LUNAR SATELLITE (BOX)
The following simplified thermal model is to be analysed for the thermal control of a small spacecraft
orbiting the Moon in a circular orbit at 300 km altitude in the Moon’s orbital plane.
The satellite is 3-axes stabilised, with a cubic main body of 0.5 m side, with four protruding flaps
protecting from sunlit the face pointing to the Moon (Fig. 1). All walls are made of aluminium 1 mm
thick, and the three faces exposed to sunlit are covered with thin solar cells with an effective area of 90%
and an electrical efficiency of 20%. Inside the box, there is a 2 mm thick aluminium plate at mid height,
with an electronics box of 40·40·20 cm3 and 10 kg at each side, centred, with an averaged thermal
capacity of 1000 J/(kg·K), holding batteries and a control system to deliver a constant electrical power all
the time.
For the thermal model, the following nodes are to be considered: one at each of the six faces of the main
box (1,2,3,4,5,6), one at each flap (7,8,9,10), and another one for the internal assembly of e-boxes and
instrument support plate (11). Thermo-optical properties of surfaces should be adequately selected.
To do:
a) Find the relative eclipse duration, and the minimum extent of the flaps to guarantee that the Sun
b)
c)
d)
e)
f)
g)
rays never fall on the face pointing to the Moon.
Find the external heat loads (solar, albedo and infrared) as a function of orbit position.
Find the thermal conductance and radiative couplings between nodes.
Establish the node equations.
Find the steady temperatures at the sub-solar and at the opposition points.
Find the orbital mean temperatures at the nodes.
Find the temperature evolution along the orbit, in the periodic state.
Fig. 1. Lunar satellite geometry.
Lunar satellite (box)
1
Solution
We start by making a compilation of relevant data for the lunar orbit (do not mix up lunar orbit of a
spacecraft around the Moon, with Moon’s orbit around the Earth). Most values (with the main exception
of thermo-optical properties) are accurate to three significant figures (we try to carry out the 3-figures
accuracy down to the results: node temperatures, which have lower uncertainties than thermo-optical
properties.
Parameter
Sun-to-Moon
distance
Moon-to-Earth
distance
Moon radius
Moon orbit period
Moon orbit
inclination
Moon axis tilt
Moon surface
temperature
Moon emissivity
Solar irradiance
Albedo
Solar to orbit
vector angle
Relative altitude
Sat orbit period
Eclipse fraction
Eclipse start angle
Table 1. Data for the lunar orbit and Moon’s orbit.
Symbol and value
Comments
9
RSM=150·10 m (1 AU) Sat to Moon and Moon to Earth distances are much smaller.
RME=385·106 m
(60 Earth diameters). Perigee 363·106 m, apogee 406·106 m.
RM=1737 km
TM=27.3 d
i=5.1º to ecliptic
(0.273 RE). Equatorial 1738 km, Polar 1736 km. g=1.62 m/s2.
Synodic period (from full-moon to full-moon) is TM=29.5 d.
Between 18º and 29º to Earth’s equator.
6.7º to Moon’s orbit plane.
Tmax=390 K at subsolar point (equatorial noon), falling as
(cos)1/4 towards the terminator, with quasi-uniform Tdark=100
K on the whole dark hemisphere, Tmin=26 K at one polar
crater (the minimum in the solar system), and Tmean=274 K.
M=0.94
From energy balance (1−ρ)EπR2=4πR2εσT4. M=Tp4=300
W/m2.
2
E=1360 W/m
With <2% annual oscillation due to Earth orbit eccentricity.
Bolometric; normal albedo is 0.07.
=0.12
Can be taken as an ecliptic orbit since sat orbit plane is in
=/2
Moon’s orbit plane, with an inclination of 5.1º to ecliptic.
h=H/R=300/1737=0.173 The Moon has no atmosphere (the Lunar Reconnaissance
Orbiter altitude is 30..70 km)
T=8470 s (2.4 h)
T=2[(Rp+H)3/(GMp)]1/2, Rp=RM, G=6.674·10--11 N·m2/kg2,
Mp=73.5·1021 kg.
Te/T=0.325
Te/T=(1/)arcsin(1/(1+h)), Te=46 min.
e=2.12 rad (121.5º).
e=arcsin(1/(1+h)).
iM=1.54º to ecliptic
TM=274 K
The Moon differs from most satellites of other planets in that its orbit is close to the plane of the ecliptic,
and not to its planet’s equatorial plane (Fig. 2). The Moon crosses the ecliptic about twice per month. If
this happens during new moon a solar eclipse occurs, and a lunar eclipse if during full moon. This was the
way the ancients could trace the ecliptic along the sky; they marked the places where eclipses could
occur.
Lunar satellite (box)
2
Fig. 2. a) Frontal sketch view along the ecliptic plane of Moon’s orbit, not to scale (Earth-Moon
barycentre is at 4641 km from Earth’s centre). b) Earth–Moon system to scale. Wiki.
We have chosen for the origin of angular positions along the orbit, , the sub-solar point (see Fig. 3,
which is an upside-down view; compare it with the frontal view in Figs. 1 and 2).
