Physics Verification
Runs canonical test cases against the bedding-plane (He-Hutchinson) and TIV (Thomsen) libraries with expected ranges. Use this as a quick smoke check when the underlying libs change.
Closed-form textbook relations with every number substituted, beside what this app's solver returns for the same inputs. Retype the arithmetic and you should land on the same answer. Fracture geometry is not here — that is graded against measured field data in the pad-scale verification suite.
- Equation
- p_burst = 2 · Y · t / OD
- Inputs
- OD 5.500 in · wall 0.361 in · yield 80,000 psi
- By hand
- 2 × 80,000 × 0.361 / 5.500 = 57,760 / 5.5 = 10,502 psi
- This app
- 10,502 psi
Exact closed form — the only difference is display rounding.
- Equation
- A = (pi/4) · (OD² − ID²), ID = OD − 2t
- Inputs
- OD 5.500 in · wall 0.361 in ⇒ ID 4.778 in
- By hand
- (pi/4) × (5.5² − 4.778²) = 0.7854 × (30.25 − 22.829) = 5.828 in²
- This app
- 5.8282 in²
Exact closed form.
- Equation
- F_yield = A · Y
- Inputs
- A 5.828 in² · yield 80,000 psi
- By hand
- 5.828 × 80,000 = 466,257 lbf ≈ 466 kips
- This app
- 466,257 lbf
Exact closed form.
- Equation
- Rs = γg · [ (p/18.2 + 1.4) · 10^(−a) ]^(1/0.83), a = 0.00091·T − 0.0125·API
- Inputs
- p 3,000 psi · T 200 °F · 35 °API · γg 0.75
- By hand
- a = 0.00091×200 − 0.0125×35 = 0.182 − 0.4375 = −0.2555; Rs = 0.75 × [(164.84 + 1.4) × 10^0.2555]^(1/0.83) = 722 scf/STB
- This app
- 721.9 scf/STB
Correlation, not a law: Standing's own scatter against his data set is roughly ±15 %. The check here only confirms the arithmetic, not the fit.
- Equation
- Bo = 0.9759 + 0.00012 · [ Rs·sqrt(γg/γo) + 1.25·T ]^1.2
- Inputs
- Rs 722 scf/STB · T 200 °F · 35 °API (γo 0.8498) · γg 0.75
- By hand
- term = 722 × sqrt(0.75/0.8498) + 1.25×200 = 678.3 + 250 = 928.3; Bo = 0.9759 + 0.00012 × 928.3^1.2 = 1.413 bbl/STB
- This app
- 1.4127 bbl/STB
Correlation; Standing quotes about ±5 % on Bo for the fluids he fitted.
- Equation
- M0_max = G · ΔV, Mw = (2/3)·(log₁₀ M0 − 9.1)
- Inputs
- ΔV 238,481 m³ (1.5 MMbbl) · G 2.0 × 10¹⁰ Pa
- By hand
- M0 = 2.0e10 × 238,481 = 4.77 × 10¹⁵ N·m; Mw = (2/3)×(15.6786 − 9.1) = 4.386
- This app
- 4.3857 Mw
Upper bound only — it assumes every injected barrel drives slip on one fault. Real events are smaller.
Kink-vs-pierce decisions at a weak laminated horizon. Uses ζ = Gc_iface / Gc_matrix and the He-Hutchinson energy bridge.
| Case | Expected | Actual | Status |
|---|---|---|---|
Weak bed, shallow dip (ζ=0.10) Gc_iface/Gc_matrix ≤ 0.25 ⇒ kink along bedding | kink | kink | PASS |
↳ energy ratio ζ Expected ≈ 0.10 | [0.090, 0.110] | 0.100 | PASS |
Tough bed, shallow dip (ζ=1.50) Gc_iface ≥ Gc_matrix ⇒ pierce through | pierce | pierce | PASS |
Mixed regime, sufficient G (ζ=0.50) G_avail ≥ Gc_iface in 0.25<ζ<1 band ⇒ kink | kink | kink | PASS |
Mixed regime, starved G (ζ=0.50) G_avail < Gc_iface in 0.25<ζ<1 band ⇒ pierce | pierce | pierce | PASS |
Steep NF (dip 60°) dip > 15° tol ⇒ defer to RP/GW | not-bedding | not-bedding | PASS |
Five-constant TIV stiffness from (Vp0, Vs0, ρ, ε, γ, δ) and Mavko et al. (2009) engineering-moduli closed forms.
| Case | Expected | Actual | Status |
|---|---|---|---|
Isotropic collapse: E_h/E_v ε=γ=δ=0 ⇒ ratio ≡ 1 (within 1e-9) | [0.999999999, 1.000000001] | 1.000000000 | PASS |
Shale E_v Typical laminated shale: 25-55 GPa | [25.00, 55.00] GPa | 28.90 GPa | PASS |
Shale E_h Stiffer in bedding plane: 35-80 GPa | [35.00, 80.00] GPa | 42.71 GPa | PASS |
Shale anisotropy E_h/E_v Expected 1.2-2.0 for ε=0.20, γ=0.25 | [1.200, 2.000] | 1.478 | PASS |
E(θ=0°) vs E_v relative deviation Closed-form must agree to <1e-6 | [0.000000000, 1.000000000e-6] | 1.320043280e-16 | PASS |
E(θ=90°) vs E_h relative deviation Closed-form must agree to <1e-6 | [0.000000000, 1.000000000e-6] | 0.000000000 | PASS |
ν_v (vertical Poisson) Physical band 0.10-0.45 | [0.100, 0.450] | 0.236 | PASS |
ν_h (in-plane Poisson) Physical band 0.05-0.45 | [0.050, 0.450] | 0.176 | PASS |