11.0 The Specific Force
The specific force function is obtained by applying the momentum principle, as shown in Fig.
(21), to a short horizontal reach of a prismatic channel, (The external force of friction and the weight effect of water can be ignored), thus:
1
C.G.
y
1
2 v
1
A
1
F
1 bed v
2
F
2
Sec 1
Fig. (21) Momentum Force in open channel
Momentum Eqn. between sections 1 & 2

F = Rate of change of momentum
F
1

F
2

 g
Q (v
2
 v )


A y
1 1
 

A y
2 2

 g
Q (v
2
 v )
1

 g
Q

Q
A
2

Q
A
1


Q gA
F

2
1


A y
1 1
Q gA
2
Q
2
 gA
2


A y


A y
2 2
= Specific force where each term of the specific force equation has the following definition:
Q
2

Momentum / unit time / Specific weight = Force / Specific weight gA
A y

Force /specific weight
It is called specific force because it is force / specific weight where
Q = Discharge
A = Cross section area
 y = Vertical depth to C.G. g = Gravity acceleration.
C.G.
A
2 y
2
Sec 2
64

12.0 Specific Force Diagram
The specific force diagram is a plot of the depth against the specific force for a given channel section and discharge. Figure (22) shows the specific force curve supplement with specific energy curve.
Fig. (22) Specific force curve supplement with specific energy curve. (a) Specific energy curve; (b) channel section; (c) specific force curve.
From Fig. (22) y
1
& y
2
are called alternate depths; y
1
& y /
2
are called conjugate depths, or sequent depths.
F

Q gA
2


A y
For minimum F dF

0 dy
and Q

constant dF dy
 
Q
2 gA
2 dA dy

 dA y

0 dy
Q
2 g A
2
 dA dy
 dA dy y
 dA dy

T

2
Q gA
T
2
 dA dy y
(1)
The term dA y dy
is define as:
64

For a change dy in the depth, the corresponding change dA y in momentum of the water is equal to: monentum at (o
/
-o
/
) - momentum (o-o).
1 st moment at o
 o

A y o
C.G.
o y
2
2 nd moment at o
/  o
/ 
A (
 y
 dy)

T dy dy
2 o /
T o / dA y = momentum at (o
/
-o
/

) - momentum (o-o).
A y dy
C.G.
y d A y

A y

Ady

T dy 2
2

 dA y

Ady (2)
By substituting equation (2) into equation (1), then we can get:
Q g

A
A
Critical flow condition
T
13.0 Rapidly Varied Flow
If the change in flow condition occurs in a short distance, the flow is called rapidly varied flow. Such as the case of:

Flow over weirs.

Flow through channel restrictions.

Flow at changes in bed slope from steep to mild slope.
Note : The hydraulic Jump is a typical example of R.V.F.
Fig. (23) shows typical forms of hydraulic jump in open channel.
64

Fig. (23) Typical forms of hydraulic jump in open channel.
13.0 Applications on Rapidly Varied Flow
13.1 Hydraulic Jump

Earliest investigation by Bidone 1815.

Hydraulic jump is a natural phenomenon in open channel. It is an abrupt reduction in flow velocity by means of a sudden increase of water depth in the downstream direction. It is used downstream irrigation structures to reduce the high velocity to prevent scour in earth canals.
64

Fig. (24) Hydraulic jump downstream sluice gate
Applying the momentum equation between sections 1 & 2

F = Rate of change of momentum
F
1

F
2

 g
Q (v
2
 v
1
)

A
1
 y
1
 
A
2
 y
2


Q g


Q
A
2

Q
A
1


Divide by

A
1
 y
1

A
2
 y
2

Q
2 g


1
A
2

1
A
1


Rearrange
Q
2

A
1 gA
1
 y
1
Q
2
 gA
2

A
2
 y
2
Specific Force at 1 = Specific Force at 2
13.1.1 Hydraulic jump for Rectangular Channels
A
1
= b y
1
, A
2
= b y
2
, Q = q b
(1)
 y
1
 y
1
,
2
From equations 1 & 2
 y
2 q
2 b
2 g y
1 b
 by
1 y
2
1 q
2 b
2
 g y
2 b

 by
2 y
2
2 y
2
2
(2)
05

q
2
 y
1
2 g y
1
2q
2 g


1 y
2
2
 q
2 g y
2


1 y
1


 y
1
2 y
2
2
2
 y
2
2
2q
2 g

 y
1 y
2
 y y
1
2



 y
1
 y
2
 y
1
 y
2

2q
2
 y (y
1
 y )
 y y
1
 y
2
2 gy
1 y
2
2  y y
2

2q
2 gy
1
= 0 y
2

 y
1
 q
2 y
1
2 
8 gy
1
2
Divide by y
2 y
1
 q
2
1 1 8 gy
1
3
2
F
1
 q y gy
1 1
 y
2 
  
1
2
2 y
1
F
1
= 1
F
1
> 1
F
1
< 1 y
1
= y
2 y
2
> y
1 y
2
< y
1
No Jump
Jump
Impossible
13.1.2 Energy loss in hydraulic jump
F
1
2 q
2
 gy
1
3
05

E
1
 y
1
 v
1
2g
E
2
 y
2

V
2
2g

E
 head loss

E
2

E
1
Length of jump L j

6y
2
13.1.3 Comparison between specific energy and specific force diagrams.
Fig. (24) Hydraulic jump interpreted by specific energy and specific force curves
Example:
A rectangular channel 5.0m wide, has an initial water depth 0.5m and velocity 5m/sec can a hydraulic jump take place? If it takes place, find the following:
1) The sequent depth;
2)
3)
The energy loss;
The length of the jump.
Solution
F
1
 v
1
 gy
1
5
9.81 * 0.5

2.25 > 1 jump will take place.
05

y
2 y
1


1

1
2

8F
1
2


1

1

8
2
* (2.25) y
2

0.5
* 2.73

1.36m

E

E
1

E
2
 y
1
 v 2
1
2g
 y
2
 v 2
2
2g
2

2.73m

0.5

5
2
2 * 9.81

1.36



5 *
0.5
1.36


2
2 * 9.81
Length of jump, L j
= 6y
2
= 6 * 1.36 = 8.16 m
For non rectangular section
F
1
= F
2
Q
2


A y
1 gA
1
Q
2
 gA
2


A y
2

0.238 m
05