Fig. 3 . Orbit view from the North Pole, to show the angular position origin along the orbit, and the
eclipse period (from =e=121.5º, to =238.5º).
We can compute and compile relevant physical parameters (geometrical and thermal) for the spacecraft.
Parameter
Main cube side
Extra panel length
Aluminium plate thickness
Cube face area (each)
Extra panel area (each)
Solar panel effective area
Solar panel efficiency
Solar panel absorptance
Solar panel emissivity
Instr. plate thickness
E-box size (each)
E-box mass (each)
E-box cp
Aluminium density
Lunar satellite (box)
Table 2. Data for the spacecraft.
Symbol and value
Comments
L=0.50 m
One axis pointing nadir (i.e. to Moon’s centre),
another to ecliptic north (appr.), and the third
one along the path direction.
Lp=0.31 m
Lp/L=tan(e)=0.613 (see Fig. 3).
=0.001 m
A=0.25 m2
A=L2.
2
Ap=0.155 m
Ap=LLp=0.5·0.31=0.155 m2.
90%
20%
=0.7
=0.7
A=0.002 m
0.4·0.4·0.2 m3
mEB=10 kg
cp=1000 J/(kg·K)
=2700 kg/m3
3
Aluminium thermal conduct.
Aluminium thermal capacity
k=200 W/(m·K)
c=900 J/(kg·K)
Notice that we neglect the mass and consequently thermal inertia of solar cells (thin technology) and
other possible coatings (e.g. MLI).
a)
Find the relative eclipse duration, and the minimum extent of the flaps to guarantee that the Sun
rays never fall on the face pointing to the Moon.
Relative to the nadir-pointing face, the Sun is going around the full 360º circumference with centre at the
face, so that the field of view cannot include any background space but the Moon; thence, the
unobstructed viewing angle from the edge of face 2, arctan(L/Lp), must equal the limb-to-centre angle of
the Moon from the satellite, arcsin(1/(1+h))=1.02 rad (58.5º), and therefore L/Lp=tan(1.02)=1.63 and
Lp=0.5/1.63=0.306 m (see Fig. 3).
On the other hand, we know that, unlike solar eclipses on Earth, which can only last a few minutes (a total
solar eclipse can last a maximum of 7 minutes, 31 seconds, moving eastward at 470 m/s along a track that
is up to 250 km wide), lunar eclipses may last several hours, with totality itself usually averaging
anywhere in between 30..90 minutes (the rest being partial eclipse and penumbral eclipse, Fig. 4). With
the parallel ray approximation, Te/T=(1/)arcsin(1/(1+h)), which, with h=RME/RE=385/6.37=60 and T=28
d gives Te=3.5 h. But with finite Sun distances and spherical bodies, when the occluding object is smaller
than the star, the length of the umbra's cone-shaped shadow, LU, is also finite and given by
LU=RSERE/(RSRE)=(1.5·109·6.37·106/(0.7·1096.37·106)=1.4·109 m, so that at Moon’s distance the
radius of the cone section has dropped to RU=RE(1RME/LU)=6.37·106(1385·106/(1.4·109))=4.6·106 m.
The Sun-Earth Lagrangian rear point, L2, where Planck and Herschel space observatories are located (and
James Webb Space Telescope will go too), is slightly beyond the reach of Earth's umbra (1.5·10 9 m
against 1.4·109 m), so that solar radiation is not completely blocked (besides, spacecraft usually follow
large-departure Lissajous orbits around the Lagrangian points), so that solar panels are used to power
them.
Fig. 4 . Umbra, penumbra and antumbra cast by Earth on the Moon, and solar irradiation fraction.
There are zero to three partial or total lunar eclipses per year (although possibly not all visible from the
same location on Earth), and not one per month, because Moon's orbit is tilted 5 degrees from Earth's
orbit. During lunar totality, the colour of the Moon takes on a dark red hue due to light scattering at the
Earth’s atmosphere. In 2010 there were only one partial lunar eclipse (on 25-Jun, of 2 h 43 m total
Lunar satellite (box)
4
duration), and one total lunar eclipse, on 21-Dec, with 3 h 29 m total duration (1 h 13 m total eclipse);
(http://eclipse.gsfc.nasa.gov/LEdecade/LEdecade2001.html).
We are not taking account of Moon eclipses on the thermal design of our lunar satellite because they are
rare events that we assume to occur outside the intended spacecraft life (less than one year for this kind of
small lunar satellites).
b)
Find the external heat loads (solar, albedo and infrared) as a function of orbit position.
The objective is to solve the energy balance of every part of the spacecraft, here performed by a finite
difference scheme of non-regular isothermal elements (nodes), to find the temperature evolution at each
point in the satellite along the orbit.
If we start with the isothermal-satellite model, i.e. one single node representative of the whole spacecraft,
Tmean(t), the energy balance per unit time (for a constant mass) can be set as:
mcmean
dTmean dEele

