What is the relative depth
to achieve maximum discharge
in a circular culvert?


Tamina Igartua and Victor M. Ponce


19 September 2015


Abstract. Using differential calculus, the relative depth y/D required to achieve maximum discharge in a circular culvert is derived. This value is shown to be y/D = 0.938. The reason for this hydraulic behavior is that when the flow depth increases beyond the value y = 0.938 D, the wetted perimeter begins to grow faster than the flow area, causing the flow rate to begin to decrease.

1.  PROBLEM STATEMENT

The maximum discharge in a circular culvert occurs, not when the culvert is full, but when the culvert is nearly full. The reason for this is that beyond a certain relative depth y/D, where y is the flow depth and D is the culvert diameter, a further increase in depth will make the wetted perimeter grow faster than the flow area, triggering a decrease in discharge.

In this article, the relative depth y/D which corresponds to maximum discharge in circular culvert flow is calculated.

2.  DERIVATION

According to the Manning equation, the discharge in SI units is (Ponce, 2014):

           1
Q  =  ____ A R 2/3 S 1/2  
           n 		(1)

in which Q = discharge, A = flow area, R = hydraulic radius, S = bottom slope, and n = Manning's n. Since R = A/P, it follows that:

           1
Q  =  ____ A 5/3 P - 2/3 S 1/2  
           n 		(2)

In terms of r and θ, the flow area and wetted perimeter are, respectively (Fig. 1):

          r 2
A  =  ____   ( θ - sin θ )  
           2 		(3)

           
P  =  r θ  
             		(4)


Fig. 1  Definition sketch.

Therefore:

  dA          r 2
_____  =  ____   ( 1 - cos θ )   dθ           2 		(5)

  dP           
_____  =  r    
  dθ             		(6)

By geometry, the relative depth is (Fig. 1):


y /D  =  (1/2) [ 1 - cos (θ/2) ]

(7)

Following differential calculus, the maximum discharge occurs when dQ/dθ = 0. Using Eq. 2, for a constant slope and Manning friction, this condition implies that:

  d       
____  ( A 5/3 P - 2/3 )  =  0
 dθ       		(8)

Operating on the derivatives:

   5                        dA            2                         dP  
____  A 2/3 P -2/3  _____   -   ____  A 5/3 P - 5/3 _____  =  0
   3                        dθ            3                         dθ
(9)

Simplifying:

           dA                    dP  
 5 P   _____   -   2 A   _____  =  0
           dθ                    dθ
(10)

Replacing Equations 3 to 6 into Equation 10 and simplifying:


 5 θ ( 1 - cos θ )  -  2 ( θ - sin θ )  =  0

(11)

Simplifying Eq. 11:


 3θ  -  5θ cos θ  +  2 sin θ = 0

(12)

Solving for θ in Eq.10: θ = 302° 25' 51.96". This angle is used in Eq. 7 to solve for relative depth, to give y/D  =  0.938 (Fig. 2).


Fig. 2   Relative depth for maximum discharge in a circular culvert..

3.  SUMMARY

The relative depth y/D required to achieve maximum discharge in a circular culvert has been derived using differential calculus. This value is shown to be: y/D = 0.938. The reason for this hydraulic behavior is that when the flow depth increases beyond the value y = 0.938 D, the wetted perimeter begins to grow faster than the flow area, causing the flow rate to decrease accordingly (Eq. 2).

REFERENCE

Ponce, V. M. 2014. Fundamentals of open-channel hydraulics. Online text.

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