We get a good estimate of R ×
X by measuring the slope of the D.O. graph shortly after aeration is stopped.
Now consider a point about midway in the curve in the region where aeration has resumed.
The tangent to the curve that is cast through that point gives the slope. This
is equal to dC/dt. The D.O. corresponding to the point is C. We look up the concentration of oxygen in water at the temperature of the process or measure it with some medium that has not been inoculated. Remember that it must be aerated with ordinary air, not pure oxygen.
Now we have everything we need for the equation and simply substitute to solve for Kla.
Graph with the tangent sketched in.
Select a value or reaeration rate that is near the middle of its scrollbar and experiment with increasing the uptake rate. Note that the line for D.O. following closure of the value is no longer straight but curves at low D.O. because the respiration is limited by oxygen. Experiments with permutations of coefficients until you get rather low D.O. concentrations. Try to avoid or ignore permutations of coefficients where the very early D.O. is not a horizontal line. Any deviation is an artifact of the way the program calculates the initial condition.
There is no problem is using the linear portion of the graph following closure of the valve to estimate R X. It is also possible to measure a slope at a point on the curve after turning the valve back on, but there is only a cramped region in which to work. If the D.O. is too close to the concentrations where oxygen is limiting, this method fails. In other words, a more voracious demand for oxygen and a lower initial D.O. would give too little room to work before approaching the critical oxygen concentration, and the method fails at low D.O. where the equations are not valid.