12.003 Atmosphere, Ocean and Climate Dynamics Lecture VI ...
Transcript of 12.003 Atmosphere, Ocean and Climate Dynamics Lecture VI ...
12.003Atmosphere, Ocean and Climate Dynamics
Lecture VIVertical Structure of the Atmosphere
Saturday, September 17, 2011
Lecture VI Outline
1. Physical properties of dry air2. Vertical distribution of temperature3. Hydrostatic balance4. Vertical distribution of pressure and density
Saturday, September 17, 2011
Physical properties of dry air (p,ρ,T)• Ideal gas law
•
ρ(z) = ρ0 e−z/H (1)ρ0 = 1.35 kg/m3 (2)H = 6.8 km (3)
Ω =vr
=2πR/day
R=
2πday
= 7.27×10−5s−1 (4)
Ve =
2GM/R (5)
Vm =
2kT/m (6)
p V = n Rg T =Mm
Rg T (7)
p = ρRg
mT = ρ RT (8)
Rg = 8.314 JK−1mol−1 (9)
R =Rg
ma= 287 Jkg−1K−1 (10)
p = ∑i
ρiRg
miT = 78%ρ
Rg
mN2
T +21%ρRg
mO2
T + · · · = ρRg
maT (11)
e = ρv Rv T (12)pd = ρd Rd T (13)p = pd + e≈ pd (14)
es = A eβT (15)
1
ρ(z) = ρ0 e−z/H (1)ρ0 = 1.35 kg/m3 (2)H = 6.8 km (3)
Ω =vr
=2πR/day
R=
2πday
= 7.27×10−5s−1 (4)
Ve =
2GM/R (5)
Vm =
2kT/m (6)
p V = n Rg T =Mm
Rg T (7)
p = ρRg
mT = ρ RT (8)
Rg = 8.314 JK−1mol−1 (9)
R =Rg
ma= 287 Jkg−1K−1 (10)
p = ∑i
ρiRg
miT = 78%ρ
Rg
mN2
T +21%ρRg
mO2
T + · · · = ρRg
maT (11)
e = ρv Rv T (12)pd = ρd Rd T (13)p = pd + e≈ pd (14)
es = A eβT (15)
1
ρ(z) = ρ0 e−z/H (1)ρ0 = 1.35 kg/m3 (2)H = 6.8 km (3)
Ω =vr
=2πR/day
R=
2πday
= 7.27×10−5s−1 (4)
Ve =
2GM/R (5)
Vm =
2kT/m (6)
p V = n Rg T =Mm
Rg T (7)
p = ρRg
mT = ρ RT (8)
Rg = 8.314 JK−1mol−1 (9)
R =Rg
ma= 287 Jkg−1K−1 (10)
p = ∑i
ρiRg
miT = 78%ρ
Rg
mN2
T +21%ρRg
mO2
T + · · · = ρRg
maT (11)
e = ρv Rv T (12)pd = ρd Rd T (13)p = pd + e≈ pd (14)
es = A eβT (15)
1
ρ(z) = ρ0 e−z/H (1)ρ0 = 1.35 kg/m3 (2)H = 6.8 km (3)
Ω =vr
=2πR/day
R=
2πday
= 7.27×10−5s−1 (4)
Ve =
2GM/R (5)
Vm =
2kT/m (6)
p V = n Rg T =Mm
Rg T (7)
p = ρRg
mT = ρ RT (8)
Rg = 8.314 JK−1mol−1 (9)
R =Rg
ma= 287 Jkg−1K−1 (10)
p = Rg78%ρmN2
T +Rg21%ρmO2
T + · · · = ρRg
maT (11)
p V = ∑i
ni Rg T = ∑i
Mi
miRg T (12)
e = ρv Rv T (13)pd = ρd Rd T (14)p = pd + e≈ pd (15)
es = A eβT (16)
1
ρ(z) = ρ0 e−z/H (1)ρ0 = 1.35 kg/m3 (2)H = 6.8 km (3)
Ω =vr
=2πR/day
R=
2πday
= 7.27×10−5s−1 (4)
Ve =
2GM/R (5)
Vm =
2kT/m (6)
p V = n Rg T =Mm
Rg T (7)
p = ρRg
mT = ρ RT (8)
Rg = 8.314 JK−1mol−1 (9)
R =Rg
ma= 287 Jkg−1K−1 (10)
p = Rg78%ρmN2
T +Rg21%ρmO2
T + · · · = ρRg
maT (11)
p V = ∑i
ni Rg T = ∑i
Mi
miRg T (12)
e = ρv Rv T (13)pd = ρd Rd T (14)p = pd + e≈ pd (15)
es = A eβT (16)
1
ρ Animation
Universal gas constant
Gas constant of a specific gas
Saturday, September 17, 2011
Physical properties of dry air• Dalton’s law of partial pressures for dry air
ρ(z) = ρ0 e−z/H (1)ρ0 = 1.35 kg/m3 (2)H = 6.8 km (3)
Ω =vr
=2πR/day
R=
2πday
= 7.27×10−5s−1 (4)
Ve =
2GM/R (5)
Vm =
2kT/m (6)
p V = n Rg T =Mm
Rg T (7)
p = ρRg
mT = ρ RT (8)
Rg = 8.314 JK−1mol−1 (9)
R =Rg
ma= 287 Jkg−1K−1 (10)
p = Rg78%ρmN2
T +Rg21%ρmO2
T + · · · = ρRg
maT (11)
p V = ∑i
ni Rg T = ∑i
Mi
miRg T (12)
e = ρv Rv T (13)pd = ρd Rd T (14)p = pd + e≈ pd (15)
