Thermal control — multi-layer insulation (MLI), heaters, heat pipes, radiators: Exercises

3.6.23 · D4 · Physics › Spacecraft Structures & Systems Engineering › Thermal control — multi-layer insulation (MLI), heaters, hea

Yeh page tumhara self-testing gym hai thermal-control topic ke liye. Har problem apna level batata hai (L1 Recognition → L5 Mastery), phir ek complete worked solution ek collapsible callout ke andar chhupaata hai. Dekhna sirf tab jab try kar lo.

Shuru karne se pehle, woh tools pin kar lete hain jinhe hum baar baar use karte rahenge, taaki koi bhi symbol bina explanation ke na aaye.

Definition Is page ke liye toolbox

$σ = 5.67 \times 1 0^{- 8} W/m^{2} K^{4}$ — yeh Stefan–Boltzmann constant hai. Yeh woh fixed number hai jo ek temperature (kelvin mein) ko radiated power per square metre mein convert karta hai. Dekho Stefan-Boltzmann Law.
$T$ absolute temperature kelvin (K) mein hai. Exactly convert karne ke liye: $T_{K} = T_{° C} + 273.15$ . Hamesha kelvin use karna padega, kyunki law tabhi kaam karta hai jab $0$ ka matlab ho "bilkul bhi thermal motion nahi". (Inn exercises mein rounded $+ 273$ wahan use kiya gaya hai jahan isse kuch farq nahi padta.)
$ε$ (epsilon) emissivity hai: $0$ aur $1$ ke beech ka ek dimensionless number jo kehta hai "yeh surface ek perfect black body ki chamak ka kitna fraction actually emit karta hai?" Ek mirror $0$ ke paas hota hai; black paint $0.9$ ke paas.
$A$ area hai m² mein. $Q$ ek power hai (heat flow ki rate) watts (W) mein — joules per second.

Level 1 — Recognition

Problem 1.1

Ek radiator surface $T = 300 K$ par hai jisme emissivity $ε = 0.85$ hai. Ek square metre cold space mein kitni power radiate karta hai? (Space ko $0 K$ maano.)

Recall Solution

Hum kya karenge: seedha $Q = ε σ A T^{4}$ mein plug karenge $A = 1 m^{2}$ ke saath. Kyun: L1 mein sirf yeh pehchaanna hota hai ki kaun si ek equation fit hoti hai — abhi koi traps nahi. $T^{4} = 30 0^{4} = 8.1 \times 1 0^{9} K^{4}$ $Q = 0.85 \times 5.67 \times 1 0^{- 8} \times 1 \times 8.1 \times 1 0^{9} \approx 390 W$ Toh 300 K par har square metre lagbhag $390 W$ reject karta hai.

Problem 1.2

Har kaam ke liye sahi thermal-control device ka naam batao: (a) eclipse mein battery ko freeze hone se bachata hai, (b) bina kisi power ke 1 m mein 1 kW heat move karta hai, (c) 30 shiny layers se radiation block karta hai, (d) waste heat space mein dump karta hai.

Recall Solution

(a) Heater (resistive) — jab part bahut thanda ho jaaye to electrical power add karta hai. (b) Heat pipe — evaporation/condensation cycle, koi moving parts nahi. Phase Change Heat Transfer aur Capillary Action par depend karta hai. (c) MLI (multi-layer insulation) — stacked reflective shields, dekho Materials Science — Kapton & Mylar. (d) Radiator — ek high-emissivity surface jo Stefan-Boltzmann Law follow karta hai.

Level 2 — Application

Problem 2.1

Ek MLI blanket mein $N = 25$ layers hain, har layer ki emissivity $ε = 0.03$ hai. Stack ki effective emissivity $ε_{MLI}$ estimate karo.

Recall Solution

Hum kya karenge: shield-network approximation $ε_{MLI} \approx \frac{ε}{N - 1}$ use karenge. Kyun $N - 1$ aur $N$ nahi: $N$ layers ke saath $N - 1$ radiation gaps hote hain unke beech; har gap ek $T^{4}$ barrier hai jo heat ko wapas re-radiate karta hai. Zyada gaps → kam leak. $ε_{MLI} \approx \frac{0.03}{25 - 1} = \frac{0.03}{24} = 0.00125$ Toh $ε_{MLI} \approx 1.25 \times 1 0^{- 3}$ — bare $ε = 0.9$ black surface se lagbhag 700× behtar.

Problem 2.2

Woh blanket use karke, $A = 3 m^{2}$ mein se heat leak $Q$ nikalo jab andar $T_{1} = 293 K$ aur bahar $T_{2} = 120 K$ ho.

Recall Solution

Hum kya karenge: do temperatures ke beech radiation fourth powers ke difference use karta hai.

