Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁: Worked examples

3.1.12 · D3 · Physics › Compressible Flow & Aerodynamics › Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂

Yeh page parent note ka drill ground hai. Wahan humne shock relations derive kiye the; yahan hum unhe use karte hain jab tak har tarah ka problem familiar na lag jaaye. Hum $γ = 1.4$ (air) lete hain jab tak koi problem aur na kahe.

The scenario matrix

Kuch bhi solve karne se pehle, chalte hain list karte hain kis tarah ki situations yeh topic de sakta hai. Neeche har worked example us cell ke saath tagged hai jisme woh aata hai.

Cell	Case class	Defining feature	Example
A	Standard moderate shock	$1 < M_{1} ≲ 3$ , plug-and-chug	Ex 1
B	Weak-shock limit	$M_{1} \to 1^{+}$ , sab kuch $\to 1$	Ex 2
C	Strong / hypersonic limit	$M_{1} \to \infty$ , density saturate ho jaati hai	Ex 3
D	Degenerate: $M_{1} = 1$ exactly	"Shock" ek sound wave hai, koi jump nahi	Ex 4
E	Forbidden direction	Diya gaya $M_{1} < 1$ — kya shock exist karta hai?	Ex 5
F	Inverse / back-solve	Ek ratio diya gaya, $M_{1}$ nikalo	Ex 6
G	Non-air gas	$γ \neq = 1.4$ (helium, combustion gas)	Ex 7
H	Real-world word problem	Dimensional data, actual $P, T, u$ nikalo	Ex 8
I	Exam twist	Stagnation-pressure loss + inlet design	Ex 9

Hamare paas sirf parent se boxed formulas chahiye. Shuru karne se pehle kuch symbols ki definition zaroori hai. $u$ = flow speed (gas kitni tezi se move karti hai, m/s mein — Mach number $M$ se alag, jo woh speed divided by local speed of sound hoti hai). $Δ s$ = entropy change shock ke across (irreversibility ka measure; positive matlab "zyada messy, kam recoverable"). $c_{p}$ = specific heat at constant pressure (joules mein energy jo 1 kg gas ko 1 K warm karne ke liye chahiye jab woh freely expand kar sake; air ke liye $c_{p} \approx 1005 J/kg\cdotK$ ). $R$ = specific gas constant (air ke liye $R = 287 J/kg\cdotK$ ); yeh dono $c_{p} = \frac{γ}{γ - 1} R$ se linked hain. Quick reference ke liye:

Intuition

$P_{02} / P_{01}$ formula kahan se aata hai ("why") Shuru karo ek ideal gas ki entropy rise se, $Δ s = c_{p} ln \frac{T _{2}}{T _{1}} - R ln \frac{P _{2}}{P _{1}}$ — jahan (jaise abhi define kiya) $c_{p}$ constant-pressure specific heat hai aur $R$ specific gas constant. Parent note dikhata hai ki stagnation pressure aur entropy $\frac{P _{02}}{P _{01}} = exp (- \frac{Δ s}{R})$ se tied hain — zyada entropy matlab kam stagnation pressure, hamesha. Ab shock ki apni $T_{2} / T_{1}$ aur $P_{2} / P_{1}$ ko $Δ s$ mein substitute karo aur simplify karo: exponential upar wale do-bracket product mein collapse ho jaata hai. Toh pehla bracket entropy ka temperature side carry karta hai aur doosra bracket pressure side; exponents $\frac{γ}{γ - 1}$ aur $\frac{1}{γ - 1}$ sirf $c_{p} / R$ aur $1/ (γ - 1)$ hain disguise mein. Tumhe ise kabhi re-derive nahi karna — lekin ab tum jaante ho ki har piece ek entropy term hai, koi magic constant nahi.

Yahan $M$ = Mach number (flow speed ÷ speed of sound), subscript $1$ = upstream (shock se pehle, hamesha supersonic), subscript $2$ = downstream (baad mein, hamesha subsonic). $P$ = static pressure, $T$ = static temperature, $ρ$ = density, $P_{0}$ = stagnation pressure. Agar inme se kuch shaky lage toh Speed of Sound and Mach Number aur Stagnation Properties T0 and P0 dekho.

