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Line Sizing — Pressure Drop vs Velocity vs Cost

Two lines, same fluid, two different answers

A centrifugal pump on a light-hydrocarbon service. 80 m³/h, ρ = 700 kg/m³. On the P&ID the discharge line is 4″ and the suction line is 6″ — same pump, same fluid, same flow, one size apart, and nobody on the review could say why.

Guess first. Is the bigger suction line (a) a drafting error, (b) a vendor's standard nozzle size, or (c) a deliberate and non-negotiable design decision?

(c). And the reason is not in any velocity table.

The belief worth killing is the one every junior engineer is handed in their first week:

"Size the line off the velocity table. 1–3 m/s for liquid, 15–30 m/s for gas."

The table is not wrong. It is a pre-solved answer to a problem you have not read. Velocity is not a design limit — nobody's pipe has ever failed because a number exceeded 3.0. Velocity is simply the variable that five real limits happen to be written in terms of:

A velocity table bundles all five into one number for one typical service, and it is right whenever your case looks like the case it was solved for. Change the density, the length, the energy price or the NPSH margin and the bundle comes apart. That is the whole lesson of the 4″ discharge and the 6″ suction: on the discharge, ΔP is bought with pump head, which is cheap per bar. On the suction, every millibar comes straight off NPSH available, and cavitation is not a cost — it is a wall.

Symbol key — every symbol on this sheet

The four judges, and the arithmetic each one uses

1 — Pressure drop. Everything starts here. The Darcy–Weisbach equation, with fittings carried as velocity heads:

ΔP = ( f · L/D  +  ΣK ) · ½ρv²

Laminar (Re < 2300):   f = 64 / Re                       — no roughness term at all
Turbulent (Colebrook): 1/√f = −2·log₁₀( ε/(3.7D) + 2.51/(Re·√f) )     — implicit, iterate
Churchill (explicit, all regimes):
    A = [ 2.457 · ln( 1 / ((7/Re)^0.9 + 0.27·ε/D) ) ]^16
    B = ( 37530 / Re )^16
    f = 8 · [ (8/Re)^12 + 1/(A+B)^1.5 ]^(1/12)

Churchill matches Colebrook to well under 1 % in the turbulent range and collapses cleanly to 64/Re in the laminar range, which is why it is the one to code.

The number that matters for sizing is ΔP per 100 m, because it is the only form that can be compared between lines of different length. Typical allowances: 0.2–0.5 bar/100 m for pump discharge, 0.05–0.1 bar/100 m for a pump suction, and whatever the elevation gives you for gravity flow.

Why ΔP falls as roughly D⁻⁵, and why that single exponent explains everything

For fixed Q: v ∝ D⁻², so ½ρv² ∝ D⁻⁴, and the L/D term adds one more power. With f varying only slowly, ΔP ∝ D⁻⁵. One standard size up therefore cuts the pressure drop by about 2.5–4× for the usual steps (6″→8″ = 3.9×, 8″→10″ = 3.1×, 10″→12″ = 2.4×) — and by 7.8× at 4″→6″, only because 5″ is skipped. Quote the step you actually mean; the exponent is 5, the ratio is not.

That exponent is the reason line sizing feels so forgiving in one direction and so brutal in the other. Guess one size too big and you waste maybe 40 % on steel. Guess one size too small and the pump you already bought is 2 bar short and cannot make rate. It is also the reason the economic optimum is sharp on the left and almost flat on the right: total cost = a·D^1.2 + b·D⁻⁵, and the D⁻⁵ term falls off a cliff. When in doubt, round up.

2 — Erosional velocity. API RP 14E, the most-quoted and most-argued-about line in the field:

V_e = C / √ρ          (USC: ft/s, lb/ft³)
V_e = 1.22 · C / √ρ   (SI: m/s, kg/m³)     C = 100 continuous · 125 intermittent

Know its limits as well as its value. It is empirical, from 1980s offshore two-phase practice. It contains no sand loading, no droplet size, no material, no geometry — yet erosion happens at bends and tees, not in straight pipe, and a single elbow with sand in it will fail long before V_e. API 14E itself now says C may be raised well above 100 for clean, corrosion-controlled, solids-free service, and that erosive service needs a proper model (DNV-RP-O501). Treat V_e as a screening number you must be able to argue about, not as a code limit.

