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首页> 外文会议>China-Japan workshop on nuclear waste management and reprocessing >The Study on Extraction Behaviour of U(VI), Pu(IV) and Am(III) by Amido Podands and Their Application in Separation of HLW
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The Study on Extraction Behaviour of U(VI), Pu(IV) and Am(III) by Amido Podands and Their Application in Separation of HLW

机译：酰胺键对U（VI），Pu（IV）和Am（III）的萃取行为及其在HLW分离中的应用研究

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By using n-octanol(40percent)-kerosen as diluent, the extraction behavior of U(VI), Pu(IV, Am(III), Eu(III) and other metal ions from nitric acid solution was studied with N, N, N'N'-tetrabutyl-3-oxa-pentanediamide (TBOPDA), N, N, N'N'-tetraisobutyl-3-oxa-pentanediamide(TiBOPDA) and N, N, N, 'N'-tetrabutyl-3, 6-dioxaoctanediamide (TBDOODA). A separation flowsheet was designed on the bassi of condition tests. A miniature mixer-settler test was carried out with 0.2mol/LTBOPDA/40percentoctanol-kerosene as extractant, to recover and separate U(VI), Pu(IV), Am(III) and Eu(III) from simualted high liquid waste solution. In coextractor(A), these elements were extracted into organic phase with the recovery ratio of 99.98percent for U(VI, >99.98 for Pu(IV), and >99.99percentAm(III) and Eu(III) respectively, witht he flow ratio in this battery O:A:S =1.5:1:0.5. In the first back-extractor(Rl), 99.99percent U, 86,2percent Pu and a fraction of Am or Eu were stripped into aqueous phase by 0.01 mol/L HNO_3+0.2mol/LAHA. In the second back-extractor (R2), the remained Pu, totally Am and Eu were stripped into aqueous phase by 0.2mol/L AHA with the flow ratio 1:1 as the same as that in battery Rl. This separation process has obvious advantages that contains no salt, extractant is completely incinerated and it is not needed to diluent or denitrate HLW.
机译：以正辛醇（40％）-煤油为稀释剂，研究了用N，N， N'N'-四丁基-3-氧杂戊二酰胺（TBOPDA），N，N，N'N'-四异丁基-3-氧杂戊二酰胺（TiBOPDA）和N，N，N，'N'-四丁基-3， 6-二氧八辛二酰胺（TBDOODA）。根据条件试验设计了分离流程图，以0.2mol / LTBOPDA / 40％辛醇-煤油为萃取剂进行了微型混合沉降试验，以回收和分离U（VI），Pu从模拟的高废液中分离出（IV），Am（III）和Eu（III）在共萃取器（A）中，这些元素被萃取到有机相中，U（VI）的回收率为99.98％，Pu（> 99.98） IV）和> 99.99％Am（III）和Eu（III），且该电池中的流量比为O：A：S = 1.5：1：0.5。在第一个反萃取器（R1）中，Um为99.99％，将86.2％的Pu和一部分Am或Eu汽提到0.01 mol /％的水相中L HNO_3 + 0.2mol / LAHA。在第二反萃取器（R2）中，将剩余的Pu，全部Am和Eu通过0.2mol / L AHA以与电池R1中相同的1：1的流量比汽提成水相。这种分离过程具有明显的优势，它不含盐，萃取剂被完全焚化，不需要稀释或反硝化HLW。

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