reel and paddle flocculator power dissipation calculation Given drag coefficient of paddle (C_D) = 1.9100; fluid density (rho) = 995.0 kg/m^3; paddle area (A) = 3.9000 m^2; paddle velocity relative to water (v_p) = 0.3400 m/s, determine the power dissipated (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• 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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for P:

   $$P = \dfrac{C_D\rho A v_p^3}{2}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.9100, fluid density (rho) = 995.0 kg/m^3, paddle area (A) = 3.9000 m^2, paddle velocity relative to water (v_p) = 0.3400 m/s.

4. Step 4 — Substitute the given values:

   $$P = \dfrac{C_D995.0 3.9000 v_p^3}{2}$$

5. Step 5 — Evaluate:

   $$P = 145.7\ \text{W}$$

6. Step 6 — Check: returning P = 145.7 W to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 291.3 — kept a factor of two that cancels in the correct rearrangement.
• 72.8279 — dropped that same factor in the other direction.
• 160.2 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle mixer design for flocculation basin Given drag coefficient of paddle (C_D) = 1.3200; fluid density (rho) = 996.5 kg/m^3; paddle velocity relative to water (v_p) = 0.5200 m/s; power dissipated (P) = 3,069 W, determine the paddle area (A) in m^2.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• Everything except A is given, so isolate A 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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for A:

   $$A = \dfrac{2P}{C_D\rho v_p^3}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.3200, fluid density (rho) = 996.5 kg/m^3, paddle velocity relative to water (v_p) = 0.5200 m/s, power dissipated (P) = 3,069 W.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{23069}{C_D996.5 v_p^3}$$

5. Step 5 — Evaluate:

   A = 33.1868\ \text{m^2}

6. Step 6 — Check: returning A = 33.1868 m^2 to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 66.3736 — kept a factor of two that cancels in the correct rearrangement.
• 16.5934 — dropped that same factor in the other direction.
• 36.5055 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle drag power for a slow-mix flocculator Given drag coefficient of paddle (C_D) = 1.0700; fluid density (rho) = 995.5 kg/m^3; paddle area (A) = 6.4500 m^2; power dissipated (P) = 4,893 W, determine the paddle velocity relative to water (v_p) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• Everything except v_p is given, so isolate v_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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for v_p:

   $$v_{p} = \sqrt[3]{\dfrac{2P}{C_D\rho A}}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.0700, fluid density (rho) = 995.5 kg/m^3, paddle area (A) = 6.4500 m^2, power dissipated (P) = 4,893 W.

4. Step 4 — Substitute the given values:

   $$v_{p} = \sqrt[3]{\dfrac{24893}{C_D995.5 6.4500}}$$

5. Step 5 — Evaluate:

   $$v_{p} = 1.1251\ \text{m/s}$$

6. Step 6 — Check: returning v_p = 1.1251 m/s to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 2.2503 — kept a factor of two that cancels in the correct rearrangement.
• 0.5626 — dropped that same factor in the other direction.
• 1.2377 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle flocculator power dissipation calculation Given drag coefficient of paddle (C_D) = 1.4700; fluid density (rho) = 995.0 kg/m^3; paddle area (A) = 1.7000 m^2; paddle velocity relative to water (v_p) = 0.2000 m/s, determine the power dissipated (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• 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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for P:

   $$P = \dfrac{C_D\rho A v_p^3}{2}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.4700, fluid density (rho) = 995.0 kg/m^3, paddle area (A) = 1.7000 m^2, paddle velocity relative to water (v_p) = 0.2000 m/s.

4. Step 4 — Substitute the given values:

   $$P = \dfrac{C_D995.0 1.7000 v_p^3}{2}$$

5. Step 5 — Evaluate:

   $$P = 9.9460\ \text{W}$$

6. Step 6 — Check: returning P = 9.9460 W to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 19.8920 — kept a factor of two that cancels in the correct rearrangement.
• 4.9730 — dropped that same factor in the other direction.
• 10.9406 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle mixer design for flocculation basin Given drag coefficient of paddle (C_D) = 1.2000; fluid density (rho) = 996.0 kg/m^3; paddle velocity relative to water (v_p) = 0.4400 m/s; power dissipated (P) = 1,915 W, determine the paddle area (A) in m^2.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• Everything except A is given, so isolate A 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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for A:

   $$A = \dfrac{2P}{C_D\rho v_p^3}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.2000, fluid density (rho) = 996.0 kg/m^3, paddle velocity relative to water (v_p) = 0.4400 m/s, power dissipated (P) = 1,915 W.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{21915}{C_D996.0 v_p^3}$$

5. Step 5 — Evaluate:

   A = 37.6184\ \text{m^2}

6. Step 6 — Check: returning A = 37.6184 m^2 to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 75.2368 — kept a factor of two that cancels in the correct rearrangement.
• 18.8092 — dropped that same factor in the other direction.
• 41.3802 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle drag power for a slow-mix flocculator Given drag coefficient of paddle (C_D) = 1.9900; fluid density (rho) = 998.0 kg/m^3; paddle area (A) = 4.0500 m^2; power dissipated (P) = 1,199 W, determine the paddle velocity relative to water (v_p) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• Everything except v_p is given, so isolate v_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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for v_p:

   $$v_{p} = \sqrt[3]{\dfrac{2P}{C_D\rho A}}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.9900, fluid density (rho) = 998.0 kg/m^3, paddle area (A) = 4.0500 m^2, power dissipated (P) = 1,199 W.

