Thrust coefficient C_F = F - (P_c A - ) — derivation: Worked examples

3.3.9 · D3 · Physics › Rocket Propulsion › Thrust coefficient C_F = F - (P_c A - ) — derivation

Yeh page thrust-coefficient derivation ke liye exhaustive drill-book hai. Hum master formula lete hain aur use har case class mein fire karte hain: perfectly matched nozzles, over-expanded ones, under-expanded ones, vacuum limit, infinite-expansion ceiling, degenerate zero-pressure inputs, ek word problem, aur ek exam twist jo ek trap chhupata hai.

Kisi bhi number se pehle, woh ek formula milte hain jis par neeche ke har example ka daromadar hai.

Formula Master formula (parent derivation se)

$C_{F} = momentum term C_{F, mom} \frac{2 γ ^{2}}{γ - 1} (\frac{2}{γ + 1})^{\frac{γ + 1}{γ - 1}} [1 - (\frac{P _{e}}{P _{c}})^{\frac{γ - 1}{γ}}] + pressure term \frac{P _{e} - P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}}$

$γ$ = ratio of specific heats (ek pure number, jaise $1.2$ ).
$P_{c}$ = chamber (stagnation) pressure, $P_{e}$ = nozzle exit pressure, $P_{a}$ = ambient pressure.
$A^{*}$ = throat area (nozzle ka sabse narrow cross-section, m² mein, jahan flow Mach 1 par choked hoti hai — dekho Choked Flow and the Throat).
$A_{e}$ = nozzle exit area (m²); ratio $A_{e} / A^{*}$ hi expansion ratio hai (exit area ÷ throat area), dekho Nozzle Expansion Ratio Ae-over-Astar.
$C_{F}$ reference force $P_{c} A^{*}$ ko multiply karke thrust deta hai: $F = C_{F} P_{c} A^{*}$ .
$C_{F, m a x}$ = sabse bada possible momentum term, tab milta hai jab gas infinitely expand kare ( $P_{e} \to 0$ ); yeh $K$ ke barabar hota hai aur sirf $γ$ par depend karta hai. Hum ise Cell E mein dekhenge.

Figure — Thrust coefficient C_F = F - (P_c A - ) — derivation

Upar ke figure mein, left panel nozzle hai: throat $A^{*}$ pinch-point hai (violet), exit $A_{e}$ chauda mouth hai (orange). Right panel teeno see-saw states stack karta hai — $(P_{e} - P_{a})$ ka sign seedha dekho ki kaun sa pan neeche hai.

The scenario matrix

Yeh har distinct case class hai jo is topic mein aapke saamne aa sakti hai. Neeche ke har worked example ko us cell ke saath tag kiya gaya hai jise woh fill karta hai.

#	Case class	Pehchaanane wali condition	Filled by
A	Perfectly matched	$P_{e} = P_{a}$ → pressure term $= 0$	Ex 1
B	Under-expanded	$P_{e} > P_{a}$ → pressure term $> 0$	Ex 2
C	Over-expanded	$P_{e} < P_{a}$ → pressure term $< 0$	Ex 3
D	Vacuum limit	$P_{a} = 0$ → pressure term hamesha $> 0$	Ex 4
E	Infinite-expansion ceiling	$P_{e} \to 0$ → $C_{F} \to C_{F, m a x}$	Ex 5
F	Back-out from data	given $F$ , find $C_{F}$	Ex 6
G	Same engine, two altitudes	pressure term ka sign flip hota hai	Ex 7
H	Word problem (design)	target ke liye $A_{e} / A^{*}$ choose karo	Ex 8
I	Exam twist (hidden trap)	student $T_{c}$ use karta hai — woh vanish ho jaana chahiye	Ex 9

Poore examples mein hum $γ = 1.2$ (typical hot rocket gas) use karte hain jab tak alag na bataya jaaye. Pehle un do building-block numbers ko pakka kar lete hain jo hum baar baar use karenge.

Cell A — Perfectly matched ( $P_{e} = P_{a}$ )

Worked example Ex 1 — matched sea-level nozzle

$γ = 1.2$ , $P_{c} = 7$ MPa, $P_{e} = P_{a} = 0.1$ MPa, $A_{e} / A^{*} = 25$ . $A^{*} = 0.01$ m² ke saath $C_{F}$ aur thrust nikalo.

Forecast: pressure term $(P_{e} - P_{a}) = 0$ hai, isliye $C_{F}$ sirf momentum term ke barabar hona chahiye — guess hai ki yeh $1.5$ – $1.6$ ke aas paas aayega.

