Learn: Open-channel specific energy
Specific energy measures open-channel energy relative to the local channel bed. It includes water depth and velocity head, but not the bed elevation above an external survey datum.
What the formula is saying
Add depth y to v²/(2g). Depth represents the pressure-and-elevation contribution measured from the bed under the hydrostatic approximation.
Read the symbols in plain language
- y
- Flow depth
Flow depth. Add water depth without adding an external bed elevation.
mMetres measure length; 1 m = 1000 mm.
- v
- Mean velocity
Mean velocity. This positive term accounts for kinetic energy relative to liquid weight.
m/sUse m/s as the base unit shown here. y and specific energy E are in m; v is m/s and g is m/s². This E is a head in metres, not energy in joules and not elastic modulus.
- g
- Gravity
Gravity. This positive term accounts for kinetic energy relative to liquid weight.
m/s²Use m/s² as the base unit shown here. y and specific energy E are in m; v is m/s and g is m/s². This E is a head in metres, not energy in joules and not elastic modulus.
- E
- Result to find
Open-channel specific energy. Add water depth without adding an external bed elevation.
m
Sort out the units first
y and specific energy E are in m; v is m/s and g is m/s². This E is a head in metres, not energy in joules and not elastic modulus.
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 E and explain the result in the stated output unit.
- y · Flow depth
- 1 m
- v · Mean velocity
- 2 m/s
- g · Gravity
- 9.81 m/s²
Calculate motion-related head
This positive term accounts for kinetic energy relative to liquid weight.
(2)^2 ÷ (2 × (9.81)) ≈ 0.2038735984 mReference the energy to the local bed
Add water depth without adding an external bed elevation.
(1) + (0.2038735984) ≈ 1.203873598 m
Does this worked answer make sense?
Specific energy cannot be smaller than y for nonnegative depth. At zero speed it equals y exactly.
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.
- y · Flow depth
- 0.6 m
- v · Mean velocity
- 3 m/s
- g · Gravity
- 9.81 m/s²
Calculate motion-related head
This positive term accounts for kinetic energy relative to liquid weight.
(3)^2 ÷ (2 × (9.81)) ≈ 0.4587155963 mReference the energy to the local bed
Add water depth without adding an external bed elevation.
(0.6) + (0.4587155963) ≈ 1.058715596 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.
- y · Flow depth
- 1.2 m
- v · Mean velocity
- 2.5 m/s
- g · Gravity
- 9.81 m/s²
Find: Learn: Open-channel specific energy
A hint, not the answer
Add depth y to v²/(2g). Depth represents the pressure-and-elevation contribution measured from the bed under the hydrostatic approximation.
y and specific energy E are in m; v is m/s and g is m/s². This E is a head in metres, not energy in joules and not elastic modulus.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Calculate motion-related head
This positive term accounts for kinetic energy relative to liquid weight.
(2.5)^2 ÷ (2 × (9.81)) ≈ 0.3185524975 mReference the energy to the local bed
Add water depth without adding an external bed elevation.
(1.2) + (0.3185524975) ≈ 1.518552497 m
Avoid the common trap
Do not add bed elevation again when asked only for specific energy. Do not forget that speed must be recalculated when depth changes at fixed discharge.
When this method applies — and when it does not
Use a hydrostatic open-channel section and velocity-head coefficient equal to one. Comparing different sections requires bed elevations and losses. A specific-energy value alone does not choose the subcritical or supercritical depth at a given discharge.
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: Open-channel specific energy. 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.