Neeche diya figure poore topic ka hamaara map hai (yeh "alt text" hai: ek labelled plot of all five downstream/upstream property ratios versus single input $M_{1}$ ). Horizontal axis upstream Mach number $M_{1}$ hai (from $1$ to $6$ ). Vertical axis kisi property ka downstream-to-upstream ratio hai. Har coloured curve ek property hai: magenta $P_{2} / P_{1}$ aur orange $T_{2} / T_{1}$ steeply climb karti hain (shock strength ke saath pressure aur temperature badhti hain); violet $ρ_{2} / ρ_{1}$ climb karti hai phir dotted density ceiling par $6$ ke paas flat ho jaati hai; dashed navy curve $M_{2}$ horizontal line ke neeche $1$ se neeche dip karti hai — proof ki flow shock ke baad hamesha subsonic hoti hai; aur teal $P_{02} / P_{01}$ curve zero ki taraf sink karti hai jaise shock strong hoti hai (entropy loss badhti hai). Ise left-to-right "shock strength badhane" ki tarah padho. Neeche har worked example is ek picture ka sirf ek vertical slice hai.

Figure — Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁

Figure 1 — Master map: $M_{1}$ ( $γ = 1.4$ ) ke functions ke roop mein saare paanch normal-shock ratios. Horizontal axis par koi $M_{1}$ chuno aur curves se har downstream property padho.

Example 1 — Cell A · Standard moderate shock

Worked example Air mein

$M_{1} = 2.5$ , $γ = 1.4$ par ek shock khada hai. $M_{2}$ , $P_{2} / P_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ , aur $P_{02} / P_{01}$ nikalo.

Forecast: Compute karne se pehle — guess karo. $M_{1} = 2.5$ ek kaafi strong shock hai. Kya $M_{2}$ $1$ ke paas hoga ya $0.4$ ke paas? Kya pressure roughly chaar guna hogi ya roughly saat guna? Apne guesses likho, phir check karo.

Step 1 — Downstream Mach number. $M_{1}^{2} = 6.25$ ke saath: $M_{2}^{2} = \frac{1 + 0.2 ( 6.25 )}{1.4 ( 6.25 ) - 0.2} = \frac{1 + 1.25}{8.75 - 0.2} = \frac{2.25}{8.55} = 0.2632$ toh $M_{2} = 0.513$ . Yeh step kyun? Har doosra ratio $M_{1}$ se anchored hai, lekin $M_{2}$ headline physical result hai — yeh confirm karta hai ki flow subsonic ho gayi, jo shock ka defining behaviour hai.

Step 2 — Pressure ratio. $M_{1}^{2} - 1 = 5.25$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2 ( 1.4 )}{2.4} (5.25) = 1 + \frac{2.8}{2.4} (5.25) = 1 + 6.125 = 7.125$ Yeh step kyun? Rankine–Hugoniot pressure jump (dekho Rankine–Hugoniot Relations) sabse zyada used engineering number hai: yeh batata hai shock kitna force exert karti hai.

Step 3 — Density ratio. $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 6.25 )}{0.4 ( 6.25 ) + 2} = \frac{15}{4.5} = 3.333$ Yeh step kyun? Density aur flow speed mass conservation se ek saath locked hain ( $ρ_{2} / ρ_{1} = u_{1} / u_{2}$ ), toh yeh ek number hume yeh bhi batata hai ki flow $3.333$ ke factor se slow ho gayi.

Step 4 — Temperature ratio ideal-gas link $T_{2} / T_{1} = (P_{2} / P_{1}) / (ρ_{2} / ρ_{1})$ se: $\frac{T _{2}}{T _{1}} = \frac{7.125}{3.333} = 2.1375$ Yeh step kyun? Hum temperature ko scratch se re-derive karne se bachte hain — ideal gas law pressure aur density ko temperature free mein de deta hai.

