普雷克斯(PUREX)水法后处理中几种重要元素行为的理论探讨

THE DISCUSSION OF BEHAVIOR OF SOME IMPORTANT ELEMENTS IN PUREX PROCESS

从热力学和动力学方面探讨了乏燃料溶解液、共去污溶剂萃取、铀钚分离中某些重要元素的化学行为。乏燃料溶解时亚硝酸浓度可从开始约３×１０－２ｍｏｌ／Ｌ降至终点时的２×１０－５ｍｏｌ／Ｌ，计算出ＮＯ－３－ＨＮＯ２电对的溶解终点电位为１．１３３伏，表明终点时溶液具有更强的氧化性。由此计算了溶解液中约含３％Ｐｕ（Ⅵ）和４７％Ｎｐ（Ⅵ）。大部分碘将以Ｉ２形式存在，只有当亚硝酸浓度极低的情况下才会生成少量ＩＯ－３。钌将以Ｒｕ（Ⅵ）形式存在，在水相硝酸中几乎不可能把Ｒｕ（Ⅳ）氧化至Ｒｕ（Ⅷ）。在共去污萃取段当ＨＮＯ３为２ｍｏｌ／Ｌ时约有２０％Ｎｐ（Ⅴ）氧化至Ｎｐ（Ⅵ），３ｍｏｌ／ＬＨＮＯ３时将有７０％的Ｎｐ（Ⅴ）氧化至Ｎｐ（Ⅵ）。铀钚分离中当用Ｕ（Ⅳ）作还原剂时，计算了Ｐｕ（Ⅳ）与Ｕ（Ⅳ）反应的浓度平衡常数为１０１９．５，求得的Ｐｕ（Ⅲ）／Ｐｕ（Ⅳ）比至少为４．７×１０８，反应速度计算表明，该反应在４．９分内即达平衡，由此指出除水力学因素外，Ｐｕ（Ⅲ）的少量萃取是铀中除钚的另一制约因素。进入铀钚分离阶段的有机相中的镎为Ｎｐ（Ⅵ），它在Ｕ（Ⅳ）和Ｐｕ（Ⅲ）的作用下很快还原至Ｎｐ（Ⅴ），然后Ｕ（Ⅳ）将把Ｎｐ（Ⅴ）慢还原至Ｎｐ（Ⅳ?

The behavior of some important elements in dissolution,co decontamination and uranium plutomium separation stages from the thermodymamic and Kinetic points of view is discussed.During the dissolution of spent fuel,the concentration of nitrous acid might decrease sharply from 3×10 -2 mol/L at beginning down to 2×10 -5 mol/L at the end.The potential of NO - 3 HNO 2 half reaction in the final dissolution solution is estimated to be 1.133V.Based on the above potential,the possible valence states of Np、 Pu、I 2、Ru in dissolution solution are predicted,the Pu(Ⅳ) and Np(Ⅵ) about 3% and 47% of its total amount respectively,the iodine existing mainly as I 2 and iodate possibly appearing only under extreme condition of very low nitrous acid concentration,the Ru existing only in the form of Ru(Ⅳ),generally impossible to be oxidized to Ru(Ⅷ) in the aqueous solution of nitric acid.Depending on the contact time in the co decontamination stage,about 20% Np(Ⅴ) could be oxidized to Np(Ⅵ) when nitric acid is 2mol/L,and 70% Np(Ⅵ) when nitric acid is 3 mol/L in the extraction section of co decontamination extractor.In the separation stage,the equilibrium constant of Pu(Ⅳ) reduction with U(Ⅳ) is calculated with K c about 10 19.5 ,the reduction process reaching equilibrium within 5 min.In the given condition the ratio of Pu(Ⅲ) to Pu(Ⅳ) is 4.7×10 8.It indicates that,beside hydrodymamic effect,the little extraction of Pu(Ⅲ) is another restriction factor for having high separation factor of Pu from Uand Pu(Ⅲ)in the Separation stage, and then, the Np(Ⅴ)is Velatively slow to be veduced to Np(Ⅳ) by U(Ⅵ),The Np(Ⅵ) coming from the first extractor could be quickly reduced to Np(Ⅴ) by U(Ⅳ) and Pu(Ⅲ),therefore, a mixture of Np(Ⅴ) and Np(Ⅳ) in the Pu product of separation stage would be obtained,the Np(Ⅳ) might be dominated if resident time is long enough.The factor causing much more consumption of U(Ⅳ) than the stoichiometric equation is discussed and suggested that the action of Tc for excess consumption of U(Ⅳ) needs more careful investigation.