic reaction1.14(a) what is the enthalpy of pure, liquid aluminum at 1000K?(b) an electric resistance furnace is used to melt pure aluminum at the rate of 100kg/h. the furnace is fed with solid aluminum at 298K. The liquid aluminum leaves the furnace at 1000K. what is the minimum electric powder rating (kW) of furnace.DATA : For aluminum : atomic weight=27g/mol, C p,s =26(J/molK), C p,L =29(J/molK), Melting point=932K, Heat of fusion=10700J/mol Solution)(28.0)(7.2793600110002727184)/(2718410700)9321000(29)298932(261000932,932298,1000,kW W P mol J H dT C dT C H SLL P S P l ==⨯⨯==+-⨯+-⨯=++=⎰⎰1.15 A waste material (dross from the melting of aluminum) is found to contain 1 wt% metallic aluminum. The rest may be assumed to aluminum oxide. The aluminum is finely divided and dispersed in the aluminum oxide; that is the two material are thermally connected.If the waster material is stored at 298K. what is the maximum temperature to which it may rise if all the metallic aluminum is oxidized by air/ the entire mass may be assumed to rise to the same temperature. Data : atomic weight Al=27g/mol, O=16g/mol, C p,s,Al =26(J/molK), C p,s, Al2O3=104J/mol, heat formation of Al 2O 3=-1676000J/molSolution;)(600)(3021041029927275.116122711676000K T K T T ==∆∆⨯⨯++⨯⨯=⨯⨯1.16 Metals exhibit some interesting properties when they are rapidly solidified from the liquid state. An apparatus for the rapid solidification of copper is cooled by water. In the apparatus, liquid copper at its melting point (1356K) is sprayed on a cooling surface, where it solidified and cools to 400K. The copper is supplied to the apparatus at the rate of one kilogram per minute. Cooling water is available at 20℃, and is not allowed to raise above 80℃. What is the minimum flow rate of water in the apparatus, in cubic meters per minute? DATA; for water: C p =4.184J/g k, Density=1g/cm 3; for copper: molecular weight=63.54g/mol C p =7cal/mol k, heat of fusion=3120 cal/molSolution:min)/(10573.2)2080(1min /min54.631000)]4001356(73120[min /33m V VQQ WaterCopper -⨯=-=⨯⨯-⨯+=1.17 water flowing through an insulated pipe at the rate of 5L/min is to be heated from 20℃ to 60℃ b an electrical resistance heater. Calculate the minimum power rating of the resistance heater in watts. Specify the system and basis for you calculation. DA TA; For water C p =4.184J/g k, Density=1g/cm 3 Solution:)(139476010005)2060(184.4W W =⨯⨯-⨯=1.18 The heat of evaporation of water at 100℃ and 1 atm is 2261J/mol (a) what percentage of that energy is used as work done by the vapor?(b)if the density of water vapor at 100℃ and 1 atm is 0.597kg/m 3 what is the internal energy change for the evaporation of water?Solution:)/(375971822613101%6.71822613101%)/(31010224.0273373101325mol J Q W U mol J V P =⨯+-=+=∆=⨯==⨯⨯=∆1.19 water is the minimum amount of steam (at 100℃ and 1 atm pressure) required to melt a kilogram of ice (at 0℃)? Use data for problem 1.20 Solution)(125,3341000)10018.42261(g m m =⨯=⨯+1.20 in certain parts of the world pressurized water from beneath the surface of the earth is available as a source of thermal energy. To make steam, the geothermal water at 180℃ is passed through a flash evaporator that operates at 1atm pressure. Two streams come out of the evaporator, liquid water and water vapor. How much water vapor is formed per kilogram of geothermal water? Is the process reversible? Assume that water is incompressible. The vapor pressure of water at 180℃ is 1.0021 Mpa( about 10 atm) Data: C P,L =4.18J/(g k), C P ,v =2.00J/(g k), △H V =2261J/g, △H m =334 J/g Solution:leirreversib g x x x )(138),1000(8018.4)8018.48022261(=-⨯⨯=⨯-⨯+The problems of the second law2.1 The solar energy flux is about 4J cm 2/min. in no focusing collector the surface temperature can reach a value of about 900℃. If we operate a heat engine using the collector as the heat source and a low temperature reservoir at 25℃, calculate the area of collector needed if the heat engine is to produce 1 horse power. Assume the engine operates at maximum efficiency.Solution)(25.6)(7466010427390)2590(24m S W tWP StQ T T T W H H L H ===⨯⨯+-=-=2.2 A refrigerator is operated by 0.25 hp motor. If the interior of the box is to be maintained at -20℃ ganister a maximum exterior temperature of 35℃, what the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume the coefficient of performance is 75% of the value for a reversible engine.Solution:)(64374625.02035202734375.0W P P T T T P Q T T T W L LL LH HHLH =⨯⨯+-⨯=-=-=2.3 suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0℃. Liquid water is taken in at 0℃ and converted to ice at 0℃. To convert 1 g of ice to 1 g liquid. △H=334J/g is required. If the temperature outside the box is 20℃, what mass of ice can be produced in one minute by a 0.25 hp motor running continuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value.Solution:)(4576033474625.020273g m M m P P T T T P L LLLH ===⨯⨯=-=2.4 under 1 atm pressure, helium boils at 4.126K. The heat of vaporization is 84 J/mol what size motor (in hp) is needed to run a refrigerator that must condense 2 mol of gaseous helium at 4.126k to liquid at the same temperature in one minute? Assume that the ambient temperature is 300K and that the coefficient of performance of the refrigerator is 50% of the maximum possible.Solution:)(52.0)(393'60284216.4216.4300'5.0%50hp W P P T T T P P Q T T T W L L L H LLLH ==⨯⨯-=-==-=2.5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical power to run a heat pump that works between 25 and 5℃, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler.(a) assume that the efficiencies are equal to the theoretical maximum values(b) assume the power plant efficiency is 70% of maximum and that coefficient of performance of the heat pumpis 10% of maximum(c) if a furnace can use 80% of the energy in fossil foe to heat the house would it be more economical in terms ofoverall fissile fuel consumption to use a heat pump or a furnace ? do the calculations for cases a and b solution:1,2,2,1,212,2,2,2,21,1,1,1,198.82527352527354050540)(H H H H H H L H H H L H P P P P P P P T T T P P T T T P a =+-=+-=-=-= .,)(6286.0)(1,2,not is b ok is a c PP b H H =2.6 calculate △U and △S when 0.5 mole of liquid water at 273 K is mixed with 0.5 mol of liquid water at 373 K and the system is allowed to reach equilibrium in an adiabatic enclosure. Assume that C p is 77J /(mol K) from 273K to 373K Solution:)/(933.0)273323ln(5.0)373323ln(5.0)ln()ln()(02211K J C C T T C n T T C n S J U P P E P E P =+=+=∆=∆2.7 A modern coal burning power plant operates with a steam out let from the boiler at 540℃ and a condensate temperature of 30℃.(a) what is the maximum electrical work that can be produced by the plant per joule of heat provided to theboiler?