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CivilSolve

Fluid Mechanics & Hydraulics

Continuity (Flow Between Pipe Sections)

Use Q = AV to find missing flow rates, velocities, or areas between pipe sections for incompressible flow, with unit conversions and worked steps.

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
section1One pipe/duct section
section2One pipe/duct section
section3One pipe/duct section

Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

contracting pipe (the motivating user problem): A1=5 ft², V1=5.1 ft/s, A2=1 ft²

Given

section1
area 5 ft^2velocity 5.1 ft/s
section2
area 1 ft^2

Assumptions

  • Incompressible steady flow with uniform (mean) velocity across each section — the volumetric flowrate is identical at every section.

Solution steps

  1. Establish the flowrate

    Every section carries the same flow; Q comes from section 1.

    Q = A · V

    Q = 25.5 cfs

    = 25.5 cfs

  2. Velocity at section 2

    The same flow through a smaller area must move faster (and slower through a larger one).

    V = Q / A

    V = 25.5 cfs / 1 ft^2 = 25.5 ft/s

    = 25.5 ft/s

Results

Volumetric flowrate Q

25.5cfs

Velocity at section 1 (given)

5.1ft/s

Velocity at section 2

25.5ft/s

Where this answer was checked
source
Continuity Q = A1·V1 = A2·V2 (NCEES FE Reference Handbook) — find V2 in the smaller pipe
verified by
hand-recomputed
derivation
Q = 5·5.1 = 25.5 cfs. V2 = Q/A2 = 25.5/1 = 25.5 ft/s.

Example 2

SI diameters: 200 mm at 2 m/s contracting to 100 mm

Given

section1
diameter 200 mmvelocity 2 m/s
section2
diameter 100 mm

Assumptions

  • Incompressible steady flow with uniform (mean) velocity across each section — the volumetric flowrate is identical at every section.
  • Sections given by diameter are circular: A = πD²/4.

Solution steps

  1. Establish the flowrate

    Every section carries the same flow; Q comes from section 1.

    Q = A · V

    Q = 0.06283 m^3/s

    = 0.06283 m^3/s

  2. Velocity at section 2

    The same flow through a smaller area must move faster (and slower through a larger one).

    V = Q / A

    V = 0.06283 m^3/s / 0.007854 m^2 = 8 m/s

    = 8 m/s

Results

Volumetric flowrate Q

0.06283m^3/s

Velocity at section 1 (given)

2m/s

Velocity at section 2

8m/s

Where this answer was checked
source
Continuity with circular sections: V2/V1 = (D1/D2)² — halving the diameter quadruples the velocity
verified by
hand-recomputed
derivation
A1 = π·0.2²/4 = 0.0314159 m²; Q = 0.0628319 m³/s. A2 = 0.00785398 m²; V2 = Q/A2 = 8 m/s exactly (ratio check: (200/100)² × 2 = 8 ✓).

Example 3

size a section for a velocity limit: Q=0.5 m³/s, V2 ≤ 4 m/s

Given

flowrate
0.5 m^3/s
section1
diameter 500 mm
section2
velocity 4 m/s

Assumptions

  • Incompressible steady flow with uniform (mean) velocity across each section — the volumetric flowrate is identical at every section.
  • Sections given by diameter are circular: A = πD²/4.

Solution steps

  1. Establish the flowrate

    Every section carries the same flow; Q comes from the given flowrate.

    Q = A · V

    Q = 0.5 m^3/s

    = 0.5 m^3/s

  2. Velocity at section 1

    The same flow through a smaller area must move faster (and slower through a larger one).

    V = Q / A

    V = 0.5 m^3/s / 0.1963 m^2 = 2.546 m/s

    = 2.546 m/s

  3. Required area at section 2

    The area follows from the flow and the velocity it must carry.

    A = Q / V

    A = 0.5 m^3/s / 4 m/s = 0.125 m^2

    = 0.125 m^2

Results

Volumetric flowrate Q

0.5m^3/s

Velocity at section 1

2.546m/s

Flow area at section 2

0.125m^2

Equivalent circular diameter at section 2

0.3989m

Where this answer was checked
source
Continuity solved for area: A = Q/V — duct sizing from a velocity limit
verified by
hand-recomputed
derivation
A2 = 0.5/4 = 0.125 m²; equivalent D = √(4·0.125/π) = √0.159155 = 0.39894 m.