Re: Drag
Today we'll ....
Explore a paradox explained by viscosity
One drag, now two
Re: Reynold's concept
Conservation of Energy: Bernoulli
What is the relationship between fluid motion and pressure?
Potential energy (PE=mgh)
Kinetic energy .... \(KE =mu^2/2\)
Mechanical work (W=Fd=PAd)
Along a streamline PE + KE + W = constant
\((P_2-P_1)/\rho+(u_2^2-u_1^2)/2=0\)
D'Alembert's Paradox
What is the relationship between fluid motion and pressure?
Along a streamline PE + KE + W = constant
- steady
- incompressible
- inviscid*
\((P_2-P_1)/\rho+(u_2^2-u_1^2)/2=0\)
No net force?
What can you say intuitively about this situation?
D'Alembert's Paradox
What is the relationship between fluid motion and pressure?
Along a streamline PE + KE + W = constant
- steady
- incompressible
- inviscid*
\((P_2-P_1)/\rho+(u_2^2-u_1^2)/2=0\)
No net force? Not when \(\mu\neq0\)
D'Alembert's Paradox
What is the relationship between fluid motion and pressure?
Along a streamline PE + KE + W = constant
- steady
- incompressible
- inviscid*
Viscosity robs fluid of its momentum. There is a shear stress exerted on the sphere and energy is dissipated by viscosity
\((P_2-P_1)/\rho+(u_2^2-u_1^2)/2=0\)
No net force? Not when \(\mu\neq0\)
D'Alembert's Paradox
What is the relationship between fluid motion and pressure?
Along a streamline PE + KE + W = constant
- steady
- incompressible
- inviscid*
Becomes even more apparent when we consider a shape that results in lots of flow changing velocity quickly (i.e., shape matters!).
\((P_2-P_1)/\rho+(u_2^2-u_1^2)/2=0\)
No net force? Not when \(\mu\neq0\)
D'Alembert's Paradox
What is the relationship between fluid motion and pressure?
Along a streamline PE + KE + W = constant
- steady
- incompressible
- inviscid*
What happens when \(P_a<P_p\)?
\((P_2-P_1)/\rho+(u_2^2-u_1^2)/2=0\)
Consequences of D'Alembert's Paradox
Flow separation
Where we would always have "skin" drag, with \(\mu\neq 0\), we now have "pressure" drag.
Consequences of D'Alembert's Paradox
Viscosity robs fluid of its momentum. There is a shear stress exerted on the sphere and energy is dissipated by viscosity.
Because of viscosity, velocity cannot increase as much as in the inviscid case.
New stagnation point where the flow separates
Reynolds number: Ratio of inertial (pressure) forces to viscous (shear) forces within a fluid which is subjected to relative internal movement due to different fluid velocities.
How much does it separate?
$$\small{\frac{\text{Pressure stress}}{\text{Shear stress}} \rightarrow \frac{P}{\tau} \rightarrow \frac{\rho u^2}{\mu u/L} \rightarrow Re= \frac{\rho uL}{\mu}}$$
Re across size and velocity scales
\(10^{-5}\) ~ Bacteria
\(10^{-4}\) ~ Spermatozoa
\(10^{-1}\) ~ Ciliate
1~Smallest Fish
\(10^2\) ~ Blood flow in brain
\(10^3\) ~ Blood flow in aorta
\(10^4\) ~ Birds flying 1
\(2 \textrm{x} 10^5\) ~ Typical pitch in Major League Baseball
\(4 \textrm{x} 10^6\) ~ Human Swimming
\(10^6\) ~ Fastest Fish
\(3 \textrm{x} 10^8\) ~Blue Whale
\(5 \textrm{x} 10^9\) ~A large ship (RMS Queen Elizabeth 2)
$$\small{\frac{\text{Pressure stress}}{\text{Shear stress}} = Re= \frac{\rho uL}{\mu}}$$
Re across size and velocity scales
The Re measures the relative importance of inertial vs viscous stresses in determining the flow. Conservation of Re implies identical flow patterns
\(10^{-4}\) ~ Spermatozoa
\(4 \textrm{x} 10^6\) ~ Human Swimming
What would it feel like to swim like a sperm?
How would you induce flow conditions to experience this?
$$\small{\frac{\text{Pressure stress}}{\text{Shear stress}} = Re= \frac{\rho uL}{\mu}}$$
