Power systems — solar arrays (I-V curve, power tracking), batteries (DoD, cycles), RTG: Exercises

3.6.22 · D4 · Physics › Spacecraft Structures & Systems Engineering › Power systems — solar arrays (I-V curve, power tracking), ba

Yeh ek self-testing ladder hai. Har problem mein sab kuch diya gaya hai jo tumhe chahiye. Pehle khud try karo, phir collapsible solution kholna. Rungs L1 Recognition (cheez ka naam bolo) se L5 Mastery tak jaati hain (poora subsystem design karo). Yahan sab kuch the parent power-systems note par build karta hai aur Semiconductor Physics, Power Electronics, Orbital Mechanics (Keplerian) aur Spacecraft Thermal Control ko touch karta hai.

Reference curve for the whole set — yeh figure ek solar cell ki I-V curve dikhata hai (blue: current voltage badhne ke saath girta hai), power curve $P = V \cdot I$ (pink: ek single hill), aur teen key points $I_{sc}$ (max current at $V = 0$ ), $V_{oc}$ (max voltage at $I = 0$ ), aur MPP (power hill ka peak):

Figure — Power systems — solar arrays (I-V curve, power tracking), batteries (DoD, cycles), RTG

Level 1 — Recognition

Problem 1.1 — Read the curve

Upar di gayi I-V curve par, teen labelled operating points ke naam batao aur ek-ek phrase mein yeh batao ki unmein kya special hai.

Recall Solution

==Short-circuit current $I_{sc}$ == — woh point jahan terminal voltage $V = 0$ hota hai (wires shorted). Current apne maximum par hoti hai, power zero hoti hai (kyunki $P = V \cdot I = 0 \cdot I_{sc} = 0$ ).
==Open-circuit voltage $V_{oc}$ == — woh point jahan current $I = 0$ hoti hai (koi load connected nahi). Voltage apne maximum par hoti hai, power phir se zero hoti hai ( $P = V_{oc} \cdot 0 = 0$ ).
Maximum power point (MPP) $(V_{m p}, I_{m p})$ — curve ka knee jahan $V \cdot I$ ka rectangle sabse bada hota hai. Yahan hum operate karna chahte hain.

Problem 1.2 — Units check

Ek battery $28 V$ , $20 Ah$ rated hai. Watt-hours mein uski total stored energy kya hai?

Recall Solution

Energy $= voltage \times charge capacity$ . Amp-hours times volts watt-hours deta hai. Stored energy ko $E_{total}$ likhte hain ( $Q$ sirf charge ke liye reserve hai, energy ke liye nahi): $E_{total} = 28 V \times 20 Ah = 560 Wh .$ Yeh kyun kaam karta hai (explicit unit chain): units ko step by step carry karo.

Amp-hours ko coulombs (charge) mein convert karo: $20 Ah \times 3600 \frac{s}{h} = 72000 A\cdots = 72000 C$ .
Charge ko volts se multiply karo (ek volt ek joule per coulomb hota hai, isliye C·V = J): $72000 C \times 28 V = 2016000 J$ .
Joules ko watt-hours mein convert karo (ek Wh $= 3600$ J): $2016000 J \div 3600 \frac{J}{Wh} = 560 Wh$ . Woh $3600$ jo step 1 mein aaya tha, step 3 mein exactly undo ho jaata hai, aur isi liye shortcut "volts × amp-hours = watt-hours" kaam karta hai — dono conversions cancel ho jaati hain.

Level 2 — Application

Problem 2.1 — Open-circuit voltage

Ek solar cell ideal single-diode equation follow karta hai $I = I_{L} - I_{0} (e^{V / (n V_{T})} - 1),$ jahan $I$ output current hai, $V$ terminal voltage, $I_{L}$ photocurrent, $I_{0}$ dark saturation current, $n$ ideality factor, aur $V_{T}$ thermal voltage. Given $I_{L} = 5 A$ , $I_{0} = 1 0^{- 12} A$ , $n = 1.5$ , $V_{T} = 0.026 V$ , open-circuit voltage $V_{oc}$ nikalo (woh voltage jis par $I = 0$ ho).

