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Yield-Stress Annular Laminar/Turbulent Friction Calculator

Calculate eccentric yield-power-law annular friction with a self-consistent local flow index, effective diameter and roughness-aware turbulent branch.

Inputs

Results

Calculate to view linked results. Default values are examples, not recommended operating inputs.

Calculation sequence

  1. Fit the complete yield-power-law rheogram first: measured yield stress, consistency K in Pa.s^n and fitted exponent n. Enter actual liquid flow, density, uniform geometry, roughness and eccentricity (center offset divided by radial clearance). The synthetic defaults are not a drilling-fluid recommendation. A nonzero yield stress must not be dropped or absorbed into a constant effective viscosity.
  2. Integrate the HB Rabinowitsch-Mooney equivalent-pipe flow curve exactly. Its local log slope n' depends on yield stress and operating rate, whereas fitted n is a material parameter. The standard stress integral uses tau squared, consistent with Maleki-Frigaard equations 5-6 and the Newtonian limit; AADE's printed equation 35 appears to contain a dimensionally inconsistent tau-cubed typographical error. This correction is our algebraic inference, not a published erratum. At each rate solve the local-slope/effective-diameter consistency problem using AADE equation 34. A 64/96-point quadrature refinement checks geometry.
  3. Use the equivalent-pipe laminar wall stress to construct generalized Reynolds number, then calculate laminar and roughness-aware turbulent Fanning factors. The larger is selected using the paper's branch rule. The stress used for this flow-curve construction is not either actual annulus-wall stress and is not a prediction of turbulent plug shape. The real hydraulic diameter D-d, not Deff, remains in the pressure-loss denominator.
  4. The 21-point sweep spans 0.5Q to 2Q, recalculating n', Deff and both friction factors at every rate. Downloaded samples retain both candidates and the selected branch. All sweep points must pass the unclamped local-index and conditioning limits; a yield-dominated point can therefore invalidate a sweep even when its central rate is acceptable. For exact concentric laminar yield surfaces, use the separate HB annulus workflow instead.

Worked check

The synthetic default uses D=0.2 m, d=0.1 m, Q=0.1 m3/s, K=0.2 Pa.s^n, n=0.8, yield=5 Pa, rho=1,200 kg/m3, e=0.5, roughness=0.00001 m and L=1,000 m. The local index is about 0.625571 rather than 0.8, and the turbulent branch is selected. Setting yield to zero recovers the existing power-law correlation, not a different roughness model.

Synthetic examples for reproducibility, not measured field results.

Assumptions and limitations

  • Stationary uniform eccentric circular annulus, incompressible time-independent HB liquid. This is the generalized empirical AADE hydraulic method, not an exact eccentric HB solution or a field-calibrated turbulence closure. No rotation, skew, cuttings bed, elasticity, gas, wall slip, local/entrance losses, hydrostatic head or temperature-dependent rheology.
  • Retain 0.3 <= d/D <= 0.9 and 0 <= e <= 0.95, with fitted 0.4 <= n <= 1 and self-consistent 0.4 <= n' <= 1. The paper's later heuristic clamp to 0.25 is deliberately not used. Extreme yield-dominated rheology is rejected, not silently reset; the original paper describes degraded prediction for such fluids.
  • Software envelopes at every sampled rate: ReG from 0.0001 to 10,000,000, epsilon/Deff <= 0.01 and friction gradient <= 2,000,000 Pa/m. The max-factor switch is not a resolved mixed/transition flow field. Compare predicted pressure losses with measured data for the actual mud and geometry before operational use.

Implemented equations

Generalized HB flow curve and local slope

τ=τy+Kγ˙nS(τw)=4τw3∫τyτwτ2γ˙(τ) dτS=8V/Deffn′=dln⁡τwdln⁡S=S4γ˙w−3S\begin{aligned}\tau&=\tau_y+K\dot\gamma^n\\S(\tau_w)&=\frac4{\tau_w^3}\int_{\tau_y}^{\tau_w}\tau^2\dot\gamma(\tau)\,d\tau\\S&=8V/D_{\rm eff}\\n'&=\frac{d\ln\tau_w}{d\ln S}\\&=\frac{S}{4\dot\gamma_w-3S}\end{aligned}

Effective diameter and selected friction

ReG=8ρV2/τwfL=16/ReGfT−1/2=−4log⁡10AA=Ar+AvAr=0.27ϵ/DeffAv=1.26(n′)−1.2[ReGfT1−n′/2](n′)−0.75Δpf=2max⁡(fL,fT)ρV2LD−d\begin{aligned}\mathrm{Re}_G&=8\rho V^2/\tau_w\\f_L&=16/\mathrm{Re}_G\\f_T^{-1/2}&=-4\log_{10}\mathcal A\\\mathcal A&=A_r+A_v\\A_r&=0.27\epsilon/D_{\rm eff}\\A_v&=\frac{1.26^{(n')^{-1.2}}}{[\mathrm{Re}_G f_T^{1-n'/2}]^{(n')^{-0.75}}}\\\Delta p_f&=\frac{2\max(f_L,f_T)\rho V^2 L}{D-d}\end{aligned}

Sources and maintenance

Published by PetroCalcHub Team. Implementation and source checks: . This does not claim independent expert review.

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