Question

Design a channel carrying a 30 cumecs. The median grain diameter is taken as 0.5 mm. The bedload concentration is 60 p.p.m by weight. Use Lacey’s Regime perimeter and Meyer-peter’s formulas.

(a) B = 26 m, S = 1/5600, y = 1.25 m

(b) B = 22 m, S = 1/5800, y = 1.55 m

(c) B = 24 m, S = 1/5500, y = 1.45 m

(d) B = 25 m, S = 1/5700, y = 1.35 m

I have been asked this question in an interview for internship.

I'm obligated to ask this question of Sediment Transport topic in chapter Sediment Transport and Irrigation Channels Design of Irrigation Engineering

Answer 1

Right choice is (d) B = 25 m, S = 1/5700, y = 1.35 m

The explanation: Quantity of bed load transported by weight = 40/10^6

Quantity of bed load transported per second = 40/10^6 (30 x 9.81 x 1000) = 11.8 N/sec

Lacey’s Regime perimeter = 4.75 x √Q = 26.03 m

Let us take channel bed width (B) as 25 m

Bed load per unit width = gb = 11.8 / 25 = 0.472 N/m/sec

Meyer Peter equation –

gb = 0.417 x [τo(n’/n) – τc]^3/2

n’ = (1/24) x (0.5)^1/6

= 0.011

n = 0.02

n’/n = 0.55

τc = 0.687 x da = 0.687 x 0.5 = 0.3435 N/m^2

gb = 0.417 [9.81 x 1000 x RS x (0.55)^3/2 – 0.3435]^3/2

RS = 0.0002

Manning’s equation –

Q = 1/n x R^2/3 x S^1/2

R^2/3 x S^1/2 = 0.6

S = 0.0002/R

S = 1/5700

R = 1.15 m

Now P = 25+y√5, A = 25y + y^2/2 for trapezoidal channel of 1/2:1 slopes

R = A/P = (25+√5y)/ (25+y^2/2) = 1.15

From this y = 1.35 m

Therefore B = 25 m, y = 1.35 m, S = 1/5700.