Learn: Bernoulli total head at a section
Total hydraulic head expresses mechanical energy per unit weight as an equivalent height of liquid. A flowing liquid can carry that energy as elevation, pressure, or motion.
What the formula is saying
Add elevation head z, pressure head p/(ρg), and velocity head v²/(2g). Dividing pressure by unit weight turns it into a liquid-column height, while squaring velocity converts motion into an energy term.
Read the symbols in plain language
- z
- Elevation head
Elevation head. Elevation, pressure head and velocity head all use metres and the same reference convention.
mMetres measure length; 1 m = 1000 mm.
- p
- Pressure
Pressure. Pressure divided by liquid weight per unit volume is an equivalent column height.
PaOne pascal is one newton per square metre. 1 MPa = 10⁶ Pa.
- ρ
- Fluid density
Mass per unit volume for the stated material and condition. This is density, not weight per volume.
kg/m³Use kg/m³ as the base unit shown here. Use z in m, p in Pa, density ρ in kg/m³, velocity in m/s and g in m/s². Every term then has unit m. A pressure of 200 kPa is 200000 Pa, not 200 Pa.
- v
- Velocity
Velocity. The squared speed divided by twice gravity expresses kinetic energy per unit weight.
m/sUse m/s as the base unit shown here. Use z in m, p in Pa, density ρ in kg/m³, velocity in m/s and g in m/s². Every term then has unit m. A pressure of 200 kPa is 200000 Pa, not 200 Pa.
- g
- Gravity
Gravity. Pressure divided by liquid weight per unit volume is an equivalent column height.
m/s²Use m/s² as the base unit shown here. Use z in m, p in Pa, density ρ in kg/m³, velocity in m/s and g in m/s². Every term then has unit m. A pressure of 200 kPa is 200000 Pa, not 200 Pa.
- H
- Result to find
Bernoulli total head at a section. Elevation, pressure head and velocity head all use metres and the same reference convention.
m
Sort out the units first
Use z in m, p in Pa, density ρ in kg/m³, velocity in m/s and g in m/s². Every term then has unit m. A pressure of 200 kPa is 200000 Pa, not 200 Pa.
Assumptions before calculating
Treat the liquid as incompressible with constant density and use section-average velocity. The velocity-head coefficient is taken as 1; all elevations and pressures must use consistent reference levels.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find H and explain the result in the stated output unit.
- z · Elevation head
- 10 m
- p · Pressure
- 200000 Pa
- ρ · Fluid density
- 1000 kg/m³
- v · Velocity
- 2 m/s
- g · Gravity
- 9.81 m/s²
Convert pressure into liquid head
Pressure divided by liquid weight per unit volume is an equivalent column height.
(200000) ÷ ((1000) × (9.81)) ≈ 20.38735984 mCalculate the velocity head
The squared speed divided by twice gravity expresses kinetic energy per unit weight.
(2)^2 ÷ (2 × (9.81)) ≈ 0.2038735984 mAdd the three compatible head terms
Elevation, pressure head and velocity head all use metres and the same reference convention.
(10) + (20.38735984) + (0.2038735984) ≈ 30.59123344 m
Does this worked answer make sense?
With no motion, the velocity term vanishes. Doubling velocity makes its head term four times as large, while shifting the elevation datum shifts all total heads by the same amount.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- z · Elevation head
- 5 m
- p · Pressure
- 150000 Pa
- ρ · Fluid density
- 1000 kg/m³
- v · Velocity
- 3 m/s
- g · Gravity
- 9.81 m/s²
Convert pressure into liquid head
Pressure divided by liquid weight per unit volume is an equivalent column height.
(150000) ÷ ((1000) × (9.81)) ≈ 15.29051988 mCalculate the velocity head
The squared speed divided by twice gravity expresses kinetic energy per unit weight.
(3)^2 ÷ (2 × (9.81)) ≈ 0.4587155963 mAdd the three compatible head terms
Elevation, pressure head and velocity head all use metres and the same reference convention.
(5) + (15.29051988) + (0.4587155963) ≈ 20.74923547 m
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- z · Elevation head
- 8 m
- p · Pressure
- 98000 Pa
- ρ · Fluid density
- 1000 kg/m³
- v · Velocity
- 1.5 m/s
- g · Gravity
- 9.81 m/s²
Find: Learn: Bernoulli total head at a section
A hint, not the answer
Add elevation head z, pressure head p/(ρg), and velocity head v²/(2g). Dividing pressure by unit weight turns it into a liquid-column height, while squaring velocity converts motion into an energy term.
Use z in m, p in Pa, density ρ in kg/m³, velocity in m/s and g in m/s². Every term then has unit m. A pressure of 200 kPa is 200000 Pa, not 200 Pa.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Convert pressure into liquid head
Pressure divided by liquid weight per unit volume is an equivalent column height.
(98000) ÷ ((1000) × (9.81)) ≈ 9.98980632 mCalculate the velocity head
The squared speed divided by twice gravity expresses kinetic energy per unit weight.
(1.5)^2 ÷ (2 × (9.81)) ≈ 0.1146788991 mAdd the three compatible head terms
Elevation, pressure head and velocity head all use metres and the same reference convention.
(8) + (9.98980632) + (0.1146788991) ≈ 18.10448522 m
Avoid the common trap
Do not add raw pressure to elevation, use mass density in place of unit weight without g, or forget to square velocity. A negative gauge pressure can be valid, but does not establish whether cavitation is avoided.
When this method applies — and when it does not
This computes head at one section. Equality of heads at two sections needs the appropriate Bernoulli assumptions; pumps, turbines and head losses must be added separately. Gauge and absolute pressure references must not be mixed.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
One idea understood. Keep going.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Bernoulli total head at a section. Water-measurement principles; for discharge devices, read the orifice/weir chapters and the installation and head-measurement conditions, not only the coefficient formula.
- U.S. Bureau of Reclamation — Water Measurement Manual, 3rd edition (1997; revised reprint 2001)
- Dawei Han, University of Bristol — Concise Hydraulics (2008, Ventus Publishing)
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.