Exercises — FEEP, MEMS thrusters — micro-propulsion

Hum kya karte hain: $80\mu \text{N}$ ko defining range se compare karo. Micro-propulsion ka matlab hai thrust $1\mu \text{N}$ se $\sim 1\text{mN}$ tak ki band mein (yaani $10^{-6}$ N se $10^{-3}$ N). Kyunki $80\mu \text{N}=8\times 10^{-5}\text{N}$ is band ke andar hai, haan, yeh micro-propulsion hai.

Speed $v_e=\sqrt{2qV/m}$ se aati hai. Current $I$ isme appear hi nahi karta. Toh $v_e$ badhane ke liye $V$ badhao. (Current control karta hai ki har second kitne ions bahar jaate hain, yaani thrust — har ek ion ki speed nahi.)

Hum kya karte hain: $I_{sp}=v_e/g_0$ apply karo (definition from Specific Impulse). $I_{sp}=\frac{1.0\times 10^5}{9.81}\approx 1.02\times 10^4\text{s}\approx 10,200\text{s}.$ Hazaaron seconds mein number electric propulsion ki signature hai — isliye yeh thrusters propellant bahut hi kam peete hain.

Hum kya karte hain: energy conservation — ion ki electrical potential energy $qV$ kinetic energy $\frac{1}{2}m{v}_e^2$ ban jaati hai, toh $v_e=\sqrt{2qV/m}$. $v_e=\sqrt{\frac{2(1.6\times 10^{-19})(8000)}{2.2\times 10^{-25}}}=\sqrt{\frac{2.56\times 10^{-15}}{2.2\times 10^{-25}}}=\sqrt{1.164\times 10^{10}}\approx 1.08\times 10^5\text{m/s}.$ Lagbhag $108\text{km/s}$.

Valid calculation (part a). Root ke andar ka fraction pehle ek clean power of ten mein laao, phir square root lo: $\frac{2mV}{q}=\frac{2(2.2\times 10^{-25})(8000)}{1.6\times 10^{-19}}=\frac{3.52\times 10^{-21}}{1.6\times 10^{-19}}=2.2\times 10^{-2}.$ $\sqrt{2mV/q}=\sqrt{2.2\times 10^{-2}}=0.148,F=I\cdot 0.148=(15\times 10^{-6})(0.148)=2.2\times 10^{-6}\text{N}\approx 2.2\mu \text{N}.$ Valid calculation (part b), independent route. $\frac{m}{q}=\frac{2.2\times 10^{-25}}{1.6\times 10^{-19}}=1.375\times 10^{-6}\text{kg/C}$. $\dot{m}=\frac{m}{q}I=(1.375\times 10^{-6})(15\times 10^{-6})=2.06\times 10^{-11}\text{kg/s}.$ $F=\dot{m}{v}_e=(2.06\times 10^{-11})(1.08\times 10^5)=2.23\times 10^{-6}\text{N}\approx 2.2\mu \text{N}.$ Dono routes $\approx 2.2\mu \text{N}$ dete hain. Direct formula aur mass-flow picture same physics hain, toh yeh agreement wahi check hai jo hum chahte the. Lesson baka hua hai: hamesha ek µN device ko $\dot{m}{v}_e$ picture se sanity-check karo, aur square root lene se pehle exponent ko root ke andar collapse karo (classic slip ke liye neeche L2 trap dekho).

$v_e=\sqrt{2qV/m}\propto 1/\sqrt{m}$: halka ion zyada fast hota hai. Toh $v_e$ mein: light ion $>$ indium $>$ caesium. (Yeh exactly wahi trade-off hai jo Ion Thrusters mein aur explore ki gayi hai.)

Hum kya karte hain: formulas se compare karo, $I$ aur $V$ fixed rakhte hue. Thrust: $F=I\sqrt{2mV/q}\propto \sqrt{m}$. Toh $\frac{F_B}{F_A}=\sqrt{\frac{m_B}{m_A}}=\sqrt{1.16}\approx 1.077$. Caesium ~7.7% zyada thrust per amp deta hai. $I_{sp}$: $v_e\propto 1/\sqrt{m}$, toh $\frac{I_{sp,B}}{I_{sp,A}}=\sqrt{\frac{m_A}{m_B}}=\frac{1}{\sqrt{1.16}}\approx 0.928$. Caesium ~7% kam $I_{sp}$ deta hai. Matlab: heavy ions thrust khareedte hain (momentum per charge) fuel economy ki keemat par. Koi bhi "better" nahi hai — yeh depend karta hai ki mission thrust-limited hai ya propellant-limited.

Pehle figure padho. Cone flanks par green inward arrows surface tension ka pull hain — woh smallest possible surface area chahti hai, toh liquid ko andar kheenchti hai. Sharp apex ke paas red outward arrows electric field $E$ se produce hone wala electrostatic pressure hai: iska size $\frac{1}{2}{\epsilon }_0{E}^2$ hai (dono $E$ aur ${\epsilon }_0$ symbol reminder mein define hain). Yellow arc half-angle mark karta hai, aur upar yellow arrows woh ions hain jo apex se fling ho rahe hain. Dashed circle apex se $r$ distance par ek test point mark karta hai jahan hum dono pulls weigh karte hain.

