Treatment plant - sewer construction, Civil Engineering

Treatment plant - sewer construction:

Problem:

How many kg of primary and secondary dry solids needs to be processed daily from the treatment plant?

Solution

An amount of solids processed from the basic sedimentation tanks equals the variance in suspended solids concentrations measured across the sedimentation tanks multiplied by the plant flow rate. Thus,

33.75 × 106 l/day (200 mg SS/l - 100 mg SS/l) × kg/1,000,000 mg = 3375 kg primary solids per day.

We are not provided with the concentration difference of suspended solids across the secondary sedimentation tanks so we can determine the amount of secondary solids produced daily in the same manner that we used for primary solids. Therefore, careful examination of the expression of solids retention time displays in which the term Qw Xw could be estimated as the following:

4 days = V =X/Qw×Xw = 5,000,000 l (4,300 mg SS/l)/Qw × Xw

Solve for Qw Xw which equals 5375 kg or say 5400 kg secondary dry solids per day.

Posted Date: 1/22/2013 6:33:01 AM | Location : United States







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