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XYLENE POWER LTD.

By Charles Rhodes, Xylene Power Ltd.

**TEMPERATURE OF AN IDEAL ROTATING SPHERICAL EARTH:**

It is helpful to investigate the temperature of an ideal isothermal and rotating spherical Earth with a partially reflecting surface. Assume that the solar reflectance and infrared emissivity emissivity of the surface are unrelated but both are nearly frequency independent. Then from the Radiation Physics section the theoretical average surface temperature Ta is given by:

**Ta = [Ho Ac (1 - Fr) / Ft Cb As]^.25**

where:

Ac = cross sectional area absorbing energy from the sun

and

As = the corresponding radiating surface area.

**NUMERICAL SOLUTION:**

Assume:

Ho = 1367 W / m^2

(Ac / As) = (1 / 4)

Fr = 0.297

Cb = 5.6697 X 10^-8 W m^-2 K^-4

Ft = 1.000

Then:

**Ta** = [Ho (1 - Fr) Ac / As Ft Cb]^.25

= [1367 W / m^2 (1 - 0.297) / (4 (1) (5.6697 X 10^-8 W m^-2 K^-4)]^0.25

= **255.14 K**

In reality due to the presence of Green House Gases in Earth's atmosphere Ft < 1. Experimental measurement of Ta in 1996 indicated that:

**Ta = 270 degrees K.**

This measurement indicates that:

**Ft** = (255.14 / 270)^4

= **0.79735**

**EARTH MODEL SHOWING DEPENDENCE OF Ta ON LATITUDE AT AN EQUINOX:**

At an equinox the geometry of a spherical Earth is sufficiently simple to permit easy calculation of the dependence of Ta on latitude.

Let R = the radius of Earth

Let L = the angle of latitude (L= 0 at the equator, L = 90 degrees = (Pi / 2) radians at the north pole)

Consider a narrow latitude band around the earth's axis that is tangent to the earth's surface.

The width of the band is (R dL)

The radius of the band is R cos(L)

The circumference of the band is 2 Pi R cos(L)

The surface area of the band is given by:

As = 2 Pi R cos(L) R dL

At an equinox the cross-sectional area of the latitude band exposed to the sun at any instant in time is given by:

Ac = 2 R cos(L) R dL cos(L)

Hence:

Ac / As = [2 R cos(L) R dL cos(L)] / [2 Pi R cos(L) R dL]

or

**Ac / As = [cos(L) / Pi]**

At an equinox numerical evaluation of (Ac / As) for a spherical Earth gives:

Ac / As = 1 / Pi = .31831 at the equator (latitude L = 0 degrees = 0 radians)

Ac / As = 3^0.5 / 2 Pi = .27566468 at latitude L = 30 degrees = (Pi / 6) radians

Ac / As = 1 / (2^0.5 Pi) = .225079 at latitude L = 45 degrees = (Pi / 4) radians

Ac / As = 1 / 2 Pi = .159155 at latitude L = 60 degrees = (2 Pi / 6) radians

Recall that:

Ta = [(Ho Ac (1 - Fr)) / (Cb As Ft)]^.25

= [(1367 (1 - Fr) / 5.6697 Ft)(Ac / As)]^.25 X 100 K

= 381.837 X (Ac / As)^.25 K

Hence, at an equinox numerical evaluation of Ta gives:

Ta = 381.837 K X (.31831)^0.25 = 286.80 K at latitude = 0 degrees = 0 radians (the equator)

Ta = 381.837 K X (.27566)^0.25 = 276.677 K at latitude = 30 degrees = (Pi / 6) radians

Ta = 381.837 K X (.22508)^0.25 = 263.003 K at latitude = 45 degrees = (Pi / 4) radians

Ta = 381.837 K X (.15916)^0.25 = 241.177 K at latitude = 60 degrees = (2 Pi / 6) radians

**Note that at latitudes < ~ 35 degrees Ta > 273.15 K and that at latitudes > ~ 35 degrees Ta < 273.15 K.**

This issue has importance in the study of thermal runaway.

This web page last updated November 6, 2016

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