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

KYA karte hain: seedha definition mein plug karo — koi nozzle geometry nahi chahiye, bas teen measured numbers. KYUN: $C_F$ ko define kiya gaya hai thrust divided by "reference force" $P_c{A}^{\ast }$ ke roop mein. Yeh kisi real test firing ko grade karne ka sabse fast tarika hai. $P_c{A}^{\ast }=5\times 10^6\times 0.008=4.0\times 10^4\text{N}=40000\text{N}$ $C_F=\frac{90000}{40000}=\boxed{2.25}$ High value hai — yeh nozzle kaafi amplifying kar raha hai.

False. $C_F=\frac{F}{P_c{A}^{\ast }}=\frac{\text{N}}{(\text{Pa})({\text{m}}^2)}=\frac{\text{N}}{({\text{N/m}}^2)({\text{m}}^2)}=\frac{\text{N}}{\text{N}}$ — force over force, isliye yeh dimensionless hai. Yeh ek pure amplification factor hai.

KYA karte hain: square root ko ek constant $K$ times bracket mein toddo, phir multiply karo. KYUN split karte hain: constant $K$ sirf $\gamma$ par depend karta hai aur kaafi problems mein reuse hota hai; bracket expansion carry karta hai. Constant: $\frac{2{\gamma }^2}{\gamma -1}=\frac{2(1.44)}{0.2}=14.4,{\left(\frac{2}{2.2}\right)}^{\frac{2.2}{0.2}}=(0.9091{)}^{11}=0.3505$ $K=14.4\times 0.3505=5.048$ Bracket (exponent $\frac{\gamma -1}{\gamma }=\frac{0.2}{1.2}=0.1667$): $1-(0.02{)}^{0.1667}=1-0.5214=0.4786$ $C_{F,\text{mom}}=\sqrt{5.048\times 0.4786}=\sqrt{2.416}=\boxed{1.554}$

KYA karte hain: momentum term mein pressure term add karo. KYUN: exit pressure $P_e$ ambient $P_a$ se alag hai, isliye exit area par unbalanced pressure contribute karta hai. $\frac{P_e-P_a}{P_c}\frac{A_e}{A^{\ast }}=\frac{0.14-0.10}{7}\times 25=\frac{0.04}{7}\times 25=+0.1429$ Yeh term positive hai kyunki $P_e>P_a$ hai (thoda under-expanded — dekho Over- and Under-Expanded Nozzles). $C_F=1.554+0.143=\boxed{1.697}$

KYA karte hain: sirf pressure term altitude ke saath change hoti hai; $C_{F,\text{mom}}$ geometry se fixed hai. (a) Sea level: $\frac{0.05-0.10}{6}\times 30=\frac{-0.05}{6}\times 30=-0.25\Rightarrow C_F=1.60-0.25=\boxed{1.35}$ (b) Vacuum: $\frac{0.05-0}{6}\times 30=\frac{0.05}{6}\times 30=+0.25\Rightarrow C_F=1.60+0.25=\boxed{1.85}$ KYUN flip hota hai: sea level par bahar ki hawa ($0.1$ MPa) exhaust exit ($0.05$ MPa) se zyada strong hai, isliye ambient pressure exit par back push karta hai — yeh ek penalty hai. Vacuum mein kuch bhi push back nahi karta, isliye wahi exit pressure ek bonus ban jaata hai. Same nozzle sirf space mein jaane se $\Delta C_F=0.50$ gain kar leta hai.

Figure kya dikhata hai: neeche wala curve is fixed nozzle ke liye ambient pressure $P_a$ ke against $C_F$ plot karta hai. Cyan line upar se neeche slope karti hai jab $P_a$ badhta hai (zyada bahar ka push-back). Dashed white line constant momentum coefficient $1.60$ mark karti hai; jahan cyan line isse cross karti hai (dotted vertical, $P_a=P_e=0.05$ MPa par) pressure term exactly zero hoti hai. Do amber dots hamare answers mark karte hain: vacuum bonus (upper-left, $C_F=1.85$) aur sea-level penalty (lower-right, $C_F=1.35$).

KYA karte hain: pressure term ko total $C_F$ se divide karo. KYUN: yeh ratio batata hai ki nozzle ka kitna score pressure bonus se aaya versus raw momentum se — yeh justify karta hai (ya nahi) ki ek bada bell carry karna worth it hai ya nahi. $\frac{\text{pressure term}}{C_F}=\frac{0.25}{1.85}=0.135=\boxed{13.5\%}$ Momentum thrust dominate karta hai (86.5%), lekin pressure bonus hi reason hai ki space engines giant bells kyon carry karte hain: vacuum mein $A_e/A^{\ast }$ area badhane se kabhi bhi sirf help hi milti hai.

