Mass Balance and Measuring Plant Performance - Basic Mathematics for Water and Wastewater Operators - page 190

Solution:
BOD in = 166 mg/L × 11.40 MGD × 8.34 lb/gal = 15,783 lb/day
BOD out = 25 mg/L × 11.40 MGD × 8.34 lb/gal = 2377 lb/day
BOD removed = 15,783 lb/day - 2377 lb/day = 13,406 lb/day
Solids produced (lb/day) = 13,406 lb/day × 0.7 lb solids/lb BOD = 9384 lb/day
Solids removed (lb/day) = 6795 mg/L × 0.15 MGD × 8.34 lb/gal = 8501 lb/day
Difference = 9384 lb/day - 8501 lb/day = 883 lb/day, or 9.4%

These results are within the acceptable range.
Note: We have demonstrated two ways in which mass balance can be used; how-
ever, it is important to note that the mass balance concept can be used for all
aspects of wastewater and solids treatment. In each case, the calculations must
take into account all of the sources of material entering the process and all of the
methods available for removal of solids.

MEASURING PLANT PERFORMANCE
To evaluate how well a plant or unit process is performing, performance efficiency
or percent removal is used. The results obtained can be compared with those listed
in the plant's operations and maintenance (O&M) manual to determine if the facility
is performing as expected. In this section, sample calculations often used to mea-
sure plant performance or efficiency are presented. The efficiency of a unit process
is its effectiveness in removing various constituents from the wastewater or water.
Suspended solids and BOD removal are therefore the most common calculations of
unit process efficiency. In wastewater treatment, the efficiency of a sedimentation
basin may be affected by such factors as the types of solids in the wastewater, the
temperature of the wastewater, and the age of the solids. Typical removal efficiencies
for a primary sedimentation basin are as follows:

Settleable solids, 90-99%
Suspended solids, 40-60%
Total solids, 10-15%
BOD, 20-50%

p lant p erforManCe /e ffiCienCy
Note: The calculation used for determining the performance (percent removal) for
a digester is different from that used for performance (percent removal) for other
processes. Care must be taken to select the correct formula:

(Influent concentrationEffluent − concentration) × 100

%Removal = (13.1)

Influent concentration

■ Example 13.3
Problem: As shown in Figure 13.3, the influent BOD 5 is 247 mg/L, and the plant
effluent BOD is 17 mg/L. What is the percent removal?