Edward J. Hickin: River Hydraulics and Channel Form
Chapter 3
The momentum equation for open-channel flow
The momentum equation and the hydraulic jump
Hydraulic jumps
Deriving the momentum equation
Application of the momentum equation to the hydraulic jump
Energy loss in hydraulic jumps
Some concluding remarks
References
We have seen that the energy equation has many useful applications to a wide variety of flow
transition problems. In all cases, however, the solutions depend on the assumption that energy is
conserved. If energy actually leaks from the system via frictional head loss the Bernoulli
equation will overstate the energy available to the flow and the related predictions of velocity
and depth will proportionately be in error. To recall our earlier strategy, we minimize this error
by considering only short reaches of channel and only gradual transitions. In certain flow
phenomena, however, we simply can no longer ignore the energy losses and we must look to
alternative ways of describing the flow.
The momentum approach and the hydraulic jump
Hydraulic jumps
One of these cases is the hydraulic jump. Hydraulic jumps mark the flow transition from
supercritical to subcritical flow. When subcritical flow accelerates into the supercritical state the
transition often is smooth with gradually increasing velocity and decreasing depth bringing about
a smooth drop in the water surface until the alternate depth is achieved. Any disturbance to the
water surface is smoothed out by the surface or gravity wave propagation mechanism discussed
earlier. In these circumstances energy losses are not great and the Bernoulli equation does a
Chapter 3: The momentum equation for open-channel flow
credible job of describing the changes to the flow.
When supercritical flow changes to
subcritical flow, however, there is no smoothing of the water surface upstream of the transition
because the high downstream velocity prevents upstream diffusion of the water-surface
deformation. As a result the transition to subcritical flow is sudden and marked by an abrupt
discontinuity, or hydraulic jump, in the water
Hydraulic
jump
with breaking
surge
An undular hydraulic
jump
F>1.0
Subcritical
Flow F<1.0
Accelerating
flow
Supercritical flow F>1.0
Subcritical
Flow F<1.0
3.1: Undular and breaking hydraulic jumps in open-channel flow
surface (Figure 3.1). The greater the difference between the alternate depths the more severe the
hydraulic jump.
Hydraulic jumps appear in the surface of rivers as standing waves or surges. In their most fully
developed state they appear as a line of breaking waves normal to the river bank which remains
stationary with respect to an observer on the bank.
Upstream of the jump the flow is
supercritical and downstream it is subcritical. Hydraulic jumps may occur as a single event in a
reach of channel or they may occur in trains as flowalternates back and forth across the two
flow states. They may be steady-state features or they may be intermittent or even periodic,
forming and collapsing, and reestablishing again, in response to the highly unsteady non-uniform
flow. As we might expect, hydraulic jumps are more likely to be found disrupting flows in steep
mountain streams rather than those in large low-slope rivers although even the latter may exhibit
this flow phenomenon at high flood discharges. Because the flow at the hydraulic jump often is
surging and highly agitated, there is a great deal of energy lost here, precisely the circumstance
we must avoid if the Bernoulli equation is the only analytical tool at our disposal.
3.2
Chapter 3: The momentum equation for open-channel flow
Deriving the momentum equation
An alternative to the energy conservation approach is one based on the momentum principle
(also known as the momentum-impulse principle). The principle is derived from Newton's
second law of motion; if we multiply equation (1.7) by time t we obtain
Ft = mat ............................................................ (3.1)
Since a =
v2 - v1
(from equation (1.1)),
t
Ft = m(v2 - v1)................................................. (3.2a)
We have in effect integrated both sides of equation (3.1) with respect to time so that in
general,
t2
t2
⌠
⌠
⌡ F dt = m ⌡ a dt = m(v2 - v1) ....................……..............(3.2b)
t1
t1
In words, the momentum equation states that the change in momentum per unit time in any body
is equal to the resultant of all the external forces acting on the body in unit time; the product Ft is
known as the impulse of the force.
Equation (3.2) can be applied to open-channel flow by reference to the control volume of channel
flow in the definitional diagram in Figure 3.2; we assume that the channel has a rectangular
cross-section.
Rearranging equation (3.2) we can say that:
F =
m(v2 - v1)
..................................................................... (3.3)
t
For the control volume, Vo, we note that m = rVo and that equation (3.3) can be written:
F=
ρVo ( v 2 −v1 )
....................…….........................………...........(3.4)
t
€
3.3
Chapter 3: The momentum equation for open-channel flow
1
water
surface
Since Vo/t = Q, we can also state that
2
F = ρQ(v2 - v1) =
L
Wsinθ
p1
V1
y V
2
p2
The net force F is the resultant of three types of force
acting on€the fluid body:
W=γVo
θ
τ0
γ
Q(v2 - v1)..................…....... (3.5)
g
the force resulting from the pressure difference at
sections 1 and 2: p1A1 - p2A2;
(a)
z1
z2
(b)
the force exerted by the downstream component of
the weight of the fluid body: γVosinϕ; and
Horizontal datum
3.2: Definitional diagram for the
momentum equation.
(c)
the force to of friction and resistance acting over the
boundary area, Ab: - toAb.
Substituting these components for the net force in equation (3.5) yields:
p1A1 - p2A2 + γVosinθ - τoAb =
γ
Q(v2 - v1)..........................…........ (3.6)
g
Equation (3.6) is the general form of the momentum equation for open channel flow. The twodimensional form of the momentum equation is obtained by dividing equation (3.6) throughout
€
by the channel width:
p1y1 - p2y2 +
€
γVo sin ϑ − τ o Ab
γ
= q(v2 - v1)............................... (3.7)
w
g
€
Application of the momentum equation to the hydraulic jump
Equation (3.7) can be further adapted to describe the particular conditions at the hydraulic jump
(see Figure 3.3). If we assume that the force of the downslope weight component of the flow
(γVosinϕ) is exactly balanced by the force of frictional resistance (τoAb), and assume further that
y
γy
the pressure terms simply are hydrostatic (the average at depth 2 , p =
), equation (3.7)
2
reduces to:
€
3.4
Chapter 3: The momentum equation for open-channel flow
γy12 γy 2 2 γ
= q(v2 - v1)
g
2
2
y12
y22
q
g (v2 - v1) = 2 - 2 .............……....................... (3.8)
or
q
Substituting y = v in the left-hand side of equation
€
€
€
(3.8)
and factoring the right-hand side, yields:
ΔH
H1
C
2
V1 /2g
q  q q 
1
=


