AcademyFluids
Academy
Viscosity and Turbulence
Level 1 - Physics topic page in Fluids.
Principle
Viscosity transfers momentum between fluid layers, and large Reynolds number promotes turbulence.
Notation
\(\eta\)
dynamic viscosity
\(\mathrm{Pa\,s}\)
\(\tau\)
shear stress
\(\mathrm{Pa}\)
\(F\)
shear force
\(\mathrm{N}\)
\(A\)
area of the sheared layer
\(\mathrm{m^{2}}\)
\(\frac{dv}{dy}\)
velocity gradient across layers
\(\mathrm{s^{-1}}\)
\(\mathrm{Re}\)
Reynolds number
1
Method
Viscosity resists relative sliding of neighboring fluid layers, while turbulence appears when inertial effects dominate over viscous smoothing.
Stress definition
\[\tau=\frac{F}{A}\]
Newtonian fluid
\[\tau=\eta\frac{dv}{dy}\]
Layer force
\[F=\eta A\frac{dv}{dy}\]
Flow regime ratio
\[\mathrm{Re}=\frac{\rho vL}{\eta}\]
Low Reynolds number favors laminar flow; large Reynolds number favors turbulence.
Laminar flow keeps smooth neighboring layers, while turbulence introduces eddies, mixing, and extra energy loss.
Rules
These are the compact viscosity and flow-regime relations.
Shear stress
\[\tau=\frac{F}{A}\]
Viscous law
\[\tau=\eta\frac{dv}{dy}\]
Layer force
\[F=\eta A\frac{dv}{dy}\]
Reynolds number
\[\mathrm{Re}=\frac{\rho vL}{\eta}\]
Examples
Question
Oil with viscosity
\[0.90\,\mathrm{Pa\,s}\]
fills a gap of \[3.0\,\mathrm{mm}\]
between plates of area \[0.20\,\mathrm{m^2}\]
If the top plate moves at \[0.60\,\mathrm{m\,s^{-1}}\]
find the shear force.Answer
Assume a linear velocity profile, so
\[\frac{dv}{dy}=\frac{0.60}{3.0\times10^{-3}}=200\,\mathrm{s^{-1}}\]
Then \[F=\eta A\frac{dv}{dy}=0.90(0.20)(200)=36\,\mathrm{N}\]
Checks
- Viscosity units are pascal-seconds, equivalent to newton-seconds per square metre.
- Larger viscosity lowers Reynolds number for the same \(\\rho\), \(v\), and \(L\).
- Turbulent flow usually produces larger losses than the ideal Bernoulli model predicts.
- Shear force acts parallel to the layer, not perpendicular to it.