Exercises — Spacecraft bus — structure, power, thermal, ADCS, C&DH, comms, propulsion

Camera payload hai — mission-specific instrument. Bus woh sab kuch hai jo ise zinda aur pointed rakhta hai. Saat bus subsystems hain:

1. Structure
2. Power (EPS)
3. Thermal (TCS)
4. ADCS (Attitude Determination & Control)
5. C&DH (Command & Data Handling)
6. Communications
7. Propulsion

Humne kya kiya: humne "asli science karta hai" (payload) ko "science support karta hai" (bus) se alag kiya. Kyun: har mass/power/cost trade is line ke kis taraf ek component hai yeh jaanne se shuru hoti hai.

• ${\sigma }_y$ ::: yield strength — woh stress jis par ek material permanently deform hone lagta hai (units: Pa). (Subscript $y$ dhyan se dekho — akela $\sigma$ Stefan–Boltzmann constant hai.)
• $\rho$ ::: density — mass per unit volume (kg/m³).
• $\alpha$ ::: absorptivity — incoming radiation ka woh fraction jo ek surface soak up karta hai (0–1, koi units nahi).
• $ϵ$ ::: emissivity — kitni efficiently ek surface heat radiate karta hai bahar (0–1).
• $S$ ::: solar constant — Earth ki distance par sunlight power per square metre, $1361$ W/m².

Yeh kyun matter karta hai: $\alpha$ aur $ϵ$ twins lagte hain lekin ulte sawaal ka jawaab dete hain — ek hai "main kitna pee leta hoon?", doosra "main kitna glow karta hoon bahar?" Inhe mix karna #1 thermal error hai (neeche dekho).

Step 1 — force. Rocket ki spine ko poore spacecraft ka weight times g-load carry karna hoga: $F=m\cdot a=4\times 98=392\text{N}$ Kyun: launch poore mission ka sabse harsh mechanical moment hai — orbit mein kuch bhi itna hard push nahi karta.

Step 2 — area. Stress force ko area par spread karna hai, limit par ${\sigma }_y=F/A$. Hum demand karte hain $F/A\le {\sigma }_y$, isliye: $A\ge \frac{F}{{\sigma }_y}=\frac{392}{240\times 10^6}=1.63\times 10^{-6}{\text{m}}^2=1.63{\text{mm}}^2$ Kyun: koi bhi choti area usi force ko kam material par concentrate karti hai, stress ko yield limit se aage push karti hai — wall permanently deform ho jaayegi. Yeh kaisa dikhta hai: shared figure ka left panel — force arrow wall cross-section par spread hota hua.

Answer: $A\ge 1.63{\text{mm}}^2$ (yeh strength minimum hai; real walls stiffness ke liye kaafi zyada moti hoti hain).

Step 1 — day/night ratio. Deficit period relatively kitni lambi hai? $\frac{T_{\text{eclipse}}}{T_{\text{sun}}}=\frac{33}{62}=0.532$ Kyun: yeh ratio sizing problem ki core hai — yeh humein batata hai ki running load ke relative, sunlight mein array ko dark stretch cover karne ke liye kitni extra energy bank karni hogi. Badi eclipse (ya choti daylight) bada surplus force karti hai.

Step 2 — sizing formula. Parent note ke energy see-saw se: $P_{\text{SA}}=P_{\text{load}}\left(1+\frac{T_{\text{eclipse}}}{T_{\text{sun}}\cdot {\eta }_{\text{charge}}{\eta }_{\text{discharge}}}\right)=50\left(1+\frac{0.532}{0.85}\right)$ $P_{\text{SA}}=50\times 1.626=81.3\text{W}$ Extra term kyun: array ko sirf sunlight mein load run nahi karna — saath mein itna bank bhi karna hai (lossy batteries ke through) ki eclipse survive ho sake.

Answer: $P_{\text{SA}}\approx 81.3$ W.

Step 1 — deciding quantity. Strength-limited part ke liye, required area $A=F/{\sigma }_y$ hai, aur mass $m=AL\rho =\frac{FL\rho }{{\sigma }_y}$ hai. Toh mass $\rho /{\sigma }_y$ ke saath scale hoti hai — equivalently, halkapan structural efficiency ke saath scale hota hai: ${\eta }_{\text{struct}}=\frac{{\sigma }_y}{\rho }$ Sirf ${\sigma }_y$ kyun nahi: ek zyada strong material kam area use karne deta hai, lekin agar woh dense bhi hai toh kuch gain nahi hota. Sirf ratio hi sach mein payoff batata hai.

