Knowledge card

Energy ODE

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

Symbol

θ ″

The second-order similarity-reduced energy equation; its formula is composed on demand from the conduction–radiation, convective, and unsteady 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.

(θ^{''} S3.4 = (\frac{( ( A ( \frac{2 θ + η θ ^{'}}{2} ) S3.1 - f θ ^{'} ) S3.2 + f ^{'} θ )}{k _{1}})_{[S3.3]})

Dimension:1Maths:𝔹 equation(·) — A statement that two sides are equal, depending on another quantityDimensionally verifiedMathematically verifiedFully reduced to base quantities

In words:equals(minusplus)divided by.

Open full derivation chain: 5 stepsDerivation

1
Open Step 1: Given / definition · The boundary-layer energy PDE [proposal eq. 3.25]: unsteady plus convective heat transport equa…
The boundary-layer energy PDE [proposal eq. 3.25]: unsteady plus convective heat transport equals the effective thermal diffusion — conduction and the radiative-flux divergence gathered as one diffusion block (the Rosseland-linearised radiation acts as an extra conduction).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.
$((\frac{\partial T}{\partial t} S5.1 + u \cdot \nabla T) S5.2 = α \partial^{2} T / \partial y^{2} - (\frac{1}{ρC p}) \partial q_{r} / \partial y)$
2
Open Step 2: Similarity transformation · Substitute the maps into EVERY block [eqs. 3.56–3.57]: T−T∞ = (T_r−T∞)θ with T_r−T∞ = ΔT₀x/(1−γ…
Substitute the maps into EVERY block [eqs. 3.56–3.57]: T−T∞ = (T_r−T∞)θ with T_r−T∞ = ΔT₀x/(1−γt)², plus the velocity and temperature-derivative maps. ΔT₀ (K/m) is independent of the stretching rate c (1/s). Because the reference excess grows with x, u∂T/∂x survives as f′θ.Similarity transformation Substitution · η
Open operand record: Substitution η
Substitution:L0 $η$ L2 $(\frac{c}{ν _{f} ( 1 - γ t )}) \sqrt ? y$
Open term-change ledger: 3 records
Open record 1: ∂T/∂t → γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2)
Consumes:L0 $\partial T / \partial t$ L1 $\frac{\partial ( T r - T \infty ) θ}{\partial t}_{?}$ Produces:L0 $γ (Δ T_{0} x / (1 - γ t)^{3}) (2 θ + η θ'/2)$ L1 $((γ i \frac{Δ T _{0} x}{( 1 - γ t ) ^{3}}) i \frac{2 θ + η θ ^{'}}{2})$ The similarity transform inserts T−T∞ = (T_r−T∞)θ(η) into ∂T/∂t; differentiation gives γΔT₀x/(1−γt)³ times 2θ+ηθ′/2.γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2)
Open record 2: u·∇T → (cΔT₀x/(1−γt)³)(f′θ−fθ′)
Consumes:L0 $u \cdot \nabla T$ L1 $u ? \frac{\partial T}{\partial x} ? + v ? \frac{\partial T}{\partial y}$ Produces:L0 $(c Δ T_{0} x / (1 - γ t)^{3}) (f' θ - f θ')$ L1 $((\frac{c Δ T _{0} x}{( 1 - γ t ) ^{3}}) f^{'} θ i - (\frac{c Δ T _{0} x}{( 1 - γ t ) ^{3}}) f θ^{'})$ The similarity transform inserts u=U_wf′ and T−T∞=(T_r−T∞)θ into both advection terms; because T_r−T∞=ΔT₀x/(1−γt)² grows with x, u∂T/∂x survives as f′θ.(cΔT₀x/(1−γt)³)(f′θ−fθ′)
Open record 3: α∂²T/∂y²−(1/ρCp)∂q_r/∂y → k̃_eff[ΔT₀x/(1−γt)²][c/(ν_f(1−γt))]θ″
Consumes:L0 $α \partial^{2} T / \partial y^{2} - (1/ ρC p) \partial q_{r} / \partial y$ L1 $k \nabla T ? - q_{r}$ Produces:L0 $\tilde{k}_{e} f f [Δ T_{0} x / (1 - γ t)^{2}] [c / (ν_{f} (1 - γ t))] θ ″$ L1 $((\frac{k}{ρC p})_{hn f} (\frac{Δ T _{0} x}{( 1 - γ t ) ^{2}}) (\frac{c}{ν _{f} ( 1 - γ t )}) θ^{''} i + (\frac{1}{( ρC p ) _{hn f}}) (\frac{16 σ * T \infty ^{3}}{3 k *}) (\frac{Δ T _{0} x}{( 1 - γ t ) ^{2}}) (\frac{c}{ν _{f} ( 1 - γ t )}) θ^{''})$ The similarity transform inserts ∂²T/∂y²=(T_r−T∞)[c/(ν_f(1−γt))]θ″ into both conduction and linearised Rosseland diffusion.k̃_eff[ΔT₀x/(1−γt)²][c/(ν_f(1−γt))]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.
$((γ (\frac{Δ T _{0} x}{( 1 - γ t ) ^{3}}) (\frac{2 θ + η θ ^{'}}{2}) S1.1 + (\frac{c Δ T _{0} x}{( 1 - γ t ) ^{3}}) (f^{'} θ - f θ^{'})) S1.2 = \tilde{k}_{e f f} [Δ T_{0} x / (1 - γ t)^{2}] [c / (ν_{f} (1 - γ t))] θ^{''})$
3
