rapid mix and flocculator design velocity gradient for coagulation Given power input (P) = 19,090 W; dynamic viscosity (mu) = 0.0011 Pa*s; basin volume (V) = 1,460 m^3, determine the velocity gradient (G) in 1/s.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except G is given, so isolate G 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for G:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

3. Step 3 — List the givens: power input (P) = 19,090 W, dynamic viscosity (mu) = 0.0011 Pa*s, basin volume (V) = 1,460 m^3.

4. Step 4 — Substitute the given values:

   $$G = \sqrt{\dfrac{19090}{0.0011 1460}}$$

5. Step 5 — Evaluate:

   $$G = 109.5\ \text{1/s}$$

6. Step 6 — Check: returning G = 109.5 1/s to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 219.1 — kept a factor of two that cancels in the correct rearrangement.
• 54.7625 — dropped that same factor in the other direction.
• 120.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

rapid mix basin design power input for a water treatment plant Given dynamic viscosity (mu) = 0.0011 Pa*s; basin volume (V) = 895.0 m^3; velocity gradient (G) = 917.0 1/s, determine the power input (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except P is given, so isolate P 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for P:

   $$P = G^2 \mu V$$

3. Step 3 — List the givens: dynamic viscosity (mu) = 0.0011 Pa*s, basin volume (V) = 895.0 m^3, velocity gradient (G) = 917.0 1/s.

4. Step 4 — Substitute the given values:

   $$P = 917.0^2 0.0011 895.0$$

5. Step 5 — Evaluate:

   $$P = 850433\ \text{W}$$

6. Step 6 — Check: returning P = 850,433 W to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 1,700,866 — kept a factor of two that cancels in the correct rearrangement.
• 425,217 — dropped that same factor in the other direction.
• 935,476 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

flocculator design velocity gradient and power requirement Given power input (P) = 6,150 W; dynamic viscosity (mu) = 0.0017 Pa*s; velocity gradient (G) = 247.0 1/s, determine the basin volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except V is given, so isolate V 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for V:

   $$V = \dfrac{P}{\mu G^2}$$

3. Step 3 — List the givens: power input (P) = 6,150 W, dynamic viscosity (mu) = 0.0017 Pa*s, velocity gradient (G) = 247.0 1/s.

4. Step 4 — Substitute the given values:

   $$V = \dfrac{6150}{0.0017 247.0^2}$$

5. Step 5 — Evaluate:

   V = 60.3622\ \text{m^3}

6. Step 6 — Check: returning V = 60.3622 m^3 to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 120.7 — kept a factor of two that cancels in the correct rearrangement.
• 30.1811 — dropped that same factor in the other direction.
• 66.3984 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

rapid mix and flocculator design velocity gradient for coagulation Given power input (P) = 4,900 W; dynamic viscosity (mu) = 0.0018 Pa*s; basin volume (V) = 1,858 m^3, determine the velocity gradient (G) in 1/s.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except G is given, so isolate G 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for G:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

3. Step 3 — List the givens: power input (P) = 4,900 W, dynamic viscosity (mu) = 0.0018 Pa*s, basin volume (V) = 1,858 m^3.

4. Step 4 — Substitute the given values:

   $$G = \sqrt{\dfrac{4900}{0.0018 1858}}$$

5. Step 5 — Evaluate:

   $$G = 38.7096\ \text{1/s}$$

6. Step 6 — Check: returning G = 38.7096 1/s to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 77.4192 — kept a factor of two that cancels in the correct rearrangement.
• 19.3548 — dropped that same factor in the other direction.
• 42.5806 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

rapid mix basin design power input for a water treatment plant Given dynamic viscosity (mu) = 0.0017 Pa*s; basin volume (V) = 491.0 m^3; velocity gradient (G) = 53.0000 1/s, determine the power input (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except P is given, so isolate P 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for P:

   $$P = G^2 \mu V$$

3. Step 3 — List the givens: dynamic viscosity (mu) = 0.0017 Pa*s, basin volume (V) = 491.0 m^3, velocity gradient (G) = 53.0000 1/s.

4. Step 4 — Substitute the given values:

   $$P = 53.0000^2 0.0017 491.0$$

5. Step 5 — Evaluate:

   $$P = 2331\ \text{W}$$

6. Step 6 — Check: returning P = 2,331 W to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 4,662 — kept a factor of two that cancels in the correct rearrangement.
• 1,165 — dropped that same factor in the other direction.
• 2,564 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

flocculator design velocity gradient and power requirement Given power input (P) = 17,610 W; dynamic viscosity (mu) = 0.0012 Pa*s; velocity gradient (G) = 454.0 1/s, determine the basin volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except V is given, so isolate V 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for V:

   $$V = \dfrac{P}{\mu G^2}$$

3. Step 3 — List the givens: power input (P) = 17,610 W, dynamic viscosity (mu) = 0.0012 Pa*s, velocity gradient (G) = 454.0 1/s.

