Problem Solved: Volatile Solids Reduction - Wastewater Math

[@burtrussel1535]

This is quite simple I suppose. After running a TSS sample on a pad at 105C, you ignite the pad at 550C then get the weight. And after calculating the TSS mg/l, divide the smaller number by the larger number to get the % of organic material. Then compare the results of the sludge fed to the digester against the results of the digested sludge that is being fed to your dewatering process. This could be applied to aerobic digestion as well. I think it would be very interesting to run these tests in an effort to try and optimize digestion in order to maximize the potential of volitile sludge reduction.

[@alexjj231]

you rock!

[@bernardgoh5307]

Thank you

[@gdubbg5306]

welcome to the wastewater world

[@godwincharity6058]

Please how to you achieve a total solid of 7.5% slurry using food wastes and cow dung as substrates?.?

[@melancholyid]

That makes no sense. According to that formula, if I had a volatile solids of 50 percent going in and 25 going out, my reduction would be 66 percent instead of 50 percent.

[@sasamiy00]

melancholyid Did you ever figure out why the denominator is in-(in*out) instead of in? I have the same question.

[@nontokozomazeka8293]

@@sasamiy00 Only the biodegradable fraction fraction of the fed volatile solids is considered at the denominator of the equation books.google.co.za/books?id=BBxRYaSaapQC&pg=PA156&lpg=PA156&dq=Why+is+the+denominator+in-(in*out)+in+Volatile+solid+removal&source=bl&ots=79aEjIfdO4&sig=ACfU3U0H3kGyn0lzmFA-qzeSKwiLLn_TtQ&hl=en&sa=X&ved=2ahUKEwi_4emvt_noAhXKZxUIHXlUB2YQ6AEwAHoECAoQAQ#v=onepage&q=Why%20is%20the%20denominator%20in-(in*out)%20in%20Volatile%20solid%20removal&f=false

[@zzytrewq]

No. There is gas production that means the weight of sludge going in and out are different.

[@christinastewart6957]

thank you