Volume-moisture relation; specific gravity

Volume-moisture relationship

Sludge consists of water and solids; when the solids are constant, the volume varies inversely with the solids percentage:

$$\frac{V_1}{V_2}=\frac{100-P_2}{100-P_1}$$

where $P$ is the moisture content (%). For example, thickening sludge from 98% to 95% moisture reduces the volume to $\dfrac{2}{5}=40\%$ of the original. The relation holds when the sludge specific gravity does not change much, and it explains why a small loss of water greatly reduces volume.

The specific gravity of sludge $S_s$ follows from the masses and specific gravities of its parts:

$$\frac{W_s}{S_s}=\frac{W_{sol}}{S_{sol}}+\frac{W_w}{S_w}$$

where $W$ is the weight fraction.

Specific gravity of the given sludge

Moisture $=95\%$, so solids $=5\%$ and water $=95\%$ by weight. Of the solids, 20% are fixed ($G=2.65$) and 80% volatile ($G=1.0$).

Specific gravity of the solids:

$$\frac{1}{G_{sol}}=\frac{0.2}{2.65}+\frac{0.8}{1.0}=0.8755\ \Rightarrow\ G_{sol}=1.142$$

Specific gravity of the sludge:

$$\frac{1}{S_s}=\frac{0.05}{1.142}+\frac{0.95}{1.0}=0.9938\ \Rightarrow\ S_s=1.0063$$

Answer: specific gravity of the sludge $\approx 1.0063$ (solids $\approx 1.142$).