(b) How many metric tons (1000kg) of coal per hour is required if the plant out put is to be 500MW (megawatts).Assume the maximum efficiency for the plant. The heat of combustion of coal is 29.0 MJ/k g(c) Electricity is used to heat a home at 25℃ when the out door temperature is 10℃ by passing a currentthrough resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied?Solution:)(3.69)(6937136005000.29)()(89.013054030540)(ton kg m T T T mb J Q T T T W a LH LH H L H ==⨯=-=+-=-=)(9.191102525273)(J Q Q T T T W c H HHLH =-+=-=2.8 an electrical resistor is immersed in water at the boiling temperature of water (100℃) the electrical energy input into the resistor is at the rate of one kilowatt(a) calculate the rate of evaporation of the water in grams per second if the water container is insulated that is noheat is allowed to flow to or from the water except for that provided by the resistor(b) at what rate could water could be evaporated if electrical energy were supplied at the rate of 1 kw to a heatpump operating between 25 and 100℃data for water enthalpy of evaporation is 40000 J/mol at 100℃; molecular weight is 18g/mol; density is 1g/cm 3solution:)(23.2,2510027310010004000018)()(45.0,10004000018)(g m m b g m ma =-+===2.9 some aluminum parts are being quenched (cooled rapidly ) from 480℃ to -20℃ by immersing them in abrine , which is maintained at -20℃ by a refrigerator. The aluminum is being fed into the brine at a rate of one kilogram per minute. The refrigerator operates in an environment at 30℃; that is the refrigerator may reject heat at 30℃. what is them minus power rating in kilowatts, of motor required to operate the refrigerator?Data for aluminum heat capacity is 28J/mol K; Molecular weight 27g/mol Solution:)(5.102)(102474202732030)20480(28271000kW W P P T T T P P L L L L H W L ==---=-=--⨯=2.10 an electric power generating plant has a rated output of 100MW. The boiler of the plant operates at 300℃. The condenser operates at 40℃(a) at what rate (joules per hour) must heat be supplied to the boiler?(b) The condenser is cooled by water, which may under go a temperature rise of no more than 10℃. Whatvolume of cooling water in cubic meters per hour, is require to operate the plant?(c) The boiler tempeture is to be raised to 540℃,but the condensed temperature and electric output will remainthe same. Will the cooling water requirement be increased, decreased, or remain the same?Data heat capacity 4.184, density 1g/cm 3 Solution: )(109.7)(102.21040300273300)(1188J t P Q W P T T T P a H H L H H H ⨯==⨯=-+=-= )(1003.1184.41010)(103.4)(34611m V Q V J Q b LL ⨯==⨯⨯⨯⨯= noW P T T T P c L H H H )(10626.11040540273540)(88⨯=-+=-= 2.11 (a) Heat engines convert heat that is available at different temperature to work. They have been several proposals to generate electricity y using a heat engine that operate on the temperature differences available at different depths in the oceans. Assume that surface water is at 20℃, that water at a great depth is at 4℃, and that both may be considered to be infinite in extent. How many joules of electrical energy may be generated for each joule of energy absorbed from surface water? (b) the hydroelectric generation of electricity use the drop height of water as the energy source. in a particular region the level of river drops from 100m above sea level to 70m above the sea level . what fraction of the potential energy change between those two levels may be converted into electrical energy? how much electrical energy ,in kilowatt-hours, may be generated per cubic meter of water that undergoes such a drop? Solution: )/(1006.136001000)()(055.0127320420)(6h kW h mg P b J Q T T T W a H H L H ⨯=⨯∆==+-=-=2.12 a sports facility has both an ice rink and a swimming pool. to keep the ice frozen during the summer requires the removal form the rink of 105 KJ of thermal energy per hour. It has