Hydraulic diameter vs hydraulic radius

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SUMMARY

The hydraulic diameter is defined as \(D_h = \frac{4A_c}{p}\) and the hydraulic radius as \(R_h = \frac{A_c}{p}\), where \(A_c\) is the cross-sectional area and \(p\) is the wetted perimeter of a pipe. The misconception that the hydraulic radius should equal \(2\frac{A_c}{p}\) arises from a misunderstanding of the definitions. The relationship between diameter and radius does not apply in this context, as the hydraulic diameter and radius are distinct concepts used in fluid mechanics.

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  • Understanding of fluid mechanics terminology
  • Knowledge of cross-sectional area and wetted perimeter
  • Familiarity with hydraulic diameter and hydraulic radius definitions
  • Basic mathematical skills for manipulating equations
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  • Study the derivation of hydraulic diameter and hydraulic radius in fluid dynamics
  • Explore applications of hydraulic diameter in pipe flow calculations
  • Learn about the significance of wetted perimeter in fluid mechanics
  • Investigate the impact of hydraulic radius on flow characteristics in various geometries
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hotjohn
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Homework Statement


i was told that the hydraulic diameter = 4Ac / p , while the hydraulic radius = Ac / p , where Ac = cross sectional area , and p = wetted perimeter of the pipe , why shouldn't the hydraulic radius = 2Ac / p ?
because as well all know , diameter = 2 Radius , so hydraulic radius = 2(hydraulic diameter ) = 2(4Ac / p) = 8Ac / p

Homework Equations

The Attempt at a Solution

 

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hotjohn said:

Homework Statement


i was told that the hydraulic diameter = 4Ac / p , while the hydraulic radius = Ac / p , where Ac = cross sectional area , and p = wetted perimeter of the pipe , why shouldn't the hydraulic radius = 2Ac / p ?

It doesn't make sense, but this is the terminology that's used. So learn to live with it.

because as well all know , diameter = 2 Radius , so hydraulic radius = 2(hydraulic diameter ) = 2(4Ac / p) = 8Ac / p
Your mathematics here is incorrect.
 

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