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By Charles Rhodes, Xylene Power Ltd.
TEMPERATURE OF AN IDEAL ROTATING SPHERICAL BODY:
It is helpful to first investigate the temperature of an ideal rotating nonconducting spherical body with a partially reflecting surface. Assume that the reflectance and emissivity of the surface are frequency independent. Then from the Radiation Physics section the theoretical surface temperature Te is given by:
Te = (Ho dAc / Cb dAs)^.25
where:
dAc = an element of cross sectional area absorbing energy from the sun
and
dAs = the corresponding element of radiating surface area.
EARTH MODEL SHOWING DEPENDENCE OF Te ON LATITUDE AT AN EQUINOX:
At an equinox the geometry of the spherical Earth is sufficiently simple to permit easy calculation of the dependence of Te on latitude.
Let R = the radius of the 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:
dAs = 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:
dAc = 2 R cos(L) R dL cos(L)
Hence:
dAc / dAs = [2 R cos(L) R dL cos(L)] / [2 Pi R cos(L) R dL]
or
dAc / dAs = [cos(L) / Pi]
At an equinox numerical evaluation of (dAc / dAs) for a spherical Earth gives:
dAc / dAs = 1 / Pi = .31831 at the equator (latitude L = 0 degrees = 0 radians)
dAc / dAs = 3^0.5 / 2 Pi = .27566468 at latitude L = 30 degrees = (Pi / 6) radians
dAc / dAs = 1 / (2^0.5 Pi) = .225079 at latitude L = 45 degrees = (Pi / 4) radians
dAc / dAs = 1 / 2 Pi = .159155 at latitude L = 60 degrees = (2 Pi / 6) radians
dAc / dAs = (.7233702 / Pi) = .230256 at latitude L = 43 40' degrees
dAc / dAs = (.7087508 / Pi) = .225603 at latitude L = 44 52' degrees
Recall that:
Te = [(Ho dAc) / (Cb dAs)]^.25
= [(1367 / 5.6697)(dAc / dAs)]^.25 X 100 K
= 394.0507 X (dAc / dAs)^.25 K
Hence, at an equinox numerical evaluation of Te gives:
Te = 295.98 K at latitude = 0 degrees = 0 radians (the equator)
Te = 285.53 K at latitude = 30 degrees = (Pi / 6) radians
Te = 271.42 K at latitude = 45 degrees = (Pi / 4) radians
Te = 248.89 K at latitude = 60 degrees = (2 Pi / 6) radians
Te = 272.96 K at 43 40' N latitude (Toronto Airport)
Te = 271.57 K at 44 52'N latitude (Halifax Airport)
EFFECTIVE AVERAGE SURFACE TEMPERATURE:
At an equinox the effective average surface temperature Tea for this ideal rotating spherical body is given by:
Tea = [Ho Ac / Cb As]^.25
= (Ho Pi R^2 / Cb 4 Pi R^2)^.25
= [Ho / (4 Cb)]^.25
= [(1367 W / m^2) / (4 X 5.6697 X 10^-8 W / m^2-K^4)]^.25
= 278.6359256 degrees K
This web page last updated October 8, 2008
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