Solid rocket Isp derivation from grain properties: Worked examples

3.3.38 · D3 · Physics › Rocket Propulsion › Solid rocket Isp derivation from grain properties

Underlying formulas ke liye dekho Nozzle Isentropic Expansion, Saint-Robert Burn Rate Law, Characteristic Velocity c-star, aur Tsiolkovsky Rocket Equation.

The scenario matrix

Is topic ka har problem neeche ke cells mein se ek (ya blend) hota hai. Har worked example ko uska cell letter diya gaya hai.

Cell	Scenario class	Kya cheez tricky banati hai	Example
A	Baseline ideal $I_{s p}$ ( $p_{e} = p_{a}$ )	master formula plug karo, koi pressure term nahi	Ex 1
B	Grain scaling ( $A_{b}$ change karo)	check karo ki $I_{s p}$ invariant hai	Ex 2
C	Mismatched nozzle ( $p_{e} \neq = p_{a}$ )	$(p_{e} - p_{a}) A_{e}$ term rakhna zaroori hai	Ex 3
D	Limiting: $p_{e} / p_{c} \to 0$ (vacuum, infinite expansion)	bracket $\to 1$ , maximum possible $v_{e}$	Ex 4
E	Degenerate: $p_{e} / p_{c} \to 1$ (no expansion)	bracket $\to 0$ , thrust $\to 0$	Ex 4
F	Molar-mass sensitivity ( $M$ half karo)	scaling $v_{e} \propto 1/ M$	Ex 5
G	Equilibrium $p_{c}$ grain + throat se, aur $n \to 1$ ka danger	exponent $1/ (1 - n)$ blow up karta hai	Ex 6
H	Real-world word problem (web se burn time)	geometry ko time se link karo, sirf rate se nahi	Ex 7
I	Exam twist (altitude: over- vs. under-expanded)	$(p_{e} - p_{a})$ ka sign flip ho jaata hai	Ex 8

Example 1 — Cell A: baseline ideal $I_{s p}$

Worked example

Ek motor ek propellant jalata hai jisme $γ = 1.2$ hai, chamber temperature $T_{c} = 3000 K$ , exhaust molar mass $M = 0.025 kg/mol$ , pressure ratio $p_{e} / p_{c} = 0.01$ , aur yeh perfectly expanded hai ( $p_{e} = p_{a}$ ). Universal gas constant $R = 8.314 J/(mol\cdotK)$ . $I_{s p}$ nikalo.

Forecast: Solid motors typically $200$ – $260 s$ tak jaate hain. Aage padhne se pehle seconds mein ek number guess karo.

Step 1 — prefactor $\frac{2 γ}{γ - 1}$ . $\frac{2 ( 1.2 )}{1.2 - 1} = \frac{2.4}{0.2} = 12$ . Yeh step kyun? Yeh factor chamber enthalpy ko kinetic energy mein convert karne se aata hai (Nozzle Isentropic Expansion); yeh "har degree mein kitni energy" wala lever hai.

Step 2 — specific gas energy $\frac{R T _{c}}{M}$ . $\frac{8.314 \times 3000}{0.025} = 9.977 \times 1 0^{5} J/kg$ . Yeh step kyun? $\frac{R}{M}$ gas constant per kilogram hai; $T_{c}$ se multiply karo toh yeh thermal energy hai jo hot gas ka har kilogram carry karta hai.

Step 3 — expansion bracket. $\frac{γ - 1}{γ} = \frac{0.2}{1.2} = 0.1667$ ; $(0.01)^{0.1667} = 0.4642$ ; bracket $= 1 - 0.4642 = 0.5358$ . Yeh step kyun? Bracket woh fraction hai jo $p_{c}$ se $p_{e}$ tak expand karne par thermal energy motion mein actually convert hoti hai. Chhota $p_{e} / p_{c}$ → bada fraction.

