Exercises — Area-Mach number relation A - A - = f(M) — isentropic flow

Figure mein U-shaped valley ka sabse neecha point dekho. Uska lowest point $A/A^{\ast }=1$ par hai, aur yeh sirf ==$M=1$== par hota hai (Mach one, sonic point). Toh $A/A^{\ast }=1$ aur $M=1$. KYU: throat woh akela spot hai jahan geometry flow ko sound ki speed tak pahunchne deti hai.

Figure mein valley ke aaar-paar height $2$ par ek horizontal line khiincho — yeh curve ko do jagah cross karti hai. Throat ke left mein ($M<1$) subsonic root hai; right mein ($M>1$) supersonic root hai. KYU unique nahi: area–Mach curve U-shaped hai (pehle neeche jaati hai phir upar), toh ek height do Mach numbers se correspond karti hai. Geometry akele nahi bata sakti kaunsa; back-pressure decide karta hai.

$a=\sqrt{\gamma RT}=\sqrt{1.4\cdot 287\cdot 300}=\sqrt{120540}\approx 347\text{m/s}$. KYU yeh formula: sound ek tiny pressure wave hai; uski speed is baat par depend karti hai ki gas kitni "stiff" aur kitni "heavy" hai, jo $\gamma RT$ se capture hoti hai. Zyada garam gas ⇒ tez sound.

KYA: hum Tool A directly use karte hain kyunki $M$ pata hai. Bracket ke andar: $1+\frac{\gamma -1}{2}{M}^2=1+0.2\cdot 4=1.8$. Factor $\frac{2}{\gamma +1}=\frac{2}{2.4}=0.8333$, toh bracket $=0.8333\cdot 1.8=1.5$. Exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.4}{0.8}=3$. $\frac{A}{A^{\ast }}=\frac{1}{2}(1.5{)}^3=\frac{3.375}{2}=\mathrm{1.6875.}$ Toh $A/A^{\ast }\approx ==1.688==$. KYU: supersonic $M$ par aap valley ki right wall par chadhte ho, toh area sonic minimum se kaafi upar hona chahiye.

KYA: ab Tool B use karo (Mach → thermodynamics). $\frac{T_0}{T}=1+0.2\cdot 4=1.8\Rightarrow T=\frac{300}{1.8}=166.7\text{K}$. $\frac{p_0}{p}=(1.8{)}^{\gamma /(\gamma -1)}=(1.8{)}^{3.5}=7.824\Rightarrow p=\frac{500}{7.824}=63.9\text{kPa}$. KYU: supersonic flow expand aur cool ho gayi hai — gas ne internal energy ko speed ke liye trade kar diya, toh $T$ aur $p$ reservoir values se kaafi neeche gir jaate hain.

Bracket: $1+0.2\cdot 0.25=1.05$; times $0.8333=0.875$. $\frac{A}{A^{\ast }}=\frac{1}{0.5}(0.875{)}^3=2\cdot 0.6699=\mathrm{1.3398.}$ $A/A^{\ast }\approx ==1.340==$. KYU: $M=0.5$ par aap valley ki left wall par ho — minimum se phir bhi upar, kyunki flow ko sonic se neeche slow karne ke liye (throat ke relative) chauda karna padega.

KYA: $A/A^{\ast }=14/8=1.75$. Tool A ko algebraically $M$ ke liye solve nahi kiya ja sakta, toh hum iterate karte hain (ya table padhte hain). KYU iterate: $f(M)$ mein $1/M$ aur $M^2$ ki power mix hoti hai — koi clean inverse nahi.

Subsonic root: guess $M=0.35\Rightarrow A/A^{\ast }=1.778$; $M=0.36\Rightarrow 1.732$. Interpolate karo, $M\approx ==0.357==$.

Supersonic root: guess $M=2.05\Rightarrow A/A^{\ast }=1.766$; $M=2.03\Rightarrow 1.744$. In dono ke beech interpolate karo, $M\approx ==2.037==$ (jo $A/A^{\ast }\approx 1.751$ deta hai). ✔

Dono $A/A^{\ast }\approx 1.75$ dete hain. KYU do answers: $1.75$ par horizontal line valley ko do baar kaatti hai — figure dekho.

