外文翻译.docx

1、外文翻译毕业设计/论文外 文 文 献 翻 译院 系 城市建设学院 专 业 班 级 环境1102 姓 名 易家喜 原 文 出 处 评 分 指 导 教 师 皮科武 华中科技大学武昌分校20 年 月 日毕业设计/论文外文文献翻译要求：1外文文献翻译的内容应与毕业设计/论文课题相关。2外文文献翻译的字数：非英语专业学生应完成与毕业设计/论文课题内容相关的不少于2000汉字的外文文献翻译任务（其中，汉语言文学专业、艺术类专业不作要求），英语专业学生应完成不少于2000汉字的二外文献翻译任务。格式按华中科技大学武昌分校本科毕业设计/论文撰写规范的要求撰写。3外文文献翻译附于开题报告之后：第一部分为译文，第

2、二部分为外文文献原文，译文与原文均需单独编制页码（底端居中）并注明出处。本附件为封面，封面上不得出现页码。4外文文献翻译原文由指导教师指定，同一指导教师指导的学生不得选用相同的外文原文。 原文： Municipal Wastewater Treatment Plant Biological Phosphorus Removal SystemAbstract: The enhanced biological phosphorus removal (EBPR) system design, the determination of sludge age and organic phosphorus

3、 concentration effect and the relationship between pairs of different levels of phosphorus goal gives the corresponding unit.Keywords: EBPR anaerobic aerobic sludge age SRT BOD5/ P phosphorus goalAs Chinas new Integrated Wastewater Discharge Standard (GB8978-1996) The promulgation and implementation

4、 of enhanced biological phosphorus removal technology (Enhanced Biological Phosphorus Removal, referred to as EBPR) in the new sewage treatment plant project received an increasingly wide range of applications. As the new effluent standards for TP with high requirements, coupled with the complexity

5、of biological phosphorus removal, so it tends to become a sewage treatment plant design and operation of the difficulty.The mechanism of biological phosphorus removal, the current consensus view is that more phosphate accumulating bacteria (PAO) the completion of the unique metabolic activities of p

6、hosphorus from the liquid (water) to solid (sludge) conversion. Conventional activated sludge phosphorus content of 1.5% 2.0% (P / VSS), while the PAO can sludge phosphorus content increased to 5% to 7%, and thus biological phosphorus removal requires the creation of suitable environment for the gro

7、wth of PAO, which so that the proliferation of PAO groups. In the craft section through aerobic anaerobic pre-set (space, such as A / O phosphorus removal process; timing, such as SBR process) to enable the PAO to obtain selective growth. PAO to obtain selective advantage due to a substantial absorp

8、tion of water in the anaerobic volatile fatty acids (VFAs), and in the body into a poly- -hydroxybutyrate (Polyhydroxybutyrate, referred to as PHB), making it no other segment in the aerobic heterotrophic bacteria compete for water and residual organic matter.Design of a sludge age1.1 aerobic sludge

9、 ageAs a heterotrophic bacteria, PAO there is a necessary condition for growth is to meet its required minimum sludge age. Because PAO under aerobic conditions in the growth and reproduction, so here is the minimum sludge age refers to the aerobic sludge age. Between the growth rate of PAO other het

10、erotrophic bacteria and autotrophic bacteria (nitrifying bacteria), with the minimum needed for growth, so the SRT is also in between. At present the design of the minimum required PAO aerobic sludge age generally based on experience in use. Figure 1 indicates that the PAO and the nitrifying bacteri

11、a needed for growth, the minimum sludge age and temperature of the relationship between the design in addition to considering the minimum sludge age should be considered outside a certain safety factor, the value of 1.5 in general.The reason is listed in Figure 1, the minimum required for the growth

12、 of nitrifying bacteria as nitrification sludge age would have a negative impact on the growth of PAO. If the section occurs in the aerobic nitrification, nitrate will be returned with the sludge into the anaerobic, making PAO denitrifiers compete with VFAs, thus undermining the selective advantage

13、of the PAO. In addition, back to the anaerobic fermentation of nitrate will interfere with response, because denitrifying bacteria under anaerobic conditions would be the direct use of easily degradable organic compounds (1mgNO3-N back into the anaerobic approximately 6mgCOD is consumed), thus under

14、mining the their reaction into VFAs by fermentation process. Thus, when the aerobic nitrification occurs when the segment to determine anaerobic sludge age should take into account the resulting additional demand for organics. As can be seen from Figure 1, when the temperature is low, PAO and nitrif

15、ication bacteria needed for growth, a larger difference between the minimum sludge age, design or run-time is easy to make the system to meet the PAO inhibit the growth of nitrifying bacteria growth, but as the temperature Both increased gradually approach (> 25 , when the two are equal). VIP, UC

