5.10. Notes on the Marshall-Palmer distribution and the Z-RR relation

A brief backgrounder in where Stull gets his equation 8.30 on page 247:

\[Z = a_3 RR^{a_4}\]

This equation works because the collision/coalescence process for rain formation produces a remarkably regular distribution of droplet sizes for different rainrates, essentially “fingerprinting” the rain rate with its drop-size distribution, which produces a unique value of the radar reflectivity.

In 1947 Marshall and Palmer published their measurements of rain drop size as a function of rain rate:

_images/marshall_size.png

These results are well fit by this equation:

\[n(D) = N_0 \exp(-\Lambda D)\]

where \(\Lambda=4.1 RR^{-0.21}\) with D in mm, \(N_D\) in \(m^{-3}\,mm^{-1}\) and RR in \(mm/hr\)

I’ve put this equation into the following function in marshall_palmer.py:

def marshall_dist(Dvec,RR):
  """
     Calcuate the Marshall Palmer drop size distribution

     Input: Dvec: vector of diameters in mm
            RR: rain rate in mm/hr
     output: n(Dvec), length of Dvec, in m^{-3} mm^{-1}
  """
  N0=8000  #m^{-3} mm^{-1}
  the_lambda= 4.1*RR**(-0.21)
  output=N0*np.exp(-the_lambda*Dvec)
  return output

You can run it with:

python -m a301.scripts.marshall_palmer

and should see the following plot:

_images/marshall_palmer.png

Here’s a 2009 paper that presents the current leading contender for why the drop-size distribution behaves this way:

Radar problems:

Will be assigned on Wednesday.

  1. Integrate \(Z=\int D^6 n(D) dD\) assuming a Marshall Palmer size distribution and show that it integrates to:

    \[Z \approx 300 RR^{1.5}\]

    with Z in \(mm^6\,m^{-3}\) and RR in mm/hr. It’s helpful to know that:

    \[\int^\infty_0 x^n \exp( -a x) dx = n! / a^{n+1}\]

Radar equation problem

  1. Notebook problem:
    1. Calculate \(P_r\) for the Nexrad radar given \(|K|^2\), Z, R, \(L_a\).