Step 4 — exhaust velocity. $v_{e} = 12 \times 9.977 \times 1 0^{5} \times 0.5358 = 6.416 \times 1 0^{6} = 2533 m/s$ . Yeh step kyun? Teeno levers (energy conversion × thermal energy × extracted fraction) multiply karo aur square root lo kyunki energy $\propto v^{2}$ hoti hai.

Step 5 — $g_{0} = 9.81$ se divide karo. $I_{s p} = 2533/9.81 = 258 s$ . Yeh step kyun? $g_{0}$ ek bookkeeping constant hai jo m/s ko $I_{s p}$ ki "seconds" definition mein convert karta hai.

Verify: Units: $J/kg = m^{2} / s^{2} = m/s$ ✓. Value $258 s$ expected $200$ – $260$ band mein hai ✓. Koi bhi grain quantity ( $ρ_{p}, A_{b}, r$ ) appear nahi hui — parent ke key insight ke saath consistent ✓.

Example 2 — Cell B: grain surface ko double karna

Worked example

Same propellant jaise Ex 1 mein (toh $I_{s p} = 258 s$ , effective exhaust speed $c = g_{0} I_{s p}$ ). Grain mein $ρ_{p} = 1800 kg/m^{3}$ , burning area $A_{b} = 0.5 m^{2}$ , burn rate $r = 0.008 m/s$ hai. (a) Thrust $F$ nikalo. (b) Ab $A_{b}$ ko $1.0 m^{2}$ par double karo. $F$ aur $I_{s p}$ ka kya hoga?

Forecast: Burning wall ko double karo — kya rocket zyada efficient ho jaata hai, ya sirf zyada powerful?

Step 1 — mass flow. $\overset{m}{˙} = ρ_{p} A_{b} r = 1800 \times 0.5 \times 0.008 = 7.2 kg/s$ . Yeh step kyun? Volume burned per second = area × recession speed; times density = kg/s. Sirf yahaan geometry enter karti hai.

Step 2 — effective exhaust speed. $c = g_{0} I_{s p} = 9.81 \times 258 = 2531 m/s$ . Yeh step kyun? Definition ke hisaab se $I_{s p} = c / g_{0}$ , toh $c$ recover karne ke liye invert karo.

Step 3 — thrust. $F = \overset{m}{˙} c = 7.2 \times 2531 = 1.822 \times 1 0^{4} N \approx 18200 N$ . Yeh step kyun? Thrust = mass flow × woh speed jis par har kilogram nikalta hai.

Step 4 — $A_{b}$ double karo. $\overset{m}{˙}^{'} = 1800 \times 1.0 \times 0.008 = 14.4 kg/s$ , toh $F^{'} = 14.4 \times 2531 = 3.645 \times 1 0^{4} N \approx 36400 N$ . $I_{s p}$ abhi bhi $= F^{'} / (\overset{m}{˙}^{'} g_{0}) = 36450/ (14.4 \times 9.81) = 258 s$ — unchanged. Yeh step kyun? $I_{s p} = F / (\overset{m}{˙} g_{0})$ mein $F$ aur $\overset{m}{˙}$ dono $A_{b}$ ke saath scale hote hain, toh cancel ho jaata hai.

Verify: $F$ double hua ( $18200 \to 36400$ ) ✓; $I_{s p}$ identical ✓. Yeh directly Mistake ① ko refute karta hai — "bigger surface → higher $I_{s p}$ ."

Example 3 — Cell C: mismatched (sea-level, over-expanded)

Worked example

Ex 2 ka motor lo ( $\overset{m}{˙} = 7.2 kg/s$ , $v_{e} = 2533 m/s$ ) lekin ab nozzle exit over-expanded hai: exit pressure $p_{e} = 0.04 MPa$ jabki ambient $p_{a} = 0.101 MPa$ (sea level). Exit area $A_{e} = 0.09 m^{2}$ hai. Thrust aur actual $I_{s p}$ nikalo.