Step 5 — Stagnation-pressure loss. $M_{1} = 2.5$ ko $P_{02} / P_{01}$ formula mein substitute karo. Pehle bracket base: $\frac{( 2.4 ) ( 6.25 ) /2}{1 + 0.2 ( 6.25 )} = \frac{7.5}{2.25} = 3.333$ , $\frac{γ}{γ - 1} = 3.5$ tak raise karo to $3.33 3^{3.5} = 67.6$ milta hai. Doosra bracket base: $\frac{2.4}{2 ( 1.4 ) ( 6.25 ) - 0.4} = \frac{2.4}{17.1} = 0.1404$ , $\frac{1}{γ - 1} = 2.5$ tak raise karo to $0.140 4^{2.5} = 0.00738$ milta hai. Product: $\frac{P _{02}}{P _{01}} = 67.6 \times 0.00738 = 0.499$ toh stagnation pressure ka lagbhag aadha destroy ho jaata hai. Yeh step kyun? Yeh entropy fingerprint hai — irreversibility numeric bana di (dekho Entropy and Second Law).

Verify: Saare static ratios 1 se zyada hain (compression ✓), $M_{2} < 1$ (subsonic ✓), $P_{02} / P_{01} < 1$ (2nd law ✓). Upar introduce ki gayi $T_{2} / T_{1}$ ki energy-conservation form se cross-check karo: $\frac{1 + 0.2 M _{1}^{2}}{1 + 0.2 M _{2}^{2}} = \frac{2.25}{1 + 0.2 ( 0.2632 )} = \frac{2.25}{1.0526} = 2.137$ ✓ — ideal-gas route (Step 4) aur energy route agree karte hain.

Example 2 — Cell B · The weak-shock limit

Worked example Air

$M_{1} = 1.05$ (barely supersonic) par flow kar rahi hai. Dikhao ki har jump choti hai — $M_{2}$ aur $T_{2} / T_{1}$ bhi — aur shock almost isentropic hai.

Forecast: $M_{1}$ $1$ se thoda sa upar hai, guess karo: kya ratios $1.0$ ke paas hain ya $1.5$ ke paas? Kya $M_{2}$ just under $1$ hoga? Aur kya stagnation-pressure loss "kuch percent" hai ya "kuch tenths of a percent"?

Step 1 — Downstream Mach number. $M_{1}^{2} = 1.1025$ : $M_{2}^{2} = \frac{1 + 0.2 ( 1.1025 )}{1.4 ( 1.1025 ) - 0.2} = \frac{1.2205}{1.3435} = 0.9084 \Rightarrow M_{2} = 0.953$ Yeh step kyun? $M_{2}$ sirf thoda sa $1$ se neeche hai — flow barely sonic line cross ki, confirm karta hai ki yeh sabse gentle possible shock hai.

Step 2 — Pressure ratio. $M_{1}^{2} - 1 = 0.1025$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (0.1025) = 1 + 0.1196 = 1.1196$ Yeh step kyun? Yeh directly measure karta hai "no-shock se kitne dur hain"; $1.12$ ki value $1$ ke paas hai — ek gentle squeeze.

Step 3 — Density ratio. $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 1.1025 )}{0.4 ( 1.1025 ) + 2} = \frac{2.646}{2.441} = 1.0840$ Yeh step kyun? Confirm karta hai ki density bhi barely move karti hai.

Step 4 — Temperature ratio ideal-gas link se: $\frac{T _{2}}{T _{1}} = \frac{1.1196}{1.0840} = 1.0328$ Yeh step kyun? Temperature sirf lagbhag $3%$ badhti hai — "tiny jump" ka claim har property ke liye hold karna chahiye, sirf pressure aur density ke liye nahi, isliye hum $T_{2} / T_{1}$ explicitly compute karte hain picture complete karne ke liye.

Step 5 — Stagnation-pressure loss. $M_{1} = 1.05$ ko two-bracket formula mein substitute karo: $\frac{P _{02}}{P _{01}} = 0.99987$ lagbhag 0.013% ka loss. Yeh step kyun? Yeh punchline hai: entropy cube $(M_{1}^{2} - 1)^{3}$ ki tarah badhti hai $M_{1} = 1$ ke paas, toh ek weak shock nearly reversible hai — yeh almost ek sound wave ki tarah behave karta hai Isentropic Flow Relations mein.

Verify: Jaise $M_{1} \to 1^{+}$ parent note kehta hai saare ratios $\to 1$ . Yahan $M_{2} = 0.953$ (just below $1$ ✓), aur $P_{2} / P_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ sab just above $1$ hain jabki $P_{02} / P_{01}$ just below $1$ hai — har jump choti hai, limit ke saath consistent. ✓

Example 3 — Cell C · Strong / hypersonic limit

Worked example Ek re-entry capsule

$M_{1} = 12$ par ek shock ke peeche hai. Density ratio nikalo aur theoretical ceiling se compare karo. $P_{2} / P_{1}$ aur $P_{02} / P_{01}$ nikalo.