Pressure ratio power. $(\frac{P _{e}}{P _{c}})^{n} = (0.1/7)^{0.1667} = (0.01429)^{0.1667} = 0.4933$ . Yeh step kyun? Yeh bracket maapti hai ki gas kitni expand hui; chhota $P_{e} / P_{c}$ matlab zyada expansion.
Momentum term. $C_{F, mom} = 5.06 (1 - 0.4933) = 5.06 \times 0.5067 = 2.564 = 1.601$ . Yeh step kyun? Yeh directed-jet contribution hai jo sirf $γ$ aur $P_{e} / P_{c}$ se set hota hai.
Pressure term. $\frac{P _{e} - P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}} = \frac{0}{7} \times 25 = 0$ . Yeh step kyun? Matched nozzle → exit gas aur atmosphere balance mein hain, kuch extra add karne ko nahi.
Total. $C_{F} = 1.601 + 0 = 1.601$ . Thrust $F = 1.601 \times 7 \times 1 0^{6} \times 0.01 = 112, 070$ N $\approx 112$ kN.

Verify: units — $C_{F}$ dimensionless hai; $P_{c} A^{*}$ = Pa·m² = N. $1.601 \times 7 MPa \times 0.01$ m² gives $\sim 1.1 \times 1 0^{5}$ N ✓. Sanity: $C_{F}$ typical $1.3$ – $2.0$ band ke andar hai ✓.

Cell B — Under-expanded ( $P_{e} > P_{a}$ )

Worked example Ex 2 — high-altitude nozzle, exit abhi bhi over-pressured

Same engine: $γ = 1.2$ , $P_{c} = 7$ MPa, $P_{e} = 0.1$ MPa, $A_{e} / A^{*} = 25$ , lekin ab aise uda rahe hain jahan $P_{a} = 0.04$ MPa.

Forecast: $P_{e} > P_{a}$ , isliye pressure term positive hai → $C_{F}$ ko matched value $1.601$ se zyada hona chahiye.

Momentum term unchanged. $P_{e} / P_{c}$ Ex 1 jaisa hi hai, isliye $C_{F, mom} = 1.601$ . Yeh step kyun? Momentum term $P_{a}$ kabhi nahi dekhta — sirf exit-to-chamber ratio.
Pressure term. $\frac{0.1 - 0.04}{7} \times 25 = \frac{0.06}{7} \times 25 = 0.2143$ . Yeh step kyun? Exit gas abhi bhi ambient se zyada push kar rahi hai $A_{e}$ area par → extra forward force.
Total. $C_{F} = 1.601 + 0.214 = 1.815$ .

Verify: pressure term positive hai jaise predict kiya, aur $1.815 > 1.601$ ✓. Kyunki $P_{a}$ $0.1$ se $0.04$ MPa tak gira, hum identical nozzle se thrust gain kar rahe hain — bilkul wahi jo Thrust Equation barhti altitude ke liye promise karta hai.

Cell C — Over-expanded ( $P_{e} < P_{a}$ )

Worked example Ex 3 — same nozzle sea level par forced down, lekin over-expanded

Ek bada vacuum-style nozzle: $γ = 1.2$ , $P_{c} = 6$ MPa, $P_{e} = 0.05$ MPa, $A_{e} / A^{*} = 30$ , sea level $P_{a} = 0.1$ MPa par test kiya.

Forecast: $P_{e} < P_{a}$ — atmosphere pushback de rahi hai — isliye pressure term negative jaayega aur $C_{F}$ ko uske momentum term se neeche khichega.

Momentum term. $(0.05/6)^{0.1667} = (0.008333)^{0.1667} = 0.4500$ . $C_{F, mom} = 5.06 (1 - 0.4500) = 5.06 \times 0.5500 = 2.783 = 1.668$ . Yeh step kyun? Ex 1 se deeper expansion (chhota $P_{e} / P_{c}$ ), isliye thoda bada momentum term.
Pressure term. $\frac{0.05 - 0.1}{6} \times 30 = \frac{- 0.05}{6} \times 30 = - 0.25$ . Yeh step kyun? Negative bracket = ambient pressure badi exit area par kamzor exit pressure ko haata deta hai → real thrust loss. Dekho Over- and Under-Expanded Nozzles.
Total. $C_{F} = 1.668 - 0.25 = 1.418$ .