 Wnet  Qnet  Qs,in  Qa,in  Qp,in  Qout
dt
dt
(1)
where m is the total spacecraft (SC) mass, cmean and Tmean are the averaged specific thermal capacity and
temperature of the SC at time t, Eele is the non-thermal energy store (we only consider electrochemical
batteries), Wnet is the net electrical and electromagnetic work rate received (no mechanical work
considered), and Qnet the net heat input rate. We recall that solar radiation is thermal radiation (only
dependent on Sun temperature), but its temperature is so high (5800 K) that it can be taken as work input
(i.e. pure exergy input), as we do here; thus, Qs,in is the total solar energy absorbed, including the part that
generates electricity in the PV panels, but not including the reflected part; and similarly for the planet
albedo input (this does not apply to planet emission input and own radiation output, both in the IR). In
this global analysis (whole SC) electrical dissipation does not appear (it is internal, does not go through
the envelop), and radioelectric exchanges for communication are neglected in the energy balance.
The total solar input power, Qs,in , is:
Qs,in  t   mean EAfrontal-lit  t 
(2)
depending on time due to changes in lit frontal projected area (including eclipse shadowing; solar
irradiance can be considered constant for a satellite, but not so for a deep probe).
Power input on the SC from albedo is modelled as follows:
Qa,in  t   mean Ap Fp,SC t  p EFa t   mean ASC FSC,p t  p EFa t 
(3)
i.e., assuming a planetary reflectance p to solar irradiance E, the maximum power reflected per unit area
of the planet (in this problem the Moon) is pE, and the average from a point of view moving along the
Lunar satellite (box)
5
orbit can be taken as pE,Fa(t), where the albedo factor usually approximated by a modified parabolic law
from the subsolar point to the entrance into eclipse like Fa()=[1(/e)2](1+cos)2/4. Thence, the fraction
of that solar reflected power that impinges on the SC is pE,Fa(t) multiplied by the area of the planet, the
view factor from the planet to the SC, and the SC average solar absorptance, meanApFp,SC, finally
applying the reciprocity relation, ApFp,SC=ASCFSC,p.
Power input on the SC from planet own emission (subindex ‘p’), assuming an average planet temperature
Tp, is modelled as follows:
Qp,in  t   IR,mean Ap Fp,SC t   p Tp4   mean ASC FSC,p t   p Tp4
(4)
where, as before, the fraction of total power emitted by the planet ( pTp4Ap) that impinges on the SC, is
pTp4ApFp,SC(t), and the reciprocity relation is applied again. Notice that we are taking an averaged value
for Moon surface temperature (274 K), but there is a great contrast with non-negligible thermal effects on
the satellite since, from the 2.4 h of orbit period, 1.2 h is flying over the lit face at some 350 K, and
another 1.2 h is flying over the unlit face at 100 K.
Finally, the power emitted by the SC itself is:
4
Qout  t   ASC mean Tmean
t 
(5)
where ASC is the envelop area of the spacecraft and Tmean its average temperature. Notice that here the
whole emission of the SC is considered, not its heat exchange with the environment, where the presence
of a close planet would substantially change the thermal radiation exchanged. Notice also that the effect
of the microwave background radiation at 2.7 K has been neglected. Substituting all this developments in
the energy balance above, provides an ordinary differential equation in Tmean(t), which can be solved if an
initial condition is known (for a periodic state, any input is valid because its effects disappear after some
periods of simulation.
When more than one node is chosen, one has to account for heat and work exchanges with the other
nodes, and the energy balance (1) takes the form:
mi ci


