Why Does the Laminar Flow Equation Use Different Coefficients?

[hotjohn]

Homework Statement

as we all know , D=4R , D=hydraulic diameter , in the equation of P= (8μLV_avg) / (R^2) , i would get (128μLV_avg)/ (D^2) , am i right ? why the author gave (32μLV_avg)/ (D^2) ?

Homework Equations

The Attempt at a Solution

 

[SteamKing]

Homework Statement

as we all know , D=4R , D=hydraulic diameter , in the equation of P= (8μLV_avg) / (R^2) , i would get (128μLV_avg)/ (D^2) , am i right ? why the author gave (32μLV_avg)/ (D^2) ?

Homework Equations

The Attempt at a Solution

It depends on the shape of the duct or tube you are using.

The hydraulic diameter is defined as D_H = 4A/P, where A is the cross sectional area of the duct or pipe and P is the wetted perimeter..

For a circular pipe, D_H = D (internal) of the pipe, or ##D_H = \frac{4 ⋅ π ⋅ \frac{D^2}{4}}{π ⋅ D} = D##

This article shows the hydraulic diameters for ducts of different shapes:

https://en.wikipedia.org/wiki/Hydraulic_diameter

 

[hotjohn]

It depends on the shape of the duct or tube you are using.

The hydraulic diameter is defined as D_H = 4A/P, where A is the cross sectional area of the duct or pipe and P is the wetted perimeter..

For a circular pipe, D_H = D (internal) of the pipe, or ##D_H = \frac{4 ⋅ π ⋅ \frac{D^2}{4}}{π ⋅ D} = D##

This article shows the hydraulic diameters for ducts of different shapes:

https://en.wikipedia.org/wiki/Hydraulic_diameter

yes , i know that . the object given is a long circular tube , so it is D=4R , right ? the notes is wrong ?

 

[haruspex]

so it is D=4R , right ? the notes is wrong ?

No, it's 2R as usual.

 

[hotjohn]

No, it's 2R as usual.

why the note gave 4R ?

 

[SteamKing]

why the note gave 4R ?

They used the wrong formula for calculating hydraulic diameter to start with

##D_H = \frac{4A}{P}##

The rest is shoddy proof-reading and failing to catch the error.

 

[Chestermiller]

They used the wrong formula for calculating hydraulic diameter to start with

##D_H = \frac{4A}{P}##

The rest is shoddy proof-reading and failing to catch the error.

Some authors define the hydraulic diameter as A/P. However, I have never liked this, and I always used 4A/P. But, if you are reading a text, you need to make sure which definition they are using.

 

[haruspex]

why the note gave 4R ?

I believe you are confusing radius with hydraulic radius.
The link from your original post mentioned R and D, with the implied relationship D=2R.
Your link in post #5 does not mention R, but R_h. This is the hydraulic radius, which is meaningful in any cross section, and does not equate to the actual radius in the case of a cylindrical pipe. Indeed, the way it is defined at https://en.m.wikipedia.org/wiki/Manning_formula would make it R/2.

Edit: to clarify, R_h=R/2 for a filled cylindrical pipe, and for a half-filled horizontal pipe.

 

[Chestermiller]

I believe you are confusing radius with hydraulic radius.
The link from your original post mentioned R and D, with the implied relationship D=2R.
Your link in post #5 does not mention R, but R_h. This is the hydraulic radius, which is meaningful in any cross section, and does not equate to the actual radius in the case of a cylindrical pipe. Indeed, the way it is defined at https://en.m.wikipedia.org/wiki/Manning_formula would make it R/2.

Oops. In my last post, I meant to say hydraulic radius, not hydraulic diameter. Thanks for catching this.

 

[hotjohn]

Oops. In my last post, I meant to say hydraulic radius, not hydraulic diameter. Thanks for catching this.

so , D=4Rh or D=2Rh ?

 

[haruspex]

so , D=4Rh or D=2Rh ?

D is the pipe diameter and is therefore twice the radius, D=2R.
R_h is defined as the cross-sectional area of flow, A, divided by the wetted perimeter, P. For a filled cylindrical pipe radius R, A=πR², P=2πR, so R_h=R/2=D/4.
A half-filled horizontal cylindrical pipe has half the wetted perimeter and half the flow cross-section, so the same R_h.
In between the two, R_h reaches a maximum somewhere.

 

[hotjohn]

D is the pipe diameter and is therefore twice the radius, D=2R.
R_h is defined as the cross-sectional area of flow, A, divided by the wetted perimeter, P. For a filled cylindrical pipe radius R, A=πR², P=2πR, so R_h=R/2=D/4.
A half-filled horizontal cylindrical pipe has half the wetted perimeter and half the flow cross-section, so the same R_h.
In between the two, R_h reaches a maximum somewhere.

i have an example from my books ,

it stated that D_h= 4A / P , why not D_h= 2A / P ? since 2 radius = diameter ?

 

[haruspex]

i have an example from my books ,

it stated that D_h= 4A / P , why not D_h= 2A / P ? since 2 radius = diameter ?

I don't know why hydraulic diameter and hydraulic radius are defined in such ways that the ratio is 4:1 instead of 2:1. My guess is that it is an accident of history and the two definitions (in terms of A and P) were made independently.

 

[hotjohn]

I don't know why hydraulic diameter and hydraulic radius are defined in such ways that the ratio is 4:1 instead of 2:1. My guess is that it is an accident of history and the two definitions (in terms of A and P) were made independently.

so , just follow D_h = 4A/ P ? R_h = A/ P ?

 

[haruspex]

so , just follow D_h = 4A/ P ? R_h = A/ P ?

Yes. I suspect that you will not usually find both R_h and D_h used in the same text/problem. Authors will tend to adhere to one or the other. I could be wrong, though.