馬曉雁,楊 帆,李青松,楊慶云,陳國(guó)元,李國(guó)新

UV/NaClO和UV/過(guò)碳酸鈉工藝降解水楊酸的對(duì)比

馬曉雁1,楊 帆1,李青松2,3*,楊慶云1,陳國(guó)元2,李國(guó)新2

(1.浙江工業(yè)大學(xué)土木工程學(xué)院,浙江 杭州 310014；2.廈門(mén)理工學(xué)院水資源環(huán)境研究所,福建 廈門(mén) 361024；3.廈門(mén)理工學(xué)院廈門(mén)市水資源利用與保護(hù)重點(diǎn)實(shí)驗(yàn)室,福建 廈門(mén) 361024)

采用UV/NaClO和UV/過(guò)碳酸鈉(SPC)工藝降解水中水楊酸(SA),對(duì)比考察了氧化劑種類(lèi)和投加量對(duì)SA去除的影響,采用淬滅法和電子順磁共振波譜儀(EPR)鑒定識(shí)別了2種工藝中的自由基,通過(guò)競(jìng)爭(zhēng)動(dòng)力學(xué)的方法計(jì)算了SA與?OH、CO3?-的二級(jí)反應(yīng)速率常數(shù)及反應(yīng)體系中不同組分的貢獻(xiàn),從環(huán)境水樣模擬、急性毒性和經(jīng)濟(jì)效益等角度比較了SA的去除效果.結(jié)果表明,UV/NaClO工藝和UV/SPC工藝降解SA的擬一級(jí)動(dòng)力學(xué)常數(shù)分別為0.4378,0.3794min-1.UV/NaClO和UV/SPC體系中分別存在?OH、Cl?和O2?-、?OH及CO3?-等自由基.SA與?OH、CO3?-的二級(jí)反應(yīng)速率常數(shù)分別為3.97′109,8′107L/(mol×s).UV/NaClO工藝中活性氯自由基(RCS)(79.91%)對(duì)SA去除起主導(dǎo)作用;而UV/SPC工藝中O2?-(51.75%)與?OH(41.42%)起主導(dǎo)作用.環(huán)境水樣中SA在UV/NaClO和UV/SPC工藝中的降解受到抑制,其反應(yīng)速率分別平均降低了67%和74%.UV/SPC工藝反應(yīng)溶液的抑制率(25%)較UV/NaClO工藝反應(yīng)溶液(63%)低38%.SA降解率達(dá)到97.5%以上時(shí)UV/SPC工藝的成本[37.1$/(m3×order)]是UV/NaClO工藝成本[4.0$/(m3×order)]的9.3倍,UV/NaClO工藝較UV/SPC工藝具有較高的經(jīng)濟(jì)效益.

UV-AOPs；自由基鑒定；自由基貢獻(xiàn)；急性毒性；經(jīng)濟(jì)效益

近年來(lái)藥物和個(gè)人護(hù)理品(PPCPs)在水體中被頻繁檢出,由于其潛在的危害與風(fēng)險(xiǎn),水環(huán)境中PPCPs的去除對(duì)于保障飲水安全具有重要的意義[1-3].目前,基于紫外的高級(jí)氧化工藝(UV-AOPs)已廣泛應(yīng)用于去除水中的PPCPs[4].UV-AOPs常見(jiàn)的氧化劑有雙氧水(H2O2)、次氯酸鈉(NaClO)、臭氧、過(guò)硫酸鹽、過(guò)碳酸鈉(SPC)等,實(shí)際應(yīng)用中可以根據(jù)水質(zhì)情況采用不同的氧化劑以達(dá)到更好的去除效果[4-6].

UV/NaClO工藝能產(chǎn)生將污染物高效去除的?OH和活性氯自由基(RCS,包括Cl?、ClO?、Cl2?-等),且去除富含電子的有機(jī)污染物時(shí),RCS比?OH具有更高的二級(jí)反應(yīng)速率常數(shù)[7-9].SPC溶于水后經(jīng)UV照射可產(chǎn)生?OH、O2?-、CO3?-等自由基,且不同pH值條件下主導(dǎo)的自由基不同,這一特點(diǎn)適用于處理不同的目標(biāo)污染物[10-11].相比于NaClO等液體氧化劑,固體氧化劑SPC具有更好的穩(wěn)定性、抗爆性和可獲得性[5,12].關(guān)于UV/NaClO和UV/SPC工藝去除水中 PPCPs 的研究已有諸多報(bào)道,然而2種工藝在相同條件下降解同一污染物時(shí)去除效果、機(jī)理與經(jīng)濟(jì)效益尚不明確.

