Solution

Problem 4.1: Estimate the Reynolds number of the flow generated a car going 55 mph, where the size of the vortices are approximately the car cross section and the velocity of hte vorticies are approximately the same as the velocity of the car. The kinematic viscosity of air at 15C is $1.48 \times 10^{-5} m^2 /s$.

We start with the definition of the Reynolds number:

$$Re = \frac{UL}{\nu}$$

We need to convert the velocity of the car to SI units:

$$U = 55 \frac{miles}{hour} \times \frac{1609.34 m}{1 mile} \times \frac{1 hour}{3600 s} = 24.587 m/s$$

We are told that the size of the vortices are approximately the car cross section. We will assume that the car is $2m$ thus $L=2m$. We are given the kinematic viscosity of air at 15C as $\nu = 1.48 \times 10^{-5} m^2/s$. Now we can compute the Reynolds number:

$$Re = \frac{UL}{\nu} = \frac{(24.587 m/s)(2m)}{1.48 \times 10^{-5} m^2/s} = 3.32 \times 10^6$$

Problem 4.2: Estimate the Reynolds number in a water bottle that you vigorously shake.

We start with the parameters of water: $\nu = 1 \times 10^{-6} m^2/s$ We assume that the size of the vortices are approximately the diameter of the water bottle. We will assume that the diameter of the water bottle is $0.1m$ thus $L=0.3m$. We will assume that you can shake the water bottle at a frequency of $2 Hz$ and the amplitude of the of the water bottle is $0.3m$. Thus, we can estimate the velocity as the maximum speed of the wave, hence:

$$U = 2\pi \times 2 Hz \times 0.3 m = 1.2 \pi m/s \approx 3.8 m/s$$

Now we can compute the Reynolds number:

$$Re = \frac{UL}{\nu} = \frac{(3.8 m/s)(0.1m)}{1 \times 10^{-6} m^2/s} = 3.8 \times 10^5$$

Problem 4.3: Estimate the Reynolds number on the commercial airplane wing. We say that the airplane is flying at $550 mph$ and the average chord length of the wing is approximately $3m$ with a kinematic viscosity of air at $1.5 \times 10^{-5} m^2/s$. We start by converting the velocity to SI units:

$$U = 550 \frac{miles}{hour} \times \frac{1609.34 m}{1 mile} \times \frac{1 hour}{3600 s} = 245.87 m/s$$

Now we can compute the Reynolds number:

$$Re = \frac{UL}{\nu} = \frac{(245.87 m/s)(3m)}{1.5 \times 10^{-5} m^2/s} = 4.92 \times 10^7$$