Examveda

Friction factor for a hydraulically smooth pipe at NRe = 2100 is f1. If the pipe is further smoothened (i.e., roughness is reduced), the friction factor at the same value of NRe, will

A. Increase

B. Decrease

C. Remain unchanged

D. Increase or decrease depending on the pipe material

Answer: Option A

Solution (By Examveda Team)

The correct answer is Option C: Remain unchanged
Here's why:
NRe stands for Reynolds Number, which is a dimensionless number that helps predict flow patterns in fluids.
It's calculated as: NRe = (density * velocity * diameter) / viscosity
At a Reynolds number of 2100, the flow is considered laminar flow.
In laminar flow, the fluid flows in smooth layers, and the friction is mainly due to the viscosity of the fluid.
The surface roughness has very little impact on friction in laminar flow because the layers of fluid slide past each other in an ordered fashion, and the irregularities of the wall are hidden under a smooth flowing layer.
Since the flow is already laminar at NRe = 2100, further smoothening the pipe (reducing roughness) won't change the flow pattern or the friction factor.
Therefore, the friction factor (f1) will remain unchanged.

This Question Belongs to Chemical Engineering >> Fluid Mechanics

Join The Discussion

Comments (2)

  1. Aditya Kumar
    Aditya Kumar:
    4 weeks ago

    Remain constant

  2. VIJITH H
    VIJITH H:
    5 years ago

    Ans is no changes, bcz there no change in Nre

Related Questions on Fluid Mechanics