Knowledge card

Substituted Convective Acceleration

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

LT⁻²

Symbol

(c^{2} x / (1 - γ t)^{2}) (f'^{2} - f f ″)

The convective acceleration after the similarity substitution — its streamwise (f′²) and wall-normal (ff″) parts grouped as ONE block, so the balance keeps the same convective-acceleration concept it had before substitution. Derives from the convective acceleration.

Main equation

Level 0 of 5

Brief — named quantities

About levels

Why the levels load one at a time

Only level 0 travels with this article; the expansion runs to level 5. The complete expansion is several megabytes, so deeper levels are fetched one at a time when you ask for them.

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{c ^{2} x}{( 1 - γ t ) ^{2}}) f^{' 2} S2.1 - (\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) f f^{''})

Dimension:LT⁻² (m s⁻²)Maths:ℝ scalar(·, ·, ·) — A single number (real numbers), depending on another quantity, another quantity, another quantityDimensionally verifiedMathematically verifiedFully reduced to base quantities

In words:minus.

Open full derivation chain: 3 stepsDerivation

1
Open Step 1: Given / definition · The source relation this block transforms: the advective transport of axial momentum carried by…
The source relation this block transforms: the advective transport of axial momentum carried by the flow itself — u∂u/∂x plus v∂u/∂y.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.
$((u S3.1 \frac{\partial u}{\partial x}) S3.3 + (v S3.2 \frac{\partial u}{\partial y}))$
2
Open Step 2: Substitution · Insert the velocity and velocity-gradient similarity maps into both advective products: u = U_w…
Insert the velocity and velocity-gradient similarity maps into both advective products: u = U_w f′(η), v = −√(cν_f/(1−γt))f(η), ∂u/∂x = (c/(1−γt))f′ (η carries no x-dependence), ∂u/∂y = U_w√(c/(ν_f(1−γt)))f″. Each product is now a raw product of substituted factors, not yet combined.Substitution Substitution · η
Open operand record: Substitution η
Substitution:L0 $η$ L2 $(\frac{c}{ν _{f} ( 1 - γ t )}) \sqrt ? y$
Open term-change ledger: 2 records
Open record 1: ∂u/∂x → U_w f′·(c/(1−γt))f′
Consumes:L0 $\partial u / \partial x$ Produces:L0 $U_{w} f' \cdot (c / (1 - γ t)) f'$ The streamwise advection u∂u/∂x becomes (U_w f′)·((c/(1−γt))f′): u = U_w f′ and ∂u/∂x = (c/(1−γt))f′, since U_w = cx/(1−γt) depends on x while η does not.U_w f′·(c/(1−γt))f′
Open record 2: ∂u/∂y → −√(cν_f/(1−γt))f·U_w√(c/(ν_f(1−γt)))f″
Consumes:L0 $\partial u / \partial y$ Produces:L0 $- \sqrt (c ν_{f} / (1 - γ t)) f \cdot U_{w} \sqrt (c / (ν_{f} (1 - γ t))) f ″$ The wall-normal advection v∂u/∂y becomes −√(cν_f/(1−γt))f · U_w√(c/(ν_f(1−γt)))f″: v = −√(cν_f/(1−γt))f and the shear-rate map ∂u/∂y = U_w√(c/(ν_f(1−γt)))f″ (the minus sign of v makes the part enter subtracted).−√(cν_f/(1−γt))f·U_w√(c/(ν_f(1−γt)))f″
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.
$(U_{w} f^{'} \cdot (\frac{c}{1 - γ t}) f^{'} S1.1 - - \sqrt (\frac{c ν _{f}}{1 - γ t}) f \cdot U_{w} \sqrt (\frac{c}{ν _{f} ( 1 - γ t )}) f^{''})$
★
Open Step 3: Simplification · Combine the dimensional factors of each product. Streamwise: U_w·(c/(1−γt)) = (cx/(1−γt))·(c/(1…
Combine the dimensional factors of each product. Streamwise: U_w·(c/(1−γt)) = (cx/(1−γt))·(c/(1−γt)) = c²x/(1−γt)², giving c²x/(1−γt)²·f′². Wall-normal: √(cν_f/(1−γt))·U_w√(c/(ν_f(1−γt))) = (c/(1−γt))·(cx/(1−γt)) = c²x/(1−γt)², giving c²x/(1−γt)²·f f″.Simplification
Open term-change ledger: 2 records
Open record 1: U_w f′·(c/(1−γt))f′ → (c²x/(1−γt)²)f′²
Consumes:L0 $U_{w} f' \cdot (c / (1 - γ t)) f'$ Produces:L0 $(c^{2} x / (1 - γ t)^{2}) f'^{2}$ Combining U_w·(c/(1−γt)) = c²x/(1−γt)² leaves the streamwise term c²x/(1−γt)²·f′².(c²x/(1−γt)²)f′²
Open record 2: −√(cν_f/(1−γt))f·U_w√(c/(ν_f(1−γt)))f″ → (c²x/(1−γt)²)ff″
Consumes:L0 $- \sqrt (c ν_{f} / (1 - γ t)) f \cdot U_{w} \sqrt (c / (ν_{f} (1 - γ t))) f ″$ Produces:L0 $(c^{2} x / (1 - γ t)^{2}) f f ″$ Combining √(cν_f/(1−γt))·U_w√(c/(ν_f(1−γt))) = c²x/(1−γt)² leaves the wall-normal term c²x/(1−γt)²·f f″.(c²x/(1−γt)²)ff″
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.
$((\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) f^{' 2} S2.1 - (\frac{c ^{2} x}{( 1 - γ t ) ^{2}}) f f^{''})$