Step 2 — har ek ke liye compute karo. ${\eta }_{\text{Al}}=\frac{240\times 10^6}{2700}=8.89\times 10^4\text{N⋅m/kg}$ ${\eta }_{\text{CFRP}}=\frac{600\times 10^6}{1600}=3.75\times 10^5\text{N⋅m/kg}$

Step 3 — ratio. $\frac{{\eta }_{\text{CFRP}}}{{\eta }_{\text{Al}}}=\frac{3.75\times 10^5}{8.89\times 10^4}=4.22$

Answer: CFRP halka choice hai; usi strength requirement ke liye ise taqreeban 4.2× kam structural mass chahiye. (Caveats: real parts aksar stiffness-limited hote hain, aur CFRP zyada costly hai aur outgass karta hai — lekin pure strength-to-weight par yeh jeet jaata hai.)

Step 1 — sun mein ek flat plate ka balance likho (argument ke liye Earth terms ignore karo). Steady state par heat in equals heat out: $\alpha A_{\text{proj}}S=ϵ\sigma A{T}^4$ jahan $\sigma =5.67\times 10^{-8}{\text{W/(m}}^2{\text{⋅K}}^4)$ Stefan–Boltzmann constant hai. Step 2 — $T$ ke liye solve karo. Plate ki sunlit aur radiating areas equal lo, $A_{\text{proj}}=A$, toh woh cancel ho jaate hain: $T={\left(\frac{\alpha }{ϵ}\cdot \frac{S}{\sigma }\right)}^{1/4}$ $A_{\text{proj}}=A$ kyun set kiya: hum coating ka effect isolate kar rahe hain, geometry ka nahi. Dono areas ko equal force karke woh equation se cancel ho jaate hain, temperature sirf material ratio $\alpha /ϵ$ par depend karti hai — exactly wahi comparison jo sawaal poochh raha hai. (Real spacecraft mein dono areas alag hote hain, lekin woh sirf ek geometric prefactor badalta hai, $\alpha /ϵ$ dependence nahi.)

Insight: $\alpha$ aur $ϵ$ alag-alag appear nahi hote — sirf unka ratio $\alpha /ϵ$ karta hai. Woh ratio hi woh single knob hai jo ek coating tumhe deta hai.

Step 3 — compare karo.

• Black: $\alpha /ϵ=0.95/0.85=1.12$
• White: $\alpha /ϵ=0.25/0.90=0.28$

Kyunki $T\propto (\alpha /ϵ{)}^{1/4}$ hai, black paint white ke absolute temperature ka $(1.12/0.28{)}^{1/4}=1.41$× par hota hai. Black kaafi zyada garam chalta hai — exactly isliye radiators aur sun-facing panels aksar white ya silvered hote hain.

Step 1 — heat in. Do sources hain: absorbed sunlight aur electronics ki apni waste heat. $Q_{\text{solar}}=\alpha A_{\text{proj}}S=0.25\times 0.01\times 1361=3.40\text{W}$ $Q_{\text{in}}=Q_{\text{solar}}+Q_{\text{dissip}}=3.40+5=8.40\text{W}$ Dissipation kyun include kiya: space mein electronics heat carry karne ke liye koi air nahi — woh sab radiation se hi nikalni hai, isliye yeh balance ke "in" side par count hoti hai.

Step 2 — heat out aur solve karo. $Q_{\text{out}}=ϵ\sigma A{T}^4=Q_{\text{in}}$ set karo: $T={\left(\frac{Q_{\text{in}}}{ϵ\sigma A}\right)}^{1/4}={\left(\frac{8.40}{0.90\times 5.67\times 10^{-8}\times 0.03}\right)}^{1/4}$ $T={\left(5.487\times 10^9\right)}^{1/4}=272.2\text{K}$

Step 3 — convert aur judge karo. $T=272.2-273.15=-0.95^{\circ }\text{C}$ Yeh $-20^{\circ }\text{C}$ to $+50^{\circ }\text{C}$ ke andar aaram se fit hai. Answer: $\approx 272$ K $\approx -1^{\circ }\text{C}$ — limits ke andar. ✅

Step 1 — end-of-life array need (2.2 se power-budget formula use karke): $P_{\text{SA,EOL}}=12\left(1+\frac{33}{62\times 0.85}\right)=12\left(1+0.626\right)=12\times 1.626=19.5\text{W}$ EOL pehle kyun: array ko year 5 mein, degrade hone ke baad bhi, load meet karna hai — isliye hum pehle worst (latest) case design karte hain.