2 (y1 + y2) (y1 - y2) .
g y2 y1
2
V2 /2
g
surge
y2
y1
V1
H2
Rearranging this expression gives
V2
1 1
 - 
q2 y2 y1
1
g y1 - y2 = 2 (y1 + y2) ........................... (3.9)
3.3: Flow characteristics at a
1 1
Noting that y1y2 y - y  = y1 - y2, and that
2
2
hydraulic jump.
q = v1y1, it follows that
v12y12 v12y1 1
gy1y2 = gy2 = 2 (y1 + y2)
y2
v12
Multiplying throughout by y yields g
1
and
v12
1
=
2
g
1 y2
= 2 y (y2 + y1)
1
y22 y1y2


 y1 + y1 
Taking y1 outside the right-hand bracket yields
v12
g
 2

1 y2
= 2  y + y2
1
y22 y2
1
= 2 y1  2 + y 
y1
1

v12 1 y2 y2
and further rearranging gives gy = 2 y y + 1 ……...............…………..................... (3.10)
1
1
1
Equation (3.10) can be cast into several different forms, including two expressions which
appropriately relate the geometry of the simple hydraulic jump to the Froude number:
v12
=
gy1
F12 =

1 y 2  y2
+ 1 ..................................................... (3.11)
2 y1  y1

- 3.5 -
Chapter 3: The momentum equation for open-channel flow
v1
= F1 =
gy1

1 y2  y2
+ 1 .................................................... (3.12)

2 y1  y1

Equation (3.12) has proved to be an acceptable description of the simple hydraulic jump and is
widely used in solving a variety of practical problems (see Sample Problems 3.1 and 3.2). It
usefully illustrates that the severity of the hydraulic jump, summarized by y2/y1, is a function of
the state of the flow specified by the Froude number.
With a little more algebraic gymnastics we can cast equation (3.12) into a form that isolates y2/y1
as a function of the Froude number only:

y2 y2
 or F 2 =
+
1
1

y1 y1
y2
  2 + y2
y1
y1
From equation (3.11) we have F12 =
1
2
and
y2
y2
2F12 = y  2 + y .................................................... (3.13)
1
1
1
2
In order to extract simple roots we multiply both sides of equation (3.13) by 4 and add 1 to both
sides, thus:
y2
y2
8F12 + 1 = 4 y  2 + 4 y + 1
1
Taking roots,
 y2

8F12 + 1 = 2 y + 1
so that
y2
y1 =
1
1
1
2
(
)
8F12 + 1 - 1
................................... (3.14)
y13
v12
We might note also that continuity gives the equality gy = F12 =  2 F22 and substituting
y2
1
Sample Problem 3.1
Problem: Water flows along a 10-m wide rectangular channel and through a hydraulic jump. If the flow depth just
before the jump is 2.0 m, and 3.0 m after it, what is the discharge through the channel?
Solution: From equation (3.12) we get
v 12
1
3.0 3.0 
9.806y1 = 2 x 2.0 2.0 +1
so that v12 = 36.7725 and v1 = 6.064 ms-1. Thus discharge, Q = wv1y1 = 10 x 6.064 x 2.0 = 121.3 m3s-1.
- 3.6 -
Chapter 3: The momentum equation for open-channel flow
Sample Problem 3.2
sluice gate
Problem: A sluice gate at the base of a large reservoir is raised
1.7 m, as shown opposite, and the water discharges through
this 5.0 m-wide rectangular orifice into a rectangular channel of
the same width. If a hydraulic jump forms in the channel, what
will be its height?
reservoir
3.0 m
hydraulic
jump
1.7 m
?
Solution: Since the reservoir is large, we can assume that the water depth will not change rapidly and that the depth
of water in the tank represents the total head E1 = 3.0 m. Thus we can write the Bernoulli equation for this case as
E1 = y2 + v22/2g
v 22
or 3.0 = 1.7 + 2 x 9.806 .
Therefore, v2 = 1.3 x 19.612 = 5.049 ms-1
5.049
The flow discharging from under the sluice gate is supercritical since F =
= 1.237
9.806 x 1.70
If a hydraulic jump forms in the channel its geometry is specified by equation (3.56) as
y2
1
y1 = 2
(
(8 x 1.2372) + 1 - 1
)
= 1.319
Since y1 = 1.70 m, y2 = 2.242 m so that the hydraulic jump must be 0.542 m high.
for F12 in equation (3.11) leads to:

