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

Answer:

Sedimentation is not a good method.

Explanation:

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

So,

$V_s= 418(G-1)d^2(\frac{3T+70}{100})$

Replacing the values,

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

Here then we calculate the time,

$t_{req}=\frac{x}{v}$

Where x= Distance, v= velocity

$t_{req}=\frac{1foot}{5.434*10^{-6}} = \frac{304.8}{5.434*10^{-6}}\\t_{req}=649.2days$

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

$S_{req}= \frac{Volume}{Distance}=\frac{34.79*10^7}{1}\\S_{req}=7987.8 Acres$

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