Max distance Between storm water inlets

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SUMMARY

The discussion focuses on calculating the maximum distance between storm water inlets on a suburban street receiving runoff at a rate of 2.0 L/m-s. Using the modified Manning's equation for a triangular gutter cross-section, the gutter flow (Q) was determined to be 0.28 m³/s. By converting the runoff rate from liters to cubic meters and applying the relationship between the gutter flow and runoff, the maximum distance between inlets was calculated to be 140 meters. This calculation is essential for effective stormwater management in urban planning.

PREREQUISITES
  • Understanding of modified Manning's equation
  • Knowledge of hydraulic principles related to stormwater management
  • Familiarity with units of flow rate and conversion between liters and cubic meters
  • Basic geometry of triangular cross-sections
NEXT STEPS
  • Study the application of Manning's equation in various gutter designs
  • Research stormwater management best practices for urban environments
  • Learn about hydraulic modeling software for stormwater systems
  • Explore the impact of different roughness coefficients on flow rates
USEFUL FOR

Civil engineers, urban planners, and environmental consultants involved in stormwater management and infrastructure design will benefit from this discussion.

Bison11
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A suburban street receives storm water runoff at rate of 2.0 L/m-s (on each side). The street has a slope of .8 percent and a cross slope of 2 percent; n = .013. If the street is 10 m wide and can be totally covered by water (depth = 0 at the midpoint), what is the max distance between inlets?

Thinking to use modified Manning's equation for triangular gutter cross section

Q = K(z/n)(s^.5)[y^(8/3)]

Q = gutter flow
z = reciprocal of the cross slope of the gutter
n = roughness coefficient
s= gutter slope
y = water depth at the curb
K = constant = .38 for metric


first I found the Q for the gutter flow
Q=?
z= 1/.02=50
n = .013
s= .008
y= 5m(.02)=.1m
K= .38

so Q = .38(50/.013)(.008^.5)[.1^(8/3)] = .28 m^3/s

at this point I am not sure how to use the runoff of 2 L/m-s and the Q = .28 m^3/s to get a max distance between inlets?
 
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may have figured this out. if I convert the L in 2 L/m-s to m^3 I get .002 m^2/s. then if I take (.28 m^3/s)/(.002m^2/s) = 140m. so basically this is a right triangle with a hypotenuse of 2L/m-s and leg1 = .28m^3/s and leg2 = length. Any thoughts or further info is welcome.
 

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