Characteristic velocity c - = P_c A - ṁ — derivation, combustion efficiency measure: Worked examples

3.3.10 · D3 · Physics › Rocket Propulsion › Characteristic velocity c - = P_c A - ṁ — derivation, combu

Yeh page Characteristic velocity $c^*$ ke liye worked-example workshop hai. Hum yahan theory re-derive nahi karte — hum use use karte hain, aur deliberately har corner mein jaate hain: normal numbers, zero inputs, $γ$ ki limiting values, ek word problem, aur ek exam trap. Agar tumne abhi tak boxed formula nahi dekha, pehle parent note padho.

Neeche sab kuch $c^{*}$ ke in do roopoon par tika hai:

Intuition Do theoretical forms same kyun hain

Yeh alag dikhte hain lekin algebraically identical hain. Right form $\frac{1}{Γ} γ R T_{c}$ se shuru karo aur $\frac{1}{Γ}$ ko alag karo. $Γ$ bracket ke andar fraction ko flip karne se exponent ka sign flip hota hai, $\frac{2}{γ + 1}$ ko $\frac{γ + 1}{2}$ mein badalta hai: $\frac{1}{Γ} = \frac{1}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} .$ Ab $γ R T_{c}$ se multiply karo: $\frac{1}{Γ} γ R T_{c} = \frac{1}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} \cdot γ R T_{c} = (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} R T_{c} .$ Dono $γ$ factors cleanly cancel ho jaate hain (ek $1/Γ$ se, ek $γ R T_{c}$ ke andar se), bachta hai $(\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} R T_{c}$ . Left form $\frac{γ R T _{c}}{γ} (\frac{γ + 1}{2})^{\dots}$ recover karne ke liye, note karo ki $\frac{γ R T _{c}}{γ} = \frac{γ R T _{c}}{γ} = \frac{R T _{c}}{γ}$ — toh left form equals $\frac{1}{γ} (\frac{γ + 1}{2})^{\dots} R T_{c}$ , yaani left form $\frac{1}{γ}$ times upar ke right-hand expression ke barabar hai. Dusre words mein, truly-equal pair yeh hai: $c_{ideal}^{*} = \frac{1}{Γ} γ R T_{c} = \frac{γ R T _{c}}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} .$ Koi stray $γ$ na appear hota hai na disappear — har ek accounted for hai. Left form tab handy hai jab $Γ$ already pata ho; right form tab handy hai jab na pata ho.

Do symbols jo kabhi mix up nahi karne:

$A^{*}$ = throat area (woh pinch, jahan flow choked hoti hai). Yeh "star" area hai.
$\overset{m}{˙}$ = mass flow rate kg/s mein (woh "m-dot" — chota dot matlab "per second").

The scenario matrix

Is topic se jo bhi question aa sakta hai woh inhi cells mein se kisi ek mein fit hota hai. Har example neeche apni cell ka tag carry karta hai, aur har cell ke worked example ke andar apna unit sanity-check hai.

Cell	Kya cheez tricky banati hai	Example
A. Plain measured	bas $P_{c} A^{*} / \overset{m}{˙}$ mein plug in karo	Ex 1
B. Plain theoretical	pehle $Γ$ banao, phir $γ R T_{c}$	Ex 2
C. Efficiency	ratio measured / ideal	Ex 3
D. Solve backwards	given $c^{*}$ , find $\overset{m}{˙}$ or $P_{c}$	Ex 4
E. Degenerate: $\overset{m}{˙} \to 0$	limiting behaviour, kyun blow up hota hai	Ex 5
F. Limiting $γ$	$γ \to 1^{+}$ vs $γ \approx 1.67$ — sanity of $Γ$	Ex 6
G. Design mantra: change $M$	propellant swap karo, $H_{2}$ kyun jeet jaata hai	Ex 7
H. Word problem	real test stand, unit soup	Ex 8
I. Exam trap	koi tumhe $A_{e}$ ya nozzle detail de	Ex 9

Hum har cell A–I cover karte hain. Units utni hi matter karti hain jitni arithmetic — zyaadatar exam marks units ki wajah se jaate hain.

