copyright © 2014 r. r. dickerson1 professor russell dickerson room 2413, computer & space...
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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
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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?
Copyright © 2010 R. R. Dickerson & Z.Q. Li
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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.
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What is the right answer?
What is the uncertainty?
Are the data Gaussian?
How can we improve the measurement?
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Experiment: Room temperature
Let’s repeat the experiment with calibrated thermometers.
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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
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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
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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
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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.
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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.
Copyright © 2013 R. R. Dickerson & Z.Q. Li
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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.
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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
Copyright © 2013 R. R. Dickerson & Z.Q. Li
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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?
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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.
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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
)(*
**
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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
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Since P = RT*
)1(
)/1(
w
wRTPand
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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 =
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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)
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Unsaturated Moist Air
Equation of state: P = RT*
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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
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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.
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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