Bracket $=1+0.2\cdot (2.037{)}^2=1+0.2\cdot 4.149=1.830$. Exponent $1/(\gamma -1)=1/0.4=2.5$. $\frac{{\rho }_0}{\rho }=(1.830{)}^{2.5}=4.531$, toh $\rho /{\rho }_0=1/4.531===0.2207==$. KYU: supersonic gas patli phail jaati hai — density reservoir density ke lagbhag ek paanchwe hisse tak gir jaati hai.

$M=0.6$ par: bracket $=1+0.2\cdot 0.36=1.072$; times $0.8333=0.8933$; $A/A^{\ast }=\frac{1}{0.6}(0.8933{)}^3=1.6667\cdot 0.7130=1.1883$. Toh $A^{\ast }=A_{\min }/1.1883=12/1.1883===10.10=={\text{cm}}^2$. Kya koi real sonic point hai? Nahi. Sabse chhoti actual area (12 cm²) abhi bhi $A^{\ast }$ (10.10 cm²) se badi hai, toh flow kabhi $M=1$ nahi pahunchti. $A^{\ast }$ yahan sirf ek mathematical yardstick ki tarah exist karta hai — yeh duct ki kisi bhi real area se chhota hai. KYU yeh matter karta hai: yeh "$A^{\ast }$ hamesha ek physical throat hai" wali galti ko steel-man karta hai — aisa nahi hai.

Step 1 (shape → Mach). $A_e/A^{\ast }=15/6=2.5$. Supersonic branch par iterate karo: $M=2.44\Rightarrow A/A^{\ast }=2.494$; $M=2.443\Rightarrow 2.502$. Toh $M_e\approx ==2.443==$.

Step 2 (Mach → $T$). Bracket $=1+0.2\cdot (2.443{)}^2=1+0.2\cdot 5.968=2.1937$. $T_e=T_0/2.1937=320/2.1937===145.9==\text{K}$.

Step 3 (Mach → $p$). $p_0/p=(2.1937{)}^{3.5}=15.63$, toh $p_e=700/15.63===44.8==\text{kPa}$.

Step 4 (velocity). Local sound speed $a_e=\sqrt{\gamma R{T}_e}=\sqrt{1.4\cdot 287\cdot 145.9}=\sqrt{58622}=242.1\text{m/s}$. Phir $V_e=M_e{a}_e=2.443\cdot 242.1===591.5==\text{m/s}$. KYU yeh chain: geometry ne $M$ diya (Tool A); $M$ ne har thermodynamic property unlock ki (Tool B); temperature ne local sound speed di, aur $V=Ma$ ne actual speed di. Yahi hai "shape → flow" master key jo parent note mein promise ki gayi thi.

KYA (a): choked throat ke liye mass flow reservoir conditions se fixed hoti hai: $\dot{m}=\frac{p_0{A}^{\ast }}{\sqrt{T_0}}\sqrt{\frac{\gamma }{R}}{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}.$ Numbers: $\frac{p_0{A}^{\ast }}{\sqrt{T_0}}=\frac{700000\cdot 6\times 10^{-4}}{\sqrt{320}}=\frac{420}{17.889}=23.48$. $\sqrt{\gamma /R}=\sqrt{1.4/287}=\sqrt{0.004878}=0.06984$. Bracket factor $(0.8333{)}^3=0.5787$. $\dot{m}=23.48\cdot 0.06984\cdot 0.5787===0.949==\text{kg/s}$.

(b) $\dot{m}\propto A^{\ast }$ (baaki sab unchanged), toh $A^{\ast }$ double karne se flow double ho jaati hai: $\dot{m}=2\cdot 0.949===1.898==\text{kg/s}$. KYU: jab throat sonic ho jaata hai, exit pressure ko kam karna aur zyada mass through nahi khiinch sakta — flow choked hai (dekho Choked Flow & Maximum Mass Flow). Ek hi lever bacha hai woh hai throat area khud. Dekho Converging-Diverging (de Laval) Nozzle.

