Substituted Unsteady Acceleration

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

LT⁻²

Symbol

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

The unsteady acceleration after the similarity substitution — the growth scale times the transient velocity group.

Main equation

Level 0 of 8

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 8. 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.

((γ S4.1 \frac{c x}{( 1 - γ t ) ^{2}}) S4.2 f^{'} + f^{''} (\frac{η}{2}))

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:timestimes.

Open full derivation chain: 5 stepsDerivation

1
Open Step 1: Given / definition · The source relation this block transforms: the rate of change of the Problem 1 axial velocity f…
The source relation this block transforms: the rate of change of the Problem 1 axial velocity function u(x, y, t) with respect to time — the transient term of the momentum balance.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.
$(\frac{\partial u}{\partial t})_{[S5.1]}$
2
Open Step 2: Substitution · Substitute the canonical Problem 1 velocity definition into the derivative before differentiati…
Substitute the canonical Problem 1 velocity definition into the derivative before differentiating: ∂u/∂t = ∂[U_w(x,t)f′(η(y,t))]/∂t.Substitution Substitution · u = U_w f′(η)
Open operand record: Substitution u = U_w f′(η)
Substitution:L0 $u = U_{w} f' (η)$ L1 $u ? = U_{w} f^{'} (η)$
Open term-change ledger: 1 records
Open record 1: u → U_w f′(η)
Consumes:L0uProduces:L0 $U_{w} f' (η)$ The canonical velocity-definition node replaces u(x,y,t) by U_w(x,t)f′(η(y,t)) inside the time derivative; dependence on both U_w and η remains explicit.U_w f′(η)
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.
$(\frac{\partial U _{w} f ^{'} ( η )}{\partial t})_{[S1.1]}$
3
Open Step 3: Product rule · Apply the product rule to U_w f′(η), and the chain rule to f′(η(y,t)): ∂u/∂t = (∂U_w/∂t)f′ + U_…
Apply the product rule to U_w f′(η), and the chain rule to f′(η(y,t)): ∂u/∂t = (∂U_w/∂t)f′ + U_w f″(∂η/∂t).Product rule
Open term-change ledger: 1 records
Open record 1: U_w f′(η) → (∂U_w/∂t)f′ + U_w f″(∂η/∂t)
Consumes:L0 $U_{w} f' (η)$ Produces:L0 $(\partial U_{w} / \partial t) f'$ L0 $U_{w} f ″ (\partial η / \partial t)$ For U_w(x,t)f′(η(y,t)), the product rule gives (∂U_w/∂t)f′ + U_w∂f′/∂t, and the chain rule gives ∂f′/∂t = f″(∂η/∂t).
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.
$((\frac{\partial U _{w}}{\partial t}) f^{'} S2.1 + U_{w} f^{''} (\frac{\partial η}{\partial t}))$
4
Open Step 4: Substitution · Evaluate the derivative definitions using the similarity variable η = y√[c/(ν_f(1−γt))] and U_w…
Evaluate the derivative definitions using the similarity variable η = y√[c/(ν_f(1−γt))] and U_w = cx/(1−γt): ∂η/∂t = γη/[2(1−γt)] and ∂U_w/∂t = γcx/(1−γt)².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_w/∂t)f′ → (γcx/(1−γt)²) f′
Consumes:L0 $(\partial U_{w} / \partial t) f'$ Produces:L0 $(γ c x / (1 - γ t)^{2}) f'$ Substituting ∂U_w/∂t = γcx/(1−γt)² gives the explicit wall-scale contribution γcx f′/(1−γt)².(γcx/(1−γt)²) f′
Open record 2: U_w f″(∂η/∂t) → (γcx/(1−γt)²) (ηf″/2)
Consumes:L0 $U_{w} f ″ (\partial η / \partial t)$ Produces:L0 $(γ c x / (1 - γ t)^{2}) (η f ″/2)$ Substituting U_w = cx/(1−γt) and ∂η/∂t = γη/[2(1−γt)] (from η = y√[c/(ν_f(1−γt))]) gives γcx(ηf″/2)/(1−γt)².(γcx/(1−γt)²) (ηf″/2)
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.
$((\frac{γ c x}{( 1 - γ t ) ^{2}}) f^{'} S3.1 + (\frac{γ c x}{( 1 - γ t ) ^{2}}) (\frac{η f ^{''}}{2}))$
★
Open Step 5: Factoring · Factor the common dimensional multiplier (γcx/(1−γt)²) from both contributions.
Factor the common dimensional multiplier (γcx/(1−γt)²) from both contributions.Factoring Multiplier · cx/(1−γt)²
Open operand record: Multiplier cx/(1−γt)²
Multiplier:L0 $c x / (1 - γ t)^{2}$ L1 $\frac{c ? x}{( 1 - γ t ) ^{2}}_{?}$
Open term-change ledger: 1 records
Open record 1: (γcx/(1−γt)²) f′ + (γcx/(1−γt)²) (ηf″/2) → cx/(1−γt)² + f′+f″(η/2)
Consumes:L0 $(γ c x / (1 - γ t)^{2}) f'$ L0 $(γ c x / (1 - γ t)^{2}) (η f ″/2)$ Produces:L0 $c x / (1 - γ t)^{2}$ L0 $f' + f ″ (η /2)$ Both evaluated terms contain γcx/(1−γt)²; factoring the common growth scale leaves exactly the deterministic unsteady velocity group f′ + (η/2)f″.f′+f″(η/2)
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.
$((γ S4.1 \frac{c x}{( 1 - γ t ) ^{2}}) S4.2 f^{'} + f^{''} (\frac{η}{2}))$

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 Unsteady Acceleration.

Domain Analysis

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

Step-by-step unit check (5 steps)

Step 1 given
Expand checked this step.
Step 2 substitution
Expand checked this step.
Step 3 product_rule
Expand checked this step.
Step 4 substitution
Expand checked this step.
Step 5 factor
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: γ(cx/(1−γt)²)(f′+f″(η/2))

“γ(cx/(1−γt)²)(f′+f″(η/2))” 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 (5 steps)

Step 1 — given

Step 2 — substitution

Step 3 — product rule

Step 4 — substitution

Step 5 — factor

Essence
S1The time derivative after substituting u(x,y,t) = U_w(x,t)f′(η).
S1.1essenceThe rate of change of U_w f′(η) with respect to t.
S2The complete product- and chain-rule expansion, with both time-dependent factors displayed.
S2.1essenceThe sum of (∂U_w/∂t)f′ and U_w f″(∂η/∂t).
S3The two evaluated contributions before their common dimensional factor is collected.
S3.1essenceThe sum of (γcx/(1−γt)²) f′ and (γcx/(1−γt)²) (ηf″/2).
S4The substituted unsteady acceleration: γcx/(1−γt)² times [f′ + (η/2)f″].
S4.1essenceThe product of γ and cx/(1−γt)².
S4.2essenceThe product of (γ · cx/(1−γt)²) and f′+f″(η/2).

Dimension I

Definition

What it is — and what it is not

What it is: S5The rate of change of the Problem 1 axial velocity function u(x, y, t) with respect to time.
S5.1essenceThe rate of change of u with respect to t.

Dimension II

In practice

How to deal with it

No practical guidance recorded yet.

vproductionmain @ 5d8445a3

Encyclopedia

Substituted Unsteady Acceleration

Equation Workspace

Domain Analysis

Mathematical Analysis