Exercises — Chamber-to-exit relation - all quantities as f(M_e, γ)

$\Phi =1+\frac{\gamma -1}{2}{M}_e^2=1+\frac{0.4}{2}(2.0{)}^2=1+0.2\cdot 4=1+0.8=\mathrm{1.8.}$

Temperature ratio bas ${\Phi }^{-1}$ hai: $\frac{T_e}{T_0}=\frac{1}{1.8}=\mathrm{0.556.}$ Toh exit gas chamber temperature (kelvin mein) ka thoda zyada aadha hai. Humne kya kiya: $\Phi$ ki definition mein substitute kiya, phir sabse chhota relation use kiya. Kyun: temperature ko bas ${\Phi }^{-1}$ chahiye, koi exponent gymnastics nahi.

Exponent hai $-\frac{\gamma }{\gamma -1}=-\frac{1.4}{0.4}=-\mathrm{3.5.}$ $\frac{P_e}{P_0}={\Phi }^{-3.5}=1.8^{-3.5}=\mathrm{0.1278.}$ Exit static pressure chamber pressure ka lagbhag 12.8% hai. Exponent kyun: isentropic link $P/P_0=(T/T_0{)}^{\gamma /(\gamma -1)}$ temperature ke ${\Phi }^{-1}$ ko pressure ke liye ${\Phi }^{-\gamma /(\gamma -1)}$ mein badal deta hai.

Exponent: $-\frac{1}{\gamma -1}=-\frac{1}{0.4}=-\mathrm{2.5.}$ $\frac{{\rho }_e}{{\rho }_0}={\Phi }^{-2.5}=1.8^{-2.5}=\mathrm{0.230.}$ Ideal gas law $\rho \propto P/T$ se check karo: $\frac{P_e/P_0}{T_e/T_0}=\frac{0.1278}{0.556}=\mathrm{0.230.}✓$ Check kyun kaam karta hai: density ka exponent $-\gamma /(\gamma -1)-(-1)=-1/(\gamma -1)$ hota hai, bilkul pressure aur temperature exponents ka difference.

$\Phi =1+\frac{0.3}{2}(2.5{)}^2=1+0.15\cdot 6.25=1+0.9375=\mathrm{1.9375.}$

Temperature: $T_e=T_0{\Phi }^{-1}=\frac{3000}{1.9375}=1548.4\text{K}.$

Pressure: exponent $-\gamma /(\gamma -1)=-1.3/0.3=-4.3333$. $P_e=P_0{\Phi }^{-4.3333}=50\cdot 1.9375^{-4.3333}=50\cdot 0.05612=2.806\text{bar}.$

Chamber density ideal gas law ${\rho }_0=P_0/(R{T}_0)$ se, jahan $R$ specific gas constant hai. $P_0$ convert karo: $50\text{bar}=50\times 10^5\text{Pa}=5\times 10^6\text{Pa}$. ${\rho }_0=\frac{5\times 10^6}{400\cdot 3000}=\frac{5\times 10^6}{1.2\times 10^6}=4.167{\text{kg/m}}^3.$ Units pehle convert kyun karein: gas-law formula ko SI pascals chahiye, bar nahi; inhe silently mix karna tumhari density ko $10^5$ se scale kar deta hai.

$V_e=M_e\sqrt{\frac{\gamma R{T}_0}{\Phi }}$ use karo. Yeh formula kyun: $V_e=M_e{a}_e$ Mach number ki definition se, aur $a_e=\sqrt{\gamma R{T}_e}$ local speed of sound hai; $T_e=T_0/\Phi$ (hamara temperature relation) substitute karne par exit temperature chamber quantities mein fold ho jaata hai, isliye humein $a_e$ alag se compute karne ki zaroorat nahi. $\gamma R{T}_0=1.3\cdot 400\cdot 3000=1.56\times 10^6.$ $\frac{\gamma R{T}_0}{\Phi }=\frac{1.56\times 10^6}{1.9375}=8.052\times 10^5.$ $V_e=2.5\cdot \sqrt{8.052\times 10^5}=2.5\cdot 897.3=2243\text{m/s}.$ Square root kya hai: $\sqrt{\gamma R{T}_e}$ local speed of sound $a_e$ hai; $M_e$ se multiply karne par "kitne sound-speeds fast" ek actual speed m/s mein ban jaata hai.

