Darcy-Weisbach Equation

Equivalent forms

h_f = f \frac{L}{D} \frac{v^2}{2g}

f = \frac{64}{Re} \text{ (laminar)}

One empirical factor f packages all wall-roughness physics into a single number.

Symbol	Name	SI unit	Dimension	Range
$f$	Friction Factor Darcy friction factor (dimensionless)	—	1	0.005 – 0.1
$L$	Pipe Length Length of the pipe	m	L	1 – 1000
$D$	Pipe Diameter Internal pipe diameter	m	L	0.01 – 1
$\rho$	Density Fluid density	kg/m³	M·L⁻³	0.1 – 14000
$v$	Velocity Mean flow velocity	m/s	L·T⁻¹	0 – 20
$\Delta P$	Pressure Dropoutput Pressure loss due to friction	Pa	M·L⁻¹·T⁻²	0 – 1000000

Unit systems

Symbol	SI	CGS	Imperial
$f$	—	—	—
$L$	m	cm	ft
$D$	m	cm	ft
$\rho$	kg/m³	g/cm³	lb/ft³
$v$	m/s	cm/s	ft/s
$\Delta P$	Pa	dyn/cm²	psi

Where it holds

Fully developed, steady, incompressible flow in a straight pipe of constant cross-section. f is read from the Moody chart for given Re and relative roughness

\varepsilon /D

. Excludes pipe fittings, valves, expansions — those use loss coefficients K.

Dimensional analysis

$[f]\cdot [L/D]\cdot [\rho v^{2}/2] = (1)(1)(M\cdot L^{-1}\cdot T^{-2}) = M\cdot L^{-1}\cdot T^{-2} \checkmark$ (pressure)

Discovery

Henry Darcy & Julius Weisbach · 1845

Weisbach proposed the form in 1845; Darcy provided extensive experimental data on the friction factor from canal-flow studies in 1857. The Moody chart later organized friction factors by Re and roughness.

Try this

Why do skyscrapers need booster pumps for water — and how is that pressure drop calculated?

Water flows at 2 m/s through a 100 m long, 0.1 m diameter pipe with friction factor 0.02. What is the pressure drop?

Research status: stable

Real-world applications

Water distribution networks and skyscraper plumbing
Oil and gas pipeline sizing
HVAC duct design
Hydraulic system engineering (brakes, presses)

Common misconceptions

Don't confuse Darcy friction factor f with Fanning friction factor $f_F = f/4$
f is NOT constant — it depends on Re and relative roughness
Darcy-Weisbach handles friction loss only; minor losses (bends, valves) need separate K coefficients

Experimental verification

Lewis Moody's 1944 chart compiled friction-factor data from Nikuradse's (1933) sand-grain roughness experiments and Colebrook's (1939) commercial-pipe measurements, and remains the standard reference. Modern CFD reproduces Moody data to within 1-2% across Re from

10^{3} to 10^{8}

.

Derivation

Apply a force balance on a cylindrical control volume of length L.

Pressure force

\Delta P\cdot (\pi D^{2}/4)

equals wall shear force

\tau _w\cdot (\pi DL)

.

The wall shear stress is parametrized

as \tau _w = f\cdot (\rho v^{2}/8)

.

Combining:

\Delta P = (4\cdot \tau _w\cdot L)/D = f\cdot (L/D)\cdot (\rho v^{2}/2)

.

The factor f is empirical — for laminar flow

f = 64/Re

, for turbulent flow it depends on Re and roughness via the Colebrook equation.

Limiting cases

v \to 0

⟶

\Delta P \to 0

No flow means no friction loss.

D \to \infty

⟶

\Delta P \to 0

A very wide pipe has negligible friction at fixed velocity.

Laminar (Re < 2300)⟶

f = 64/Re

In laminar flow, Darcy-Weisbach reduces to Poiseuille's law.

What if…

What if you double the pipe diameter?

$\Delta P$ drops by $32\times$ (factor of 1/D plus $v^{2}$ halved by continuity at same Q means another 1/16). Bigger pipes are pump savings.

What if the pipe is very rough?

f increases substantially. In fully rough turbulent flow, f depends only $on \varepsilon /D$ , not Re — there's a permanent friction penalty.

What if flow is laminar?

$Use f = 64/Re$ , and the formula reduces exactly to Poiseuille's law.

1

Water through a long pipe

Given ·

f:: 0.02
L:: 100
D:: 0.1
\rho:: 1000
v:: 2

Find · \Delta P

Steps

$\Delta P = f\cdot (L/D)\cdot (\rho v^{2}/2)$
$L/D = 1000$
$\rho v^{2}/2 = 1000 \times 4 / 2 = 2000\,\mathrm{Pa}$
$\Delta P = 0.02 \times 1000 \times 2000 = 40\,\mathrm{kPa}$

Result ·

\Delta P = 0.02 \times (100/0.1) \times (1000 \times 4/2) = 0.02 \times 1000 \times 2000 = 40000\,\mathrm{Pa}

2

Doubling velocity quadruples loss

Given ·

f:: 0.02
L:: 100
D:: 0.1
\rho:: 1000
v:: 4

Find · \Delta P

Steps

$\Delta P$ scales as $v^{2}$
Doubling v (from 2 to 4 m/s) multiplies $\Delta P$ by 4
$\Delta P = 40\,\mathrm{kPa} \times 4 = 160\,\mathrm{kPa}$

Result ·

\Delta P = 0.02 \times 1000 \times (1000 \times 16 / 2) = 160\,\mathrm{kPa}

Poiseuille's Law→Reynolds Number→Bernoulli's Equation→