水楊酸(SA)是地表水中檢出頻次較高的典型PPCP,地表水中其濃度高達(dá)2014.4ng/L,對(duì)水生生物構(gòu)成威脅[2-3].本文采用UV/NaClO和UV/SPC工藝降解水中的SA,從氧化劑投加量、自由基種類(lèi)與貢獻(xiàn)率、環(huán)境水樣應(yīng)用、急性毒性和經(jīng)濟(jì)效益等角度對(duì)比考察2種工藝的差異,以期為去除水中PPCPs工藝選擇提供參考.

1 材料與方法

1.1 實(shí)驗(yàn)材料與儀器

SA(純度99.9%,德國(guó)Dr.Ehrenstorfer);硝基苯(NB,AR,阿法埃莎化學(xué)有限公司);N,N—二甲基苯胺(DMA,純度399%,上海麥克林生化科技有限公司);三氯甲烷(CF,HPLC,美國(guó)Anaqua).過(guò)碳酸鈉(SPC, CP)、對(duì)氯苯甲酸(pCBA,GC)、5,5-二甲基-1-氧化吡咯啉(DMPO,AR)等購(gòu)于上海阿拉丁生化科技股份有限公司;五水硫代硫酸鈉(Na2S2O3·5H2O, AR)、叔丁醇(TBA,HPLC)、乙腈(C2H3N,HPLC)、苯酚(PhOH,AR)等購(gòu)于安譜實(shí)驗(yàn)科技股份有限公司. NaClO、H2O2、NaOH、HCl、NaHCO3、CH3COOH等購(gòu)于國(guó)藥集團(tuán)化學(xué)試劑有限公司,除NaClO為化學(xué)純外其余均為分析純.實(shí)驗(yàn)用水為Milli-Q超純水.

高效液相色譜儀(HPLC)(LC-20A,日本Shimadzu)、電子順磁共振波譜儀(EPR) (SN0253,德國(guó)Bruker)、便攜式余氯計(jì)(CL200,上海三信儀表廠)、pH計(jì)(ST2100,常州奧豪斯儀器有限公司)、磁力攪拌器(HJ-6A,江蘇金壇崢嶸儀器)、純水機(jī)(Milli-Q,美國(guó)Milipore)、發(fā)光細(xì)菌毒性檢測(cè)儀(LumiFox 6800,深圳朗石科學(xué)儀器有限公司).

1.2 降解實(shí)驗(yàn)方法

實(shí)驗(yàn)在一個(gè)置于磁力攪拌器上的圓柱形容器(容積500mL)中進(jìn)行,光源為低壓紫外汞燈(0.18mW/cm2,波長(zhǎng)254nm),汞燈外套有石英套管.實(shí)驗(yàn)溶液為濃度500μg/L的SA溶液(300mL),實(shí)驗(yàn)前用0.1mol/L的NaOH溶液或0.1mol/L的HCl溶液調(diào)節(jié)pH值至7.實(shí)驗(yàn)開(kāi)始時(shí)首先投加一定濃度的NaClO或SPC,同時(shí)開(kāi)啟磁力攪拌器與汞燈開(kāi)始計(jì)時(shí),在特定時(shí)間取樣經(jīng)0.22μm濾膜過(guò)膜后進(jìn)行HPLC分析.進(jìn)樣瓶在實(shí)驗(yàn)開(kāi)始前先加入10μL濃度為0.01mol/L的Na2S2O3·5H2O以確保完全淬滅剩余氧化劑.所有實(shí)驗(yàn)重復(fù)3次并取平均值.

1.3 分析方法

SA的濃度采用HPLC進(jìn)行檢測(cè).HPLC方法:流動(dòng)相A:0.1%乙酸溶液;流動(dòng)相B:乙腈;A:B=65:35;流速為1mL/min;紫外檢測(cè)器波長(zhǎng)為292nm;柱溫為40℃;進(jìn)樣體積20μL;

1.4 自由基捕獲實(shí)驗(yàn)方法

鑒定UV/NaClO工藝中Cl?和?OH時(shí),分別調(diào)節(jié)pH=7和pH=12.5,UV/SPC工藝中?OH和CO3?-鑒定時(shí)調(diào)節(jié)溶液pH=7.由于DMPO-O2?-在水溶液中的存活時(shí)間極短,鑒定O2?-在二甲基亞砜溶液中進(jìn)行[13].降解實(shí)驗(yàn)開(kāi)始后,在1min時(shí)抽取1mL反應(yīng)液立即與0.1mL 100mmol/L的DMPO溶液混合,迅速用50μL毛細(xì)管吸取至一定高度后封口,立即進(jìn)行EPR檢測(cè).

