1.

6 Specific Energy
 The amount of energy expressed in terms of head or is
the energy per unit weight of liquid at any section of
the channel measured with respect to the channel
bottom.
 Specific energy is the energy at a cross-section of an
open channel flow with respect to the channel bed.
 The concept of specific energy first introduced in 1932
and it was very useful in analysis of open channel flow
 Total energy (datum), H= Z+ Ycosθ + αV2/2g
 Specific energy (E) : if datum coincides with channel
bed
 E= Ycosθ + αV2/2g
 When cosθ =1 and α =1, the equation of specific energy is
further simplified as , E= Y + V2/2g.
 It indicates that the specific energy is equal to the
sum of the depth of water and the velocity head.
 Therefore, we defined:
 Specific energy is the energy at a cross-section of an open
channel flow with respect to the channel bed.
Specific energy is the height of the energy grade line above
the channel bottom.
 In other respect, since V=Q/A, the equation of specific
energy may be written as:
Substitute v=Q/A, E= Y + Q2/2gA2
Here, cross-sectional area A depends on water
depth y and can be defined as, A = f(y) and also there
is a functional relation between the three variables as, f
( E , y , Q ) relates E, Y and Q. In a general case, all three
quantities vary, but particular interest attaches to those
instances in which Q is constant while Y and E vary, and those
in which E is constant while Y and Q vary.
 In order to examine the functional relationship on the
plane, two cases are introduced:
1. Constant discharge: Q = Q1 E = f (y, Q). Variation of
the specific energy with the water depth at a cross-
section for a given discharge Q1.
2. Constant Specific Energy E = E1 E = f (y, Q). Variation
of the discharge with the water depth at across-section
for a given specific energy, E .
Generally, most of time in open channel flow
(Canal), we have regulated flow. Which implies
Discharge, Q is constant.

E = y + Q2/2gA2
1. Constant discharge situation

Fig. 2.1 Specific Energy Diagram / Curve

 Specific energy
 For a channel of known geometry, E=f(Q, Y), Keeping Q =
constant, the variation of E with y is represented by a cubic
parabola. (Figure 2.1) is called specific energy diagram.
 It is seen that there are two positive roots for the equation E1
indicating that any particular discharge Q1 can be passed in a
given channel at two depths and still maintain the same specific
energy E1.
 PR = y1 or PR` = y1`, two possible depths having the same
specific energy are known as alternate depths
 For a given Q, as the specific energy is increased the
difference between the two alternate depths increases. On
the other hand, if E is decreased, the difference (y1` – y1)
will decrease and a certain value E = Ec, the two depths will
merge with each other (at point C in Fig. 2.1). No value for
y can be obtained when E < Ec denoting that the flow
under the given conditions is not possible in this region.
Thus this condition of minimum specific energy is known as
the critical flow condition.
 The flow condition at point C is called the critical flow the
corresponding depth is called critical depth. Therefore, the
critical flow condition also corresponds to the condition
for maximum discharge in the channel for a fixed specific
energy.
 1.6.1 Critical Depth
 At critical depth, the specific energy is minimum, if
differentiating the equation of specific energy with
respect to y (keeping Q constant) and equating to zero,

It is the basic equation governing the critical flow conditions in open

channel flow.
 Critical Depth
It may be noted that the critical flow condition is governed
solely by the channel geometry and discharge. Other channel
characteristics such as the bed slope and roughness do not
influence the critical flow condition for any given Q.
 The critical flow
corresponds to the
minimum specific energy
and at this condition the
Froude number of the
flow is unity.
 Critical Depth
Computation of Critical Depth
A) Rectangular Channel
 Critical Depth
B. Triangular channel

For triangular channel having side slope of m or (H:V= m: 1)

The Froude number for a triangular

channel be
 Critical depth
C. Circular Channel
Let D be the diameter of a circular channel and 2θ be the angel in radians subtended by the
water surface at the center.
Critical Depth
 Examples
1.12. A rectangular channel 2.50 m wide has a specific
energy of 1.50 m when carrying a discharge of 6.48 m3
/sec. Calculate the alternate depths, corresponding
Froude numbers, critical depth and critical energy .

1.13. A flow of 5.0 m3 /sec is passing at a depth of 1.50 m

through a rectangular channel of width 2.50 m. What
is the specific energy of the flow? What is the value of
the alternate depth to the existing depth?
 Example 1.14
Calculate the critical depth and corresponding specific
energy for a discharge of 5.0m3/sec in the following
channels:
a) Rectangular channel, B = 2.0 m
b) Triangular channel, m = 0.5
c) Trapezoidal Channel, B = 2.0 m, m = 1.5
d) Circular channel, D = 2.0 m and θ = 600
 1.7. Flow in Channel Transitions
(Application of Bernoulli equation)
 Steady uniform flow is interrupted by a raised bed level as
shown. If the upstream depth and discharge are known, we can
use Bernoulli’s equation and the continuity equation to get the
velocity and depth of flow over the raised hump.

Fig. Uniform flow interrupted by raised hump

 Applying Bernoulli’s equation between sections 1 and 2
(assuming a horizontal rectangular channel z1 = z2 )

 Applying specific energy equation at section 1 and 2, taking

datum as bed of the channel:
Es1 = Es2 +Δz
Example 1.15
A rectangular channel with a flat bed has width 5m and a
discharge of 10 m3/sec. The normal depth is 1.25m. What is the
depth of flow in a section in which the bed rises 0.2m.
 Example 1.16

A rectangular channel has 5m width and 3m

water depth. If the bed slope of the channel is 1 in
1200, find the maximum height of the hump
without changing the water level in the upstream.
Take manning’s n = 0.02.
 1.8. Hydraulic Jump
 A hydraulic jump occurs when a super-critical flow
transit to a sub-critical flow. The jump is the mechanism
for the two surfaces to join. They join in an extremely
turbulent manner which causes large energy losses.
 when supercritical flow has its velocity reduced to sub
critical, there is sudden rise in water level at the point
where hydraulic jump occurs. At this moment it
dissipates much of the destructive energy of the high –
velocity water, thereby reducing downstream erosion.
 The main importance of hydraulic jump
 To dissipate energy in water flowing over
dams, weirs, sluice gate etc and thus prevent
scouring downstream from structures;
 To recover head or raise the water level on
the downstream and thus divert water for
irrigation or water distribution purposes;
 To increase the water depth (weight) in apron
to counteract the uplift pressure;
 To mix chemicals used for water purification,
etc; and
 To aerate water for city water supplies
General equations used in hydraulic jump
 The two alternative / conjugated depths are obtained
by:

 Energy dissipated by hydraulic jump between section 1

and 2:
 Example 1.17
 A hydraulic jump occurs on downstream side of the
under sluice of rectangular cross section and the
following values are obtained.
Initial depth = 0.5m
Velocity in upstream, v1 = 7 m/sec
Froude number, Fr1 = 2
Determine the sequent depth, velocity, Froude
number after the jump, and the energy losses.
 Exercise
 Water flow in a wide rectangular channel at q = 10 m2/sec
and depth of flow at section 1 is 1.25m. If the flow under goes
a hydraulic jump, determine:
A. Sequent depth D. Head loss / energy dissipated
B. Velocity in downstream E. % energy dissipated
C. Froude number in downstream F. The power dissipated per unit
width