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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

rays never fall on the face pointing to the Moon. b) Find the external heat loads (solar, albedo and infrared) as a function of orbit position. c) Find the thermal conductance and radiative couplings between nodes .d) Establish the node equations. e) Find the steady temperatures at the sub-solar and at the opposition points. f) Find the orbital mean temperatures at the nodes. g) 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.

Table 1. Data for the lunar orbit and Moon’s orbit.Parameter Symbol and value Comments

Sun-to-Moon distance

RSM=150·109 m (1 AU) Sat to Moon and Moon to Earth distances are much smaller.

Moon-to-Earth distance

RME=385·106 m (60 Earth diameters). Perigee 363·106 m, apogee 406·106 m.

Moon radius RM=1737 km (0.273 RE). Equatorial 1738 km, Polar 1736 km. g=1.62 m/s2.Moon orbit period TM=27.3 d Synodic period (from full-moon to full-moon) is TM=29.5 d.Moon orbit inclination

i=5.1º to ecliptic Between 18º and 29º to Earth’s equator.

Moon axis tilt iM=1.54º to ecliptic 6.7º to Moon’s orbit plane.Moon surface temperature

TM=274 K 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.

Moon emissivity M=0.94 From energy balance (1−ρ)EπR2=4πR2εσT4. M=Tp4=300

W/m2.Solar irradiance E=1360 W/m2 With <2% annual oscillation due to Earth orbit eccentricity.Albedo =0.12 Bolometric; normal albedo is 0.07.Solar to orbit vector angle

=/2 Can be taken as an ecliptic orbit since sat orbit plane is in Moon’s orbit plane, with an inclination of 5.1º to ecliptic.

Relative altitude h=H/R=300/1737=0.173

The Moon has no atmosphere (the Lunar Reconnaissance Orbiter altitude is 30..70 km)

Sat orbit period 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.

Eclipse fraction Te/T=0.325 Te/T=(1/)arcsin(1/(1+h)), Te=46 min.Eclipse start angle e=2.12 rad (121.5º). e=arcsin(1/(1+h)).

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.

Table 2. Data for the spacecraft.Parameter Symbol and value Comments

Main cube side 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.

Extra panel length Lp=0.31 m Lp/L=tan(e)=0.613 (see Fig. 3).Aluminium plate thickness =0.001 mCube face area (each) A=0.25 m2 A=L2.Extra panel area (each) Ap=0.155 m2 Ap=LLp=0.5·0.31=0.155 m2.Solar panel effective area 90%Solar panel efficiency 20%Solar panel absorptance =0.7Solar panel emissivity =0.7Instr. plate thickness A=0.002 mE-box size (each) 0.4·0.4·0.2 m3

E-box mass (each) mEB=10 kgE-box cp cp=1000 J/(kg·K)Aluminium density =2700 kg/m3


https://en.wikipedia.org/wiki/Orbit_of_the_Moon

http://upload.wikimedia.org/wikipedia/commons/4/43/Earth-Moon.PNG

Aluminium thermal conduct. k=200 W/(m·K)Aluminium thermal capacity 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/(RSRE)=(1.5·109·6.37·106/(0.7·1096.37·106)=1.4·109 m, so that at Moon’s distance the radius of the cone section has dropped to RU=RE(1RME/LU)=6.37·106(1385·106/(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 (1.5·109 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


http://en.wikipedia.org/wiki/File:Umbra01.svg

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:

11\*MERGEFORMAT ()

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), is the net electrical and electromagnetic work rate received (no mechanical work considered), and 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, 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, , is:

22\*MERGEFORMAT ()

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:

33\*MERGEFORMAT ()


http://eclipse.gsfc.nasa.gov/LEdecade/LEdecade2001.html

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 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:

44\*MERGEFORMAT ()

where, as before, the fraction of total power emitted by the planet (pTp4Ap) that impinges on the SC, is

pTp4ApFp,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:

55\*MERGEFORMAT ()

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:

66\* MERGEFORMAT ()

where Eele is the non-thermal energy store (only for batteries), is the electrical dissipation in the node (all active nodes have electrical input, but only two have electrical output: solar Lunar satellite (box) 6

panels and batteries. Notice that in 6 is the total solar absorptance, including the part used to generate 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 may refer just to the latter, as done below.

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:

77\*MERGEFORMAT ()

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:

88\* MERGEFORMAT ()

Thermal energy balance (we only detail solar input):

99\* MERGEFORMAT ()

with the following solar, albedo, and planet inputs (only the thermal part):


1010\* MERGEFORMAT ()



with peak values of =(0.70.2·0.9)·1360·0.25=177 W and =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 divided by panel area A) have been introduced. The orbit averaged external heat input and electrical input are:


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.




