Environmental testing — thermal vacuum (TVAC), vibration, acoustic, EMC - EMI: Exercises

3.6.33 · D4 · Physics › Spacecraft Structures & Systems Engineering › Environmental testing — thermal vacuum (TVAC), vibration, ac

Shuru karne se pehle, woh do symbols dobara explain karte hain jinpar hum sabse zyada rely karte hain, taaki aap unhe kabhi unfamiliar na paao.

Definition Do workhorse laws (zero se dobara bataye gaye)

Radiated power per unit area: $q = σ ε T^{4}$ .

$q$ = watts jo ek warm surface ke har square metre se light/infrared ke roop mein nikal rahe hain.
$σ = 5.67 \times 1 0^{- 8} W/(m^{2} K^{4})$ nature ka ek fixed constant hai.
$ε$ (emissivity) ek dial hai $0$ se (shiny, glow karne se inkar) lekar $1$ tak (perfect glower).
$T$ temperature hai kelvin (K) mein: kelvin matlab celsius $+ 273.15$ , toh $0$ K ka matlab hai "bilkul bhi thermal wiggling nahi".

Natural frequency of a mass on a spring: $f_{n} = \frac{1}{2 π} \frac{k}{m}$ .

$m$ = mass kg mein (kitna mushkil hai dhakka dena).
$k$ = stiffness N/m mein (spring kitna hard pull back karta hai per metre stretch pe).
$f_{n}$ = woh frequency, cycles-per-second (hertz, Hz) mein, jis par cheez bounce karna chahti hai agar aap use ek baar flick karo.

Level 3 se pehle neeche diya figure zaroor dekho — yahi picture har vibration problem ke peeche hai.

Space mein koi air nahi hai. Teen heat-transfer routes mein se — conduction, convection, radiation — orbit par kaun sa route band ho jaata hai, aur kaun sa route woh akeela tarika ban jaata hai jisse spacecraft waste heat ko deep space mein dump karta hai?

Recall Solution — L1.1

WHAT: Band hone wala route aur survivor batao. WHY: Convection ke liye ek fluid chahiye (air, water) jo heat ko physically move karke le jaaye. Vacuum mein koi fluid move karne ke liye hota hi nahi, toh convection khatam. Conduction abhi bhi heat ko solid structure ke through craft ke dusre hisson tak move karta hai, lekin woh heat eventually craft ko chhodni padti hai — aur empty space mein heat shed karne ka ek hi tarika hai: use radiation ke roop mein glow kar do. Answer: Convection band hoti hai; radiation deep space ka akeela route hai.

L1.2

Law $q = σ ε T^{4}$ mein, kaun si ek quantity hai jise double karne par $q$ sabse zyada badlega, aur $q$ kitne factor se badlega jab woh double ho?

Recall Solution — L1.2

WHAT: Fourth-power term dhundo. WHY: $σ$ fixed hai. $ε$ first power par aata hai, toh use double karne se $q$ double ho jaata hai. Lekin $T$ , $T^{4}$ ke roop mein aata hai, toh $T$ double karne par $q$ , $2^{4} = 16$ se multiply ho jaata hai. Answer: ==Temperature $T$ ==; use double karne par radiated power $16$ guna ho jaati hai.

Level 2 — Application

L2.1

Ek radiator plate $T = 300 K$ par hai jiska emissivity $ε = 0.85$ aur area $A = 0.5 m^{2}$ hai. Space ke $3$ K background ko ignore karte hue, yeh kitni power (watts) radiate karta hai?

Recall Solution — L2.1

WHAT: $Q = A σ ε T^{4}$ mein plug karo. WHY background drop kiya gaya hai: deep space $\approx 3$ K hai, aur $3^{4} = 81$ versus $30 0^{4} = 8.1 \times 1 0^{9}$ — cold side ka contribution ek sau million mein ek part jaisa hai, toh hum use ignore karte hain. $Q = 0.5 \times 5.67 \times 1 0^{- 8} \times 0.85 \times 30 0^{4}$ $Q = 0.5 \times 5.67 \times 1 0^{- 8} \times 0.85 \times 8.1 \times 1 0^{9} \approx 195.2 W$ Answer: $\approx 195 W$ .

L2.2

Ek component jiska mass $m = 0.5 kg$ hai, total stiffness $k = 2 \times 1 0^{6} N/m$ wale mounts se bolt kiya gaya hai. Uski natural frequency $f_{n}$ Hz mein nikalo.

