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Lecture 6

Lecture 6. Adiabatic Processes. Definition. Process is adiabatic if there is no exchange of heat between system and environment, i.e., dq = 0. Work and Temperature (General). First law: dU = dQ – dW Adiabatic process: dU = -dW If system does work (dW > 0), dU < 0  system cools

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Lecture 6

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  1. Lecture 6 Adiabatic Processes

  2. Definition • Process is adiabatic if there is no exchange of heat between system and environment, i.e., dq = 0

  3. Work and Temperature (General) • First law: dU = dQ – dW • Adiabatic process: dU = -dW • If system does work (dW > 0), dU < 0  system cools • If work is done on system (dW < 0), dU > 0  system warms

  4. Work and Temperature (Ideal Gas) • Adiabatic process: cvdT = - pd • Expansion (d > 0)  dT < 0 (cooling) • Contraction (d < 0)  dT > 0 (warming)

  5. Relationship between T and p [Eq. (4) from last lecture.] dq = 0  But,

  6. Thus

  7. Adiabatic Transition • Suppose system starts in state with thermodynamic “coordinates” (T0, p0) • System makes an adiabatic transition to state with coordinates (T1, p1)

  8. Integrate (1)

  9. Continue (1)   (2)

  10. Review

  11. Thus, (2)becomes (3)

  12. Thus, • (3) becomes (4) Poisson’s Equation

  13. Dry Air

  14. Exercise • p0 = 989 hPa • T0 = 276 K • p1 = 742 hPa • T1 = ? • Answer: T1 = 254 K

  15. Exercise • p0 = 503 hPa • T0 = 230 K • p1 = 1000 hPa • T1 = ? • Answer: 280 K

  16. Potential Temperature • Let p0 = 1000 hPa • Remove the subscripts from p1 and T1 • Denote T0 by   is called the potential temperature

  17. Physical Meaning • Initial state: (T, p) • Suppose system makes an adiabatic transition to pressure of 1000 hPa • New temperature =  • Potential temperature is the temperature a parcel would have if it were to expand or compress adiabatically from its present pressure and temperature to a reference pressure level. Po = 1000 mb.

  18. Physical Meaning • Removes adiabatic temperature changes experienced during vertical motion • ºC and K are interchangeable; best to convert it to K when making calculations such as differences. • is invariant along an adiabatic path • adiabatic behavior of individual air parcels is a good approximation for many atmospheric applications…from small parcels to larger convection.

  19. Adiabats • Let  be given • Re-write last equation: In the T-p plane, this describes a curve. Curve is called a dry adiabat.

  20. Dry Adiabats

  21.  = 290 K Initial state: T = 290.0 K, p = 1000 hPa

  22. Reduce pressure to 900 hPa New temperature: T  281 K

  23. Exercise • Calculate T to nearest tenth of a degree

  24. Reduce pressure to 700 hPa New temperature: T  262 K

  25. Exercise • Calculate new T to nearest tenth of a degree • Answer: 261.9 K

  26. Adiabatic Processes • In the T-p plane, an adiabatic process can be thought of as a point moving along an adiabat.

  27. The Parcel Model • An air parcel is a hypothetical volume of air that does not mix with its surroundings • Parcel is a closed system. • Parcel moves adiabatically if there is no exchange of heat with surroundings. • 1 and 2  parcel is isolated • Parcel doesn’t interact with surroundings.

  28. Rising & Sinking Parcels • If a parcel rises adiabatically, its pressure decreases  parcel cools • If a parcel sinks adiabatically, its pressure increases  parcel warms

  29. Movie: “The Day After Tomorrow” • Premise: global warming produces gigantic storms • In these storms, cold air from upper troposphere is brought down to surface, causing sudden cooling • (People freeze in their tracks!)

  30. Problem With Premise • As air sinks, it WARMS! • Suppose air at height of 10 km sinks rapidly to surface • Pressure at 10 km  260 hPa • Temperature  220 K • If surface pressure  1000 hPa, what is air temperature upon reaching surface?

  31. Solution

  32. Height Dependence of T Start with *

  33. Parcel Temperature • Consider a parcel with pressure p and temperature T. • Assume the parcel rises adiabatically   is constant • Goal: Determine dT/dz. • Method: logarithmic differentiation

  34. Step 1 • Take log of both sides of * Constants

  35. Step 2: Differentiate **

  36. Step 3: Hydrostatic Equation Substitute into **

  37. Step 4

  38. Exercise • Simplify the expression for dT/dz.

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