Reynolds Number

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

20 formulas

10 exam-style examples

~60 min

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Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

K and n are defined in the Stress, Pressure, and Viscosity section.
The critical Reynolds number (Re)c is defined to be the minimum Reynolds number at which a flow will turn turbulent.
Flow through a pipe is generally characterized as laminar for Re < 2,100 and fully turbulent for Re > 10,000, and transitional
The velocity distribution for laminar flow in circular tubes or between planes is
The shear stress distribution is
where τ and τw are the shear stresses at radii r and R, respectively.

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

Reynolds number — solve for Reynolds number — Reynolds Number

the Reynolds number for water flowing through a pipe Given density (rho) = 1,070 kg/m^3; velocity (V) = 2.2500 m/s; pipe diameter (D) = 0.4250 m; dynamic viscosity (mu) = 0.0006 Pa*s, determine the Reynolds number (Re).

Given

$density (rho) = 1,070 kg/m^3$
$velocity (V) = 2.2500 m/s$
$pipe diameter (D) = 0.4250 m$
$dynamic viscosity (mu) = 0.0006 Pa*s$

Find

Reynolds number (Re)

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
Everything except Re is given, so isolate Re 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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 1 — schematic for Reynolds number — solve for Reynolds number — Reynolds Number

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that Re stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 1,070 kg/m^3, velocity (V) = 2.2500 m/s, pipe diameter (D) = 0.4250 m, dynamic viscosity (mu) = 0.0006 Pa*s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$Re = 1705313$
Step 6 — Check: returning Re = 1,705,313 to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

Re = 1705313

Why the other options are there

3,410,625 — kept a factor of two that cancels in the correct rearrangement.
852,656 — dropped that same factor in the other direction.
1,875,844 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 2

Reynolds number — solve for velocity — Reynolds Number (2)

the Reynolds number of oil in a pipeline Given density (rho) = 840.0 kg/m^3; pipe diameter (D) = 0.2750 m; dynamic viscosity (mu) = 0.0014 Pa*s; Reynolds number (Re) = 275,770, determine the velocity (V) in m/s.

Given

$density (rho) = 840.0 kg/m^3$
$pipe diameter (D) = 0.2750 m$
$dynamic viscosity (mu) = 0.0014 Pa*s$
$Reynolds number (Re) = 275,770$

Find

velocity (V), in m/s

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 2 — schematic for Reynolds number — solve for velocity — Reynolds Number (2)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that V stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 840.0 kg/m^3, pipe diameter (D) = 0.2750 m, dynamic viscosity (mu) = 0.0014 Pa*s, Reynolds number (Re) = 275,770.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$V = 1.6713\ \text{m/s}$
Step 6 — Check: returning V = 1.6713 m/s to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

V = 1.6713\ \text{m/s}

Why the other options are there

3.3427 — kept a factor of two that cancels in the correct rearrangement.
0.8357 — dropped that same factor in the other direction.
1.8385 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 3

Reynolds number — solve for pipe diameter — Reynolds Number (3)

the Reynolds number used to check laminar versus turbulent flow Given density (rho) = 1,100 kg/m^3; velocity (V) = 3.1500 m/s; dynamic viscosity (mu) = 0.0006 Pa*s; Reynolds number (Re) = 116,230, determine the pipe diameter (D) in m.

Given

$density (rho) = 1,100 kg/m^3$
$velocity (V) = 3.1500 m/s$
$dynamic viscosity (mu) = 0.0006 Pa*s$
$Reynolds number (Re) = 116,230$

Find

pipe diameter (D), in m

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
Everything except D is given, so isolate D 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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 3 — schematic for Reynolds number — solve for pipe diameter — Reynolds Number (3)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that D stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 1,100 kg/m^3, velocity (V) = 3.1500 m/s, dynamic viscosity (mu) = 0.0006 Pa*s, Reynolds number (Re) = 116,230.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$D = 0.0184\ \text{m}$
Step 6 — Check: returning D = 0.0184 m to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

D = 0.0184\ \text{m}

Why the other options are there

0.0369 — kept a factor of two that cancels in the correct rearrangement.
0.0092 — dropped that same factor in the other direction.
0.0203 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 4

Reynolds number — solve for Reynolds number (case 2) — Reynolds Number (4)

the Reynolds number for water flowing through a pipe Given density (rho) = 830.0 kg/m^3; velocity (V) = 3.2500 m/s; pipe diameter (D) = 0.2750 m; dynamic viscosity (mu) = 0.0010 Pa*s, determine the Reynolds number (Re).