1.5 急性毒性實(shí)驗(yàn)方法

實(shí)驗(yàn)前混合復(fù)蘇稀釋液與發(fā)光細(xì)菌凍干粉形成細(xì)菌液,并活化15min,將空白樣、待測(cè)樣與滲透壓調(diào)節(jié)液以900μL:100μL混合并搖勻.實(shí)驗(yàn)開(kāi)始時(shí)空白樣和待測(cè)樣中分別加入50μL細(xì)菌液混勻,測(cè)量初始發(fā)光強(qiáng)度后送入儀器樣本區(qū),培養(yǎng)30min后再進(jìn)行發(fā)光強(qiáng)度檢測(cè).急性毒性的結(jié)果可以通過(guò)相對(duì)抑制率來(lái)表示水樣急性毒性強(qiáng)度.

2 結(jié)果與討論

2.1 UV/NaClO和UV/SPC工藝對(duì)SA的降解

由圖1可知,NaClO與SA的物質(zhì)的量比從3.7增加至18.5,SA的擬一級(jí)動(dòng)力學(xué)常數(shù)()由0.0842min-1增加至0.4378min-1,相同反應(yīng)時(shí)間內(nèi)相應(yīng)SA的去除從57.3%增加至100%;SPC與SA的物質(zhì)的量比由7增加至35時(shí),SA的從0.1239min-1增加至0.3794min-1,對(duì)應(yīng)的SA去除率從71.5%提高到97.8%;2種工藝處于同一量級(jí),氧化劑與SA的物質(zhì)的量比小于14時(shí)UV/SPC工藝較大,氧化劑與SA的物質(zhì)的量比大于14時(shí)UV/NaClO工藝較大.SA的與去除率均隨著氧化劑投加量的增大而增加,原因是NaClO的增加可以生成更多的?OH和RCS(式(2～7)),SPC的增加可以生成更多的?OH、O2?-和CO3?-等(式(8～12))[7,10,14].UV/NaClO和UV/SPC工藝氧化劑物質(zhì)的量投加比增加至5倍時(shí),分別增加至5.2和3.1倍,UV/NaClO工藝的增幅是UV/SPC工藝增幅的1.4倍;當(dāng)SPC投加的物質(zhì)的量比超過(guò)14時(shí),的增幅變緩,可能原因是溶液中的CO32-和HCO3-會(huì)消耗?OH(式(10,13～14))[9],這與Yan等[11]采用Fe2+活化SPC降解磺胺甲惡唑得到的規(guī)律相似.

圖1 UV/NaClO和UV/SPC對(duì)SA的去除

[SA]0=3.6μmol/L, [NaClO]: [SA]0=3.7、7.4、11.1、14.8、18.5,[SPC] :[SA]0=7、14、21、28、35, pH=(7.0±0.2)

2.2 UV/NaClO和UV/SPC工藝中自由基的鑒定

TBA可用于淬滅?OH和Cl?[15];PhOH可用于淬滅?OH和CO3?-(PhOH-?OH=6×108L/(mol×s),PhOH-CO3?-= 2.2×107L/(mol×s))[12,16];CF可用于淬滅O2?-(CF-O2?-= 3.0×1010L/(mol×s))[10];加入自由基清掃劑TBA、PhOH和CF后2種工藝SA的去除見(jiàn)圖2.

UV/NaClO工藝中SA的去除率隨TBA的增加而減小,表明體系中含有?OH和Cl?.加入過(guò)量TBA (20mM)后SA的去除率仍大于單獨(dú)NaClO(4.5%)和UV(13%)的去除率,說(shuō)明體系中存在其他RCS的貢獻(xiàn).UV/SPC工藝中加入CF、TBA和PhOH后,SA的去除率均不同程度降低,表明體系中含有O2?-、?OH和CO3?-.Yue等[10]用UV/SPC工藝降解辣椒素同樣證明了體系中存在O2?-、?OH和CO3?-.

