CHEM-5151_S05_L4_Transport2

# CHEM-5151_S05_L4_Transport2 - Lecture 4 Chemical Transport...

This preview shows pages 1–6. Sign up to view the full content.

1 Suggested Reading: SP Chapter 17 Atmospheric Chemistry CHEM-5151 / ATOC-5151 Spring 2005 Prof. Brian Toon (PAOS) Lecture 4: Chemical Transport in the Atmosphere The Aerosol Continuity Equation A. transport The change in the concentration of a chemical often can be written as C t = P LC • Here L is the loss rate, P is the production rate, and C is the species concentration (per unit volume).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 How do we account for the effects of transport on C? x y z UC u u U C dd Wind direction δ δ δ An observer standing at a fixed point in space measures changing concentrations. The observer must account for the chemical sources and sinks as well as for the motion of the air. This is called an Eulerian measurement since it is at a point. x y z u u U C Wind direction δ δ δ The flux of material into the upwind side of the box is U u C u (particles per cm 2 per sec.) The total number of particles being added per second is U u C u dydz (particles per second), where dydz is the area of the open face of the box. Therefore, considering that material is also leaving the box on the downwind side, the total amount of material added to the box per second, divided by the volume of the box so that we have the particles added cm -3 s -1 , is C t = ( U u C u U d C d )/ dx =− dUC / dx
3 Considering all three directions and adding the sources and sinks, we arrive at the flux form of the continuity equation C t =− (UC) x (VC) y (WC) z + P-LC This is often called the total derivative in dC dt = P LC Example of flux based Eulerian transport: 10 13 mol cm -3 9x10 12 mol cm -3 L=10 13 mol cm -3 hr -1 U=10 km/hr 1 km Assume the wind speed is constant at 10 km hr -1 . Assume the concentration declines by 10 12 molecules cm -3 km -1 in the wind direction. Assume the concentration declines by 10 13 molecules cm -3 hr -1 due to a chemical sink. The rate of change in the concentration at the fixed downwind position is C t =− 10 13 mol hr (10 km hr )( 9x10 12 10 13 1 mol km )=0 Hence, in this example the advection by the wind completely masks the ongoing chemical loss of the material.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4 The Lagrangian form of the continuity equation U U U d u An observer moving with the wind, so that the same air parcel is always observed, only has to account for physical and chemical changes within the air parcel, and not for air motions, to understand how the mixing ratio varies. The Lagrangian form of the continuity equation Since neither air molecules, nor the species being observed can be lost from within the parcel, the ratio of the species concentration to the air density is not changed no matter how winds distort the volume of the air parcel. d(C/ ρ ) dt = (P LC)/ ρ
5 A view of aerosol transport , There were large fires in Russia prior to this time period, and dust storms in Africa. Can you tell the source and sink regions just by glancing at the distributions and knowing the winds?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 19

CHEM-5151_S05_L4_Transport2 - Lecture 4 Chemical Transport...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online