Thermodynamic energy eqn. (conservation of energy) Conservation of energy xyz

1. Energy conservation for air 2. Internal energy?

3. What is work?

Thermodynamic energy eqn. (conservation of energy)

From uspowerandlight.com

Q = ΔU + W

c

DT

Dt − α Dp Dt = J

Conservation of energy

x

y z

Q = ΔU + W

m m m ₍ ^c

ΔT + pΔα = q ₎ ⁽ α = V/m ⁾ c

ΔT + pΔv = q

What is work?

W = F ⋅ Δx

From khan academy

Conservation of energy

x

y z

Q = ΔU + W ( )

( )

c

ΔT + pΔα = q α = V/m c

ΔT + pΔv = q

Conservation of energy

x

y z

Q = ΔU + W

m m m ₍ ^c

ΔT + pΔα = q ₎ ⁽ α = V/m ⁾

c

ΔT + pΔv = q

Conservation of energy

Q = ΔU + W

m m m ( )

c

ΔT + pΔα = q c

ΔT + pΔv = q

c

= 717.6 J K

kg

(specific heat at constant volume)

* Temperature is simply 


a measure of internal energy

Conservation of energy

Q = ΔU + W

m m m ₍ ^c

ΔT + pΔα = q ₎ c

ΔT + pΔv = q

(from www.dlt.ncssm.edu)

N number of molecules m’ mass of molecule (kg)

n amount of molecules (in moles) R* gas constant (=8.314 J K⁻¹mol⁻¹) (N/n = NA, Avogadro’s number)

Internal energy

without rotation

N [ 1

2 m′v

]

= 3 2 nR*T

c

= 717.6 J K

kg

(specific heat at constant volume)

Conservation of energy

Q = ΔU + W

m m m ( )

c

ΔT + pΔα = q c

ΔT + pΔv = q

(from www.dlt.ncssm.edu)

N number of molecules m’ mass of molecule (kg)

n amount of molecules (in moles) R* gas constant (=8.314 J K⁻¹mol⁻¹) (N/n = NA, Avogadro’s number)

R specific gas constant (=287 J K−1 kg−1) of dry air
 (= R*/M; M is moler mass of dry air)

Internal energy

without rotation

[KE]

m = 3 2 RT

c

= 717.6 J K

kg

(specific heat at constant volume)

Conservation of energy

Q = ΔU + W

m m m ₍ ^c

ΔT + pΔα = q ₎ c

ΔT + pΔv = q

(from www.dlt.ncssm.edu)

N number of molecules m’ mass of molecule (kg)

n amount of molecules (in moles) R* gas constant (=8.314 J K⁻¹mol⁻¹) (N/n = NA, Avogadro’s number)

R specific gas constant (=287 J K−1 kg−1) of dry air
 (= R*/M; M is moler mass of dry air)

Internal energy

with rotation (two atom molecules)

[KE]

m = 5 2 RT

c

= 717.6 J K

kg

(specific heat at constant volume)

First law of thermodynamics For a closed system,

ΔU = Q - W (or Q = ΔU + W)


Atmospheric version (ideal gas, per unit mass) c

ΔT + pΔα = q

Conservation of energy

Heat supplied
 from outside

change in


Internal energy

Work done 
 by the system

c

DT

Dt + p D α Dt = J c

DT

Dt − α ^Dp Dt = J

popular form ( c

ΔT − αΔp = q )

ΔU/m = c

T


= (5/2) RT
 c

= 717 J K

kg

(specific heat at constant volume)

c

= 1004 J K

kg

(specific heat at constant pressure)

Dry adiabatic lapse rate ( ) Γ _d

For a rising air parcel, temperature decreases ~ 10 K/km

How do we know?

Γ

= 10 K/km

c

ΔT − αΔp = q

10 K/

km

Potential temperature (θ)

Temperature


(January, mid-latitude)

Pressure (hPa)

Temperature (K)

11 km

5 km 16 km

3 km

c

ΔT − αΔp = q

Potential temperature (θ)

Temperature


(January, mid-latitude)

Pressure (hPa)

Temperature (K)

11 km

5 km 16 km

3 km

c

ΔT − αΔp = q

Potential temperature

Temperature (T), potential temperature (θ)

Temperature


(January, mid-latitude)

Pressure (hPa)

Temperature (K)

Temperature


(January, mid-latitude) Potential

temperature

T=218 K

θ = 345 K

= 218*(1000/200)

Dry adia bat

Temperature (T), potential temperature (θ)

Pressure (hPa)

Temperature (K)

Temperature


(January, mid-latitude) Potential

temperature

T=218 K

θ = 345 K

= 218*(1000/200)

Dry adia bat (g

/c_p

~ 10K /km)

Temperature (T), potential temperature (θ)

Pressure (hPa)

Temperature (K)

11 km

5 km

Temperature (T), potential temperature (θ)

Shading: zonal wind (m/s) Contour: temperature (K)
 Thick line: tropopause

Global transport processes


(stratosphere-troposphere exchange process; STE )

(Stohl et al. 2003)

Transport along isentrope


(constant potential temperature surface)