Pressure drop in gas pipes: formulas and a worked example

Why squared pressure

In liquids the head loss is written as a difference of pressures: the fluid is incompressible, density is constant, Darcy-Weisbach suffices. Not so with gas: as it compresses, its density grows with pressure, so along the same pipe the gas «slows down» where pressure is high and «speeds up» where it falls. Integrating the flow law along the pipe, this dependence naturally produces a drop expressed as the difference of the squared absolute pressures:

P₁² − P₂² = R(L, D, s, T) · f · z · Q²

with R the pipe's base resistance, f the Darcy friction from Colebrook-White, z the compressibility and Q the standard flow. At low pressure, where P₁ ≈ P₂ ≈ 1 atm, the approximation P₁²−P₂² ≈ 2·P·ΔP holds and the linearised form becomes acceptable — but the quadratic form holds always, which is why a serious calculation engine works in P² from low pressure to transmission mains.

What drives the loss

Flow: Q² — doubling the flow quadruples the drop (up to the slight variation of f with Reynolds).
Internal diameter: 1/D⁵ — the dominant factor. One catalogue step changes the drop by tens of percent; doubling the diameter cuts it by more than thirty times.
Length: L — linear, no surprises.
Relative density: s — a denser gas loses more at equal volumetric flow (it matters for LPG).
Friction: f — depends on the pipe roughness (PE ≈ 0.007 mm, commercial steel ≈ 0.046 mm) and the Reynolds number; computed with Colebrook-White, which converges in a few fixed-point iterations.
Compressibility: z — practically 1 in distribution, dropping significantly at high pressures (0.89 at 50 bar for natural gas).

And elevation? It adds a height term with a surprising sign: for natural gas, climbing gives back pressure — that is the altitude correction.

A worked example

Case 1 of the validation dossier, where every digit is developed in full and compared with the calculation engine:

Data: a 100 m pipe in PE 100 DE63 SDR 11 (internal 51.4 mm, ε = 0.007 mm), 30 Sm³/h of natural gas (s = 0.591), upstream at 25 mbar gauge.

Upstream absolute pressure: P₁ = 1.01325 + 0.025 = 1.03825 bar.
Reynolds: Re ≈ 13,568 → turbulent; Colebrook-White gives f = 0.0288.
Compressibility at the mean pressure: z = 0.9972 (≈ 1, as expected in LP).
Quadratic drop: P₁² − P₂² = R·f·z·Q² = 6.61·10⁻³ bar².
Downstream: P₂ = √(P₁² − 6.61·10⁻³) = 1.03506 bar → 21.81 mbar gauge, i.e. 3.19 mbar lost over 100 m.

With the next catalogue diameter (DE75, internal 61.4 mm) the same flow would lose about 60% less: the effect of the fifth power of the diameter.

The Reynolds → Colebrook-White → compressibility → drop chain is the same for any pipe, from low pressure to a 50 bar trunk main: what changes is only how hard z and the friction «work». The validation dossier applies the full chain to six cases, replicable with a calculator.

Frequently asked questions

Why do gas network calculations use squared pressure?

Because gas is compressible: density changes along the pipe with pressure, and integrating the flow law over the length the drop comes out expressed as the difference of the squared absolute pressures, P₁² − P₂². At low pressure the linearised ΔP form is an acceptable approximation; the quadratic form always holds.

What drives the pressure drop in a gas pipe?

The flow squared, the length linearly, the internal diameter with the fifth power (the dominant factor), the gas relative density, the friction factor (roughness and Reynolds number) and the compressibility at higher pressures.

How much does a larger diameter reduce the drop?

Enormously: the drop scales with 1/D⁵ at equal flow. Moving from a 51.4 mm to a 61.4 mm internal diameter (a single PE catalogue step) cuts the loss by about 60%; doubling the diameter cuts it by more than thirty times. That is why sizing is decided almost entirely by the diameter choice.

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