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

3.6.21 · Physics › Spacecraft Structures & Systems Engineering

1. Structure Subsystem

Purpose

Primary structure provide karta hai:

Baaki saare subsystems ke liye mechanical support
Launch ke dौران load path (high g-forces, vibration)
Launch vehicle se attachment payload adapter ke zariye

Key Design Drivers

Worked example Example: CubeSat Structure Sizing

Diya hua hai:

Total mass: $m = 4$ kg (3U CubeSat)
Launch acceleration: $a = 10 g = 98$ m/s²
Aluminum 6061-T6: $σ_{y} = 240$ MPa, $ρ = 2700$ kg/m³ Dhundna hai: Ek 10 cm × 10 cm square cross-section frame ke liye minimum wall thickness.

Step 1: Axial load calculate karo. $F = 4 \times 98 = 392 N$ Yeh step kyun? Launch sabse harsh mechanical environment hota hai; structure ko bina buckle kiye survive karna chahiye.

Step 2: Load-bearing walls ka required cross-sectional area nikalo. $A_{req} = \frac{392}{240 \times 1 0 ^{6}} = 1.63 \times 1 0^{- 6} m^{2} = 1.63 mm^{2}$ Yeh step kyun? Yield stress ke neeche rehne ke liye yeh minimum material area hai.

Step 3: 4 walls mein distribute karo (har ek 10 cm lambi). $A_{per wall} = \frac{1.63}{4} = 0.41 mm^{2}$ Wall thickness: $t = \frac{0.41}{100} = 0.004$ mm Reality check: Yeh bilkul absurdly thin hai! Real CubeSats 1–2 mm walls use karte hain kyunki:

Vibration modes ke liye stiffness chahiye (sirf strength nahi)
Handling, thermal cycling, aur safety factors requirement ko 10–20× multiply kar dete hain

2. Electrical Power System (EPS)

Function

Saare subsystems ko electrical power generate, store, regulate, aur distribute karna.

Components

Power Budget

Formula Orbit Mein Energy Balance

Derivation: LEO mein ek spacecraft day/night cycles experience karta hai. Ek orbit period $T_{orbit}$ ke upar energy balance karna zaroori hai.

Daytime phase (duration $T_{sun}$ ):

Solar array power generate karta hai $P_{SA}$
Load consume karta hai $P_{load}$
Excess battery charge karta hai: $P_{charge} = P_{SA} - P_{load}$

Eclipse phase (duration $T_{eclipse} = T_{orbit} - T_{sun}$ ):

Battery $P_{load}$ par discharge hoti hai

Poore orbit par energy balance — daylight mein battery mein store ki gayi energy (charging losses ke baad) eclipse ke dौران draw ki gayi energy ke barabar honi chahiye (discharge losses ke saath): $E_{charge} = E_{discharge}$

jahan:

$E_{charge} = P_{charge} \cdot T_{sun} \cdot η_{charge}$ = battery mein daali gayi usable energy
$E_{discharge} = \frac{P _{load} \cdot T _{eclipse}}{η _{discharge}}$ = battery se nikalni padhne wali energy (discharge inefficiency including)

Barabar set karne par: $P_{charge} \cdot T_{sun} \cdot η_{charge} = \frac{P _{load} \cdot T _{eclipse}}{η _{discharge}}$

Ek common simplification assume karti hai ki array isliye sized hota hai ki daytime generation sirf load aur recharge dono cover kare, jo standard sizing form deta hai: $P_{SA} = P_{load} (1 + \frac{T _{eclipse}}{T _{sun} \cdot η _{charge} \cdot η _{discharge}})$

Typical Li-ion efficiencies: $η_{charge} = 0.9$ , $η_{discharge} = 0.95$

Worked example Example: LEO Satellite ke liye Power Sizing

Diya hua hai:

Orbit: 500 km altitude, $T_{orbit} = 95$ min
Eclipse fraction: 35% → $T_{eclipse} = 33$ min, $T_{sun} = 62$ min
Average load: $P_{load} = 50$ W
Battery efficiency: round-trip 85% ( $η_{charge} \times η_{discharge} = 0.85$ ) Dhundna hai: Required solar array power.

Step 1: Charge/discharge ratio calculate karo. $\frac{T _{eclipse}}{T _{sun}} = \frac{33}{62} = 0.532$ Yeh step kyun? Array ko sunlight ke dौران itna surplus generate karna hai ki eclipse ka deficit cover ho sake.

Step 2: Energy balance apply karo. $P_{SA} = 50 (1 + \frac{0.532}{0.85}) = 50 \times 1.626 = 81.3 W$ Yeh step kyun? Yeh battery losses aur unequal day/night durations ko account karta hai.