Recall Solution — L2.2

WHAT: $f_{n} = \frac{1}{2 π} k / m$ use karo. WHY yahi formula aur sirf $k$ nahi: mass acceleration ko resist karta hai, toh heavier part slower bounce karta hai; spring pull back karta hai, toh stiffer mount faster bounce karaata hai. Ratio $k / m$ dono ko capture karta hai, aur square root "energy-like" quantities ko frequency mein convert karta hai. $f_{n} = \frac{1}{2 π} \frac{2 \times 1 0 ^{6}}{0.5} = \frac{1}{2 π} 4 \times 1 0^{6} = \frac{2000}{2 π} \approx 318.3 Hz$ Answer: $\approx 318 Hz$ .

Level 3 — Analysis

Figure — Environmental testing — thermal vacuum (TVAC), vibration, acoustic, EMC - EMI

Upar wala curve transmissibility $T (r)$ hai: input ki tulna mein shaken output kitna bada hai, frequency ratio $r = ω / ω_{n}$ ke function ke roop mein. Peak $r = 1$ par hai (resonance); dashed line $r = 2$ wahan hai jahan amplification khatam hoti hai aur isolation shuru hoti hai.

L3.1

Ek structure mein damping ratio $ζ = 0.02$ hai. Resonance par amplification factor $Q = \frac{1}{2 ζ}$ hota hai. Agar shaker input $1.0 g$ hai, toh resonance par component actually kitna acceleration feel karta hai?

Recall Solution — L3.1

WHAT: $Q$ compute karo, input se multiply karo. WHY $Q = 1/ (2 ζ)$ : damping $ζ$ woh fraction hai jitni "bounce energy" har cycle mein bleed hoti hai. Tiny damping ( $ζ = 0.02$ ) matlab almost kuch bhi bleed nahi hota, toh energy cycle-by-cycle pile up hoti rehti hai jab tak response bahut bada na ho jaaye — pile-up factor exactly $1/ (2 ζ)$ hota hai. $Q = \frac{1}{2 \times 0.02} = 25$ $a_{felt} = Q \times a_{in} = 25 \times 1.0 = 25 g$ Answer: == $25 g$ == — ek benign $1$ g test input part par ek violent $25$ g ban jaata hai. Isliye hi $f_{n}$ locate karne ke liye sine sweep low level par ki jaati hai.

L3.2

$T (r) = \frac{1}{( 1 - r ^{2} ) ^{2} + ( 2 ζ r ) ^{2}}$ use karte hue $ζ = 0.02$ ke saath, $T$ evaluate karo (a) $r = 0.5$ par, (b) $r = 2$ par, aur confirm karo ki har ek region kaunsa represent karta hai (amplification, unity-crossover, isolation).

Recall Solution — L3.2

WHAT: Dono ratios substitute karo. (a) $r = 0.5$ (resonance se neeche): $T = \frac{1}{( 1 - 0.25 ) ^{2} + ( 2 \cdot 0.02 \cdot 0.5 ) ^{2}} = \frac{1}{0.5625 + 0.0004} \approx 1.333$ Thoda $1$ se upar — mild amplification, jaise resonance se neeche expected hai. (b) $r = 2$ (crossover): $T = \frac{1}{( 1 - 2 ) ^{2} + ( 2 \cdot 0.02 \cdot 2 ) ^{2}} = \frac{1}{1 + 0.0032} \approx 0.998 \approx 1$ WHY $r = 2$ special hai: damping ignore karte hue, wahan $(1 - r^{2})^{2} = 1$ hota hai, toh $T = 1$ exactly — mount na amplify karta hai na protect karta hai. Iske neeche aap zyada shake hote ho; iske upar mass fast shaking ke saath "keep up nahi kar paata" aur quieter ride karta hai (isolation). Answer: (a) $T \approx 1.33$ (amplification region); (b) $T \approx 1.0$ (crossover — isolation yahan se aage shuru hoti hai).

Level 4 — Synthesis

L4.1

Ek flat solar panel jiska area $A = 1 m^{2}$ hai, Sun ki taraf face karta hai. Yeh sunlight ko absorptivity $α = 0.90$ ke saath solar constant $S = 1368 W/m^{2}$ se absorb karta hai, aur emissivity $ε = 0.80$ ke saath dono faces se radiate karta hai. Uska equilibrium temperature nikalo.

Recall Solution — L4.1

WHAT: Power in ko power out ke saath balance karo aur $T$ ke liye solve karo. WHY balance: equilibrium par panel na warm ho raha hai na cool, toh watts absorbed = watts radiated. Power in (sirf sunlit face collect karta hai): $P_{in} = α S A = 0.90 \times 1368 \times 1 = 1231.2 W$ . Power out (dono faces glow karte hain, isliye factor $2$ ): $P_{out} = 2 A σ ε T^{4}$ . Equal set karo: $1231.2 = 2 \times 1 \times 5.67 \times 1 0^{- 8} \times 0.80 \times T^{4}$ $T^{4} = \frac{1231.2}{2 \times 5.67 \times 1 0 ^{- 8} \times 0.80} = \frac{1231.2}{9.072 \times 1 0 ^{- 8}} = 1.357 \times 1 0^{10}$ $T = (1.357 \times 1 0^{10})^{1/4} \approx 341.4 K \approx 68. 3^{\circ} C$ Answer: $\approx 341 K (\approx 6 8^{\circ} C)$ — itna hot ki TVAC ko confirm karna padega ki coatings/heat-paths use safe band mein rakhen.