Exit par kinetic energy per kg: $\frac{1}{2}{V}_e^2=\frac{1}{2}(2243{)}^2=2.516\times 10^6\text{J/kg}.$

Chamber thermal energy per kg (enthalpy) $h_0=c_p{T}_0$: $c_p=\frac{1.3\cdot 400}{0.3}=1733.3\text{J/(kg⋅K)},h_0=1733.3\cdot 3000=5.200\times 10^6\text{J/kg}.$ Fraction $=\frac{2.516\times 10^6}{5.200\times 10^6}=0.484\approx 48.4\%.$ Sanity: yeh $1-T_e/T_0=1-1/1.9375=0.484$ ke barabar hona chahiye, kyunki energy conservation kehti hai ki khoyi enthalpy haasil ki kinetic energy ban jaati hai. Match karta hai. ✓

${\Phi }_A=1+\frac{0.20}{2}(9)=1+0.9=1.9,T_e/T_0{|}_A=1/1.9=\mathrm{0.5263.}$ ${\Phi }_B=1+\frac{0.25}{2}(9)=1+1.125=2.125,T_e/T_0{|}_B=1/2.125=\mathrm{0.4706.}$

Nozzle A (kam $\gamma$) exit par hotter hai. Ratios ka ratio: $\frac{0.5263}{0.4706}=\mathrm{1.118.}$ A ka exit temperature ratio lagbhag 12% zyada hai. Kam $\gamma$ kyun hotter rehta hai: ek chhota $\gamma -1$ kinetic term $\frac{\gamma -1}{2}{M}_e^2$ ko shrink karta hai, isliye kam thermal energy motion mein drain hoti hai — parent ki us remark ke saath consistent hai ki H₂-rich (low $\gamma$) exhausts preferred hote hain.

$f(M)={\Phi }^{-n}$ likho jahan $n=\gamma /(\gamma -1)=3.5$ aur $\Phi =1+0.2M^2$.

Yahan derivative kyun: hum chahte hain "output kitna wiggle karta hai jab input thoda wiggle karta hai" — yeh bilkul derivative $df/dM$ ka kaam hai, local slope.

$\frac{df}{dM}=-n{\Phi }^{-n-1}\cdot \frac{d\Phi }{dM}=-n{\Phi }^{-n-1}\cdot (0.4M).$ Fractional sensitivity $\frac{d\ln f}{d\ln M}=\frac{M}{f}\frac{df}{dM}=-n\cdot \frac{0.4M^2}{\Phi }.$ $M=2$ par: $\Phi =1.8$, toh $\frac{d\ln f}{d\ln M}=-3.5\cdot \frac{0.4\cdot 4}{1.8}=-3.5\cdot \frac{1.6}{1.8}=-\mathrm{3.111.}$ Toh $M_e$ mein $+1\%$ change se $P_e/P_0$ mein lagbhag $-3.11\%$ change hoti hai. Ise padhna: pressure ratio Mach number par bahut leveraged hai — chhote Mach gains bade pressure drops khareedte hain (yahi thrust banata hai).

Density ratio $={\Phi }^{-1/(\gamma -1)}$ aur temperature ratio $={\Phi }^{-1}$. Toh ${\left(\frac{T_e}{T_0}\right)}^{1/(\gamma -1)}={\left({\Phi }^{-1}\right)}^{1/(\gamma -1)}={\Phi }^{-1/(\gamma -1)}=\frac{{\rho }_e}{{\rho }_0}.✓$ Numbers ($\gamma =1.4$, toh $1/(\gamma -1)=2.5$): $T_e/T_0=1/1.8=0.5556$; $0.5556^{2.5}=0.2299$, jo L1.3 ke $0.230$ se match karta hai. Yeh isentropic $\rho \propto T^{1/(\gamma -1)}$ kyun hai: yeh density-temperature form hai "adiabatic, reversible" ki — wahi physics, alag variables mein.

$\frac{A_e}{A^{\ast }}=\frac{1}{M_e}{\left[\frac{2\Phi }{\gamma +1}\right]}^{\frac{\gamma +1}{2(\gamma -1)}}.$ $\Phi =1+0.2(9)=\mathrm{2.8.}$ Exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.4}{0.8}=\mathrm{3.0.}$ $\frac{2\Phi }{\gamma +1}=\frac{2\cdot 2.8}{2.4}=\frac{5.6}{2.4}=\mathrm{2.3333.}$ $\frac{A_e}{A^{\ast }}=\frac{1}{3.0}\cdot 2.3333^3=\frac{1}{3.0}\cdot 12.70=\mathrm{4.235.}$ Diverging cone ko throat area ka lagbhag 4.2× tak kholna hoga (supersonic branch lete hue, kyunki hum $M_e=3>1$ chahte hain). Figure kya dikhata hai: neeche, throat (jahan $M=1$) pinch hai; area grow karta hai $M_e=3$ reach karne ke liye.