Verify: pressure term ka sign negative hai, aur $1.418 < 1.668$ ✓. Yeh the classic over-expansion penalty hai — giant vacuum bell sea level par wasteful hai.

Cell D — Vacuum limit ( $P_{a} = 0$ )

Worked example Ex 4 — same bada nozzle ab space mein

Ex 3 ka engine vacuum mein le jao: $γ = 1.2$ , $P_{c} = 6$ MPa, $P_{e} = 0.05$ MPa, $A_{e} / A^{*} = 30$ , $P_{a} = 0$ .

Forecast: $P_{a} = 0$ ke saath pressure term $\frac{P _{e}}{P _{c}} \frac{A _{e}}{A ^{*}}$ hai, hamesha positive — yeh is $P_{e}$ par nozzle ka best $C_{F}$ hona chahiye.

Momentum term. Recompute (Ex 3 jaisa $P_{e} / P_{c}$ ): $(0.05/6)^{0.1667} = 0.4500$ , isliye $C_{F, mom} = 5.06 (1 - 0.4500) = 1.668$ . Yeh step kyun? Momentum term $P_{a}$ ignore karta hai; sirf $P_{e} / P_{c}$ matter karta hai, isliye Ex 3 se match karta hai.
Pressure term. $\frac{0.05 - 0}{6} \times 30 = \frac{0.05}{6} \times 30 = 0.25$ . Yeh step kyun? Koi atmosphere pushback nahi, isliye $A_{e}$ par poora exit pressure pure bonus thrust ban jaata hai.
Total. $C_{F, vac} = 1.668 + 0.25 = 1.918$ .

Verify: Ex 3 se compare karo — same nozzle $1.418$ (sea level) se $1.918$ (vacuum) tak gayi, exactly $0.50$ ka jump, jo $\frac{P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}} = \frac{0.1}{6} \times 30 = 0.50$ ke barabar hai ✓. Isliye space stages huge expansion ratios rakhte hain.

Cell E — Infinite-expansion ceiling ( $P_{e} \to 0$ )

Worked example Ex 5 —

$γ$ se set hoti hard maximum Expansion ko theoretical limit tak push karo: $P_{e} \to 0$ , abhi bhi $γ = 1.2$ . $C_{F, m a x}$ nikalo.

Forecast: jab $P_{e} \to 0$ , bracket $[1 - (P_{e} / P_{c})^{n}] \to 1$ , isliye $C_{F, mom}$ apni ceiling $K$ tak pahunchega. Guess $5.06 \approx 2.25$ .

Bracket limit. $(P_{e} / P_{c})^{n} \to 0^{0.1667} = 0$ , isliye $[1 - 0] = 1$ . Yeh step kyun? Infinite expansion saari available enthalpy extract kar leti hai — pressure ratio mein kuch nahi bachta.
Ceiling value. $C_{F, m a x} = K = 5.06 = 2.249$ . Yeh step kyun? Yeh sirf $γ$ par depend karta hai — ek fundamental limit jo koi nozzle nahi tod sakta, Isentropic Nozzle Flow se juda hua.
Interpret. $P_{a} = 0$ wale real engines $2.25$ ke paas pahunchte hain lekin kabhi nahi pahunte kyunki finite nozzle $P_{e} > 0$ chhodti hai.

Verify: Ex-4 numbers plug karo — uska $C_{F, vac} = 1.918$ safely $2.249$ ke neeche hai ✓. Bada $A_{e} / A^{*}$ lo (chhota $P_{e}$ ) aur $C_{F}$ $2.25$ ki taraf creep karta hai lekin kabhi cross nahi karta, bilkul ceiling ki demand ke anusaar.

Neeche ka figure dikhata hai ki $C_{F}$ vacuum mein expansion ratio ke saath kaise climb karta hai, is ceiling ki taraf flatten hote hue.

Us plot mein horizontal axis expansion ratio $A_{e} / A^{*}$ hai aur vertical axis $C_{F}$ hai. Magenta curve full vacuum $C_{F}$ hai, dashed violet curve sirf momentum term hai, aur dotted orange line ceiling $K = 2.25$ hai — notice karo ki magenta curve uski taraf bend karti hai lekin kabhi touch nahi karti.

Cell F — Measured data se back-out

Worked example Ex 6 — test-stand reading se engine grade karo

Sea-level test $F = 90$ kN, $P_{c} = 5$ MPa, $A^{*} = 0.008$ m² read karta hai. $C_{F}$ nikalo.