Equation Workspace

Set known values, calculate unknowns, inspect dependencies, and run numerical solves.

Calculator

Equation workspace

Set the values you know, then calculate the remaining quantities in Substituted Convective Acceleration.

Domain Analysis

Unit status, dimensional trails, per-step checks, and custom unit expressions.

Step-by-step unit check (3 steps)

Step 1 given
Expand checked this step.
Step 2 substitution
Expand checked this step.
Step 3 simplify
Expand checked this step.

Try your own — the unit calculator

Ever wondered whether a formula “adds up”? This tool answers one plain question: do the units match? Pick a couple of quantities (distance, time, a speed…), join them with ×, ÷, +, or =, and press Run. It works out the resulting unit — for example distance ÷ time gives a speed, LT⁻¹ (metres per second) — and flags anything that can’t be right, like adding a length to a time.

Reads as: (c²x/(1−γt)²)(f′²−ff″)

“(c²x/(1−γt)²)(f′²−ff″)” 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

Object type, lawful operations, conditions, comparisons, and derivation-step checks.

Check one step at a time (3 steps)

Step 1 — given

Step 2 — substitution

Step 3 — simplify

Essence
S1The convective products after the similarity maps are inserted, before their dimensional factors are combined: (U_w f′)(c/(1−γt)f′) − √(cν_f/(1−γt))f · U_w√(c/(ν_f(1−γt)))f″.
S1.1essenceU_w f′·(c/(1−γt))f′ reduced by −√(cν_f/(1−γt))f·U_w√(c/(ν_f(1−γt)))f″.
S2The substituted convective acceleration: the convective scale c²x/(1−γt)² times f′² (streamwise) minus the same scale times f f″ (wall-normal).
S2.1essence(c²x/(1−γt)²)f′² reduced by (c²x/(1−γt)²)ff″.

Dimension I

Definition

What it is — and what it is not

What it is: S3The advective transport of axial momentum carried by the flow itself: u∂u/∂x + v∂u/∂y.
S3.1essenceThe product of u and ∂u/∂x.
S3.2essenceThe product of v and ∂u/∂y.
S3.3essenceThe sum of (u · ∂u/∂x) and (v · ∂u/∂y).

Dimension II

In practice

How to deal with it

No practical guidance recorded yet.

vproductionmain @ 5d8445a3

Encyclopedia

Substituted Convective Acceleration

Equation Workspace

Domain Analysis

Mathematical Analysis