Worked examples — Thermal control — multi-layer insulation (MLI), heaters, heat pipes, radiators

3.6.23 · D3 · Physics › Spacecraft Structures & Systems Engineering › Thermal control — multi-layer insulation (MLI), heaters, hea

Yeh page drill hall hai. Parent note ne tumhe chaar tools diye the (MLI, heaters, heat pipes, radiators). Yahan hum un tools pe har tarah ke numbers throw karte hain — hot sides, cold sides, zero temperatures, infinite layers, gravity-hostile pipes, aur ek exam trap — taaki test mein koi bhi problem tumhe surprise na kare.

Neeche sab kuch ek master law pe tikaa hai, isliye koi bhi symbol aane se pehle use plain words mein ek baar pin kar lete hain.

Master radiation formula, zor se padho: heat leaving = throwing-strength × how-black × area × (hot⁴ minus cold⁴).

$Q=\epsilon \sigma A(T_{\text{hot}}^4-T_{\text{cold}}^4)$

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The scenario matrix

Is topic pe har possible problem inhi cells mein se ek hai. Neeche ke examples tagged hain us cell ke saath jo unhe hit hoti hai.

#	Cell class	Kya cheez alag banati hai	Kis example mein
C1	Hot side dominates	$T_{\text{hot}}\gg T_{\text{cold}}$, cold term negligible	Ex 1, Ex 6
C2	Both terms matter	$T_{\text{cold}}$ $T_{\text{hot}}$ ka real fraction hai, dono rakhte hain	Ex 2
C3	Degenerate: cold side = 0 K	deep space $\approx 3$ K, cold term vanishes	Ex 1, Ex 6
C4	Limiting: layers $N\to \infty$	MLI diminishing returns, idealisation kahan toot ti hai	Ex 3
C5	Zero / no-source case	eclipse, no Sun, component khud se cool hoti hai	Ex 4
C6	Time-dependent (rate)	kisi limit tak pahunchne mein kitna time lagega, $dT/dt$ use hoti hai	Ex 4
C7	Phase-change transport	heat pipe, latent heat, koi $T^4$ nahi	Ex 5
C8	Sign / direction flip	gravity wick ki help karta hai ya ladta hai	Ex 5b
C9	Real-world word problem	full mission sizing, khud tool chunna padta hai	Ex 6
C10	Exam twist	ek hidden unit ya "which is bigger" comparison	Ex 7

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Ex 1 — Hot radiator into cold void · cells C1, C3

Forecast: Aage padhne se pehle nearest 100 W tak answer guess karo. Kya 3 K ka koi fark padta hai?

1. Net law likho. $Q=\epsilon \sigma A(T_r^4-T_s^4)$. Yeh step kyun? Vacuum mein radiation hi ek maatra heat path hai; dono surfaces glow karti hain, isliye jo space humari taraf glow karta hai woh subtract karte hain.
2. Check karo ki cold term survive karta hai ya nahi. $T_r^4=300^4=8.1\times 10^9$, jabki $T_s^4=3^4=81$. Yeh step kyun? Calculate karne se pehle dekho ki koi term negligible toh nahi — isse arithmetic bachti hai aur physics samajh aati hai. Yahan $81$ vs $8.1\times 10^9$ ka ratio $10^{-8}$ hai: void kuch contribute nahi karta. Isliye hum hamesha $T_s\to 0$ approximate karte hain (cell C3).
3. Compute karo. $Q=0.85\times 5.67\times 10^{-8}\times 1.2\times 8.1\times 10^9\approx 468\text{W}$. Yeh step kyun? Baache hue terms ko multiply karo.

Verify: Units: $[{\text{W/m}}^2{\text{K}}^4]\cdot [{\text{m}}^2]\cdot [{\text{K}}^4]=\text{W}$. ✓. Figure dekho — red outgoing arrow bahut bada hai, aur space se cyan return arrow ek whisker jaisa hai. Yeh $10^{-8}$ ratio se match karta hai.

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Ex 2 — Both temperatures matter (MLI leak) · cell C2

Forecast: Kya $180$ K "ignore karne ke liye kaafi chhota" hai jaise 3 K tha? Guess karo haan/nahi.