Given

$density (rho) = 830.0 kg/m^3$
$velocity (V) = 3.2500 m/s$
$pipe diameter (D) = 0.2750 m$
$dynamic viscosity (mu) = 0.0010 Pa*s$

Find

Reynolds number (Re)

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
Everything except Re is given, so isolate Re 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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 4 — schematic for Reynolds number — solve for Reynolds number (case 2) — Reynolds Number (4)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that Re stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 830.0 kg/m^3, velocity (V) = 3.2500 m/s, pipe diameter (D) = 0.2750 m, dynamic viscosity (mu) = 0.0010 Pa*s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$Re = 780855$
Step 6 — Check: returning Re = 780,855 to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

Re = 780855

Why the other options are there

1,561,711 — kept a factor of two that cancels in the correct rearrangement.
390,428 — dropped that same factor in the other direction.
858,941 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 5

Reynolds number — solve for velocity (case 2) — Reynolds Number (5)

the Reynolds number of oil in a pipeline Given density (rho) = 930.0 kg/m^3; pipe diameter (D) = 0.1000 m; dynamic viscosity (mu) = 0.0011 Pa*s; Reynolds number (Re) = 197,390, determine the velocity (V) in m/s.

Given

$density (rho) = 930.0 kg/m^3$
$pipe diameter (D) = 0.1000 m$
$dynamic viscosity (mu) = 0.0011 Pa*s$
$Reynolds number (Re) = 197,390$

Find

velocity (V), in m/s

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 5 — schematic for Reynolds number — solve for velocity (case 2) — Reynolds Number (5)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that V stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 930.0 kg/m^3, pipe diameter (D) = 0.1000 m, dynamic viscosity (mu) = 0.0011 Pa*s, Reynolds number (Re) = 197,390.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$V = 2.2286\ \text{m/s}$
Step 6 — Check: returning V = 2.2286 m/s to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

V = 2.2286\ \text{m/s}

Why the other options are there

4.4572 — kept a factor of two that cancels in the correct rearrangement.
1.1143 — dropped that same factor in the other direction.
2.4515 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 6

Reynolds number — solve for pipe diameter (case 2) — Reynolds Number (6)

the Reynolds number used to check laminar versus turbulent flow Given density (rho) = 1,000 kg/m^3; velocity (V) = 1.5500 m/s; dynamic viscosity (mu) = 0.0008 Pa*s; Reynolds number (Re) = 261,750, determine the pipe diameter (D) in m.

Given

$density (rho) = 1,000 kg/m^3$
$velocity (V) = 1.5500 m/s$
$dynamic viscosity (mu) = 0.0008 Pa*s$
$Reynolds number (Re) = 261,750$

Find

pipe diameter (D), in m

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
Everything except D is given, so isolate D 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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 6 — schematic for Reynolds number — solve for pipe diameter (case 2) — Reynolds Number (6)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that D stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 1,000 kg/m^3, velocity (V) = 1.5500 m/s, dynamic viscosity (mu) = 0.0008 Pa*s, Reynolds number (Re) = 261,750.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$D = 0.1267\ \text{m}$
Step 6 — Check: returning D = 0.1267 m to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

D = 0.1267\ \text{m}

Why the other options are there

0.2533 — kept a factor of two that cancels in the correct rearrangement.
0.0633 — dropped that same factor in the other direction.
0.1393 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 7

Reynolds number — solve for Reynolds number (case 3) — Reynolds Number (7)

the Reynolds number for water flowing through a pipe Given density (rho) = 890.0 kg/m^3; velocity (V) = 0.3000 m/s; pipe diameter (D) = 0.4350 m; dynamic viscosity (mu) = 0.0016 Pa*s, determine the Reynolds number (Re).