es = A eβT (16)
1
ρ(z) = ρ0 e−z/H (1)ρ0 = 1.35 kg/m3 (2)H = 6.8 km (3)
Ω =vr
=2πR/day
R=
2πday
= 7.27×10−5s−1 (4)
Ve =
2GM/R (5)
Vm =
2kT/m (6)
p V = n Rg T =Mm
Rg T (7)
p = ρRg
mT = ρ RT (8)
Rg = 8.314 JK−1mol−1 (9)
R =Rg
ma= 287 Jkg−1K−1 (10)
p = Rg78%ρmN2
T +Rg21%ρmO2
T + · · · = ρRg
maT = ρ R T (11)
p V = ∑i
ni Rg T = ∑i
Mi
miRg T (12)
e = ρv Rv T (13)pd = ρd Rd T (14)p = pd + e≈ pd (15)
es = A eβT (16)
1
ρ(z) = ρ0 e−z/H (1)ρ0 = 1.35 kg/m3 (2)H = 6.8 km (3)
Ω =vr
=2πR/day
R=
2πday
= 7.27×10−5s−1 (4)
Ve =
2GM/R (5)
Vm =
2kT/m (6)
p V = n Rg T =Mm
Rg T (7)
p = ρRg
mT = ρ RT (8)
Rg = 8.314 JK−1mol−1 (9)
R =Rg
ma= 287 Jkg−1K−1 (10)
p = Rg78%ρmN2
T +Rg21%ρmO2
T + · · · = ρRg
maT = ρ R T (11)
p V = ∑i
ni Rg T = ∑i
Mi
miRg T (12)
e = ρv Rv T (13)pd = ρd Rd T (14)p = pd + e≈ pd (15)
es = A eβT (16)
1
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Dry air gas constant
Saturday, September 17, 2011
Atmospheric temperature variations • Atmospheric temperature varies in horizontal and vertical• Vertical variations are similar everywhere
pattern set by radiation budget (thermodynamics)• Horizontal variations depend on height
pattern set atmospheric transport (dynamics)
Saturday, September 17, 2011
Vertical variations of temperature Three hot spots:
1) thermopause =>
2) stratopause =>
3) surface =>Saturday, September 17, 2011
Thermosphere• Named from the Greek θερμός (heat) σφαιρα (sphere), begins about 90 km above Earth’s surface• Molecules are dissociated and electrons are ionized (no triatomic molecules)• Absorbs high-energy UV (λ < 0.1 μm) - heated from above• Temperature increases with height
Saturday, September 17, 2011
Mesosphere• Named from the Greek μεσό (middle) σφαιρα (sphere), located from about 50 km to 80-90 km above Earth's surface• Composition is similar to troposphere and stratosphere (N2, O2, Ar, CO2) • Absorbs decreasing amounts of low-energy UV through ozone• Temperature decreases with height from 0oC to -100oC
Saturday, September 17, 2011
Stratosphere• Named from the Greek στράτος (layered) σφαιρα (sphere), located from about 10 km to 50 km above Earth's surface• Contains most of the ozone in the atmosphere• Ozone absorbs medium wavelength UV (0.1 < λ < 0.35 μm) - heated from above• Temperature increases with height from -40oC to 0oC
Saturday, September 17, 2011
Ozone layer• Ozone is byproduct of photodissociation of molecular oxygen• Ozone absorbs medium wavelength UV (important for life on Earth)
Saturday, September 17, 2011
Troposhere• Named from the Greek τρόπος (turn) σφαιρα (sphere), located within 10 km of the Earth's surface• Contains 85% of the atmospheric mass• Absorbs infrared radiation through H2O and CO2 - heated from below• Temperature decreases with height from to 15oC -40oC
Saturday, September 17, 2011
Troposphere• Atmospheric layer most dynamically active
weather we experience is confined to troposphere• Transport modifies temperature profiles
