Recently I am reading aqchem.F. I notice the description of DTW (cloud chemistry timestep) in aqchem.F:
DTW( 1 ) = -0.05D0 * MIN( TH2O2, TSIV ) / DSIVDT( 1 )
DTW( 2 ) = -0.01D0 * MIN( TO3, TSIV ) / DSIVDT( 2 )
DTW( 3 ) = -0.1D0 * TSIV / DSIVDT( 3 )
DTW( 4 ) = -0.1D0 * MIN( TMHP, TSIV ) / DSIVDT( 4 )
DTW( 5 ) = -0.1D0 * MIN( TPAA, TSIV ) / DSIVDT( 5 )
I wonder why the coefficients multiplied in each of the expressions are different.
Much appreciated for your answer!
In many cases, the aqueous reactions of S(IV) with H2O2 and O3 dominate the aqueous production of S(VI). These reactions can quickly change the system pH (especially under high pH conditions). In order to reduce error associated with the separation of mass transfer and chemical kinetic calculations (and in order to properly represent the contribution of different oxidation pathways), smaller timesteps are appropriate when the system is changing rapidly. Under high pH conditions, the O3 reaction rate is often fastest, followed by H2O2, which is essentially pH-independent for pH > 2 (see Figure 1 from Walcek and Taylor, 1986 (https://doi.org/10.1175/1520-0469(1986)043<0339:ATMFCV>2.0.CO;2) or Figure 2A from Cheng et al., 2016 (DOI: 10.1126/sciadv.1601530) for reference), and I think that is why those reactions have smaller coefficients in their associated timestep calculations.
Sorry for my late reply! I am very appreciated for your help! Your answer is very helpful for me!