Question

two sedimentation tanks have identical surface areas and flow rates. tank a depth = 3 m, tank b depth = 6 m. which statement is most correct?

• tank b removes all particles removed by tank a plus additional particles
• both tanks theoretically remove the same discrete particles under ideal conditions.
• tank b has a lower overflow rate
• tank a has longer retention time

Explanation:

🆕 New Concept Discovered: Sedimentation Tank Design
Depth does not affect ideal discrete particle removal.

Step 1: Analyze Overflow Rate

The surface overflow rate (or hydraulic loading rate) \( v_0 \) is defined as:

$$ v_0 = \frac{Q}{A_s} $$

where \( Q \) is the flow rate and \( A_s \) is the surface area.

Since both tanks have identical surface areas \( A_s \) and flow rates \( Q \), their surface overflow rates are identical:

$$ v_{0, A} = v_{0, B} $$

This rules out the statement "Tank B has a lower overflow rate".

Step 2: Analyze Retention Time

The hydraulic retention time (HRT) \( \theta \) is given by:

$$ \theta = \frac{V}{Q} = \frac{A_s \cdot H}{Q} $$

where \( H \) is the depth of the tank.

Since Tank B is deeper (\( H_B = 6\text{ m} \)) than Tank A (\( H_A = 3\text{ m} \)), Tank B has a larger volume and therefore a longer retention time:

$$ \theta_B > \theta_A $$

This rules out the statement "Tank A has longer retention time".

Step 3: Analyze Particle Removal Efficiency

For ideal, discrete particle settling (Type 1 sedimentation), the removal efficiency of particles with a settling velocity \( v_s \) depends solely on the surface overflow rate \( v_0 \):

• Particles with \( v_s \ge v_0 \) are \( 100\% \) removed.
• Particles with \( v_s < v_0 \) are removed in the ratio \( \frac{v_s}{v_0} \).

Because depth \( H \) does not enter into the removal efficiency equations for ideal discrete settling, both tanks theoretically remove the exact same discrete particles under ideal conditions.

Answer:

Both tanks theoretically remove the same discrete particles under ideal conditions.