Contents preface (VII) introduction 1—37

Fig. 9.1. Control volume for hydraulic jump
With the assumptions that θ is small (i.e., sin θ ≅ 0), and β₁ = β₂ = 1, Eq. (9.1) becomes
ρ g z₁ A₁ – ρ g z₂ A₂ – F_f = ρQ (u₂ – u₁) (9.2) Equation (9.2) can be rewritten as

Ff

F

Q^2

I

F

Q^2

I

= G

+

A1

^z1J

− G

+ A2

^z2 J

ρ g

gA2

H

gA1

K

H

K

or

^Ff

= M – M

2

ρ g

1

where,

M =

Q2

+ A z

gA

and M is termed the specific momentum or force function or specific force.

(9.3)

(9.4)

(9.5)

SURFACE AND SUBSURFACE FLOW CONSIDERATIONS FOR DESIGN OF CANAL STRUCTURES

319

If the jump occurs in a horizontal channel and is not assisted by any other means, such as baffle blocks, then F_f ~₌ 0, and Eq. (9.4) yields

		M1 = M2		(9.6)
or	Q^2	+ A1	z1	=	Q^2	+ A2 z2	(9.7)
gA1	gA2

9.2.1. Hydraulic Jump in Rectangular Channels
A hydraulic jump formed in a smooth, wide, and horizontal rectangular channel is termed classical hydraulic jump. For rectangular channel of width B,

Q = u1 A1 = u2 A2
A1 = B1 h1	and	A2 = B2h2
z1 = h1/2	and	z2 = h2/2

Substituting these values into Eq. (9.7) one can, after simplification, obtain

1	h h (h	+ h ) =	q^2
2	g
		1 2 1	2

where, q = Q/B. Equation (9.8) has the following solutions :

	^h2 =	^1 ( 1 + 8F ^2	− 1)
	h1		2	1
and	h1	=		^1 ( 1 + 8F	2	− 1)
	h2		2	2

(9.8)

(9.9)
(9.10)

in which, F₁ and F₂ are the Froude numbers at sections 1 and 2, respectively. Froude number




F equals u/ gD in which, D is the hydraulic depth and equals A/T where, T is the top width of flow. Equations (9.9) and (9.10) are the well-known Belanger’s momentum equations.
9.2.2. Energy Loss in Hydraulic Jump in a Rectangular Channel
In a horizontal rectangular channel with the channel bed chosen as the datum, the total energies at sections 1 and 2, (Fig. 9.1 with θ = 0) are equal to the specific energies E₁ and E₂ at sections 1 and 2, respectively, i.e.,

q^2
E1 ^{= h}1 ⁺ _{2 g h1}2




q^{2 E}2 ^{= h}2 ⁺ _{2g h2}2

so that energy loss, ∆E = E₁ – E₂

q<sup>2

= h</sup>1 ^{– h}2 ⁺ _{2 g}

1 = (h₁ – h₂) + ₄

L	1	1	O
M		−		P
h ^2	h ^2
M	1	2	P
N					Q

	F h ^2	− h	^2 I
h1h2(h1	+ h2) G	2		1	J
h1	2	2
		H		h2			K

(9.11)

(9.12)
(9.13)



320 IRRIGATION AND WATER RESOURCES ENGINEERING

					4h		^2 h − 4h	^2h	+ h h ^2	− h ^3	+ h ^3	− h	^2h
	=	1	2		2	1	1	2			1	2	1	2
											4h1 h2
					h ^3		− h	^3 + 3h ^2 h − 3h h	2
	=		2		1				1	2	1		2
								4h1 h2
					(h		− h )^3
∴	∆E =	2		1															(9.14)
		4h1 h2
	∆E		[ h	+ (u	^2 /2 g )] − [ h +	(u	2	/2 g)]
Also,		=	1		1				2	2								(9.15)
						[ h1 + u1^2 /2 g]
	^E1
	=			h [ 1 − ( h	/ h )] +		( q ^2 /2 g h	^2 ) [ 1 − ( h	/ h ) ^2	]
		1			2	1					1		1	2
											( h /2) [ 2	+ F	2 ]
											1				1
	∆E2 − 2 ( h	/ h ) + F	^2 [ 1 − ( h		/ h ) ^2	]
∴		=					2		1	1		1		2					(9.16)
								2 + F1^2
	^E1
Combining Eq. (9.8), (9.11) and (9.14), one can obtain
	∆E	8 F	^4 + 20 F ^2	− (8 F ^2 + 1) ^3 /2 − 1
		=	1				1		1										(9.17)
							8 F1^2 ( 2 + F1^2 )
	^E1

Hence, for a supercritical Froude number F₁ equal to 20, the energy loss ∆E is equal to 0.86 E₁. This means that 86 per cent of the initial specific energy is dissipated. Because of this energy dissipating capability, hydraulic jump is widely used as an energy dissipator for spillways and other hydraulic structures. In a hydraulic jump, the mean kinetic energy is first converted into turbulence and then dissipated through the action of viscosity. Equation (9.17) can be rewritten as

E		(8 F	^2 + 1) ^3	^/2 − 4 F	2	+ 1
2	=	1		1			...(9.18)
E1		8 F1^2 ( 2 + F1^2 )

The term E₂/E₁ is called the efficiency of the jump.