Question

Question 7 Consider flow in an open channel with slope 1:500 and a Manning's n of 0.030. The channel has a rectangular cross-section with a width of 5 m (a) Uniform flow in this channel has a depth of 1.6 m. What is the flowrate in the channel? (4 marks) (b) Suppose that this uniform flow (at point 1 in the figure below) encounters an obstacle on the channel bed (at point 2). How big would the obstacle need to be to cause a transition to supercritical flow downstream? (5 marks) (c) If a transition to supercritical flow does occur, show that the depth after the transition (at point 3) is approximately 48 cm, and thus determine the flow velocity after the hydraulic jump that forms (at point 4) (6 marks) 1 4 2

Answer 1

Ans We know,

Q = (A/N) R^2/3 S^1/2

where, Q = flow rate

A = cross sectional area of channel

R = hydraulic radius

S = slope of bed

N = Manning roughness coefficient

Also, R = area / wetted perimeter

= BD/B + 2D

= 5(1.6) / (5 + 3.2) = 0.976 m

Putting values,

Q = (8/ 0.030) (0.976)^0.667 (0.002)^0.50

= 11.73 m³/s

Ans b) We know,

Q = A x V

=> V = Q/A = 11.73/8

= 1.466 m/s

To change flow to supercritical , Froude number must be more than 1 ,

Fr = V² / (g D)^0.50

Take limiting value Fr = 1,

1 = 1.466² / (9.81 x D)^0.50

=> D = 0.47 m

Earlier , depth of flow = 1.6 m

Now, depth of flow = 0.47 m

=> Height of obstacle = 1.60 - 0.47 = 1.13 m

Ans) Since, the flow is now super critical,

  Fr = V₂² / (g d)^0.50

^=> gd = V₂⁴  

=> gd = (1.466)⁴

=>d = 0.48 m or 48 cm

Ans iii) We know ,

y2 = 0.50 y₁ ( 1 + 8Fr² )^1/2 - 1

Fr = V3² / (gd)^0.50

From part ii) V3 = 0.689/d = 0.689/0.48 = 1.48 m/s

=> Fr = 1.48² / (9.81 x 0.48)^0.50

= 1.01

=> y2 = 0.50 x 0.48 ( 1+ 8(1.01)²)^0.5 -1

= 1.65 m

Using continuity equation ,

A3 V3 = A4 V4

=> V4 = 0.48 x 5 x 1.48 / ( 5 x 1.6)

=> V4 = 0.444 m/s