dTi dEele,i

 Wele,in,i  Wele,out,i  Qs,in,i  Qa,in,i  Qp,in,i   Qcond,j ,i   Qconv,j ,i   Qrad,j ,i   Qout,i
dt
dt
j
j
 j

(6)
where Eele is the non-thermal energy store (only for batteries), Wele,in,i  Wele,out,i  Wele,dis,i is the electrical
dissipation in the node (all active nodes have electrical input, but only two have electrical output: solar
panels and batteries. Notice that Qs,in,i in (6) is the total solar absorptance, including the part used to
generate Wele,out,i in the solar cells; and the same for the albedo contribution; however, it is usual to
decouple the whole energy balance in an electrical part and a thermal part, and then Qs,in,i may refer just
to the latter, as done below.
Lunar satellite (box)
6
With the node notation shown in Fig. 5, face 2 is permanently facing the Moon and under shadow from
the Sun. Faces 5 (North) and 6 (South) are assumed to be parallel to Sun rays (so no solar lit) on the
approximation of ecliptic orbit (really between 5.1º and +5.1º). The relation between orbit angle and
time (from subsolar point) is:
t


,
T 2
with T  8470 s
(7)
Fig. 5. Node notation.
Node 1. It gets direct solar radiation at its external face when looking at the Sun, and never gets albedo
nor infrared from the Moon. We decouple the general energy balance (6) into:
 Electrical energy balance:
0  Wele,out,1  Fp EAfrontal,1

(8)
Thermal energy balance (we only detail solar input):
m1c1


dT1
 1  Fp  EA1 cos   Qa,in,1  Qp,in,1   Qcond,j ,1   Qconv,j ,1   Qrad,j ,1   Qout,1
dt
j
j
 j

(9)
with the following solar, albedo, and planet inputs (only the thermal part):
Qs ,1  1 EA1 cos 



Qs ,1,th  1   Fp  EA1 cos  


 if    
2
2
Ws ,1,out   Fp EA1 cos 



( 0 otherwise)
(10)
Qa,1  0
(11)
Qp,1  0
(12)
with peak values of Qs,1,th =(0.70.2·0.9)·1360·0.25=177 W and Ws ,1,out =62 W, and where the electrical
efficiency of solar cells, defined by (VI)max/(EA), and the packaging factor for the cells, Fp (cell area
Lunar satellite (box)
7
divided by panel area A) have been introduced. The orbit averaged external heat input and electrical input
are:


1  Fp  EA  57 W
Qs ,1,th,mean 
2
 EA 
1

(13)
Qs ,1,mean 
1 EA cos  d  1 

2 
 

F
EA
0.2

0.9

1370

0.25
p


 19.6 W
2
W1,out,mean  

Node 2. Always pointing nadir, it gets no direct solar radiation, but gets albedo when not under eclipse,
and infrared from the Moon all along the orbit. Notice that from the centre of face 2 one can only see a
large ‘square’ of the Moon surface (no background), but from the centre of each of its four borders, the
opposite Moon limb is just in the limit of the field of view.
Qs,2  0
Qa ,2   2 AM FM,2 M aM
(14)
 1  cos  
  2 A2 F2,M M a   2 A2 F2,M  M E 

2 

2
   2 
1    
  e  
Qp,2   IR,2 AM FM,2 M M   2 A2 F2,M M TM4
(15)
(16)
where a modified parabolic dependence with orbit angle of albedo input is assumed (from maximum at
subsolar point to 0 at eclipse).
The view factor from face 2 to the Moon, approximated as the view factor from face 2 to a virtual closing
panel at the rim of the protruding panels can be found from Table of view factors: vie factor between two
identical parallel square plates of side L and separation H, with a=L/H:

