Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
456 views
in General by (108k points)
closed by
In turbulent flows through rough pipes, the ration of the maximum velocity to the mean velocity is 
1. 2
2. \(\frac{4}{3}\)
3. 1.1
4. Dependent on the friction factor 

1 Answer

0 votes
by (101k points)
selected by
 
Best answer
Correct Answer - Option 4 : Dependent on the friction factor 

Concept:

Common mean velocity distribution for turbulent flow,

\(\frac{{u - \bar u}}{{{V^*}}} = 5.75{\log _{10}}\frac{y}{R} + 3.75\)

Where, V* = shear velocity, u = Flow velocity, u̅ = avg. velocity

\({V^*} = \bar u \times \sqrt {\frac{f}{8}} \)

f = friction factor

At y = R, u = umax

\(\frac{{{u_{max}} - \bar u}}{{\bar u \times \sqrt {\frac{f}{8}} }} = 3.78\)

\(\frac{{{u_{max}}}}{{\bar u}} - 1 = \frac{{3.75}}{{\sqrt 8 }} \times \sqrt f \)

\(\frac{{{u_{max}}}}{{\bar u}} = 1 + 1.33\;\sqrt f \)

\(\frac{{{u_{max}}}}{{\bar u}} = 1 + 1.43\;\sqrt f \)

So, from the above equations, we can conclude that, Ratio of maximum velocity to mean velocity is dependent on friction factor.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...