Appendix D: Compost Losses

Assume that organic matter has three components: carbon, nitrogen, and a residue:

O = C + N + R

The residue is hydrogen and oxygen (maybe also sulfur), which disappear along with the carbon; and minerals, most of which remain, except for what may leach out (such as potassium).

The presence of the minerals requires an additional parameter and complicates the calculation. Since they constitute a small part of the residues, a convenient solution is simply to ignore them. This introduces an error, but the goal is to obtain only an estimate of the loss of organic matter.

What then remains of the organic matter is:

O = C(1 + γ + N)

O = N[(C/N)(1 + γ) + 1]

where γ = R/C is the ratio of volatiles to carbon.

Define the initial and final carbon/nitrogen ratios:

η0 = (C/N)0

η1 = (C/N)1

The ratio of the final to initial quantity of organic matter is then:

 O1     N11(1 + γ) + 1]
———— = ——————————————————
 O0     N00(1 + γ) + 1]

Set β = N1/N0 (the fraction of nitrogen remaining.

 O1         η1(1 + γ) + 1
———— = β * ——————————————
 O0         η0(1 + γ) + 1
  

Since both carbon/nitrogen ratios are much larger than 1:

 O1         η1
———— = β * ————
 O0         η0
  

The total amount of material in the compost pile is

T = M + O

where M is the soil (and mineral) content. The ratio of the final to initial material is:

 T1     M + O1
———— = ————————
 T0     M + O0
  

Let α = M / O, the ratio of soil to the initial quantity of organic matter. Then:

 T1     α * O0 + O1
———— = ————————————
 T0     α * O0 + O0
  
 T1     α + O1 / O0
———— = ————————————
 T0     α + 1
  
 T1     α + β + η1 / η0
———— = ————————————————
 T0     α + 1
  

To summarize:

The last statement is perhaps trivial, since it does no more than confirm what we might expect intuitively: the fraction of remaining organic matter is proportional to the fraction of remaining nitrogen. But it does lend credibility to the graphs in figure 2. Compost Losses

© 2013 Robert Parnes

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