Handbook of Civil Engineering Calculations

71.2 = 1QQ — V= 18.51 103 ft^3 (523.8 m^3 )

1+0.0561 /2,2002 Ib/d

V F(2.08)

Using the given depth of 7 feet (2.1 m), and a circular trickling filter, the area and diame-
ter of the first stage filter are

Area=^^ 18.51 xl 0 3ft3=2>644ft2 diameter=5g-02 ft(177m)

depth 7 ft

3. Analyze the second stage filter
   The BOD 5 loading for the second stage trickling filter is calculated using

W = (I-E 1 )W

where W = BOD 5 loading to the second stage filter
W=(I- 0.712)(2,002 Ib/d) = 577 Ib BOD 5 /d (261.9 kg/d)

The NRC equation for a second stage trickling filter is

^IT 2 100 _

| 0.0561 W

1-E 1 ^VF

Using terms defined previously and values calculated above, the volume of the second
stage trickling filter is

100

E 2 =. V= 64.33 103 ft^3 (1.82 m^3 )

| 0.0561 /577 Ib/d

1-0.712 V F(2.08)

The area and diameter of the second stage filter are

Area =^64 '^33 *^1 ^ ft3 = 9,190 ft^2 diameter - 108.17 ft (32.97 m)

7 ft

4. Compute the BOD 5 loading and hydraulic loading to each filter
   The BOD 5 (organic) loading to each filter is calculated by dividing the BOD 5 loading by
   the volume of the filter in 103 ft^3 (m^3 ):

First stage filter: BOD 5 loading = Jffig^ = 108.2 -^

(1.74kg/m^3 -d)

Second stage filter: BOD 5 loading = J™^ = 8.97 -^

(0.14kg/m

3

-d)