horizontal velocities through a rectangular sedimentation basin Given flow rate (Q) = 17,830 m^3/day; basin width (W) = 2.6000 m; basin depth (H) = 5.0000 m, determine the horizontal velocity (v_h) in m/day.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except v_h is given, so isolate v_h 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for v_h:

   $$v_{h} = \dfrac{Q}{WH}$$

3. Step 3 — List the givens: flow rate (Q) = 17,830 m^3/day, basin width (W) = 2.6000 m, basin depth (H) = 5.0000 m.

4. Step 4 — Substitute the given values:

   $$v_{h} = \dfrac{17830}{W5.0000}$$

5. Step 5 — Evaluate:

   $$v_{h} = 1372\ \text{m/day}$$

6. Step 6 — Check: returning v_h = 1,372 m/day to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

• 2,743 — kept a factor of two that cancels in the correct rearrangement.
• 685.8 — dropped that same factor in the other direction.
• 1,509 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities check to prevent scour in a settling tank Given flow rate (Q) = 48,430 m^3/day; basin depth (H) = 4.0000 m; horizontal velocity (v_h) = 2,793 m/day, determine the basin width (W) in m.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except W is given, so isolate W 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for W:

   $$W = \dfrac{Q}{v_hH}$$

3. Step 3 — List the givens: flow rate (Q) = 48,430 m^3/day, basin depth (H) = 4.0000 m, horizontal velocity (v_h) = 2,793 m/day.

4. Step 4 — Substitute the given values:

   $$W = \dfrac{48430}{v_h4.0000}$$

5. Step 5 — Evaluate:

   $$W = 4.3349\ \text{m}$$

6. Step 6 — Check: returning W = 4.3349 m to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

• 8.6699 — kept a factor of two that cancels in the correct rearrangement.
• 2.1675 — dropped that same factor in the other direction.
• 4.7684 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities computed from basin cross-sectional area and flow Given flow rate (Q) = 39,170 m^3/day; basin width (W) = 3.9000 m; horizontal velocity (v_h) = 4,402 m/day, determine the basin depth (H) in m.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except H is given, so isolate H 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for H:

   $$H = \dfrac{Q}{v_hW}$$

3. Step 3 — List the givens: flow rate (Q) = 39,170 m^3/day, basin width (W) = 3.9000 m, horizontal velocity (v_h) = 4,402 m/day.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{39170}{v_h3.9000}$$

5. Step 5 — Evaluate:

   $$H = 2.2816\ \text{m}$$

6. Step 6 — Check: returning H = 2.2816 m to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

• 4.5632 — kept a factor of two that cancels in the correct rearrangement.
• 1.1408 — dropped that same factor in the other direction.
• 2.5098 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities through a rectangular sedimentation basin Given flow rate (Q) = 8,180 m^3/day; basin width (W) = 12.5000 m; basin depth (H) = 4.5000 m, determine the horizontal velocity (v_h) in m/day.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except v_h is given, so isolate v_h 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for v_h:

   $$v_{h} = \dfrac{Q}{WH}$$

3. Step 3 — List the givens: flow rate (Q) = 8,180 m^3/day, basin width (W) = 12.5000 m, basin depth (H) = 4.5000 m.

4. Step 4 — Substitute the given values:

   $$v_{h} = \dfrac{8180}{W4.5000}$$

5. Step 5 — Evaluate:

   $$v_{h} = 145.4\ \text{m/day}$$

6. Step 6 — Check: returning v_h = 145.4 m/day to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

• 290.8 — kept a factor of two that cancels in the correct rearrangement.
• 72.7111 — dropped that same factor in the other direction.
• 160.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities check to prevent scour in a settling tank Given flow rate (Q) = 38,000 m^3/day; basin depth (H) = 2.6000 m; horizontal velocity (v_h) = 3,628 m/day, determine the basin width (W) in m.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except W is given, so isolate W 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for W:

   $$W = \dfrac{Q}{v_hH}$$

3. Step 3 — List the givens: flow rate (Q) = 38,000 m^3/day, basin depth (H) = 2.6000 m, horizontal velocity (v_h) = 3,628 m/day.