$\Phi =1+\frac{0.25}{2}(3.2{)}^2=1+0.125\cdot 10.24=1+1.28=\mathrm{2.28.}$

Exit velocity $V_e=M_e\sqrt{\frac{\gamma R{T}_0}{\Phi }}$ se. Yeh formula phir kyun: yeh bas $V_e=M_e{a}_e$ hai jahan $a_e=\sqrt{\gamma R{T}_e}$ aur $T_e=T_0/\Phi$ substitute hai, toh chamber quantities mein ek single expression poora kaam karta hai. $\gamma R{T}_0=1.25\cdot 480\cdot 3400=2.04\times 10^6.$ $V_e=3.2\sqrt{\frac{2.04\times 10^6}{2.28}}=3.2\sqrt{8.947\times 10^5}=3.2\cdot 945.9=3027\text{m/s}.$

Area ratio: exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.25}{0.5}=\mathrm{4.5.}$ $\frac{2\Phi }{\gamma +1}=\frac{2\cdot 2.28}{2.25}=\frac{4.56}{2.25}=\mathrm{2.0267.}$ $\frac{A_e}{A^{\ast }}=\frac{1}{3.2}\cdot 2.0267^{4.5}=\frac{1}{3.2}\cdot 25.10=\mathrm{7.844.}$ One-pass summary: $V_e\approx 3027\text{m/s}$, area ratio $\approx 7.84$ (supersonic branch). Zyada $M_e$ ke liye L4.1 ke $M_e=3$ case se bahut wider exit chahiye — exponent ise amplify karta hai.

$A_e=7.844\cdot 0.010=0.07844{\text{m}}^2.$

Diameter ratio: kyunki $A\propto d^2$, $\frac{d_e}{d^{\ast }}=\sqrt{A_e/A^{\ast }}=\sqrt{7.844}=\mathrm{2.801.}$ Toh exit diameter throat diameter ka lagbhag 2.8× hai. Square root kyun: area diameter ke square ke saath scale karta hai, isliye 7.8× area increase sirf 2.8× diameter increase hai — nozzles area ratio se kam dramatic dikhte hain.

Hum $P_e/P_0={\Phi }^{-\gamma /(\gamma -1)}$ ko invert karte hain. Exponent $n=1.3/0.3=4.3333$. $\Phi ={\left(\frac{P_e}{P_0}\right)}^{-1/n}=0.010^{-1/4.3333}=0.010^{-0.23077}=\mathrm{2.899.}$ Phir $\Phi =1+\frac{0.3}{2}{M}_e^2=1+0.15M_e^2$, toh $M_e^2=\frac{\Phi -1}{0.15}=\frac{1.899}{0.15}=12.66,M_e=\mathrm{3.559.}$ Humne kya kiya: har formula ulta chalaya — $\Phi$ recover karne ke liye log-style root liya, phir $M_e$ mein quadratic piece solve kiya. Yeh design workflow kyun hai: practice mein tum ek target expansion specify karte ho (altitude se set pressure ratio) aur Mach number back out karte ho, phir area ratio.

$\Phi =2.899$ (L5.1 se). Exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.3}{0.6}=\mathrm{3.8333.}$ $\frac{2\Phi }{\gamma +1}=\frac{2\cdot 2.899}{2.3}=\frac{5.798}{2.3}=\mathrm{2.5209.}$ $\frac{A_e}{A^{\ast }}=\frac{1}{3.559}\cdot 2.5209^{3.8333}=\frac{1}{3.559}\cdot 33.86=\mathrm{9.514.}$ Nozzle ko lagbhag 9.5 ka expansion ratio chahiye (supersonic branch) — ek bada, high-altitude bell. Link: deeper vacuum (kam ambient $P_e$ target) hamesha $M_e$ aur $A_e/A^{\ast }$ dono upar push karta hai.

$M_e=1$ set karo: tab $\Phi =1+\frac{\gamma -1}{2}=\frac{\gamma +1}{2}.$ Toh $\frac{2\Phi }{\gamma +1}=\frac{2}{\gamma +1}\cdot \frac{\gamma +1}{2}=1.$ Kuch bhi kisi bhi power par $1$ hi rehta hai, aur $1/M_e=1/1=1$, isliye $A/A^{\ast }=1$ milta hai. Kyun hona chahiye: throat hai $M=1$ station definition se, isliye uska area trivially $A^{\ast }$ ke barabar hai. Numerically: $\gamma =1.4\Rightarrow \frac{2\cdot 1.2}{2.4}=1$, exponent $3$, result $1$; $\gamma =1.2\Rightarrow \frac{2\cdot 1.1}{2.2}=1$, result $1$. ✓ Yeh self-consistency ek accha exam sanity check hai — aur yeh woh single point hai jahan subsonic aur supersonic branches milti hain.

Saare exit ratios ek single factor $\Phi$ ki powers hain — ise aur uske teen exponents name karo. ::: $\Phi =1+\frac{\gamma -1}{2}{M}_e^2$; temperature ${\Phi }^{-1}$, pressure ${\Phi }^{-\gamma /(\gamma -1)}$, density ${\Phi }^{-1/(\gamma -1)}$. $\Phi \ge 1$ kyun hona chahiye? ::: Kyunki $M_e^2\ge 0$ aur $\gamma >1$, toh $\frac{\gamma -1}{2}{M}_e^2\ge 0$ aur $\Phi =1+(\text{woh})\ge 1$. Ek diye gaye $A_e/A^{\ast }>1$ ke liye, area relation kitne Mach numbers solve karta hai, aur hum kaun sa choose karte hain? ::: Do — ek subsonic, ek supersonic; rocket jet ke liye hum supersonic ($M_e>1$) branch choose karte hain.