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McKay, Donald. "Front matter" Multimedia Environmental Models ...

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where<br />

Figure 8.5 QWASI model: steady-state and unsteady-state solutions.<br />

©2001 CRC Press LLC<br />

QWASI Equations<br />

Steady-state solutions (i.e., derivatives are equal to zero)<br />

Since Sum of all input rates = sum of all output rates<br />

The sediment mass balance is<br />

fW(DD + DT) = fS(DR + DT + DS + DB) and can be rewritten as<br />

fS = fW (DD + DT)/(DR + DT + DS + DB) The water mass balance is<br />

EW + fI(DI + DX) + fA(DV + DQ + DC + DM) + fS(DR + DT) = fW(DV + DW + DJ + DY + DD + DT) and can be rewritten as<br />

EW + f I ( Di + DX ) + f A ( DV + DQ + DC + DM ) + f S ( DR + DT )<br />

f W = ------------------------------------------------------------------------------------------------------------------------------------------------------------<br />

DV + DW + DJ + DY + DD + DT To solve water mass balance, eliminate fS. EW + f I ( Di + DX ) + f A ( DV + DQ + DC + DM )<br />

f W = --------------------------------------------------------------------------------------------------------------------<br />

( DD + DT ) ( DS + DB )<br />

DV + DW + DJ + DY + -----------------------------------------------------<br />

DR + DT + DS + DB The sediment fugacity can then be calculated.<br />

Unsteady-state analytical solutions<br />

Since VZdf/dt = (total input rate – total output rate)<br />

the sediment differential equation is<br />

V SZ BSdfS -------------------------- = f<br />

dt W ( DD + DT ) – f s ( DR + DT + DS + DB )<br />

The water differential equation is<br />

V W ZBWdfW -------------------------------- = E<br />

dt<br />

W + f I ( DI + DX ) + f A ( DV + DQ + DC + DM ) + f s ( DR + DT ) – f W ( DV + DW + DJ + DY + DD + DT )<br />

Here, subscript B refers to the bulk or total phase including dissolved and sorbed material.<br />

This pair of equations can be written more compactly as<br />

dfW ----------- = I<br />

dt 1 + I2 f S – I3 f W<br />

dfS --------- = I<br />

dt 4 f W – I5 f S<br />

where<br />

EW + f I ( DI + DX ) + f A ( DV + DQ + DC + DM )<br />

I1 = ---------------------------------------------------------------------------------------------------------------------<br />

V W ZBW DR + DT I2 = ---------------------<br />

V W ZBW DV + DW + DJ + DY + DD + DT I3 = -------------------------------------------------------------------------------<br />

V W ZBW DD + DT I4 = ---------------------<br />

V SZ BS<br />

DR + DT + DS + DB I5 = ------------------------------------------------<br />

V SZ BS<br />

The solution with the initial conditions fSO and fWO and final conditions fS• and fW• is<br />

f W = f W • + I8 exp [ – ( I6 – I7 )t]<br />

+ I9 exp [ – ( I6 + I7 ) t ]<br />

( I3 – I6 + I7 )I8exp [ – ( I6 – I7 )t]<br />

+ ( I3 – I6 – I7 )I9exp [ – ( I6 + I7 )t]<br />

f S =<br />

f S • + --------------------------------------------------------------------------------------------------------------------------------------------------------------------<br />

I2 f W• = I 1I 5/(I 3I 5 – I 2I 4), as in the steady-state solution above<br />

f S• = I 1I 4/(I 3I 5 – I 2I 4), as in the steady-state solution above<br />

I 6 = (I 3 + I 5)/2<br />

I 7 = [(I 3 – I 5) 2 + 4I 2I 4] 0.5 /2<br />

I 8 = [–I 2(f S• – f SO) + (I 3 – I 6 – I 7)(f W• – f WO)]/2I 7<br />

I 9 = [+I 2(f S• – f SO) – (I 3 – I6 + I 7)(f W• – f WO)]/2I 7

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