1 
x4
a
2
F12  2  ln

4
ay
 , with x  1  a and y  x arctan  arctan a
2
 a  1  2a
x

(17)
which, for a=0.5/0.31=1.6, yields x=1.9, y=0.33, and F2M=0.35. The view factor from face 2 to the
protruding panels, due to symmetry, are all equal: F2i=(1F2M)/4=0.16. Notice the shielding effect of the
shroud of panels; without it, the view factor from 2 to the planet would be more than double,
F12=1/h2=0.72.
The orbit peak input values in (15) and (16) are Qa ,2,m =14 W and Qp,2,m =26 W, and the averaged external
heat input:
2
2

1 e
 1  cos       
Qa ,2,mean 
 AF2,M  M E 
 1    d  4.4 W
2 e
2    e  



4
Qp,2,mean   A2 F2,M M TM  1 0.25  0.36  0.94  5.67 108  2744  26 W
Node 3.
Node 3 has solar panels that work with direct solar radiation and planet-reflected (albedo).
Lunar satellite (box)
8

Qs ,3,th   3   Fp  EA3 sin  

(18)

 if 0    e ( 0 otherwise)
W


F
EA
sin



s
,3,out
p
3


2
2

1  cos       

Qa ,3,th   3   Fp A3 F3,M  M E 
 1    
2


   e  
Qa ,3   3 AM FM,3 M a   3 A3 F3,M M a 
(19)
2
2 


 
 1  cos  
Wa ,3,out   Fp A3 F3,M  M E 
 1    
2

   e  

(20)
Qp,3   IR AM FM,3M M   3 A3 F3,M M TM4
Qs ,3   3 EA3 sin 


The view factor from face 3 to the Moon can be found from Table of view factors: view factor from a
small planar plate at an altitude H, tilted an angle =/2, to a sphere of radius R, with hH/R:
F12 
1
1 x 
2
 arctan  2  with x  h  1

x h 
(21)
so that, with h=1.173 (Table 1), x=0.613, and F12=F3,M=0.18.
Node 3 has then the peak and mean values collected in Table 3:
Table 3. Node 3 peak and mean values.
Peak Mean
Solar thermal input
178
43
Solar electr. output
62
15
Albedo thermal input 3.9
0.6
Albedo electr. output 1.4
0.2
Planetary input
9.6
9.6
Node 4
Node 4 is similar to node 3, changing the domain 0<<e to e<<2. Results are those in Table 3.
Nodes 5 and 6
Node 5 is facing Nord, and node 6 is facing South; they have no solar input, and those of albedo and
planet are the same as for node 3 or 4.
Node 7 and 8
Node 7 is exactly as node 3 except for the smaller area, and node 8 is like 4 except for the smaller area.
Nodes 9 and 10
Node 9 is facing Nord, and node 10 is facing South; they have no solar input, and those of albedo and
planet are the same as for node 7 or 8.
Lunar satellite (box)
9
Node 11. It contains the batteries, which receive a time-dependent electrical input from the solar panels in
other nodes, and delivers a constant electrical power that we assume to be consumed all within the same
node 11. We decouple the general energy balance (6) into:
 Electrical energy balance:
dEele,11
dt

 Wele,out,1  Wele,out,3  Wele,out,4  Wele,out,7  Wele,out,8  Wele,dis,11
(22)
Thermal energy balance:
m11c11
dT11
 Wdis ,11   Qcond,j ,11   Qrad,j ,11
dt
j
j
From the electrical balance (assumed periodic) one gets Wdis ,11 
(23)
1
2
 W
ele,out,i
 69 W , and, integrating
(22):
Fig. 6 Accumulated electrical energy (in [kJ]), and comparison with net input rate (in [W]).
A summary of external inputs is compiled in Table 4.
Table 4. Summary of external inputs (peak, mean) values [W].
Node
Solar thermal Solar electr. Albedo thermal
Albedo electr.
Planetary
1
178, 57
62, 20
(zenith)
(zenith)
(zenith)
2
(nadir+shield) (nadir+shield)
14, 4.4
(no solar cells)
26, 26
3
178, 43
62, 15
3.9, 0.6
1.4, 0.2
10, 10
4
178, 43
62, 15
3.9, 0.6
1.4, 0.2
10, 10
5
(North)
(North)
3.9, 0.6
(no solar cells)
10, 10
6
(South)
(South)
3.9, 0.6
(no solar cells)
10, 10
7
109, 35
38, 9
2.5, 0.6
0.9, 0.2
6, 6
8
109, 35
38, 9
2.5, 0.6
0.9, 0.2
6, 6
9
(North)
(North)
2.5, 0.6
(no solar cells)
6, 6
10
(South)
(South)
2.5, 0.6
(no solar cells)
6, 6
11
(interior)
120, 69
(interior)
(see Solar electr.) (interior)
(alb. added)
Whole SC
340, 196
37, 24
(see Solar electr.)
88, 88
51 (69),0
(alb. added)
The global mean external input is then 196+(69)+24+88=376 W, which must be balanced by outgoing
thermal radiation towards the 4 steradians (i.e. including planet and Sun directions).
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c)
Find the thermal conductance and radiative couplings between nodes.
Conductive couplings
We want now the conductive couplings Cj,i that enter into the heat input to node i from node j:
Qcond,j ,i  C j ,i (Tj  Ti )
(24)
which can be computed using a quasi-one-dimensional model of heat conduction along two materials of
different geometrical and material properties (generic nodes 1 and 2) in series:
Q  k1 A1
T1  Tjoint
L1
 k2 A2
Tjoint  T2
L2