Surface tension $1/r$ se kyun scale karti hai. Surface tension ek inward pressure produce karti hai jo $\gamma$ (tension) times surface ki curvature ke barabar hai. Ek cone par, apex se slant distance $r$ par surface ek circular cross-section ke around curve karti hai jiska radius $r$ ke proportional hai (cone tip se door jaate jaate linearly choda hota hai). Curvature us radius ka ulta hai, toh curvature $\propto 1/r$ hai. Isliye inward surface-tension pressure $\propto \gamma /r$ hai. Tip ke paas (chhota $r$) surface zyada sharply curved hai, toh tension zyada squeeze karti hai.

Electrostatic pull bhi $1/r$ se kyun scale karta hai. Ek charged conductor apna field wahan concentrate karta hai jahan surface sabse sharp ho. Conical tip ke paas electric field apex ke approach karte waqt $E\propto r^{-1/2}$ ki tarah badhta hai (conical point wale conductor ka ek standard result). Outward electrostatic pressure $\frac{1}{2}{\epsilon }_0{E}^2$ hai, aur $r^{-1/2}$ ko square karne par $E^2\propto r^{-1}$ milta hai. Toh outward pressure $\propto \frac{1}{2}{\epsilon }_0/r$ — yeh bhi tip ki taraf $1/r$ se blow up karta hai.

Specifically cone kyun? Inward pull ($\propto \gamma /r$) aur outward pull ($\propto \frac{1}{2}{\epsilon }_0{E}^2\propto 1/r$) dono mein same $1/r$ dependence hai. Yeh special hai: iska matlab hai agar woh ek radius par balance karte hain, toh har radius par simultaneously balance karte hain. Cone ek unique self-similar shape hai jahan red arrows aur green arrows tip tak poore raaste cancel karte hain. Woh balanced shape hi Taylor cone hai, half-angle $\approx 49.3^{\circ }$ (yellow arc; aur detail Taylor Cone mein). Sphere ek curvature deta hai jo field ke $1/r$ growth se match nahi karti, toh field tip par jeetta rehta hai aur use aur kheenchta hai (unstable); flat film mein koi sharp point nahi, toh field kabhi concentrate nahi hota aur kuch emit nahi hota. Sirf cone har jagah balance karta hai — aur iska apex woh jagah hai jahan ions (yellow) ripped off hote hain.

$Re$ kya measure karta hai: $Re=\rho vL/\mu$ hai (flow ki inertia)/(viscous friction). Ek chhota $Re$ (yahan 40) matlab friction dominate karta hai. (Contrast: ek airliner wing ka $Re\sim 10^7$ hota hai.) Consequence: ek tiny channel mein walls ke against sticky boundary layer poore flow cross-section ka ek bada fraction hoti hai. Woh friction directed kinetic energy ko heat mein convert karta hai aur gas ko slow kar deta hai. Toh real exhaust speed 700 m/s se kam hai, aur efficiency ideal-expansion prediction se neeche jaati hai. Yeh fundamental MEMS penalty hai — Reynolds Number dekho.

(a) Propellant. Kyunki $\Delta v\ll v_e$, Rocket Equation ka small-$x$ approximation use karo, $e^{-x}\approx 1-x$: $m_p=m_0\left(1-e^{-\Delta v/v_e}\right)\approx m_0\frac{\Delta v}{v_e}=4\cdot \frac{12}{10^5}=4.8\times 10^{-4}\text{kg}=0.48\text{g}.$ Aadha gram se bhi kam — bade $v_e$ ka payoff. (b) Firing time. Zaroori total impulse $J=m_0\Delta v=4\times 12=48\text{kg⋅m/s}$ hai (propellant negligible hai toh $m_0$ use karo). Constant $F$ par fire karte hue: $t=J/F$. $t=\frac{48}{2.0\times 10^{-6}}=2.4\times 10^7\text{s}=\frac{2.4\times 10^7}{86400}\approx 278\text{days}.$ Matlab: micro-thrust matlab mission ke zyaatar time almost lagaataar, dheere dheere fire karo — exactly wahi "eyedropper, not firehose" picture jo parent note mein thi.

Hum kya karte hain: $F=\dot{m}{v}_e=(m/q)I{v}_e$ ko invert karke $I$ solve karo. $I=\frac{Fq}{m{v}_e}=\frac{(2.0\times 10^{-6})(1.6\times 10^{-19})}{(1.9\times 10^{-25})(1.0\times 10^5)}.$ Numerator: $3.2\times 10^{-25}$. Denominator: $1.9\times 10^{-20}$. $I=\frac{3.2\times 10^{-25}}{1.9\times 10^{-20}}=1.68\times 10^{-5}\text{A}\approx 17\mu \text{A}.$ ~17-microamp beam ~2 µN deta hai — aur hum woh current electronically nanoamps tak dial down kar sakte hain, isliye FEEP ko µN thrust resolution milti hai.