Step 2 — beginning-of-life nikalo. Hum degradation ko linear model karte hain: har saal original output ka $2.5\%$ khota hai, toh 5 saalon mein total loss $2.5\%\times 5=12.5\%$ hai aur array apni start value ka $1-0.125=0.875$ retain karta hai. End mein $19.5$ W bacha ke rakhne ke liye: $P_{\text{SA,BOL}}=\frac{P_{\text{SA,EOL}}}{0.875}=\frac{19.5}{0.875}=22.3\text{W}$

Answer: array ko beginning of life par $\approx 22.3$ W ke liye design karo taaki 5 saalon baad bhi $\approx 19.5$ W deliver ho.

Assumption flag — linear vs. compound: humne ek linear model use kiya, $\text{retained}=1-pN$ with $p=0.025$, $N=5$. Bahut se space-solar analyses ek compound (exponential) model use karte hain, $\text{retained}=(1-p{)}^N=0.975^5=0.881$, kyunki har saal ki degradation already-degraded output par act karti hai, original par nahi. Dono yahan $0.875$ vs $0.881$ dete hain — almost identical 5 saalon mein, lekin lambi missions ke liye gap badh jaata hai. Hamesha batao tumne kaun sa use kiya; hum is page par arithmetic simplicity ke liye linear use karte hain.

Part (a) — structure. Launch force: $F=ma=6\times (12\times 9.8)=6\times 117.6=705.6\text{N}$. Strength-limited area: $A=F/{\sigma }_y=705.6/(240\times 10^6)=2.94\times 10^{-6}{\text{m}}^2$. Structural mass: $m_{\text{struct}}=AL\rho =2.94\times 10^{-6}\times 0.3\times 2700=2.38\times 10^{-3}\text{kg}\approx 2.4$ g. Reality note: parent CubeSat example ki tarah, yeh pure-strength number tiny hai — real stiffness/vibration/safety-factor requirements ise 10–20× multiply karte hain. Yeh floor set karta hai, design nahi.

Part (b) — power. EOL array: $P_{\text{SA,EOL}}=40\left(1+\frac{35}{60\times 0.85}\right)=40\left(1+0.686\right)=40\times 1.686=67.5\text{W}$. Linear degradation over 3 yr: $1-0.025\times 3=0.925$ retain karta hai. BOL: $P_{\text{SA,BOL}}=67.5/0.925=72.9\text{W}$.

Part (c) — thermal. Array apna BOL power generate karta hai (waste heat ke liye worst case), aur har 8 W generated par $1$ W waste heat produce karta hai: $Q_{\text{array}}=\frac{P_{\text{SA,BOL}}}{8}=\frac{72.9}{8}=9.1\text{W}$ Payload ki internal dissipation add karo, jo (L4 trap ki tarah) bhi radiate karni hogi: $Q_{\text{reject}}=Q_{\text{array}}+Q_{\text{payload}}=9.1+6=15.1\text{W}$ Answer: structure floor $\approx 2.4$ g; BOL array $\approx 72.9$ W; thermal system ko $\approx 15.1$ W reject karna hoga.

Humne abhi kya kiya — L5 ka poora point: ek 6 kg number teen subsystems ke through ripple kiya. Launch load ne walls size ki; usi spacecraft ki electronics ne array size kiya; array ki apni inefficiency aur payload ne thermal load feed kiya. Woh coupling — jahan har subsystem ka answer doosre ka input ban jaata hai — exactly wahi hai jo systems engineering hai.

• Woh quantity jo decide karti hai strength-limited part ke liye kaun sa material halka hai woh ==${\sigma }_y/\rho$ (structural efficiency)== hai.
• Sunlight mein equilibrium temperature ratio ==$\alpha /ϵ$== par depend karta hai, akele $\alpha$ par nahi.
• Internal dissipation power budget aur thermal budget dono mein appear karta hai.
• Tum array ko pehle end-of-life conditions ke liye size karte ho, phir retained fraction se divide karke beginning-of-life nikalte ho.
• Akela $\sigma$ is page par Stefan–Boltzmann constant hai, jabki ${\sigma }_y$ yield strength hai.