Step 1 (Mach → area ratio). Bracket $=1+0.2\cdot 9=2.8$; times $0.8333=2.3333$. $A_e/A^{\ast }=\frac{1}{3}(2.3333{)}^3=\frac{1}{3}\cdot 12.70=4.2346$. Step 2 (throat ke liye solve karo). $A^{\ast }=A_e/4.2346=45/4.2346===10.63=={\text{cm}}^2$. Step 3 ($T$, $p$). $T_0/T=2.8\Rightarrow T=500/2.8===178.6==\text{K}$. $p_0/p=(2.8{)}^{3.5}=36.73\Rightarrow p=1000/36.73===27.2==\text{kPa}$. KYU: design chain ko ulta chalata hai — target Mach area ratio fix karta hai, jo given exit ke liye throat size karta hai. Phir thermodynamics $M_e$ se nikal aati hai.

Step 1. Downstream stagnation pressure: $p_{0,2}=0.499\cdot 900===449.1==\text{kPa}$. Step 2 (interpretation). Ek shock isentropic nahi hota — entropy badhta hai, toh $p_0$ girta hai. Kyunki choked mass flow $\dot{m}\propto p_0{A}^{\ast }$ shock ke across conserved rehta hai jabki $p_0$ girta hai, sonic reference area badhna chahiye: $A_2^{\ast }/A_1^{\ast }=p_{0,1}/p_{0,2}=1/0.499=2.004$. Toh $A^{\ast }$ shock ke across roughly double ho jaata hai. KYU yeh matter karta hai: $A^{\ast }$ sirf ek isentropic run ke along constant rehta hai. Ek shock cross karo aur yardstick khud jump kar jaata hai — yeh ek key subtlety hai jab nozzle sections match karo (dekho Normal Shock Waves).

Jab $M\to 0$: bracket $\to {\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma +1}{2(\gamma -1)}}$, ek finite positive constant, lekin prefactor $1/M\to \infty$. Toh $A/A^{\ast }\to \infty$. Physical: fixed mass flow par gas ko near-rest tak slow karne ke liye bahut bada area chahiye (bheed infinitely thin aur slow phail jaati hai).

Jab $M\to \infty$: bracket $M^2$ ki tarah grow karta hai jise $\frac{\gamma +1}{2(\gamma -1)}$ tak raise kiya gaya, matlab $M^{\frac{\gamma +1}{\gamma -1}}$ ki tarah; $M$ se divide karne par bhi $M^{\frac{\gamma +1}{\gamma -1}-1}=M^{2/(\gamma -1)}\to \infty$ bachta hai. Physical: hypersonic flow mein density negligible hoti hai, toh wahi mass pass karne ke liye area blow up karna padta hai. Dono ends → $\infty$; single minimum $M=1$ par baitha hai — yahi exactly U-valley shape hai.

$M=1$ par $A/A^{\ast }$ kya hai? ::: Exactly 1 — U-valley ka minimum. Air ke liye AREA relation mein kaunsa exponent hota hai? ::: $\frac{\gamma +1}{2(\gamma -1)}=3$ (na ki $\gamma /(\gamma -1)=3.5$, woh pressure relation ka hai). Choked-flow problems $\dot{m}$ ke liye back-pressure kyun ignore karte hain? ::: Sonic throat density×velocity ko cap kar deta hai, toh $\dot{m}=\frac{p_0{A}^{\ast }}{\sqrt{T_0}}(\dots )$ reservoir conditions se fixed hai. Normal shock ke across $A^{\ast }$ ka kya hota hai? ::: Yeh increase hota hai ($p_{0,1}/p_{0,2}$ se), kyunki $p_0$ girta hai jabki $\dot{m}\propto p_0{A}^{\ast }$ conserved rehta hai. Jab $M$ pata ho toh velocity kaise nikaalte hain? ::: $V=Ma$ ke saath $a=\sqrt{\gamma RT}$, $T$ stagnation relation se lene ke baad.