v22
1 y1 y1
 .................................…........... (3.15)
2=
=
F
+
1
2

2 y2 y2
gy2
Furthermore, from equation (3.15) it can be shown readily that :
y1
y2 =
1
2
(
8F22 + 1 - 1
) ..................................……............. (3.16)
a sort of mirror image of equation (3.14). Obviously equation (3.16) will be very useful when only
the subcritical flow conditions downstream of the hydraulic jump are known.
Energy loss in hydraulic jumps
To obtain the energy loss across a hydraulic jump in a rectangular channel, we have to compute
E1-E2 when M1 = M2. It will be found that unless F12 and y2/y1 are quite large, the difference E1E2 is substantially smaller than either E1 or E2, making it difficult to calculate the difference with
reasonable accuracy.
- 3.7 -
Chapter 3: The momentum equation for open-channel flow
This is a common problem in computing work; the obvious but clumsy remedy is to calculate E1
and E2 to a higher degree of precision than is required for the difference E1-E2. A much better
way out of this difficulty, one suggested by the influential river engineer Professor Frank
Henderson (1966), is outlined below. He recommended rearranging the algebra so that E1-E2 is
expressed in a form involving products rather than differences.
We have
E1-E2 = (y1 +
v 12
2g
) - (y2 +
v 22
2g
) = (y1 - y2) +
q2  1
1 
 2 − 2
2g  y1
y2 
In order to express (E1-E2) as a product, it will be useful to factorize the expression as far as
possible.
So, making the substitution F12 =
q2
gy13
, we take out the factor (y1 - y2) leading to

F 2 y1(y1 + y2 ) 
 ...............……....................... (3.17)
y 22

E1- E2 = (y1 - y2) 1− 1
2

Now we impose the condition M1 = M2 by using the hydraulic jump equation:
F1 2 =

1 y 2  y2
+ 1 .

2 y1  y1

Setting r =
y2
and simplifying the expression within the bracket yields:
y1


E1- E2 = (y1 - y2) 1−
or
and finally,
r(r + 1) (r + 1) 
...............................................……….................(3.18)
4
r 2 

E1 − E 2
(r + 1)2 
= (1− r)1−

y1
4r 

E1 − E 2 (r − 1)3
=
...................................................................................(3.19)
y1
4r
Thus the energy loss at a hydraulic jump becomes a simple function of the relative flow depths in
question (see Sample Problem 3.3). Of course this resulting increased ease of computation
offered by the Henderson method does not obviate the user from the task of assessing the
significance of the result in the light of the precision of the primary measurements.
- 3.8 -
Chapter 3: The momentum equation for open-channel flow
Sample Problem 3.3
For example, in a hydraulic jump where y1 = 1.000 and y2 = 1.100 (r =
E1- E2 = (1.000)
y2
= 1.100), equation (3.19) yields:
y1
(1.100− 1)3
= 0.00023 m.
4(1.100)
In other words, energy loss through this mild hydraulic jump amounts to about two fifths of one millimetre of head.
In the case of severe hydraulic jumps, energy losses may be very large indeed (see Sample
Problem 3.4).
Sample Problem 3.4
In a hydraulic jump where y1 = 1.000 and y2 = 2.000 (r =
E1- E2 = (1.000)
y2
= 2.0), equation (3.61) yields:
y1
(2.000 − 1.000)3
= 0.125 m.
4(2.00)
Equation (3.14) yields F1=1.73 for these flow conditions, associated with a head loss amounting to 12.5% of the
energy of the approach flow. As we noted earlier, hydraulic jumps can be very efficient dissipaters of flow energy
and are sometimes designed by engineers to form in outlet flows where the approach flow might otherwise cause
damaging scour to the channel bed.
Some concluding remarks
The momentum equation is conveniently brought to bear on problems such as the task of
describing the hydraulic jump where the flow is far less than perfectly energy conservative as the
component DH in Figure 3.3 indicates. Indeed, all real flows in nature dissipate energy in
overcoming frictional resistance and our discussion of flow has reached a point where it is no
longer useful to ignore these energy losses. Ways of incorporating these sources of resistance
into our thinking about river flows is the subject of the next chapter.
References
Henderson, F.M., 1966, Open Channel Flow, Macmillan, New York.
- 3.9 -