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

Figure (upar): warm inside wall (orange) bahar ki taraf ek rate par glows karti hai jo

T_{1}^{4}

se set hoti hai — woh thick magenta arrow hai jo violet MLI stack ko cross karta hai. Cold outside wall (navy) wapas andar ki taraf bahut weak rate par glows karti hai jo

T_{2}^{4}

se set hoti hai — woh thin navy arrow hai. Jo actually leak hota hai woh un dono glows ka difference hai, figure ke bottom par

Q \propto (T_{1}^{4} - T_{2}^{4})

print hua hai. Kyun difference: hot side

\propto T_{1}^{4}

power bahar bhejtaa hai, cold side

\propto T_{2}^{4}

wapas bhejtaa hai; net leak unke beech ka gap hai.

Q = ε_{MLI} σ A (T_{1}^{4} - T_{2}^{4})

T_{1}^{4} = 29 3^{4} = 7.37 \times 1 0^{9}, T_{2}^{4} = 12 0^{4} = 2.07 \times 1 0^{8}

T_{1}^{4} - T_{2}^{4} = 7.16 \times 1 0^{9}

Q = 1.25 \times 1 0^{- 3} \times 5.67 \times 1 0^{- 8} \times 3 \times 7.16 \times 1 0^{9} \approx 1.52 W

Toh sirf lagbhag

1.5 W

leak hota hai — yehi MLI ka poora point hai. Yeh Heat Transfer in Vacuum mein pure radiation hai.

Level 3 — Analysis

Problem 3.1

Ek radiator ko $Q = 500 W$ reject karna hai jisme $ε = 0.85$ hai. (a) $T = 300 K$ aur (b) $T = 350 K$ par required area compare karo. Area kitne factor se shrink hoti hai?

Recall Solution

Hum kya karenge: $Q = ε σ A T^{4}$ ko rearrange karke area solve karenge: $A = \frac{Q}{ε σ T ^{4}}$ . Kyun rearrange karein: yahan $Q$ fixed hai (electronics 500 W dump karte hain chahe kuch bhi ho), toh area free variable hai.

Figure (upar): magenta curve required area

A

ko radiator temperature

T

ke against plot karti hai, sab ke liye same fixed 500 W. Yeh steeply plunge karti hai kyunki

A

1/ T^{4}

ki tarah girta hai. Do dashed drop-lines hamare do cases mark karti hain: orange point

T = 300 K

par

A \approx 1.28 m^{2}

chahta hai, jabki violet point

T = 350 K

par sirf

A \approx 0.69 m^{2}

chahta hai — bilkul same heat load ke liye visibly chhota radiator. (a)

T = 300

T^{4} = 8.10 \times 1 0^{9}

A_{a} = \frac{500}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 8.10 \times 1 0 ^{9}} = \frac{500}{390} \approx 1.28 m^{2}

(b)

T = 350

T^{4} = 1.50 \times 1 0^{10}

A_{b} = \frac{500}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 1.50 \times 1 0 ^{10}} = \frac{500}{723} \approx 0.69 m^{2}

Factor:

A_{a} / A_{b} = 1.28/0.69 \approx 1.85

. Kyun:

T

ko 300 se 350 tak raise karna

350/300 = 1.167

ka ratio hai; fourth power tak yeh

1.16 7^{4} \approx 1.85

hai. Area exactly usi factor se shrink hoti hai. Zyada garam chalane se chhota radiator use kar sakte ho.

Problem 3.2

Ek single ammonia heat pipe $\overset{m}{˙} = 8 \times 1 0^{- 4} kg/s$ flow rate par vapour le jaata hai, latent heat $L_{v} = 1.37 \times 1 0^{6} J/kg$ . Yeh kitni power transport karta hai? Same $A = 7.85 \times 1 0^{- 5} m^{2}$ , length $1 m$ , conductivity $k = 400 W/m\cdotK$ , $Δ T = 10 K$ across wale ek solid copper rod se compare karo.

Recall Solution

Heat pipe heat ko moving vapour ki latent heat ki tarah carry karta hai: $Q = \overset{m}{˙} L_{v}$ (Phase Change Heat Transfer use karta hai). $Q_{pipe} = 8 \times 1 0^{- 4} \times 1.37 \times 1 0^{6} = 1096 W \approx 1.1 kW$ Copper rod conduction se heat carry karta hai (Fourier): $Q = \frac{k A Δ T}{L}$ . $Q_{Cu} = \frac{400 \times 7.85 \times 1 0 ^{- 5} \times 10}{1} = 0.314 W$ Ratio: $Q_{pipe} / Q_{Cu} = 1096/0.314 \approx 3500$ . Heat pipe saade teen hazaar times se jeet jaata hai per unit cross-section mein — kyunki latent heat ek slow conductive trickle se kahin zyada energy carry karta hai.

