Question

Consider a single bacterial cell as a discrete particle with a diameter of 1x10-6m and a specific gravity of 1.01. Assuming laminar flow conditions, calculate how long would it take for the cell the settle 1 foot in water at 20 oC?For a flow rate of 4MGD, what surface area would be required to settle individual cells? Is sedimentation a practical method for removing individual bacterial cells?

Answer 1

Sedimentation is not a good method.

Step-by-step explanation:

We need to apply Stoke laws and assume that is valid here.

So,

`V_s= 418(G-1)d^2((3T+70)/(100))`

Replacing the values,

`V_s= (418)(1.01-1)(10^-3)^3*((3*20+70)/(100))\\V_s=5.434*10^-6mm/s`

Here then we calculate the time,

`t_(req)=(x)/(v)`

Where x= Distance, v= velocity

`t_(req)=(1foot)/(5.434*10^(-6)) = (304.8)/(5.434*10^(-6))\\t_(req)=649.2days`

To calculate the surface required we need first to calculate the volume through the volume,

So,

`Q=4MGD=4*10^6*0.135 (ft^3/day)`

Then,

`V_(req) = Q*t\\V_(req)=4*10^6*0.134*649.2\\V_(req)=34.79*10^7ft^3`

Here we can calculate the surface

`S_(req)= (Volume)/(Distance)=(34.79*10^7)/(1)\\S_(req)=7987.8 Acres`

So, the requeriment of Area of tank and settlement time is huge, it's not a practical method.