Sludge Drying Water Removal

Mass balance during drying

If water is removed while dry solids remain constant, final product mass follows directly. First multiply initial wet mass by its solids fraction. Divide that dry mass by the target solids fraction. The difference between initial and final total mass is the water that must leave.

Example and limits

1000 kg of sludge at 20 percent solids contains 200 kg of dry matter. At a target of 80 percent solids, the product weighs 250 kg and 750 kg of water must be removed. The target is not an operating recommendation. Solids loss, added material or loss of volatile compounds requires a balance with additional streams.

Distinguish water removal from product loss

Start with 600 kg of sludge at 15 percent solids. Dry solids are 600 × 0.15 = 90 kg. At a target of 45 percent solids, final product is 90 / 0.45 = 200 kg. Water removed is 600 − 200 = 400 kg and water remaining is 200 − 90 = 110 kg. Initial water is 510 kg, which must also equal removed plus remaining water.

A decrease in total mass does not imply dry solids loss. This model preserves solids and removes water only. Do not confuse target solids percentage with moisture percentage. A drying target cannot be below the starting solids fraction. Equal fractions require zero water removal.

The result is water mass to remove, not required heat or electricity. Energy needs drying conditions and efficiency as separate inputs. If additives such as polymer change dry solids, or solids escape in filtrate, do not apply this single inlet balance directly. Check that initial and target analyses use the same definition of dry solids.

Frequently asked questions

How do you calculate sludge weight after drying?

For 1000 kg of wet sludge at 20 percent solids dried to 80 percent solids, final product weight is 250 kg and water removal is 750 kg. This preliminary mass balance conserves dry solids, so the example excludes solids loss, added material and loss of volatile matter.

Calculate with these values