Level 4 — Synthesis

Problem 4.1

Ek $5 kg$ battery ( $c = 1000 J/kg\cdotK$ , toh heat capacity $C = 5000 J/K$ ) $20° C$ par start hoti hai aur $95$ -minute eclipse mein $Q_{loss} = 6 W$ lose karti hai. (a) Koi heater nahi to final temperature kya hogi? (b) Kya yeh apni $0° C$ limit se upar rehti hai? (c) Exactly $10° C$ par rakhne ke liye steady heater power kya chahiye?

Recall Solution

(a) Bina heater ke cooling. Battery ki stored heat ek steady rate par drain hoti hai: $\frac{d T}{d t} = - \frac{Q _{loss}}{C} = - \frac{6}{5000} = - 1.2 \times 1 0^{- 3} K/s$ Kyun $C$ se divide karein: heat capacity kehta hai "1 K temperature change ke liye kitne joules chahiye". $Q_{loss}$ joules per second drain ho rahe hain, toh temperature $Q_{loss} / C$ per second neeche slide karti hai. Eclipse time $t = 95 min = 5700 s$ : $Δ T = - 1.2 \times 1 0^{- 3} \times 5700 = - 6.84 K$ Final temperature $= 20 - 6.84 = 13.16° C$ . (b) $13.16° C > 0° C$ → haan, yeh is eclipse mein survive karti hai bina heater ke. Battery ki apni thermal mass buffer hai. (c) $10° C$ hold karne ke liye heater. Temperature constant rakhne ke liye, power in aur power out equal honi chahiye. Toh heater exactly loss replace karta hai: $P_{heater} = Q_{loss} = 6 W$ Yeh heater power Spacecraft Power Systems se li jaati hai, isliye engineers usually margin ke liye duty cycle par $10$ – $12 W$ heater specify karte hain.

Problem 4.2

Same eclipse, lekin ab battery $2° C$ par start hoti hai aur loss $10 W$ hai. Dikhao ki bina heater ke yeh freeze ho jaayegi, aur woh heater power nikalo jo isse $5700 s$ eclipse mein $0° C$ se neeche girne se roknے ke liye chahiye.

Recall Solution

Bina heater ke: $Δ T = - \frac{10}{5000} \times 5700 = - 11.4 K$ , landing $2 - 11.4 = - 9.4° C$ par — freezing se kaafi neeche, fail. Heater ke saath: starting temperature hold karne ke liye (worst case: $0° C$ se neeche mat giro, aur yeh $2° C$ par start hoti hai, toh sabse simple yeh hai ki koi net cooling na ho), heater ko kam se kam puri loss supply karni chahiye: $P_{heater} = 10 W$ Refinement: hum $0° C$ tak girne ki ijazat de sakte hain, $2 K$ ka drop buffer ke roop mein available hai. Buffer $C Δ T = 5000 \times 2 = 1 0^{4} J$ absorb karta hai $5700 s$ mein, yaani yeh $1 0^{4} /5700 = 1.75 W$ offset kar sakta hai. Toh minimum heater hai: $P_{heater,min} = 10 - 1.75 = 8.25 W$ Kyun subtraction: 2 K ka "temperature room" free stored energy hai; heater ko sirf baaki ki loss cover karni hai.

Level 5 — Mastery

Problem 5.1

Ek spacecraft bay continuously $Q = 450 W$ generate karta hai. Yeh $30$ -layer MLI ( $ε = 0.03$ per layer, wall area $A_{wall} = 4 m^{2}$ , andar $293 K$ , bahar $150 K$ ) se wrapped hai aur apni waste heat ek radiator ( $ε_{r} = 0.88$ , $0 K$ space dekh raha hai) ke through vent karta hai jo $310 K$ par rakha gaya hai. (a) MLI se kitni heat bahar leak hoti hai? (b) Baaki heat reject karne ke liye kitni radiator area chahiye? (c) Ek heat pipe 450 W ko radiator tak ammonia ke $\overset{m}{˙} = 3.5 \times 1 0^{- 4} kg/s$ ke saath deliver karta hai — kya yeh flow kaafi hai?

