Knowledge card

Momentum Equation

Mathematical Modelling — Fluid Dynamics & Heat TransferContinuum mechanics; boundary-layer similarity theory

LT⁻²

Symbol

ΣF

The boundary-layer momentum balance combining unsteady, convective, Powell–Eyring, buoyancy, and Lorentz terms.

Main equation

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To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.

((\frac{\partial u}{\partial t} S5.1 + u \cdot \nabla u) S5.4 = ((τ S5.2 + ρ g β) S5.3 - J \times B))

Dimension:LT⁻² (m s⁻²)Maths:𝔹 equation(·, ·, ·, ·) — A statement that two sides are equal, depending on another quantity, another quantity, another quantity, another quantityDimensionally verifiedMathematically verifiedFully reduced to base quantities

In words:plusequalsplusminus.

Open full derivation chain: 3 stepsDerivation

1
Open Step 1: Given / definition · Start from the x-momentum boundary-layer balance after the standard thin-shear-layer assumption…
Start from the x-momentum boundary-layer balance after the standard thin-shear-layer assumptions: local inertia plus convective inertia equals density-scaled shear-stress divergence, inclined thermal buoyancy, and magnetic damping. This is the conceptual momentum PDE before substituting a constitutive law.Given / definition
Level 0 · more available
Brief — named quantities
About levels
Why the levels load one at a time
Only level 0 travels with this article. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them — and how deep it goes is not known until you get there.
To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.
How many levels remain is not yet known — the depth is discovered one level at a time.
$((\frac{\partial u}{\partial t} S4.1 + u \cdot \nabla u) S4.4 = ((\frac{( \frac{1}{ρ _{hn f}} ) \partial τ _{x y}}{\partial y} S4.2 + ρ g β) S4.3 - J \times B))$
2
Open Step 2: Substitution · Substitute the Taylor-truncated Powell–Eyring shear stress into the same balance: τ_PE = (μ + 1…
Substitute the Taylor-truncated Powell–Eyring shear stress into the same balance: τ_PE = (μ + 1/(βς))∂u/∂y − (1/(6βς³))(∂u/∂y)³. The only change is replacing τ_xy by its constitutive expression; the equality is unchanged.Substitution Substitution · Powell–Eyring Rheology
Open term-change ledger: 1 records
Open record 1: (1/ρ_hnf)∂τ_xy/∂y → (1/ρ_hnf)∂τ_PE/∂y
Consumes:L0 $(1/ ρ_{h} n f) \partial τ_{x} y / \partial y$ Produces:L0 $(1/ ρ_{h} n f) \partial τ_{P} E / \partial y$ The generic shear stress τ_xy is replaced by its equal, the truncated Powell–Eyring constitutive expression τ_PE = (μ + 1/(βς))∂u/∂y − (1/(6βς³))(∂u/∂y)³; the density-scaled divergence keeps the same place in the balance.(1/ρ_hnf)∂τ_PE/∂y
Level 0 · more available
Brief — named quantities
About levels
Why the levels load one at a time
Only level 0 travels with this article. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them — and how deep it goes is not known until you get there.
To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.
How many levels remain is not yet known — the depth is discovered one level at a time.
$((\frac{\partial u}{\partial t} S2.1 + u \cdot \nabla u) S2.4 = ((\frac{( \frac{1}{ρ _{hn f}} ) \partial τ _{P E}}{\partial y} S2.2 + ρ g β) S2.3 - J \times B))$
★
Open Step 3: Chain rule · Differentiate the truncated stress with respect to y and divide by ρ_hnf: the linear part gives…
Differentiate the truncated stress with respect to y and divide by ρ_hnf: the linear part gives (ν_hnf + 1/(ρ_hnf βς))∂²u/∂y², while d[(∂u/∂y)³]/dy = 3(∂u/∂y)²∂²u/∂y² changes 1/6 into 1/2. This produces the explicit Momentum PDE used as equation (3.24).Chain rule
Open term-change ledger: 1 records
Open record 1: (1/ρ_hnf)∂τ_PE/∂y → τ
Consumes:L0 $(1/ ρ_{h} n f) \partial τ_{P} E / \partial y$ Produces:L0 $τ$ Differentiating the truncated stress in y: the linear part yields (ν_hnf + 1/(ρ_hnf βς))∂²u/∂y², and d[(∂u/∂y)³]/dy = 3(∂u/∂y)²·∂²u/∂y² converts the 1/6 coefficient into 1/2 — the explicit Powell–Eyring diffusion and retardation terms.τ
Level 0 · more available
Brief — named quantities
About levels
Why the levels load one at a time
Only level 0 travels with this article. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them — and how deep it goes is not known until you get there.
To go deeper still, open any quantity in the reading as its own concept card — it carries its own derivation and its own depth ladder.
How many levels remain is not yet known — the depth is discovered one level at a time.
$((\frac{\partial u}{\partial t} S5.1 + u \cdot \nabla u) S5.4 = ((τ S5.2 + ρ g β) S5.3 - J \times B))$

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Step-by-step unit check (3 steps)

Step 1 given
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Step 2 substitution
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Step 3 chain_rule
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Reads as: ΣF

“ΣF” is analysed by expanding its full defining equation (its children and their relations), not as a bare symbol. Use Check this concept’s units above for the same result.

Mathematical Analysis

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Check one step at a time (3 steps)

Step 1 — given

Step 2 — substitution

Step 3 — chain rule

Equation

Essence
S1Unsteady + convective acceleration balanced by Powell–Eyring viscous diffusion, buoyancy, and the Lorentz force. [2.1]
S2The same momentum balance after replacing the generic shear stress by the truncated Powell–Eyring stress.
S2.1essenceThe sum of ∂u/∂t and u·∇u. [2.1]
S2.2essenceThe sum of (1/ρ_hnf)∂τ_PE/∂y and ρgβ. [2.1]
S2.3essence((1/ρ_hnf)∂τ_PE/∂y + ρgβ) reduced by J×B. [2.1]
S2.4essence(∂u/∂t + u·∇u) set equal to (((1/ρ_hnf)∂τ_PE/∂y + ρgβ) − J×B). [2.1]

Dimension I

Definition

What it is — and what it is not

What it is: S3The streamwise momentum balance in the boundary layer.
S4The conceptual x-momentum boundary-layer balance: inertia equals shear-stress divergence plus buoyancy minus Lorentz damping.
S4.1essenceThe sum of ∂u/∂t and u·∇u. [2.1]
S4.2essenceThe sum of (1/ρ_hnf)∂τ_xy/∂y and ρgβ. [2.1]
S4.3essence((1/ρ_hnf)∂τ_xy/∂y + ρgβ) reduced by J×B. [2.1]
S4.4essence(∂u/∂t + u·∇u) set equal to (((1/ρ_hnf)∂τ_xy/∂y + ρgβ) − J×B). [2.1]
Wisdoms: S5The final explicit Momentum PDE: inertia balances Powell–Eyring diffusion and retardation, buoyancy, and Lorentz damping.
S5.1essenceThe sum of ∂u/∂t and u·∇u. [2.1]
S5.2essenceThe sum of τ and ρgβ. [2.1]
S5.3essence(τ + ρgβ) reduced by J×B. [2.1]
S5.4essence(∂u/∂t + u·∇u) set equal to ((τ + ρgβ) − J×B). [2.1]

Dimension II

In practice

How to deal with it

Consequences: S6The buoyancy term couples it to the energy equation through the temperature θ.

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