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TERMINAL SETTLING VELOCITY (ft/s)

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics
1 formulas
10 exam-style examples
~47 min
All Fluid Mechanics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • FACTOR IS INCLUDED FOR FINE

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s)

A pipe reduces from 14 in. to 7 in. diameter while carrying 9.0 cfs. Find the velocity in each section.

Given

  • D1=14inD_{1} = 14 in
  • D2=7inD_{2} = 7 in
  • Q=9.0cfsQ = 9.0 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=14D₂=7v₁ = 8.42 ft/sv₂ = 33.68 ft/s

Figure 1 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s)

Step-by-step solution

  1. Area 1

    A1=(π/4)(1.167)2=1.0690ft2A_{1} = (\pi/4)(1.167)^{2} = 1.0690 ft^{2}

  2. Area 2

    A2=(π/4)(0.583)2=0.2673ft2A_{2} = (\pi/4)(0.583)^{2} = 0.2673 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=9.0/1.0690=8.42ft/sv_{1} = 9.0/1.0690 = 8.42 ft/s

  5. Velocity 2

    v2=9.0/0.2673=33.68ft/sv_{2} = 9.0/0.2673 = 33.68 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=4.00✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 4.00 ✓
Answer:

v₁ ≈ 8.42 ft/s, v₂ ≈ 33.68 ft/s

Why the other options are there

  • v₂ = 16.84 ft/s (diameter ratio not squared)
  • v₂ = 0.234 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 2
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (2)

A pipe reduces from 14 in. to 6 in. diameter while carrying 1.5 cfs. Find the velocity in each section.

Given

  • D1=14inD_{1} = 14 in
  • D2=6inD_{2} = 6 in
  • Q=1.5cfsQ = 1.5 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=14D₂=6v₁ = 1.40 ft/sv₂ = 7.64 ft/s

Figure 2 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (2)

Step-by-step solution

  1. Area 1

    A1=(π/4)(1.167)2=1.0690ft2A_{1} = (\pi/4)(1.167)^{2} = 1.0690 ft^{2}

  2. Area 2

    A2=(π/4)(0.500)2=0.1963ft2A_{2} = (\pi/4)(0.500)^{2} = 0.1963 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=1.5/1.0690=1.40ft/sv_{1} = 1.5/1.0690 = 1.40 ft/s

  5. Velocity 2

    v2=1.5/0.1963=7.64ft/sv_{2} = 1.5/0.1963 = 7.64 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=5.44✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 5.44 ✓
Answer:

v₁ ≈ 1.40 ft/s, v₂ ≈ 7.64 ft/s

Why the other options are there

  • v₂ = 3.27 ft/s (diameter ratio not squared)
  • v₂ = 0.053 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 3
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (3)

A pipe reduces from 17 in. to 4 in. diameter while carrying 7.5 cfs. Find the velocity in each section.

Given

  • D1=17inD_{1} = 17 in
  • D2=4inD_{2} = 4 in
  • Q=7.5cfsQ = 7.5 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=17D₂=4v₁ = 4.76 ft/sv₂ = 85.94 ft/s

Figure 3 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (3)

Step-by-step solution

  1. Area 1

    A1=(π/4)(1.417)2=1.5763ft2A_{1} = (\pi/4)(1.417)^{2} = 1.5763 ft^{2}

  2. Area 2

    A2=(π/4)(0.333)2=0.0873ft2A_{2} = (\pi/4)(0.333)^{2} = 0.0873 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=7.5/1.5763=4.76ft/sv_{1} = 7.5/1.5763 = 4.76 ft/s

  5. Velocity 2

    v2=7.5/0.0873=85.94ft/sv_{2} = 7.5/0.0873 = 85.94 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=18.06✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 18.06 ✓
Answer:

v₁ ≈ 4.76 ft/s, v₂ ≈ 85.94 ft/s

Why the other options are there

  • v₂ = 20.22 ft/s (diameter ratio not squared)
  • v₂ = 0.597 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 4
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (4)

A pipe reduces from 6 in. to 5 in. diameter while carrying 3.5 cfs. Find the velocity in each section.

