Spacecraft bus — structure, power, thermal, ADCS, C&DH, comms, propulsion: Worked examples

3.6.21 · D3 · Physics › Spacecraft Structures & Systems Engineering › Spacecraft bus — structure, power, thermal, ADCS, C&DH, comm

Hum sirf wahi do ideas reuse karte hain jo the parent bus note mein hain:

Recall Jin do pillars par hum build karte hain

Steady-state balance ::: $Q_{in} = Q_{out}$ — andar aane wali heat bahar jaane wali heat ke barabar hoti hai, isliye temperature change hona band ho jaata hai. Radiated heat ::: $Q_{out} = ε σ A T^{4}$ — ek garam surface energy glow karke bahar bhejti hai; jitni zyada hot hogi, utni zyada glow hogi (fourth power).

Kisi bhi number se pehle, aao har symbol ko dobara samjhein taaki koi naya student kabhi confused na ho.

Definition Har symbol, seedhi zabaan mein

$S$ = solar constant, Earth ki distance par raw sunlight ki power per square metre, $S = 1361 W/m^{2}$ . Socho ek $1 m \times 1 m$ ka square Sun ke saamne pakda hua hai: usse 1361 watts guzar rahi hain.
$α$ = absorptivity, arriving sunlight ka woh fraction jo surface nigal leta hai (0 se 1 tak). Black paint lagbhag sab nigal leti hai; ek mirror lagbhag kuch nahi nigalta.
$ε$ = emissivity, ideal glow ka woh fraction jo surface actually radiate karta hai (0 se 1 tak). Ek accha radiator high $ε$ wala hota hai.
$A_{proj}$ = projected area, woh shadow jo spacecraft ek peeche wali wall par dalta hai — woh flat silhouette jo Sun actually dekhta hai. Sirf yahi sunlight pakadta hai.
$A$ = total radiating area, poori skin jo heat bahar glow kar sakti hai.
$A_{Earth}$ = Earth-facing area, skin ka woh hissa jo Earth ki taraf pointed hai; sirf yahi bounced sunlight aur Earth ki apni glow pakadta hai.
$Q_{solar}$ = direct solar heat, woh sunlight jo surface actually nigal leta hai. Likha jaata hai $Q_{solar} = α A_{proj} S$ .
$Q_{dissip}$ = internal dissipation, waste heat jo electronics sirf chalne se paida karte hain (jaise phone garam ho jaata hai).
$Q_{albedo}$ = albedo heat, woh sunlight jo Earth reflect karke spacecraft par wapas bhejti hai. Likha jaata hai $Q_{albedo} = α A_{Earth} a S$ .
$a$ = Earth's albedo, sunlight ka woh fraction jo Earth bounce karta hai, $a \approx 0.3$ (lagbhag ek tehai).
$Q_{Earth}$ = Earth-IR heat, infrared glow jo garam Earth khud upar bhejti hai. Likha jaata hai $Q_{Earth} = ε A_{Earth} E_{Earth}$ .
$E_{Earth}$ = Earth-IR flux, us Earth glow ki power per square metre, $\approx 237 W/m^{2}$ .
$σ$ = Stefan–Boltzmann constant, $5.67 \times 1 0^{- 8} W/(m^{2} K^{4})$ — woh fixed number jo nature temperature ko glow mein convert karne ke liye use karti hai.
$T$ = surface temperature kelvin (K) mein. Kelvin bas Celsius $+ 273.15$ hai; hum ise isliye use karte hain kyunki glow temperature par depend karta hai jo absolute zero se measure hoti hai, paani ke freezing point se nahi.

The scenario matrix

Is topic ka har thermal problem inhi cells mein se ek hoga. Neeche ke examples mein har ek ke saath ek [Cell n] tag hai taaki tum dekh sako ki poori space cover ho gayi hai.

Cell	Case class	Kya special hai	Example
1	Sunlit, broadside	full solar input, hot case	Ex. 1
2	Eclipse (zero Sun)	$S = 0$ , cold limiting case	Ex. 2
3	Edge-on ( $A_{proj} \to 0$ )	degenerate projected area	Ex. 3
4	Coating sign-flip ( $α / ε$ )	ratio hot vs cold decide karta hai	Ex. 4
5	Sare charon inputs on	albedo + Earth-IR shaamil	Ex. 5
6	Limit $ε \to 0$	near-zero radiator, runaway heat	Ex. 6
7	Real-world word problem	target hit karne ke liye coating chunna	Ex. 7
8	Exam twist: area ke liye solve karo	unknown $A$ hai, $T$ nahi	Ex. 8

Example 1 — Sunlit, broadside (hot case) · [Cell 1]

Worked example The baseline

Ek CubeSat panel jiska total radiating area $A = 0.03 m^{2}$ hai, Sun ke saamne broadside face karta hai jiska projected area $A_{proj} = 0.01 m^{2}$ hai. Black paint: $α = 0.95$ , $ε = 0.85$ . Internal dissipation $Q_{dissip} = 5 W$ . Abhi ke liye albedo aur Earth-IR ignore karo. Equilibrium temperature nikalo.