Forecast: $p_{e} < p_{a}$ hone par, kya pressure term bonus hai ya penalty?

Step 1 — momentum thrust. $F_{mom} = \overset{m}{˙} v_{e} = 7.2 \times 2533 = 1.824 \times 1 0^{4} N$ . Yeh step kyun? Pehla thrust term hamesha ejected gas ka raw momentum hota hai.

Step 2 — pressure thrust. $(p_{e} - p_{a}) A_{e} = (40000 - 101000) \times 0.09 = (- 61000) \times 0.09 = - 5490 N$ . Yeh step kyun? Jab exit pressure ambient se neeche hota hai, atmosphere exit plane par wapas push karta hai — yeh term genuinely negative hai.

Step 3 — total thrust. $F = 18240 - 5490 = 12750 N$ . Yeh step kyun? Full thrust equation $F = \overset{m}{˙} v_{e} + (p_{e} - p_{a}) A_{e}$ ; yahaan doosra term DROP nahi kar sakte (Mistake ④).

Step 4 — actual $I_{s p}$ . $I_{s p} = F / (\overset{m}{˙} g_{0}) = 12750/ (7.2 \times 9.81) = 180.5 s$ . Yeh step kyun? $I_{s p}$ hamesha real thrust use karta hai, toh pressure penalty ise ideal $258 s$ se neeche kheeench leti hai.

Verify: Over-expanded ( $p_{e} < p_{a}$ ) ne penalty di ✓; actual $180.5 s < 258 s$ ideal ✓. $(p_{e} - p_{a})$ ka sign negative hai jaisa chahiye ✓.

Example 4 — Cells D & E: expansion bracket ki do limits

Figure — Solid rocket Isp derivation from grain properties

Worked example

Ex 1 ki chemistry use karo ( $\frac{2 γ}{γ - 1} \frac{R T _{c}}{M} = 12 \times 9.977 \times 1 0^{5} = 1.197 \times 1 0^{7}$ ), bracket $B = 1 - (p_{e} / p_{c})^{(γ - 1) / γ}$ ke extreme cases evaluate karo. (D) Vacuum limit $p_{e} / p_{c} \to 0$ . (E) No-expansion limit $p_{e} / p_{c} \to 1$ .

Forecast: Ek limit sabse bada possible exhaust speed deti hai; doosri zero thrust deti hai. Kaun si kaun si hai?

Figure padhna: horizontal axis pressure ratio $p_{e} / p_{c}$ hai jo $0$ (left) se $1$ (right) tak jaati hai; vertical axis $I_{s p}$ seconds mein hai. Yellow curve hai $I_{s p} = 1.197 \times 1 0^{7} B / g_{0}$ . Ise left-to-right follow karo: yeh upar se shuru hoti hai (vacuum), pink dot se guzarti hai (Ex 1, $p_{e} / p_{c} = 0.01$ , $258 s$ ), aur right edge par origin tak girti hai.

Step 1 — Cell D bracket. Jaise $p_{e} / p_{c} \to 0$ , $(0)^{0.1667} \to 0$ , toh $B \to 1$ . Yeh step kyun? Zero pressure tak expand karna saari available thermal energy extract karta hai — physical ceiling yahi hai.

Step 2 — Cell D max speed. $v_{e, m a x} = 1.197 \times 1 0^{7} \times 1 = 3460 m/s$ ; $I_{s p, m a x} = 3460/9.81 = 353 s$ . Yeh step kyun? Yeh theoretically best hai jo yeh chemistry kabhi deliver kar sakti hai — yeh top-left ka blue dot hai jahan yellow curve left axis ke against flatten hoti hai.

Step 3 — Cell E bracket. Jaise $p_{e} / p_{c} \to 1$ , $(1)^{0.1667} = 1$ , toh $B \to 0$ . Yeh step kyun? Agar gas chamber pressure par exit karti hai toh usne kabhi accelerate nahi kiya — koi energy convert nahi hui.

