Settleable solids; first-stage BOD

Practical approach to determine solids

1. Total solids (TS): a measured volume (100 mL) of well-mixed sample is evaporated in a pre-weighed dish on a water bath and dried at 103-105°C to constant weight. TS = (final − empty weight) / volume.

2. Suspended and dissolved solids: another sample is filtered through a weighed glass fibre filter. The dried residue on the filter is the suspended solids (SS); $TDS = TS - SS$.

3. Volatile and fixed solids: the dried residue is ignited in a muffle furnace at 550°C. The weight lost is the volatile (organic) solids; the ash remaining is the fixed (inorganic) solids.

4. Settleable solids: 1 litre of mixed sample is poured into an Imhoff cone and left undisturbed for 1 hour (after 45 min, the sides are gently stirred). The volume of sludge settled in the tip is read in mL/L. (The gravimetric way: settleable solids = SS of the raw sample − SS of the clear supernatant left after 1 h of settling.)

5. Non-settleable solids: the SS of the supernatant after 1 hour of settling, which are mainly colloids and fine particles; i.e.

$$\text{Non-settleable solids} = \text{Total suspended solids} - \text{Settleable solids}$$

Derivation of first-stage BOD expression

The first stage (carbonaceous) BOD is the oxygen used to oxidise organic carbon. It is a first order reaction: the rate of oxidation at any time is proportional to the organic matter still remaining (not yet oxidised).

Let $L_t$ be the oxygen equivalent of organic matter remaining at time $t$ and $K$ the rate constant (day⁻¹).

$$\frac{dL_t}{dt}=-K\,L_t$$

Separate variables and integrate between $t=0$ ($L_t=L$, the ultimate first-stage BOD) and time $t$:

$$\int_{L}^{L_t}\frac{dL_t}{L_t}=-K\int_0^t dt\ \Rightarrow\ \ln\frac{L_t}{L}=-Kt$$ $$L_t=L\,e^{-Kt}=L\cdot10^{-k t}\quad(k=K/2.303)$$

BOD exerted up to time $t$ is the amount used up:

$$y_t = L - L_t = L\left(1-e^{-Kt}\right)=L\left(1-10^{-kt}\right)$$

For example, with $k=0.1$/day, $BOD_5=0.68L$. The rate constant depends on temperature as $K_T=K_{20}(1.047)^{T-20}$.