Stokes' law settling velocity for fine sand particles in a clarifier Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,920 kg/m^3; fluid density (rho) = 996.5 kg/m^3; particle diameter (d) = 0.0000 m; dynamic viscosity (mu) = 0.0010 Pa*s, determine the settling velocity (v_s) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for v_s:

   $$v_{s} = \dfrac{g(\rho_p-\rho)d^2}{18\mu}$$

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) = 996.5 kg/m^3, particle diameter (d) = 0.0000 m, dynamic viscosity (mu) = 0.0010 Pa*s.

4. Step 4 — Substitute the given values:

   $$v_{s} = \dfrac{9.8100(\rho_p-996.5)0.0000^2}{180.0010}$$

5. Step 5 — Evaluate:

   $$v_{s} = 0.0004\ \text{m/s}$$

6. Step 6 — Check: returning v_s = 0.0004 m/s to

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

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

Why the other options are there

• 0.0009 — kept a factor of two that cancels in the correct rearrangement.
• 0.0002 — dropped that same factor in the other direction.
• 0.0005 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law applied to discrete particle settling in water treatment Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,990 kg/m^3; fluid density (rho) = 998.5 kg/m^3; dynamic viscosity (mu) = 0.0012 Pa*s; settling velocity (v_s) = 0.0040 m/s, determine the particle diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for d:

   $$d = \sqrt{\dfrac{18\mu v_s}{g(\rho_p-\rho)}}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 1,990 kg/m^3, fluid density (rho) = 998.5 kg/m^3, dynamic viscosity (mu) = 0.0012 Pa*s, settling velocity (v_s) = 0.0040 m/s.

4. Step 4 — Substitute the given values:

   $$d = \sqrt{\dfrac{180.0012 v_s}{9.8100(\rho_p-998.5)}}$$

5. Step 5 — Evaluate:

   $$d = 0.0001\ \text{m}$$

6. Step 6 — Check: returning d = 0.0001 m to

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

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

Why the other options are there

• 0.0002 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0001 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law used to size a grit chamber for small particles Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,430 kg/m^3; fluid density (rho) = 991.0 kg/m^3; particle diameter (d) = 0.0001 m; settling velocity (v_s) = 0.0009 m/s, determine the dynamic viscosity (mu) in Pa*s.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• Everything except mu is given, so isolate mu 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for mu:

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

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,430 kg/m^3, fluid density (rho) = 991.0 kg/m^3, particle diameter (d) = 0.0001 m, settling velocity (v_s) = 0.0009 m/s.

4. Step 4 — Substitute the given values:

   $$\mu = \dfrac{9.8100(\rho_p-991.0)0.0001^2}{18v_s}$$

5. Step 5 — Evaluate:

   $$\mu = 0.0075\ \text{Pa*s}$$

6. Step 6 — Check: returning mu = 0.0075 Pa*s to

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

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

Why the other options are there

• 0.0150 — kept a factor of two that cancels in the correct rearrangement.
• 0.0038 — dropped that same factor in the other direction.
• 0.0083 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law settling velocity for fine sand particles in a clarifier Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,490 kg/m^3; fluid density (rho) = 991.5 kg/m^3; particle diameter (d) = 0.0001 m; dynamic viscosity (mu) = 0.0014 Pa*s, determine the settling velocity (v_s) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for v_s:

   $$v_{s} = \dfrac{g(\rho_p-\rho)d^2}{18\mu}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,490 kg/m^3, fluid density (rho) = 991.5 kg/m^3, particle diameter (d) = 0.0001 m, dynamic viscosity (mu) = 0.0014 Pa*s.

4. Step 4 — Substitute the given values:

   $$v_{s} = \dfrac{9.8100(\rho_p-991.5)0.0001^2}{180.0014}$$

5. Step 5 — Evaluate:

   $$v_{s} = 0.0038\ \text{m/s}$$

6. Step 6 — Check: returning v_s = 0.0038 m/s to

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

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

Why the other options are there

• 0.0076 — kept a factor of two that cancels in the correct rearrangement.
• 0.0019 — dropped that same factor in the other direction.
• 0.0042 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law applied to discrete particle settling in water treatment Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,490 kg/m^3; fluid density (rho) = 993.5 kg/m^3; dynamic viscosity (mu) = 0.0013 Pa*s; settling velocity (v_s) = 0.0024 m/s, determine the particle diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for d:

   $$d = \sqrt{\dfrac{18\mu v_s}{g(\rho_p-\rho)}}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,490 kg/m^3, fluid density (rho) = 993.5 kg/m^3, dynamic viscosity (mu) = 0.0013 Pa*s, settling velocity (v_s) = 0.0024 m/s.

4. Step 4 — Substitute the given values:

   $$d = \sqrt{\dfrac{180.0013 v_s}{9.8100(\rho_p-993.5)}}$$

5. Step 5 — Evaluate:

   $$d = 0.0001\ \text{m}$$

6. Step 6 — Check: returning d = 0.0001 m to

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

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

Why the other options are there

• 0.0001 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0001 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law used to size a grit chamber for small particles Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,760 kg/m^3; fluid density (rho) = 994.5 kg/m^3; particle diameter (d) = 0.0000 m; settling velocity (v_s) = 0.0048 m/s, determine the dynamic viscosity (mu) in Pa*s.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• Everything except mu is given, so isolate mu 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for mu:

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

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 1,760 kg/m^3, fluid density (rho) = 994.5 kg/m^3, particle diameter (d) = 0.0000 m, settling velocity (v_s) = 0.0048 m/s.