T1  T2
L1
L
 2
k1 A1 k2 A2
 C12 
1
L1
L
 2
k1 A1 k2 A2
(25)
where A stands for areas transversal to the heat flow, and L for the characteristic lengths along the flow
path, for each material. The procedure in (25) can be extended to accommodate other more involved
configurations, e.g. to add a thermal joint resistance for un-welded metal unions, or consider several heat
paths in parallel.
Node 1 is conductively connected to nodes 3, 4, 5, 6, and 11, with the first four, the conductive couplings
Cj,1 are the same: Cj,1=kAlAC/L=0.2 W/K, with thermal conductivity kAl=200 W/(m·K) for aluminium
alloys, the contact area AC=L, with L=0.5 m plate edge, and =1 mm plate thickness, and L=0.5 m the
total node-to-node thermal path. We neglect the thermal effect of the solar cells, typically several tenths
of millimetre thin, either made of silicon (=2330 kg/m3, cp=703 J/(kg·K), k=150 W/(m·K)), of gallium
arsenide (=5300 kg/m3, cp=330 J/(kg·K), k=55 W/(m·K)), or a combination of both as in new
multijunction solar cells. As for C11,1, the material is the same, but the contact area is double (=2 mm
plate thickness), and the distance between node 1 and node 11 is L/2, so that C11,1=0.8 W/K.
Node 2 is conductively connected to nodes 3, 4, 5, 6, 7, 8, 9, 10, and 11. With the first four, the
conductive couplings Cj,2 are (see above) Cj,2=kAlAC/L=0.2 W/K. With the second four, the conductive
couplings Cj,2 change because now the two half-paths are not equal to L/2, but one is L/2 and the other
Lp/2, what yields Cj,2=kAlAC/(L/2+Lp/2)=0.25 W/K. Last, C11,2=0.8 W/K as before.
Node 3 is conductively connected to nodes 1, 2, 5, 6, 7, and 11. With the first four, the conductive
couplings are (see above) Cj,3=0.2 W/K. With node 7 is C7,3=0.25 W/K. Last, C11,3=0.8 W/K as before.
Node 4 is conductively connected to nodes 1, 2, 5, 6, 8, and 11. With the first four, the conductive
couplings are (see above) Cj,4=0.2 W/K. With node 8 is C8,4=0.25 W/K. Last, C11,4=0.8 W/K as before.
Node 5 is conductively connected to nodes 1, 2, 3, 4, 9, and 11. With the first four, the conductive
couplings are (see above) Cj,5=0.2 W/K. With node 9 is C9,5=0.25 W/K. Last, C11,5=0.8 W/K as before.
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Node 6 is conductively connected to nodes 1, 2, 3, 4, 10, and 11. With the first four, the conductive
couplings are (see above) Cj,6=0.2 W/K. With node 10 is C10,6=0.25 W/K. Last, C11,6=0.8 W/K as before.
Node 7 is conductively connected to nodes 2, 3, 9, and 10. With the first two, the conductive couplings
are (see above, L/2+Lp/2) Cj,7=0.25 W/K. With the second two (Lp/2+Lp/2) Cj,7=0.32 W/K.
Node 8 is conductively connected to nodes 2, 4, 9, and 10. With the first two, the conductive couplings
are (see above, L/2+Lp/2) Cj,8=0.25 W/K. With the second two (Lp/2+Lp/2) Cj,8=0.32 W/K.
Node 9 is conductively connected to nodes 2, 5, 7, and 8. With the first two, the conductive couplings are
(see above, L/2+Lp/2) Cj,9=0.25 W/K. With the second two (Lp/2+Lp/2) Cj,9=0.32 W/K.
Node 10 is conductively connected to nodes 2, 6, 7, and 8. With the first two, the conductive couplings
are (see above, L/2+Lp/2) Cj,10=0.25 W/K. With the second two (Lp/2+Lp/2) Cj,10=0.32 W/K.
Node 11 is conductively connected to nodes 1, 2, 3, 4, 5, and 6, all of them with Cj,11=0.2 W/K.
Nodes
1
2
3
4
5
6
7
8
9
10
11
1
0.2
0.2
0.2
0.2
0.2
2
0.2
0.2
0.2
0.2
0.25
0.25
0.25
0.25
0.2
Table 5. Conductive couplings between nodes, in [W/K].
3
4
5
6
7
8
9
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.25
0.25
0.25
0.2
0.2
0.25
0.2
0.2
0.25
0.2
0.2
0.25
0.2
0.2
0.25
0.32
0.25
0.32
0.25
0.32
0.32
0.25
0.32
0.32
0.2
0.2
0.2
0.2
-
10
0.25
0.25
0.32
0.32
-
11
0.8
0.8
0.8
0.8
0.8
0.8
-
Radiative couplings
We want to find now the radiative couplings Rj,i that enter into the heat input from node j to node i:
Qrad,j ,i   R j ,i (Tj4  Ti 4 )
(26)
which, in the restricted case where surfaces are isothermal, opaque, and diffusively radiating, can be
worked out analytically in terms of view factors and emissivities:

