Handbook of Civil Engineering Calculations

The increase in MLVSS is computed using the following equation:

P* = Y 0 bsQ(So - S)(8-34 Ib/Mgal/mg/L)

where Px is the net waste activated sludge produced each day in (Ib VSS/d).
Using values defined above:

f

* = (°-

41 J2I

S <

4

'

0 M

^ (

24

° ^ - 1°

1

x \ mgBOD 5 )^ & J\ L LT- )\Y«-34 mg/L ^T) )

= 3146 Ib VSS/d (1428.3 kg VSS/d)

This represents the increase of volatile suspended solids (organics) in the reactor. Of
course the total increase in sludge mass will include fixed suspended solids (inorganics)
as well. Therefore, the increase in the total mass of mixed liquor suspended solids
(MLSS) = Px(ss} = (3146 Ib VSS/d)/(0.8) = 3933 Ib SS/d (1785.6 kg SS/d). This represents
the total mass of sludge that must be wasted from the system each day.

4. Compute the oxygen requirements based on ultimate
   carbonaceous oxygen demand (BOD 1 J
   The theoretical oxygen requirements are calculated using the BOD 5 of the wastewater and
   the amount of organisms (Px) wasted from the system each day. If all BOD 5 were convert-
   ed to end products, the total oxygen demand would be computed by converting BOD 5 to
   ultimate BOD (BODL), using an appropriate conversion factor. The "Quantity of Sludge
   Wasted" calculation illustrated that a portion of the incoming waste is converted to new
   cells which are subsequently wasted from the system. Therefore, if the BODL of the wast-
   ed cells is subtracted from the total, the remaining amount represents the amount of oxy-
   gen that must be supplied to the system. From stoichiometry, it is known that the BODL
   of one mole of cells is equal to 1.42 times the concentration of cells. Therefore, the theo-
   retical oxygen requirements for the removal of the carbonaceous organic matter in waste-
   water for an activated-sludge system can be computed using the following equation:

Ib O 2 /d = (total mass of BODL utilized, Ib/d)

• 1.42 (mass of organisms wasted, Ib/d)

Using terms that have been defined previously where/= conversion factor for con-
verting BOD 5 to BODL (0.68 is commonly used):

flM"*^

1

)

Ib O 2 Id = •—

:

'- - (1 .42X*,)

Using the above quantities:

Ib02/d= (4.0 Mgd)(240 ^-^10 mg/L)(8.34) _ ^^ ^

= 6816 Ib O 2 At (3094.5 kg O 2 AI)

This represents the theoretical oxygen requirement for removal of the influent BOD 5.
However, to meet sustained peak organic loadings, it is recommended that aeration equip-