[SA]0=3.6μmol/L, [NaClO]=40μmol/L,[SPC]=75.6μmol/L,pH=(7.0±0.2)

[NaClO]=40μmol/L, [SPC]=75.6μmol/L, [DMPO]=100mmol/L, 未特殊說(shuō)明pH=(7.0±0.2)

采用DMPO作為自由基捕獲劑檢測(cè)UV/NaClO和UV/SPC工藝中產(chǎn)生的自由基,測(cè)定EPR圖譜如圖3.強(qiáng)堿性條件時(shí)UV/NaClO體系中?OH(EPR圖譜四線強(qiáng)度比1:2:2:1)更明顯,而弱堿性時(shí)DMPO將被氧化為DMPOX(EPR圖譜七線強(qiáng)度比1:2:1:2:1:2:1),其為?OH和Cl?共同氧化的結(jié)果[17-18].而UV/SPC體系中檢測(cè)到?OH和O2?-(EPR圖譜四線強(qiáng)度比1:1: 1:1)[19].CO3?-作為UV/SPC體系中重要的自由基實(shí)驗(yàn)中未檢測(cè)到,這可能是體系中產(chǎn)生的CO3?-量較少,以至EPR無(wú)法檢出.

2.3 UV/NaClO和UV/SPC工藝降解SA過(guò)程中各組分貢獻(xiàn)

2.3.1 SA與?OH、CO3?-的二級(jí)反應(yīng)速率常數(shù) 二級(jí)反應(yīng)速率常數(shù)用來(lái)描述自由基與化合物之間的反應(yīng)快慢[16].通過(guò)UV/H2O2工藝降解SA與pCBA (pCBA-?OH=5′109L/(mol×s))進(jìn)行競(jìng)爭(zhēng)動(dòng)力學(xué)實(shí)驗(yàn)計(jì)算SA與?OH的二級(jí)反應(yīng)速率常數(shù)(式(15～16))[15](圖4).聯(lián)立式(15～16)得式(17),可求出SA與?OH的二級(jí)反應(yīng)速率常數(shù)為3.94′109L/(mol×s).該結(jié)果略小于Peralta等[20]報(bào)道的SA與?OH的二級(jí)反應(yīng)速率常數(shù)5′109L/(mol×s),但仍為同一數(shù)量級(jí).

圖4 SA和pCBA在UV/H2O2工藝中的降解

[SA]0=[pCBA]0=3.6μmol/L, [H2O2]=60μmol/L, pH=(7.0±0.2)

式中:obs,UV/H2O2,SA和obs,UV/H2O2,pCBA為降解SA、pCBA的擬一級(jí)動(dòng)力學(xué)常數(shù);obs,UV,SA和obs,UV,pCBA為單獨(dú)UV降解SA、pCBA的擬一級(jí)動(dòng)力學(xué)常數(shù);HO?-SA和HO?-pCBA為?OH與SA、pCBA的二級(jí)反應(yīng)速率常數(shù);[HO?]UV/H2O2,SS表示?OH的穩(wěn)態(tài)濃度.

選用TBA與CF分別作為?OH與O2?-的清除劑[10,15],通過(guò)SA和PhOH在UV/SPC工藝中的降解求出SA與CO3?-的二級(jí)反應(yīng)速率常數(shù)(式(18～ 19))[12],結(jié)果見(jiàn)圖5.

據(jù)此簡(jiǎn)化式(18～19)為式(20～21),聯(lián)立式(20～21)得到式(22),可求出SA與CO3?-的二級(jí)反應(yīng)速率常數(shù)為1.66′107L/(mol×s).此前Wojnárovits等[16]報(bào)道CO3?-與有機(jī)分子的二級(jí)反應(yīng)速率常數(shù)在102～109L/(mol×s)范圍內(nèi),實(shí)驗(yàn)結(jié)果與該范圍相符.

圖5 SA和PhOH在UV/SPC工藝中的降解

[SA]0=[PhOH]0=3.6μmol/L, [SPC]=75.6μmol/L, [TBA]=10mmol/L,[CF]=10mmol/L, pH=(7.0±0.2)

