CT Pressure Loss
Hydraulics
CT Pressure Loss
Frictional pressure loss for fluids pumped through coiled tubing. Choose the rheology model, enter the fluid, the tubing and the flow rate, in oilfield or metric units.
Inputs
Result
Check the inputs
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Total friction pressure loss
– psi
- Off the reel (straight)
- – psi
- On the reel (curved)
- – psi
- Velocity
- – ft/min
- Reynolds number
- –
- Equivalent flow index n′
- –
- Equivalent consistency K
- – lbf·sⁿ/100 ft²
- Dean number
- –
Flow regime: –
Flow regime: –
Method
The loss is calculated separately for the part of the string off the reel (straight) and the part on the reel (curved), and the two are added.
- Rheology: the model gives the wall shear stress τ at a wall shear rate γ. Newtonian: τ = μγ. Bingham plastic: τ = YP + PV·γ. Power law: τ = Kγⁿ. Herschel–Bulkley: τ = τ₀ + Kγⁿ. Empirical: the measured Fann points joined by straight lines on log–log axes. Fann speed N rpm gives γ = 1.7023 N 1/s and τ = 1.0678 θ lbf/100 ft².
- Equivalent power law: at the flow in question the pseudo shear rate is Γ = 8V/D. The wall shear rate γw = Γ(3n′+1)/(4n′) is found by iterating with n′, the local slope of the flow curve, and the model's K and n′ are those that give the same wall stress and slope. A power-law fluid returns its own n and K.
- Fann fit: Bingham plastic PV = θ600 − θ300 and YP = θ300 − PV; Herschel–Bulkley τ₀ = 2τ3 − τ6 and a log–log fit for n and K; power law a log–log fit; Newtonian a fit through the origin. The deviation is the root-mean-square relative error of the model against the readings.
- Reynolds number (Metzner–Reed): Re = 31.042 ρ Q^(2−n) / (9.804^n K ID^(4−3n)) · (n/(1+3n))^n, with ρ in lb/gal, Q in gal/min, ID in inches and K in lbf·sⁿ/100 ft².
- Straight tubing: laminar below Re = 3470 − 1370n and turbulent above 4270 − 1370n. Laminar f = 16/Re. Turbulent f = a/Re^b, with a = (log n + 3.93)/50 and b = (1.75 − log n)/7.
- Coiled tubing: the laminar factor is multiplied by ψ·De^φ, where De = Re_g·(ID/D)^0.5 is the Dean number, ψ = 47.969 − 153.8n + 166.22n² − 60.132n³ and φ = 0.875n − 0.515. Turbulent: f = 1.069·a/Re^(0.8b)·(ID/D)^0.1. The laminar limit is (4n/(3n+1))^n · [2100 ψ (ID/D)^(0.5φ)]^(1/(1−φ)) and the turbulent limit is 800 above it.
- Pressure loss: ΔP = f ρ Q² L / (154.63 ID⁵) in oilfield units. Between the two limits the loss is interpolated linearly in Re, from the laminar loss at the lower limit to the turbulent loss at the upper limit.
Assumptions
- Wall roughness enters only the turbulent friction factor, as the ratio of the Haaland factor with roughness to the one without, so a roughness of 0 gives the smooth-tube result. For n other than 1 this is an estimate.
- For models with a yield stress the equivalent power law is evaluated at the flow rate entered. This is sound for laminar flow; turbulent and transitional values for such fluids are estimates, and the critical Reynolds numbers are those of a power-law fluid.
- A single-phase power-law fluid without solids.
- Smooth tubing and fully developed flow, with fluid properties constant along the string.
- The curvature is that of the reel diameter you enter; tubing off the reel is straight.
- The correlations cover a limited range of n and ID/D. Check results outside it.
Based on published correlations for power-law fluids in coiled tubing (Zhou and Shah), with the Metzner–Reed Reynolds number and Dodge–Metzner turbulent factors. The calculation reproduces a reference worksheet for its example case. Check the original papers before you rely on a result.
The tools are aids for engineers. Check the inputs and the results: responsibility for any decision stays with the user.