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Section 2.4 Vanadium crossover If we take a look at reaction (2-18), we see that each VO2+ ion reacts with two V2+ ions to three V3+ ions. Thereby, all entered VO2+ ions are converted. This means that the effective ionic flux of VO2+ ions, J5Cross−, is zero. After all calculations are carried out for a particular time step, VO2+ ions are no longer present in the negative half-cell. Simultaneously, a negative flux arises for the V2+ ions and a positive flux arises for the V3+ ions. The ionic flux of the V2+ ions is caused by reaction (2-18) and is 2c5C c5CD5Mem. Thus, it is twice the ionic flux of the VO2+ ions penetrating the negative half-cell due to the vanadium crossover. The ionic flux of the V3+ ions is three times the ionic flux of the VO2+ ions. Now, we take a look at reaction (2-19). Each present VO2+ ion reacts with one V2+ ion to two V3+ ions. Hence, the VO2+ ions penetrating the negative half-cell through the membrane are now converted into V3+ ions. Thus, the effective ionic flux of VO2+ ions, J4Cross−, is zero as well. The number of V2+ ions is reduced by the number of VO2+ ions. Hence, the ionic flux of V2+ ions due to the reaction (2-19) is c4C c4CD4Mem. The ionic flux of V3+ ions is positive with twice the magnitude. Effectively, the ionic flux of V2+ ions because of the self-discharge reactions is c4C c4CD4Mem 2c5C c5CD5Mem. The number of V3+ ions in the negative half-cell is increased by the self-discharge reactions. The entered VO2+ ions each react to three V3+ ions in the reaction (2-18). Further, each entered VO2+ ion reacts to two additional V3+ ions due to reaction (2-19). In total, the V3+ flux due to the self-discharge reactions is 2c4C c4CD4Mem 3c5C c5CD5Mem. Together with the fluxes of V2+ and V3+ ions leaving the negative half-cell and diffusing into the positive half-cell, the fluxes can be put into a matrix form, as shown in Eq. (2-26), which allows for an efficient implementation in MATLAB. J2Cross J3Cross J4Cross J5Cross (2-26) 2.4.4 Ionic flux resulting from vanadium crossover and the self-discharging reactions in the positive half-cell The aforementioned considerations are conducted in an analogue manner for the positive half-cell. The self-discharge reaction (2-20) yields V3+ ions, which further react with VO2+ ions in the self-discharge reaction (2-22). Hence, reaction (2-20) should not be considered at last. Consequently, there are two reasonable orders of the reactions, namely (2-20)→(2-21)→(2-22) and (2-21)→(2-20)→(2-22). A c2C c2CD2Mem c4C c4CD4Mem 2c5C c5CD5Mem Mem c3C c3CD3Mem 2c4C c4CD4Mem 3c5C c5CD5Mem Mem 0 0 23PDF Image | Model-based Design Vanadium Redox Flow Batteries
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