Open Step 3: Rearrange terms · Move every term to one side and set the equation to zero [eq. 3.60], the standard homogeneous f…
Move every term to one side and set the equation to zero [eq. 3.60], the standard homogeneous form: subtracting the effective-diffusion block collects the whole thermal balance against zero — the SAME three blocks, now equated to nought.Rearrange terms
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.
$(((γ (\frac{Δ T _{0} x}{( 1 - γ t ) ^{3}}) (\frac{2 θ + η θ ^{'}}{2}) S2.1 + (\frac{c Δ T _{0} x}{( 1 - γ t ) ^{3}}) (f^{'} θ - f θ^{'})) S2.2 - \tilde{k}_{e f f} [Δ T_{0} x / (1 - γ t)^{2}] [c / (ν_{f} (1 - γ t))] θ^{''}) S2.3 = 0)$
4
Open Step 4: Non-dimensionalisation · Divide by the thermal scale cΔT₀x/(1−γt)³, multiply the homogeneous balance by −1, and identify…
Divide by the thermal scale cΔT₀x/(1−γt)³, multiply the homogeneous balance by −1, and identify γ/c ≡ A and k₁ = (1/Pr)(1/φ_Cp)(φ_k + 4Rd/3). The canonical result is k₁θ″ + fθ′ − f′θ − A(2θ+ηθ′/2) = 0.Non-dimensionalisation Divisor · cΔT₀x/(1−γt)³
Open operand record: Divisor cΔT₀x/(1−γt)³
Divisor:L0 $c Δ T_{0} x / (1 - γ t)^{3}$ L1 $\frac{c ? Δ T _{0} ? x}{( 1 - γ t ) ^{3}}_{?}$
Open term-change ledger: 3 records
Open record 1: γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2) → A(2θ+ηθ′/2)
Consumes:L0 $γ (Δ T_{0} x / (1 - γ t)^{3}) (2 θ + η θ'/2)$ L1 $((γ i \frac{Δ T _{0} x}{( 1 - γ t ) ^{3}}) i \frac{2 θ + η θ ^{'}}{2})$ Produces:L0 $A (2 θ + η θ'/2)$ L1 $(A i \frac{2 θ + η θ ^{'}}{2})$ Dividing γΔT₀x(2θ+ηθ′/2)/(1−γt)³ by cΔT₀x/(1−γt)³ leaves γ/c=A.A(2θ+ηθ′/2)
Open record 2: (cΔT₀x/(1−γt)³)(f′θ−fθ′) → fθ′−f′θ
Consumes:L0 $(c Δ T_{0} x / (1 - γ t)^{3}) (f' θ - f θ')$ L1 $((\frac{c Δ T _{0} x}{( 1 - γ t ) ^{3}}) f^{'} θ i - (\frac{c Δ T _{0} x}{( 1 - γ t ) ^{3}}) f θ^{'})$ Produces:L0 $f θ' - f' θ$ L1 $(f θ^{'} i - f^{'} θ)$ After division and the final sign reversal, the canonical advection block is fθ′−f′θ; expanding this block exposes both terms retained directly by the ODE root.fθ′−f′θ
Open record 3: k̃_eff[ΔT₀x/(1−γt)²][c/(ν_f(1−γt))]θ″ → k₁θ″
Consumes:L0 $\tilde{k}_{e} f f [Δ T_{0} x / (1 - γ t)^{2}] [c / (ν_{f} (1 - γ t))] θ ″$ L1 $((\frac{k}{ρC p})_{hn f} (\frac{Δ T _{0} x}{( 1 - γ t ) ^{2}}) (\frac{c}{ν _{f} ( 1 - γ t )}) θ^{''} i + (\frac{1}{( ρC p ) _{hn f}}) (\frac{16 σ * T \infty ^{3}}{3 k *}) (\frac{Δ T _{0} x}{( 1 - γ t ) ^{2}}) (\frac{c}{ν _{f} ( 1 - γ t )}) θ^{''})$ Produces:L0 $k_{1} θ ″$ L1 $(k_{1} i θ^{''})$ Dividing the effective-diffusion block by the thermal scale and applying the definitions of Pr and Rd GATHERS conduction and radiation into the single verified coefficient k₁ = (1/Pr)(1/φ_Cp)(φ_k+4Rd/3).k₁θ″
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.
$(((k_{1} θ^{''} S6.1 + f θ^{'} - f^{'} θ) S6.2 - A (\frac{2 θ + η θ ^{'}}{2})) S6.3 = 0)$
★
Open Step 5: Separation of variables · Isolate the highest derivative: factor θ″ out of the thermal-diffusion term and divide the bala…
Isolate the highest derivative: factor θ″ out of the thermal-diffusion term and divide the balance by its coefficient, giving the explicit normal form θ″ = … that the numerical solver marches.Separation of variables Divisor · k₁
Open term-change ledger: 2 records
Open record 1: k₁θ″ → k₁
Consumes:L0 $k_{1} θ ″$ Produces:L0k₁The thermal-diffusion term k₁θ″ separates into the common factor θ″ (isolated on the left as the highest derivative) and its coefficient k₁, by which the isolated balance is divided.
Open record 2: fθ′−f′θ → fθ′ + f′θ
Consumes:L0 $f θ' - f' θ$ Produces:L0 $f θ'$ L0 $f' θ$ The heat-advection block fθ′−f′θ, moved to the right and subtracted, is written out as its two terms −fθ′+f′θ so the explicit form stays unambiguous.
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.
$(θ^{''} S3.4 = (\frac{( ( A ( \frac{2 θ + η θ ^{'}}{2} ) S3.1 - f θ ^{'} ) S3.2 + f ^{'} θ )}{k _{1}})_{[S3.3]})$

