Fluid mechanics: finding friction factor, should be simple

1. Apr 10, 2012

musicmar

1. The problem statement, all variables and given/known data
Given: water at 20°C flows through a 3cmID smooth brass tube at 0.002 m^3/s.

2. Relevant equations
V=Q/A
f=64/Re
Re=VD/(nu)

3. The attempt at a solution
V=Q/A=(0.002 m^3/s)/((∏/4)*(0.03 m)^2) = 2.829 m/s

Re=VD/(nu)
= (2.829 m/s)*(0.03 m)/(1x10^-6 m^2/s)=84882.64

f=64/Re=0.0075

I know this is wrong for two reasons:
1. The numeric answer is in the back of the book: 0.0185
2. Looking at the Moody Diagram: there is no f value on the chart below 0.008.

Any help in finding my mistake would be appreciated.
Thank you.

2. Apr 11, 2012

Staff: Mentor

Out of interest I've had a look at this, but I've not touched fluid flow since I was a student.

Here is a very similar example: http://answers.yahoo.com/question/index?qid=20071223185604AAiTXv9

How does your working compare? I see you haven't explicitly used density. (BTW, you've lost one of the 0's from your ƒ.)

Here's a Moody diagram http://www.engineeringtoolbox.com/moody-diagram-d_618.html

https://www.physicsforums.com/images/icons/icon2.gif [Broken] It seems to indicate that for Reynold's numbers up to about 2,500 the flow is laminar, the graph is a straight line so the Darcy–Weisbach friction factor formula ƒ=Re/64 applies. But for Re exceeding about 2,500 the flow is turbulent and you must consult the Moody diagram. If you do, I think you'll get your 0.018 figure.

How have I done?

Last edited by a moderator: May 5, 2017
3. Apr 11, 2012

musicmar

We had another problem that explicitly told us to use the Moody Diagram, so I guess I didn't realize it would be necessary, but you are right. I didn't use density because it is included in the kinematic viscosity, nu(didn't bother to find a symbol for it). I did get the right answer from the Moody diagram.
Thank you.