Given

$density (rho) = 890.0 kg/m^3$
$velocity (V) = 0.3000 m/s$
$pipe diameter (D) = 0.4350 m$
$dynamic viscosity (mu) = 0.0016 Pa*s$

Find

Reynolds number (Re)

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
Everything except Re is given, so isolate Re 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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 7 — schematic for Reynolds number — solve for Reynolds number (case 3) — Reynolds Number (7)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that Re stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 890.0 kg/m^3, velocity (V) = 0.3000 m/s, pipe diameter (D) = 0.4350 m, dynamic viscosity (mu) = 0.0016 Pa*s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$Re = 72591$
Step 6 — Check: returning Re = 72,591 to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

Re = 72591

Why the other options are there

145,181 — kept a factor of two that cancels in the correct rearrangement.
36,295 — dropped that same factor in the other direction.
79,850 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 8

Reynolds number — solve for velocity (case 3) — Reynolds Number (8)

the Reynolds number of oil in a pipeline Given density (rho) = 900.0 kg/m^3; pipe diameter (D) = 0.0800 m; dynamic viscosity (mu) = 0.0015 Pa*s; Reynolds number (Re) = 127,420, determine the velocity (V) in m/s.

Given

$density (rho) = 900.0 kg/m^3$
$pipe diameter (D) = 0.0800 m$
$dynamic viscosity (mu) = 0.0015 Pa*s$
$Reynolds number (Re) = 127,420$

Find

velocity (V), in m/s

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 8 — schematic for Reynolds number — solve for velocity (case 3) — Reynolds Number (8)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that V stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 900.0 kg/m^3, pipe diameter (D) = 0.0800 m, dynamic viscosity (mu) = 0.0015 Pa*s, Reynolds number (Re) = 127,420.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$V = 2.6546\ \text{m/s}$
Step 6 — Check: returning V = 2.6546 m/s to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

V = 2.6546\ \text{m/s}

Why the other options are there

5.3092 — kept a factor of two that cancels in the correct rearrangement.
1.3273 — dropped that same factor in the other direction.
2.9200 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 9

Reynolds number — solve for pipe diameter (case 3) — Reynolds Number (9)

the Reynolds number used to check laminar versus turbulent flow Given density (rho) = 850.0 kg/m^3; velocity (V) = 1.4000 m/s; dynamic viscosity (mu) = 0.0015 Pa*s; Reynolds number (Re) = 276,150, determine the pipe diameter (D) in m.

Given

$density (rho) = 850.0 kg/m^3$
$velocity (V) = 1.4000 m/s$
$dynamic viscosity (mu) = 0.0015 Pa*s$
$Reynolds number (Re) = 276,150$

Find

pipe diameter (D), in m

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
Everything except D is given, so isolate D 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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 9 — schematic for Reynolds number — solve for pipe diameter (case 3) — Reynolds Number (9)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that D stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 850.0 kg/m^3, velocity (V) = 1.4000 m/s, dynamic viscosity (mu) = 0.0015 Pa*s, Reynolds number (Re) = 276,150.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$D = 0.3365\ \text{m}$
Step 6 — Check: returning D = 0.3365 m to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

D = 0.3365\ \text{m}

Why the other options are there

0.6730 — kept a factor of two that cancels in the correct rearrangement.
0.1682 — dropped that same factor in the other direction.
0.3701 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Example 10

Reynolds number — solve for Reynolds number (case 4) — Reynolds Number (10)

the Reynolds number for water flowing through a pipe Given density (rho) = 1,100 kg/m^3; velocity (V) = 3.5500 m/s; pipe diameter (D) = 0.0400 m; dynamic viscosity (mu) = 0.0017 Pa*s, determine the Reynolds number (Re).

