# Homework Help: Pressure drop in laminar flow

1. Feb 3, 2016

### hotjohn

1. The problem statement, all variables and given/known data
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) ?

2. Relevant equations

3. The attempt at a solution

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2. Feb 3, 2016

### SteamKing

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

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

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

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

3. Feb 3, 2016

### hotjohn

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

4. Feb 3, 2016

### haruspex

No, it's 2R as usual.

5. Feb 3, 2016

### hotjohn

why the note gave 4R ?

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6. Feb 3, 2016

### SteamKing

Staff Emeritus
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.

7. Feb 3, 2016

### Staff: Mentor

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.

8. Feb 3, 2016

### haruspex

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 Rh. 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, Rh=R/2 for a filled cylindrical pipe, and for a half-filled horizontal pipe.

Last edited: Feb 4, 2016
9. Feb 3, 2016

### Staff: Mentor

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

10. Feb 4, 2016

### hotjohn

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

11. Feb 4, 2016

### haruspex

D is the pipe diameter and is therefore twice the radius, D=2R.
Rh 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 Rh=R/2=D/4.
A half-filled horizontal cylindrical pipe has half the wetted perimeter and half the flow cross-section, so the same Rh.
In between the two, Rh reaches a maximum somewhere.

12. Feb 4, 2016

### hotjohn

i have an example from my books ,

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

13. Feb 4, 2016

### haruspex

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.

14. Feb 4, 2016

### hotjohn

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

15. Feb 4, 2016

### haruspex

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