Answer / kang chuen tat (malaysia - pen

Answer 68 – (a) BOD = (8.8 - 5.9) (300 / 30) = 29 mg / L.

(b) BOD = (8.8 - 4.2) (600 / 100) = 27.6 mg / L.

(c) BOD = (29 27.6) / 2 = 28.3 mg / L.

(d) NBOD = 45 - 28.3 = 16.7 mg / L.

The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS 61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

Question 68 – Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).

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