式中:T&C表示溶液中加入足量TBA與CF作為自由基淬滅劑;obs,UV/SPC,T&C,SA和obs,UV/SPC,T&C,PhOH為降解SA與PhOH的擬一級(jí)動(dòng)力學(xué)常數(shù);obs,UV,T&C,SA和obs,UV,T&C,PhOH為單獨(dú)UV降解SA、PhOH的擬一級(jí)動(dòng)力學(xué)常數(shù);obs,SPC,T&C,SA和obs,SPC,T&C,PhOH為單獨(dú)SPC降解SA、PhOH的擬一級(jí)動(dòng)力學(xué)常數(shù);obs,O2?-,T&C,SA和obs,O2?-,T&C,PhOH為O2?-降解SA、PhOH擬一級(jí)動(dòng)力學(xué)常數(shù),假定加入的CF完全淬滅O2?-,此處取0;HO?-PhOH為?OH與PhOH的二級(jí)反應(yīng)速率常數(shù);CO3?--SA和CO3?--PhOH為CO3?-與SA、PhOH的二級(jí)反應(yīng)速率常數(shù);[HO?]SS和[CO3?-]SS為該體系下?OH與CO3?-的穩(wěn)態(tài)濃度,設(shè)定加入的TBA完全淬滅?OH,此處[HO?]SS取0.

2.3.2 UV/NaClO和UV/SPC不同組分對(duì)SA去除的貢獻(xiàn)對(duì)比 選用NB、DMA分別作為?OH和CO3?-的探針(NB-?OH=3.9′109L/(mol×s),CO3?--DMA=1.8′109L/(mol×s)),采用探針?lè)ㄍㄟ^(guò)NB和DMA在UV/ NaClO和UV/SPC工藝中的降解(式(23～25))求出?OH和CO3?-的穩(wěn)態(tài)濃度(圖6)[21-22],進(jìn)而求出不同組分對(duì)SA去除的貢獻(xiàn).將擬合所得到的擬一級(jí)動(dòng)力學(xué)常數(shù)代入式(23～25)得出UV/NaClO工藝中?OH的穩(wěn)態(tài)濃度為4.21′10-12mol/L,UV/SPC工藝中?OH和CO3?-的穩(wěn)態(tài)濃度分別為2.72′10-11,8.20′10-11mol/L.

圖6 NB和DMA在UV/NaClO和UV/SPC工藝中的降解

[NB]0=[DMA]0=3.6μmol/L, [NaClO]= 40μmol/L, [SPC]=75.6μmol/L, pH=(7.0±0.2)

式中:obs,UV/NaClO,NB、obs,UV/SPC,NB和obs,UV/SPC,DMA為降解NB、DMA的擬一級(jí)動(dòng)力學(xué)常數(shù);obs,UV,NB和obs,UV,DMA分別為單獨(dú)UV降解NB、DMA的擬一級(jí)動(dòng)力學(xué)常數(shù);HO?-NB和CO3?--DMA分別為NB與?OH、DMA與CO3?-的二級(jí)反應(yīng)速率常數(shù).

由SA與?OH、CO3?-的二級(jí)反應(yīng)速率常數(shù)和?OH與CO3?-的穩(wěn)態(tài)濃度,可求得2個(gè)工藝下?OH和CO3?-降解SA的擬一級(jí)動(dòng)力學(xué)常數(shù)(式(26)～(28)).

式中:obs,UV/NaClO,HO?-SA為UV/NaClO工藝下?OH降解SA的擬一級(jí)動(dòng)力學(xué)常數(shù);obs,UV/SPC,HO?-SA和obs,UV/SPC,CO3?--SA分別為UV/SPC工藝下?OH、CO3?-降解SA的擬一級(jí)動(dòng)力學(xué)常數(shù).

圖7 不同物種在UV/NaClO和UV/SPC工藝中降解SA的貢獻(xiàn)

[SA]0=3.6μmol/L, [NaClO]=40μmol/L,[SPC]=75.6μmol/L,pH=(7.0±0.2)

SA在2種工藝中的降解可表示為式(29～30),由式(29)可求得UV/NaClO工藝中RCS降解SA的擬一級(jí)動(dòng)力學(xué)常數(shù)為0.1384min-1.由式(30)可求得UV/SPC工藝下O2?-降解SA的擬一級(jí)動(dòng)力學(xué)常數(shù)為0.1339min-1.

