Thermal analysis — conduction in structures, thermal stress

3.6.14 · Physics › Spacecraft Structures & Systems Engineering

Overview

Spacecraft orbit mein extreme temperature gradients experience karte hain — sunlight mein +120°C se shadow mein -150°C tak. Structural members ke through thermal conduction temperature distributions create karta hai jo thermal stress induce karte hain, jisse solar panels warp ho sakte hain, antennas misalign ho sakte hain, ya materials crack bhi kar sakte hain. Yeh note governing equations ko first principles se derive karta hai aur dikhata hai ki spacecraft structures mein temperature fields aur stress fields dono ko kaise predict karein.

Core Concepts

Intuition Spacecraft ke liye Thermal Analysis Kyun Zaroori Hai

Ek metal beam imagine karo: ek end Sun ki taraf hai (+100°C), doosra end deep space ki taraf (-100°C). Hot end expand karna chahta hai, cold end contract karna chahta hai. Agar beam constrained hai (rigid frame mein bolt ki gayi hai), toh woh freely move nahi kar sakti — isliye internal forces build up hote hain. Yeh forces hi thermal stresses hain. Spacecraft mein:

Yeh optical benches ko micrometers se bend kar sakte hain → telescope pointing errors
Yeh thousands of thermal cycles mein joints ko fatigue karte hain (har 90 min orbit)
Structures ko sahi size karne ke liye design ke dauran inhe predict karna zaroori hai

Key insight: Heat flow (conduction) temperature distribution set karta hai. Temperature distribution expansion drive karti hai. Constrained expansion stress create karti hai.

Derivation: 1D Steady-State Conduction

Goal: Length $L$ ki ek rod mein temperature distribution $T (x)$ find karo, jahan $x = 0$ par $T_{1}$ aur $x = L$ par $T_{2}$ fixed temperatures hain.

Step 1: Conservation of Energy

Position $x$ par thickness $d x$ ka ek thin slice consider karo. Steady state mein (koi time variation nahi), energy in = energy out:

$q (x) \cdot A = q (x + d x) \cdot A$

jahan $A$ cross-sectional area hai. $q (x + d x) \approx q (x) + \frac{d q}{d x} d x$ expand karne par:

$q (x) = q (x) + \frac{d q}{d x} d x ⟹ \frac{d q}{d x} = 0$

Yeh step kyun? Steady state ka matlab hai slice mein koi energy accumulation nahi. Flux ka derivative zero hona chahiye.

Step 2: Fourier's Law Apply Karo

1D mein, $q = - k \frac{d T}{d x}$ . Substitute karo:

$\frac{d}{d x} (- k \frac{d T}{d x}) = 0$

Constant $k$ ke liye:

$\frac{d ^{2} T}{d x ^{2}} = 0$

Kyun? Second derivative ka zero hone ka matlab hai $T (x)$ , $x$ mein linear hai.

Step 3: Integrate Karo

$\frac{d T}{d x} = C_{1} ⟹ T (x) = C_{1} x + C_{2}$

Step 4: Boundary Conditions

$x = 0$ par: $T (0) = T_{1} ⟹ C_{2} = T_{1}$
$x = L$ par: $T (L) = T_{2} ⟹ C_{1} L + T_{1} = T_{2} ⟹ C_{1} = \frac{T _{2} - T _{1}}{L}$

Derivation: Constrained Rod mein Thermal Stress

Setup: Length $L$ ki ek rod, uniformly $T_{0}$ se $T_{0} + Δ T$ tak heat ki gayi, dono ends fixed hain (move nahi kar sakti).

Step 1: Free Thermal Expansion

Agar unconstrained hoti, toh rod itni elongate hoti:

$Δ L = α L Δ T$

Kyun? Zyaadatar materials temperature ke saath linearly expand karte hain: $\frac{Δ L}{L} = α Δ T$ ( $α$ ki definition).

Step 2: Constraint Force

Rod ko original length $L$ par hold kiya gaya hai. Expansion $Δ L$ ko rokne ke liye, humein ek compressive force $F$ apply karni hogi jo elastic strain cause kare:

$ε_{elastic} = \frac{Δ L}{L} = α Δ T$

Lekin elastic strain $ε = \frac{σ}{E}$ hai (Hooke's Law). Toh:

$\frac{σ}{E} = α Δ T$

Step 3: Non-Uniform Temperature

Agar $T = T (x)$ hai, toh har slice ka expansion alag hoga. Local thermal strain:

$ε_{thermal} (x) = α [T (x) - T_{0}]$

Agar rod constrained hai, toh total strain $ε_{total} = ε_{elastic} + ε_{thermal}$ hai. Statically determinate structure mein:

$ε_{total} = 0 ⟹ ε_{elastic} = - ε_{thermal}$

$σ (x) = E ε_{elastic} = - E α [T (x) - T_{0}]$

Linear temperature profile $T (x) = T_{1} + (T_{2} - T_{1}) x / L$ ke liye:

$σ (x) = - E α [\frac{( T _{2} - T _{1} ) x}{L} + (T_{1} - T_{0})]$

$σ (0) = - E α (T_{1} - T_{0})$ se $σ (L) = - E α (T_{2} - T_{0})$ tak linear stress distribution.

