01/25/20061 cveen7920 hydrometeorology heat and radiation naoki mizukami

01/25/200601/25/2006 11CVEEN7920 HydrometeorologyCVEEN7920 Hydrometeorology

Heat and RadiationHeat and Radiation

Naoki Mizukami

01/25/200601/25/2006 CVEEN7920 HydrometeorologyCVEEN7920 Hydrometeorology 22

OutlineOutline1. Heat

Definition of energy and heat Conduction Advection Latent heat/Sensible heat Surface heat budget

2. 1st law of thermodynamics Definition Thermodynamics equation Potential Temperature Lapse rate

3. Radiation Electromagnetic wave Blackbody radiation Interaction of EM energy with materials


1. Heat 1. Heat – energy & heat– energy & heat

Heat: energy transferred between two objects (or system) associated with temperature -> unit [J]

Energy: the ability to work on object

measured in an unit of work [Joule = kg·m2·s-2]

1 [J] = work done by 1 [N] of force to move 1 kg of object by 1 m distance

1 [cal] = 4.185 [J]

rate of work [w =J/s]

Types of energy:

• Potential energy: energy associated with position

e.g. gravitational (mgh), magnetic, electric field

• Kinetic energy: energy due to motion of object

motion of body (macroscopic) -> ke = 1/2*mv2

motion of molecule (microscopic) -> Thermal energy

Temperature: directly proportional to molecular or atomic motion, measure of thermal energy


1. Heat 1. Heat – energy & heat– energy & heat Heat transfer modes

• Conduction: transfer within a substance via molecular bouncing

• Advection (convection): transfer by movement of medium substance (fluid)

• Radiation: transfer by electromagnetic (EM) waves

radiation

convection

conduction


1. Heat1. Heat – energy & heat– energy & heat Conduction (W·m-2)

z

Tk

y

Tk

x

TkTkqc ,,

Where

qc: heat flux [W·m-2]

k: thermal conductivity [W·m-1·K-1] For air, kair = 0.03 [W·m-1·K-1] )

T: Temperature [K]

Flux: quantity transferred into/out of unit area per unit time

e.g. energy flux (J·m-2·s-1 = W·m-2)

Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere?

Ans. 300 [W·m-2] = -0.03 [W·m-1·K-1] x ΔT[K] / 0.001[m]

ΔT[K] = -10 [K]

This temperature gradient within air mostly happens just above the ground on sunny day

W C

0x

T

x

qc >0


1. Heat1. Heat – energy & heat– energy & heat Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere?

W

C z

1mm


1. Heat1. Heat – energy & heat– energy & heat Q. what is the vertical temperature difference is needed to conduct 300 W·m-2 of heat flux across 1 mm depth of atmosphere?

Ans. 300 [W·m-2] = -0.03 [W·m-1·K-1] x ΔT[K] / 0.001[m]

ΔT[K] = -10 [K]

This temperature gradient within air mostly happens just above the ground on sunny day

W

C z


1. Heat1. Heat – energy & heat– energy & heat

z

Tw

y

Tv

x

TuTVTadv

w c

x

u >0

0x

T

e.g. warm advection

Warm advection -> positive

Cold advection -> negative

Advection (convection) [K· s-1]

where

u, v and w: wind speed [m·s-1]

T: Temperature [K]

This is temperature change rate. Can convert heat transfer rate using specific heat

Advection: horizontal transfer (x and y direction)

Convection: vertical transfer (z direction)



w

c

x

Q. Air temperature at a point 50 km north of a station is 3 ° C cooler than station. Wind is blowing 60 ° from north at 20 m s-1. What is temperature change due to advection?

y

Wind = 20ms-1

u = ??

v = ????y

T



w

c

x

??y

T

Q. Air temperature at a point 50 km north of a station is 5 ° C cooler than station. Wind is blowing 60 ° from north at 20 m s-1 for 1 hr. What is temperature change due to advection?

y

u = ??

v = ??