Step 4 — Cell E speed. $v_{e} = 1.197 \times 1 0^{7} \times 0 = 0 m/s$ , toh $I_{s p} = 0 s$ . Yeh step kyun? Koi pressure drop nahi → koi push nahi. Yeh degenerate case hai — off-white dot jo exactly figure ke origin (right edge) par baitha hai.

Verify: $B$ monotonically $0$ (no expansion) se $1$ (vacuum) tak range karta hai ✓; $I_{s p}$ $0$ se $353 s$ tak ✓; Ex 1 case ( $p_{e} / p_{c} = 0.01$ , $258 s$ ) dono limits ke beech mein hai ✓.

Example 5 — Cell F: molar mass half karna

Worked example

Ex 1 ki sab cheezein fixed rakho lekin lighter-molecule exhaust par switch karo: $M$ $0.025$ se half karke $0.0125 kg/mol$ karo. Poora calculation dobaara kiye bina naya $I_{s p}$ predict karo, phir confirm karo.

Forecast: Parent note ka Forecast-then-Verify block $2$ factor claim karta hai. Sach hai?

Step 1 — $M$ dependence isolate karo. $v_{e} = \frac{2 γ}{γ - 1} \frac{R T _{c}}{M} B$ mein, sirf $M$ badalta hai; baaki sab constant $k$ hai. Toh $v_{e} = k / M$ , yaani $v_{e} \propto M^{- 1/2}$ . Yeh step kyun? Untouched terms factor out karo taaki scaling law akela khada ho.

Step 2 — scaling apply karo. $M$ half karne se $1/ M$ $2$ se multiply hoti hai, toh $v_{e}$ (aur isliye $I_{s p}$ ) $2 = 1.414$ se multiply hota hai. Yeh step kyun? Kyunki $v_{e}$ $1/ M$ ke square root par depend karta hai, factor-2 change factor- $2$ change ban jaata hai.

Step 3 — new value. $I_{s p, new} = 258 \times 1.414 = 365 s$ . Yeh step kyun? Known baseline scale karo — recompute karne se faster aur kam error-prone.

Verify: Direct recompute: $\frac{R T _{c}}{M} = \frac{8.314 \times 3000}{0.0125} = 1.995 \times 1 0^{6}$ ; $v_{e} = 12 \times 1.995 \times 1 0^{6} \times 0.5358 = 3582 m/s$ ; $I_{s p} = 3582/9.81 = 365 s$ ✓. Scaling prediction se match ✓. (Isliye $H_{2}$ / $H_{2} O$ -rich exhausts itne pasand kiye jaate hain.)

Example 6 — Cell G: equilibrium chamber pressure aur $n \to 1$ cliff

Worked example

Ek grain mein $a = 5 \times 1 0^{- 5}$ (SI), $ρ_{p} = 1800 kg/m^{3}$ , characteristic velocity $c^{*} = 1500 m/s$ , area ratio $A_{b} / A_{t} = 200$ hai. Equilibrium chamber pressure $p_{c}$ nikalo (a) $n = 0.35$ ke liye aur (b) $n = 0.9$ se contrast karo. Phir explain karo ki $n \geq 1$ kyun forbidden hai.

Forecast: Same grain, zyada exponent — kya $p_{c}$ thoda badhta hai ya explode karta hai?

Figure padhna: horizontal axis pressure exponent $n$ hai ( $0$ se $1$ ki taraf); vertical axis solution exponent $\frac{1}{1 - n}$ hai. Yellow curve pehle dheere chadhti hai, phir $n \to 1$ hote hi rocket ki tarah upar jaati hai; pink dashed line forbidden wall $n = 1$ par mark karti hai jahan curve infinity tak shoot karta hai. Blue dot $n = 0.35$ par hamare stable case ko mark karta hai.

