ATOC 4710/5710 Atmospheric Physics
Problem Solving 7
April 4, 2003

1. A needle shaped ice crystal is growing by vapor deposition in a relative humidity of 100% at a temperature of -5 oC and pressure of 800 hPa.

a) Calculate the supersaturation with respect to ice.

Supersaturation with respect to ice is defined as Si = (e-esi)/esi = e/esi - 1

where esi is the saturation vapor pressure over an ice surface. We can find the ambient vapor pressure e from the relative humidity, since RH = 100% implies e = es. Thus, for a tenperatuee of - 5oC, the supersaturation with respect to ice is

Si = 4.214 hPa /4.014 hPa - 1 = 0.050 = 5.0%

b) The needle has length L=400 mm and diameter D = 40 m m. Modeling the needle as a prolate spheroid, the capacitance for D << L is

 


C = (1/2L) / [ln(2L/D)].

The growth constant is Gi = 4 p D rsi =1.1x10-6 kg m-1 s-1. Calculate the mass growth rate by vapor deposition (dM/dt). For comparison the mass of the needle is M = rip(D/2)2 L = 4.6 x 10-10 kg (ri = 916 kg m-3).

The mass growth rate equation for vapor diffusion/deposition is

dM/dt = 4 pC D r si Si = C Gi Si

dM/dt = (66.8 x 10-6 m) (1.1 x 10-6 kg m-1 s-1) (0.050) = 3.67 x 10-12 kg s-1

Comparing to the mass of the ice crystal

(1/M) dM/dt = 1/125 s-1

meaning that it would take just 12 seconds to increase the mass by 10%. Note the difference compared to what we found for growth of a liquid drop.

 

2. The needle crystal in 1 is falling at 0.20 m s-1 in a cloud with liquid water content of W=0.5g m-3. Calculate the mass growth rate by riming (dM/dt).

The collection equation for riming is
 

dM/dt = A vt W Ec

where A is the horizontal cross sectional area of the ice crystal, vt is the terminal velocity (fallspeed), W is the liquid water content of cloud droplets, and Ec is the collection efficiency. We will assume that Ec = 1. Due to aerodynamic forces, the needle falls with its long axis in the horizontal plane. Therefore, the area is
 

A = LD = (400x10-6 m) (40 x 10-6 m) = 1.6 x 10-8 m2

The mass growth rate is

dM/dt = (1.6 x 10-8 m2) (0.20 m s-1) (0.5 g m-3) (1.0) = 1.6 x 10-12 kg s-1

The growth rate by riming for this needle is actually smaller than the growth rate by vapor deposition. This would never be the case for a similarly sized liquid droplet.