Fluid Mechanics Open Channel Problem

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SUMMARY

The discussion revolves around solving a fluid mechanics problem involving a trapezoidal open channel with a bottom width of 6.10m and a Manning's roughness coefficient (n) of 0.030. The goal is to determine the slope required to achieve an average velocity of 0.76 m/s for a flow rate of 5.66 m³/s. The solution involves using the Manning's equation, V = (1.486/n)R^(2/3) * S^(1/2), where the slope (S) is the unknown variable. The correct slope calculated is 0.000675.

PREREQUISITES
  • Understanding of Manning's equation for open channel flow
  • Knowledge of trapezoidal channel geometry
  • Familiarity with hydraulic radius (R) calculations
  • Basic principles of fluid mechanics
NEXT STEPS
  • Study the derivation and application of Manning's equation in different channel shapes
  • Learn how to calculate hydraulic radius for various channel geometries
  • Explore methods for determining flow rates in open channels
  • Investigate the impact of channel slope on flow velocity and discharge
USEFUL FOR

This discussion is beneficial for civil engineers, hydraulic engineers, and students studying fluid mechanics, particularly those focused on open channel flow design and analysis.

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The problem: A canal cut in rock (n= 0.030) is a trapezoidal in section with a bottom width of 6.10m and the side slopes of 1 on 1. The allowable average velocity is 0.76 m/s. What slope will produce 5.66 m^3/s? The books answer is 0.000675

from the mannings equation

V=(1.486/n)R^2/3 x S^1/2 i get... 0.76=(1.486/0.030)? I am confused on where to go
 
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Use of the Manning formula necessitates knowing the slope. But that is what is sought. So how did you do justify you did?

Hint:
Start out by determining the area of the open channel. Then use the Manning formula to determine the slope.
 

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