Step 1 — base quantity. base $= ρ_{p} a c^{*} (A_{b} / A_{t}) = 1800 \times 5 \times 1 0^{- 5} \times 1500 \times 200 = 2.7 \times 1 0^{4}$ . Yeh step kyun? Mass generated ( $ρ_{p} A_{b} r$ ) ko mass vented ( $p_{c} A_{t} / c^{*}$ ) ke equal set karna, $r = a p_{c}^{n}$ ke saath, is base term ko ek power tak raised karta hai (Characteristic Velocity c-star).

Step 2 — $n = 0.35$ ke liye exponent. $\frac{1}{1 - n} = \frac{1}{0.65} = 1.538$ — yeh figure mein blue dot ki height hai. Yeh step kyun? $p_{c} = (base) p_{c}^{n}$ ko $p_{c}$ ke liye solve karne par $p_{c} = (base)^{1/ (1 - n)}$ milta hai.

Step 3 — evaluate karo. $p_{c} = (2.7 \times 1 0^{4})^{1.538} = 6.7 \times 1 0^{6} Pa = 6.7 MPa$ . Yeh step kyun? Yeh woh pressure hai jahan gas made = gas vented — ek stable balance point.

Step 4 — bade $n$ ka danger. $n = 0.9$ ke liye: $\frac{1}{1 - 0.9} = 10$ , toh $p_{c} = (2.7 \times 1 0^{4})^{10} \approx 2.06 \times 1 0^{44} Pa$ — physically absurd. Yeh step kyun? Exponent $1/ (1 - n)$ $n \to 1$ hote hi diverge karta hai (figure mein vertical pink wall). $n \geq 1$ ke liye pressure bump burn rate ko throat ke vent karne se zyada tezi se badhata hai → runaway. Stable motors ko $n < 1$ chahiye.

Verify: $(2.7 \times 1 0^{4})^{1.538} \approx 6.7 MPa$ ✓ (ek plausible solid-motor chamber pressure). $n = 0.9$ pressure nonphysically huge hai, instability argument (Mistake ③) confirm karta hai ✓.

Example 7 — Cell H: real-world word problem (burn time)

Worked example

Ek cylindrical end-burning grain (sirf ek flat face par jalta hai) mein web thickness (jitni length burn through karni hai) $w = 0.60 m$ hai aur constant $r = 0.008 m/s$ par jalta hai. Face area $A_{b} = 0.5 m^{2}$ , $ρ_{p} = 1800 kg/m^{3}$ hai, aur motor $I_{s p} = 258 s$ deliver karta hai. Nikalo (a) burn time, (b) total propellant mass, (c) total impulse.

Forecast: $\sim 18 kN$ (Ex 2 se) ke constant thrust ke saath, roughly kitne seconds fire karta hai, aur kya total impulse $I_{s p}$ par depend karta hai ya total mass par?

Step 1 — burn time. Face $r$ par recede karta hai, toh $t_{b} = w / r = 0.60/0.008 = 75 s$ . Yeh step kyun? Burn time = web thickness divided by recession speed — pure kinematics.

Step 2 — total propellant mass. Volume $= A_{b} \times w = 0.5 \times 0.60 = 0.30 m^{3}$ ; $m_{p} = ρ_{p} \times 0.30 = 1800 \times 0.30 = 540 kg$ . Yeh step kyun? Saara propellant woh face area hai jo poore web mein sweep hoti hai.

Step 3 — total impulse. $J = I_{s p} g_{0} m_{p} = 258 \times 9.81 \times 540 = 1.367 \times 1 0^{6} N\cdots$ . Yeh step kyun? Definition $I_{s p} = J / (g_{0} m_{p})$ se, total impulse $I_{s p} g_{0}$ times burned mass hai.