Forecast: Black paint sunlight pee leti hai — andaza hai: bahut uncomfortable hot, 20 °C se kaafi zyada?

Step 1 — Solar input. $Q_{solar} = α A_{proj} S = 0.95 \times 0.01 \times 1361 = 12.93 W$ Yeh step kyun? Sirf projected silhouette sunlight pakadta hai, aur sirf fraction $α$ nigla jaata hai.

Step 2 — Total heat in. $Q_{in} = 12.93 + 5 = 17.93 W$ Yeh step kyun? Steady state mein jo bhi andar aata hai woh bahar jaana chahiye; pehle har source ko sum karo.

Step 3 — Temperature ke liye fourth-root. $T = (\frac{17.93}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 330 K = 57 °C$ Yeh step kyun? $T^{4}$ glow law ko undo karo taaki pucha ja sake "kaun sa temperature 17.93 W radiate karta hai?"

Verify: $T = 330$ K wapas plug karo: $0.85 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 33 0^{4} = 17.9 W$ ✓ — $Q_{in}$ se match karta hai. Units: $W/m^{2} K^{4} \cdot m^{2} \cdot K^{4} = W$ ✓. Electronics ke liye bahut hot — hamare forecast se match karta hai.

Neeche ki picture yeh balance visually dikhati hai: do heat arrows andar aa rahe hain (orange solar, magenta dissipation) aur violet arrows bahar glow ho rahe hain.

Figure 1 — Example 1 ka steady state: $12.93 W$ solar plus $5 W$ dissipation andar equal karta hai $17.9 W$ radiated bahar $T = 330$ K par.

Example 2 — Eclipse: Sun switch off ho jaata hai · [Cell 2]

Worked example Zero-input limiting case

Wahi CubeSat, ab Earth ki shadow mein: $S = 0$ , koi albedo nahi, koi Earth-IR nahi. Sirf $Q_{dissip} = 5 W$ ise garam rakhti hai. Coating $ε = 0.85$ , $A = 0.03 m^{2}$ . Temperature nikalo.

Forecast: Sunlight gone hone ke baad, yeh cold hona chahiye — freezing se neeche?

Step 1 — Solar term ko zero set karo. $Q_{solar} = α A_{proj} \cdot 0 = 0$ Yeh step kyun? Eclipse degenerate input $S = 0$ hai; equation phir bhi hold karti hai, ek term gayab ho jaata hai.

Step 2 — Sirf dissipation bachti hai. $Q_{in} = 5 W$ Yeh step kyun? Jab har solar aur Earth term switch off ho, electronics ki waste heat akela cheez hai jo radiate honi hai — isliye poora $Q_{in}$ collapse hokar $Q_{dissip}$ ban jaata hai.

Step 3 — Fourth-root. $T = (\frac{5}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 243 K = - 30 °C$ Yeh step kyun? Glow tabtak girega jab tak woh sirf woh 5 W radiate kare jo electronics banate hain.

Verify: $0.85 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 24 3^{4} = 5.0 W$ ✓. Cold, jaise forecast kiya tha — aur yahi woh exact wajah hai kyun heaters exist karte hain: hot case (57 °C) aur cold case (−30 °C) survival range bracket karte hain.

Example 3 — Sun ke saamne Edge-on ( $A_{proj} \to 0$ ) · [Cell 3]

Worked example Degenerate projected area

Wahi panel, lekin edge-on moda gaya taaki Sun ek razor-thin sliver dekhe: $A_{proj} = 0.0005 m^{2}$ . Black paint, $Q_{dissip} = 5 W$ . $T$ nikalo.

Forecast: Lagbhag koi sunlight catch nahi hui — temperature eclipse value ke paas rehna chahiye, thoda sa zyada garm.

Step 1 — Tiny solar input. $Q_{solar} = 0.95 \times 0.0005 \times 1361 = 0.65 W$ Yeh step kyun? Projected area woh shadow hai jo object cast karta hai; edge-on, woh shadow zero ki taraf shrink ho jaata hai.