Recall Solution

$V_{oc}$ woh voltage hai jahan current vanish ho jaati hai. Ideal cell equation $I = I_{L} - I_{0} (e^{V / n V_{T}} - 1)$ mein $I = 0$ set karke solve karo: $V_{oc} = n V_{T} ln (\frac{I _{L}}{I _{0}} + 1) .$ Logarithm kyun? Diode current voltage ke saath exponentially badhta hai; ek fixed photocurrent cancel karne ke liye humein woh exponential undo karna hoga — $e^{x}$ ka inverse $ln x$ hota hai. Values plug in karo: $V_{oc} = 1.5 \times 0.026 \times ln (\frac{5}{1 0 ^{- 12}} + 1) = 0.039 \times ln (5 \times 1 0^{12}) \approx 0.039 \times 29.25 \approx 1.14 V .$

Problem 2.2 — Fill factor

Usi cell ke liye, numerically found MPP hai $V_{m p} = 1.00 V$ , $I_{m p} = 4.90 A$ . Fill factor $F F$ aur $P_{ma x}$ compute karo.

Recall Solution

Maximum power: $P_{ma x} = V_{m p} I_{m p} = 1.00 \times 4.90 = 4.90 W .$ Fill factor yeh batata hai ki curve kitna "square" hai — MPP rectangle divided by ideal $V_{oc} I_{sc}$ rectangle: $F F = \frac{V _{m p} I _{m p}}{V _{oc} I _{sc}} = \frac{4.90}{1.14 \times 5} = \frac{4.90}{5.70} \approx 0.86.$ Yeh typical GaAs-space-cell band (0.85–0.88) mein hai.

Problem 2.3 — Sun-angle scaling

Sunlight ek panel par surface normal se $β = 3 0^{\circ}$ off angle par pad rahi hai. Full-sun intensity $1367 W/m^{2}$ hai. Panel tak kitni intensity pahunchti hai, aur $I_{sc}$ roughly kitne fraction se drop hoti hai?

Recall Solution

Sirf light ka woh component jo surface normal ke along hai power deliver karta hai, aur woh projection ek cosine hota hai (light-angle triangle par adjacent over hypotenuse): $S = 1367 cos 3 0^{\circ} = 1367 \times 0.866 = 1184 W/m^{2} .$ Photocurrent absorbed photon flux ke proportional hoti hai, isliye $I_{sc}$ intensity ke saath scale karta hai: $\frac{I _{sc}}{I _{sc, 0}} = cos 3 0^{\circ} = 0.866,$ matlab 13.4% drop.

Level 3 — Analysis

Problem 3.1 — Derive the MPP condition

$P (V) = V I (V)$ se start karke, dikhao ki maximum power point par $\frac{d I}{d V} = - \frac{I}{V}$ hota hai.

Recall Solution

WHAT: hum power hill ka top chahte hain. WHY derivative: slope $d P / d V$ peak par zero hota hai — yeh smooth maximum ki definition hai. $\frac{d P}{d V} = \frac{d}{d V} (V I) = I + V \frac{d I}{d V} .$ (Product rule: $V$ differentiate karo $I$ chhod ke, plus $V$ times derivative of $I$ .) MPP par zero set karo: $I + V \frac{d I}{d V} = 0 ⟹ \frac{d I}{d V} = - \frac{I}{V} .$ Iska matlab: incremental conductance $d I / d V$ exactly bulk conductance $I / V$ cancel karta hai. Yahi identity incremental-conductance MPPT controller watch karta hai. Dekho Power Electronics.

Problem 3.2 — Cycle-life scaling

Li-ion cycle life follow karta hai $N \approx N_{0} (DoD / DoD_{0})^{- k}$ jahan $N_{0} = 5000$ cycles at $DoD_{0} = 50%$ aur $k = 1.5$ hai. $80%$ DoD par kitne cycles? $30%$ DoD par?

Recall Solution

Formula kya kehta hai: zyada deep discharge ( $↑$ DoD) matlab kam cycles — negative exponent ratio ko flip kar deta hai. $80%$ par: $N = 5000 (\frac{80}{50})^{- 1.5} = 5000 \times (1.6)^{- 1.5} = 5000 \times 0.4941 \approx 2470 cycles .$ $30%$ par (shallower — zyada cycles hone chahiye): $N = 5000 (\frac{30}{50})^{- 1.5} = 5000 \times (0.6)^{- 1.5} = 5000 \times 2.152 \approx 10760 cycles .$ Trend sahi hai: shallower cycling life double se zyada kar deta hai.