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Step-by-step unit check (5 steps)

Step 1 given
Expand checked this step.
Step 2 similarity_transform
Expand checked this step.
Step 3 rearrange
Expand checked this step.
Step 4 non_dimensionalise
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Step 5 separation_of_variables
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Reads as: θ″

“θ″” 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 — similarity transform

Step 3 — rearrange

Step 4 — non dimensionalise

Step 5 — separation of variables

Equation

Essence
S1The SAME balance after substitution: substituted unsteady plus substituted convective heat transport equals the substituted effective diffusion.
S1.1essenceThe sum of γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2) and (cΔT₀x/(1−γt)³)(f′θ−fθ′). [1.1]
S1.2essence(γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2) + (cΔT₀x/(1−γt)³)(f′θ−fθ′)) set equal to k̃_eff[ΔT₀x/(1−γt)²][c/(ν_f(1−γt))]θ″. [1.1]
S2The substituted thermal balance in homogeneous form: substituted unsteady plus substituted convective heat transport minus the substituted effective diffusion equals zero.
S2.1essenceThe sum of γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2) and (cΔT₀x/(1−γt)³)(f′θ−fθ′). [1.1]
S2.2essence(γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2) + (cΔT₀x/(1−γt)³)(f′θ−fθ′)) reduced by k̃_eff[ΔT₀x/(1−γt)²][c/(ν_f(1−γt))]θ″. [1.1]
S2.3essence((γ(ΔT₀x/(1−γt)³)(2θ+ηθ′/2) + (cΔT₀x/(1−γt)³)(f′θ−fθ′)) − k̃_eff[ΔT₀x/(1−γt)²][c/(ν_f(1−γt))]θ″) set equal to 0. [1.1]
S3Isolate the highest derivative: factor θ″ out of the thermal-diffusion term and divide the balance by its coefficient, giving the explicit normal form θ″ = … that the numerical solver marches.
S3.1essenceA(2θ+ηθ′/2) reduced by fθ′.
S3.2essenceThe sum of (A(2θ+ηθ′/2) − fθ′) and f′θ.
S3.3essence((A(2θ+ηθ′/2) − fθ′) + f′θ) divided by k₁.
S3.4essenceθ″ set equal to (((A(2θ+ηθ′/2) − fθ′) + f′θ) / k₁).

Dimension I

Definition

What it is — and what it is not

What it is: S4The similarity-reduced energy equation.
S5The thermal balance: unsteady plus convective heat transport equals the effective thermal diffusion.
S5.1essenceThe sum of ∂T/∂t and u·∇T. [1.1]
S5.2essence(∂T/∂t + u·∇T) set equal to α∂²T/∂y²−(1/ρCp)∂q_r/∂y. [1.1]
Wisdoms: S6The corrected reduced energy ODE: k₁θ″ + fθ′ − f′θ − A(2θ+ηθ′/2) = 0.
S6.1essenceThe sum of k₁θ″ and fθ′−f′θ. [1.1]
S6.2essence(k₁θ″ + fθ′−f′θ) reduced by A(2θ+ηθ′/2). [1.1]
S6.3essence((k₁θ″ + fθ′−f′θ) − A(2θ+ηθ′/2)) set equal to 0. [1.1]
Attributes: S7Balances effective conduction and thermal radiation against convective heating and unsteadiness; the equation is composed on demand. [1.1]

Dimension II

In practice

How to deal with it

Consequences: S8The convective term carries f, coupling it to the momentum ODE.

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