Qrad , j ,i  Ai Fi , j T j4  Ti 4


Ri , j   i Ai  j  Fj ,k Fj ,k
(27)
where:
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 1 F13
...
1  1F11  1F12

  F

1   2 F22   2 F21
...
2 21


F
  3 F31

 3 F32 1  3 F33
...


...
...
1   N FNN 
 ...
(28)
and =1. If we further take all internal surfaces as blackbodies, then (26) reduces to:

Qrad , j ,i  Ai Fi , j T j4  Ti 4


Ri , j  Ai Fi , j
(29)
The internal face of node 1 can be approximated as a planar surface of area A1=0.52=0.25 m2 facing an
equal planar area at node 11 at a separation of 50 mm; in fact, one face of node 11 has A11=0.42=0.16 m2,
and the internal side of node 1 also sees the supporting aluminium plate in the middle (2 mm thick), and a
slant view of part of the internal faces of nodes 3, 4, 5, and 6, but we think the simplification is good for
the purpose of this simulation.
Thus, we only have to deal with the nodal faces exposed to the space environment.
 The external faces of nodes 1, 3, 4, 5, 6, 7, 8, 9, and 10, all have a view factor of unity towards the
background envelop; the temperature of the latter, 2.7 K, will be neglected in the analysis.
 The external face of node 2 has a view factor F2,∞=0.35 (obtained above to compute planet
inputs), and, since the other four lateral faces (7,8,9,10) are all equal and complete an enclosure,
F2,7=F2,8=F2,9=F2,10=(1F2,∞)/4=0.16.

The ‘internal’ face of each of the panel nodes (7,8,9,10) present three different view factors.
Taken node 7 as wild-card node:
o With their lateral square sections: F7,∞=F7,2=0.27 (from A7F7,2=A2F2,7).
o With their lateral nodes: F7,9=F7,10=0.16 (from view factor data).
o With its frontal node: F7,8=0.14 (from view factor data). One may check that
2*0.266+2·0.164+0.138=1.
Table 6. Non-trivial view factors from node i to node j.
Nodes j=2
7
8
9
10 Env.
i=2
0.16 0.16 0.16 0.16 0.35
7
0.27
0.14 0.16 0.16 0.27
8
0.27 0.14
0.16 0.16 0.14
9
0.27 0.16 0.16
0.14 0.14
10
0.27 0.16 0.16 0.14
0.14
Env. 0.35 0.16 0.16 0.16 0.16
-
Notice that a nodal matrix containing all thermal couplings might be established by taking advantage of
the symmetry in node interaction. For instance:
 Radiative couplings between node i and node j in the upper triangular side of the matrix.
 Thermal capacities of each node i in the diagonal of the matrix.
 Conductive couplings between node i and node j in the lower triangular side of the matrix.
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d)
Establish the node equations.
The nodal equations are:
mi ci


dTi dEele,i

 Wele,in,i  Wele,out,i  Qs,in,i  Qa,in,i  Qp,in,i   Qcond,j ,i   Qrad,j ,i   Qout,i (30)
dt
dt
j
 j

