In water treatment plant, we need to be evaporated water or Reactivated sludge (MLSS) and Chemical sludge.In this case we should know the Heat energy calculation for Sludge drying.
It is help to analyze quondam of Heat Energy require to evaporate such a medium.
Ok, for example. we’ll take. A Sludge containing 80% water is to be dried at 100°C down to moisture content of 10%. If the initial temperature of the sludge is 21°C, calculate the quantity of heat energy required per unit weight of the original material, for drying under atmospheric pressure.
The latent heat of vaporization of water at 100°C and at standard atmospheric pressure is 2257 kJ kg-1. The specific heat capacity of the sludge is 4.18 kJ kg-1 °C-1 and of water is 4.186 kJ kg-1 °C-1. (Specific heat of sludge, equal to the specific heat of water (cpw=4.187 kJ/ kgK)Find also the energy requirement/kg water removed.
Calculating for 1 kg Sludge Initial moisture = 80%
800 g moisture are associated with 200 g dry matter.
Final moisture = 10 %, 100 g moisture are associated with 900 g dry matter,
Therefore (100 x 200)/900 g = 22.2 g moisture are associated with 200 g dry matter.
1kg of original matter must lose (800 – 22) g moisture = 778 g = 0.778 kg moisture.
Heat energy required for 1kg original material
= heat energy to raise temperature to 100°C + latent heat to remove water
= (100 – 21) x 4.18 + 0.778 x 2257
= 330.2 + 1755.9
= 2086 kJ.
Energy/kg water removed, as 2056 kJ are required to remove 0.778 kg of water,
= 2681 kJ.
Now we can select the sources of heat generator like boiler, electric power or solar energy.