Problem 3.3 — Temperature shift of MPP

Ek cell string mein 40 series cells hain. Har cell ka $V_{oc}$ drift karta hai $- 2.2 mV/^{\circ} C$ se, aur har cell ki short-circuit current heat ke saath slightly rise karti hai $+ 0.05% /^{\circ} C$ of $I_{sc}$ . Har cell $V_{oc} = 1.14 V$ se start karta hai. Panel $- 1 0^{\circ} C$ se $+ 6 0^{\circ} C$ tak warm hota hai. String $V_{oc}$ kitna move karta hai, per-cell percentage voltage drop kya hai, $I_{sc}$ kitna change hoti hai, aur MPPT ko operating voltage kis direction mein push karni chahiye?

Recall Solution

Temperature change: $Δ T = 60 - (- 10) = 7 0^{\circ} C$ . Voltage side (large, negative): per-cell drop $- 2.2 mV/^{\circ} C \times 70 = - 154 mV$ . 40 series cells mein voltages add hote hain: $Δ V_{oc, string} = 40 \times (- 0.154 V) = - 6.16 V .$ Per-cell percentage drop ( $1.14 V$ start ke relative): $\frac{- 0.154}{1.14} \times 100% = - 13.5% per cell.$ Current side (tiny, positive): heat mildly bandgap narrow karta hai isliye thode zyada photons absorb hote hain, $I_{sc}$ ko up nudge karta hai: $\frac{Δ I _{sc}}{I _{sc}} = + 0.05% /^{\circ} C \times 70 = + 3.5%.$ Dono effects compare karo: voltage $13.5%$ per cell se girta hai jabki current sirf $+ 3.5%$ badhta hai. Voltage loss dominate karta hai, isliye panel heat hone par $V_{m p}$ fall karta hai aur MPPT ko peak par rehne ke liye operating voltage lower karni chahiye. (Hotter carriers → zyada diode leakage $I_{0}$ → lower voltage; dekho Semiconductor Physics aur Spacecraft Thermal Control.)

Level 4 — Synthesis

Problem 4.1 — Eclipse energy and battery sizing

Ek LEO spacecraft ka $90$ -minute orbit hai jisme $35$ -minute eclipse hai. Eclipse-load $500 W$ hai. Battery $28 V$ , $20 Ah$ hai. Agar mission rules DoD $80%$ par cap karte hain, toh kya battery ek eclipse cover karti hai, aur kitne margin ke saath?

Recall Solution

Step 1 — energy needed: eclipse minutes ko hours mein convert karo: $35 min = 0.583 h$ . $E_{need} = 500 W \times 0.583 h = 291.7 Wh .$ Step 2 — usable energy at 80% DoD: $E_{usable} = (28 \times 20) \times 0.80 = 560 \times 0.80 = 448 Wh .$ Step 3 — compare. $448 > 291.7$ , isliye haan, ek eclipse cover hoti hai. Margin: $448 - 291.7 = 156.3 Wh$ , yaani $\frac{156.3}{291.7} \times 100% \approx 53.6% spare.$ Equivalently eclipse sirf $291.7/560 = 52.1%$ DoD use karti hai, jo 80% cap se comfortably under hai.

Problem 4.2 — Mission lifetime in eclipse cycles

4.1 se actual DoD ( $52.1%$ ) aur 3.2 se cycle-life law ( $N_{0} = 5000$ at $50%$ , $k = 1.5$ ) use karke, battery ki cycle life aur mission years estimate karo, given ki har $90$ -min orbit mein ek eclipse hoti hai.

Recall Solution

Cycle life at 52.1% DoD: $N = 5000 (\frac{52.1}{50})^{- 1.5} = 5000 \times (1.042)^{- 1.5} = 5000 \times 0.9403 \approx 4701 cycles .$ Orbits per year: har orbit mein ek cycle; ek saal $365.25 \times 24 \times 60 = 525960$ minutes ka hota hai, isliye orbits per year $= 525 960/90 = 5844$ . Mission years: $\frac{4701}{5844} \approx 0.80 years .$ Interpretation: ~52% DoD par battery life-limiter hai — eclipse cycling ka ek saal se kam. Multi-year life ke liye ya toh DoD kam karo (badi battery) ya scheduled capacity fade accept karo. Yeh LEO power design mein classic tension hai.

Level 5 — Mastery

Problem 5.1 — RTG power budget over mission life

Ek radioisotope thermoelectric generator $^{238} Pu$ use karta hai (half-life $T_{1/2} = 87.7 yr$ ). Launch par uska thermal output $4400 W_{th}$ hai; thermoelectric conversion efficiency $6.5%$ hai. $t$ ko mission elapsed time measured in years since launch maano. (a) Launch par electrical power? (b) $t = 14$ years baad electrical power, fuel decay aur $0.8%$ /year converter degradation dono treat karke? Assume karo ki converter efficiency $(1 - 0.008)^{t}$ se multiply hoti hai jahan $t$ years mein hai.

