acalculator

What is the pressure drop in my pipe?

Type the flow rate or velocity, the inside diameter and length of the pipe, and pick the pipe material and fluid. The pressure drop calculator gives the friction loss in psi, kPa and feet of head, with the friction factor, Reynolds number and flow regime.

Your numbers

Units
Flow given as
Pressure drop
0.91197 psi

The pressure drop is 0.91197 psi (2.1074 ft of head), with f = 0.0240783 at Re = 47,277.8.

Pressure drop (Pa)
6,287.8
Pressure drop (psi)
0.91197
Head loss (m)
0.64233
Head loss (ft)
2.1074
Velocity (m/s)
0.93383
Flow rate (L/s)
1.8927
Reynolds number
47,277.8
Flow
Turbulent
Friction factor f
0.0240783
Colebrook friction factor
0.0239359
Relative roughness ε ÷ D
0.0009
Density used (kg/m³)
998.21

Pressure drop: 0.91197 psi. The pressure drop is 0.91197 psi (2.1074 ft of head), with f = 0.0240783 at Re = 47,277.8.

How to calculate

Pressure drop calculator for pipe flow: the Darcy-Weisbach pressure and head loss from the flow rate or velocity, pipe size, length and material, with the friction factor, Reynolds number and flow regime.

Example with the default inputs (Flow given as Flow rate, Flow rate (Q) 30, Flow unit US gal/min (gpm), Inside diameter (D) 2 in, Pipe length (L) 100 ft, Pipe material Commercial steel or wrought iron (0.00015 ft), Fluid Water at 20 °C (68 °F)): The pressure drop is 0.91197 psi (2.1074 ft of head), with f = 0.0240783 at Re = 47,277.8.

Method: Δp = f × (L ÷ D) × ρv² ÷ 2; h = Δp ÷ ρg; Re = ρvD ÷ μ; f = 64 ÷ Re for Re < 2,000; f = 0.25 ÷ [log10(ε ÷ 3.7D + 5.74 ÷ Re^0.9)]² for Re > 4,000; Dunlop’s cubic between (EPANET).

