Question

A settling tank treating 5.5x10*6 L of water per day has dimensions as follows: length, 12.2m; width, 7.0m; depth, 3.5m. Calculate the detention time of water in the tank? Calculate the minimum particle (expressed as diameter of spheres) size that could settle the tank?

Answer 1

Water flow rate Q = 5.5*10^6 L /day is to be treated by settling tank.

Tank dimensions:

L= 12.2 m

W = 7 m

H = 3.5 m

Total volume of the settling tank , V = L*W*H = 12.2m*7m*3.5m = 298.9 m3

Q = 5.5*10^6 L/day = 5.5*10^6 * 10^-3 m3/3600s = 1.527m3/s

As we know that detention time is defined as residence time of incoming treated fluid.

Td = V/Q = 298.9m3/(1.527m3/s) = 195.64 sec

detention time of water in the tank Td = 195.64 sec

ans : 195.64 sec

minimum size of the particle that could settle in the tank.

That particle having settling velocity less than critical velocity of the solid, those particles are settle down.

Flow velocity of water, V = Q/W.H =( 1.527m3/s)/(7m*3.5m) = 0.0623 m/s

For settling tank,

V/Vs = L/H

Vs = V*(H/L) = 0.0623 *(3.5/12.2) = 0.0178 m/s

Settling velocity of the particles Vs = 0.0178 m/s

Let us assume settling velocity formula for stokes law, where Re < 1 for laminar flow

Vs = gdp^2(G - 1)/18mu

g = 9.8 m/s2

dp - particle size

G = specific gravity of particle

mu - viscosity of water = 10^-3 Pa.s

0.0178 = 9.8*dp^2(G -1)/18*10^-3

dp = 0.0057/√(G-1)

Here dp is the minimum diameter of particle that could settle down .

Here G is not given for particle solid.

If given then we calculate minimum diameter of particle and then verify

Reynolds number Re < 1 then our assumption is correct.

If not then we have to assume turbulent flow.