Step 2 — Total in. $Q_{in} = 0.65 + 5 = 5.65 W$ Yeh step kyun? Tiny surviving solar input ko hamesha present dissipation mein add karo taaki poora $Q_{in}$ mile jo radiate hona hai.

Step 3 — Temperature. $T = (\frac{5.65}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 250 K = - 23 °C$ Yeh step kyun? Woh temperature dhundho jiska glow exactly $5.65$ W le jaaye, uske liye fourth root lo.

Verify: $25 0^{4} \times 0.85 \times 5.67 \times 1 0^{- 8} \times 0.03 = 5.65 W$ ✓. Jaise hi $A_{proj} \to 0$ hota hai, hum smoothly eclipse case (243 K) recover karte hain — limiting behaviour continuous hai, koi surprise nahi.

Example 4 — Coating sign-flip: white vs black · [Cell 4]

Worked example

$α / ε$ ratio sab kuch decide karta hai Broadside sunlit case ( $A_{proj} = 0.01$ , $A = 0.03$ , $Q_{dissip} = 5$ W). Black ( $α = 0.95, ε = 0.85$ ) ko white ( $α = 0.25, ε = 0.90$ ) se compare karo. Dono temperatures nikalo.

Forecast: White sunlight reflect karta hai phir bhi accha radiate karta hai — andaza hai yeh black se kaafi cooler hoga.

Step 1 — Driving ratio compute karo. $\frac{α}{ε}_{black} = \frac{0.95}{0.85} = 1.12, \frac{α}{ε}_{white} = \frac{0.25}{0.90} = 0.28$ Yeh step kyun? Purely sunlit plate ke liye, $T^{4} \propto α / ε$ : high ratio → hot, low ratio → cold. Yeh akela number outcome predict karta hai.

Step 2 — White paint temperature. $Q_{solar} = 0.25 \times 0.01 \times 1361 = 3.40 W, Q_{in} = 3.40 + 5 = 8.40 W$ $T = (\frac{8.40}{0.90 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 272 K = - 1 °C$ Yeh step kyun? Reduced solar swallow ko dissipation ke saath sum karo $Q_{in}$ pane ke liye, phir fourth-root lo — wahi master move, ab white paint ke low $α$ ke saath jo kaafi kam sunlight andar aane deta hai.

Step 3 — Black se compare karo (57 °C from Ex. 1). Sirf ek paint choice se 58 °C ka swing.

Verify: $0.90 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 27 2^{4} = 8.4 W$ ✓. Low $α / ε$ ⇒ cold, exactly jaisa ratio ne warn kiya tha. Yahi sign-flip hai: $α / ε = 1$ cross karna "runs hot" aur "runs cold" ko separate karta hai.

Neeche ka curve equilibrium temperature ko ratio $α / ε$ ke against plot karta hai, jisme black aur white points mark hain — tum 58 °C gap directly read off kar sakte ho.

Figure 2 — Example 4 ka sign-flip: ek sunlit plate ka temperature $α / ε$ ke saath badhta hai; black paint (ratio 1.12) hot rehti hai, white paint (ratio 0.28) cold rehti hai.

Example 5 — Charon heat inputs on · [Cell 5]

Worked example Full LEO environment

Low Earth orbit mein ek flat radiator, ek face Sun aur Earth ki taraf. $A = 0.03 m^{2}$ (radiating), $A_{proj} = 0.01 m^{2}$ (Sun ki taraf), Earth ki taraf face area $A_{Earth} = 0.01 m^{2}$ . White paint $α = 0.25, ε = 0.90$ . Albedo $a = 0.3$ ; Earth-IR flux $E_{Earth} = 237 W/m^{2}$ . $Q_{dissip} = 5 W$ . $T$ nikalo.

Forecast: Sun ke upar Earth ki warmth bhi add ho rahi hai — pure-sunlit white case (−1 °C) se thodi zyada hot?

Step 1 — Solar. $Q_{solar} = 0.25 \times 0.01 \times 1361 = 3.40 W$ . Kyun: direct sunlight, pehle ki tarah.

Step 2 — Albedo (Earth se bounce hui sunlight). $Q_{albedo} = α A_{Earth} a S = 0.25 \times 0.01 \times 0.3 \times 1361 = 1.02 W$ Kyun: Earth sunlight ka ek fraction $a$ reflect karti hai; surface jo dekhta hai uska fraction $α$ absorb karta hai.