由圖7可知,UV/NaClO工藝中各組分貢獻(xiàn)分別為:RCS(79.91%)＞?OH(9.58%)＞UV(7.85%)＞NaClO(2.66%);UV/SPC工藝中各組分貢獻(xiàn)分別為:O2?-(51.75%)＞?OH(41.42%)＞UV(5.25%)＞SPC(1.04%)＞CO3?-(0.54%).UV/NaClO工藝中占主導(dǎo)的自由基是RCS,其貢獻(xiàn)是?OH的8.3倍,原因是生成的?OH轉(zhuǎn)化為RCS(式(2～4)),該結(jié)果與李博強(qiáng)等[23]研究UV-LED/NaClO去除對(duì)乙酰氨基酚所得到的規(guī)律一致.而UV/SPC工藝中占主導(dǎo)的自由基是O2?-和?OH,兩者貢獻(xiàn)了SA去除的93.17%,O2?-的貢獻(xiàn)超過(guò)50%,造成該結(jié)果的原因可能是CO32-的存在促進(jìn)了O2?-的生成(式(10～12)).實(shí)際應(yīng)用中可以加強(qiáng)自由基的生成與轉(zhuǎn)化進(jìn)而增強(qiáng)SA的降解效果.UV對(duì)兩種工藝降解SA的貢獻(xiàn)相差不大,這可能是因?yàn)?種工藝使用同一光源.?OH在2種工藝中的貢獻(xiàn)率相差31.84%,這是因?yàn)閁V/SPC工藝中?OH穩(wěn)態(tài)濃度(2.72′10-11mol/L)是UV/NaClO工藝下?OH穩(wěn)態(tài)濃度(4.21′10-12mol/L)的6.5倍.UV/SPC工藝中CO3?-在所有組分中的貢獻(xiàn)最低,該結(jié)果為EPR無(wú)法檢出CO3?-提供了佐證.原因是?OH與CO32-反應(yīng)(式(10))的二級(jí)反應(yīng)速率常數(shù)為3.9′108L/(mol×s),低于SA與?OH的二級(jí)反應(yīng)速率常數(shù)3.94′109L/(mol×s)[9].

2.4 UV/NaClO和UV/SPC工藝環(huán)境水樣中SA的降解

考察環(huán)境水樣(廈漳地區(qū)北溪和江東泵站)、自來(lái)水及實(shí)驗(yàn)室超純水(水質(zhì)參數(shù)見(jiàn)表1)中對(duì)SA降解的影響,不同水體中SA的降解速率如圖8.

表1 環(huán)境水樣的主要水質(zhì)參數(shù)

UV/NaClO和UV/SPC工藝在江東泵站、北溪、超純水和自來(lái)水水樣中的反應(yīng)速率分別為0.0517、0.0253、0.2174、0.1354min-1和0.0452、0.0213、0.2588、0.1383min-1.SA的降解速率為超純水>自來(lái)水>江東泵站>北溪,這是因?yàn)橐环矫嫠畼拥臐岫炔煌?影響UV的穿透;另一方面水體由有機(jī)物、無(wú)機(jī)物、浮游生物和微生物等復(fù)雜基質(zhì)構(gòu)成,這些物質(zhì)在降解時(shí)也參與了反應(yīng)并消耗了氧化劑及產(chǎn)生的自由基[24-25].Parastoo等[25]應(yīng)用超聲/臭氧工藝時(shí)得出濁度會(huì)消耗氧化劑.環(huán)境水樣中UV/NaClO工藝降解SA的反應(yīng)速率平均降低了67%,UV/SPC工藝中SA的反應(yīng)速率平均降低了74%,UV/SPC工藝較UV/NaClO工藝易受自然水體的抑制,可能的原因是由于?OH的無(wú)選擇性幾乎可以與所有有機(jī)分子發(fā)生反應(yīng)[6],而UV/SPC工藝中?OH的貢獻(xiàn)率大于UV/NaClO工藝中?OH的貢獻(xiàn)率,故其反應(yīng)速率受自然水體的抑制更大.

圖8 環(huán)境水樣對(duì)UV/NaClO和UV/SPC工藝降解SA的影響

[SA]0=3.6μmol/L, [NaClO]=40μmol/L,[SPC]=75.6μmol/L,pH=(7.0±0.2)

2.5 UV/NaClO和UV/SPC工藝降解SA過(guò)程中溶液急性毒性變化

圖9 UV/NaClO和UV/SPC工藝降解SA過(guò)程中急性毒性變化

[SA]0=3.6μmol/L, [NaClO]=40μmol/L,[SPC]=75.6μmol/L,pH=(7.0±0.2)