Worked Examples

Worked example Example 1: Sun-Shadow Transition mein Aluminum Beam

Problem: Ek aluminum beam ( $k = 200$ W/m·K, $α = 23 \times 1 0^{- 6}$ /K, $E = 70$ GPa) ki length $L = 1$ m hai, jo ek sunlit panel ( $T_{1} = 100° C$ ) ko shadowed radiator ( $T_{2} = - 50° C$ ) se connect karti hai. Find karo:

Midpoint par temperature
Heat flux
Maximum thermal stress (dono ends fixed) Solution:
Midpoint temperature: $T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x$ use karte hue: $T (0.5) = 100 + \frac{- 50 - 100}{1} (0.5) = 100 - 75 = 25° C$ Yeh step kyun? Linear profile → midpoint endpoints ka average hai.
Heat flux: $q = - k \frac{T _{2} - T _{1}}{L} = - 200 \frac{- 50 - 100}{1} = 200 \times 150 = 30, 000 W/m^{2} = 30 kW/m^{2}$ Positive kyun? Heat hot (x=0) se cold (x=L) ki taraf flow karta hai, +x direction mein.
Thermal stress: $x = 0$ par (sabse hot), agar hum $T_{0} = 25° C$ (reference) lete hain: $σ (0) = - E α (T_{1} - T_{0}) = - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times (100 - 25)$ $= - 1.61 \times 1 0^{6} \times 75 = - 121 MPa (compression)$

$x = L$ par (sabse cold): $σ (L) = - 70 \times 1 0^{9} \times 23 \times 1 0^{- 6} \times (- 50 - 25) = + 121 MPa (tension)$

Kyun? Hot end expand karne ki koshish karta hai → compression. Cold end contract karne ki koshish karta hai → tension.

Worked example Example 2: Spacecraft Strut Thermal Cycle

Problem: Ek titanium strut ( $α = 8.6 \times 1 0^{- 6}$ /K, $E = 110$ GPa, length 0.5 m) har 90 minutes mein -100°C aur +50°C ke beech cycle karta hai. Agar fully constrained hai, toh stress range kya hai?

Solution:

Stress range $Δ σ = E α Δ T$ : $Δ T = 50 - (- 100) = 150 K$ $Δ σ = 110 \times 1 0^{9} \times 8.6 \times 1 0^{- 6} \times 150 = 142 MPa$

Yeh kyun important hai? Stress range fatigue life drive karta hai. Aluminum alloys typically ~100 MPa range par $1 0^{6}$ cycles ke baad fail ho jaate hain. Yeh strut har orbit mein ek thermal cycle experience karta hai. 90-minute orbit ke saath: $\frac{1440 min/day}{90 min/cycle} = 16$ cycles/day → $16 \times 365 \approx 5, 840$ cycles/year. 10-year mission ke liye: $\approx 5.8 \times 1 0^{4}$ cycles → yeh phir bhi ek significant fatigue concern hai. Designer ko chahiye:

Low-CTE materials use karo (CFRP: $α \approx 0$ )
Compliance add karo (flexible mounts)
Pre-stress karo taaki stress hamesha compressive rahe (koi tension reversal nahi)

Common Mistakes

Active Recall Questions

#flashcards/physics

What is Fourier's Law of heat conduction? :: $q = - k \nabla T$ , jahan $q$ heat flux hai, $k$ thermal conductivity hai, aur $\nabla T$ temperature gradient hai. Heat hot se cold ki taraf flow karta hai.

Why does thermal stress arise in a heated, constrained rod?

Rod

α L Δ T

se expand karna chahti hai, lekin constraints is expansion ko prevent karte hain. Expansion ko suppress karne ke liye, internal forces develop hote hain, jo stress

σ = E α Δ T

create karte hain.

For a rod with ends at

T_{1}

and

T_{2}

, what is the steady-state temperature distribution?

T (x) = T_{1} + \frac{T _{2} - T _{1}}{L} x

(linear). Steady state mein constant thermal conductivity ke saath

\frac{d ^{2} T}{d x ^{2}} = 0

se follow karta hai.