L4.2

Ek random-vibration spectrum $50$ – $800$ Hz band par flat hai jiska PSD $G = 0.04 g^{2} / Hz$ hai. Sirf is band mein poore level ko flat approximate karte hue, RMS acceleration $g_{rms} = \int G df$ estimate karo.

Recall Solution — L4.2

WHAT: Flat PSD ke liye, integral sirf level $\times$ bandwidth hai. WHY square root: PSD $G$ ke units $g^{2} / Hz$ hain — yeh power (amplitude-squared) per hertz hai. Band par power add karne se total mean-square ( $g^{2}$ ) milta hai; RMS woh square root hai jo hame wapas $g$ mein le jaata hai. Bandwidth $= 800 - 50 = 750 Hz$ . $\int G df = 0.04 \times 750 = 30 g^{2}$ $g_{rms} = 30 \approx 5.48 g$ Answer: $\approx 5.5 g$ (sloped ends wala full GEVS spectrum $\approx 7.7$ g total hota hai; flat plateau akela hi zyaadatar energy contribute karta hai).

Level 5 — Mastery

L5.1

Ek cubesat electronics box ka weight $m = 0.8 kg$ hai. Mission rules require karti hain ki uski pehli natural frequency kam se kam $f_{min} = 140 Hz$ ho (taaki woh low-frequency launch loads se clear rahe). (a) Minimum total mounting stiffness $k$ kitni chahiye? (b) Us $f_{n}$ par, damping $ζ = 0.05$ ke saath, resonant amplification $Q$ kitna hai? (c) Agar us frequency ke paas flat random input $0.04 g^{2} / Hz$ hai, toh kya resonance ko flat band ke andar rakhna achha idea hai ya bura — $Q$ aur transmissibility use karke argue karo.

Recall Solution — L5.1

(a) WHAT: $f_{n} = \frac{1}{2 π} k / m$ ko $k$ ke liye invert karo. WHY: hume frequency batayi gayi hai aur hum woh stiffness chahte hain jo use guarantee kare. $k = m (2 π f_{n})^{2} = 0.8 \times (2 π \times 140)^{2} = 0.8 \times (879.6)^{2} \approx 6.19 \times 1 0^{5} N/m$ Answer (a): $k ≳ 6.19 \times 1 0^{5} N/m$ .

(b) WHAT: $Q = 1/ (2 ζ)$ . $Q = \frac{1}{2 \times 0.05} = 10$ Answer (b): $Q = 10$ .

(c) WHAT: judge karo ki resonance ka flat band mein baithna ( $140$ Hz, $50$ – $800$ Hz ke andar hai) acceptable hai ya nahi. WHY / reasoning: $Q = 10$ ke saath, resonance par box local input amplitude ka $10 \times$ feel karta hai, aur kyunki woh frequency flat $0.04 g^{2} / Hz$ plateau ke andar hai, shaker $140$ Hz par seedha full broadband energy daalega. Yahi intended philosophy hai: aap chahte ho ki test resonance ko hit kare taaki koi weak solder joint ground par hi fail ho jaaye. Mission ka $140$ Hz floor test dodge karne ke liye nahi hai — yeh ensure karne ke liye hai ki $f_{n}$ launch vehicle ke own low-frequency transients se upar rahe, jabki $Q = 10$ ( $ζ = 0.05$ ki wajah se, $0.02$ nahi) amplification ko survivable rakhta hai. Toh: resonance ko band mein rakhna theek aur deliberate hai, provided damping $Q$ ko modest rakhti ho. Answer (c): Acceptable hai — band placement guarantee karta hai ki test mode ko excite kare; design safety adequate damping se aati hai ( $Q = 10$ ) aur is baat se ki $f_{n}$ launch transients se clear karne ke liye itna high hai, resonance chhupane se nahi.

Connections: Spacecraft Thermal Control Systems · Structural Mechanics · Launch Vehicle Dynamics · Reliability Engineering · Quality Assurance in Aerospace · Electromagnetic Wave Propagation

Exercises — Environmental testing — thermal vacuum (TVAC), vibration, acoustic, EMC - EMI

Level 1 — Recognition

L1.1