Contents preface (VII) introduction 1—37

283


Example 8.2 Design a lined channel to carry a discharge of 300 m³/s through an alluvium whose angle of repose is 31°. The bed slope of the channel is 7.75 × 10^–5 and Manning’s n for the lining material is 0.016.
Solution: Since Q > 55 m³/s, trapezoidal section with rounded corners, Fig. 8.2, is to be designed. Here,

Side slope	θ = 31° = 0.541 radians
∴	cot θ = 1.664
∴	θ + cot θ = 2.205
∴	A = Bh + 2.205 h^2
	P = B + 4.41h
Adopting	U = 2 m/s
	A = ^300 = 150 m2
	2

∴Bh + 2.205h² = 150

and	R =	F Un I ^3/2	F	2 × 0.016	I 3/2
G		J	= G			−5	J= 6.93 m
			G	7.75	× 10		J
		H	S K	H				K

∴ 6.93 (B + 4.41h) = 150
or B = 21.645 – 4.41h
∴ 21.645h – 4.41h² + 2.205h² = 150

or	2.205h^2 – 21.645h + 150 = 0
∴	h =	21645. ± (21645.)^2	− 4 × 2.205 × 150
4.41

Obviously, the roots of h are imaginary. Using the criterion, Eq. (8.4), one gets,

L

QS3/2

O1/4

U ≤ M

P

3

N 4n

(θ + cot θ) Q

L

300 × (7.75 ×

10

−5 )3 /2 O^1/4

≤ M

P

4 (0.016)

3

( 2.205)

N

Q

≤ 1.543 m/s

∴ Adopt

U = 1.5 m/s

∴

A = 200 m^2

F

15. × 0.016

I 3/2

R = G

J

= 4.50 m

G

7.75 × 10

−5

J

H

K

4.50(B + 4.41h) = 200
B = 44.44 – 4.41h

284			IRRIGATION AND WATER RESOURCES ENGINEERING
Again, using	Bh + 2.205h^2 = A, one gets,
	44.44h – 4.41h^2 + 2.205h^2 = 200
or	2.205h^2 – 44.44h + 200 = 0
∴		h =	44.44 ±	(44.44)^2 − 4 × 2.205 × 200
			4.41
		= 13.37 m	or 6.784 m
∴		B = 44.44 – 4.41h
		= 14.52 m for h = 6.784 m
Other value of h (= 13.37 m) gives negative value of B which is meaningless.
∴		B = 14.52 m	and h = 6.784 m.

8.3. RIGID BOUNDARY CHANNELS CARRYING SEDIMENT–LADEN WATER
These channels are to be designed in such a way that the sediment in suspension does not settle on the channel boundary. The design is, therefore, based on the concept of minimum permissible velocity which will prevent both sedimentation as well as growth of vegetation. In general, velocities of 0.7 to 1.0 m/s will be adequate for this purpose if the sediment load is small. If the sediment concentration is large, Fig. 8.3 can be used to ensure that the sediment does not deposit. In Fig. 8.3, C_s is the concentration of sediment in ppm (by volume), f_b the friction factor of the channel bed, D_o the central depth, T the top width and S_c equals S/(∆ρ_s/ρ). If the designed channel section is not able to carry the specified sediment load, the slope S of the channel is increased.
Example 8.3 A rectangular channel 5 m wide is to carry 2.5 m³/s on a slope of 1 in 2000 at a depth of 0.75 m. It is expected that fine silt of 0.04 mm size will enter the channel. What is the maximum concentration of this sediment that can be allowed into the channel without causing objectionable deposition ? Assume that the fall velocity of the given sediment in water is 1.5 mm/s, kinematic viscosity of water is 10^–6 m²/s, and specific gravity of the sediment is 2.65.
Solution: Since the wall and the bed of the channel are of the same material, the friction factor for the channel bed f_b is given as

	fb =		8 gRS
			U2
Since	U =	2.5	= 0.7143 m/s
5 × 0.75
and	R =			5 × 0.75	= 0.5769 m
5 + 2 × 0.75
∴	fb =	8 × 9.81 × 0.5769 × (12000/		)		= 0.0444
(0.7143)^2
	Sc = S/(∆ρs/ρ) =	1	=				1
	2000 × 165.	3300

DESIGN OF STABLE CHANNELS

285

Now

qS 2.5

1

F

A I ^2

c

G

J

2

( w0 d/ ν)

0.6

νf b

H

TDo K

F

× 0.75I

(2.5 / 5.0)(1 / 3300)^2.5

5

2

=

G

J

= 2.2

(10

− 6

)(0.0444)

2

(1.5

× 10

−3

× 0.4 ×

10

−3

/ 10

−6

)

0.6

H 5

× 0.75K

Therefore, from Fig. 8.3, maximum concentration for no deposition,
C_s = 500 ppm

4

10

3

10

(ppm)

s C

2

10

1

2×10

0.4

Deposition

No deposition

— Data for different sizes of sand and coal particles
size range : 0.013–0.27 mm

1.0								10	40
2.5
	q Sc		1		A	2
			×			× (			)
2			0.6	TD0
	fb		(w0d/ )

Fig. 8.3 Relation for the limiting concentration of suspended sediment in channels (4)