Verify (cross-check via thrust×time): $\overset{m}{˙} = ρ_{p} A_{b} r = 7.2 kg/s$ ; $F = \overset{m}{˙} g_{0} I_{s p} = 7.2 \times 9.81 \times 258 = 18220 N$ ; $J = F t_{b} = 18220 \times 75 = 1.367 \times 1 0^{6} N\cdots$ ✓. Aur $\overset{m}{˙} \times t_{b} = 7.2 \times 75 = 540 kg = m_{p}$ ✓. Do independent routes agree karte hain.

Example 8 — Cell I: exam twist (altitude, sign flip)

Worked example

Ek motor design kiya gaya hai taki $p_{e} = 0.101 MPa$ (sea level par matched), $A_{e} = 0.09 m^{2}$ , $\overset{m}{˙} = 7.2 kg/s$ , $v_{e} = 2533 m/s$ . Actual $I_{s p}$ compute karo (a) sea level par $p_{a} = 0.101 MPa$ aur (b) high altitude par $p_{a} = 0.010 MPa$ . Kaun sa bada hai aur kyun?

Forecast: Jaise rocket chadhta hai aur hawa patli hoti hai, $I_{s p}$ upar jaata hai ya neeche?

Step 1 — sea level, matched. $(p_{e} - p_{a}) = 0$ , toh $F = \overset{m}{˙} v_{e} = 7.2 \times 2533 = 18240 N$ ; $I_{s p} = 18 240/ (7.2 \times 9.81) = 258 s$ . Yeh step kyun? Matched nozzle pressure term ko khatam kar deta hai — yeh clean ideal case hai.

Step 2 — altitude, under-expanded. Ab $p_{e} > p_{a}$ , toh $(p_{e} - p_{a}) A_{e} = (101000 - 10000) \times 0.09 = 91000 \times 0.09 = + 8190 N$ — ek bonus. Yeh step kyun? Jab exit pressure ambient se zyada hota hai, exit gas patli hawa ke against abhi bhi baahir push kar raha hai; sign ab positive hai (Ex 3 ke opposite).

Step 3 — altitude par thrust aur $I_{s p}$ . $F = 18240 + 8190 = 26430 N$ ; $I_{s p} = 26 430/ (7.2 \times 9.81) = 374 s$ . Yeh step kyun? Same mass flow ke liye zyada thrust → zyada $I_{s p}$ .

Verify: Altitude $I_{s p}$ ( $374 s$ ) $>$ sea-level ( $258 s$ ) ✓ — isliye rockets patli upper atmosphere/space mein zyada efficient hote hain. Pressure-term ka sign $-$ (Ex 3, over-expanded) se flip hokar $+$ (yahaan, under-expanded) ho gaya, exactly woh exam trap jo Mistake ④ warn karta hai ✓.

Recall Main kaun se cell mein hoon? (quick decision tree)

Kya nozzle matched hai ( $p_{e} = p_{a}$ )? ::: Agar haan → Cell A, use karo $I_{s p} = v_{e} / g_{0}$ . Agar nahi → $(p_{e} - p_{a}) A_{e}$ rakho: negative agar $p_{e} < p_{a}$ (over-expanded, Ex 3), positive agar $p_{e} > p_{a}$ (under-expanded, Ex 8). Unhone grain ( $A_{b}, ρ_{p}, r$ ) change kiya lekin $I_{s p}$ pucha? ::: Trap — $I_{s p}$ unchanged rehta hai (Cell B). Sirf thrust aur burn time move karte hain. Unhone ratio ko extreme tak push kiya ( $p_{e} / p_{c} \to 0$ ya $\to 1$ )? ::: Cell D/E limits: bracket $\to 1$ (max $I_{s p}$ ) ya $\to 0$ (zero thrust). Unhone $a, n, c^{*}, A_{b} / A_{t}$ diye? ::: Cell G — equilibrium pressure $p_{c} = (ρ_{p} a c^{*} A_{b} / A_{t})^{1/ (1 - n)}$ ; $n \to 1$ blow-up watch karo.