Specific Energy

E = y + \frac{V ^{2}}{2 g}

Specific energy is the energy head relative to the channel bottom: depth plus velocity head. For a given discharge, E(y) has a minimum at critical depth (Fr = 1). Subcritical and supercritical alternate depths share the same E.

$E$ (m): Specific energy
$y$ (m): Flow depth
$V$ (m/s): Mean velocity
$g$ (m/s²): Gravity acceleration

Step-by-step solved example

Rectangular flow: y = 0.80 m, V = 3.0 m/s. Find E. (g = 9.81 m/s²)

E = y + V²/2g. Alternate depths sit on the same E curve.

1. Velocity head
V²/(2g) = 9.00/19.62 = 0.459 m.
2. Add depth
E = 0.80 + 0.459 = 1.259 m.
$E = 1.26 m$

Answer: E = 1.26 m

10 practice questions

0/10 correct

1.If V = 0, specific energy equals

2.For a wide rectangle at critical flow, E_min =

3.Two depths with the same E and Q are called

4.y = 2.0 m, V = 3.0 m/s. E is nearest

5.A mild-slope sluice-gate contraction typically moves the flow from

6.Choking occurs when the required E

7.E is measured relative to the

8.Increasing Q at fixed width shifts the E–y curve

9.g = 9.81 m/s², y = 1.2 m, V = 0. Velocity head is

10.Specific energy is NOT the same as