  • Steady flow of a liquid (incompressible) filling a straight round pipe; fittings, valves and height changes are not included.
  • Between Re = 2,000 and 4,000 the flow is unstable and f is uncertain (Hydraulic Institute); the page interpolates as EPANET does.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Flow given as Velocity, Velocity (v) 1, Inside diameter (D) 1.969 in, Pipe length (L) 32.81 ft, Pipe material Drawn tubing: glass, brass, plastic (0.000005 ft), Fluid Other: type the density and viscosity, Density (ρ) 56.19 lb/ft³, Viscosity (μ) 0.1, Viscosity unit Pa·s gives Reynolds number 450, Friction factor f 0.142222, Pressure drop (Pa) 12,800, Flow Laminar.Source: Hydraulic Institute, HI Data Tool, Pipe Frictional Losses (Eq. 3.A.1 h_f = f × (L ÷ D) × v² ÷ 2g; Eq. 3.A.2 Colebrook; Eq. 3.A.3 explicit approximation; Eq. 3.A.5 f = 64 ÷ Re below 2,000; Table 3.A.1 roughness), https://datatool.pumps.org/fluid-flow-iii/general (retrieved 2026-10-18) (f = 64 ÷ Re); OpenStax, University Physics Volume 1, §14.7 Viscosity and Turbulence (Poiseuille’s law, Equation 14.19: Q = (p₂ − p₁)πr⁴ ÷ 8ηl), https://openstax.org/books/university-physics-volume-1/pages/14-7-viscosity-and-turbulence (retrieved 2026-10-18): Δp = 8ηlv ÷ r² = 8 × 0.1 × 10 × 1 ÷ 0.025² = 12,800 Pa, the same as Darcy-Weisbach with f = 64 ÷ 450
  2. Flow given as Flow rate, Flow rate (Q) 30, Flow unit US gal/min (gpm), Inside diameter (D) 2 in, Pipe length (L) 100 ft, Pipe material Commercial steel or wrought iron (0.00015 ft), Fluid Water at 20 °C (68 °F) gives Velocity (v) 0.933826, Reynolds number 47,277.787126, Friction factor f 0.024078, Pressure drop (psi) 0.911971, Colebrook friction factor 0.023936.Source: Hydraulic Institute, HI Data Tool, Pipe Frictional Losses (Eq. 3.A.1 h_f = f × (L ÷ D) × v² ÷ 2g; Eq. 3.A.2 Colebrook; Eq. 3.A.3 explicit approximation; Eq. 3.A.5 f = 64 ÷ Re below 2,000; Table 3.A.1 roughness), https://datatool.pumps.org/fluid-flow-iii/general (retrieved 2026-10-18); US EPA, EPANET Users Manual, 13. Analysis Algorithms, §13.1 (Hagen-Poiseuille below Re = 2,000, Swamee and Jain above 4,000, Dunlop’s cubic interpolation between), https://epanet-manual.readthedocs.io/en/latest/analysis_algorithms.html (retrieved 2026-10-18); NIST Chemistry WebBook, SRD 69, Isobaric Properties for Water at 0.101325 MPa (20 °C: 998.21 kg/m³, 0.0010016 Pa·s), https://webbook.nist.gov/cgi/fluid.cgi?Action=Load&ID=C7732185&Type=IsoBar&Digits=5&P=0.101325&THigh=100&TLow=10&TInc=10&RefState=DEF&TUnit=C&PUnit=MPa&DUnit=kg%2Fm3&HUnit=kJ%2Fkg&WUnit=m%2Fs&VisUnit=Pa*s&STUnit=N%2Fm (retrieved 2026-10-18)
  3. Flow given as Velocity, Velocity (v) 0.3, Inside diameter (D) 0.3937 in, Pipe length (L) 3.281 ft, Pipe material Drawn tubing: glass, brass, plastic (0.000005 ft), Fluid Water at 20 °C (68 °F) gives Reynolds number 2,989.846246, Flow Transitional (critical zone), Friction factor f 0.033043.Source: US EPA, EPANET Users Manual, 13. Analysis Algorithms, §13.1 (Hagen-Poiseuille below Re = 2,000, Swamee and Jain above 4,000, Dunlop’s cubic interpolation between), https://epanet-manual.readthedocs.io/en/latest/analysis_algorithms.html (retrieved 2026-10-18) (Dunlop cubic interpolation)
  4. Flow given as Velocity, Velocity (v) 2, Inside diameter (D) 7.874 in, Pipe length (L) 3,281 ft, Pipe material Cast iron, uncoated (0.00085 ft), Fluid Water at 10 °C (50 °F) gives Reynolds number 306,210.276438, Friction factor f 0.021877, Colebrook friction factor 0.021731.Source: Hydraulic Institute, HI Data Tool, Pipe Frictional Losses (Eq. 3.A.1 h_f = f × (L ÷ D) × v² ÷ 2g; Eq. 3.A.2 Colebrook; Eq. 3.A.3 explicit approximation; Eq. 3.A.5 f = 64 ÷ Re below 2,000; Table 3.A.1 roughness), https://datatool.pumps.org/fluid-flow-iii/general (retrieved 2026-10-18) (Colebrook and the explicit form); US EPA, EPANET Users Manual, 13. Analysis Algorithms, §13.1 (Hagen-Poiseuille below Re = 2,000, Swamee and Jain above 4,000, Dunlop’s cubic interpolation between), https://epanet-manual.readthedocs.io/en/latest/analysis_algorithms.html (retrieved 2026-10-18); NIST Chemistry WebBook, SRD 69, Isobaric Properties for Water at 0.101325 MPa (20 °C: 998.21 kg/m³, 0.0010016 Pa·s), https://webbook.nist.gov/cgi/fluid.cgi?Action=Load&ID=C7732185&Type=IsoBar&Digits=5&P=0.101325&THigh=100&TLow=10&TInc=10&RefState=DEF&TUnit=C&PUnit=MPa&DUnit=kg%2Fm3&HUnit=kJ%2Fkg&WUnit=m%2Fs&VisUnit=Pa*s&STUnit=N%2Fm (retrieved 2026-10-18)