Saturday, September 17, 2011
Hydrostatic balanced pdz
=−gρ (1)
FgGravitational force
+ FTTop pressure force
+ FBBottom pressure force
= 0 (2)
κ|∇T |2 (3)
uuueTe · ∇T (4)
(∂t +uuug · ∇) T +ΓT v+N2T w = DT (5)
(∂t +uuug · ∇) S +ΓS v+N2S w = DS (6)
(∂t +uuug · ∇) ρ +Γρ v+N2ρ w = Dρ (7)
(∂t +uuug · ∇) C +ΓC v = DC (8)
ρ = ρ0 [1−α (T −T0)+β (S−S0)] (9)
C = ρ0
1−
N2S
N2ρ
α (T −T0)+N2
TN2
ρβ (S−S0)
(10)
(11)
1
d pdz
=−gρ (1)
FgGravitational force
+ FTTop pressure force
+ FBBottom pressure force
= 0 (2)
κ|∇T |2 (3)
uuueTe · ∇T (4)
(∂t +uuug · ∇) T +ΓT v+N2T w = DT (5)
(∂t +uuug · ∇) S +ΓS v+N2S w = DS (6)
(∂t +uuug · ∇) ρ +Γρ v+N2ρ w = Dρ (7)
(∂t +uuug · ∇) C +ΓC v = DC (8)
ρ = ρ0 [1−α (T −T0)+β (S−S0)] (9)
C = ρ0
1−
N2S
N2ρ
α (T −T0)+N2
TN2
ρβ (S−S0)
(10)
(11)
1
Saturday, September 17, 2011
Vertical distribution of pressure• Isothermal atmosphere
• Non-isothermal atmosphere
d p
dz= −gρ (1)
FgGravitational force
+ FTTop pressure force
+ FBBottom pressure force
= 0 (2)
p(z) = ps exp− z
H
H =
RT0
g(3)
p(z) = ps exp−
z
0
dz
H(z)
H(z) =
RT (z)g
(4)
uuueTe · ∇T (5)
(∂t +uuug · ∇) T +ΓT v+N2T
w = DT (6)(∂t +uuug · ∇) S +ΓS v+N
2S
w = DS (7)
(∂t +uuug · ∇) ρ +Γρ v+N2ρ w = Dρ (8)
(∂t +uuug · ∇) C +ΓC v = DC (9)
ρ = ρ0 [1−α (T −T0)+β (S−S0)] (10)
C = ρ0
1−
N2S
N2ρ
α (T −T0)+N
2T
N2ρ
β (S−S0)
(11)
(12)
1
d p
dz= −gρ (1)
FgGravitational force
+ FTTop pressure force
+ FBBottom pressure force
= 0 (2)
p(z) = ps exp− z
H
H =
RT0
g(3)
p(z) = ps exp−
z
0
dz
H(z)
H(z) =
RT (z)g
(4)
uuueTe · ∇T (5)
(∂t +uuug · ∇) T +ΓT v+N2T
w = DT (6)(∂t +uuug · ∇) S +ΓS v+N
2S
w = DS (7)
(∂t +uuug · ∇) ρ +Γρ v+N2ρ w = Dρ (8)
(∂t +uuug · ∇) C +ΓC v = DC (9)
ρ = ρ0 [1−α (T −T0)+β (S−S0)] (10)
C = ρ0
1−
N2S
N2ρ
α (T −T0)+N
2T
N2ρ
β (S−S0)
(11)
(12)
1
Saturday, September 17, 2011
Vertical distribution of density• Isothermal atmosphere
• Non-isothermal atmosphere
d p
dz= −gρ (1)
FgGravitational force
+ FTTop pressure force
+ FBBottom pressure force
= 0 (2)
p(z) = ps exp− z
H
H =
RT0
g(3)
p(z) = ps exp−
z
0
dz
H(z)
H(z) =
RT (z)g
(4)
ρ(z) =ps
RT0exp
− z
H
H =
RT0
g(5)
ρ(z) =ps
RT (z)exp
−
z
0
dz
H(z)
H(z) =
RT (z)g
(6)
uuueTe · ∇T (7)
(∂t +uuug · ∇) T +ΓT v+N2T
w = DT (8)(∂t +uuug · ∇) S +ΓS v+N
2S
w = DS (9)
(∂t +uuug · ∇) ρ +Γρ v+N2ρ w = Dρ (10)
(∂t +uuug · ∇) C +ΓC v = DC (11)
ρ = ρ0 [1−α (T −T0)+β (S−S0)] (12)
C = ρ0
1−
N2S
N2ρ
α (T −T0)+N
2T
N2ρ
β (S−S0)
(13)
(14)
1
d p
dz= −gρ (1)
FgGravitational force
+ FTTop pressure force
+ FBBottom pressure force
= 0 (2)
p(z) = ps exp− z
H
H =
RT0
g(3)
p(z) = ps exp−
z
0
dz
H(z)
H(z) =
RT (z)g
(4)
ρ(z) =ps
RT0exp
− z
H
H =
RT0
g(5)
ρ(z) =ps
RT (z)exp
−
z
0
dz
H(z)
H(z) =
RT (z)g
(6)
uuueTe · ∇T (7)
(∂t +uuug · ∇) T +ΓT v+N2T
w = DT (8)(∂t +uuug · ∇) S +ΓS v+N
2S
w = DS (9)
(∂t +uuug · ∇) ρ +Γρ v+N2ρ w = Dρ (10)
(∂t +uuug · ∇) C +ΓC v = DC (11)
ρ = ρ0 [1−α (T −T0)+β (S−S0)] (12)
C = ρ0
1−
N2S
N2ρ
α (T −T0)+N
2T
N2ρ
β (S−S0)
(13)
(14)
1
Saturday, September 17, 2011