Wind = 20ms-1

Ans.For x direction;

u = 20*sin30° [ms-1], dT/dx = 0

so ΔTx = 0 [° C/sec]

For y direction;

v = 20*cos30 ° = 10 [ms-1], dT/dy = -3/50e+3[° C/m]

So ΔTy = -10* (-5/50e+3) = 0.001[° C /sec]

So 0.001*60*60 = 3.6 [° C] increase

For total,

0+ 3.6 [° C] = 3.6 [° C] increase


1. Heat1. Heat – Sensible heat vs. Latent heat– Sensible heat vs. Latent heat

Latent heat :

Heat released or absorbed by a substance causes phase change of substance, not changing temperature.

Sensible heat:

Heat released or absorbed by a substance cause temperature change of a substance, not changing phase.

-> Can calculate how much heat is added or released by knowing heat capacity of substance and temperature change (next slide)

Bowen ratio :

Ratio of energy available for sensible heat to energy available for latent heat


1. Heat1. Heat – Heat capacity and Specific heat– Heat capacity and Specific heat

Heat capacity [J·K-1]: The amount of change in thermal energy due to the corresponding temperature change

Specific heat [J·kg-1·K-1]: Heat capacity per unit mass

Specific heat of some substances

Substance Specific heat

Water 4186

Ice 2030

Dry air const pressure (cp) 1004

Dry air const volume (cv) 717

Granite 790

Q. Which substance undergoes the least temperature changes given the same heat

If you know temperature change in a system, can know how much thermal energy increases


1. Heat1. Heat – Latent heat– Latent heat

ice watervapor

qin

melting

qout

fusion

qout

condensation

qin

evaporation

qin

sublimation

qout

deposition

Latent heat [J·kg-1] : the amount of heat released or absorbed by a substance per unit mass due to its phase change

Latent heat (H2O)

Phase change Latent heat [J·kg-1]

melting (fusion) 3.34e+5

evaporation (condensation) 2.50e+6

sublimation (deposition) 2.83e+6


1. Heat1. Heat – Latent heat– Latent heat

Assume that in dry air (mass is mair [kg] ), mlquid [kg] of liquid water is produced due to condensation during Δt [sec].

Computing air temperature increase due to latent heat release

1. The amount of heat released: [J]

liquidv mLq 1

2. This heat raises temperature of dry air. Use dry air heat capacity, cp [J·kg-1·K-1] to convert heat [J] to temperature change [K]:

airp

liquidv

airp mc

mL

mc

qT

1

3. Divided by Δt to find rate of temperature change

tmc

mL

t

T

airp

liquidv


1. Heat1. Heat – Surface heat balance– Surface heat balance

Heat flux (W·m-2) on the surface

• Radiation (FR)

• Heat transfer into ground via conduction (FG)

• Latent heat transfer (FL)

• Sensible heat transfer into air via conduction and convection (FS)

Sum of all the fluxes is zero (+ : flux upward, - flux downward)

FR<0

FS>0 FL>0

FG<0

Daytime, moist surface Nighttime, moist surface

FR>0

FS<0 FL<0

FG>0

Daytime, dry surface

FR<0FS>0

FG<0


2. 12. 1stst law of thermodynamics law of thermodynamics - description- description

Conservation of energy of system : Change in thermal energy of the system (e.g. air parcel) is equal to heat added to the system and work done by external forces.

If heat is added to air parcel, the energy is used following way.

