Why is fluid velocity unaffected by a change of pipe roughness here?

Click For Summary
SUMMARY

The discussion centers on the relationship between fluid velocity and pipe roughness in a single pipe system. It concludes that fluid velocity (V) remains constant when flow rate (Q) is unchanged, as expressed by the equation Q = V * A, where A is the cross-sectional area. The participant grapples with understanding how roughness affects pressure but not velocity, ultimately recognizing that the flow rate must remain constant for velocity to be stable, regardless of pipe roughness. The Darcy–Weisbach equation is referenced to illustrate the complexity of friction and roughness coefficients while maintaining velocity as a constant.

PREREQUISITES
  • Understanding of fluid dynamics principles
  • Familiarity with the equation Q = V * A
  • Knowledge of the Darcy–Weisbach equation
  • Basic concepts of flow rate and pressure in fluid systems
NEXT STEPS
  • Study the Darcy–Weisbach equation in detail
  • Explore the impact of pipe roughness on pressure loss
  • Investigate flow rate stability in various pipe configurations
  • Learn about computational fluid dynamics (CFD) simulations for practical applications
USEFUL FOR

Students and professionals in fluid mechanics, engineers designing piping systems, and anyone interested in the principles of fluid flow and pressure dynamics.

cdux
Messages
187
Reaction score
0
I had no luck in the coursework forums though I guess the question becomes simpler if I state it the way I did here.

Assuming no complex network, or just a single pipe first feeding the network, why do the speeds in the pipe and even the rest of the network pipes remain unaffected if I change the roughness of that first pipe? The pressures do change.

I guess I need the basic answer which I'm sure is simple so I can then expand to a very rational explanation that will make me remember it because the way I think of it now, it's a bit unclear how roughness can keep velocities unaffected (in a non-part-of-a-branch, single pipe scenario).
 
Physics news on Phys.org
OK I got an answer from somewhere, but I still can't get it to hold still in my head.

The answer is basically that since Q = V * A then V remains the same since Q is the same (and A).

Now, how can I make that make sense when also thinking of roughness as irrelevant?

edit: Basically how can Q remain the same? hrm.. I'm probably trapped in a circular logic that is totally wrong but I'm not sure exactly how.
 
Now I'm thinking of Darcy–Weisbach et al. complex equations and I can't figure out how while they do include coefficients for friction/roughness, they just take velocity for granted as stable (if the above is true).
 
Ah, I think I got somewhere. In the simulations I was running, I guess whatever output is entered, "it must be supplied", hence Q is taken as a hard-constant no matter what hence V is satisfied to a constant based on that, provided same cross section.

Or at least that's how far I got.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 48 ·
2
Replies
48
Views
5K
Replies
4
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
Replies
9
Views
6K
  • · Replies 20 ·
Replies
20
Views
11K