Hydraulic diameter vs hydraulic radius

[hotjohn]

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

 

[Chestermiller]

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.

 

[haruspex]

See posts #13 onwards in https://www.physicsforums.com/threads/pressure-drop-in-laminar-flow.855413/#post-5366220