4. Step 4 — Substitute the given values:

   $$V = \dfrac{17610}{0.0012 454.0^2}$$

5. Step 5 — Evaluate:

   V = 72.4045\ \text{m^3}

6. Step 6 — Check: returning V = 72.4045 m^3 to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 144.8 — kept a factor of two that cancels in the correct rearrangement.
• 36.2023 — dropped that same factor in the other direction.
• 79.6450 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

rapid mix and flocculator design velocity gradient for coagulation Given power input (P) = 11,380 W; dynamic viscosity (mu) = 0.0018 Pa*s; basin volume (V) = 1,055 m^3, determine the velocity gradient (G) in 1/s.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except G is given, so isolate G 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for G:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

3. Step 3 — List the givens: power input (P) = 11,380 W, dynamic viscosity (mu) = 0.0018 Pa*s, basin volume (V) = 1,055 m^3.

4. Step 4 — Substitute the given values:

   $$G = \sqrt{\dfrac{11380}{0.0018 1055}}$$

5. Step 5 — Evaluate:

   $$G = 77.6280\ \text{1/s}$$

6. Step 6 — Check: returning G = 77.6280 1/s to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 155.3 — kept a factor of two that cancels in the correct rearrangement.
• 38.8140 — dropped that same factor in the other direction.
• 85.3908 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

rapid mix basin design power input for a water treatment plant Given dynamic viscosity (mu) = 0.0012 Pa*s; basin volume (V) = 269.0 m^3; velocity gradient (G) = 863.0 1/s, determine the power input (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except P is given, so isolate P 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for P:

   $$P = G^2 \mu V$$

3. Step 3 — List the givens: dynamic viscosity (mu) = 0.0012 Pa*s, basin volume (V) = 269.0 m^3, velocity gradient (G) = 863.0 1/s.

4. Step 4 — Substitute the given values:

   $$P = 863.0^2 0.0012 269.0$$

5. Step 5 — Evaluate:

   $$P = 234401\ \text{W}$$

6. Step 6 — Check: returning P = 234,401 W to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 468,802 — kept a factor of two that cancels in the correct rearrangement.
• 117,201 — dropped that same factor in the other direction.
• 257,841 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

flocculator design velocity gradient and power requirement Given power input (P) = 16,330 W; dynamic viscosity (mu) = 0.0009 Pa*s; velocity gradient (G) = 488.0 1/s, determine the basin volume (V) in m^3.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except V is given, so isolate V 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for V:

   $$V = \dfrac{P}{\mu G^2}$$

3. Step 3 — List the givens: power input (P) = 16,330 W, dynamic viscosity (mu) = 0.0009 Pa*s, velocity gradient (G) = 488.0 1/s.

4. Step 4 — Substitute the given values:

   $$V = \dfrac{16330}{0.0009 488.0^2}$$

5. Step 5 — Evaluate:

   V = 73.7333\ \text{m^3}

6. Step 6 — Check: returning V = 73.7333 m^3 to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 147.5 — kept a factor of two that cancels in the correct rearrangement.
• 36.8666 — dropped that same factor in the other direction.
• 81.1066 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design

rapid mix and flocculator design velocity gradient for coagulation Given power input (P) = 3,890 W; dynamic viscosity (mu) = 0.0017 Pa*s; basin volume (V) = 767.0 m^3, determine the velocity gradient (G) in 1/s.

Start with the thinking

• The governing relation printed in this handbook section is Rapid Mix and Flocculator Design.
• Everything except G is given, so isolate G 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.
• Rapid mix and flocculator design uses the velocity gradient G to set power input for a coagulation-flocculation basin.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

2. Step 2 — Rearrange symbolically for G:

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

3. Step 3 — List the givens: power input (P) = 3,890 W, dynamic viscosity (mu) = 0.0017 Pa*s, basin volume (V) = 767.0 m^3.

4. Step 4 — Substitute the given values:

   $$G = \sqrt{\dfrac{3890}{0.0017 767.0}}$$

5. Step 5 — Evaluate:

   $$G = 54.6201\ \text{1/s}$$

6. Step 6 — Check: returning G = 54.6201 1/s to

   $$G = \sqrt{\dfrac{P}{\mu V}}$$

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

Why the other options are there

• 109.2 — kept a factor of two that cancels in the correct rearrangement.
• 27.3101 — dropped that same factor in the other direction.
• 60.0821 — rounded an intermediate value before the final step.

Reference: FE Handbook — Rapid Mix and Flocculator Design