The examples

Worked example Example 1 — Cell A: plain measured

$c^{*}$ Test stand read karta hai $P_{c} = 6.0 MPa$ , $A^{*} = 0.008 m^{2}$ , $\overset{m}{˙} = 20 kg/s$ . $c^{*}$ find karo.

Forecast: Order of magnitude guess karo — chamber pressures $c^{*}$ ko roughly $1500$ – $2500$ m/s dete hain. Yeh kahan land karega?

Measured definition likho: $c^{*} = \frac{P _{c} A ^{*}}{m ˙}$ . Yeh step kyun? Koi nozzle info nahi diya gaya, aur zaroorat bhi nahi — yahi toh $c^{*}$ ke chamber quantity hone ka poora point hai.
MPa ko Pa mein convert karo: $6.0 MPa = 6.0 \times 1 0^{6} Pa$ . Yeh step kyun? Sirf SI units, warna end mein m/s meaningless ho jaata hai.
Plug in karo: $c^{*} = \frac{( 6.0 \times 1 0 ^{6} ) ( 0.008 )}{20} = \frac{48000}{20} = 2400 m/s$ . Yeh step kyun? Units clean hone ke baad pure arithmetic hai.

Verify: Units: $\frac{Pa \cdot m ^{2}}{kg/s} = \frac{N/m ^{2} \cdot m ^{2}}{kg/s} = \frac{N}{kg/s} = \frac{kg\cdotm/s ^{2}}{kg/s} = m/s$ . ✓ Expected band mein land kiya.

Worked example Example 2 — Cell B: theoretical

$c^{*}$ chemistry se $γ = 1.20$ , $T_{c} = 3200 K$ , $M = 22 g/mol$ . $c_{ideal}^{*}$ find karo.

Forecast: Theoretical $c^{*}$ usually real stand measurement se thoda zyaada hota hai (real engines kuch percent lose karte hain). Expect ~ $1500$ – $1600$ m/s.

Gas constant: $R = R_{u} / M$ . Pehle units fix karo: $R_{u} = 8314 J/(kmol\cdotK)$ aur $M = 22 g/mol = 22 kg/kmol$ , toh $R = 8314/22 = 378 J/(kg\cdotK)$ . Yeh step kyun? Formula ko specific gas constant $R$ (per kg) chahiye. $R_{u}$ ko J/(kmol·K) mein aur $M$ ko kg/kmol mein match karne se prefixes consistent rehte hain.
Exponent: $\frac{γ + 1}{2 ( γ - 1 )} = \frac{2.2}{0.4} = 5.5$ . Yeh step kyun? Yeh single power density ratio ( $\frac{1}{γ - 1}$ ) aur isentropic relations se sound-speed ko package karta hai.
$Γ$ build karo $= 1.2 (\frac{2}{2.2})^{5.5} = 1.0954 \times (0.9091)^{5.5} = 1.0954 \times 0.5875 = 0.6436$ . Yeh step kyun? $Γ$ saari $γ$ -dependence ek factor mein collect karta hai, toh formula ka baaki hissa sirf thermochemistry hai.
Thermochemical part: $γ R T_{c} = 1.2 \times 378 \times 3200 = 1.452 \times 1 0^{6} = 1205 m/s$ . Yeh step kyun? Yeh "khub jalaao, halki gas banao" ka core hai — $γ R T_{c}$ carry karta hai taaki hum seedha $Γ$ se divide kar sakein.
Divide karo: $c_{ideal}^{*} = \frac{γ R T _{c}}{Γ} = \frac{1205}{0.6436} = 1872 m/s$ . Yeh step kyun? Clean identity $c_{ideal}^{*} = \frac{1}{Γ} γ R T_{c}$ dono pieces ko ek division mein assemble karta hai.