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:


http://webserver.dmt.upm.es/~isidoro/tc3/Radiation%20View%20factors.pdf


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=(1F2M)/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 =14 W and =26 W, and the averaged external heat input:

Node 3. Node 3 has solar panels that work with direct solar radiation and planet-reflected (albedo).




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 hH/R:

with 2121\* MERGEFORMAT ()

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:


http://webserver.dmt.upm.es/~isidoro/tc3/Radiation%20View%20factors.pdf

Table 3. Node 3 peak and mean values.Peak

Mean

Solar thermal input 178 43Solar electr. output 62 15Albedo thermal input 3.9 0.6Albedo electr. output 1.4 0.2Planetary input 9.6 9.6

Node 4Node 4 is similar to node 3, changing the domain 0<<e to e<<2. Results are those in Table 3.

Nodes 5 and 6Node 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 8Node 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 10Node 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.

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:


Thermal energy balance:


From the electrical balance (assumed periodic) one gets , 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, 104 178, 43 62, 15 3.9, 0.6 1.4, 0.2 10, 105 (North) (North) 3.9, 0.6 (no solar cells) 10, 106 (South) (South) 3.9, 0.6 (no solar cells) 10, 107 109, 35 38, 9 2.5, 0.6 0.9, 0.2 6, 68 109, 35 38, 9 2.5, 0.6 0.9, 0.2 6, 69 (North) (North) 2.5, 0.6 (no solar cells) 6, 610 (South) (South) 2.5, 0.6 (no solar cells) 6, 611 (interior) 120, 69

(alb. added)(interior) (see Solar electr.) (interior)

Whole SC 340, 196 51 (69),0(alb. added)

37, 24 (see Solar electr.) 88, 88

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).

c) Find the thermal conductance and radiative couplings between nodes.Conductive couplingsWe want now the conductive couplings Cj,i that enter into the heat input to node i from node j:


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:



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 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 11 is conductively connected to nodes 1, 2, 3, 4, 5, and 6, all of them with Cj,11=0.2 W/K.

Table 5. Conductive couplings between nodes, in [W/K].Nodes 1 2 3 4 5 6 7 8 9 10 11

1 - - 0.2 0.2 0.2 0.2 - - - - 0.82 - - 0.2 0.2 0.2 0.2 0.25 0.25 0.25 0.25 0.83 0.2 0.2 - - 0.2 0.2 0.25 - - - 0.84 0.2 0.2 - - 0.2 0.2 - 0.25 - - 0.85 0.2 0.2 0.2 0.2 - - - - 0.25 - 0.86 0.2 0.2 0.2 0.2 - - - - - 0.25 0.87 - 0.25 0.25 - - - - - 0.32 0.32 -8 - 0.25 - 0.25 - - - - 0.32 0.32 -9 - 0.25 - - 0.25 - 0.32 0.32 - - -10 - 0.25 - - - 0.25 0.32 0.32 - - -11 0.2 0.2 0.2 0.2 0.2 0.2 - - - - -

Radiative couplingsWe want to find now the radiative couplings Rj,i that enter into the heat input from node j to node i:


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:


where:



and =1. If we further take all internal surfaces as blackbodies, then 26 reduces to:


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=(1F2,∞)/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.Node

sj=2 7 8 9 10 Env.

i=2 - 0.16 0.16 0.16 0.16 0.357 0.27 - 0.14 0.16 0.16 0.278 0.27 0.14 - 0.16 0.16 0.149 0.27 0.16 0.16 - 0.14 0.1410 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.

d) Establish the node equations.The nodal equations are:



Thermal capacities of nodal plates 1, 3 and 4 are the same: mc=AlL2AlcAl+SiL2SicSi= 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=AlL2AlcAl=2700·0.52·0.001·900=610 J/K. Nodes 7 and 8 have mc=AlLLpAlcAl+SiLLpSicSi= 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=AlLLpAlcAl=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 mm supporting plate: m11c11=2mc+AlL2AlcAl= 2·10·1000+2700·0.52·0.002·900=21 200 J/K.

All terms in 30 have been worked out before, except the last one, which is simply 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()/d t=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.


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 2412 219 268 279 265 2443 238 236 294 259 2454 237 236 292 259 2455 238 239 253 251 2486 238 239 253 251 2487 226 260 267 259 2468 224 260 267 259 2469 226 262 269 260 24710 226 262 269 260 24711 254 253 270 268 265


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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