general spherical particle settling velocity in a sedimentation basin Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,050 kg/m^3; fluid density (rho) = 991.5 kg/m^3; particle diameter (d) = 0.0027 m; drag coefficient (C_D) = 0.8300, determine the settling velocity (v_s) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except v_s is given, so isolate v_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for v_s:

   $$v_{s} = \sqrt{\dfrac{4g(\rho_p-\rho)d}{3C_D\rho}}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,050 kg/m^3, fluid density (rho) = 991.5 kg/m^3, particle diameter (d) = 0.0027 m, drag coefficient (C_D) = 0.8300.

4. Step 4 — Substitute the given values:

   $$v_{s} = \sqrt{\dfrac{49.8100(\rho_p-991.5)0.0027}{3C_D991.5}}$$

5. Step 5 — Evaluate:

   $$v_{s} = 0.2135\ \text{m/s}$$

6. Step 6 — Check: returning v_s = 0.2135 m/s to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.4270 — kept a factor of two that cancels in the correct rearrangement.
• 0.1068 — dropped that same factor in the other direction.
• 0.2349 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical settling equation with a drag coefficient for larger particles Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,170 kg/m^3; fluid density (rho) = 997.0 kg/m^3; drag coefficient (C_D) = 4.2600; settling velocity (v_s) = 0.4575 m/s, determine the particle diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for d:

   $$d = \dfrac{v_s^2\,3C_D\rho}{4g(\rho_p-\rho)}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,170 kg/m^3, fluid density (rho) = 997.0 kg/m^3, drag coefficient (C_D) = 4.2600, settling velocity (v_s) = 0.4575 m/s.

4. Step 4 — Substitute the given values:

   $$d = \dfrac{v_s^2\,3C_D997.0}{49.8100(\rho_p-997.0)}$$

5. Step 5 — Evaluate:

   $$d = 0.0579\ \text{m}$$

6. Step 6 — Check: returning d = 0.0579 m to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.1159 — kept a factor of two that cancels in the correct rearrangement.
• 0.0290 — dropped that same factor in the other direction.
• 0.0637 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical settling velocity used for grit chamber design Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,670 kg/m^3; fluid density (rho) = 996.0 kg/m^3; particle diameter (d) = 0.0020 m; settling velocity (v_s) = 0.8567 m/s, determine the drag coefficient (C_D).

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except C_D is given, so isolate C_D symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for C_D:

   $$C_{D} = \dfrac{4g(\rho_p-\rho)d}{3v_s^2\rho}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 1,670 kg/m^3, fluid density (rho) = 996.0 kg/m^3, particle diameter (d) = 0.0020 m, settling velocity (v_s) = 0.8567 m/s.

4. Step 4 — Substitute the given values:

   $$C_{D} = \dfrac{49.8100(\rho_p-996.0)0.0020}{3v_s^2996.0}$$

5. Step 5 — Evaluate:

   $$C_{D} = 0.0238$$

6. Step 6 — Check: returning C_D = 0.0238 to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0475 — kept a factor of two that cancels in the correct rearrangement.
• 0.0119 — dropped that same factor in the other direction.
• 0.0261 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical particle settling velocity in a sedimentation basin Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,570 kg/m^3; fluid density (rho) = 996.0 kg/m^3; particle diameter (d) = 0.0039 m; drag coefficient (C_D) = 2.4700, determine the settling velocity (v_s) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except v_s is given, so isolate v_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for v_s:

   $$v_{s} = \sqrt{\dfrac{4g(\rho_p-\rho)d}{3C_D\rho}}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,570 kg/m^3, fluid density (rho) = 996.0 kg/m^3, particle diameter (d) = 0.0039 m, drag coefficient (C_D) = 2.4700.

4. Step 4 — Substitute the given values:

   $$v_{s} = \sqrt{\dfrac{49.8100(\rho_p-996.0)0.0039}{3C_D996.0}}$$

5. Step 5 — Evaluate:

   $$v_{s} = 0.1816\ \text{m/s}$$

6. Step 6 — Check: returning v_s = 0.1816 m/s to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.3632 — kept a factor of two that cancels in the correct rearrangement.
• 0.0908 — dropped that same factor in the other direction.
• 0.1997 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical settling equation with a drag coefficient for larger particles Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,580 kg/m^3; fluid density (rho) = 997.5 kg/m^3; drag coefficient (C_D) = 3.4400; settling velocity (v_s) = 0.0915 m/s, determine the particle diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for d:

   $$d = \dfrac{v_s^2\,3C_D\rho}{4g(\rho_p-\rho)}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 1,580 kg/m^3, fluid density (rho) = 997.5 kg/m^3, drag coefficient (C_D) = 3.4400, settling velocity (v_s) = 0.0915 m/s.

4. Step 4 — Substitute the given values:

   $$d = \dfrac{v_s^2\,3C_D997.5}{49.8100(\rho_p-997.5)}$$

5. Step 5 — Evaluate:

   $$d = 0.0038\ \text{m}$$

6. Step 6 — Check: returning d = 0.0038 m to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0075 — kept a factor of two that cancels in the correct rearrangement.
• 0.0019 — dropped that same factor in the other direction.
• 0.0041 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical settling velocity used for grit chamber design Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,010 kg/m^3; fluid density (rho) = 991.5 kg/m^3; particle diameter (d) = 0.0017 m; settling velocity (v_s) = 0.2500 m/s, determine the drag coefficient (C_D).