Verify (doosra form, step by step): Left form $c_{ideal}^{*} = \frac{γ R T _{c}}{γ} (\frac{γ + 1}{2})^{5.5}$ use karo.

$\frac{γ R T _{c}}{γ} = \frac{1205}{1.2} = 1004$ .
$(\frac{γ + 1}{2})^{5.5} = (1.1)^{5.5} = 1.8642$ .
Product: $1004 \times 1.8642 = 1872 m/s$ . ✓

Dono forms $1872$ m/s exactly dete hain (rounding tak). Note karo yeh $1500$ – $1600$ ke forecast se thoda upar hai kyunki $γ = 1.2$ is $R$ ke saath thoda higher-energy case hai mental estimate se — forecast rough guess tha, aur math arbiter hai.

Worked example Example 5 — Cell E: degenerate input

$\overset{m}{˙} \to 0$ $c^{*} = P_{c} A^{*} / \overset{m}{˙}$ kya karta hai jab tum flow ko zero ki taraf throttle karte ho ( $P_{c}$ , $A^{*}$ hold karte hue)? Kya yeh physical hai?

Forecast: Formula $\overset{m}{˙}$ se divide karta hai. Guess karo "almost nothing" se divide karne par kya hota hai.

Limit lo: jaise $\overset{m}{˙} \to 0^{+}$ , $c^{*} = \frac{P _{c} A ^{*}}{m ˙} \to + \infty$ . Yeh step kyun? Yeh $1/ \overset{m}{˙}$ curve ka mathematical behaviour hai — yeh explode karta hai.
Poochho ki kya premise survive karti hai: koi propellant flow kiye bina fixed $P_{c}$ hold karna impossible hai. Real chambers ko $\overset{m}{˙}$ aur choking condition ( $M = 1$ , yaani gas throat pe speed of sound pe chal rahi ho — upar $M$ ki definition dekho) dono chahiye taaki tidy formula bilkul bhi hold ho sake. Yeh step kyun? Cell E sikhata hai ki formula ki math aur uski physical validity alag sawaal hain — limit real math hai lekin forbidden physics hai.
Honest physical limit: ek fixed propellant ke liye, $c^{*}$ ek constant hai ( $\propto T_{c} / M$ wala stuff). Agar $\overset{m}{˙}$ genuinely drop kare, toh $P_{c}$ bhi saath mein drop karta hai taaki ratio same rahe. Choked rahte hue tum ek ko doosre ke bina nahi hila sakte.

Verify: Ex-1 numbers se numerically, $\overset{m}{˙}$ ko $20$ se $10$ tak halving karte hue aur $P_{c}$ force karke fixed rakho toh apparent $c^{*}$ $2400$ se $4800$ m/s tak double ho jaata hai — physically flag karta hai ki tumne choked coupling violate kiya. ✓ Neeche ka figure yeh concrete banata hai.

Figure padhna: horizontal axis mass flow $\overset{m}{˙}$ (kg/s) hai, vertical axis apparent $c^{*}$ (km/s) hai, dono plot par explicitly labelled hain. Coral curve $c^{*} = P_{c} A^{*} / \overset{m}{˙}$ hai jisme $P_{c}$ artificially fixed rakha gaya hai. Ise leftward follow karo: jaise $\overset{m}{˙}$ buttery-shaded danger zone mein shrink hota hai, yeh sky ki taraf rocket karta hai — woh vertical wall $\overset{m}{˙} \to 0$ blow-up hai step 1 ka, figure par annotated hai. Mint dot (bhi labelled) ek hi honest reading hai: woh real operating point jahan $P_{c}$ aur $\overset{m}{˙}$ choking se lock hain, sensible $2400$ m/s deta hai jaise Ex 1 mein. Figure ka lesson: blow-up ek artefact hai yeh pretend karne ka ki $P_{c}$ stable rehta hai — nature tumhe us dot ke left pe kabhi nahi baithne deta.

