Question

A circular concrete culvert has a diameter D = 4 ft, bottom slope S = 0.01,
and length L = 100 ft. The culvert inlet is grooved with a headwall and is
not mitered. Determine the maximum discharge this culvert can convey
under inlet control conditions if the headwater depth, HW, is not to exceed
6.0 ft.

Answer 1

Given that

• Slope, S = 0.01
• Length, L = 100 ft
• Diameter, D = 4 ft
• Head water depth (HW) = 6 ft

We know, according to Manning equation,

• V = (1/N) R2/3 S1/2

where, V = velocity

S = Bed slope = 0.01

R = hydraulic depth = D/4 = 4/4 = 1ft

N = Manning roughness coefficient of concrete = 0.012

• V = (1 /0.012) (1)2/3 (0.01)1/2
• V = 8.33 ft/s

Now , for inlet control culvert,

HW = (0.429 Qp0.5 / D0.26) + V2 /2g + D [ C1 ( 4 Qp / D5/2)C2 + C3S)

C1 , C2 and C3 = Manning roughness coefficients for grooved inlet with head wall

• C1 = 0.0078 ,
• C2 = 2
• C3 = -0.5

where, Qp = peak discharge

Putting values,

6 =  (0.429 Qp0.5 / 40.26) + 8.332 /2g + 4 [ 0.00787 ( 4 Qp / 45/2)2 -0.5 (0.01)]

• 6 = 0.3 Qp0.5 + 1.07 + 4 [ 0.0016 Qp2 - 0.005]
• 6 = 0.3 Qp0.5 + 1.07 +0.0064 Qp2 - 0.02
• 4.95 = 0.3 Qp0.5 + 0.0064 Qp2

On solving above equation , we get peak discharge ,

• Qp = 23.38 cfs

Conclusion

Hence, this culvert can carry maximum discharge of 23.38 cfs under inlet control conditions.