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Fan versus system curve

IQ-CAL-004 · Rev 0 Indicative · 11/11/2026 · Not a certificate

INPUTS + RESULTSUNITS AS TABULATED

IQ Engineering Group · Sydney · IQ-CAL-004 · Fan versus system curve · millimetres, kN, kNm, MPa unless noted

DISCLAIMER

Fans + air

Tick up to three catalogue fans. Curves are at the selected frequency and air density.

ρ = 1.204 kg/m³ · p = 101.3 kPa · ν = 1.51e-5 m²/s
Normal (20 °C, 0 m): ρₙ = 1.2 kg/m³. Qₙ = Q × ρ/ρₙ.

What the curves are
Fan curve
Catalogue static pressure vs actual volume, scaled by speed (Q∝n, P∝n²) and by ρ/1.20. Chart Y is fitted to the fan so a steep system does not squash it.
Design takeoff
ΣΔP of the pipes and fittings below, evaluated at each Q. ΔP ∝ Q² (Darcy and K both go with V²). Equipment ΔP is entered at Qdes and scaled (Q/Qdes)².
0.8× / 1.25×
Same takeoff × 0.8 (open damper, clean filter) or × 1.25 (restricted). Duty is the intersection of the fan with the ticked 1.0× curve, unless Q is locked.
Y-scale
Pinned to the fan. If a system curve runs off the top, the takeoff at that Q is above the fan — the fan cannot deliver it.

Curves

07501500225030000.000.751.502.253.00Volume flow Q (m³/s actual)Static pressure ΔP (Pa)System off scale — takeoff above the fan at high Q
Backward-curved 450 mm @ 50 HzDesign takeoffOpen / clean 0.8×Restricted 1.25×Duty

System ΔP = Σ f(L/D)ρV²/2 + Σ K ρV²/2 + equipment. 0.8× open / clean; 1.25× restricted. Chart Y is the fan — a steep system going off the top means the fan cannot make that Q.

Operating point

Qdes actual — m³/s1.2
Q* actual — m³/s1.3
Q* normal — m³/s1.3
ΔP* — Pa1,633
V (ref D) — m/s15.4
ρ / ρₙ1.00
Air — kW2.1
Shaft — kW2.9
Motor — kW (×1.15)3.4

Catalogue fan pressure is scaled by ρ/1.20. Volume Q is actual. Mass flow is ρQ. Qₙ = Q × ρ/1.20 at 20 °C sea level.

System takeoff · add pipes and fittings

Straight uses Darcy with Haaland f from the selected surface. Knees 90° / 45° (r/D = 1.5) and mitred knees use K. Equipment ΔP is entered at Qdes and scales (Q/Qdes)².

ComponentnD mmL mSurfaceK / ΔPdes
——
f from Moody
——
f from Moody
——
——
——
——
——

Loss schedule at Qdes = 1.20 m³/s

Same Q as the design flow (or locked Q). Moody on the right holds f. Re and f live there to keep this short.

ItemnØVΔP Pa
Sharp entry131515.471
Straight pipe / duct131515.4121
90° elbow r/D = 1.5331515.4116
Straight pipe / duct131515.4121
90° elbow r/D = 1.5331515.4116
45° elbow r/D = 1.5231515.446
Damper (open)131515.429
Equipment ΔP (at Qdes)131515.4650
Discharge / exit131515.4143
TOTAL1,412

Moody chart · f vs Re

10^310^410^510^610^710^80.010.0150.020.030.040.060.08ε/D smoothε/D 0.0002ε/D 0.001ε/D 0.002ε/D 0.01Reynolds number ReDarcy friction factor flaminar 64/Retransition

Re = VD/ν = 3.22e+5 · ε/D = 0.00048 · f = 0.0179 · Galvanised duct ε 0.15 mm · 20 °C · 0 m

Calculation · one component at Qdes

DARCY–WEISBACH · HAALAND f · AIR STATE
  1. 01

    Air at the stated temperature and altitude. ISA pressure. Density from p / RT. Normal condition is 20 °C, sea level, ρₙ = 1.20 kg/m³.

    T = 20 °C = 293.2 K z = 0 m p = 101,325 Pa ρ = 1.204 kg/m³ ν = 1.51e-5 m²/s Qₙ = Q × ρ/ρₙ
  2. 02

    Velocity and Reynolds number in the reference diameter at Qdes (actual).

    D = 315 mm A = πD²/4 = 0.0779 m² V = Q/A = 15.40 m/s Re = VD/ν = 3.22e+5 ρV²/2 = 142.8 Pa
  3. 03

    Friction factor. Laminar 64/Re below Re 2300. Turbulent Haaland. Transition blended 2300–4000.

    ε = 0.15 mm ε/D = 0.00048 1/√f = −1.8 log10[ (ε/D / 3.7)^1.11 + 6.9/Re ] f = 0.0179
  4. 04

    Straight pipe — Darcy–Weisbach. ΔP = f (L/D) (ρV²/2).

    L = 15.0 m × n=1 f = 0.0179 ΔP_straight = 121 Pa
  5. 05

    Sum the takeoff. Fan catalogue pressure is scaled by n² and by ρ/1.20. Intersect with the selected fans.

    ΣΔP(Qdes) = 1,412 Pa Q* = 1.291 m³/s actual Qₙ* = 1.295 m³/s ΔP* = 1,633 Pa air kW = 2.11

IQEG · Sydney · info@iqeg.com.au · 0433 597 205 · Investigate. Understand. Solve.

Rev 0 indicative calculation. Not certified engineering advice. Check mill tables and the current Standard.