Engineering · Fluid mechanics

Navier–Stokes equation for an incompressible Newtonian fluid

The momentum balance with Newton’s viscosity law describes an incompressible fluid of constant density and viscosity.

AdvancedUniversity/collegeBachelor foundationsSubject specialization

Create flashcards or a quiz I already have an account

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Formula

ρ (\partial_{t} v + (v \cdot \nabla) v) = - \nabla p + μ \nabla^{2} v + ρ g, \nabla \cdot v = 0

ρ in kg/m³; p in Pa; μ in Pa·s; v in m/s; g in m/s²

Variables & units

Variables & units
Symbol	Meaning	Unit
$ρ$	Constant fluid density	kg/m³
$v$	Velocity field	m/s
$p$	Pressure field	Pa
$μ$	Dynamic viscosity	Pa·s
$g$	Gravitational acceleration vector	m/s²

LaTeX

\rho(\partial_t\vec v+(\vec v\cdot\nabla)\vec v)=-\nabla p+\mu\nabla^2\vec v+\rho\vec g,\quad\nabla\cdot\vec v=0

QUANTA STUDYquanta-study.de

Derivation & background

This form assumes an incompressible Newtonian fluid with constant ρ and μ; also impose ∇·v=0. It combines local momentum conservation with a linear stress–strain-rate relation. Variable viscosity, compressibility, non-Newtonian behavior or additional body forces require adjusted terms or constitutive models. Boundary and initial conditions are essential; the PDE alone does not determine a unique flow field.

Exam blueprint

Validity range

Derivation steps

The momentum balance with Newton’s viscosity law describes an incompressible fluid of constant density and viscosity.

1State the model assumptions and list quantities with units.
2Substitute into ρ Dv/Dt = −∇p + μ∇²v + ρg; convert quantities to SI first.
3Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh. For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa.

Rearrangements

Base equation for the target quantity

\rho(\partial_t\vec v+(\vec v\cdot\nabla)\vec v)=-\nabla p+\mu\nabla^2\vec v+\rho\vec g,\quad\nabla\cdot\vec v=0

Preserve signs, boundary conditions and units when rearranging.

Task variant

Work through the numerical example for “Navier–Stokes equation for an incompressible Newtonian fluid” and check its unit and magnitude.

Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh.
For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa.

Common mistakes

Substituting a related quantity that shares the same symbol.

Use variable definitions and check: ρ in kg/m³; p in Pa; μ in Pa·s; v in m/s; g in m/s².

Exam context

Place the governing equation for viscous flow, CFD, pipes and boundary layers in context.

Formula cluster

Fluid mechanics

The momentum balance with Newton’s viscosity law describes an incompressible fluid of constant density and viscosity.

Re = ρvL/η = vL/ν

Reynolds number

Q = πr⁴Δp/(8ηL)

Hagen–Poiseuille law for laminar pipe flow

p₁/ρg + α₁v₁²/2g + z₁ + hp = p₂/ρg + α₂v₂²/2g + z₂ + ht + hL

Extended Bernoulli energy equation with head loss

Worked example

Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh. For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa.

Applications

Place the governing equation for viscous flow, CFD, pipes and boundary layers in context.

How do you calculate with Navier–Stokes equation for an incompressible Newtonian fluid?

Here is how to work through a typical Navier–Stokes equation for an incompressible Newtonian fluid (ρ Dv/Dt = −∇p + μ∇²v + ρg) task step by step:

1
Task
Work through the numerical example for “Navier–Stokes equation for an incompressible Newtonian fluid” and check its unit and magnitude.
Solution path
1. Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh.
2. For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa.

Frequently asked questions about Navier–Stokes equation for an incompressible Newtonian fluid

What does the convective term v·∇v represent?+

A fluid particle can change velocity while moving through a spatially varying velocity field even when the field is steady in time. The term v·∇v captures that change along the particle path. Steady flow therefore does not imply zero acceleration or zero inertia. Assumptions: constant density and viscosity, incompressible Newtonian fluid; initial and boundary conditions are part of the problem. Application: Place the governing equation for viscous flow, CFD, pipes and boundary layers in context. The numerical example Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh. For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa anchors the idea in a concrete measurement case.

Why is continuity also required?+

The incompressible momentum equation must be coupled to ∇·v=0. This constrains the velocity field and couples pressure to velocity. Without continuity, a necessary mass-conservation condition is missing; momentum alone is not a complete flow model. Assumptions: constant density and viscosity, incompressible Newtonian fluid; initial and boundary conditions are part of the problem. The corresponding units are ρ in kg/m³; p in Pa; μ in Pa·s; v in m/s; g in m/s². Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh. For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa. This checks that the reference quantities are compatible.

