A rectangular clarifier is 25 m long and 8 m wide, treating 6,000 m³/day. Find the surface overflow rate and, for a 3.0 m depth, the detention time.

Start with the thinking

• Overflow rate uses plan area only — depth does not appear.
• Detention time uses the full volume.

Step-by-step solution

1. Plan area

   $$A = 25(8) = 200 m^{2}$$

2. Overflow rate

   $$v_o = Q/A = 6,000/200 = 30.0 m/day$$

3. Volume

   $$V = 200(3.0) = 600 m^{3}$$

4. Detention time

   $$t = V/Q = 600/6,000 = 0.100 day$$

5. Result — v_o = 30 m³/(m²·day), t = 2.4 hours

Why the other options are there

• 10 m/day (depth divided in)
• t = 24 h (day-to-hour conversion missed)

Reference: FE Reference Handbook — Environmental — Sedimentation

A circular clarifier 14 m in diameter and 4.5 m deep treats 24,359 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 80 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(14)^{2} = 153.9 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 24359/153.9 = 158.2 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 692.7/24359 = 0.68 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 24359/(π × 14) = 553.8 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(8.0e-5)^{2}/(18 \times 0.001) = 497.2 m/d$$

10. Compare

    $$v_s = 497.2 m/d > SOR = 158.2 m/d \to particle is removed$$

Why the other options are there

• SOR = 123.1 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings

A circular clarifier 21 m in diameter and 4.5 m deep treats 20,902 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 100 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(21)^{2} = 346.4 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 20902/346.4 = 60.35 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 1,559/20902 = 1.79 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 20902/(π × 21) = 316.8 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(1.0e-4)^{2}/(18 \times 0.001) = 777.0 m/d$$

10. Compare

    $$v_s = 777.0 m/d > SOR = 60.35 m/d \to particle is removed$$

Why the other options are there

• SOR = 70.4 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings

A circular clarifier 18 m in diameter and 3.0 m deep treats 4,503 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 30 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(18)^{2} = 254.5 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 4503/254.5 = 17.70 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 763.4/4503 = 4.07 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 4503/(π × 18) = 79.6 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(3.0e-5)^{2}/(18 \times 0.001) = 69.93 m/d$$

10. Compare

    $$v_s = 69.93 m/d > SOR = 17.70 m/d \to particle is removed$$

Why the other options are there

• SOR = 26.5 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings

A circular clarifier 26 m in diameter and 5.0 m deep treats 9,923 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 20 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(26)^{2} = 530.9 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 9923/530.9 = 18.69 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 2,655/9923 = 6.42 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 9923/(π × 26) = 121.5 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(2.0e-5)^{2}/(18 \times 0.001) = 31.08 m/d$$

10. Compare

    $$v_s = 31.08 m/d > SOR = 18.69 m/d \to particle is removed$$

Why the other options are there

• SOR = 24.3 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings

A circular clarifier 24 m in diameter and 3.5 m deep treats 16,281 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 30 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(24)^{2} = 452.4 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 16281/452.4 = 35.99 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 1,583/16281 = 2.33 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 16281/(π × 24) = 215.9 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(3.0e-5)^{2}/(18 \times 0.001) = 69.93 m/d$$

10. Compare

    $$v_s = 69.93 m/d > SOR = 35.99 m/d \to particle is removed$$

Why the other options are there

• SOR = 61.7 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings

A circular clarifier 34 m in diameter and 4.5 m deep treats 19,123 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 100 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(34)^{2} = 907.9 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 19123/907.9 = 21.06 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 4,086/19123 = 5.13 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 19123/(π × 34) = 179.0 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(1.0e-4)^{2}/(18 \times 0.001) = 777.0 m/d$$

10. Compare

    $$v_s = 777.0 m/d > SOR = 21.06 m/d \to particle is removed$$

Why the other options are there

• SOR = 39.8 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings

A circular clarifier 40 m in diameter and 3.5 m deep treats 19,583 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 20 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(40)^{2} = 1,257 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 19583/1,257 = 15.58 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 4,398/19583 = 5.39 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 19583/(π × 40) = 155.8 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(2.0e-5)^{2}/(18 \times 0.001) = 31.08 m/d$$

10. Compare

    $$v_s = 31.08 m/d > SOR = 15.58 m/d \to particle is removed$$

Why the other options are there

• SOR = 44.5 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings

A circular clarifier 23 m in diameter and 3.5 m deep treats 5,180 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 80 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(23)^{2} = 415.5 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 5180/415.5 = 12.47 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 1,454/5180 = 6.74 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 5180/(π × 23) = 71.7 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(8.0e-5)^{2}/(18 \times 0.001) = 497.2 m/d$$

10. Compare

    $$v_s = 497.2 m/d > SOR = 12.47 m/d \to particle is removed$$

Why the other options are there

• SOR = 20.5 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings

A circular clarifier 34 m in diameter and 3.5 m deep treats 20,891 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 50 µm particle (SG = 2.65) settles out by Stokes' law.

Start with the thinking

• An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
• Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

1. Surface area

   $$A = (\pi/4)(34)^{2} = 907.9 m^{2}$$

2. Formula

   $$SOR = Q/A$$

3. Substituting — SOR = 20891/907.9 = 23.01 m³/m²·d

4. Formula — τ = V/Q = A·h/Q

5. Substituting

   $$\tau = 3,178/20891 = 3.65 h$$

6. Formula

   $$weir loading = Q/(\pi D)$$

7. Substituting — 20891/(π × 34) = 195.6 m³/m·d

8. Formula

   $$v_s = g(\rho_s - \rho)d^{2}/(18 \mu)$$

9. Substituting

   $$v_s = 9.81(1650)(5.0e-5)^{2}/(18 \times 0.001) = 194.2 m/d$$

10. Compare

    $$v_s = 194.2 m/d > SOR = 23.01 m/d \to particle is removed$$

Why the other options are there

• SOR = 55.9 m/d (side area used)
• v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Weir Loadings