How it works

  • Velocity from a flow rate: v = Q ÷ (πD² ÷ 4).
  • Reynolds number Re = ρvD ÷ μ, and the relative roughness ε ÷ D.
  • Friction factor f (the EPANET method):
    • Re < 2,000 (laminar): f = 64 ÷ Re.
    • Re > 4,000 (turbulent): f = 0.25 ÷ [log10(ε ÷ 3.7D + 5.74 ÷ Re^0.9)]² (Swamee and Jain; HI Eq. 3.A.3).
    • 2,000 ≤ Re ≤ 4,000 (transitional): with y = ε ÷ 3.7D + 5.74 ÷ 4,000^0.9, Y3 = −2 log10(y), FA = 1 ÷ Y3², FB = FA × (2 − (3.6 ÷ ln 10) × (5.74 ÷ 4,000^0.9) ÷ (y × Y3)) and R = Re ÷ 2,000: f = X1 + R × (X2 + R × (X3 + R × X4)), with X1 = 7FA − FB, X2 = 0.128 − 17FA + 2.5FB, X3 = −0.128 + 13FA − 2FB, X4 = 0.032 − 3FA + 0.5FB. This is EPANET's cubic: it equals 64 ÷ Re at 2,000 and the turbulent formula, with the same slope, at 4,000. (The EPANET manual writes y with the actual Re; its source code, which this page follows, takes it at 4,000.)
  • Pressure drop Δp = f × (L ÷ D) × ρv² ÷ 2, and the head loss h = Δp ÷ (ρ × 9.80665 m/s²).
  • Colebrook factor, for comparison when Re > 4,000: the root of 1 ÷ √f = −2 log10(ε ÷ 3.7D + 2.51 ÷ (Re√f)) (HI Eq. 3.A.2). When ε ÷ 3.7D is 1 or more (the roughness is at least 3.7 times the diameter), the right side is below 0 for every f, so there is no root and the page leaves the Colebrook factor out. The page compares ε and 3.7D on the typed values, so 0.37 mm in a 0.1 mm pipe counts as exactly 1.

Pipe roughness ε (Hydraulic Institute Table 3.A.1, in feet, 1 ft = 0.3048 m): drawn tubing (glass, brass, plastic) 0.000005; brass or copper tubing 0.000005; fiberglass 0.0000164; rubber, smoothed 0.000033; stainless steel 0.00005; commercial steel or wrought iron 0.00015; cast iron, asphalt dipped 0.0004; galvanized iron 0.0005; cast iron, uncoated 0.00085; concrete 0.001 (smooth end) to 0.01 (rough end). Or type your own ε.

Water at 1 atm (NIST Chemistry WebBook): 10 °C, 999.70 kg/m³ and 0.0013059 Pa·s; 20 °C, 998.21 and 0.0010016; 40 °C, 992.22 and 0.00065273; 60 °C, 983.20 and 0.00046604; 80 °C, 971.79 and 0.00035405. Or type your own density and viscosity (1 mPa·s = 1 cP = 0.001 Pa·s; 1 P = 0.1 Pa·s).

Flow units: 1 US gal/min = 0.003785411784 ÷ 60 m³/s; 1 L/s = 0.001 m³/s; 1 L/min = 0.001 ÷ 60 m³/s; 1 m³/h = 1 ÷ 3,600 m³/s; 1 ft³/s = 0.028316846592 m³/s.

Rules

  • Flow rate from 10⁻⁹ to 10⁹ in its unit; velocity from 10⁻⁶ to 100 m/s; diameter from 0.1 mm to 10 m; length from 1 mm to 10,000 km; roughness from 0 to 0.1 m; a typed density from 0.01 to 30,000 kg/m³ and viscosity from 10⁻⁷ to 10⁶ in its unit.
  • A pressure drop that is 0 or not finite has no answer.