• Air parcel expand (change volume), meaning air parcel does work against pressure force (external force)

• Air parcel gains thermal energy inside the parcel

air parcel

heat


2. 12. 1stst law of thermodynamics law of thermodynamics - equation- equation

Thermodynamic energy equation

Jdt

dp

dt

dTcv

where

cv: specific heat at constant volume = 717 [J·kg-1·K-1]

p: pressure [Pa]

α: specific volume of dry air [m3·kg-1]

Rate of work done by air parcel against pressure force

Heat input into the systemRate of change in thermal energy


2. 12. 1stst law of thermodynamics law of thermodynamics - equation- equation

Entropy form of thermodynamic energy equation

where

cp: specific heat at constant pressure = 1004 [J·kg-1·K-1]

cp = cv + R

R: dry air gas constant = 287 [J·kg-1·K-1]

dt

dTR

dt

dp

dt

dp

Equation of state (Ideal gas law)

Substitute into thermodynamic equation

Jdt

dpR

dt

dTcp

dt

ds

T

J

dt

pdR

dt

Tdcp

lnln

Divided by T and again use equation of state (verify it)


2. 12. 1stst law of thermodynamics law of thermodynamics – potential temperature– potential temperature

pcRs

p

pT

Definition:

Temperature, Θ [K], that a dry air parcel at pressure, p [Pa], and temperature, T [K], would have if it were moved adiabatically to a standard pressure, ps [Pa], typically sea level pressure (~ 1000 [hPa])

0)lnln(lnln pRTcdpdRTdc pp

From entropy form of thermodynamics equationAdiabatic -> J = 0Rewrite in differential form

Integrate from a state (some height) where pressure is p [Pa], and temperature is T [K], to a state of surface where pressure is ps [Pa], and temperature is Θ [K]. (verify it)


2. 12. 1stst law of thermodynamics law of thermodynamics – potential temperature– potential temperature

T

J

dt

dcp

ln

Take logarithm of potential temperature equation and then differentiate with time

(verify it)

• From this equation, Adiabatic process (no heat exchange, J = 0) -> no potential temperature change

•This rule is used to find dry adiabatic lapse rate (next few slides)


2. 12. 1stst law of thermodynamics law of thermodynamics – Lapse rate– Lapse rate

Lapse rate: the rate of air temperature decrease with height

• Environmental lapse rate: actual lapse rate measured by sounding etc.

• Process lapse rate: lapse rate determined by physical processes

-> Dry adiabatic lapse rate: lapse rate determined by moving dry air

parcel without any heat exchange from the environment

-> Moist adiabatic lapse rate: lapse rate determined by moving

saturated air parcel without any heat exchange from the environment,

but latent heat of condensation released


2. 12. 1stst law of thermodynamics law of thermodynamics – Lapse rate– Lapse rate

pc

g

z

T

z

T

dpc

g

z

T

z

T

pc

g

z

T

z

T

Take logarithm of potential temperature equation and differentiate with height, z

And then use hydrostatic equation (dp = - ρgdz) and Ideal gas law (verify)

-> Environmental lapse rate

For adiabatic process, the left side of the equation is zero

-> dry adiabatic lapse rate (~9.8 K/1000m)


Dry adiabatic lapse rate

Dew pointsActual temperature

Moist adiabatic lapse rate


3. Radiation3. Radiation – Electromagnetic energy– Electromagnetic energy

c: speed of light 3e+6 [m·s-1]

λ: Wavelength [m]

ν: frequency [s-1]

c

Electromagnetic wave

c

λ

A

A: amplitude - energy is proportional to A2

T: period [s] –time required to travel by full wavelength, ν = 1/T



1exp

125

2

Tkhc

hcTE

Blackbody: a body that emit the maximum energy it can emit or absorb given by body temperature [K]

Plank’s law: EM energy flux, Eλ(T) [W·m-2·μm-1], emitted from blackbody whose temperature is T [K] as a function of λ [μm]

->

Any objects above zero absolute temperature ( 0K = -273 °C) emit EM energy

whereh: plank’s constant 6.624e-34 [J·s]c: speed of light 3e+6 [m·s-1]k: Boltzmann constant 1.381e-23 [J·K-1]


3. Radiation3. Radiation – Electromagnetic energy– Electromagnetic energyBlackbody radiation curve

10-1

100

101

102

103

10-4

10-2

100

102

104

106

108

Blackbody radiation curve

Wavelength [m]

B(T

) [W

m-2s-1

m-1

]

6000 [K]

500 [K]

250 [K]



Electromagnetic spectral

Peak of solar radiation Peak of earth radiation


3. Radiation3. Radiation – Electromagnetic wave– Electromagnetic wave

4TTE

Stefan-Boltzmann law:

Total energy flux, E(T) [W·m-2], emitted from blackbody whose absolute temperature is T [K]

σ: Stefan-Boltzmann constant (5.67e-8 [W·m-2·K-4])



Q: How does total energy flux change if temperature increases by 1 K for earth (275 [K]) and sun (6000 [K])?