Step 3 — Earth ka apna infrared glow. $Q_{Earth} = ε A_{Earth} E_{Earth} = 0.90 \times 0.01 \times 237 = 2.13 W$ Kyun: garam Earth IR radiate karti hai; surface ise apni IR efficiency $ε$ se absorb karta hai.

Step 4 — Sum karo aur solve karo. $Q_{in} = 3.40 + 1.02 + 2.13 + 5 = 11.55 W$ $T = (\frac{11.55}{0.90 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 293 K = 20 °C$ Yeh step kyun? Ab jab charon inputs on hain, har ek ko ek single $Q_{in}$ mein add karo aur fourth root lo — wahi master move, bas koi bhi term switch off nahi hai.

Verify: $0.90 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 29 3^{4} = 11.55 W$ ✓. Sunlit-only white case (−1 °C) se zyada warm, exactly jaisa forecast kiya tha — Earth ka albedo aur IR ~21 °C add karte hain.

Example 6 — Limit $ε \to 0$ : runaway radiator · [Cell 6]

Example 7 — Word problem: 25 °C hit karne ke liye coating chunna · [Cell 7]

Worked example Design, sirf analysis nahi

Ek sunlit instrument box ( $A_{proj} = 0.01 m^{2}$ , $A = 0.03 m^{2}$ , $Q_{dissip} = 5 W$ ) exactly $T = 298 K$ (25 °C) par rehna chahiye. Hum paints mix karke emissivity tune kar sakte hain; $α = 0.30$ fix hai. Kaun sa $ε$ chahiye?

Forecast: Hum na hot chahte hain na cold — moderate $ε$ , shayad 0.9 ke paas?

Step 1 — $ε$ unknown ke saath balance likho. $Q_{solar} + Q_{dissip} = ε σ A T^{4}$ Yeh step kyun? Ab coating unknown hai aur temperature fixed hai — design inverse.

Step 2 — Left side ek known number hai. $Q_{in} = 0.30 \times 0.01 \times 1361 + 5 = 4.08 + 5 = 9.08 W$ Yeh step kyun? Heat-in side ki har cheez ab numeric hai, isliye unknown $ε$ ko isolate karne se pehle ise ek value mein collapse karo.

Step 3 — $ε$ ke liye solve karo. $ε = \frac{Q _{in}}{σ A T ^{4}} = \frac{9.08}{5.67 \times 1 0 ^{- 8} \times 0.03 \times 29 8 ^{4}} = 0.677$ Yeh step kyun? $ε$ ke siwa sab kuch divide out karo; target temperature $T^{4}$ se enter karta hai.

Verify: $0.677 \times 5.67 \times 1 0^{- 8} \times 0.03 \times 29 8^{4} = 9.08 W$ ✓ $= Q_{in}$ . Ek physically realisable $ε \in (0, 1)$ — ek real grey paint kaam kar sakti hai.

Example 8 — Exam twist: radiator area ke liye solve karo · [Cell 8]

Worked example Unknown

$A$ hai, $T$ nahi Ek electronics box eclipse mein (koi Sun nahi) $Q_{dissip} = 40 W$ waste heat dump karta hai. Uske radiator ( $ε = 0.90$ ) ko $T = 313 K$ (40 °C) se zyada nahi jaana chahiye. Minimum radiating area $A$ kitna chahiye?

Forecast: Bahut saari heat, moderate temperature cap — area tiny nahi ho sakta.

Step 1 — $A$ unknown ke saath balance. $Q_{dissip} = ε σ A T^{4} \Rightarrow A = \frac{Q _{dissip}}{ε σ T ^{4}}$ Yeh step kyun? Temperature limit par, radiator just itna bada hona chahiye ki poore 40 W glow away ho sakein.

Step 2 — Numbers plug karo. $A = \frac{40}{0.90 \times 5.67 \times 1 0 ^{- 8} \times 31 3 ^{4}} = 0.0824 m^{2}$ Yeh step kyun? Bada $Q$ ya lower allowed $T$ ek bada radiator force karta hai — area $1/ T^{4}$ se scale karta hai, isliye cool cap area mein expensive hai.

Verify: $0.90 \times 5.67 \times 1 0^{- 8} \times 0.0824 \times 31 3^{4} = 40.0 W$ ✓. Units: $W / (W/m^{2} K^{4} \cdot K^{4}) = m^{2}$ ✓. Poore CubeSat panel se bada — isliye high-power satellites deployable radiators carry karte hain.