been suggested that this task be performed by a thermodynamic machine, which would be use the swimming pool as the high temperature reservoir. The ice in the rink is to be maintain at a temperature of –15℃, and the swimming pool operates at 20℃, (a) what is the theoretical minimum power, in kilowatts, required to run the machine? (b) how much heat , in joule per hour , would be supplied t the pool by this machine?Solution: )(1014.1101527320273)()(77.33600/10152731520)(555kJ Q b kW P T T T P a H L L L H ⨯=-+==-+=-=2.13solution:)/(81.6810ln 314.877.45277.6282.4)/(152940)()/(67.4977.45277.6282.4)()/(152940)(22)(2molK cal S mol cal H d molK cal S c mol cal H b AlNN Al a -=+-⨯-⨯=∆=∆-=-⨯-⨯=∆=∆=+ 2.14 solution:)/(2257412000)27340273ln 184.4273336263273ln1.2()(400,010,K J dT T C T H dT T C m S WATER P m m ICE P =+++=+∆+=∆⎰⎰-2.15)(70428)(2896100077773002J W J Q T T T W L L L H ==-=-= 2.16)(4.3719))2.4300(314.85.13.83(3002.4300)(7.58663.832.42.4300J Q T T T W J Q T T T W H H L H L L L H =-⨯+-=-==-=-=2.17 yes d Q c K J P P nR S b J pdV n W Q OU T a )(0)()/(1.1910ln 314.81ln)()(570410ln 298314.810)(0==⨯⨯==∆=⨯⨯=-=-==∆=∆⎰ 2.18)(1222335273020********g m m m T T T L L H =-=-=⨯ Property Relations 1. At -5︒C, the vapor pressure of ice is 3.012mmHg and that of supercooled liquid water is 3.163mmHg. The latent heat of fusion of ice is 5.85kJ/mol at -5︒C. Calculate ∆G and ∆S per mole for the transition of from water to ice at -5︒C. (3.2, 94) Solution: mol J P P RT G waterO H iceO H /9.1089523.0ln 268314.8163.3012.3ln )5273(314.8ln ,,22-=⨯⨯=-⨯==∆mol J H /1085.53⨯=∆)/(23.22268)9.108(5850K mol J T G H S S T H G ⋅=--=∆-∆=∆∴∆-∆=∆ 2. (1) A container of liquid lead is to be used as a calorimeter to determine the heat of mixing of two metals, A andB. It has been determined by experiment that the “heat capacity ” of the bath is 100cal/︒C at 300︒C. With the bath originally at 300︒C, the following experiments are performed;(2) A mechanical mixture of 1g of A and 1g of B is dropped into the calorimeter. A and B were originally at 25︒C. When the two have dissolved, the temperature of the bath is found to have increased 0.20︒C. 2. Two grams of a 50:50(wt.%) A-B alloy at 25︒C is dropped similarly into the calorimeter. The temperature decreases 0.40︒C. (a) What is the heat of mixing of the 50:50 A-B alloy (per gram of alloy)? (b) To what temperature does it apply ? (3.5, 94)Solution: mol J K cal C bath P /418/100,==(a) g cal T C Q bath P /102/2.01002/,=⨯=∆=This is the heat of mixing.(b) The heat capacity of C P, alloy : )/(072.06.27424.0100)254.0300(2,,K g cal TC C bath P alloy P ⋅=⨯⨯=--⨯∆⨯=Assuming that the calorimeter can be applied to the maximum of T ︒C, the for mixing to form 1 gram of alloy:10)'300(,1+-=T C Q bath P , )'(,2T T C Q alloy P -⋅=, 21Q Q =)'(10)'300(,,T T C T C alloy P bath P -=+-3. The equilibrium freezing point of water is 0︒C. At that temperature the latent heat of fusion of ice (the heat required to melt the ice) is 6063J/mol. (a) What is the entropy of fusion of ice at 0︒C ? (b) What is the change of Gibbs free energy for ice →water at 0︒C?(c) What is the heat of fusion of ice at -5︒C ? C P(ice) = 0.5 cal/(g. ︒C); C P(water) = 1.0 cal/(g. ︒C). (d) Repeat parts a and b at -5︒C. (3.6, p94)Solution: (a) At 0︒C, ∆G =0, ∴ T m ∆S = ∆H)./(09.222736030K mol J T H S m ==∆=∆(b) At 0︒C, ∆G =0© )./(62.37)./(1818.45.0)./(5.0,K mol J K mol J K g cal C ice P =⨯⨯==)./(24.75)./(1818.40.1)./(0.1,K mol J K mol J K g cal C water P =⨯⨯==a reversible process can be designed as follows to do the calculation:。