Sludge volume before and after digestion

Method

Sludge volume $=$ volume of water $+$ volume of volatile solids $+$ volume of fixed solids, with solids $=600$ kg dry. Water is 1000 kg/m³; volatile solids $G=1.0$ (1000 kg/m³); fixed solids $G=2.5$ (2500 kg/m³).

Primary sludge (6% solids, 65% volatile)

$$\begin{aligned} VM &= 0.65\times600=390\ \text{kg},\quad FM=210\ \text{kg}\\ \text{Mass of wet sludge} &= \frac{600}{0.06}=10000\ \text{kg}\ \Rightarrow\ \text{water}=9400\ \text{kg}=9.400\ \text{m}^3\\ \text{Volume of solids} &= \frac{390}{1000}+\frac{210}{2500}=0.4740\ \text{m}^3\\ V_{primary} &= 9.400+0.4740=9.874\ \text{m}^3 \end{aligned}$$

Digested sludge (12% solids, 65% of volatile matter removed)

$$\begin{aligned} VM_d &= 390\times(1-0.65)=136.5\ \text{kg},\quad FM=210\ \text{kg}\ (\text{unchanged})\\ \text{Dry solids} &= 136.5+210=346.5\ \text{kg}\\ \text{Mass of wet sludge} &= \frac{346.5}{0.12}=2887.5\ \text{kg}\ \Rightarrow\ \text{water}=2541.0\ \text{kg}=2.541\ \text{m}^3\\ \text{Volume of solids} &= \frac{136.5}{1000}+\frac{210}{2500}=0.2205\ \text{m}^3\\ V_{digested} &= 2.541+0.2205=2.761\ \text{m}^3 \end{aligned}$$

Percentage reduction

$$\frac{9.874-2.761}{9.874}\times100=72.0\%$$

Answer: volume before digestion $=9.874$ m³, after digestion $=2.761$ m³, reduction $=72.0\%$.