4. Step 4 — Substitute the given values:

   $$\mu = \dfrac{9.8100(\rho_p-994.5)0.0000^2}{18v_s}$$

5. Step 5 — Evaluate:

   $$\mu = 0.0000\ \text{Pa*s}$$

6. Step 6 — Check: returning mu = 0.0000 Pa*s to

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

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

Why the other options are there

• 0.0000 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law settling velocity for fine sand particles in a clarifier Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,740 kg/m^3; fluid density (rho) = 995.5 kg/m^3; particle diameter (d) = 0.0001 m; dynamic viscosity (mu) = 0.0013 Pa*s, determine the settling velocity (v_s) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for v_s:

   $$v_{s} = \dfrac{g(\rho_p-\rho)d^2}{18\mu}$$

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) = 995.5 kg/m^3, particle diameter (d) = 0.0001 m, dynamic viscosity (mu) = 0.0013 Pa*s.

4. Step 4 — Substitute the given values:

   $$v_{s} = \dfrac{9.8100(\rho_p-995.5)0.0001^2}{180.0013}$$

5. Step 5 — Evaluate:

   $$v_{s} = 0.0028\ \text{m/s}$$

6. Step 6 — Check: returning v_s = 0.0028 m/s to

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

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

Why the other options are there

• 0.0056 — kept a factor of two that cancels in the correct rearrangement.
• 0.0014 — dropped that same factor in the other direction.
• 0.0031 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law applied to discrete particle settling in water treatment Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,660 kg/m^3; fluid density (rho) = 994.5 kg/m^3; dynamic viscosity (mu) = 0.0008 Pa*s; settling velocity (v_s) = 0.0032 m/s, determine the particle diameter (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for d:

   $$d = \sqrt{\dfrac{18\mu v_s}{g(\rho_p-\rho)}}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,660 kg/m^3, fluid density (rho) = 994.5 kg/m^3, dynamic viscosity (mu) = 0.0008 Pa*s, settling velocity (v_s) = 0.0032 m/s.

4. Step 4 — Substitute the given values:

   $$d = \sqrt{\dfrac{180.0008 v_s}{9.8100(\rho_p-994.5)}}$$

5. Step 5 — Evaluate:

   $$d = 0.0001\ \text{m}$$

6. Step 6 — Check: returning d = 0.0001 m to

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

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

Why the other options are there

• 0.0001 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0001 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law used to size a grit chamber for small particles Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 1,650 kg/m^3; fluid density (rho) = 992.0 kg/m^3; particle diameter (d) = 0.0000 m; settling velocity (v_s) = 0.0027 m/s, determine the dynamic viscosity (mu) in Pa*s.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• Everything except mu is given, so isolate mu 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for mu:

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

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 1,650 kg/m^3, fluid density (rho) = 992.0 kg/m^3, particle diameter (d) = 0.0000 m, settling velocity (v_s) = 0.0027 m/s.

4. Step 4 — Substitute the given values:

   $$\mu = \dfrac{9.8100(\rho_p-992.0)0.0000^2}{18v_s}$$

5. Step 5 — Evaluate:

   $$\mu = 0.0000\ \text{Pa*s}$$

6. Step 6 — Check: returning mu = 0.0000 Pa*s to

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

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

Why the other options are there

• 0.0000 — kept a factor of two that cancels in the correct rearrangement.
• 0.0000 — dropped that same factor in the other direction.
• 0.0000 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law

Stokes' law settling velocity for fine sand particles in a clarifier Given gravitational acceleration (g) = 9.8100 m/s^2; particle density (rho_p) = 2,440 kg/m^3; fluid density (rho) = 994.5 kg/m^3; particle diameter (d) = 0.0000 m; dynamic viscosity (mu) = 0.0015 Pa*s, determine the settling velocity (v_s) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Stokes' Law.
• 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.
• Stokes' law describes the settling velocity of small spherical particles in the laminar (viscous) settling regime.

Step-by-step solution

1. Step 1 — State the governing relation:

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

2. Step 2 — Rearrange symbolically for v_s:

   $$v_{s} = \dfrac{g(\rho_p-\rho)d^2}{18\mu}$$

3. Step 3 — List the givens: gravitational acceleration (g) = 9.8100 m/s^2, particle density (rho_p) = 2,440 kg/m^3, fluid density (rho) = 994.5 kg/m^3, particle diameter (d) = 0.0000 m, dynamic viscosity (mu) = 0.0015 Pa*s.

4. Step 4 — Substitute the given values:

   $$v_{s} = \dfrac{9.8100(\rho_p-994.5)0.0000^2}{180.0015}$$

5. Step 5 — Evaluate:

   $$v_{s} = 0.0003\ \text{m/s}$$

6. Step 6 — Check: returning v_s = 0.0003 m/s to

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

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

Why the other options are there

• 0.0005 — kept a factor of two that cancels in the correct rearrangement.
• 0.0001 — dropped that same factor in the other direction.
• 0.0003 — rounded an intermediate value before the final step.

Reference: FE Handbook — Stokes' Law