copyright © 2014 r. r. dickerson1 professor russell dickerson room 2413, computer & space...

Copyright © 2014 R. R. Dickerson 1

Professor Russell Dickerson Room 2413, Computer & Space Sciences Building Phone(301) [email protected] web site www.meto.umd.edu/~russ

AOSC 620 Lecture 2PHYSICS AND CHEMISTRY

OF THE ATMOSPHERE I

Copyright © 2013 R. R. Dickerson & Z.Q. Li

22

Experiment: Room temperatureWhat is the temperature of the room?

We’ll get 22 answers.

How well do they agree?

Does location matter?

Does observer bias play a role?


3

Experiment: Room temperature

Last class

Average oC

Stdev oC

Max oC

Min oC

TMean = +/- °C (+/- °F)

Measured with uncalibrated or no thermometers at n locations.


4

What is the right answer?

What is the uncertainty?

Are the data Gaussian?

How can we improve the measurement?


5

Experiment: Room temperature

Let’s repeat the experiment with calibrated thermometers.


6

Lecture 2. Thermodynamics of Air, continued – water vapor.

Objective: To find some useful relationships among air temperature, volume, and pressure.

ReviewIdeal Gas Law: PV = nRT

Pα = R’T

First Law of Thermodynamics: đq = du + đw

W = ∫ pdα


7

Review (cont.)

Definition of heat capacity:

cv = du/dT = Δu/ΔT

cp = cv + R

Reformulation of first law for unit mass of an ideal gas:

đq = cvdT + pdα

đq = cpdT − αdp


8

Review (cont.)For an isobaric process:

đq = cpdT

For an isothermal process:đq = − αdp = pdα = đw

For an isosteric process:

đq = cvdT = du

For an adiabatic process:

cvdT = − pdα and cpdT = αdp


9

Review (cont.)For an adiabatic process:

cvdT = − pdα and cpdT = αdp du = đw

(T/T0) = (p/p0)K

Where K = R’/cp = 0.286

(T/θ) = (p/1000)K

Define potential temperature:θ = T(1000/p)K

• Potential temperature, θ, is a conserved quantity in an adiabatic process.


10

Review (cont.)The Second Law of Thermodynamics is the definition of φ as entropy.

dφ ≡ đq/T

ჶ dφ = 0Entropy is a state variable.

Δφ = cpln(θ/θ0)

In a dry, adiabatic process potential temperature doesn’t change, thus entropy is conserved.


11

Useful idea - a perfect or exact differential:

If z = f(x,y), dz is a perfect differential iff:

∂2f/∂x∂y = ∂2f/∂y∂x dz = 0ჶ

For example, v = f(T,p)

dv = (∂v/∂p)T dp + (∂v/∂T)p dT

This is true for dU, dH, dG, but not đw or đq.


12

Various Measures of Water Vapor Content

• Vapor pressure• Vapor density –

absolute humidity• Mixing ratio, w (g/kg)• Specific humidity• Relative humidity

• Virtual temperature

(density temp)• Dew point temperature• Wet bulb temperature• Equivalent temperature• Isentropic Condensation

Temperature•Potential temperature•Wet-bulb potential temperature•Equivalent potential temperature


13

Virtual Temperature: Tv or T*

Temperature dry air would have if it had thesame density as a sample of moist air at the same pressure.

Question: should the virtual temperature be higher orlower than the actual temperature?


14

Consider a mixture of dry air and water vapor. Let

Md = mass of dry airMv = mass of water vapormd = molecular weight of dry airmv = molecular weight of water.


15

Dalton’s law: P = pi

V

MM

V

M

volumeVm

M

m

M

V

TRP

Tm

RT

m

RepP

vd

v

v

d

d

vv

ddd

)(*

**


16

Combine P and to eliminate V:

)1(

)/1(

1

1)/1(

1

11

1

*

*

*

w

wRTP

wwT

m

R

wm

w

mTR

MMm

M

m

MTRP

d

vd

vdv

v

d

d


17

Since P = RT*

)1(

)/1(

w

wRTPand


18

Alternate derivation: Sinceproportional toMwt

xxm

molegm

m

mTetemperaturvirtualT

w

d

w

d

18)1(29

/29

)(

)(*

Where w is the mass mixing ratio and x (molar or volume mixing ratio) =

[H2O] = w/0.62

T* = T (29/(29-11[H2O]))

e.g., [H2O] = 1% then T* = T(1.004)

If , [H2O] = 1% then w = 0.01*.62 = 0.062 T* = 1.01/1.0062 =


19

Where w is the mass mixing ratio and x (molar or volume mixing ratio) =

[H2O] = w/0.62

T* = T (29/(29-11[H2O]))

e.g., [H2O] = 1% then T* = T(1.004)

Test: if [H2O] = 1% then w(18/29) = 0.01*.62 = 0.0062

T* = T (1 + w/)/(1+w) =

T (1 + 0.01)/(1+0.0062) = T(1.004)


20

Unsaturated Moist Air

Equation of state: P = RT*


21

Specific Heats for Moist Air

Let mv = mass of water vapor md = mass of dry air

To find the heat flow at constant volume:

96.1

)1()1(

11

)(

v

vv

v

vd

vvvvd

d

vd

vvvvddv

ccrwith

rwdTcqdw

cm

cmdTcm

m

mm

dTcmdTcmqdmmQd


22

For constant pressure )9.01( wcc ppm

So Poisson’s equation becomes

)2.01(

1000

wc

R

c

Rk

k

P

mbT

ppm

m

(1+0.6w)/(1+0.9w) => (1-0.2w) due to rounding error.


23

Water Vapor Pressure

Equation of state for water vapor: ev = v Rv Twhere ev is the partial pressure of water vapor

622.0

/

**

*

d

v

d

v

d

dd

d

v

v

v

v

m

m

TR

Tm

m

m

RT

m

m

m

RTRe

m

RR

v

copyright © 2014 r. r. dickerson1 professor russell dickerson room 2413, computer & space...

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