Forecast: sirf definition $C_{F} = F / (P_{c} A^{*})$ invert karo; koi $γ$ ki zaroorat nahi. Guess "thoda $2$ se upar".

Reference force. $P_{c} A^{*} = 5 \times 1 0^{6} \times 0.008 = 40, 000$ N. Yeh step kyun? Yeh coefficient ka denominator hai — "force yardstick".
Divide. $C_{F} = \frac{90 , 000}{40 , 000} = 2.25$ . Yeh step kyun? $C_{F}$ raw thrust reading ko nozzle ke liye size-independent grade mein convert karta hai. Companion combustion grade ke liye dekho Characteristic Velocity c-star.

Verify: units — N/(Pa·m²) = N/N = dimensionless ✓. $2.25$ ki value sea level ke liye suspiciously high hai — yeh ek bahut well-expanded, mildly under-pressured engine flag karta hai (ya measurement double-check ke liye).

Cell G — Same engine, do altitudes (sign flip)

Worked example Ex 7 — pressure term ka sign flip hote dekho

Engine: $γ = 1.2$ , $P_{c} = 6$ MPa, $P_{e} = 0.05$ MPa fixed, $A_{e} / A^{*} = 30$ . Sea level aur vacuum par $C_{F}$ compute karo, aur change bhi.

Forecast: momentum term dono altitudes par identical hai, isliye poora difference pressure term ka hai. Sea level ( $P_{a} = 0.1$ MPa) se vacuum ( $P_{a} = 0$ ) tak, $P_{a}$ $0.1$ MPa gir jaata hai; swing $\frac{Δ P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}} = \frac{0.1}{6} \times 30 = 0.50$ ke barabar honi chahiye. Guess: exactly $0.50$ ka rise.

Momentum term. Recompute: $(0.05/6)^{0.1667} = 0.4500$ , isliye $C_{F, mom} = 5.06 (1 - 0.4500) = 1.668$ . Yeh step kyun? Ex 3/Ex 4 jaisa $P_{e} / P_{c}$ → same momentum term; yeh altitude-blind hai.
Sea level ( $P_{a} = 0.1$ MPa). Pressure term $= \frac{0.05 - 0.1}{6} \times 30 = - 0.25$ → $C_{F} = 1.668 - 0.25 = 1.418$ . Yeh step kyun? Ground par over-expanded — atmosphere pushback de rahi hai.
Vacuum ( $P_{a} = 0$ ). Pressure term $= \frac{0.05}{6} \times 30 = + 0.25$ → $C_{F} = 1.668 + 0.25 = 1.918$ . Yeh step kyun? Koi back-pressure nahi — poora exit pressure thrust ban jaata hai.
Swing. $Δ C_{F} = 1.918 - 1.418 = 0.50$ . Yeh step kyun? Momentum term cancel ho jaata hai, isliye poora difference $\frac{Δ P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}}$ hai.

Verify: $\frac{0.1}{6} \times 30 = 0.50$ swing se exactly match karta hai ✓. Yeh Ex 3/Ex 4 jaisa hi engine hai, isliye numbers cross-check ho jaate hain.

Cell H — Word problem (design choice)

Worked example Ex 8 — thrust target ke liye vacuum nozzle size karo

Mission spec: vacuum ( $P_{a} = 0$ ) mein $F = 150$ kN deliver karo $P_{c} = 7$ MPa, $A^{*} = 0.01$ m² ke saath. Ground rules dete hain $γ = 1.2$ aur matched-at- $P_{e} = 0.1$ MPa design jisse $C_{F, mom} = 1.601$ (Ex 1 se). Humein kaisa expansion ratio $A_{e} / A^{*}$ chahiye?

Forecast: hume $C_{F} = F / (P_{c} A^{*})$ chahiye; phir momentum term subtract karo taaki isolate ho sake ki pressure term (isliye area ratio) ko kya supply karna hai.

Required $C_{F}$ . $C_{F} = \frac{150 , 000}{7 \times 1 0 ^{6} \times 0.01} = \frac{150 , 000}{70 , 000} = 2.143$ . Yeh step kyun? Definition sets karta hai total coefficient jo hume hit karna hai.
Required pressure term. $C_{F} - C_{F, mom} = 2.143 - 1.601 = 0.542$ . Yeh step kyun? Jo momentum term provide nahi kar sakta, woh vacuum pressure term ko karna hai.
Area ratio solve karo. Vacuum: $\frac{P _{e}}{P _{c}} \frac{A _{e}}{A ^{*}} = 0.542$ , isliye $\frac{A _{e}}{A ^{*}} = \frac{0.542}{0.1/7} = 0.542 \times 70 = 37.9$ . Yeh step kyun? $P_{e} / P_{c} = 0.1/7$ matched design se fixed hai, isliye sirf $A_{e} / A^{*}$ free hai.

