Electric propulsion — thrust, power, Isp trade-off

3.3.40 · Physics › Rocket Propulsion

1. Electric propulsion KYA hai?

YEH KYUN exist karta hai: Chemical rockets apne fuel ki chemical energy se limited hote hain, jis se exhaust speed $v_{e} \sim 4.5 km/s$ par cap lag jaati hai. EP energy source (electricity) ko reaction mass (propellant) se decouple karta hai, is liye hum $v_{e} \sim 20 - 50 km/s$ tak pahunch sakte hain. Rocket equation $Δ v = v_{e} ln (m_{0} / m_{f})$ se, zyada $v_{e}$ ⇒ propellant ke har kg se enormously zyada $Δ v$ .

The catch: har ion ko itni badi speed dene ke liye aapko energy pump in karni padti hai, aur power limited hoti hai. Is liye EP bahut tiny thrust produce karta hai (millinewtons) — deep-space cruising ke liye perfect, launch ke liye bilkul useless.

2. Core relations ko SCRATCH SE derive karna

Hum exhaust beam ko track karte hain. Maan lo thruster propellant ko mass flow rate $\overset{m}{˙}$ (kg/s) par exhaust speed $v_{e}$ (m/s) se eject karta hai.

2.1 Thrust

Time $d t$ mein hum $d m = \overset{m}{˙} d t$ mass ko $v_{e}$ speed par eject karte hain, jo $d p = v_{e} d m$ momentum carry karta hai. Force = momentum ejection ki rate:

F = \frac{d p}{d t} = v_{e} \overset{m}{˙}

2.2 Beam (jet) power

Mass $d m$ ko jo kinetic energy di jaati hai woh hai $\frac{1}{2} d m v_{e}^{2}$ . Power = energy per second:

P_{j e t} = \frac{d}{d t} (\frac{1}{2} m v_{e}^{2}) = \frac{1}{2} \overset{m}{˙} v_{e}^{2}

Real thrusters perfect nahi hote: electrical input $P_{in}$ zyada hota hai. Efficiency define karo $η = P_{j e t} / P_{in}$ , to

P_{in} = \frac{m ˙ v _{e}^{2}}{2 η} .

2.3 KEY relation — thrust per watt

$\overset{m}{˙}$ ko eliminate karo $F = \overset{m}{˙} v_{e} \Rightarrow \overset{m}{˙} = F / v_{e}$ use karke:

P_{j e t} = \frac{1}{2} \frac{F}{v _{e}} v_{e}^{2} = \frac{1}{2} F v_{e}

2.4 Specific impulse

$F = \overset{m}{˙} v_{e}$ se:

I_{s p} = \frac{m ˙ v _{e}}{m ˙ g _{0}} = \frac{v _{e}}{g _{0}} \Rightarrow v_{e} = g_{0} I_{s p}

Yeh step kyun? $I_{s p}$ sirf exhaust velocity ko rescale kiya hua hai — ek fuel-economy score. High $I_{s p}$ = thoda propellant use karta hai.

Trade-off mein substitute karke:

F = \frac{2 η P _{in}}{g _{0} I _{s p}}

To high $I_{s p}$ ⇒ same power ke liye low thrust. Yahi trade-off hai, phir se stated.

Figure — Electric propulsion — thrust, power, Isp trade-off

3. Worked examples

4. Common mistakes (Steel-manned)

Recall Feynman: ek 12-year-old ko samjhao

Socho tum ek skateboard par baith ke khud ko aage push karne ke liye tennis balls throw kar rahe ho. Agar balls super fast throw karo, to har ball tumhe zor se dhakelta hai — lekin ek ball ko itna fast banana bahut saari arm energy leta hai, to tum sirf kuch hi throw kar sakte ho. Agar balls gently throw karo, to tum bahut saare throw kar sakte ho, aur bahut saare gentle throws ek steady push mein add up hote hain. Ek electric rocket electricity ko "arm" ki tarah use karta hai. Ek fixed-size arm (fixed power) ke saath, tumhe choose karna padta hai: few fast balls (great mileage, weak push) ya many slow balls (poor mileage, stronger push). Same choti battery se strong push AUR great mileage dono nahi ho sakti.

5. Active-recall flashcards

\overset{m}{˙}

aur

v_{e}

ke terms mein rocket ka thrust?

F = \overset{m}{˙} v_{e}

(momentum ejected per second).

\overset{m}{˙}, v_{e}

ke terms mein jet (beam) power?

P_{j e t} = \frac{1}{2} \overset{m}{˙} v_{e}^{2}

Power, thrust, aur exhaust speed ke beech relation?

P_{j e t} = \frac{1}{2} F v_{e}

, isliye

F = 2 η P_{in} / v_{e}

Fixed power par high thrust AUR high

I_{s p}

dono kyun nahi ho sakte?

Kyunki fixed power par

F \propto 1/ v_{e}

hai;

v_{e}

(hence

I_{s p}

) badhane se thrust kam hota hai.

I_{s p}

ko exhaust velocity mein convert karo?

v_{e} = g_{0} I_{s p}

jahan

g_{0} = 9.81 m/s^{2}

Thruster efficiency

η

ki definition?

η = P_{j e t} / P_{in}

; input power beam power se zyada supply karta hai.

I_{s p}

ki units aur kyun?

Seconds; yeh propellant weight per unit impulse hai (

F / \overset{m}{˙} g_{0}

EP launch ke liye useless lekin deep space ke liye great kyun hai?

Thrust tiny hota hai (mN) — gravity ko jaldi beat nahi kar sakta — lekin huge

v_{e}

long time mein enormous

Δ v

deta hai.

Wasted power

P_{in} - P_{j e t}

ka physical meaning?

Heat jo spacecraft ko radiate karni padti hai.

6. Connections

Tsiolkovsky Rocket Equation — $Δ v = v_{e} ln (m_{0} / m_{f})$ ; explain karta hai kyun high $v_{e}$ matter karta hai.
Specific Impulse — fuel-economy metric $I_{s p} = v_{e} / g_{0}$ .
Chemical vs Electric Propulsion — energy-source vs reaction-mass decoupling.
Ion and Hall Thrusters — hardware jo high $v_{e}$ achieve karta hai.
Spacecraft Power Systems — solar arrays / RTGs $P_{in}$ set karte hain.
Thermal Control (radiators) — wasted $(1 - η) P_{in}$ kahan jaata hai.
Newton's Third Law — $F = \overset{m}{˙} v_{e}$ ki root.