4. Step 4 — Substitute the given values:

   $$v_{p} = \sqrt[3]{\dfrac{21199}{C_D998.0 4.0500}}$$

5. Step 5 — Evaluate:

   $$v_{p} = 0.6680\ \text{m/s}$$

6. Step 6 — Check: returning v_p = 0.6680 m/s to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 1.3361 — kept a factor of two that cancels in the correct rearrangement.
• 0.3340 — dropped that same factor in the other direction.
• 0.7348 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle flocculator power dissipation calculation Given drag coefficient of paddle (C_D) = 1.8600; fluid density (rho) = 998.5 kg/m^3; paddle area (A) = 3.0000 m^2; paddle velocity relative to water (v_p) = 0.2500 m/s, determine the power dissipated (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• 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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for P:

   $$P = \dfrac{C_D\rho A v_p^3}{2}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.8600, fluid density (rho) = 998.5 kg/m^3, paddle area (A) = 3.0000 m^2, paddle velocity relative to water (v_p) = 0.2500 m/s.

4. Step 4 — Substitute the given values:

   $$P = \dfrac{C_D998.5 3.0000 v_p^3}{2}$$

5. Step 5 — Evaluate:

   $$P = 43.5284\ \text{W}$$

6. Step 6 — Check: returning P = 43.5284 W to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 87.0567 — kept a factor of two that cancels in the correct rearrangement.
• 21.7642 — dropped that same factor in the other direction.
• 47.8812 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle mixer design for flocculation basin Given drag coefficient of paddle (C_D) = 1.8400; fluid density (rho) = 994.5 kg/m^3; paddle velocity relative to water (v_p) = 0.8100 m/s; power dissipated (P) = 1,679 W, determine the paddle area (A) in m^2.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• Everything except A is given, so isolate A 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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for A:

   $$A = \dfrac{2P}{C_D\rho v_p^3}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.8400, fluid density (rho) = 994.5 kg/m^3, paddle velocity relative to water (v_p) = 0.8100 m/s, power dissipated (P) = 1,679 W.

4. Step 4 — Substitute the given values:

   $$A = \dfrac{21679}{C_D994.5 v_p^3}$$

5. Step 5 — Evaluate:

   A = 3.4531\ \text{m^2}

6. Step 6 — Check: returning A = 3.4531 m^2 to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 6.9061 — kept a factor of two that cancels in the correct rearrangement.
• 1.7265 — dropped that same factor in the other direction.
• 3.7984 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle drag power for a slow-mix flocculator Given drag coefficient of paddle (C_D) = 1.0500; fluid density (rho) = 998.0 kg/m^3; paddle area (A) = 0.9000 m^2; power dissipated (P) = 4,091 W, determine the paddle velocity relative to water (v_p) in m/s.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• Everything except v_p is given, so isolate v_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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for v_p:

   $$v_{p} = \sqrt[3]{\dfrac{2P}{C_D\rho A}}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.0500, fluid density (rho) = 998.0 kg/m^3, paddle area (A) = 0.9000 m^2, power dissipated (P) = 4,091 W.

4. Step 4 — Substitute the given values:

   $$v_{p} = \sqrt[3]{\dfrac{24091}{C_D998.0 0.9000}}$$

5. Step 5 — Evaluate:

   $$v_{p} = 2.0548\ \text{m/s}$$

6. Step 6 — Check: returning v_p = 2.0548 m/s to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

• 4.1096 — kept a factor of two that cancels in the correct rearrangement.
• 1.0274 — dropped that same factor in the other direction.
• 2.2603 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reel and Paddle Flocculator

reel and paddle flocculator power dissipation calculation Given drag coefficient of paddle (C_D) = 1.7700; fluid density (rho) = 992.5 kg/m^3; paddle area (A) = 6.0000 m^2; paddle velocity relative to water (v_p) = 0.4500 m/s, determine the power dissipated (P) in W.

Start with the thinking

• The governing relation printed in this handbook section is Reel and Paddle.
• 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.
• Reel and paddle flocculators dissipate power in proportion to paddle drag, area, and relative velocity through the water.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

2. Step 2 — Rearrange symbolically for P:

   $$P = \dfrac{C_D\rho A v_p^3}{2}$$

3. Step 3 — List the givens: drag coefficient of paddle (C_D) = 1.7700, fluid density (rho) = 992.5 kg/m^3, paddle area (A) = 6.0000 m^2, paddle velocity relative to water (v_p) = 0.4500 m/s.

4. Step 4 — Substitute the given values:

   $$P = \dfrac{C_D992.5 6.0000 v_p^3}{2}$$

5. Step 5 — Evaluate:

   $$P = 480.2\ \text{W}$$

6. Step 6 — Check: returning P = 480.2 W to

   $$P = C_D \rho A \dfrac{v_p^3}{2}$$

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

Why the other options are there

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

Reference: FE Handbook — Reel and Paddle Flocculator