If a titanium strut (

α = 8.6 \times 1 0^{- 6}

/K,

E = 110

GPa) is heated by 100K while fully constrained, what is the thermal stress?

σ = E α Δ T = 110 \times 1 0^{9} \times 8.6 \times 1 0^{- 6} \times 100 = 94.6

MPa (compression).

How does thermal stress scale with material properties?

σ \propto E

(stiffness),

σ \propto α

(CTE),

σ \propto Δ T

(temperature change). Stress reduce karne ke liye: low-

E

materials use karo (flexible), low-

α

materials (CFRP, Invar), ya

Δ T

reduce karo (insulation, coatings).

What is the difference between steady-state and transient thermal conduction?

Steady-state:

\frac{\partial T}{\partial t} = 0

\nabla^{2} T = 0

(Laplace). Transient:

\frac{\partial T}{\partial t} \neq = 0

\frac{\partial T}{\partial t} = \frac{k}{ρ c _{p}} \nabla^{2} T

(diffusion equation). Spacecraft thermal cycling transient hoti hai.

Why is thermal stress a fatigue concern in spacecraft?

Spacecraft har orbit (~90 min) mein thermal cycles experience karte hain. Stress range

Δ σ = E α Δ T

100+ MPa ho sakta hai. Tens of thousands of cycles ke baad, yeh fatigue crack growth aur eventual failure cause karta hai.

Memory Aids

Recall Ek 12-Saal-Ke Bacche Ko Explain Karo

Ek metal ruler imagine karo jo table par rakhi hai. Tum ek end ko candle se heat karte ho aur doosre end ko ice se thanda karte ho. Hot end lambi hona chahti hai (atoms zyaada vibrate karte hain, alag ho jaate hain), cold end chhoti hona chahti hai. Lekin ruler ek piece hai — woh aadhi-aadhi nahi ho sakti!

Toh hot end cold end ke against push karta hai, aur cold end wapas pull karta hai. Yeh push aur pull hi thermal stress hai. Agar tum ruler ko bahut zyaada heat karo, toh woh in internal forces se bend ya crack bhi ho sakti hai.

Ek spacecraft mein, imagine karo ek metal beam jo ek solar panel (sunlight mein super hot) ko ek radiator (shadow mein super cold) se connect karti hai. Waahi cheez hoti hai: beam ek saath expand aur contract karne ki koshish kar rahi hoti hai, jo stress create karta hai. Engineers ko ensure karna hota hai ki beam itni strong ho, ya woh special materials use karte hain jo zyaada expand nahi karte (jaise carbon fiber), taaki stress kam rahe.

Key baat: Heat cheezein expand karti hai. Agar tum unhe freely expand nahi karne dete, toh woh literally stressed ho jaate hain!

Connections

Thermal Environment in Orbit — solar flux, albedo, Earth IR, $T_{1}, T_{2}$ ke liye boundary conditions define karte hain
Material Selection for Spacecraft — $α$ , $k$ , $E$ thermal-structural performance drive karte hain
Finite Element Analysis — real structures ki 3D geometry hoti hai; FEA $\nabla \cdot (k \nabla T) = 0$ aur coupled thermal-stress solve karta hai
Thermal Control Subsystems — heaters, radiators, MLI $Δ T$ manage karne aur thermal stress reduce karne ke liye use hote hain
Structural Dynamics — thermal stress natural frequencies shift kar sakta hai, jo vibration response affect karta hai
Fatigue and Fracture Mechanics — thermal cycling high-cycle fatigue cause karta hai; crack growth predict karo
Composite Materials in Spacecraft — CFRP ka near-zero ya negative $α$ hota hai → "athermal" structures
Deployable Structures — hinges aur booms high thermal gradients experience karte hain; stress mechanisms jam kar sakta hai

Summary

Spacecraft structures mein thermal conduction Fourier's Law follow karta hai, temperature gradients produce karta hai jo thermal stress $σ = E α Δ T$ drive karte hain jab expansion constrained hoti hai. 1D steady-state ke liye, temperature boundaries ke beech linear hoti hai. Stress wahan maximum hota hai jahan $Δ T$ sabse zyaada hota hai. Key design levers: material CTE ( $α$ ), stiffness ( $E$ ), thermal isolation ( $Δ T$ reduce karo), aur compliance (expansion allow karo). Thermal cycling fatigue cause karta hai — long-duration missions ke liye critical. Hamesha constraint conditions check karo: free structures bina stress ke expand karte hain, fixed structures mein high stress develop hoti hai.