Question

An engineer has recently driven his car into a culvert during a storm and has completely damaged his car. In an effort to ensure that this does not happen again, he wants to petition to the town’s council to redesign the culvert to appropriately accommodate the storm design flowrate of 4 m3/s. The headwater depth should be 2 m and the tailwater depth should be no greater than the height of his car (1 m), the length of culvert is 25 m and the slope Is 2%. Determine the required diameter of the culvert. (assume reinforced concrete pipe material)

Answer 1

Ans) Given,

Flow rate(Q) = 4 m³/s

Slope(s) = 2%

Length of culvert (L)= 25 m

Headwater depth (Hw) = 2m

Tailwater depth (Tw) = 1 m

We know, headloss (H),

H = Hw - Tw + Ls

H = 2 - 1 + 25(0.02)

H = 1.5 m

Headloss through culvert can also be given by,

H =( 29 N² L / R^1.33 ) V²/2g

where, N = Manning roughness coefficient ( 0.013 for reinforced concrete pipe)

   R = Hydraulic depth ( D/4 for circular pipe)

   V = flow velocity  

V = Q / (/4) D²

V = 4 / 0.785 D²

V = 5.01 /D²

Putting values,

1.5 = [29(0.013)² (25) / (D/4)^1.33] ( 26 / D⁴ x 19.62)

1.5 = ( 0.774 / D^1.33 ) 1.325 / D⁴

1.5 = 1.0255 / D^5.33

D^5.33 = 0.6837

D = 0.6837 ^1/5.33

D = 0.930 m or 930 mm