Wastewater from a small town is treated in a trickling filter which is preceded by a screen, sand-removal and primary clarification. The maximum hydraulic capacity, or peak flow, of the treatment plant is 1,500 m³/h. The dry weather flow (Qdwf) amounts 7,500 m³/day, with a BOD concentration of 300 mg/l.

The maximum scouring velocity or slip velocity is 0.3 m/s and the maximum surface load or Hazen velocity is 40 m/h.

Assume 4 channels. Take a W:L ratio of 1:15

a. Calculate the vertical surface area Avertical in m²:
b.Calculate the horizontal surface area Ahorizontal in m²:

Qdesing= (1500m3/s)/(3600s/h)= 0.42m3/s.

Qchannel= (0.42m3/s)/4channels = 0.104m3/s

assumptions: Depth = 0.7 and settling velocity of smallest grit particle V= 0.012 (derived by stoke´s law).

-Detention time:
t= Depth/vel ; t=(0.7m)/(0.012m/s) = 60s

-theorical length:
L= Vh*t ; L=(0.3m/s)*(60s)= 18m

-channel volume:
Vol= Q*t ; Q=(0.104m3/s)*(60s)= 6.24m3

-cross section area:
A= Vol/Length ; A= (6.24m3)/(18m)= 0.35m2

-channel width:
W= A/D ; W= (0.35m2)/(0.7m)= 0.5m

-And now we can calculate the vertical and horizontal area
Vertical area = 0.5m*0.7 = 0.35m2
Horizontal area = 0.5m*18m = 9m2

:)

To calculate the vertical surface area (Avertical):

Given: Maximum scouring velocity (Vs) = 0.3 m/s
Maximum surface load or Hazen velocity (Vh) = 40 m/h
Peak flow rate (Qpeak) = 1,500 m³/h

Step 1: Convert the Hazen velocity from meters per hour to meters per second.
Vh = 40 m/h = 40/3600 m/s = 0.0111 m/s

Step 2: Calculate the vertical surface area.
Avertical = Qpeak / (Vs + Vh)

Substituting the values:
Avertical = 1,500 m³/h / (0.3 m/s + 0.0111 m/s)
Avertical = 1,500 m³/h / 0.3111 m/s
Avertical ≈ 4,818.98 m²

Therefore, the vertical surface area (Avertical) is approximately 4,818.98 m².

To calculate the horizontal surface area (Ahorizontal):

Given: Width to length (W:L) ratio = 1:15

Step 1: Determine the width and length of the trickling filter using the W:L ratio.
Let W = width
Let L = length

W/L = 1/15

Step 2: Make an assumption for the width or length.
Assuming L = 1 unit

Step 3: Calculate the width.
W = (W/L) * L
W = (1/15) * 1
W = 1/15

Step 4: Calculate the horizontal surface area.
Ahorizontal = W * L

Substituting the values:
Ahorizontal = (1/15) * 1
Ahorizontal = 1/15

Therefore, the horizontal surface area (Ahorizontal) is approximately 1/15 m².

To calculate the vertical surface area (Avertical) of the trickling filter, you will need the following information:

1. Number of channels: In this case, there are 4 channels.

2. Width-to-length ratio (W:L): Given as 1:15.

To calculate Avertical, you can use the formula:

Avertical = Number of channels x Width x Length

1. Width:
Width = (W:L) x Length

Given the W:L ratio as 1:15 and the Length as L, the Width can be calculated as:
Width = (1/15) x L

2. Length:
Length can be determined based on the maximum hydraulic capacity (peak flow) of the treatment plant.

Peak flow = 1,500 m³/h = (1,500 x 24) m³/day
Peak flow = 36,000 m³/day

Length = (Peak flow / Dry weather flow)
Length = (36,000 m³/day) / (7,500 m³/day)
Length = 4.8 days

Now, substitute the calculated Width and Length into the formula for Avertical:

Avertical = Number of channels x Width x Length
Avertical = 4 x [(1/15) x L] x L
Avertical = 4/15 x L²

Finally, substitute the value of Length:
Avertical = 4/15 x (4.8 days)²
Avertical = 4/15 x 23.04 days²
Avertical ≈ 2.944 m²

So, the vertical surface area (Avertical) of the trickling filter is approximately 2.944 m².

To calculate the horizontal surface area (Ahorizontal) of the trickling filter, you can use the formula:

Ahorizontal = Avertical / W:L

Ahorizontal = 2.944 m² / (1/15)
Ahorizontal = 2.944 m² x 15
Ahorizontal ≈ 44.16 m²

Therefore, the horizontal surface area (Ahorizontal) of the trickling filter is approximately 44.16 m².

Its wrong!