Worked example Example 6 — Cell F:

$γ$ ki limiting values $γ$ ki poori physical range mein theoretical formula ko probe karo: soft limit $γ \to 1^{+}$ (heavy, many-atom exhaust) aur hard limit $γ \approx 1.67$ (monatomic gas jaise helium). $R T_{c} = 1.21 \times 1 0^{6} m^{2} / s^{2}$ hold karo (Ex-2 value) aur left form $c_{ideal}^{*} = \frac{γ R T _{c}}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}} = \frac{R T _{c}}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}}$ use karo.

Forecast: Konsa $γ$ bada $c^{*}$ deta hai — soft low- $γ$ gas ya stiff monatomic wala? Computing se pehle guess karo.

Hum $c_{ideal}^{*} = \frac{1}{Γ} γ R T_{c} = \frac{R T _{c}}{γ} (\frac{γ + 1}{2})^{\frac{γ + 1}{2 ( γ - 1 )}}$ evaluate karte hain chaar points par: $γ = 1.10$ , $γ = 1.13$ , $γ = 1.30$ , aur monatomic extreme $γ = 1.67$ . Yahan $R T_{c} = 1.21 \times 1 0^{6} = 1100$ . Yeh step kyun? Chaar points poori physical range bracket karte hain aur hume actually dikhne dete hain ki $Γ$ kaise behave karta hai.
$γ = 1.10$ par: exponent $= \frac{2.1}{0.2} = 10.5$ ; $Γ = 1.1 (2/2.1)^{10.5} = 1.0488 \times 0.6035 = 0.6329$ ; phir $c^{*} = \frac{γ R T _{c}}{Γ} = \frac{1.1 \cdot 1100}{0.6329} = \frac{1153.7}{0.6329} = 1823 m/s$ . Yeh step kyun? "Soft gas" edge, explicitly computed.
$γ = 1.13$ par: exponent $= \frac{2.13}{0.26} = 8.192$ ; $Γ = 1.13 (2/2.13)^{8.192} = 1.0630 \times 0.6062 = 0.6444$ . $γ = 1.30$ par: exponent $= \frac{2.3}{0.6} = 3.833$ ; $Γ = 1.3 (2/2.3)^{3.833} = 1.1402 \times 0.5254 = 0.5991$ ; $c^{*} = \frac{1.3 \cdot 1100}{0.5991} = \frac{1254.2}{0.5991} = 2094 m/s$ . Yeh step kyun? $Γ$ rise karta hai $0.6329$ ( $γ = 1.10$ ) se $0.6444$ ( $γ = 1.13$ ) tak phir fall karta hai $0.5991$ ( $γ = 1.30$ ) tak — ek genuine hump. Note karo $c^{*}$ khud badhta rehta hai kyunki $γ$ factor grow karta hai; hump $Γ$ mein hai, $c^{*}$ mein nahi.
Monatomic extreme $γ = 1.67$ : exponent $= \frac{2.67}{1.34} = 1.993$ ; $Γ = 1.67 (2/2.67)^{1.993} = 1.2923 \times 0.5619 = 0.7262$ ; $c^{*} = \frac{1.67 \cdot 1100}{0.7262} = \frac{1421.5}{0.7262} = 1958 m/s$ . Yeh step kyun? High- $γ$ end close karta hai: stiff monatomic gas kam $c^{*}$ deta hai $γ = 1.30$ peak se lekin softest gas se phir bhi zyaada — $γ$ (rising) aur $Γ$ -bracket (humped) ka interplay $γ \approx 1.3$ ke paas ek broad maximum produce karta hai.
Poori sweep mein $Γ$ ki sanity: $Γ$ $\approx [0.60, 0.73]$ mein bounded rehta hai aur $γ \in (1, 1.67]$ ke liye kabhi $0$ nahi touch karta ya blow up nahi hota. Yeh step kyun? Ek formula tabhi trustworthy hai jab apne poore domain par well-behaved ho — yeh check karna ki $Γ$ mein koi poles ya zeros nahi hain confirm karta hai ki $c_{ideal}^{*}$ har real propellant ke liye finite aur positive hai.

