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Unformatted text preview: PROBLEM :A SATELLITE IN SUNSHINE In this problem you will calculate the temperature of a
space satellite. The satellite is assumed to be a Sphere with
a diameter of 1 m. All of the satellite remains at a uniform
temperature. All of the spherical surface of the satellite is
coated with the same kind of coating. The satellite is located
near the earth but is not in the earth's shadow. The surface temperature of the sun (its blackbody
temperature) '1'“In  6000 K and its radius is 6.96 x 103 m. The
distance between the sun and the earth is 1.5 x 10H m. The
sunlight heats the satellite to .a temperature at which the
blackbody emission from the satellite equals the power absorbed
from the sunlight. The p0wer per unit area emitted by a black
body is given by StefanBoltzmann's law P = aT‘ where a is the
universal constant 5.67 x 10“ wqu“. In the first
approximation, you can assume that both the sun and the satellite
absorb all electromagnetic radiation incident upon them. 1) Find an expression for the temperature T of the satellite.
what is the numerical value of this temperature? 2) The blackbody radiation spectrum u(T,f) of a body at
temperature T obeys Planck's radiation law ‘
11(T,f)df==_§££Ll: lﬁiﬂl
c3h3 e"1 where n = hf/kT and u(T,f)df is the energy density of the
electromagnetic radiation in a frequency interval [f, f + df].
In the equation h = 6.6 x 104‘ Js is Planck's constant, k = 1.4
x 1043 JKd is Boltzmann's constant, and c = 3.0 x 10‘ ms4 is
the speed of light. The blackbody spectrum, integrated over all frequencies f and
directions of emission, gives the total radiated power per unit
area P = oT‘ as expressed in the StefanBoltzmann law given above.
Znsk‘
1502133
The figure shows the normalized spectrum
c3113 u(T.f)
Bnk‘ T‘
as a function of n. a: In many applications it is necessary to keep the satellite as
cool as possible. To cool the satellite, engineers use a
reflective coating that reflects light above a cutoff frequency
but does not prevent heat radiation at lower frequency from
escaping. Assume that this (sharp) cutoff frequency corresponds
to hf/k  1200 K. xxm IPHO JULY92 What is the new equilibrium temperature of the satellite? The
exact answer is not needed. Therefore, do not perform any
tedious integrations; make approximations where necessary. The
integral over the entire range is 00 3 J 17 dn 7r4 0 Cn—lui—g and the maximum of nJ/(en  1) occurs at n = 2.82. For small n
you can expand the exponential function as e1 = 1 + n. 3) If we now have a real satellite, with extending solar panels
that generate electricity, the dissipated heat in the electronics
inside the satellite acts as an extra source of heat. Assuming
that the power of the internal heat source is 1 kw, what is the
equilibrium temperature of the satellite in case 2 above? 4) A manufacturer advertises a special paint in the following
way: "This paint will reflect more than 90% of all incoming radiation
(both visible light and infrared) but it will radiate at all
frequencies (visible light and infrared) as a black body, thus
removing lots of heat from the satellite. Thus the paint will
help keep the satellite as cool as possible." Can such paint exist? Why or why not?
5) What properties should a coating have in order to raise the temperature of a spherical body similar to that of the satellite
considered here above the temperature calculated in 1? 1.6
c3h3 u(T, f)
8114s:4 T4
1.2
0.8
0.4 0 2 4 6 8 10 77 = hf/kT ESPOO FINLAND SOLUTION : PROBLEM 1. Over the whole surface of the sun, the emitted energy is 4112122  a'Tgun. All this Sun
energy passes through a. spherical shell at earth’s distance R, where the intensity now is 4‘IrR2  organ/4sz. Sun
The satellite is a. circular object absorbing 2 4:7er2  0T;1m Sun m 
"1 ll‘rrR2 but a spherical object emitting 4min  0T4 aat'
Equating this absorption and emission We get Tut : Tguﬂ/ﬂfﬁu giving Tm, =
289 K = 16 °C. . We have to calculate what part of the absorbed power comes from the part of the
spectrum below 1200 K. _ 1200 K
“mm” “ 6000 K =0.2<<1 This fraction of power is Now, the satellite is cold with respect to 1200 K so we ignore that a small part of the
satellite blackbody emission will be retained. The energy balance is now 2 47rR2  UT;1m Sun sat ' 47rR2 2 4 _
47rrwt  a'T — 6  11'7' sat
by which the new satellite temperature is the previous corrected by a factor 61/4 TW = (4.1 10—4)1/4 289 K = 41 K . The whole absorbed energy is 2 47TR2  aTgun Sun . = . W
sat 41rR2 67rr which is small (ignorable) compared to Pinter“; = 1 kW. Thus the energy balance
becomes 2 4
Pinternal : 47rrgat I aTaat xxm IPHO JULY92 PROBLEM : A SATELLITE IN SUNSHINE giving T,“ = 274 K (77 = 4.38). Note: strictly speaking, this is not accurate, because
for a blackbody radiation of 274 K, some 33 % of the emitted power lies above the
1200 K cutoff ! This means that the satellite has to be hotter, to emit all of the 1 kW in frequencies below the cutoff. The resulting integral equation is 1? )4=/” 77de ii
4.38 e’T—l 15 which can be solved numerically by iteration. The true solution is r] = 3.80 corres
ponding to a temperature of 316 K. 4. The paint cannot exist, because it would violate the second law of thermodynamics.
The physics textbook explanation is the principle of detailed balance, which means
that for equilibrium to exist, the emission and absorption in a given frequency interval
must match exactly. This should not be confused with the fact that reﬂection and absorption can be quite different. If the manufacturer’s paint existed, one could create
a temperature difference between two bodies in a closed system, and hence a perpetum mobile. 5. The coating should be transparent for high frequencies (in the range of the peak or
tail of the sun’s radiation), but reﬂective and hence insulating at low frequencies (the satellite temperature). ESPOO FINLAND ...
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 Spring '11
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