Greenhouse effect Masatsugu Sei Suzuki Department of...

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Greenhouse effect Masatsugu Sei Suzuki Department of Physics (Date: September 21, 2018) 1. Solar constant ( 0 p ) The solar constant is a flux density measuring mean solar electromagnetic radiation (solar irradiance) per unit area. It is measured on a surface perpendicular to the rays, one astronomical unit (AU) from the Sun (roughly the distance from the Sun to the Earth). It is measured by satellite as being 1361 W/m² at solar minimum and approximately 0.1% greater (roughly 1362 W/m²) at solar maximum. Solar constant is the energy flux from the Sun to the Earth, energy per unit area per unit time. The luminosity of the Sun (which is the total energy per unit time from the Sun) is given by 6 4 2 0 2 2 3.828 10 4 4 sun sun B sun L R T dp W. So we have 2 4 0 2 4 sun B sun R T p d =1361.17 W/m 2 The luminosity of the Earth is 2 4 4 earth E B E L R T which is equal to 2 0 E R p ; 2 2 4 0 4 earth E E B E L R p R T or 4 0 0 4 B E p J T The real temperature of the Earth is determined from the relation 2 2 4 0 0 (1 )( ) 4 E E B r R p R T

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Greenhouse effect

Masatsugu Sei Suzuki

Department of Physics

(Date: September 21, 2018)

1. Solar constant ( 0p )

The solar constant is a flux density measuring mean solar electromagnetic radiation (solar

irradiance) per unit area. It is measured on a surface perpendicular to the rays, one astronomical

unit (AU) from the Sun (roughly the distance from the Sun to the Earth). It is measured by satellite

as being 1361 W/m² at solar minimum and approximately 0.1% greater (roughly 1362 W/m²) at

solar maximum.

Solar constant is the energy flux from the Sun to the Earth, energy per unit area per unit time.

The luminosity of the Sun (which is the total energy per unit time from the Sun) is given by

64 2

0

22 3.828 104 4sun sun B sunL R T d p W.

So we have

2 4

0 2

4 sun B sunR Tp

d

=1361.17 W/m2

The luminosity of the Earth is

2 44earth E B EL R T

which is equal to 2

0ER p ;

2 2 4

0 4earth E E B EL R p R T

or

400

4B E

pJ T

The real temperature of the Earth is determined from the relation

2 2 4

0 0(1 )( ) 4E E Br R p R T

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or

400(1 )

4B

pr T

where r is the reflection constant. When r = 0.3, we have 0 255T K.

The solar constant is not really constant. There is a small variation, typically less than 1W/m2,

due to the change of the solar output over an 11-year solar cycle, as well as variations on

cosmological time scales. The solar constant is defined as the amount of solar radiation received

outside the Earth’s atmosphere at the Earth’s mean distance from the Sun. The Earth’s orbit is

slightly eccentric, which leads to a variation of solar irradiance from 1412W/m2 in early January

to 1321W/ m2 in early July.

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2. Example:

6-12

Ralph Baierlein; Thermal Physics

(a)

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4 4( )u SB u lJ T T

(b)

4 4

1 ( )SB u mJ T T , 4 4

2 ( )SB m lJ T T

Tu

Tl

J

SBTu4

SBTl4

Tu

Tl

Tm

J1

J2

SBTm4

SBTl4

SBTu4

SBTm4

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The third black plane is allowed to come to a steady state temperature,

1 2J J

or

4 4 4 4( ) ( )SB u m SB m lT T T T

or

1/4

4 4

2

u lm

T TT

Thus we have

4 4

1 2

1( )

2SB u lJ J T T

(c)

4 4

1 1( )SB uJ T T ,

4 4

2 1 2( )SBJ T T

4 4

3 2 3( )SBJ T T

……………………………

4 4

1( )N SB N NJ T T

4 4

1 ( )N SB N lJ T T

Note that 1 2 1... NJ J J J for the steady flow of energy. Addition of both sides of these

equations leads to

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4 41( )

1SB u lJ T T

N

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3. Example-II

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6-13 Ralph Baierlein

See the solution of Homework.