16、T process due to return fluid to remove NO3-N Er Shi PAO has obvious advantages.To meet the minimum sludge age where the system phosphorus removal efficiency will be increased as the sludge age decreases, which is due to: sludge age reduces the rate of increase in soil lead to production, so that by

17、 excluding the number of phosphorus removing sludge reduction; PAO activities in order to maintain his life led to the decomposition of phosphorus secondary phosphorus release. Therefore, the biological phosphorus removal system to meet the PAO growth under the conditions required for a shorter slud

18、ge age should be used. 1.2 anaerobic sludge ageThe completion of VFAs in the anaerobic PAO adsorption, transfer and release of phosphorus in vivo. It is absorbed phosphorus in the aerobic section with the number of VFAs in the anaerobic adsorption the number of closely related. If the water in a lar

19、ge number of VFAs exist, PAO adsorption of VFAs can quickly be completed at this time requires less anaerobic sludge age; and if only part of the influent VFAs, you need to carry out the fermentation of organic matter in anaerobic reactions produce VFAs, due to fermentation reaction was slow, and th

20、erefore will become the anaerobic fermentation reaction to the size of the controlling factors of SRT. At 20 in the presence of water, if needed VFAs, while anaerobic sludge age could be as short 0.5d; if the water does not contain VFAs, but it contains easily degradable organic matter produced by f

21、ermentation reactions required for sufficient VFAs, while anaerobic sludge age of about 1.5d (20 ); if the water contains some of VFAs (still need part of the fermentation) is sludge age of 0.5 1.5d. On the other hand, if the insufficient number of easily degradable substances, then the slow degrada

22、tion of organic compounds still need to first hydrolyzed, and then through the fermentation reaction of VFAs, at this time of the anaerobic sludge age longer (2.5 3d). At this point, if the fermentation products (as ever sludge fermentation products) into the anaerobic pond, it will improve the phos

23、phorus removal.Urban sewage will be determined according to the influent COD concentration of anaerobic MLSS quantity and the number of the total system of MLSS ratio (Table 1). MLSS concentration in the case of unification of the ratio of the number of anaerobic sludge age and the ratio of total sy

24、stem sludge age.Table 1, the proportion of anaerobic biomass and the relationship between the influent CODInfluent COD (mg / L)Anaerobic MLSS quantity accounts for the proportion of the total system of the MLSS"4000.20 0.25Four hundred to 7000.15 0.20"7000.10 0.15Two phosphorus removal eff

25、ect and the relationship between organic matterPhosphorus removal and COD / TP, BOD5/TP and SBOD5/TP the value of the relevant, while it varies with the process. Minimum required for biological phosphorus removal leads to the concept of organic carbon (organic matter) to limit sewage and effluent ph

26、osphorus limits. Carbon limits the number of sewage is not sufficient to remove all water and organic phosphorus, and the results of a high concentration of phosphorus in water, its concentration by the corresponding influent concentrations of phosphorus concentration and organic matter to determine

27、. Phosphorus limit of organic wastewater effluent is required for organic phosphorus content is more than the number of the effluent phosphorus concentration is very low at this time. So when in need of a good water quality, the raw water quality is a phosphorus effluent limit is expected.Through th

28、e understanding of carbon-limited water gives a variety of EBPR processes required for the proportion of organic matter with P. The ratio is under carbon constraints determined through testing and the actual project, while it also shows that the phosphorus removal capacity of the system. The most co

29、mmonly used ratio is the ratio of BOD5 the same to the phosphorus (BOD5/ P), its calculation formula is:BOD5/ P = influent BOD5 / water TP-water solubility of P The use of water soluble phosphorus in the formula because of a particulate phosphorus in water is usually outflow from the secondary sedim

30、entation tank, which is a function of the efficiency of the secondary settling tank, but has nothing to do with the biological processes. Table 2 provides a variety of biological phosphorus removal process required for a typical BOD5, COD and the ratio of P. Table 2 EBPR processes required for BOD5

31、and COD and the ratio of PEBPR processBOD5/ P (mgBOD5/mgP)COD / P (mgCOD / mgP)In addition to P and efficient processes (such as nitrite-free A / O-oriented, VIP, UCT)15 2026 34Medium Efficiency In addition to P processes (such as with nitrification and A / O and A2 / O)20 2534 43In addition to low

32、efficiency of P processes (such as the oxidation ditch)"25"43As can be seen from Table 2, when the value of the smaller BOD5/ P need efficient phosphorus removal process, because such processes require less removal of the organic units of phosphorus. Efficient phosphorus removal efficiency is hig