1. Stack ki effective emissivity. ${\epsilon }_{\text{MLI}}\approx \frac{\epsilon }{N-1}=\frac{0.03}{29}\approx 1.03\times 10^{-3}$. Yeh step kyun? 30 shiny Mylar layers mein se har ek re-radiate karti hai, resistances ki ek chain banati hai; poora stack ek bahut low-emissivity surface ki tarah behave karta hai.
2. Cold term check karo. $T_1^4=293^4=7.37\times 10^9$; $T_2^4=180^4=1.05\times 10^9$. Yeh step kyun? Ab $1.05\times 10^9$, $7.37\times 10^9$ ka $14\%$ hai — negligible nahi. Yeh genuine cell-C2 problem hai; $T_2$ drop karne se leak $14\%$ zyada dikhti.
3. Compute karo. $Q=1.03\times 10^{-3}\times 5.67\times 10^{-8}\times 2\times (7.37\times 10^9-1.05\times 10^9)$. $=1.03\times 10^{-3}\times 5.67\times 10^{-8}\times 2\times 6.32\times 10^9\approx 0.74\text{W}$.

Verify: Agar hum galti se $T_2=0$ set karte: $Q^{\prime }\approx 0.86\text{W}$, jo $16\%$ zyada hai — isse confirm hota hai ki cold term rakhna worth it tha. Ex 1 se compare karo jahan rakhne se koi fark nahi pada. Lesson: pehle hamesha $T_{\text{cold}}^4$ ko $T_{\text{hot}}^4$ se compare karo.

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Ex 3 — The limit $N\to \infty$: jahan MLI tumhe jhooth bolta hai · cell C4

Forecast: Formula kehta hai ${\epsilon }_{\text{MLI}}\to 0$. Guess karo ki kya 100-layer blanket practice mein 30-layer wale ko beat karta hai?

1. Plug in karo. $N=2:0.03/1=0.030$; $N=11:0.03/10=0.0030$; $N=31:0.03/30=0.0010$; $N=101:0.03/100=0.0003$. Yeh step kyun? Returns ki shape dekhne ke liye.
2. Limit lo. Jab $N\to \infty$, ${\epsilon }_{\text{MLI}}=\frac{0.03}{N-1}\to 0$. Yeh step kyun? Mathematically insulation perfect lagti hai. Idealisation kehti hai infinite layers = zero leak.
3. Dekho kahan model toot ta hai. $N=31$ se $N=101$ tak humne 70 extra layers lagaaye aur $\epsilon$ sirf $0.0010$ se $0.0003$ tak gayi — sirf $3.3$ ka factor, bahut bade mass cost par. Isse bhi bura, real layers launch load ke neeche compress hoti hain aur touch karti hain, solid-conduction shortcuts bana deti hain jo formula ignore karta hai. Yeh step kyun? Cell C4 is baare mein hai ki kahan tumhara equation sach kehna band kar deta hai. Figure mein curve flat ho jaata hai: yeh diminishing returns hai.

Verify: Ratio check: $\epsilon (31)/\epsilon (101)=0.0010/0.0003=3.33$ jabki layer count $\times 3.26$ badha — almost one-for-one, yaani koi leverage nahi bachi. Isliye flight blankets 10–30 layers par ruke rehte hain.

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Ex 4 — Eclipse cooldown: the zero-source, time-dependent case · cells C5, C6

Forecast: Cooldown rate °C per minute mein guess karo, phir 0°C tak pahunchne ka time.

