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Eccentric Power-Law Annular Pressure Loss Calculator

Estimate eccentric annular laminar/turbulent friction using power-law rheology, effective diameter and wall roughness, with both branch losses.

Inputs

Results

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

Learn, compare and practice

Worked explanations, controlled studies and printable exercises connected to this calculation.

Calculation sequence

  1. Enter actual annular volume flow, stationary circular geometry, measured power-law K and n, liquid density, absolute roughness and interval length. Eccentricity is the center offset divided by (D-d)/2: zero is concentric. K has dimensions Pa.s^n and is not a constant Newtonian viscosity unless n equals one. Do not discard a nonzero fitted HB yield stress to use this power-law tool. The example parameters are synthetic, not a recommended drilling-fluid formulation.
  2. The concentric laminar power-law solution retains curvature and balances velocity from both walls. Homogeneous scaling gives its gradient at the entered rate. Multiply this laminar gradient by the published eccentricity ratio R, then recover the effective diameter by enforcing fL*ReG=16 and equation 26. This is algebraically equivalent to the paper's power-law effective diameter, rather than applying a Newtonian effective viscosity to an eccentric gap.
  3. Solve the roughness-extended Dodge-Metzner equation for the turbulent Fanning factor, and select the larger laminar or turbulent factor as prescribed by AADE. The result warning identifies the selected branch. The effective diameter controls Reynolds number and relative roughness; the actual hydraulic diameter D-d still controls the pressure-loss denominator. Multiplying the final turbulent loss directly by the laminar ratio R would be incorrect.
  4. The flow sweep covers 0.25 to 2 times the entered rate at fixed geometry, rheology and length. Its CSV includes both candidate Fanning factors, the selected factor and the branch code. The entire sweep must pass the numerical envelope. No radial velocity field is claimed in the eccentric or turbulent geometry; use the exact concentric laminar tool only when its separate assumptions hold.

Worked check

The synthetic default has D=0.2 m, d=0.1 m, Q=0.03 m3/s, K=0.2 Pa.s^n, n=0.6, rho=1,200 kg/m3, e=0.5, roughness=0.00001 m and L=1,000 m. Both branch losses are reported separately; the larger determines friction loss. Setting e=0 restores the concentric correlation. In laminar flow, reducing Q by a factor of two reduces loss by 2^n, not necessarily by two.

Synthetic examples for reproducibility, not measured field results.

Assumptions and limitations

  • Power-law, time-independent incompressible liquid in a stationary uniform eccentric annulus. No yield stress/Herschel-Bulkley extrapolation, pipe rotation, cuttings bed, wall slip, gas, thermal feedback, local/entrance losses or hydrostatic head. This is friction-only, not a complete standpipe-pressure, ECD, hole-cleaning or well-control design.
  • Retains the published correction's ranges: 0.3 <= d/D <= 0.9, 0.4 <= n <= 1 and 0 <= e <= 0.95. Pipe-wall contact at e=1 is deliberately excluded. Do not interpret the paper's approximately 5% laminar correction-fit statement as a 5% guarantee for turbulent pressure predictions or field muds. No fitted exponent is silently clamped.
  • Additional software envelopes apply to every sweep point: 0.0001 <= ReG <= 10,000,000, epsilon/Deff <= 0.01 and gradient <= 2,000,000 Pa/m. The max-factor crossover is an empirical branch rule, not a resolution of mixed laminar/turbulent flow around the annulus. Use measured pressure-loss data to assess field applicability, particularly close to transition or for polymer elasticity.

Implemented equations

Eccentric laminar correction and effective Reynolds number

R=1−0.072enσ0.8454−1.5e2n σ0.1852+0.96e3n σ0.2527σ=d/D,Dh=D−dReG=DeffnV2−nρ8n−1K[(3n+1)/(4n)]n\begin{aligned}R&=1-0.072\frac{e}{n}\sigma^{0.8454}\\&\quad-1.5e^2\sqrt n\,\sigma^{0.1852}\\&\quad+0.96e^3\sqrt n\,\sigma^{0.2527}\\\sigma&=d/D,\quad D_h=D-d\\\mathrm{Re}_G&=\frac{D_{\rm eff}^{n}V^{2-n}\rho}{8^{n-1}K[(3n+1)/(4n)]^n}\end{aligned}

Laminar/turbulent Fanning factors and friction loss

fL=16/ReGfT−1/2=−4log⁡10AA=Ar+AvAr=0.27ϵ/DeffAv=1.26n−1.2[ReGfT1−n/2]n−0.75f=max⁡(fL,fT)Δpf=2fρV2LDh\begin{aligned}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}}}\\f&=\max(f_L,f_T)\\\Delta p_f&=\frac{2f\rho V^2L}{D_h}\end{aligned}

Sources and maintenance

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

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