14.8 Problem Set

  1. What is the Reynolds number for an elephant in a 0.1 and a 10.0 m s-1 wind? Assume \(\nu = 1.5\times10^{-5}\). (Make your own estimate for \(d\).)

  2. Given the set of equations below, solve for the unknowns \(x\), \(y\), \(z\). \[x + 2y + 3z = 10\] \[4x - z = 2\]

  3. Refer to the discussion at the end of the section titled “Diffusivity” concerning diffusive resistance to blood uptake. It was stated there that eddy size (\(d\)) and rate of diffusion are inversely proportional. It has also been observed that as the Reynolds number for the flow increases, the mixing rate of the flow increases while at the same time so does the eddy size. To resolve this apparent contradiction, complete the following table and comment on your results in light of the conflict.

Re 3000 6000
Mean flow velocity \(\overline{u}\) 1.6 cm s-1 —
Typical eddy size 0.8 mm 1.0 mm
Relative velocity fluctuation \(u'/\overline{u}\) 0.0242 0.0229
Eddy time scale \(T\) — —

Notice that \(u'\) represents the RMS (root-mean-square) value of the velocity fluctuations about the mean \(\overline{u}\). As such, it can be interpreted as the rotational velocity, on the average, of the eddies.

  1. Solve (14.10) in the text for \(\overline{u}\) as a function of \(z\). Use as limits of integration \(z=0\) and \(z=z\). What are the corresponding limits for velocity? (Hint: remember the “no slip” condition.) What problem arises at the lower limit (\(z=0\))? To remedy this, assume that \(u=0\) at \(z=z_0\) , where \(z_0\) is a small distance above the surface. Assuming that \(k=0.4\), \(z_0=1 cm\) and \(u_*=0.2 m s^{-1}\), plot \(u\) (horizontal axis) versus \(\ln z\). What is the potential significance of a plot such as this?

  2. Given the following conditions, solve for the heat transfer from a leaf using (14.14), (14.15) and Fig. 14.10. \(Re=\frac{\overline{u}d}{v}\)
    \(v=151\times10^{-7}m^2s^{-1}\)(kinematic viscosity of air)
    \(d\) = characteristic length = 0.042m
    \(\overline{u}\) = average velocity = 3 ms\(^{-1}\)
    \(k\) = thermal conductivity of air = 26 mW m\(^{-1}\)°C\(^{-1}\)
    Leaf temperature = 30°C
    Air temperature = 20°C