Output format

The pressure drop shows to 5 significant figures in psi (kPa in Metric) and also in Pa and psi (1 psi = 6,894.757 Pa); the head loss in metres and feet, the velocity, flow rate, relative roughness and density to 5; Re and the friction factors to 6. Every result is a 64-bit float (logarithms and powers). Flow regime: Laminar below 2,000, Transitional (critical zone) from 2,000 to 4,000, Turbulent above 4,000.

Worked examples by hand

Laminar oil. ρ = 900 kg/m³, μ = 0.1 Pa·s, v = 1 m/s, D = 50 mm, L = 10 m: Re = 900 × 1 × 0.05 ÷ 0.1 = 450, laminar; f = 64 ÷ 450 = 0.142222; Δp = 0.142222 × (10 ÷ 0.05) × 900 × 1² ÷ 2 = 12,800 Pa. Poiseuille's law gives the same: Δp = 8ηLv ÷ r² = 8 × 0.1 × 10 × 1 ÷ 0.025² = 12,800 Pa.

Water in steel pipe. 30 gpm of water at 20 °C in 100 ft of 2 in commercial steel: Q = 0.00189271 m³/s, A = 0.00202683 m², v = 0.933826 m/s; Re = 998.21 × 0.933826 × 0.0508 ÷ 0.0010016 = 47,277.8; ε ÷ D = 0.00015 ÷ (2 ÷ 12) = 0.0009; f = 0.25 ÷ [log10(0.0009 ÷ 3.7 + 5.74 ÷ 47,277.8^0.9)]² = 0.0240783; Δp = 0.0240783 × (30.48 ÷ 0.0508) × 998.21 × 0.933826² ÷ 2 = 6,287.8 Pa = 0.911971 psi, or 0.642329 m (2.11 ft) of head. Colebrook gives f = 0.0239359.

Transitional flow. Water at 20 °C at 0.3 m/s in a 10 mm drawn tube: Re = 998.21 × 0.3 × 0.01 ÷ 0.0010016 = 2,989.85, so the cubic applies: f = 0.0330428, and over 1 m Δp = 148.43 Pa.

Old cast iron main. Water at 10 °C at 2 m/s in 1 km of 200 mm uncoated cast iron: Re = 999.7 × 2 × 0.2 ÷ 0.0013059 = 306,210; ε ÷ D = 0.00129540; f = 0.0218767 (Colebrook 0.0217312); Δp = 0.0218767 × 5,000 × 999.7 × 4 ÷ 2 = 218,702 Pa = 31.72 psi.

Other questions people ask

How do you calculate pressure drop in a pipe?

Use the Darcy-Weisbach equation: Δp = f × (L ÷ D) × ρv² ÷ 2, with f the friction factor, L the pipe length, D the inside diameter, ρ the density and v the mean velocity. 30 gpm of water in 100 ft of 2 in steel pipe loses about 0.91 psi.

How is the friction factor found?

From the Reynolds number Re = ρvD ÷ μ and the relative roughness ε ÷ D. In laminar flow (Re below 2,000), f = 64 ÷ Re. In turbulent flow (above 4,000) the page uses the Swamee-Jain formula 0.25 ÷ [log10(ε ÷ 3.7D + 5.74 ÷ Re^0.9)]², as the EPA’s EPANET does, and shows the Colebrook value beside it.

What happens between Re = 2,000 and 4,000?

The flow is unstable and can switch between laminar and turbulent, so the friction factor is uncertain. The page follows EPANET and joins the two curves with a smooth cubic, so the answer does not jump.

What is head loss?

The pressure drop written as a height of the flowing fluid: h = Δp ÷ ρg. A head loss of 2.11 ft of water is the same as about 0.91 psi.

How rough are common pipes?

From the Hydraulic Institute: plastic and drawn tubing 0.000005 ft (0.0015 mm), stainless steel 0.00005 ft, commercial steel 0.00015 ft (0.046 mm), galvanized iron 0.0005 ft, uncoated cast iron 0.00085 ft (0.26 mm), and concrete 0.001 to 0.01 ft. The HI adds that commercial installations may need 15% more.

Does the pressure drop depend on the fluid?

Yes. Density and viscosity set the Reynolds number and the friction factor. Warm water is thinner than cold water, so it loses a little less pressure at the same flow.