Thermal capacities of nodal plates 1, 3 and 4 are the same: mc=AlL2AlcAl+SiL2SicSi=
2700·0.52·0.001·900+2300·0.52·0.0005·700=810 J/K. Those of nodes 2, 5, and 6 have no solar cells, and
their thermal capacities are just mc=AlL2AlcAl=2700·0.52·0.001·900=610 J/K. Nodes 7 and 8 have
mc=AlLLpAlcAl+SiLLpSicSi= 2700·0.5·0.31·0.001·900+2300·0.5·0.31·0.0005·700=510 J/K., Nodes 9
and 10 have mc=AlLLpAlcAl=2700·0.5·0.31·0.001·900=380 J/K. Finally, for node 11 we have to add the
two electronic boxes and the 2
2·10·1000+2700·0.52·0.002·900=21 200 J/K.
mm
supporting
plate:
m11c11=2mc+AlL2AlcAl=
All terms in (30) have been worked out before, except the last one, which is simply Qout,i  Ai i Ti 4 for
the external faces of nodes 1, 3, 4, 5, 6, 7, 8, 9, and 10.
e)
Find the steady temperatures at the sub-solar and at the opposition points.
This point is solved using the full dynamic routine to be developed later at point g), fixing the orbital
angle but letting the time run until a steady state is reached. We might try to solve it analytically, i.e. by
solving the 11 node temperatures from the set of 11 nodal steady energy balances (i.e. all d()/dt=0), but
the system is non-linear due to the T- and T4-terms, and numerical solution is required anyway. The
results are in Table 7, below.
f)
Find the orbital mean temperatures at the nodes.
The intention here was to take the orbit-average of every term in (30) and solve for the 11
mean-temperatures unknowns, but, as in the previous point, it yields a set of 11 non-linear equations, and
this point is better solved using the full dynamic routine, as before. The results are in Table 7, below.
g)
Find the temperature evolution along the orbit, in the periodic state.
A numeric code has been set, using Maple, for solving the 11 nodal equations (30) as a function of time
(orbit angle, really, just to avoid large numbers); for academic purpose, it is more instructive to develop a
personal code than to use a closed-package (of course, for routine analysis of complex ever-changing
geometries one has to resort to standard professional programs like ESATAN). The transients after an
artificial choice for initial conditions (we choose Ti=300 K), is presented in Fig. 7 (first 5 orbits), and the
steady periodic solution edited in Fig. 8.
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Fig. 7. Transient temperature evolution, in [K] (five first orbits, in [rad]).
Fig. 8. Periodic temperature evolution at nodes 1 to 11, vs. orbit angle  in [rad] (origin at the subsolar
point).
With the same numerical code, but now fixing the orbit angle at =0 or at =, and letting time to run
freely instead of as t=8470/(2) [s], we solve question e) above, completing Table 7.
Table 7. Summary of steady-point and orbit-averaged temperatures [K].
Node Steady subsolar Steady opposite Orbit-max Orbit-mean Orbit-min
1
282
231
291
260
241
2
219
268
279
265
244
3
238
236
294
259
245
4
237
236
292
259
245
5
238
239
253
251
248
6
238
239
253
251
248
7
226
260
267
259
246
8
224
260
267
259
246
9
226
262
269
260
247
10
226
262
269
260
247
11
254
253
270
268
265
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Notice that the periodic maximum in temperatures goes over the values found in the steady subsolar case,
in spite of the smoothing due to thermal inertia; the reason is that a cube box gets more solar radiation
when skewed than when frontal to the Sun.
The internal sides of the flaps (nodes 7,8,9,10) may be covered with a low-emissivity coating (e.g. a
material with =0.2, =0.1) to minimise thermal inputs to face 2.
The thermal modelling of the satellite interior with just one node is a very crude approach. In reality,
concentrated heat sources may force to implement specific heat paths (e.g. heat pipes).
Conclusion: the thermal design must be improved (e.g. by using low emissivity coatings on surfaces 5
and 6), to decrease emission and get internal temperatures (T11) around 10 ºC or 15 ºC, instead of around
the 5 ºC seen in Fig. 7.
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