Bod

D 10/300

2.1.15 Nitrification in the BOD Test

The general profile of oxygen consumption in a BOD test for a waste containing oxygen-consuming constituents is shown in Figure 2.1. As mentioned previously, because the nitrifiers cannot easily compete with the carbonaceous bacteria, it takes about 5 days or so for them to develop. Thus, after about 5 days the curve abruptly rises due to the nitrogenous oxygen demand, NBOD. If the nitrifiers are abundant in the beginning of the test, however, the nitrogen portion can be exerted immediately as indicated by the dashed line after a short lag. This figure shows the necessity of inhibiting the nitrifiers if the carbonaceous oxygen demand, CBOD, is the one desired in the BOD test.

The reactions in the nitrification process are mediated by two types of autotrophic bacteria: Nitrosomonas and Nitrobacter. The ammonia comes from the nitrogen content of any organic substance, such as proteins, that contains about 16% nitrogen. As soon as the ammonia has been hydrolyzed from the organic substance, Nitrosomonas consumes it and in the process also consumes oxygen according to

FIGURE 2.1 Exertion of CBOD and NBOD.

the following reactions:

Adding Eqs. (2.3) and (2.4) produces j-NH+ + J-O2 ^ I-H2O + 1-NO-+ 3-H+ (2.5)

Equation (2.4) is called an electron acceptor reaction. Equation (2.3) is an elector donor reaction, that is, it provides the electron for the electron acceptor reaction. Together, these two reactions produce energy for the Nitrosomonas.

The NO- produced in Equation (2.5) serves as an electron source for another genus of bacteria, the Nitrobacter. The chemical reactions when Nitrobacter uses the nitrite are as follows:

As with Nitrosomonas, the previous reactions taken together provide the energy needed by Nitrobacter. The combined reactions for the destruction of the ammonium ion, NH+, (or the ammonia, NH3) can be obtained by adding Eqs. (2.5) and (2.8). This will produce

From Equation (2.9), 1.0 mg/L of NH4-N is equivalent to 4.57 mg/L of dissolved oxygen.

2.1.16 Mathematical Analysis of BOD Laboratory Data

The ultimate carbonaceous oxygen demand may be obtained by continuing the incubation period beyond five days up to 20 to 30 days. To do this, the nitrifiers should be inhibited by adding the appropriate chemical in the incubation bottle. The other way of obtaining CBOD is through a mathematical analysis.

In the incubation process, let y represent the cumulative amount of oxygen consumed (oxygen uptake) at any time t, and let Lc represent the CBOD of the original waste. The rate of accumulation of the cumulative amount of oxygen, dy/dt, is proportional to the amount of CBOD left to be consumed, Lc - y. Thus, ddyy = y = kc(Lc - y) (2.10)

where kc is a proportionality constant called deoxygenation coefficient.

In the previous equation, if the correct values of kc and Lc are substituted, the left-hand side should equal the right-hand side of the equation; otherwise, there will be a residual R such that

At each equal interval of time, the values of y may be determined. For n intervals, there will also be n values of y. The corresponding Rs for each interval may have positive and negative values. If these Rs are added, the result may be zero which may give the impression that the residuals are zero. On the other hand, if the residuals are squared, the result of the sum will always be positive. Thus, if the sum of the squares is equal to zero, there is no ambiguity that the residuals are, in fact, equal to zero.

The n values of y corresponding to n values of time t will have inherent in them one value of kc and one value of Lc. Referring to Equation (2.11), these values may be obtained by partial differentiation. From the previous paragraph, when the sum of the squares of R is equal to zero, it is certain that the residual is zero. This means that when the sum of the squares is zero, the partial derivative of the sum of the squares must also be zero. Consequently, the partial derivatives of the sum of R with respect to kcLc and kc are zero. Thus, to obtain kc and Lc, the latter partial derivative of the sum of the squares must be equated to zero to force the solutions. The method just described is called the method of least squares, because equating the partial derivatives to zero is equivalent to finding the minimum of the squares. The corresponding equations are derived as follows:

Solving for kc,

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