Q: How does total energy flux change if temperature increases by 1 K for earth (275 [K]) and sun (6000 [K])?

Ans:

Write Stefan-Boltzmann equation in derivative form

TTTE d3d 3

For earth, dE = 3 x 5.67e-8 x 275^3 x 1 = 2.22 [W·m-2]

For sun, dE = 3 x 5.67e-8 x 6000^3 x 1 = 2.31e4 [W·m-2]


3. Radiation3. Radiation – Electromagnetic wave– Electromagnetic waveBlackbody emits theoretically maximum EM energy. But Actual emission, I, is less than blackbody emission.

TE

TI

Emissivity ε: a fraction blackbody radiation that is actual emitted depends on wavelength and substances (0 ~1)

TE

TI

Emissivity at a particular wavelength

Emissivity over all range of wavelength

Infrared emissivity of some surface

surface εinfrared

asphalt 0.95

Snow, fresh 0.99

Snow, old 0.82

sand 0.98

Human skin 0.95


3. Radiation3. Radiation – interaction of radiation with materials– interaction of radiation with materials

Absorption (absorptivity):

In vacuum, characteristics of EM wave (energy) is intact.

if EM wave encounter the matter (air, water, other terrestrial materials), it is partitioned into three processes

Reflection (reflectivity):

Transmission (transmissivity):

incident

refrected

E

Er

incident

absorbed

E

Ea

incident

dtransmitte

E

Et

All the coefficients depends on wavelength

Kirchhoffs law: ε λ = aλ

Objects that absorb EM energy of a particular wavelength emit EM energy of the same wavelength in the same efficiency

1 tar


3. Radiation3. Radiation – interaction of radiation with materials– interaction of radiation with materialsAbsorption in atmosphere


Reference TextsReference Texts

A first course in atmospheric Radiation, G.W. Petty

Meteorology today, C.D. Ahrens

Introduction to Dynamic meteorology, J.R. Holton


1. Drive the equation for this using rainfall rate, P [ms-1], liquid water density, ρliquid [kg·m-3], dry air density, ρair [kg·m-3] and h [m].

2. Using this equation, find average latent heat heating for 5 [mm·h-1]. Use the following parameters

h = 11 km, ρair = 0.689 [kg·m-3] , ρliquid = 1025 [kg·m-3], Lv = 2.5e+6 [J·kg-1], and cp = 1004 [J·kg-1·K-1]

Q1 Latent heat

Consider an air column of which height is h [m]. The average of condensation heating rate [K/s] over the air column is estimated by measuring rainfall rate on the ground.

Hint; See slide 9


Q.2. (Lapse rate)

Wind blowing from west to east is pushing air up the mountain. On the windward side of the mountain, air temperature decrease with height at 5 °C per 1000 m. On the leeward side of the mountain, air temperature increase with decreasing height at dry adiabatic lapse rate. Surface air temperature was measured as 15 °C at A. What is the surface temperature at B ??

wind

A, 15 °C

B, ?? °C

1400 m

2000 m

Top: 3000 m


Q.3. (Radiation)

Suppose air polluted air reflects 30% and transmits 50% of the incoming solar radiation (Assume 1368 [W·m-2] ). A) How much is absorbed by this air? B) How much is emitted by this air, assuming Tair = 270 [K] (hint: use Kirchhoffs law to find emissivity of this air)

01/25/20061 cveen7920 hydrometeorology heat and radiation naoki mizukami

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