基于自由基处理难降解有机废水的研究

发布时间：2018-04-18 01:30

本文选题：自由基 + 紫外光　；参考：《大连工业大学》2015年硕士论文

【摘要】：基于自由基处理难降解有机废水是一种效率较高的新型高级氧化技术,近年来受到国内外广泛的关注。自由基的种类很多(如羟基自由基、硫酸根自由基等),不同的活化方式和反应条件对有机废水的处理效果也各不相同。在提高有机废水的处理效率方面,研究不同的自由基及其活化方法的选择成为国内外研究的热点方向。本文采用微波、紫外的活化方式,以较为常见的有机污染物苯酚为处理对象,分析对比反应前后的苯酚和COD的浓度变化,研究硫酸根自由基,氯自由基,以及羟基自由基在不同反应条件下处理苯酚的降解效果。具体研究内容如下:(1)微波辅助活性炭活化过硫酸盐产生硫酸根自由基降解苯酚,研究结果表明:在微波功率560 W下,辐射5 mi n,活性炭的投加量为0.17 g,m(K2S2O8)/m(苯酚)=0.5,pH=4时,苯酚的降解效果较好,去除率可达到86%,COD去除率达到85%。(2)通过紫外光照次氯酸盐产生氯自由基降解苯酚,通过改变反应条件得出苯酚的优化处理条件:在紫外光照射2 h下,加入有效氯为5 g/L的次氯酸钠溶液,使m(有效氯)/m(苯酚)=0.7,调节pH=12,可使苯酚的降解率达到60%,COD的去除率达到51%。(3)过碳酸根在紫外光的诱导下产生具有强氧化性的羟基自由基降解苯酚,通过改变光照时间、过碳酸钠的投加量、pH,得出苯酚降解的优化条件:在紫外光照下3 h,加入过碳酸钠0.035 g,使m(活性氧)/m(苯酚)=1,pH在4~11之间时,苯酚的去除率可达85%以上,COD的去除率可达80%以上。在酸性条件下,加入Fe2+,考察紫外光照时间、过碳酸钠投加量(以活性氧计)、Fe2+投加量以及pH的变化对苯酚去除率的影响。结果表明:Fe2+的存在加速了苯酚的降解,在紫外光下照射20 min,m(活性氧)/m(苯酚)=0.6,m(Fe2+)/m(苯酚)=0.06,pH=5,苯酚的去除率就可达到85%。(4)在微波的诱导下,过碳酸根产生羟基自由基降解苯酚。实验表明:在微波功率800 W下消解3 mi n,加入过碳酸钠,使m(活性氧)/m(苯酚)=1.6,pH=4~9之间时,苯酚的去除率可达85%以上,COD的去除率可达到84%。
[Abstract]:The treatment of refractory organic wastewater based on free radical is a new type of advanced oxidation technology with high efficiency, which has attracted wide attention at home and abroad in recent years.There are many kinds of free radicals (such as hydroxyl radical, sulfate radical and so on). Different activation methods and reaction conditions have different effects on the treatment of organic wastewater.In order to improve the treatment efficiency of organic wastewater, the choice of different free radicals and their activation methods has become a hot research direction at home and abroad.In this paper, microwave and UV activation methods were used to treat phenol, a common organic pollutant. The concentration of phenol and COD were compared before and after the reaction, and the sulphate radical and chlorine free radical were studied.And the degradation effect of phenol treated by hydroxyl radical under different reaction conditions.The specific contents of the study are as follows: microwave-assisted activated carbon activates persulfate to produce sulfate radical to degrade phenol. The results show that when microwave power is 560 W, radiation time is 5 min, the dosage of activated carbon is 0.17 g / m K2S _ 2O _ 8 / m (pH = 4).The degradation efficiency of phenol was better, and the removal rate of COD reached 850.The phenol was degraded by chlorinated hypochlorite by UV irradiation. The optimum treatment conditions of phenol were obtained by changing the reaction conditions: under UV irradiation for 2 h.By adding 5 g / L sodium hypochlorite solution with available chlorine, the degradation rate of phenol can reach 60% and COD removal rate can reach 51% by adding 5 g / L sodium hypochlorite solution. The hydroxyl radical can degrade phenol by hydroxyl radical with strong oxidation ability induced by ultraviolet light.By changing the illumination time and the dosage of sodium percarbonate (pH), the optimum conditions for phenol degradation were obtained: under ultraviolet light irradiation for 3 h, adding sodium percarbonate 0.035 g, the pH value of m (Ros / m) (Phenol) was between 4 ~ 11,The removal rate of phenol can reach more than 85% and the removal rate of COD can reach more than 80%.Under acidic conditions, the effects of UV irradiation time, sodium percarbonate dosage (Fe2 addition by Ros) and pH on phenol removal rate were investigated by adding Fe2.The results showed that the presence of% Fe _ 2 accelerated the degradation of phenol. Under microwave irradiation, hydroxyl radical was produced by percarbonate to produce hydroxyl radical to degrade phenol under the irradiation of 20 min (reactive oxygen species / m (phenol ~ (0.6) / m) Fe ~ (2 +) / m (pH ~ (5)). The removal rate of phenol could reach 85 路L ~ ((4)) under microwave irradiation.The experimental results show that the removal rate of phenol can reach more than 85% and the removal rate of COD can reach 84% when the microwave power is 800W and sodium percarbonate is added.
【学位授予单位】：大连工业大学
【学位级别】：硕士
【学位授予年份】：2015
【分类号】：X783

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