Forecast: Parent note kehta hai density $\frac{γ + 1}{γ - 1} = 6$ par saturate hoti hai. $M_{1} = 12$ par, guess karo: kya $ρ_{2} / ρ_{1}$ already $5.9$ par hai, ya sirf $4$ par? Aur kya pressure jump tens mein hai ya hundreds mein?

Step 1 — Density ratio. $M_{1}^{2} = 144$ : $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 144 )}{0.4 ( 144 ) + 2} = \frac{345.6}{59.6} = 5.799$ Yeh step kyun? Saturation claim test karta hai; $5.799$ ki value ceiling $6$ se $3.4%$ ke andar hai — density essentially maxed out hai.

Step 2 — Pressure ratio. $M_{1}^{2} - 1 = 143$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (143) = 1 + 166.83 = 167.83$ Yeh step kyun? Contrast dikhata hai: pressure bina bound ke climb karti rehti hai jabki density flat ho jaati hai. Extra energy ke paas sirf $P$ aur $T$ mein jaane ki jagah hai.

Step 3 — Stagnation-pressure loss. $M_{1} = 12$ ko two-bracket formula mein substitute karo: $\frac{P _{02}}{P _{01}} = 0.001645$ — stagnation pressure ka 99.8% se zyada chala gaya. Yeh step kyun? Yahi reason hai ki hypersonic inlets kabhi ek single normal shock use nahi karte (dekho Supersonic Inlets and Diffusers).

Verify: Ratio $ρ_{2} / ρ_{1} < 6$ ceiling ke neeche hai ✓, aur $M_{1}$ badhne par $6$ ke paas aana chahiye; value $5.799 < 6$ hai aur parent ke Example 2 ki $M_{1} = 5$ value $5.0$ se badi hai ✓. Aur $P_{02} / P_{01} ≪ 1$ ✓.

Example 4 — Cell D · Degenerate case $M_{1} = 1$

Worked example Bilkul

$M_{1} = 1$ par shock kya karti hai? $M_{2}$ , $P_{2} / P_{1}$ , $ρ_{2} / ρ_{1}$ compute karo.

Forecast: $M_{1} = 1$ "shock possible" aur "shock forbidden" ke beech ki boundary hai. Guess karo: kya $M_{2}$ $1$ aata hai, ya kuch $< 1$ ? Kya ratios exactly $1$ equal hote hain?

Step 1 — $M_{2}$ . $M_{1}^{2} = 1$ : $M_{2}^{2} = \frac{1 + 0.2 ( 1 )}{1.4 ( 1 ) - 0.2} = \frac{1.2}{1.2} = 1 \Rightarrow M_{2} = 1$ Yeh step kyun? Boundary par shock equation ke do roots merge ho jaate hain: "shock branch" aur "no-shock branch" same point ban jaate hain. Koi jump nahi hota.

Step 2 — Pressure ratio. Yahan $M_{1}^{2} - 1 = 0$ : $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (0) = 1$ Yeh step kyun? Zero pressure jump confirm karta hai — "shock" infinitely weak hai.

Step 3 — Density ratio. $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 1 )}{0.4 ( 1 ) + 2} = \frac{2.4}{2.4} = 1$ Yeh step kyun? Density bhi unchanged hai.

Verify: Saare ratios exactly $1$ hain aur $M_{2} = M_{1} = 1$ . Yeh degenerate "sound wave" case hai — physically matlab koi shock nahi bilkul, sirf ek isentropic sonic point. Example 2 ke weak-shock limit ke saath consistent. ✓

Example 5 — Cell E · The forbidden direction

Worked example Ek student claim karta hai ki ek normal shock subsonic air ko

$M_{1} = 0.6$ se kisi $M_{2}$ tak decelerate karta hai. Formula use karke check karo ki kya yeh physically allowed hai.

Forecast: Guess karo: kya formula ek valid $M_{2}$ deta hai, koi nonsense value, ya ek warning sign (jaise $> 1$ ki value, jo 2nd law ek compression shock ke liye forbid karta hai)?