(a) Momentum term. $K=5.048$ (2.1 se, same $\gamma$). Exponent $0.1667$; $P_e/P_c=0.1/7=0.01429$. KYUN bracket: yeh measure karta hai ki expansion ne gas ki kitni energy speed mein convert ki — chhota $P_e/P_c$ matlab fuller expansion aur bracket 1 ke paas. $1-(0.01429{)}^{0.1667}=1-0.4926=0.5074,C_{F,\text{mom}}=\sqrt{5.048\times 0.5074}=\sqrt{2.561}=1.600$ (b) Pressure term (vacuum, $P_a=0$). KYUN $P_a=0$: vacuum mein kuch bhi exit par push back nahi karta, isliye poora exit pressure $P_e$ thrust bonus ban jaata hai. $\frac{P_e-0}{P_c}\frac{A_e}{A^{\ast }}=\frac{0.1}{7}\times 25=0.3571$ (c) Donon contributions add karo — full coefficient ki definition momentum term plus pressure term hai: $C_F=1.600+0.357=\boxed{1.957}$ (d) $F=C_F{P}_c{A}^{\ast }=1.957\times (7\times 10^6)\times 0.01=\boxed{1.370\times 10^5\text{N}\approx 137\text{kN}}$. KYUN (d) kaam karta hai: $C_F$ by construction reference force $P_c{A}^{\ast }$ par multiplier hai. Multiply karo aur thrust wapas mil jaata hai — $T_c$, $R$, ya $\dot{m}$ jaanne ki zaroorat nahi. $T_c$ se yeh independence hi poora reason hai ki $C_F$ exist karta hai (dekho Characteristic Velocity c-star us part ke liye jo actually $T_c$ carry karta hai).

KYA karte hain: donon relations se $\dot{m}$ eliminate karo. $I_{sp}=\frac{F}{\dot{m}{g}_0}=\frac{C_F{c}^{\ast }\dot{m}}{\dot{m}{g}_0}=\frac{C_F{c}^{\ast }}{g_0}$ $I_{sp}=\frac{1.85\times 1700}{9.81}=\frac{3145}{9.81}=\boxed{320.6\text{s}}$ KYUN itna cleanly factor hota hai: exhaust speed $=C_F{c}^{\ast }$ hai; $g_0$ se divide karne par seconds milte hain. Yeh clean split dikhata hai — nozzle score $C_F$ times combustion score $c^{\ast }$ equals performance $I_{sp}$. Dekho Specific Impulse Isp.

KYA karte hain: bracket $[1-(P_e/P_c{)}^{(\gamma -1)/\gamma }]\to 1$ set karo (kyunki $P_e\to 0$), sirf $\sqrt{K}$ bachta hai. $C_{F,\max }=\sqrt{K}=\sqrt{5.048}=\boxed{2.247}$ KYUN yeh ceiling hai: ek infinitely lamba nozzle bhi gas ki poori available energy sirf ek baar extract kar sakta hai. Jo bachta hai woh sirf $\gamma$ se set hota hai — gas ki apni thermodynamic springiness. Koi bhi geometry $\sqrt{K}$ ko beat nahi kar sakti; softer gas (lower $\gamma$) ki ceiling higher hoti hai kyunki uski zyada energy directed motion mein convert ho sakti hai.

Argument. Nozzle ko lamba karna (bada $A_e/A^{\ast }$) $P_e$ ko kam karta hai. Yeh momentum term badhata hai (bada bracket) lekin pressure term $\frac{P_e-P_a}{P_c}\frac{A_e}{A^{\ast }}$ negative ho jaata hai jab $P_e<P_a$ ho jaata hai. Yeh dono effects trade off karte hain, aur calculus dikhata hai ki total $C_F$ tab peak karta hai jab exactly $P_e=P_a$ ho — wahan pressure term zero cross karta hai. Physically: koi bhi exit pressure mismatch energy waste karta hai (ya to un-expanded pressure bacha rehta hai, ya over-expansion sideways push karta hai). 3.1 ke engine ke liye sea level par numeric confirmation ($P_a=0.1$ MPa). Agar hum nozzle re-tune kar sakein taaki $P_e=P_a=0.1$ MPa ho, toh pressure term vanish ho jaata hai aur $C_F=C_{F,\text{mom}}$ us naye (bade) matched expansion par hota hai. Diye gaye fixed geometry mein $P_e=0.05<P_a$ hai, engine over-expanded hai aur $0.25$ lose karta hai (3.1a se), jo confirm karta hai ki un-matched case matched optimum se neeche hai: $C_{F,\text{sea, actual}}=1.35<C_{F,\text{mom, matched}}=\mathrm{1.60.}$

Bracket at $P_e/P_c=0.10$: $1-(0.10{)}^{0.1667}=1-0.6813=0.3187$, toh $C_{F,\text{mom}}=\sqrt{5.048\times 0.3187}=\sqrt{1.609}=1.269$. Bracket at $P_e/P_c=0.01$: $1-(0.01{)}^{0.1667}=1-0.4642=0.5358$, toh $C_{F,\text{mom}}=\sqrt{5.048\times 0.5358}=\sqrt{2.705}=1.645$. Factor: $1.645/1.269=\boxed{1.296}$ — exit pressure mein tenfold drop se momentum coefficient mein lagbhag 30% gain. Diminishing lekin real hai; isliye vacuum bells itne bade hote hain.

$C_F$ chamber temperature $T_c$ se independent kyun hai? ::: Kyunki mass flow se $\sqrt{R{T}_c}$ aur exit velocity se $\sqrt{R{T}_c}$ $P_c{A}^{\ast }$ se divide karne par cancel ho jaate hain — $C_F$ sirf ratios ($P_e/P_c$, $P_a/P_c$) aur geometry ($A_e/A^{\ast }$) dekhta hai, absolute temperature kabhi nahi.