Given

  • D1=6inD_{1} = 6 in
  • D2=5inD_{2} = 5 in
  • Q=3.5cfsQ = 3.5 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=6D₂=5v₁ = 17.83 ft/sv₂ = 25.67 ft/s

Figure 4 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (4)

Step-by-step solution

  1. Area 1

    A1=(π/4)(0.500)2=0.1963ft2A_{1} = (\pi/4)(0.500)^{2} = 0.1963 ft^{2}

  2. Area 2

    A2=(π/4)(0.417)2=0.1364ft2A_{2} = (\pi/4)(0.417)^{2} = 0.1364 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=3.5/0.1963=17.83ft/sv_{1} = 3.5/0.1963 = 17.83 ft/s

  5. Velocity 2

    v2=3.5/0.1364=25.67ft/sv_{2} = 3.5/0.1364 = 25.67 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=1.44✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 1.44 ✓
Answer:

v₁ ≈ 17.83 ft/s, v₂ ≈ 25.67 ft/s

Why the other options are there

  • v₂ = 21.39 ft/s (diameter ratio not squared)
  • v₂ = 0.178 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 5
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (5)

A pipe reduces from 8 in. to 10 in. diameter while carrying 11.5 cfs. Find the velocity in each section.

Given

  • D1=8inD_{1} = 8 in
  • D2=10inD_{2} = 10 in
  • Q=11.5cfsQ = 11.5 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=8D₂=10v₁ = 32.95 ft/sv₂ = 21.08 ft/s

Figure 5 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (5)

Step-by-step solution

  1. Area 1

    A1=(π/4)(0.667)2=0.3491ft2A_{1} = (\pi/4)(0.667)^{2} = 0.3491 ft^{2}

  2. Area 2

    A2=(π/4)(0.833)2=0.5454ft2A_{2} = (\pi/4)(0.833)^{2} = 0.5454 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=11.5/0.3491=32.95ft/sv_{1} = 11.5/0.3491 = 32.95 ft/s

  5. Velocity 2

    v2=11.5/0.5454=21.08ft/sv_{2} = 11.5/0.5454 = 21.08 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=0.64✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 0.64 ✓
Answer:

v₁ ≈ 32.95 ft/s, v₂ ≈ 21.08 ft/s

Why the other options are there

  • v₂ = 26.36 ft/s (diameter ratio not squared)
  • v₂ = 0.146 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 6
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (6)

A pipe reduces from 12 in. to 7 in. diameter while carrying 2.0 cfs. Find the velocity in each section.

Given

  • D1=12inD_{1} = 12 in
  • D2=7inD_{2} = 7 in
  • Q=2.0cfsQ = 2.0 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=12D₂=7v₁ = 2.55 ft/sv₂ = 7.48 ft/s

Figure 6 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (6)

Step-by-step solution

  1. Area 1

    A1=(π/4)(1.000)2=0.7854ft2A_{1} = (\pi/4)(1.000)^{2} = 0.7854 ft^{2}

  2. Area 2

    A2=(π/4)(0.583)2=0.2673ft2A_{2} = (\pi/4)(0.583)^{2} = 0.2673 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=2.0/0.7854=2.55ft/sv_{1} = 2.0/0.7854 = 2.55 ft/s

  5. Velocity 2

    v2=2.0/0.2673=7.48ft/sv_{2} = 2.0/0.2673 = 7.48 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=2.94✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 2.94 ✓
Answer:

v₁ ≈ 2.55 ft/s, v₂ ≈ 7.48 ft/s

Why the other options are there

  • v₂ = 4.37 ft/s (diameter ratio not squared)
  • v₂ = 0.052 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 7
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (7)

A pipe reduces from 10 in. to 9 in. diameter while carrying 5.0 cfs. Find the velocity in each section.

Given

  • D1=10inD_{1} = 10 in
  • D2=9inD_{2} = 9 in
  • Q=5.0cfsQ = 5.0 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=10D₂=9v₁ = 9.17 ft/sv₂ = 11.32 ft/s

Figure 7 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (7)

Step-by-step solution

  1. Area 1

    A1=(π/4)(0.833)2=0.5454ft2A_{1} = (\pi/4)(0.833)^{2} = 0.5454 ft^{2}

  2. Area 2

    A2=(π/4)(0.750)2=0.4418ft2A_{2} = (\pi/4)(0.750)^{2} = 0.4418 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=5.0/0.5454=9.17ft/sv_{1} = 5.0/0.5454 = 9.17 ft/s

  5. Velocity 2

    v2=5.0/0.4418=11.32ft/sv_{2} = 5.0/0.4418 = 11.32 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=1.23✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 1.23 ✓
Answer:

v₁ ≈ 9.17 ft/s, v₂ ≈ 11.32 ft/s

Why the other options are there

  • v₂ = 10.19 ft/s (diameter ratio not squared)
  • v₂ = 0.079 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 8
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (8)

A pipe reduces from 7 in. to 10 in. diameter while carrying 1.5 cfs. Find the velocity in each section.