本文考察了2種工藝降解純水中SA的溶液急性毒性變化,如圖9,隨著降解進(jìn)行,UV/NaClO工藝中反應(yīng)溶液的相對(duì)抑制率先增加到100%再降至63%,可能的原因是在產(chǎn)物中發(fā)現(xiàn)的典型消毒副產(chǎn)物三氯甲烷導(dǎo)致了反應(yīng)初期毒性升高,隨著反應(yīng)的進(jìn)行有毒產(chǎn)物進(jìn)一步分解從而相對(duì)抑制率逐漸下降[26].而UV/SPC工藝中反應(yīng)溶液相對(duì)抑制率保持在25%左右波動(dòng),UV/SPC工藝較UV/NaClO工藝低38%的抑制率、表現(xiàn)出較低的急性毒性.Zupanc等[27]人研究表明SA與?OH會(huì)生成PhOH和2,5-二羥基苯甲酸(2,5-DHBA)等產(chǎn)物.通過(guò)基于QSAR的ECOSAR軟件模擬評(píng)估了這兩種產(chǎn)物的急性毒性[28],結(jié)果表明產(chǎn)物急性毒性(PhOHEC50=2.40mg/L, 2,5-DHBAEC50=2.92mg/L,EC50數(shù)值越小代表毒性越大)均高于SA(SAEC50=11.35mg/L),故UV/SPC工藝降解SA過(guò)程中溶液相對(duì)抑制率升高可能是由于生成PhOH和2,5-DHBA造成.推測(cè)隨著降解的進(jìn)行產(chǎn)物將會(huì)與生成的自由基進(jìn)一步發(fā)生反應(yīng),最終相對(duì)抑制率進(jìn)一步降低.

2.6 UV/NaClO和UV/SPC工藝的經(jīng)濟(jì)效益比較

為從經(jīng)濟(jì)效益角度比較2個(gè)工藝,采用單位去除能耗(EE/OUV)表示m3水中降解1個(gè)數(shù)量級(jí)SA所需的電能(式(31)).單個(gè)工藝總成本(Cost/Ototal)可表示為電能成本(Cost/OUV)與氧化劑成本(Cost/ Ooxidant)之和(式(32～34)),其中工藝總成本單位為$/ (m3×order),表示m3水中降解1個(gè)數(shù)量級(jí)的SA所需要的成本[26].

式中:表示UV汞燈輸入功率,為2×10-3kW;表示反應(yīng)時(shí)間,h;表示反應(yīng)溶液的體積,L;obs表示擬一級(jí)動(dòng)力學(xué)常數(shù),min-1;0表示SA的初始濃度,為500μg/L;C表示SA在反應(yīng)時(shí)刻的濃度,μg/L; [Oxidant]0表示氧化劑的投加量,mg/L; electricity cost表示電費(fèi),這里取0.1$/kWh; price[Ox]表示氧化劑的單價(jià),NaClO為0.13$/kg; SPC為0.3$/kg[12,26].

如圖10所示,單獨(dú)UV降解SA的成本為18.9$/(m3×order),UV/NaClO工藝成本隨著投加量的增加呈下降趨勢(shì),從最高8.7$/(m3×order)降至4.0$/ (m3×order).Ox/SA=3.7～7.4時(shí)UV/NaClO工藝效益曲線發(fā)生了轉(zhuǎn)折,原因是NaClO投加比較低時(shí)整體降解效率隨著投加比的增加提升不明顯,但NaClO增加一倍導(dǎo)致氧化劑成本增加1倍,故表現(xiàn)為Ox/SA= 3.7的總體降解成本低于Ox/SA=7.4時(shí). UV/SPC工藝成本隨著投加量的增加從24.4$/ (m3×order)上升至37.1$/(m3×order),且高于單獨(dú)UV降解SA的成本.SA降解率達(dá)到97.5%以上時(shí), UV/SPC工藝成本是UV/NaClO工藝的9.3倍.從經(jīng)濟(jì)效益角度UV/NaClO工藝較UV/SPC工藝更節(jié)約成本,該結(jié)果與Lu等[26]比較UV、UV/NaClO和UV/過(guò)硫酸鹽的經(jīng)濟(jì)效益得到的結(jié)果相似.

圖10 UV/NaClO和UV/SPC工藝降解SA的經(jīng)濟(jì)效益評(píng)價(jià)

[SA]0=500μg/L, pH=(7.0±0.2)

3 結(jié)論

3.1 兩種工藝對(duì)SA均能有效降解,反應(yīng)速率均隨著NaClO、SPC投加量的增加而增加,氧化劑摩爾投加比超過(guò)7.4后UV/NaClO工藝比UV/SPC工藝有更高的反應(yīng)速率與降解率,UV/NaClO工藝和UV/SPC工藝反應(yīng)速率最高分別達(dá)0.4378, 0.3794min-1.

