Lecture 6 Adiabatic Processes

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

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

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

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

Thus

Adiabatic Transition Suppose system starts in state with thermodynamic “coordinates” (T 0, p 0 ) System makes an adiabatic transition to state with coordinates (T 1, p 1 )

Integrate (1)

Continue (1)   (2)

Review

Thus, (2) becomes (3)

Thus, (3) becomes Poisson’s Equation (4)

Dry Air 

Exercise p 0 = 989 hPa T 0 = 276 K p 1 = 742 hPa T 1 = ? Answer: T 1 = 254 K

Exercise p 0 = 503 hPa T 0 = 230 K p 1 = 1000 hPa T 1 = ? Answer: 280 K

Potential Temperature Potential Temperature Let p 0 = 1000 hPa Remove the subscripts from p 1 and T 1 Denote T 0 by   is called the potential temperature

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.

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.

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

Dry Adiabats

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

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

Exercise Calculate T to nearest tenth of a degree

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

Exercise Calculate new T to nearest tenth of a degree Answer: K

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

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

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

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!) (People freeze in their tracks!)

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?

Solution

Height Dependence of T * Start with

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

Step 1 Take log of both sides of * Constants

Step 2: Differentiate **

Step 3: Hydrostatic Equation Substitute into **

Step 4 

Exercise Simplify the expression for dT/dz.

Result Constant!

z Parcel temperature

Conclusion A parcel that rises adiabatically cools at the rate g/c p A parcel that sinks adiabatically warms at the rate g/c p Exercise: Calculate g/c p (in K  km -1 ) Answer: 9.8 K  km -1 Answer: 9.8 K  km -1

Dry-Adiabatic “Lapse Rate”

Important Note Previous discussion assumed that there was no condensation. We will look at the effect of water vapor next.