Hello,

I am trying to work out the sludge production. Basically the wastewater is from a sulphur station facility and the wastewater is generated from washing down various equipments, such as: scrubbers, rail cars etc. Thus the sulphur fines are the predominant contaminant in the washdown water.

The info. i have is a stream of 15m3/hr has a suspended solids content of around 15% sulphur fines. Solids will be removed by a settling basin.

My question is, i am trying to work out the suldge quantity produced and thus need to figure out how much mg/L of sulphur fines there is in the wastewater. Now, i have calculated it based on a Sulphur density of 2070kg/m3 but the answer doesnt seem to make sense.

Following is what i did:

15% sulphur fines with S.G. 2.07 = 150,000mg/L / 2.07
= 72,463mg/L
therefore, (72463mg/L * 15m3/hr * 24h/d)/1000 = 26,087kg/d???

Please advice.

Regards,

Your assumption of 15% "sulfur fines" seems (at least to me) to be extremely high.  Was this determined from testing of the wash water?  And, from at least several tests?  At 15% solids the wash water itself would be "sludge."

From my experience with sulfur facilities, your wash water will have a low pH, say in the 5 to 6 range.  I would, in the absence of actual test results, assume a TSS level of more like 500 mg/l.

The following types of screens are very effective in screening out particulates to a dry solid:

http://www.parkson.com/products/rotostrainer

http://www.lenntech.com/parabolic_screen.htm