Question

A treatment plant being designed for Cynusoidal City requires an equalization basin to even out flow and BOD variations. The average daily flow is 0.400 m^3/s. The following flows and BOD_5 have been found to be typical of the average variation over a day. What size equalization basin, in cubic meters, is required to provide for a uniform outflow equal to the average daily flow

Answer 1

Answer:

The size of equalization basin is 6105.6 m³

Explanation:

The average flow is:

flow = ∑flow/n = 9.788/24 = 0.408 m³/s

Where n is the number or observations.

The inflow volume is:

$V_{inflow} =Q*t$

where t is the time interval

$V_{inflow} =0.446(1*3600)=1605m^{3}$

in the same way it is calculated the inflow volume for each observation

The outflow volume is:

$V_{out} =Q_{out} *t=0.4*(1*3600)=1400m^{3}$

The volume of flow is:

$ds=V_{inflow} -V_{out} =1605-1400=205m^{3}$

in the same way it is calculated the volume of flow for each observation. According to the file attach, the highest volume is 6105.6 m³

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