Verify: $C_{F}$ rebuild karo $= 1.601 + \frac{0.1}{7} \times 37.9 = 1.601 + 0.542 = 2.143$ ✓, aur $F = 2.143 \times 70, 000 = 150, 000$ N ✓. $38$ ke aas paas expansion ratio upper stage ke liye realistic hai — dekho Specific Impulse Isp ki vacuum mein bade ratios $I_{s p}$ kyun raise karte hain.

Cell I — Exam twist ( $T_{c}$ trap)

Worked example Ex 9 — deliberately useless temperature

Exam deta hai: $γ = 1.2$ , $P_{c} = 7$ MPa, $P_{e} = P_{a} = 0.1$ MPa, $A_{e} / A^{*} = 25$ , aur $T_{c} = 3200$ K, $R = 350$ J/(kg·K). Yahan $T_{c}$ chamber temperature hai aur $R$ specific gas constant hai — exhaust ke per kilogram gas constant, units J/(kg·K) ke saath. $C_{F}$ nikalo. (Ex 1 jaisa hi physical case, lekin extra numbers thunse hue hain.)

Forecast: trap yeh hai ki $T_{c}$ aur $R$ aapko exit velocity compute karne ke liye tempt karte hain. Lekin $C_{F}$ $T_{c}$ aur $R$ se independent hai — woh derivation mein cancel ho jaate hain. Answer Ex 1 ka $1.601$ hona chahiye.

Cancellation spot karo. Parent ke Step 4 mein, $\overset{m}{˙} \propto \frac{1}{R T _{c}}$ aur $u_{e} \propto R T_{c}$ , isliye $\overset{m}{˙} u_{e} / (P_{c} A^{*})$ mein $R T_{c}$ cancel ho jaata hai. Yeh step kyun? $C_{F}$ sirf nozzle geometry aur ratios measure karta hai — kabhi absolute temperature ya specific gas constant nahi.
$T_{c}, R$ ignore karo; master formula use karo. Ex 1 se identical: $(0.1/7)^{0.1667} = 0.4933$ , isliye $C_{F, mom} = 5.06 (1 - 0.4933) = 1.601$ , aur pressure term $= 0$ (matched). Yeh step kyun? Same $γ$ , $P_{e} / P_{c}$ , $A_{e} / A^{*}$ , aur matched exit → same coefficient.
Total. $C_{F} = 1.601 + 0 = 1.601$ . Yeh step kyun? Extra data $T_{c}$ aur $R$ kabhi enter nahi hui — prove karta hai ki woh $C_{F}$ question ke liye decoration hain.

Verify: Ex 1 se unchanged, extra $T_{c} = 3200$ K aur $R = 350$ ke bawajood ✓. Agar aapne galti se unhe "use" kiya hota toh aap exit velocity compute karte $u_{e} \approx \frac{2 γ R T _{c}}{γ - 1} [1 - (P_{e} / P_{c})^{n}] = \frac{2 ( 1.2 ) ( 350 ) ( 3200 )}{0.2} (0.5067) \approx 2860$ m/s — ek real, meaningful number, lekin woh $c^{*}$ aur $I_{s p}$ ka hai, $C_{F}$ ka nahi.

Recall Quick self-test

Matched nozzle pressure term kiske barabar hota hai? ::: Zero (kyunki $P_{e} = P_{a}$ ). Over-expanded ka matlab kaun si inequality hai, aur pressure term ka sign kya hoga? ::: $P_{e} < P_{a}$ , negative term (thrust penalty). Sea level se vacuum jaane par $C_{F}$ ka gain kiske barabar hota hai? ::: $\frac{Δ P _{a}}{P _{c}} \frac{A _{e}}{A ^{*}}$ . Infinite-expansion ceiling $C_{F, m a x}$ sirf kisi par depend karti hai? ::: $γ$ par (for $γ = 1.2$ yeh $\approx 2.25$ hai). Kya $C_{F}$ chamber temperature $T_{c}$ ya specific gas constant $R$ par depend karta hai? ::: Nahi — dono cancel ho jaate hain; woh $\overset{m}{˙}$ aur $c^{*}$ mein rehte hain.