Given

$density (rho) = 1,100 kg/m^3$
$velocity (V) = 3.5500 m/s$
$pipe diameter (D) = 0.0400 m$
$dynamic viscosity (mu) = 0.0017 Pa*s$

Find

Reynolds number (Re)

Start with the thinking

The governing relation printed in this handbook section is Reynolds number.
Everything except Re is given, so isolate Re 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.
The Reynolds number classifies flow through a pipe as laminar or turbulent.

Figure 10 — schematic for Reynolds number — solve for Reynolds number (case 4) — Reynolds Number (10)

Step-by-step solution

Step 1 — State the governing relation:
$Re = \dfrac{\rho V D}{\mu}$
Step 2 — Rearrange the relation so that Re stands alone on the left-hand side.
Step 3 — List the givens: density (rho) = 1,100 kg/m^3, velocity (V) = 3.5500 m/s, pipe diameter (D) = 0.0400 m, dynamic viscosity (mu) = 0.0017 Pa*s.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$Re = 91882$
Step 6 — Check: returning Re = 91,882 to
$Re = \dfrac{\rho V D}{\mu}$
reproduces the given quantities, and both sides carry the same units.

Answer:

Re = 91882

Why the other options are there

183,765 — kept a factor of two that cancels in the correct rearrangement.
45,941 — dropped that same factor in the other direction.
101,071 — rounded an intermediate value before the final step.

Reference: FE Handbook — Reynolds Number

Reynolds Number

Handbook notes for this section

Core formulas for this FE topic

Re=µ=υRe = µ = \upsilonRe=µ=υ

Rel=nRel = nRel=n

3n+1−h3n + 1 ^ - h3n+1−h

v=fluidvelocityv = fluid velocityv=fluidvelocity

ρ=massdensity\rho = mass densityρ=massdensity

D=diameterofthepipe,dimensionofthefluidstreamline,orcharacteristiclengthD = diameter of the pipe, dimension of the fluid streamline, or characteristic lengthD=diameterofthepipe,dimensionofthefluidstreamline,orcharacteristiclength

µ=dynamicviscosityµ = dynamic viscosityµ=dynamicviscosity

y=kinematicviscosityy = kinematic viscosityy=kinematicviscosity

Re=Reynoldsnumber(Newtonianfluid)Re = Reynolds number (Newtonian fluid)Re=Reynoldsnumber(Newtonianfluid)

Re′=Reynoldsnumber(Powerlawfluid)Re' = Reynolds number (Power law fluid)Re′=Reynoldsnumber(Powerlawfluid)

vrh=vmax=1−cRmGv ^ r h = vmax =1 - c R m Gvrh=vmax=1−cRmG

r=distance(m)fromthecenterliner = distance (m) from the centerliner=distance(m)fromthecenterline

R=radius(m)ofthetubeorhalfthedistancebetweentheparallelplanesR = radius (m) of the tube or half the distance between the parallel planesR=radius(m)ofthetubeorhalfthedistancebetweentheparallelplanes

v=localvelocity(m/s)atrv = local velocity (m/s) at rv=localvelocity(m/s)atr

vmax=velocity(m/s)atthecenterlineoftheductvmax = velocity (m/s) at the centerline of the ductvmax=velocity(m/s)atthecenterlineoftheduct

vmax=1.18v,forfullyturbulentflowvmax = 1.18v, for fully turbulent flowvmax=1.18v,forfullyturbulentflow

vmax=2v,forcirculartubesinlaminarflowandvmax = 2v, for circular tubes in laminar flow andvmax=2v,forcirculartubesinlaminarflowand

vmax=1.5v,forparallelplanesinlaminarflow,wherevmax = 1.5v, for parallel planes in laminar flow, wherevmax=1.5v,forparallelplanesinlaminarflow,where

v=averagevelocity(m/s)intheductv = average velocity (m/s) in the ductv=averagevelocity(m/s)intheduct

τw = RNeeds technical review

Worked exam-style examples