Recall Solution

(a) Launch electrical power: $P_{e, 0} = 4400 W_{th} \times 0.065 = 286 W_{e} .$ (b) Fuel decay exponential radioactive decay follow karta hai — activity har half-life mein half ho jaati hai, isliye thermal power $(1/2)^{t / T_{1/2}}$ se multiply hoti hai jahan $t$ aur $T_{1/2}$ dono years mein hain (isi liye exponential use karte hain: per unit time constant fractional decay exactly $e^{- λ t}$ hai, equivalently $2^{- t / T_{1/2}}$ ). $t = 14 yr$ par: $fuel factor = (\frac{1}{2})^{14/87.7} = 2^{- 0.1596} = 0.8953.$ Converter degradation yearly compound karta hai, usi $t = 14 yr$ ke saath: $conv factor = (1 - 0.008)^{14} = 0.99 2^{14} = 0.8935.$ Combine (dono independent losses multiply karo): $P_{e} (14) = 286 \times 0.8953 \times 0.8935 \approx 286 \times 0.7999 \approx 228.8 W_{e} .$ Toh RTG year 14 par approximately $229 W_{e}$ deliver karta hai — launch power ka roughly $80%$ . Deep-space loads ko is monotone, predictable fade ke around schedule karna hoga (solar se contrast karo, jo radiation-driven degradation suffer karta hai lekin distance aur geometry par depend karta hai).

Problem 5.2 — Choose the architecture

Ek probe $9.5 AU$ par (Saturn ke paas) $12$ years ke liye operate karegi, steady $200 W_{e}$ chahiye. Solar intensity $1/ d^{2}$ se fall karti hai. $1 AU$ par solar $1367 W/m^{2}$ deta hai; assume karo $30%$ end-of-life array efficiency. Required solar-array area compute karo, isse 5.1 ke RTG option se compare karo, aur ek architecture recommend karo.

Recall Solution

Step 1 — intensity at $9.5 AU$ (inverse-square: distance double ho toh light quarter ho jaati hai): $S = \frac{1367}{9. 5 ^{2}} = \frac{1367}{90.25} = 15.15 W/m^{2} .$ Step 2 — array area needed $30%$ efficiency par $200 W_{e}$ banana. Electrical power out $P_{e} = η S A$ hai, isliye area $A$ ke liye solve karo: $A = \frac{P _{e}}{η S} = \frac{200}{0.30 \times 15.15} = \frac{200}{4.545} \approx 44.0 m^{2} .$ Yeh ek bada, deployment-risky array hai (approximately $6.6 m \times 6.6 m$ square) jo permanent cold aur dark mein feeble power deliver karta hai, long-duration radiation damage se aur kharab ho jaata hai. Step 3 — RTG comparison: 5.1 ka year-14 RTG already $\approx 229 W_{e}$ yield karta hai, jo full 12-year mission ke liye required $200 W_{e}$ se zyada hai, distance, sun angle, ya eclipse par koi dependence nahi aur koi $44 m^{2}$ deployment nahi. Step 4 — recommendation: RTG choose karo. $9.5 AU$ par solar flux Earth ke mukable $1367/15.15 \approx 90 \times$ weaker hai, jo modest power ke liye impractically huge $\sim 44 m^{2}$ array force karta hai jo phir bhi planet ki shadow mein fail ho jaata hai; RTG ka fade gentle hai, distance-independent hai, aur uska $\approx 229 W_{e}$ output comfortably $200 W_{e}$ budget meet karta hai. Isi liye outer-planet missions RTGs fly karte hain.

Recall

Kis operating point par maximum power hai, $I_{sc}$ , $V_{oc}$ , ya MPP? ::: MPP (the knee) par, kyunki power product $V \cdot I$ hai aur dono endpoints par ek factor zero ke barabar hota hai. Jab light dim hoti hai, kaun zyada change hota hai, $I_{sc}$ ya $V_{oc}$ ? ::: $I_{sc}$ (intensity mein linear); $V_{oc}$ sirf logarithmically move karta hai. RTGs Saturn par solar ko kyun beat karte hain? ::: Solar flux $1/ d^{2}$ se girta hai; $9.5 AU$ par yeh $\sim 90 \times$ weaker hai, isliye solar ko bada array chahiye jabki RTG output distance-independent hota hai.