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 close to being full. The reason for this behavior is that a certain relative depth y/D, where y is the flow depth and D is the culvert diameter, the wetted perimeter begins to grow faster than the flow area and, consequently, the discharge starts to decrease.

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


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)

In Eq. 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 3-6 into Equation 9 and simplifying:


 5 θ ( 1 - cos θ )  -  2 ( θ - sin θ )  =  0
(11)

Simplifying Eq. 11:


 3θ  -  5θ cos θ  +  2sin θ = 0
(12)

In Equation 12, solving for θ by trial and error:  θ = 302° 25' 51.96".

This angle is used in Eq. 7 to solve for relative depth, to give y/D  =  0.938.


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 begin to decrease.


References

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


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