Setup. Fixed $I,V,q$ par: thrust $F=I\sqrt{2mV/q}\propto \sqrt{m}$ (heavy wins thrust), aur $v_e=\sqrt{2qV/m}\propto 1/\sqrt{m}$ (light wins $I_{sp}$).

(a) Thrust. $\sqrt{2mV/q}$ compute karo (root ke andar exponent pehle collapse karo), phir $I=20\times 10^{-6}$ se multiply karo.

• Indium: $\frac{2(1.9\times 10^{-25})(7000)}{1.6\times 10^{-19}}=\frac{2.66\times 10^{-21}}{1.6\times 10^{-19}}=1.6625\times 10^{-2}$; $\sqrt{}=0.1289$; $F=20\times 10^{-6}\times 0.1289=2.58\times 10^{-6}\text{N}\approx 2.6\mu \text{N}$.
• Caesium: $\frac{2(2.2\times 10^{-25})(7000)}{1.6\times 10^{-19}}=\frac{3.08\times 10^{-21}}{1.6\times 10^{-19}}=1.925\times 10^{-2}$; $\sqrt{}=0.1387$; $F=20\times 10^{-6}\times 0.1387=2.77\times 10^{-6}\text{N}\approx 2.8\mu \text{N}$. Caesium zyada thrust deta hai (~2.8 vs ~2.6 µN), jaisa $F\propto \sqrt{m}$ se expect tha.

(b) $I_{sp}$. $v_e=\sqrt{2qV/m}$, phir $I_{sp}=v_e/g_0$ jahan $g_0=9.81$.

• Indium: $v_e=\sqrt{\frac{2(1.6\times 10^{-19})(7000)}{1.9\times 10^{-25}}}=\sqrt{1.179\times 10^{10}}=1.086\times 10^5\text{m/s}$; $I_{sp}=1.086\times 10^5/9.81=1.107\times 10^4\text{s}$.
• Caesium: $v_e=\sqrt{\frac{2(1.6\times 10^{-19})(7000)}{2.2\times 10^{-25}}}=\sqrt{1.018\times 10^{10}}=1.009\times 10^5\text{m/s}$; $I_{sp}=1.009\times 10^5/9.81=1.029\times 10^4\text{s}$. Indium zyada $I_{sp}$ (~11,100 vs ~10,300 s) deta hai, jaisa $v_e\propto 1/\sqrt{m}$ se expect tha.

(c) $\Delta v=15\text{m/s}$, $m_0=5\text{kg}$ ke liye propellant, small-$x$ form $m_p\approx m_0\Delta v/v_e$ use karke:

• Indium: $m_p=5\cdot \frac{15}{1.086\times 10^5}=6.91\times 10^{-4}\text{kg}\approx 0.69\text{g}$.
• Caesium: $m_p=5\cdot \frac{15}{1.009\times 10^5}=7.43\times 10^{-4}\text{kg}\approx 0.74\text{g}$.

Decision. Propellant ka difference ~0.05 g hai — 5 kg satellite par bilkul negligible. Thrust ka difference bhi tiny hai (~0.2 µN). Toh choice yeh numbers decide nahi karte; yeh engineering decide karti hai: indium ~157 °C par melt hota hai aur optics ko chemically contaminate nahi karta jaisa caesium (highly reactive, low work-function, surfaces par plate ho jaata hai) karta hai. Indium fly karo — almost identical performance with far less contamination risk. Lesson: jab do designs equations par tie ho jaayein, tie-breaker woh physics hoti hai jo equations bhool gayi.

Hum kya karte hain: minimum thrust $F_{\min }=(m/q)I_{\min }{v}_e$, phir impulse bit $J_{\min }=F_{\min }\Delta t$. $F_{\min }=(1.19\times 10^{-6})(100\times 10^{-9})(1.0\times 10^5)=1.19\times 10^{-8}\text{N}\approx 12\text{nN}.$ $J_{\min }=F_{\min }\Delta t=(1.19\times 10^{-8})(1\times 10^{-3})=1.19\times 10^{-11}\text{N⋅s}\approx 12\text{pN⋅s}.$ Yeh headline number kyun hai: ek ~$10^{-11}\text{N⋅s}$ impulse bit hi ek CubeSat ko arc-second precision tak point karne deta hai — electronically-tunable beam current hi aisi absurdly small, repeatable "taps" possible banata hai. Woh precision, raw thrust nahi, hi woh poora reason hai jis ke liye FEEP exist karta hai.

Woh ek sentence jo har level ko ek saath jodta hai ::: Voltage exhaust speed aur $I_{sp}$ set karta hai; current thrust set karta hai; heavy ions $I_{sp}$ ko thrust-per-amp ke liye trade karte hain; aur jab numbers tie ho jaayein, real-world material properties decide karte hain.