3 — Noise, vibration and surge. Gas lines above roughly 0.3 Mach get loud; 25–30 m/s is a common working ceiling in process gas, and control-valve and orifice noise adds to it. Two-phase lines have a separate problem — flow regime. A line sized on average velocity can sit unknowingly in the slug-flow region of a Baker or Taitel–Dukler map, and slugs deliver momentum pulses at every elbow (that is a dynamic-load problem: see the DLF topic). Liquid lines that are closed quickly get Joukowsky surge, ΔP = ρ·c·Δv — low velocity is the cheapest mitigation there is.

4 — NPSH. On a pump suction, ΔP is not an operating cost; it is subtracted directly from NPSHa. A suction line is therefore sized short, straight and slow — 1.0–1.2 m/s is normal practice — with an eccentric reducer flat-side-up at the nozzle so no vapour pocket forms. The economic optimum on a suction line is almost always smaller than what you actually install, and you overrule it deliberately.

Slide all four judges at once and watch which one is binding — the same fluid, five services, and a pass/fail matrix for every standard size: ▶ open the interactive: process line sizing calc

And the fifth judge, the one that decides when the other four are silent: money.

Total life cost (D) = installed pipe cost + lifetime pumping energy
                    ≈ a·D^1.2 · L     +     (ΔP(D)·Q/η) · hours · years · price
                      ~ D^1.2                        ~ D⁻⁵

Because one term rises as D^1.2 and the other falls as D⁻⁵, there is a genuine minimum. Its position moves with the energy price, the running hours and the economic life — and with nothing else. A line that runs 500 h/yr wants to be small; the same line at 8760 h/yr wants to be a size or two bigger. This is exactly why an intermittent transfer line and a continuous circulating header, carrying the same fluid at the same rate, are honestly different sizes.

Worked example — 100 m³/h of cooling water, 200 m of pipe

Water, ρ = 1000 kg/m³, μ = 1 cP, ε = 0.046 mm, 200 m straight, ΣK ≈ 8, 8000 h/yr, $0.10/kWh, 20 years, η = 0.70. Try 6″ Sch 40, bore 154.1 mm:

A  = π/4 × 0.1541²            = 0.018650 m²
v  = (100/3600) / 0.018650    = 1.489 m/s
Re = 1000 × 1.489 × 0.1541 / 0.001   = 229 500          (fully turbulent)
ε/D = 0.046 / 154.1           = 2.99 × 10⁻⁴
f  (Churchill)                = 0.01750      (Colebrook gives 0.01744 — 0.3 % apart)
ΔP/100 m = f · (100/D) · ½ρv² = 0.0175 × 648.9 × 1109 = 12 600 Pa = 0.126 bar/100 m
ΔP total = 0.0175 × (200/0.1541) × 1109 + 8 × 1109     = 34 100 Pa = 0.341 bar
Power    = 34 100 × 0.02778 / 0.70                     = 1.35 kW
V_e      = 1.22 × 100 / √1000                          = 3.86 m/s   →  v/V_e = 0.39, fine

Now the economics. Installed pipe on an illustrative basis of $55·(D/25.4)^1.2 per metre gives $479/m → $95 700 of steel. Pumping: 1.35 kW × 8000 h × 20 yr × $0.10 = $21 600 of energy. Sweep the diameter and the total bottoms out at 151 mm — which is 6″ Sch 40 to within 2 %.

So on this service the velocity table, the ΔP allowance and the economics all say 6″, and the table earns its reputation. Now change one thing at a time and watch them separate:

Watch the same duty in two diameters side by side — flow animated, pressure drawn as a colour ramp along the wall, and the 20-year cost of each: ▶ open the interactive: process line sizing 3d. That model makes one point that a spreadsheet hides: the colour fades continuously along the straight run, not in jumps at the elbows. On a 200 m line the two bends and the gate valve are about 5 % of the loss (ΣK = 1.1 against f·L/D = 22.7). Fittings matter on short, congested runs; on long ones, the wall wins — but note the worked example above carries ΣK = 8, where they are a quarter of it.

Common pitfalls

Outcome

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