Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

Air: nr=1.00006, ni=1e-10 Substance: nr=1.67, ni=0.2

I0

0 x

Given flux I0 incident on the air:substance boundary.Calculate the Flux Transmitted to Point X:

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

Given flux I0 incident on the air:substance boundary.Calculate the Flux Transmitted to Point X:

Air: nr=1.00006, ni=1e-10 Substance: nr=1.67, ni=0.2

I0

0 x

Answer:

1R

1-R=T

Propagator from 0 to x.

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

0 x

What if we divide the substance into particles?Calculate the Flux Transmitted to Point X:

Air: nr=1.00006, ni=1e-10 Substance: nr=1.67, ni=0.2

I0

N identical particles / volumev = particle volumea = average particle projected area.ext = a Qext = Single Particle Extinction Cross Section. Qext=Extinction Efficiency.

ext = abs+sca , ext= ext(nr,ni,)

Pext = I0 ext = Power (watts) removed by a single particle from I0 by extinction.

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

0 x

What if we divide the substance into particles?Calculate the Flux Transmitted to Point X:

Air: nr=1.00006, ni=1e-10 Substance: nr=1.67, ni=0.2

I0

N particles / volumev = particle volumea = average projected area for each particle.ext = a Qext = Particle Extinction Cross Section. Qext=Extinction Efficiency.

ext=N ext=Extinction Coefficient. Assume sca= N sca x <<1 .(single scattering assumption).

I(x)=I0 exp(- ext x)Otherwise, use multiple scattering theory (to

be developed soon)

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

Single Particle Perspective: Assume ext ≈ abs , sca ≈ 0 (particle size much less than the wavelength, deep in the Rayleigh

range. Size parameter << 1.)

absorption

Gross, Special Purpose, ad-hoc Approximation:

abs = a[1-exp(-Deq/)]. Let Deq=v/a. =/(4ni)=skin depth.

Limits:

Deq<< , (1-e-small)≈small, abs = 4niv/

Deq>> , (1-e-large)≈1, abs = a .

v = particle volumea = particle projected area

Deq

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

0 x

What if we divide the substance into particles?Calculate the Flux Transmitted to Point X:

Air: nr=1.00006, ni=1e-10 Substance: nr=1.67, ni=0.2

I0

N particles / volumev = particle volumea = average projected area for each particle.

abs= N abs x. Deq<< , abs = 4niv/

I(x)=I0 exp(- 4nivNx/ )vN=C=(Particle Volume)/Volume

C=Concentration (e.g. ppmv)I(x)=I0 exp(- 4nixC/ )

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

Air: nr=1.00006, ni=1e-10 Substance: nr=1.67, ni=0.2

I0x

0

Compare

(Assumes no particle scattering,dilute (C<<1), weak absorption).

C=volumetric concentration.

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

Homework: Compare Mie Theory for Spheres with the simple model for absorption below.

Gross Special Purpose Approximation:

abs = a[1-exp(-Deq/)]. Let Deq=v/a. =/(4ni)=skin depth.

v=4r3/3. a=average projected area=r2 for a sphere. D=2r. Deq=2D/3.

Cases in a 3 matrices for fixed nr and variable D and ni:(calculate the percentage error of the model and Mie theory.)= 0.5 um.

nr=1, nr=1.33, nr=1.5

D=0.01 um, 0.1 um, 1 um, 10 um.

ni=0.001, ni=0.01, ni=0.1, ni=1.

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

Table for Homework (one for each real refractive index, 1.0, 1.333, and 1.5). Fill each empty table with a

percentage error as defined below.

D (microns)

ni

0.01 0.1 1 10

0.001

0.01

0.1

1

Pat Arnott, ATMS 749 Atmospheric Radiation Transfer

Air: nr=1.00006, ni=1e-10

Bulk Substance: nr=1.67, ni=0.2

I0x

0

Compare: Bulk Substance, Gas, and Particles

(Assumes no particle scattering,dilute (C<<1), weak absorption).C=volumetric concentration.

Gas

Particles