Recall Solution

(a) MLI leak (ek bonus heat path, chhoti). Hum kya karenge: blanket ko ek effective emissivity maano aur two-temperature radiation law use karo. Kyun: MLI warm inside (293 K) aur cold outside (150 K) ke beech baitha hai, toh iska leak Problem 2.2 ki tarah same difference-of-fourth-powers form follow karta hai. $ε_{MLI} = \frac{0.03}{30 - 1} = \frac{0.03}{29} = 1.034 \times 1 0^{- 3}$ $T_{1}^{4} - T_{2}^{4} = 29 3^{4} - 15 0^{4} = 7.37 \times 1 0^{9} - 5.06 \times 1 0^{8} = 6.87 \times 1 0^{9}$ $Q_{MLI} = 1.034 \times 1 0^{- 3} \times 5.67 \times 1 0^{- 8} \times 4 \times 6.87 \times 1 0^{9} \approx 1.61 W$ Toh MLI sirf lagbhag $1.6 W$ leak karta hai — 450 W ke next negligible. (b) Radiator area. Hum kya karenge: single-surface radiation law $Q = ε_{r} σ A T_{r}^{4}$ ko $A = Q / (ε_{r} σ T_{r}^{4})$ par rearrange karo. Kyun yahan $Q = ε σ A T^{4}$ apply karein: radiator ek hot surface hai jo 0 K space mein glow kar raha hai, toh plain single-surface form (koi back-radiation term nahi) exactly sahi tool hai. Isse net heat reject karni hai jo pipe deliver karta hai. MLI leak (1.6 W) already thoda chhota amount shed kar leta hai, toh radiator $450 - 1.6 \approx 448.4 W$ handle karta hai. (Hum conservatively pura 450 W use kar sakte hain.) $T_{r}^{4} = 31 0^{4} = 9.24 \times 1 0^{9}$ $A_{r} = \frac{448.4}{0.88 \times 5.67 \times 1 0 ^{- 8} \times 9.24 \times 1 0 ^{9}} = \frac{448.4}{461} \approx 0.97 m^{2}$ Toh $A_{r} \approx 0.97 m^{2}$ . (c) Heat-pipe check. Hum kya karenge: pipe ki carrying power $Q = \overset{m}{˙} L_{v}$ se compute karo aur 450 W load se compare karo. Kyun: pipe heat ko flowing ammonia vapour ki latent heat ki tarah transport karta hai, toh iska capacity sirf mass flow times latent heat hai. $L_{v} = 1.37 \times 1 0^{6} J/kg$ ke saath: $Q_{pipe} = \overset{m}{˙} L_{v} = 3.5 \times 1 0^{- 4} \times 1.37 \times 1 0^{6} = 479.5 W$ Kyunki $479.5 W > 450 W$ hai, haan — flow load carry karta hai lagbhag $30 W$ ke margin ke saath.

Problem 5.2 — Reasoning

Ground par, ek engineer yahi ammonia heat pipe vertically test karta hai, condenser evaporator ke $0.5 m$ neeche ke saath. Wick maximum capillary pressure $Δ P_{cap} = 1500 Pa$ pump kar sakta hai. Ammonia liquid density $ρ \approx 600 kg/m^{3}$ , $g = 9.81 m/s^{2}$ . Kya pipe is orientation mein kaam karega?

Recall Solution

Hum kya karenge: wick ki maximum pumping pressure ko gravity head se compare karo jo returning liquid ko climb karna padega. Kyun: liquid ko condenser se wapas evaporator tak push karna padta hai. Ek hi pump hai capillary suction; yahan ek hi opposing force hai gravity. Agar gravity ka pressure demand wick ki supply se jeet jaaye, toh wick dry out ho jaata hai. Gravity head height $h = 0.5 m$ par: $Δ P_{grav} = ρ g h = 600 \times 9.81 \times 0.5 = 2943 Pa$ Compare: $Δ P_{cap} = 1500 Pa < Δ P_{grav} = 2943 Pa$ Verdict: kyunki capillary pumping ( $1500 Pa$ ) gravity head ( $2943 Pa$ ) se kam hai, wick liquid ko wapas upar nahi utha sakta. Evaporator starve ho jaata hai, pipe is orientation mein dry out ho ke fail ho jaata hai. $The pipe does NOT work.$ Orbit mein kyun kaam karta hai: microgravity mein $g \approx 0$ hota hai, toh $Δ P_{grav} \to 0$ aur modest capillary pressure (Capillary Action se) kisi bhi orientation mein aasaani se liquid wapas le aata hai. Yehi wajah hai ki ground testing mein gravity ka dhyan rakhna zaroori hai.

Recall Quick self-check reveals

300 K, ε=0.85 par Radiator power per m² ::: lagbhag 390 W 25-layer MLI (ε=0.03) ki effective emissivity ::: 0.03/24 ≈ 0.00125 Kyun hum $T_{1}^{4} - T_{2}^{4}$ use karte hain na ki $(T_{1} - T_{2})^{4}$ ::: har surface apne $T^{4}$ ke proportional power radiate karta hai; net leak un powers ka difference hai Heat pipe jo heat carry karta hai ::: $Q = \overset{m}{˙} L_{v}$ (mass flow × latent heat) Kyun heat pipes ground par fail ho sakte hain ::: gravity head $ρ g h$ wick ki capillary pressure se zyada ho sakta hai