Verify: Chaar $Γ$ values $0.6329, 0.6444, 0.5991, 0.7262$ (sab appendix mein checked) $γ \approx 1.13$ ke paas $Γ$ mein hump aur bounded range confirm karte hain. Resulting $c^{*}$ : $1823, 2094, 1958$ m/s for $γ = 1.10, 1.30, 1.67$ — ek broad peak, monotone slide nahi. ✓

Worked example Example 7 — Cell G: propellant swap karo (design mantra)

$T_{c} = 3200 K$ , $γ = 1.20$ rakho lekin exhaust $M = 22$ g/mol se hydrogen-rich $M = 10$ g/mol mein switch karo. $c^{*}$ ka kya hota hai?

Forecast: Mantra kehta hai "halki gases banao." $M$ ko aadhe se zyaada kam karne se kitna faayda hoga?

$c_{ideal}^{*} \propto γ R T_{c} = γ R_{u} T_{c} / M$ , toh $c^{*} \propto 1/ M$ fixed $T_{c}, γ$ par. Yeh step kyun? Baaki sab ( $Γ$ , $T_{c}$ , $γ$ ) unchanged hai, toh sirf $1/ M$ factor move karta hai.
Ratio: $\frac{c _{10}^{*}}{c _{22}^{*}} = \frac{22}{10} = 2.2 = 1.483$ . Yeh step kyun? Direct proportionality se $Γ$ re-compute karna skip kar sakte hain.
New value: $c_{10}^{*} = 1872 \times 1.483 = 2776 m/s$ . Yeh step kyun? Combustion score mein $48%$ jump sirf exhaust halka karne se — exactly yahi reason hai ki $H_{2}$ propellants high-performance stages mein dominate karte hain.

Verify: Scratch se recompute karo. $R = 8314/10 = 831.4$ ; $γ R T_{c} = 1.2 \times 831.4 \times 3200 = 3.193 \times 1 0^{6} = 1787$ ; $Γ = 0.6436$ (unchanged); $c^{*} = 1787/0.6436 = 2776 m/s$ . ✓ Ratio method se match karta hai. Ratio ke units: dimensionless (g/mol over g/mol). ✓

Worked example Example 8 — Cell H: full word problem (real test stand)

Ek engineer $12$ -second burn log karta hai. Chamber $P_{c} = 3.5 MPa$ steady raha, throat diameter $d^{*} = 90 mm$ , aur propellant tanks ne burn ke dauraan $216 kg$ mass lose kiya. Is propellant ke liye predicted (ideal) $c^{*}$ $1720 m/s$ hai. $c_{measured}^{*}$ aur combustion efficiency report karo.

Forecast: Efficiency healthy $0.9$ – $0.99$ range mein land honi chahiye agar engine theek hai.

Mass flow: $\overset{m}{˙} = \frac{216 kg}{12 s} = 18 kg/s$ . Yeh step kyun? $\overset{m}{˙}$ hai total mass burned over burn time — tanks directly yeh batate hain.
Diameter se throat area: $A^{*} = π (d^{*} /2)^{2} = π (0.045)^{2} = 6.362 \times 1 0^{- 3} m^{2}$ . Yeh step kyun? Tumhe mm mein diameter diya gaya tha — m mein convert karna aur area mein turn karna zaroori hai, warna units collapse ho jaati hain.
Measured $c^{*} = \frac{P _{c} A ^{*}}{m ˙} = \frac{( 3.5 \times 1 0 ^{6} ) ( 6.362 \times 1 0 ^{- 3} )}{18} = \frac{22267}{18} = 1237 m/s$ . Yeh step kyun? Measured definition — yahan har quantity chamber/throat se hai, nozzle se kuch bhi nahi.
Efficiency: $η_{c^{*}} = 1237/1720 = 0.719 = 72%$ . Yeh step kyun? Real-over-ideal. Yeh kam hai — ek red flag.