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4. Radiative transfer

A vacuum or Dewar flask (sometimes called thermo flask) consists of a cylindrical container

surrounded by a thin evacuated jacket. The entire surface of the evacuated jacket with a material

such as silver, which is opaque and has a high reflectivity. The reflectivity is r, and the

temperatures inside and outside the flask are Ta and Tb, respectively. What is the energy flux (the

rate of radiative energy transfer per unit area)?

Since the separation between the walls is much less than the diameter of the cylinder, they can

be treated as plane parallel surfaces. By the Kirchhoff and Stefan-Boltzmann law, the inner surface

radiates at a rate 4(1 ) B ar T per square meter, and the outer one at 4(1 ) B br T . Each surface

reflects a fraction r of the radiation falling on it. Let the total ingoing flux be J and the total

outgoing flux be 'J . Then

4' (1 ) B aJ rJ r T , 4' (1 ) B bJ rJ r T

The net ingoing flux is thus

4 41' ( )

1B a b

rJ J T T

r

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If r = 0.98, the net flux is reduced to about 1% of the value it would have if the surfaces were black

bodies.

REFERENCES

M.D. Sturge, Statistical and Thermal Physics: Fundamentals and Applications (A K Peters, 2003).

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5. Greenhouse effect

Schroeder: Thermal Physicsp.307

The equilibrium temperature of 255 K applies (roughly) to the atmosphere, while the surface

below is heated both by the incoming sunlight and by the atmospheric “blanket.” If we model the

atmosphere as a single layer that is transparent to visible light but opaque to infrared, we get the

situation in Fig. as shown below. Equilibrium requires that the energy of the incident sunlight

(minus what is reflected) be equal to the energy emitted upward by the atmosphere, which in turn

is equal to the energy radiated downward by the atmosphere. Therefore, the earth’s surface receives

twice as much energy (in the simplified model) as it would from sunlight alone. This mechanism

raises the surface temperature by a factor of 1/42 , to 303 K. This is a bit high, but then, the

atmosphere is not just a single perfectly oparaque layer. By the way, this mechanism is called the

greenhouse effect, even though most greenhouses depend primarily on a different mechanism

(namely, limiting convective cooling).

((Problem))

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Suppose that the concentration of infrared-absorbing gases in earth’s atmosphere were to double,

effectively creating a second “blanket” to warm the surface. Estimate the equilibrium surface

temperature of the earth that would result from this catastrophe.

((Energy conservation))

On the surface of the Earth:

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4 4

0 0 1 0B B EJ rJ T T (1)

At the atmosphere (1)

4 4

12 B B ET T (3)

At the top of the atmosphere (1)

4

0 0 1BJ rJ T (4)

From these equations, we get

1/4 1/4

1 0 0

(1 ) (1 )[ ] [ ] 255

4B B

r rT J p

K

1/4

1 12 1.1892 303.2ET T T K

6. Greenhouse effect (II)

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((Energy conservation))

On the surface of the Earth:

4 4

0 0 2 0B B EJ rJ T T (1)

At the atmosphere (2):

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4 4

22 B B ET T (2)

At the atmosphere (1)

4 4

1 22 B BT T (3)

At the top of the atmosphere (2)

4

0 0 1BJ rJ T (4)

From these equations, we get

1/4

1 0

(1 )[ ] 255

4 B

rT p

K

1/4

1 13 1.31607ET T T = 335 K,

1/4

1 1

31.10668

2ET T T

= 282.2 K

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REFERENCES

M.H.P. Ambaum, Thermal Physics of the Atmosphere (Wiley-Blackwell, 2010).

M.D. Sturge, Statistical and Thermal Physics: Fundamentals and Applications (A K Peters, 2003).

APPENDIX Temperature of the Earth

The radius of the Earth:

66.372 10EarthR m

The radius of the Sun:

86.9599 10SunR m

The distance between the Sun and the Earth:

d = AU=1.49597870 x 1011 m

The Stefan Boltzmann constant

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85,670400 10B W/m2

The temperature of the Earth

4 2 4 2( )( ) ( )(4 )B sun Sun B E ET R d T R

where the solid angle is

2 2

Ed R

Thus we have

2

4 2 4 2

2( )( ) ( )(4 )E

B sun Sun B E E

RT R T R

d

or

2

4 4

2( )

4

SunE sun

RT T

d

or

2

SunE sun

RT T

d = 282.15 K.