When may I use μ∇²v?+

For constant μ and incompressible flow, the divergence of Newtonian viscous stress simplifies to μ∇²v. With variable viscosity, compressibility or a different rheology, this compact form does not remain unchanged. The viscous stress term must then be derived from the appropriate constitutive law. Assumptions: constant density and viscosity, incompressible Newtonian fluid; initial and boundary conditions are part of the problem. Scope: This form assumes an incompressible Newtonian fluid with constant ρ and μ; also impose ∇·v=0. It combines local momentum conservation with a linear stress–strain-rate relation. Variable viscosity, compressibility, non-Newtonian behavior or additional body forces require adjusted terms or constitutive models. Boundary and initial conditions are essential; the PDE alone does not determine a unique flow field. A practical application is Place the governing equation for viscous flow, CFD, pipes and boundary layers in context.

How do I check the result and units for Navier–Stokes equation for an incompressible Newtonian fluid?+

Write the quantity and unit from the variable list next to every value: ρ in kg/m³; p in Pa; μ in Pa·s; v in m/s; g in m/s². Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh. For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. Assumptions: constant density and viscosity, incompressible Newtonian fluid; initial and boundary conditions are part of the problem.

Which model limit matters for Navier–Stokes equation for an incompressible Newtonian fluid?+

The background states this starting condition: This form assumes an incompressible Newtonian fluid with constant ρ and μ; also impose ∇·v=0. It combines local. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. Assumptions: constant density and viscosity, incompressible Newtonian fluid; initial and boundary conditions are part of the problem.

Quanta exam set

Curated exam set for "Navier–Stokes equation for an incompressible Newtonian fluid":

Question (front)

Which formula describes Navier–Stokes equation for an incompressible Newtonian fluid?

Answer in your set

Question (front)

How do you rearrange Navier–Stokes equation for an incompressible Newtonian fluid for Base equation for the target quantity?

Answer in your set

Question (front)

Which common mistake happens with Navier–Stokes equation for an incompressible Newtonian fluid?

Answer in your set

+ 8 more cards: units, variables, derivation, example, exam task

The formula set contains 11 cards. After importing, review them and start practicing; your ratings determine later review intervals.

Create flashcards or a quiz I already have an account

Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]MIT OpenCourseWare, 2.20 Marine Hydrodynamics, Lecture 4: Navier–Stokes equations for incompressible Newtonian fluid.

More formulas: Fluid and flow engineering

Q=A\bar v,\qquad A_1\bar v_1=A_2\bar v_2

Volume flow rate and continuity for incompressible flow

\dot m=\rho A\bar v

Mass flow rate through a section

D_h=\frac{4A}{P_\mathrm{benetzt}}

Hydraulic diameter

h_f=f_D\frac{L}{D}\frac{\bar v^2}{2g}

Darcy–Weisbach equation for pipe-friction loss

Spaced-repetition apps

Compare scheduling workflows

Study evidence

Read methods, results and limits

Compare flashcard apps

FSRS support and study features

Create a formula exam set

From Essential in Quanta

Retain Navier–Stokes equation for an incompressible Newtonian fluid for exams

Create a curated FSRS exam set for ρ Dv/Dt = −∇p + μ∇²v + ρg: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Create flashcards or a quiz Already registered? Continue directly

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Quanta Study editorial team

Navier–Stokes equation for an incompressible Newtonian fluid

The momentum balance with Newton’s viscosity law describes an incompressible fluid of constant density and viscosity.

AdvancedUniversity/collegeBachelor foundationsSubject specialization

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

ρ

Constant fluid density

kg/m³

v

Velocity field

m/s

p

Pressure field

μ

Dynamic viscosity

Pa·s

g

Gravitational acceleration vector

m/s²

Derivation & background

Exam blueprint

Validity range

Derivation steps

The momentum balance with Newton’s viscosity law describes an incompressible fluid of constant density and viscosity.

1State the model assumptions and list quantities with units.
2Substitute into ρ Dv/Dt = −∇p + μ∇²v + ρg; convert quantities to SI first.
3Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh. For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa.

How do you calculate with Navier–Stokes equation for an incompressible Newtonian fluid?

Here is how to work through a typical Navier–Stokes equation for an incompressible Newtonian fluid (ρ Dv/Dt = −∇p + μ∇²v + ρg) task step by step:

Task

Work through the numerical example for “Navier–Stokes equation for an incompressible Newtonian fluid” and check its unit and magnitude.

Solution path

Hydrostatic check with z upward, v=0 and g⃗=−g e_z gives −∂p/∂z−ρg=0, hence Δp=ρgΔh.
For water over 2.0 m, Δp≈1000·9.81·2=19.6 kPa.

Frequently asked questions about Navier–Stokes equation for an incompressible Newtonian fluid

What does the convective term v·∇v represent?+

Why is continuity also required?+

When may I use μ∇²v?+

How do I check the result and units for Navier–Stokes equation for an incompressible Newtonian fluid?+

Which model limit matters for Navier–Stokes equation for an incompressible Newtonian fluid?+

Retain Navier–Stokes equation for an incompressible Newtonian fluid for exams

Create a curated FSRS exam set for ρ Dv/Dt = −∇p + μ∇²v + ρg: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential