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Engineering formulas & units

Fluid mechanics formulas with units and examples

Ten fluid mechanics formulas, from hydrostatic pressure to pump power and drag, each with symbols, SI units, assumptions and a worked example.

On this page
  1. 1. Hydrostatic pressure
  2. 2. Pascal’s law and the hydraulic press
  3. 3. Buoyancy (Archimedes’ principle)
  4. 4. Continuity equation
  5. 5. Bernoulli’s equation
  6. 6. Reynolds number
  7. 7. Friction loss (Darcy–Weisbach)
  8. 8. Laminar friction factor
  9. 9. Pump power
  10. 10. Drag force

Fluids are in almost every machine: oil in a hydraulic cylinder, coolant in an engine, air over a car. These ten formulas cover pressure, flow and losses, the quantities you need to size a cylinder, a pump or a duct. Water is taken as ρ = 1000 kg/m³ and air as ρ = 1.2 kg/m³, with g = 9.81 m/s².

ΔhV₁V₂FlowSection 1large area A₁, slow, high p₁Section 2 (throat)small area A₂, fast, low p₂p₁/ρgp₂/ρgSame flow rate Q everywhere
Water flowing through a narrowing (a venturi). Continuity makes it speed up in the throat, and Bernoulli's equation says the pressure must drop there, which the lower water column in tube 2 shows. Δh measures the flow rate.

1. Hydrostatic pressure

p=ρ g hp = \rho\,g\,h
SymbolMeaningSI unit
ppGauge pressure at depth hhPa
ρ\rhoFluid densitykg/m³
ggGravitational accelerationm/s²
hhDepth below the free surfacem

Assumptions: the fluid is at rest and incompressible. Add atmospheric pressure (101.3 kPa) for absolute pressure.

Worked example

Find the gauge pressure 10 m below the surface of a water tank.

p=1000×9.81×10=98 100 Pap = 1000 \times 9.81 \times 10 = 98\,100\ \text{Pa}
Answerp = 98.1 kPa (about 1 bar)

2. Pascal’s law and the hydraulic press

F1A1=F2A2⇒F2=F1A2A1\frac{F_1}{A_1} = \frac{F_2}{A_2} \quad\Rightarrow\quad F_2 = F_1 \frac{A_2}{A_1}
SymbolMeaningSI unit
F1,F2F_1, F_2Forces on the small and large pistonsN
A1,A2A_1, A_2Piston areasm²

Assumptions: a confined fluid at rest, pistons at the same height, no friction or leakage. The force is multiplied but the large piston moves less: A1x1=A2x2A_1 x_1 = A_2 x_2.

Worked example

A hydraulic press has a 20 mm diameter input piston and a 200 mm diameter output piston. The operator pushes with 200 N.

F2=200×(20020)2=20 000 NF_2 = 200 \times \left(\frac{200}{20}\right)^2 = 20\,000\ \text{N}
AnswerF₂ = 20 kN

3. Buoyancy (Archimedes’ principle)

FB=ρf g VsubF_B = \rho_f\,g\,V_{sub}
SymbolMeaningSI unit
FBF_BUpward buoyant forceN
ρf\rho_fDensity of the fluidkg/m³
VsubV_{sub}Submerged volume of the bodym³

Assumptions: fluid at rest with uniform density. A body floats when the buoyant force equals its weight before it is fully submerged.

Worked example

An aluminium block (ρ = 2700 kg/m³) of 2 litres is hung under water. Find the buoyant force and its apparent weight.

FB=1000×9.81×0.002=19.6 N,W=2700×9.81×0.002=53.0 NF_B = 1000 \times 9.81 \times 0.002 = 19.6\ \text{N}, \qquad W = 2700 \times 9.81 \times 0.002 = 53.0\ \text{N}
AnswerF_B = 19.6 N; apparent weight 53.0 − 19.6 = 33.4 N

4. Continuity equation

Q=A1V1=A2V2Q = A_1 V_1 = A_2 V_2
SymbolMeaningSI unit
QQVolume flow ratem³/s
AAFlow aream²
VVAverage velocitym/s

Assumptions: steady flow of an incompressible fluid with no leaks. For gases with large density changes use ρ1A1V1=ρ2A2V2\rho_1 A_1 V_1 = \rho_2 A_2 V_2.

Worked example

Water flows at 2 m/s in a 100 mm bore that narrows to 50 mm.

V2=V1(D1D2)2=2×4=8 m/s,Q=π×0.124×2=0.0157 m3/sV_2 = V_1 \left(\frac{D_1}{D_2}\right)^2 = 2 \times 4 = 8\ \text{m/s}, \qquad Q = \frac{\pi \times 0.1^2}{4} \times 2 = 0.0157\ \text{m}^3/\text{s}
AnswerV₂ = 8 m/s; Q = 15.7 L/s

5. Bernoulli’s equation

p1+12ρV12+ρgz1=p2+12ρV22+ρgz2p_1 + \tfrac{1}{2}\rho V_1^2 + \rho g z_1 = p_2 + \tfrac{1}{2}\rho V_2^2 + \rho g z_2
SymbolMeaningSI unit
ppStatic pressurePa
ρ\rhoDensitykg/m³
VVVelocitym/s
zzHeight above a datumm

Assumptions: steady, incompressible, frictionless flow along a streamline, with no pump or turbine between the two points. Real flow loses some pressure to friction; add a head loss term ρghL\rho g h_L to the right-hand side.