Step 3: Degradation ke liye margin add karo. Solar cells LEO mein radiation damage se ~2.5% per year degrade hoti hain. 5-year mission ke liye: $P_{SA,BOL} = \frac{81.3}{( 1 - 0.025 \times 5 )} = \frac{81.3}{0.875} = 92.9 W$ BOL = Beginning of Life

3. Thermal Control System (TCS)

Purpose

Saare components ko operational temperature limits ke andar rakho (typically electronics ke liye −20°C se +50°C).

Heat Sources & Sinks

Formula Thermal Balance Equation

Derivation: Space mein spacecraft ke paas convection nahi hoti. Sirf radiation aur internal dissipation matter karti hai.

Steady state par energy balance: $Q_{in} = Q_{out}$

Heat inputs:

Solar flux: $Q_{solar} = α \cdot A_{proj} \cdot S$
- $α$ = absorptivity (0–1)
- $A_{proj}$ = Sun ki taraf projected area
- $S$ = solar constant (1361 W/m² at Earth)
Albedo (Earth se reflected sunlight): $Q_{albedo} = α \cdot A \cdot a \cdot S$
- $a$ = Earth ka albedo ≈ 0.3
Earth IR emission: $Q_{Earth} = ϵ \cdot A \cdot E_{Earth}$
- $E_{Earth}$ ≈ 237 W/m²
Internal dissipation: $Q_{dissip}$ (electronics se waste heat)

Heat outputs: $Q_{out} = ϵ \cdot σ \cdot A \cdot T^{4}$ jahan:

$ϵ$ emissivity (0–1)
$σ$ = Stefan-Boltzmann constant = $5.67 \times 1 0^{- 8}$ W/(m²·K⁴)
$T$ = spacecraft temperature (Kelvin)

Key insight: Surface coatings choose karke hum $α$ (kitni heat absorb karein) aur $ϵ$ (kitni radiate karein) control karte hain. Hot environments ke liye low $α / ϵ$ ratio chahiye, cold ke liye high ratio.

Worked example Example: Thermal Coating Selection

Scenario: LEO mein CubeSat ko electronics 20°C par rakhne chahiye. Internal dissipation = 5 W. Exposed area = 0.03 m².

Option A: Black paint ( $α = 0.95$ , $ϵ = 0.85$ ) Option B: White paint ( $α = 0.25$ , $ϵ = 0.90$ )

Step 1: Solar input calculate karo (sun-facing side, 0.01 m² projected). Option A: $Q_{solar} = 0.95 \times 0.01 \times 1361 = 12.9$ W Option B: $Q_{solar} = 0.25 \times 0.01 \times 1361 = 3.4$ W Yeh step kyun? High absorptivity matlab zyada solar heating.

Step 2: Reject karne ke liye total heat. Option A: $Q_{total} = 12.9 + 5 = 17.9$ W Option B: $Q_{total} = 3.4 + 5 = 8.4$ W

Step 3: Required equilibrium temperature. $Q_{out} = ϵ σ A T^{4} = Q_{total}$ Option A: $T = (\frac{17.9}{0.85 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 330$ K = 57°C ❌ Bahut zyada hot! Option B: $T = (\frac{8.4}{0.90 \times 5.67 \times 1 0 ^{- 8} \times 0.03})^{1/4} = 281$ K = 8°C ✓ Conclusion: White paint spacecraft ko cooler rakhta hai kyunki low $α / ϵ$ ratio heat rejection favor karta hai.

Passive vs Active Control

Passive: Coatings, multi-layer insulation (MLI), heat pipes
Active: Heaters, louvers, fluid loops (bade spacecraft ke liye)

4. Attitude Determination & Control System (ADCS)

Purpose

Yeh jaanna ki spacecraft kahan point kar raha hai (determination) aur us orientation ko change/maintain karna (control).

Components

Reaction Wheels ki Physics

Worked example Example: Slew Maneuver Calculation

Diya hua hai:

Spacecraft moment of inertia: $I_{sc} = 10$ kg·m²
Reaction wheel: $I_{wheel} = 0.01$ kg·m², max speed = 6000 RPM
Desired rotation: 90° in 60 seconds Dhundna hai: Required wheel acceleration aur check karo ki saturation hota hai ya nahi.

Step 1: Required spacecraft angular acceleration calculate karo. Pehli aadhi ke liye constant acceleration par $θ = \frac{1}{2} α t^{2}$ use karte hue, phir deceleration: $α_{sc} = \frac{4 θ}{t ^{2}} = \frac{4 \times ( π /2 )}{6 0 ^{2}} = 1.75 \times 1 0^{- 3} rad/s^{2}$ Yeh step kyun? Hume mid-point tak accelerate karna hai, phir symmetrically decelerate karna hai.

Step 2: Required wheel acceleration nikalo (opposite direction). $I_{sc} α_{sc} = I_{wheel} α$ $α_{wheel} = \frac{10 \times 1.75 \times 1 0 ^{- 3}}{0.01} = 1.75 rad/s^{2}$ Yeh step kyun? Angular momentum ka conservation yeh ratio dictate karta hai.