3.2 UV/NaClO體系存在?OH和Cl?, UV/SPC體系中存在O2?-、?OH和CO3?-.SA與?OH、CO3?-的二級(jí)反應(yīng)速率常數(shù)分別為3.97′109,8′107L/(mol×s). UV/ NaClO工藝下?OH的穩(wěn)態(tài)濃度為4.21′10-12mol/ L,UV/SPC工藝下?OH、CO3?-的穩(wěn)態(tài)濃度分別為2.72′10-11,8.20′10-11mol/L.自由基在SA的降解中起主導(dǎo)作用,UV/NaClO工藝中以RCS為主導(dǎo)的各組分貢獻(xiàn)順序?yàn)?RCS>?OH>UV>NaClO; UV/SPC工藝中以O(shè)2?-和?OH為主導(dǎo)的各組分貢獻(xiàn)順序?yàn)?O2?->?OH>UV>SPC>CO3?-.

3.3 兩種工藝在環(huán)境水樣中降解SA時(shí)反應(yīng)速率均受到了抑制,UV/SPC工藝較UV/NaClO工藝受自然水體的抑制更明顯.從急性毒性角度,UV/SPC工藝反應(yīng)溶液較UV/NaClO工藝反應(yīng)溶液低了38%的抑制率.從經(jīng)濟(jì)效益角度,UV/NaClO工藝較UV/SPC工藝具有較高的經(jīng)濟(jì)效益.

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Comparison of UV/NaClO and UV/ Sodium percarbonate processes for degradation of salicylic acid.

MA Xiao-yan1, YANG Fan1, LI Qing-song2,3*, YANG Qing-yun1, CHEN Guo-yuan2, LI Guo-xin2

(1.College of Civil Engineering, Zhejiang University of Technology, Hangzhou 310014, China；2.Water Resource and Environment Institute, Xiamen University of Technology, Xiamen 361024, China；3.Key Laboratory of Water Resources Utilization and Protectionof Xiamen, Xiamen University of Technology, Xiamen 361024, China)., 2022,42(3)：1182～1190

The degradation of salicylic acid (SA) in aqueous solution by UV/NaClO and UV/SPC processes was investigated. The effects of oxidizer types and dosage on SA removal were compared. The free radicals in the two processes were identified by quenching method and electron paramagnetic resonance(EPR) spectroscopy. The second-order rate constants of ?OH and CO3?-with SA and the contributions of different components in the reaction system were determined by competitive kinetics. The removal of SA was compared in terms of environmental water samples simulation, acute toxicity and economic benefit. The pseudo-first-order kinetic rate constants of UV/NaClO and UV/SPC processes were 0.4378 and 0.3794min-1, respectively. ?OH and Cl? were detected in UV/NaClO process, while O2?-, ?OH and CO3?-were detected in UV/SPC process. The second-order rate constants of ?OH and CO3?-with SA were calculated to be 3.97′109and 8′107L/(mol×s), respectively. Reactive chlorine species (RCS) (79.91%) functioned a dominant role in the removal of SA in UV/NaClO process, while O2?-(51.75%) and ?OH (41.42%) functioned a dominant role in UV/SPC process. The degradation of SA in environmental water samples by UV/NaClO and UV/SPC processes was inhibited, and the removal rates were reduced by 67% and 74%, respectively. The inhibition rate of UV/SPC process (25%) was 38% lower than that of UV/NaClO process(63%). The cost of UV/SPC process[37.1$/(m3×order)] was 9.3 times higher than that of UV/NaClO process[4.0$/(m3×order)] when SA degradation rate was above 97.5%. UV/NaClO process had higher economic benefit than UV/SPC process.

UV-AOPs；radical identify；contribution of free radicals；acute toxicity；economic benefit

X703

A

1000-6923(2022)03-1182-09

馬曉雁(1978-),女,山東萊州人,教授,博士,主要研究方向?yàn)轱嬘盟⒘坑袡C(jī)污染物控制.發(fā)表論文30余篇.

2021-08-18

國(guó)家自然科學(xué)基金資助項(xiàng)目(51878582,51978618);福建省科技計(jì)劃引導(dǎo)性資助項(xiàng)目(2021Y0041);福建省自然科學(xué)基金資助項(xiàng)目(2020J01256);福建省高校新世紀(jì)優(yōu)秀人才支持計(jì)劃項(xiàng)目(JA14227);廈門(mén)理工學(xué)院科研攀登計(jì)劃項(xiàng)目(XPDKT19026)

*責(zé)任作者, 研究員, leetsingsong@sina.com