Worked example

In the reducer from example 4, the pressure upstream is 200 kPa and the pipe is horizontal. Find the pressure in the 50 mm section.

p2=p1+12ρ (V12−V22)=200 000+500×(4−64)=170 000 Pap_2 = p_1 + \tfrac{1}{2}\rho\,(V_1^2 - V_2^2) = 200\,000 + 500 \times (4 - 64) = 170\,000\ \text{Pa}
Answerp₂ = 170 kPa

A full walk-through with a diagram is in Bernoulli’s equation explained with a solved numerical.

6. Reynolds number

Re=ρVDμ=VDνRe = \frac{\rho V D}{\mu} = \frac{V D}{\nu}
SymbolMeaningSI unit
ReReReynolds numbernone
VVAverage velocitym/s
DDInside diameter (or hydraulic diameter)m
μ\muDynamic viscosityPa·s
ν\nuKinematic viscosity, μ/ρ\mu/\rhom²/s

Assumptions: the limits below are for flow in a round, full bore. Below about 2300 the flow is laminar; above about 4000 it is turbulent; in between it can be either.

Worked example

Water at 20 °C (μ = 1.0 × 10⁻³ Pa·s) flows at 2 m/s in a 100 mm bore.

Re=1000×2×0.11.0×10−3=200 000Re = \frac{1000 \times 2 \times 0.1}{1.0 \times 10^{-3}} = 200\,000
AnswerRe = 200 000: turbulent

7. Friction loss (Darcy–Weisbach)

hf=f LD V22g,Δp=ρghfh_f = f\,\frac{L}{D}\,\frac{V^2}{2g}, \qquad \Delta p = \rho g h_f
SymbolMeaningSI unit
hfh_fHead lost to frictionm
ffDarcy friction factornone
LLLengthm
DDInside diameterm
VVAverage velocitym/s

Assumptions: fully developed flow in a straight, round, full bore. For turbulent flow, ff comes from the Moody chart or the Colebrook equation; it depends on Re and wall roughness. Bends and valves add further losses.

Worked example

Water flows at 2 m/s through 50 m of 100 mm bore with f = 0.02.

hf=0.02×500.1×222×9.81=2.04 m,Δp=1000×9.81×2.04=20.0 kPah_f = 0.02 \times \frac{50}{0.1} \times \frac{2^2}{2 \times 9.81} = 2.04\ \text{m}, \qquad \Delta p = 1000 \times 9.81 \times 2.04 = 20.0\ \text{kPa}
Answerh_f = 2.04 m, Δp = 20 kPa

8. Laminar friction factor

f=64Re⇒Δp=32 μLVD2f = \frac{64}{Re} \quad\Rightarrow\quad \Delta p = \frac{32\,\mu L V}{D^2}
SymbolMeaningSI unit
ffDarcy friction factornone
μ\muDynamic viscosityPa·s
LL, DDLength and diameterm
VVAverage velocitym/s

Assumptions: laminar flow (Re below about 2300) of a Newtonian fluid in a round bore. This is the Hagen–Poiseuille result; it is exact for laminar flow and does not depend on roughness.

Worked example

Oil (ρ = 900 kg/m³, μ = 0.1 Pa·s) flows at 0.5 m/s through a 20 mm bore, 10 m long.

Re=900×0.5×0.020.1=90,Δp=32×0.1×10×0.50.022=40 000 PaRe = \frac{900 \times 0.5 \times 0.02}{0.1} = 90, \qquad \Delta p = \frac{32 \times 0.1 \times 10 \times 0.5}{0.02^2} = 40\,000\ \text{Pa}
AnswerLaminar (Re = 90); Δp = 40 kPa

9. Pump power

P=ρ g Q HηP = \frac{\rho\,g\,Q\,H}{\eta}
SymbolMeaningSI unit
PPShaft power neededW
QQFlow ratem³/s
HHTotal head the pump adds (lift plus losses)m
η\etaPump efficiencynone

Assumptions: steady flow of an incompressible liquid. ρgQH\rho g Q H alone is the hydraulic power given to the fluid.

Worked example

A pump delivers 10 L/s of water against a total head of 30 m at 70 % efficiency.

P=1000×9.81×0.01×300.7=4200 WP = \frac{1000 \times 9.81 \times 0.01 \times 30}{0.7} = 4200\ \text{W}
AnswerP ≈ 4.2 kW

10. Drag force

FD=12 ρ V2 CD AF_D = \tfrac{1}{2}\,\rho\,V^2\,C_D\,A
SymbolMeaningSI unit
FDF_DDrag forceN
ρ\rhoFluid densitykg/m³
VVSpeed relative to the fluidm/s
CDC_DDrag coefficient (from tests or tables)none
AAReference area, usually frontal aream²

Assumptions: CDC_D is roughly constant over the speed range, which holds for bluff bodies at high Reynolds number. The power to overcome drag is FDVF_D V, so it rises with the cube of speed.

Worked example

A car with a frontal area of 2.2 m² and C_D = 0.30 drives at 100 km/h (27.8 m/s).

FD=0.5×1.2×27.82×0.30×2.2=306 N,P=306×27.8=8.5 kWF_D = 0.5 \times 1.2 \times 27.8^2 \times 0.30 \times 2.2 = 306\ \text{N}, \qquad P = 306 \times 27.8 = 8.5\ \text{kW}
AnswerF_D ≈ 306 N, about 8.5 kW to push the air aside

Key points

  • Continuity and Bernoulli together explain most flow problems: where the area falls, speed rises and pressure drops.
  • Check the Reynolds number first; laminar and turbulent flow need different friction factors.
  • Pressure losses grow with the square of velocity, so a slightly larger bore can save a lot of pump power.
  • Drag force grows with the square of speed and drag power with the cube.

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