Given

  • D1=7inD_{1} = 7 in
  • D2=10inD_{2} = 10 in
  • Q=1.5cfsQ = 1.5 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=7D₂=10v₁ = 5.61 ft/sv₂ = 2.75 ft/s

Figure 8 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (8)

Step-by-step solution

  1. Area 1

    A1=(π/4)(0.583)2=0.2673ft2A_{1} = (\pi/4)(0.583)^{2} = 0.2673 ft^{2}

  2. Area 2

    A2=(π/4)(0.833)2=0.5454ft2A_{2} = (\pi/4)(0.833)^{2} = 0.5454 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=1.5/0.2673=5.61ft/sv_{1} = 1.5/0.2673 = 5.61 ft/s

  5. Velocity 2

    v2=1.5/0.5454=2.75ft/sv_{2} = 1.5/0.5454 = 2.75 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=0.49✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 0.49 ✓
Answer:

v₁ ≈ 5.61 ft/s, v₂ ≈ 2.75 ft/s

Why the other options are there

  • v₂ = 3.93 ft/s (diameter ratio not squared)
  • v₂ = 0.019 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 9
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (9)

A pipe reduces from 7 in. to 4 in. diameter while carrying 6.5 cfs. Find the velocity in each section.

Given

  • D1=7inD_{1} = 7 in
  • D2=4inD_{2} = 4 in
  • Q=6.5cfsQ = 6.5 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=7D₂=4v₁ = 24.32 ft/sv₂ = 74.48 ft/s

Figure 9 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (9)

Step-by-step solution

  1. Area 1

    A1=(π/4)(0.583)2=0.2673ft2A_{1} = (\pi/4)(0.583)^{2} = 0.2673 ft^{2}

  2. Area 2

    A2=(π/4)(0.333)2=0.0873ft2A_{2} = (\pi/4)(0.333)^{2} = 0.0873 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=6.5/0.2673=24.32ft/sv_{1} = 6.5/0.2673 = 24.32 ft/s

  5. Velocity 2

    v2=6.5/0.0873=74.48ft/sv_{2} = 6.5/0.0873 = 74.48 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=3.06✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 3.06 ✓
Answer:

v₁ ≈ 24.32 ft/s, v₂ ≈ 74.48 ft/s

Why the other options are there

  • v₂ = 42.56 ft/s (diameter ratio not squared)
  • v₂ = 0.517 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

Example 10
Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (10)

A pipe reduces from 18 in. to 5 in. diameter while carrying 1.5 cfs. Find the velocity in each section.

Given

  • D1=18inD_{1} = 18 in
  • D2=5inD_{2} = 5 in
  • Q=1.5cfsQ = 1.5 cfs

Find

v₁ and v₂

Start with the thinking

  • Q is the same in both sections — only the area changes.
  • Convert inches to feet before computing area.
D₁=18D₂=5v₁ = 0.85 ft/sv₂ = 11.00 ft/s

Figure 10 — schematic for Continuity through a pipe contraction — TERMINAL SETTLING VELOCITY (ft/s) (10)

Step-by-step solution

  1. Area 1

    A1=(π/4)(1.500)2=1.7671ft2A_{1} = (\pi/4)(1.500)^{2} = 1.7671 ft^{2}

  2. Area 2

    A2=(π/4)(0.417)2=0.1364ft2A_{2} = (\pi/4)(0.417)^{2} = 0.1364 ft^{2}

  3. Continuity

    Q=A1v1=A2v2Q = A_{1}v_{1} = A_{2}v_{2}

  4. Velocity 1

    v1=1.5/1.7671=0.85ft/sv_{1} = 1.5/1.7671 = 0.85 ft/s

  5. Velocity 2

    v2=1.5/0.1364=11.00ft/sv_{2} = 1.5/0.1364 = 11.00 ft/s

  6. Ratio check

    v2/v1=(D1/D2)2=12.96✓v_{2}/v_{1} = (D_{1}/D_{2})^{2} = 12.96 ✓
Answer:

v₁ ≈ 0.85 ft/s, v₂ ≈ 11.00 ft/s

Why the other options are there

  • v₂ = 3.06 ft/s (diameter ratio not squared)
  • v₂ = 0.076 ft/s (inches left unconverted)

Reference: FE Reference Handbook — Fluid Mechanics → TERMINAL SETTLING VELOCITY (ft/s)

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