Step 3: Saturation check karo. Maneuver ke dौran maximum wheel speed ( $t = 30$ s par): $ω_{wheel,max} = α_{wheel} \times 30 = 52.5 rad/s = 501 RPM$ Yeh max speed (6000 RPM) ka sirf 8.4% hai → ✓ Koi saturation nahi

5. Command & Data Handling (C&DH)

Function

Onboard computer jo:

Ground se commands receive karta hai (comms subsystem ke zariye)
Command sequences execute karta hai (autonomy)
Saare subsystems se telemetry collect karta hai
Science/housekeeping data store karta hai
Onboard processing karta hai (compression, fault detection)

Architecture

Radiation-hardening kyun? Space radiation (cosmic rays, solar protons) single-event upsets (bit flips) cause karta hai. Rad-hard chips larger transistors, error-correcting memory, aur triple-modular redundancy use karte hain.

6. Communications Subsystem

Purpose

Ground stations ke saath radio link ke liye:

Uplink: Commands (low data rate, high reliability)
Downlink: Telemetry aur science data (high data rate)

Link Budget

Formula Friis Transmission Equation

Derivation: Receiver par kitni power pahunchti hai?

Transmitter: Power $P_{t}$ (watts) radiate karta hai $G_{t}$ gain (dimensionless) wale antenna ke zariye.

Free-space spreading: Signal $4 π d^{2}$ area wale sphere par spread hota hai, jahan $d$ distance hai.

Receiver par power density: $Power density = \frac{P _{t} G _{t}}{4 π d ^{2}}$

Receiver antenna: Power collect karta hai apne effective area $A_{eff}$ ke proportional: $A_{eff} = \frac{λ ^{2} G _{r}}{4 π}$ jahan $G_{r}$ receiver antenna gain hai, $λ$ wavelength hai.

Received power: $P_{r} = \frac{P _{t} G _{t}}{4 π d ^{2}} \cdot \frac{λ ^{2} G _{r}}{4 π}$

Simplify karne par: $P_{r} = P_{t} G_{t} G_{r} (\frac{λ}{4 π d})^{2}$

Decibels mein (dBW = $10 lo g_{10} (P /1 W)$ ): $P_{r} [dBW] = P_{t} [dBW] + G_{t} [dBi] + G_{r} [dBi] - L_{path} [dB]$

jahan path loss: $L_{path} = 20 lo g_{10} (\frac{4 π d}{λ}) = 20 lo g_{10} (d) + 20 lo g_{10} (f) - 147.55$ ( $d$ meters mein, $f$ Hz mein)

Worked example Example: CubeSat ke liye Downlink Budget

Diya hua hai:

Ground station tak range $d \approx 2000$ km (500 km orbit se max slant range)
Frequency: 437 MHz (UHF) → $λ = 0.686$ m
Transmit power: $P_{t} = 2$ W = 3 dBW
Spacecraft antenna gain: $G_{t} = 2$ dBi (dipole)
Ground station antenna gain: $G_{r} = 15$ dBi (Yagi array)
System losses: $L_{sys} = 3$ dB
Receiver noise temperature: $T_{sys} = 200$ K

Dhundna hai: bit-error-rate (BER) = 10⁻⁶ ke saath achievable data rate.

Step 1: Path loss calculate karo. $L_{path} = 20 lo g_{10} (2 \times 1 0^{6}) + 20 lo g_{10} (437 \times 1 0^{6}) - 147.55$ $= 126.0 + 172.8 - 147.55 = 151.3 dB$ Yeh step kyun? Yeh dominant loss term hai—radio waves 3D space mein spread hoti hain.

Step 2: Received power. $P_{r} = 3 + 2 + 15 - 151.3 - 3 = - 134.3 dBW$ Yeh step kyun? Yeh woh signal power hai jo ground station par pahunchi hai.

Step 3: Receiver noise spectral density. $N_{0} = k T_{sys} = 1.38 \times 1 0^{- 23} \times 200 = 2.76 \times 1 0^{- 21} W/Hz = - 205.6 dBW/Hz$ jahan $k = 1.38 \times 1 0^{- 23}$ J

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

Spacecraft Bus Kya Hota Hai?

1. Structure Subsystem

Purpose

Key Design Drivers

2. Electrical Power System (EPS)

Function

Components

Power Budget

3. Thermal Control System (TCS)

Purpose

Heat Sources & Sinks

Passive vs Active Control

4. Attitude Determination & Control System (ADCS)

Purpose

Components

Reaction Wheels ki Physics

5. Command & Data Handling (C&DH)

Function

Architecture

6. Communications Subsystem

Purpose

Link Budget

Concept Map