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except C_D is given, so isolate C_D symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for C_D:

   $$C_{D} = \dfrac{4g(\rho_p-\rho)d}{3v_s^2\rho}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,010 kg/m^3, fluid density (rho) = 991.5 kg/m^3, particle diameter (d) = 0.0017 m, settling velocity (v_s) = 0.2500 m/s.

4. Step 4 — Substitute the given values:

   $$C_{D} = \dfrac{49.8100(\rho_p-991.5)0.0017}{3v_s^2991.5}$$

5. Step 5 — Evaluate:

   $$C_{D} = 0.3569$$

6. Step 6 — Check: returning C_D = 0.3569 to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.7137 — kept a factor of two that cancels in the correct rearrangement.
• 0.1784 — dropped that same factor in the other direction.
• 0.3926 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical particle settling velocity in a sedimentation basin Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,000 kg/m^3; fluid density (rho) = 993.0 kg/m^3; particle diameter (d) = 0.0041 m; drag coefficient (C_D) = 2.1300, determine the settling velocity (v_s) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except v_s is given, so isolate v_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for v_s:

   $$v_{s} = \sqrt{\dfrac{4g(\rho_p-\rho)d}{3C_D\rho}}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,000 kg/m^3, fluid density (rho) = 993.0 kg/m^3, particle diameter (d) = 0.0041 m, drag coefficient (C_D) = 2.1300.

4. Step 4 — Substitute the given values:

   $$v_{s} = \sqrt{\dfrac{49.8100(\rho_p-993.0)0.0041}{3C_D993.0}}$$

5. Step 5 — Evaluate:

   $$v_{s} = 0.1596\ \text{m/s}$$

6. Step 6 — Check: returning v_s = 0.1596 m/s to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.3192 — kept a factor of two that cancels in the correct rearrangement.
• 0.0798 — dropped that same factor in the other direction.
• 0.1756 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical settling equation with a drag coefficient for larger particles Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,740 kg/m^3; fluid density (rho) = 990.5 kg/m^3; drag coefficient (C_D) = 4.1700; settling velocity (v_s) = 0.6852 m/s, determine the particle diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except d is given, so isolate d symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for d:

   $$d = \dfrac{v_s^2\,3C_D\rho}{4g(\rho_p-\rho)}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 1,740 kg/m^3, fluid density (rho) = 990.5 kg/m^3, drag coefficient (C_D) = 4.1700, settling velocity (v_s) = 0.6852 m/s.

4. Step 4 — Substitute the given values:

   $$d = \dfrac{v_s^2\,3C_D990.5}{49.8100(\rho_p-990.5)}$$

5. Step 5 — Evaluate:

   $$d = 0.1978\ \text{m}$$

6. Step 6 — Check: returning d = 0.1978 m to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.3956 — kept a factor of two that cancels in the correct rearrangement.
• 0.0989 — dropped that same factor in the other direction.
• 0.2176 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical settling velocity used for grit chamber design Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,030 kg/m^3; fluid density (rho) = 995.0 kg/m^3; particle diameter (d) = 0.0004 m; settling velocity (v_s) = 0.3463 m/s, determine the drag coefficient (C_D).

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except C_D is given, so isolate C_D symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for C_D:

   $$C_{D} = \dfrac{4g(\rho_p-\rho)d}{3v_s^2\rho}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,030 kg/m^3, fluid density (rho) = 995.0 kg/m^3, particle diameter (d) = 0.0004 m, settling velocity (v_s) = 0.3463 m/s.

4. Step 4 — Substitute the given values:

   $$C_{D} = \dfrac{49.8100(\rho_p-995.0)0.0004}{3v_s^2995.0}$$

5. Step 5 — Evaluate:

   $$C_{D} = 0.0408$$

6. Step 6 — Check: returning C_D = 0.0408 to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.0817 — kept a factor of two that cancels in the correct rearrangement.
• 0.0204 — dropped that same factor in the other direction.
• 0.0449 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling

general spherical particle settling velocity in a sedimentation basin Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,920 kg/m^3; fluid density (rho) = 992.5 kg/m^3; particle diameter (d) = 0.0035 m; drag coefficient (C_D) = 3.7100, determine the settling velocity (v_s) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is General Spherical (settling).
• Everything except v_s is given, so isolate v_s symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• The general spherical particle settling equation gives settling velocity for particles outside the Stokes' law range.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

2. Step 2 — Rearrange symbolically for v_s:

   $$v_{s} = \sqrt{\dfrac{4g(\rho_p-\rho)d}{3C_D\rho}}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 1,920 kg/m^3, fluid density (rho) = 992.5 kg/m^3, particle diameter (d) = 0.0035 m, drag coefficient (C_D) = 3.7100.

4. Step 4 — Substitute the given values:

   $$v_{s} = \sqrt{\dfrac{49.8100(\rho_p-992.5)0.0035}{3C_D992.5}}$$

5. Step 5 — Evaluate:

   $$v_{s} = 0.1078\ \text{m/s}$$

6. Step 6 — Check: returning v_s = 0.1078 m/s to

   $$v_s = \sqrt{\dfrac{4 g (\rho_p - \rho) d}{3 C_D \rho}}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 0.2157 — kept a factor of two that cancels in the correct rearrangement.
• 0.0539 — dropped that same factor in the other direction.
• 0.1186 — rounded an intermediate value before the final step.

Reference: FE Handbook — General Spherical Particle Settling