1. Heat capacity. $C=mc=5\times 1000=5000\text{J/K}$. Yeh step kyun? $C$ batata hai ki battery ko 1 K badlane ke liye kitne joules chahiye — thermal "inertia".
2. Cooling rate. Koi source nahi, energy balance: $C\frac{dT}{dt}=-Q_{\text{loss}}$, isliye $\frac{dT}{dt}=-\frac{Q_{\text{loss}}}{C}=-\frac{5}{5000}=-1.0\times 10^{-3}\text{K/s}=-0.06\text{K/min}.$ Yeh step kyun? Hum rate (ek derivative) use karte hain kyunki question time ke baare mein hai, kisi steady value ke baare mein nahi. Yahi ise cell C6 banata hai. Vacuum mein bina Sun ke (Orbital Thermal Environment) source term exactly zero hoti hai — cell C5.
3. Limit tak pahunchne ka time. Allowed drop $\Delta T=20\text{K}$ ($20°$ se $0°$ tak). Time $t=\frac{\Delta T}{0.06\text{K/min}}=\frac{20}{0.06}\approx 333\text{min}$. Yeh step kyun? Constant rate ⇒ time = drop ÷ rate.
4. Eclipse length se compare karo. $333\text{min}\gg 90\text{min}$, isliye 90-min eclipse ise sirf $90\times 0.06=5.4\text{K}$ cool karega, aur temperature $14.6°\text{C}$ ho jayegi — safe, is eclipse ke liye koi heater nahi chahiye.

Verify: $90\text{min}\times (-0.06\text{K/min})=-5.4\text{K}$; $20-5.4=14.6°\text{C}>0$. ✓. Agar eclipse $>333$ min hoti, toh ek heater (Spacecraft Power Systems) jo $\ge 5$ W draw kare, mandatory hota.

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Ex 5 — Heat pipe capacity: phase change, no $T^4$ · cell C7

Forecast: Order-of-magnitude guess karo: copper se kitne times better?

1. Heat pipe transport. $Q=\dot{m}{L}_v=0.001\times 1.37\times 10^6=1370\text{W}$. Yeh step kyun? Ek heat pipe energy ko latent heat ke roop mein carry karta hai (phase-change): fluid hot end par boil karte waqt $L_v$ joules per kg absorb karta hai aur cold end par condense karte waqt dump karta hai. Koi $T^4$ nahi aata — yeh transport hai, rejection nahi.
2. Copper conduction. $Q_{\text{cu}}=\frac{kA\Delta T}{L}=\frac{400\times 7.85\times 10^{-5}\times 10}{1}=0.314\text{W}$. Yeh step kyun? Solids heat ko conduction se move karte hain, Fourier's law — honest competitor.
3. Ratio. $1370/0.314\approx 4360\times$. Yeh step kyun? Dikhata hai kyun har bade spacecraft mein heat pipes use hoti hain: same pencil-thin cross-section, thousands of times throughput.

Verify: Units: $[\text{kg/s}]\cdot [\text{J/kg}]=\text{J/s}=\text{W}$ ✓; $[\text{W/m⋅K}]\cdot [{\text{m}}^2]\cdot [\text{K}]/[\text{m}]=\text{W}$ ✓.

Ex 5b — Sign flip: gravity wick ki help karta hai ya ladta hai · cell C8

1. Gravity head. $\Delta P_g=680\times 9.81\times 1=6670\text{Pa}\approx 6.67\text{kPa}$. Yeh step kyun? Yeh woh pressure hai jo gravity 1 m liquid ke across impose karta hai.
2. Case (a): condenser upar, liquid evaporator ki taraf neeche girna chahiye. Gravity assists return: capillary pressure $1$ kPa plus gravity — wick easily jeet jaata hai. Gravity term ka sign: $+$.
3. Case (b): condenser neeche, liquid ko 1 m gravity ke against chadna padega. Ab wick ka $1\text{kPa}$, $6.67\text{kPa}$ overcome nahi kar sakta — yeh nahi kar sakta (Capillary Action pump too weak). Sign: $-$, aur $|-6.67|>1$ ⇒ dry-out, pipe fail.
4. Microgravity resolution. Orbit mein $g\approx 0$, isliye $\Delta P_g\to 0$ aur dono orientations kaam karti hain — poora sign problem khatam ho jaata hai.

Verify: $1\text{kPa}<6.67\text{kPa}$ ⇒ uphill case mein ground failure; $1\text{kPa}>0$ ⇒ orbit success. Isliye ground tests mein pipe ko "condenser-up" orient karna zaroori hai.

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Ex 6 — Real-world sizing: reject 500 W · cells C1, C3, C9

Forecast: Area m² mein guess karo. Phir guess karo: allowed temperature 20 K badhane se area thoda shrink hoga ya bahut zyada?

