1001Ferramentas
🛢️ Calculators

Hedström Number

Calculates the Hedström number (He), a dimensionless group combining density, yield stress, pipe diameter and plastic viscosity of a Bingham fluid. Used together with the Bingham Reynolds number, it locates the laminar-to-turbulent transition for drilling fluids, mineral slurries and cement pastes. Enter the four quantities.

Result

The dimensionless group that delays turbulence in Bingham flow

Drilling mud, mineral slurry and cement paste refuse to move until wall stress overcomes the yield stress. That pushes the laminar-to-turbulent transition well away from the Reynolds 2100 everyone memorised at university, and sizing a pump on the Newtonian criterion gets the pressure drop wrong in an awkward direction: designing for turbulent flow a line that still runs laminar, or the reverse. The Hedström number pins down where the transition actually landed.

He = ρ·τ₀·D² / μₚ², with ρ the fluid density in kg/m³, τ₀ the Bingham yield stress in Pa, D the pipe inner diameter in metres and μₚ the plastic viscosity in Pa·s. Output has no unit and climbs fast: doubling the diameter multiplies He by four. The figures already on screen, 1250 kg/m³ mud, 8.5 Pa yield stress, a 4-inch pipe at 0.1016 m and 35 cP plastic viscosity, return He = 89532.4. Under the Hanks correlation that value lifts the critical Bingham Reynolds number to roughly 6600, so the line stays laminar at flow rates where a Newtonian fluid would already have gone turbulent.

The expression assumes Bingham plastic rheology, a fair description of bentonite mud and dense slurry, yet wrong for anything following power-law or Herschel-Bulkley behaviour; those have a generalised Hedström number with a different definition. Unit slips cause most of the trouble: plastic viscosity in centipoise needs dividing by 1000, yield point read in lbf/100 ft² multiplies by 0.4788 to reach Pa, mud weight in lb/gal by 119.83, and diameter in inches by 0.0254. Leaving μₚ in cP inflates He by a factor of a million.

Frequently asked questions

Which He value counts as high enough to change the design?
No single cut-off exists, since He only carries meaning alongside the Bingham Reynolds number. As a guide from the Hanks correlation: below one thousand the fluid behaves almost Newtonian and the critical Reynolds sits near 2300; at ten thousand it rises to about 3300; at the 89532 produced by the default inputs, to roughly 6600; and at one million, past 15000. That shift is the whole point — the larger the He, the greater the flow rate a line tolerates while remaining laminar.
Can I enter viscosity in centipoise and diameter in inches?
No, the fields are strictly SI. Divide centipoise by 1000 (35 cP becomes 0.035 Pa·s) and multiply inches by 0.0254 (4 in becomes 0.1016 m). Since μₚ sits squared in the denominator, leaving 35 where 0.035 belongs drops He by a factor of a million, and the page still shows a result with no warning: it rejects input only when μₚ, ρ or D reaches zero or below, or when τ₀ turns negative. Sanity-check the order of magnitude before using the figure.
Does the tool work out the pressure drop as well?
It does not. What comes back here is the Hedström group alone, one half of the transition criterion; the other half being the Bingham Reynolds number, Re = ρ·v·D/μₚ, which depends on velocity and therefore falls outside this screen. With both in hand you place the point on the Hanks diagram, settle the flow regime, and only then apply the matching friction factor to reach a pressure drop. Working the other way round, estimating head loss without checking the regime, is exactly the mistake the He/Re pair guards against.

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The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.