Solving the Haaland Equation for Re

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SUMMARY

The discussion focuses on solving the Haaland equation to determine the Reynolds number (Re) using given parameters: friction factor λ = 0.000032 m, absolute roughness Ks = 1.5 x 10-6 m, and internal bore diameter d = 0.04 m. Participants express confusion regarding the algebraic manipulation required to isolate Re, particularly in handling logarithmic functions. The equation is presented as 1/√λ = -1.8log[(6.9/Re) + [(Ks/d)/3.7]1.11]. Understanding the inverse of logarithmic functions is crucial for progressing in the solution.

PREREQUISITES
  • Understanding of logarithmic functions and their inverses
  • Basic algebra skills for manipulating equations
  • Familiarity with fluid dynamics concepts, specifically the Reynolds number
  • Knowledge of the Haaland equation and its application in calculating friction factors
NEXT STEPS
  • Study the properties and applications of the Haaland equation in fluid mechanics
  • Learn about logarithmic identities and how to manipulate them in equations
  • Explore the concept of Reynolds number and its significance in fluid flow
  • Practice solving equations involving logarithms and exponential functions
USEFUL FOR

Students in engineering or physics, particularly those studying fluid dynamics, as well as educators looking to clarify the application of logarithmic functions in real-world equations.

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Homework Statement



Using the Haaland equation shown below, determine the Reynold number Re

Friction factor λ = 0.032mm = 0.000032m
Absolute roughness Ks = 1.5 x 10-6m
Internal bore diameter d= 0.04m

1/√ λ = -1.8log[(6.9/Re) + [(ks/d)/3.7]1.11]

I have tried to solve the above equation but haven't got a clue as I have not been taught to this level. Consequently it's a bit unfair. Someone tried to help me but confused me even more. I do know the answer but need to show how I got the answer.

Any help would be appreciated.
 
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It seems a bit surprising you'd be working with an equation like this if you haven't mastered basic algebra yet. Where are you getting stuck?
 
vela said:
It seems a bit surprising you'd be working with an equation like this if you haven't mastered basic algebra yet. Where are you getting stuck?

Basic algebra yes but other people are speaking about anti logs? The left hand side of the equation is easy but I am struggling with the first part of the right hand side.

1/√0.032=-1.8log[(6.9/Re)+[( 1.5 x 10-6/0.04)/3.7]1.11]

Any help would be appreciated.
 
The inverse of the log function is the exponential function ex.
 
Well, more generally, the inverse of the log base n function is the exponential nx. So, logn(nx) = x.
 

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