Verify: $72%$ healthy $0.92$ – $0.99$ band se bahut neeche hai → is engine mein serious incomplete combustion hai ya throat/flow galat measure kiya gaya. Math check karta hai ( $0.719 \times 1720 = 1237$ ✓), lekin engineering verdict hai "investigate." ✓ Sab units m/s aur dimensionless mein resolve hote hain.

Worked example Example 9 — Cell I: exam trap (galat area)

Exam deta hai: $P_{c} = 5.0 MPa$ , throat area $A^{*} = 0.010 m^{2}$ , exit area $A_{e} = 0.080 m^{2}$ , $\overset{m}{˙} = 25 kg/s$ . $c^{*}$ compute karo. Trap kya hai?

Forecast: Do areas latkaaye gaye hain. $c^{*}$ mein konsa wala jaata hai — aur trap se galat answer kya nikalta hai?

Trap identify karo: $c^{*}$ throat area $A^{*}$ use karta hai, kyunki flow wahan choked hoti hai ( $M = 1$ ). Exit area $A_{e}$ nozzle se related hai aur sirf thrust coefficient $C_F$ ko affect karta hai. Yeh step kyun? Badi "obvious" area uthaa lena $c^{*}$ ki sabse common error hai.
Correct: $c^{*} = \frac{( 5.0 \times 1 0 ^{6} ) ( 0.010 )}{25} = 2000 m/s$ . Yeh step kyun? Same as parent Ex 1 — sirf throat.
Trap answer ( $A_{e}$ use karte hue): $\frac{( 5.0 \times 1 0 ^{6} ) ( 0.080 )}{25} = 16000 m/s$ — physically absurd (aath guna zyaada; koi chemistry yeh nahi deti). Yeh step kyun? Galat number dekhna hi tumhe train karta hai ki use sight pe reject karo.

Verify: Correct $c^{*} = 2000$ m/s (parent se match karta hai). Trap $16000$ m/s deta hai = $8 \times$ zyaada, exactly ratio $A_{e} / A^{*} = 8$ . ✓ Sanity: chemical propellants ke liye koi bhi $c^{*}$ ~ $3000$ m/s se upar ho toh suspicion trigger hona chahiye.

Recall Which area, and why?

$c^{*} = P_{c} A^{*} / \overset{m}{˙}$ mein, area throat hi kyun honi chahiye, exit nahi? ::: Kyunki flow choked (sonic, $M = 1$ ) hoti hai throat par — wahi par $\overset{m}{˙}$ $(P_{c}, T_{c}, A^{*})$ se lock hota hai. Exit area sirf $C_{F}$ (nozzle expansion) govern karta hai.

Recall Degenerate

$\overset{m}{˙}$ Jaise $\overset{m}{˙} \to 0$ formula $\to \infty$ — kya yeh real prediction hai? ::: Nahi. $P_{c}$ fixed rakhte hue $\overset{m}{˙} \to 0$ karna choked coupling violate karta hai; physically $P_{c}$ $\overset{m}{˙}$ ke saath girta hai toh $c^{*}$ ek constant rehta hai jo $T_{c}, M, γ$ se set hota hai.

Recall Design mantra check

Exhaust molecular mass $M$ ko half karne se $c^{*}$ kis factor se multiply hota hai? ::: $2 \approx 1.41$ , kyunki $c^{*} \propto 1/ M$ .

Recall Which

$γ$ wins? Fixed $R T_{c}$ par $γ \in (1, 1.67]$ across, kya peak $c^{*}$ soft end par hai, stiff end par, ya middle mein? ::: Middle mein — $c^{*}$ broad maximum tak badha $γ \approx 1.3$ ke paas ( $2094$ m/s) aur dono $γ = 1.10$ ( $1823$ ) aur $γ = 1.67$ ( $1958$ ) par kam tha.