4. Step 4 — Substitute the given values:

   $$W = \dfrac{38000}{v_h2.6000}$$

5. Step 5 — Evaluate:

   $$W = 4.0285\ \text{m}$$

6. Step 6 — Check: returning W = 4.0285 m to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

• 8.0570 — kept a factor of two that cancels in the correct rearrangement.
• 2.0142 — dropped that same factor in the other direction.
• 4.4313 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities computed from basin cross-sectional area and flow Given flow rate (Q) = 36,250 m^3/day; basin width (W) = 8.0000 m; horizontal velocity (v_h) = 1,401 m/day, determine the basin depth (H) in m.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except H is given, so isolate H 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for H:

   $$H = \dfrac{Q}{v_hW}$$

3. Step 3 — List the givens: flow rate (Q) = 36,250 m^3/day, basin width (W) = 8.0000 m, horizontal velocity (v_h) = 1,401 m/day.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{36250}{v_h8.0000}$$

5. Step 5 — Evaluate:

   $$H = 3.2343\ \text{m}$$

6. Step 6 — Check: returning H = 3.2343 m to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

• 6.4686 — kept a factor of two that cancels in the correct rearrangement.
• 1.6171 — dropped that same factor in the other direction.
• 3.5577 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities through a rectangular sedimentation basin Given flow rate (Q) = 10,660 m^3/day; basin width (W) = 9.2000 m; basin depth (H) = 3.0000 m, determine the horizontal velocity (v_h) in m/day.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except v_h is given, so isolate v_h 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for v_h:

   $$v_{h} = \dfrac{Q}{WH}$$

3. Step 3 — List the givens: flow rate (Q) = 10,660 m^3/day, basin width (W) = 9.2000 m, basin depth (H) = 3.0000 m.

4. Step 4 — Substitute the given values:

   $$v_{h} = \dfrac{10660}{W3.0000}$$

5. Step 5 — Evaluate:

   $$v_{h} = 386.2\ \text{m/day}$$

6. Step 6 — Check: returning v_h = 386.2 m/day to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

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

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities check to prevent scour in a settling tank Given flow rate (Q) = 30,670 m^3/day; basin depth (H) = 4.9000 m; horizontal velocity (v_h) = 4,423 m/day, determine the basin width (W) in m.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except W is given, so isolate W 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for W:

   $$W = \dfrac{Q}{v_hH}$$

3. Step 3 — List the givens: flow rate (Q) = 30,670 m^3/day, basin depth (H) = 4.9000 m, horizontal velocity (v_h) = 4,423 m/day.

4. Step 4 — Substitute the given values:

   $$W = \dfrac{30670}{v_h4.9000}$$

5. Step 5 — Evaluate:

   $$W = 1.4151\ \text{m}$$

6. Step 6 — Check: returning W = 1.4151 m to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

• 2.8303 — kept a factor of two that cancels in the correct rearrangement.
• 0.7076 — dropped that same factor in the other direction.
• 1.5567 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities computed from basin cross-sectional area and flow Given flow rate (Q) = 12,450 m^3/day; basin width (W) = 19.5000 m; horizontal velocity (v_h) = 1,310 m/day, determine the basin depth (H) in m.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except H is given, so isolate H 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for H:

   $$H = \dfrac{Q}{v_hW}$$

3. Step 3 — List the givens: flow rate (Q) = 12,450 m^3/day, basin width (W) = 19.5000 m, horizontal velocity (v_h) = 1,310 m/day.

4. Step 4 — Substitute the given values:

   $$H = \dfrac{12450}{v_h19.5000}$$

5. Step 5 — Evaluate:

   $$H = 0.4874\ \text{m}$$

6. Step 6 — Check: returning H = 0.4874 m to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

• 0.9748 — kept a factor of two that cancels in the correct rearrangement.
• 0.2437 — dropped that same factor in the other direction.
• 0.5361 — rounded an intermediate value before the final step.

Reference: FE Handbook — Horizontal Velocities (sedimentation)

horizontal velocities through a rectangular sedimentation basin Given flow rate (Q) = 32,330 m^3/day; basin width (W) = 19.6000 m; basin depth (H) = 4.4000 m, determine the horizontal velocity (v_h) in m/day.

Start with the thinking

• The governing relation printed in this handbook section is Horizontal Velocities.
• Everything except v_h is given, so isolate v_h 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.
• Horizontal velocities through a sedimentation basin must remain low enough to avoid resuspension of settled particles.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_h = \dfrac{Q}{W H}$$

2. Step 2 — Rearrange symbolically for v_h:

   $$v_{h} = \dfrac{Q}{WH}$$

3. Step 3 — List the givens: flow rate (Q) = 32,330 m^3/day, basin width (W) = 19.6000 m, basin depth (H) = 4.4000 m.

4. Step 4 — Substitute the given values:

   $$v_{h} = \dfrac{32330}{W4.4000}$$

5. Step 5 — Evaluate:

   $$v_{h} = 374.9\ \text{m/day}$$

6. Step 6 — Check: returning v_h = 374.9 m/day to

   $$v_h = \dfrac{Q}{W H}$$

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

Why the other options are there

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

Reference: FE Handbook — Horizontal Velocities (sedimentation)