1. Law ko area ke liye rearrange karo. $T_s\approx 0$ ke saath (cell C3): $A=\frac{Q}{\epsilon \sigma T^4}$. Yeh step kyun? Humein heat aur temperature limit pata hai; area unknown hai.
2. 313 K par plug in karo. $313^4=9.60\times 10^9$; $A=\frac{500}{0.85\times 5.67\times 10^{-8}\times 9.60\times 10^9}=\frac{500}{462.6}\approx 1.08{\text{m}}^2$. Yeh step kyun? Denominator $462.6{\text{W/m}}^2$ hai — har square metre 313 K par kitna radiate karta hai.
3. Limit 333 K tak badhao. $333^4=1.23\times 10^{10}$; per-m² emission $=0.85\times 5.67\times 10^{-8}\times 1.23\times 10^{10}=593{\text{W/m}}^2$; $A=500/593\approx 0.84{\text{m}}^2$. Yeh step kyun? $T^4$ leverage dikhata hai: sirf +20 K lekin area $\sim 22\%$ shrink ho jaata hai.

Verify: Areas ka ratio $=(313/333{)}^4=0.94^4=0.78$, aur indeed $0.84/1.08=0.78$ ✓. $T^4$ dependence ka matlab hai hotter run karna radiator size par ek strong lever hai — classic thermal trade.

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Ex 7 — Exam twist: hidden unit / "which is bigger" · cell C10

Forecast: Student ki galti ka size guess karo — factor of 2? 10? 1000?

1. Trap pakdo. $T^4$ ko kelvin chahiye, celsius nahi. $40°\text{C}=313\text{K}$, na ki $40$. Yeh step kyun? Stefan–Boltzmann law absolute zero se define hota hai; celsius ka origin galat hai.
2. Student ka galat number. $\sigma \times 40^4=5.67\times 10^{-8}\times 2.56\times 10^6=0.145{\text{W/m}}^2$. Yeh step kyun? Dekho yeh slip kitni catastrophic hai.
3. Sahi number. $\sigma \times 313^4=5.67\times 10^{-8}\times 9.60\times 10^9=544{\text{W/m}}^2$. Yeh step kyun? Correct emission.
4. Comparison twist. $-40°\text{C}=233\text{K}$; uski emission $=\sigma \times 233^4=5.67\times 10^{-8}\times 2.95\times 10^9=167{\text{W/m}}^2$. Usse double $334{\text{W/m}}^2$ hai. Kyunki $544>334$, $40°\text{C}$ radiator double cold wale se zyada beat karta hai. Yeh step kyun? Exam ko "is it more than double?" bahut pasand hai — $T^4$ curve ka matlab hai 80 K rise emission ko double se zyada kar deta hai.

Verify: Error factor $=544/0.145\approx 3750\times$ — ek mission-killing mistake. Aur $544/167=3.26>2$ ✓, confirm karta hai ki hot radiator cold wale se double se zyada hai. Cryogenics intuition se cross-check: cold surfaces barely radiate, exactly wahi $167$ vs $544$ gap hai.

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Recall Is page ko close karne se pehle self-test karo

Kisi matrix cell mein derivative kyun chahiye? ::: C6 (time-dependent), Ex 4 — cooling rate $dT/dt=-Q/C$. 3 K space term kab drop karna safe hai? ::: Jab $T_{\text{cold}}^4\ll T_{\text{hot}}^4$; yahan $3^4=81$ vs $\sim 10^9$, $10^{-8}$ ratio (Ex 1). MLI returns ~30 layers ke baad kyun flatten ho jaate hain? ::: ${\epsilon }_{\text{MLI}}=\epsilon /(N-1)$ $1/N$ ki tarah girta hai (weak leverage) plus compression conduction shorts (Ex 3). Ground heat-pipe rule of thumb? ::: Condenser ko evaporator ke upar rakho taaki gravity return assist kare; wick akela (~1 kPa) 1 m ammonia column (~6.7 kPa) se haar jaata hai (Ex 5b). Radiator "at 40°" — pehle kya convert karna hai? ::: Celsius → kelvin (313 K); bhoolne par ~3750× ki galti hoti hai (Ex 7).