Substituted Unsteady Heating

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

T⁻¹Θ

Symbol

γ (Δ T_{0} x / (1 - γ t)^{3}) (2 θ + η θ'/2)

The unsteady heating after substitution — the independent heating amplitude and transient rate times the temperature 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{Δ T _{0} x}{( 1 - γ t ) ^{3}}) S4.2 \frac{2 θ + η θ ^{'}}{2})

Dimension:T⁻¹Θ (s⁻¹ K)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 temperature fiel…
The source relation this block transforms: the rate of change of the Problem 1 temperature field T(x, y, t) with respect to time — the transient term of the thermal balance.Given / definition
$(\frac{\partial T}{\partial t})_{[S5.1]}$
2
Open Step 2: Substitution · Substitute T=T∞+(T_r−T∞)θ(η), where T_r−T∞=ΔT₀x/(1−γt)²; then ∂T/∂t=∂[(T_r−T∞)θ]/∂t.
Substitute T=T∞+(T_r−T∞)θ(η), where T_r−T∞=ΔT₀x/(1−γt)²; then ∂T/∂t=∂[(T_r−T∞)θ]/∂t.Substitution Substitution · θ
Open operand record: Substitution θ
Substitution:L0 $θ$ L1 $\frac{T - T \infty}{T r - T \infty}_{?}$
Open term-change ledger: 1 records
Open record 1: T → (Tr−T∞)θ
Consumes:L0TProduces:L0 $(T r - T \infty) θ$ The corrected temperature map leaves (T_r−T∞)θ after the constant T∞ differentiates to zero.(Tr−T∞)θ
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.
$(\frac{\partial ( T r - T \infty ) θ}{\partial t})_{[S1.1]}$
3
Open Step 3: Product rule · Apply product and chain rules: ∂T/∂t=[∂(T_r−T∞)/∂t]θ+(T_r−T∞)θ′∂η/∂t.
Apply product and chain rules: ∂T/∂t=[∂(T_r−T∞)/∂t]θ+(T_r−T∞)θ′∂η/∂t.Product rule
Open term-change ledger: 1 records
Open record 1: (Tr−T∞)θ → (∂(Tr−T∞)/∂t)θ + (Tr−T∞)θ′(∂η/∂t)
Consumes:L0 $(T r - T \infty) θ$ Produces:L0 $(\partial (T r - T \infty) / \partial t) θ$ L0 $(T r - T \infty) θ' (\partial η / \partial t)$ Both the prescribed reference excess and η depend on time, so both contributions are required.
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.
$((\frac{\partial ( T r - T \infty )}{\partial t}) θ S2.1 + (T r - T \infty) θ^{'} (\frac{\partial η}{\partial t}))$
4
Open Step 4: Substitution · With η=y√[c/(ν_f(1−γt))] and T_r−T∞=ΔT₀x/(1−γt)², evaluate ∂η/∂t=γη/[2(1−γt)] and ∂(T_r−T∞)/∂t=…
With η=y√[c/(ν_f(1−γt))] and T_r−T∞=ΔT₀x/(1−γt)², evaluate ∂η/∂t=γη/[2(1−γt)] and ∂(T_r−T∞)/∂t=2γΔT₀x/(1−γt)³.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: (∂(Tr−T∞)/∂t)θ → (γΔT₀x/(1−γt)³)(2θ)
Consumes:L0 $(\partial (T r - T \infty) / \partial t) θ$ Produces:L0 $(γ Δ T_{0} x / (1 - γ t)^{3}) (2 θ)$ The reference-excess derivative contributes [γΔT₀x/(1−γt)³]2θ.(γΔT₀x/(1−γt)³)(2θ)
Open record 2: (Tr−T∞)θ′(∂η/∂t) → (γΔT₀x/(1−γt)³)(ηθ′/2)
Consumes:L0 $(T r - T \infty) θ' (\partial η / \partial t)$ Produces:L0 $(γ Δ T_{0} x / (1 - γ t)^{3}) (η θ'/2)$ Substituting the reference excess and ∂η/∂t contributes [γΔT₀x/(1−γt)³](ηθ′/2).(γΔT₀x/(1−γt)³)(ηθ′/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.
$((\frac{γ Δ T _{0} x}{( 1 - γ t ) ^{3}}) (2 θ) S3.1 + (\frac{γ Δ T _{0} x}{( 1 - γ t ) ^{3}}) (\frac{η θ ^{'}}{2}))$
★
Open Step 5: Factoring · Factor γΔT₀x/(1−γt)³ from both contributions.
Factor γΔT₀x/(1−γt)³ from both contributions.Factoring Multiplier · ΔT₀x/(1−γt)³
Open operand record: Multiplier ΔT₀x/(1−γt)³
Multiplier:L0 $Δ T_{0} x / (1 - γ t)^{3}$ L1 $\frac{Δ T _{0} ? x}{( 1 - γ t ) ^{3}}_{?}$
Open term-change ledger: 1 records
Open record 1: (γΔT₀x/(1−γt)³)(2θ) + (γΔT₀x/(1−γt)³)(ηθ′/2) → ΔT₀x/(1−γt)³ + 2θ+ηθ′/2
Consumes:L0 $(γ Δ T_{0} x / (1 - γ t)^{3}) (2 θ)$ L0 $(γ Δ T_{0} x / (1 - γ t)^{3}) (η θ'/2)$ Produces:L0 $Δ T_{0} x / (1 - γ t)^{3}$ L0 $2 θ + η θ'/2$ Factoring the shared γΔT₀x/(1−γt)³ leaves 2θ+(η/2)θ′.2θ+ηθ′/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.
$((γ S4.1 \frac{Δ T _{0} x}{( 1 - γ t ) ^{3}}) S4.2 \frac{2 θ + η θ ^{'}}{2})$

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Domain Analysis

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Step-by-step unit check (5 steps)

Step 1 given
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Step 2 substitution
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Step 3 product_rule
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Step 4 substitution
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Step 5 factor
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Reads as: γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2)

“γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/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 acts on (T_r−T∞)θ.
S1.1essenceThe rate of change of (Tr−T∞)θ with respect to t.
S2The complete product- and chain-rule expansion, with both time-dependent factors displayed.
S2.1essenceThe sum of (∂(Tr−T∞)/∂t)θ and (Tr−T∞)θ′(∂η/∂t).
S3The two evaluated contributions before their common thermal growth scale is collected.
S3.1essenceThe sum of (γΔT₀x/(1−γt)³)(2θ) and (γΔT₀x/(1−γt)³)(ηθ′/2).
S4The substituted unsteady heating: [γΔT₀x/(1−γt)³][2θ+(η/2)θ′].
S4.1essenceThe product of γ and ΔT₀x/(1−γt)³.
S4.2essenceThe product of (γ · ΔT₀x/(1−γt)³) and 2θ+ηθ′/2.

Dimension I

Definition

What it is — and what it is not

What it is: S5The rate of change of the Problem 1 temperature field T(x, y, t) with respect to time.
S5.1essenceThe rate of change of T with respect to t.

Dimension II

In practice

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Substituted Unsteady Heating

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