Step 1 — $M_{2}$ formula blindly apply karo. $M_{1}^{2} = 0.36$ : $M_{2}^{2} = \frac{1 + 0.2 ( 0.36 )}{1.4 ( 0.36 ) - 0.2} = \frac{1.072}{0.504 - 0.2} = \frac{1.072}{0.304} = 3.526$ toh $M_{2} = 1.878$ . Yeh step kyun? Algebra khushi se ek answer deta hai — lekin answer $M_{2} > 1$ ka matlab hai flow subsonic se supersonic jaayegi.

Step 2 — Entropy / pressure direction check karo. $M_{1} = 0.6$ ke saath, pressure "ratio" formula deta hai $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (0.36 - 1) = 1 + \frac{2.8}{2.4} (- 0.64) = 1 - 0.7467 = 0.2533$ ek pressure drop. Yeh step kyun? Is tarah ka rarefaction jump $Δ s < 0$ require karega — flow spontaneously zyada ordered ho jaaye. Second Law yeh forbid karta hai (dekho Entropy and Second Law).

Step 3 — Conclusion. Aisa koi shock exist nahi karta. Math ka ek mathematically valid root hai, lekin woh unphysical branch hai. Shocks require karte hain $M_{1} > 1$ aur hamesha $M_{2} < 1$ produce karte hain. Yeh step kyun? Yeh classic trap hai: formulas physics nahi jaante; tumhe 2nd-law filter apply karna hai.

Verify: Kisi bhi $M_{1} < 1$ ke liye, formula $M_{2} > 1$ aur $P_{2} / P_{1} < 1$ deta hai — forbidden expansion shock ka signature. Numerically confirmed: $M_{2}^{2} = 3.526 > 1$ aur $P_{2} / P_{1} = 0.2533 < 1$ . ✗ physically. ✓ trap ka demonstration.

Example 6 — Cell F · Inverse problem (back-solve $M_{1}$ )

Worked example Ek pressure gauge ek shock ke across

$P_{2} / P_{1} = 4.5$ air mein read karta hai. Upstream Mach number kya tha? (Yeh parent ka Example 1 hai, ulta chalaya.)

Forecast: Parent ke forward example ne $M_{1} = 2$ se $P_{2} / P_{1} = 4.5$ nikala tha. Guess karo: kya formula invert karne par exactly $M_{1} = 2$ recover hoga?

Step 1 — Pressure formula ko $M_{1}^{2}$ ke liye rearrange karo. Shuru karo $\frac{P _{2}}{P _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1),$ solve karo: $M_{1}^{2} = 1 + \frac{γ + 1}{2 γ} (\frac{P _{2}}{P _{1}} - 1)$ Yeh step kyun? Pressure ratio Mach number $M_{1}$ ka ek monotonic function hai, toh ise uniquely invert kiya ja sakta hai — ek gauge reading exactly ek upstream Mach number pin down karta hai.

Step 2 — Plug in karo. $\frac{P _{2}}{P _{1}} - 1 = 3.5$ : $M_{1}^{2} = 1 + \frac{2.4}{2.8} (3.5) = 1 + 3 = 4 \Rightarrow M_{1} = 2.0$ Yeh step kyun? Confirm karta hai ki inverse forward answer recover karta hai.

Verify: Recovered $M_{1} = 2$ ko forward formula mein substitute karo: $1 + \frac{2.8}{2.4} (4 - 1) = 1 + 3.5 = 4.5$ ✓ — round trip exactly close hota hai.

Example 7 — Cell G · Non-air gas ( $γ \neq = 1.4$ )

Worked example

Helium ( $γ = 5/3$ , monatomic) mein $M_{1} = 2.0$ par ek shock banta hai. $M_{2}$ , $ρ_{2} / ρ_{1}$ , aur maximum possible density ratio nikalo. Air se compare karo.

Forecast: Helium ka $γ = 1.667$ air ke $1.4$ se bada hai. Density ceiling $\frac{γ + 1}{γ - 1}$ hai. Guess karo: kya helium ki ceiling air ki $6$ se zyada hai ya kam?

Step 1 — Density ceiling. $\frac{ρ _{2}}{ρ _{1}}_{m a x} = \frac{γ + 1}{γ - 1} = \frac{2.667}{0.667} = 4.0$ Yeh step kyun? Physical limit establish karta hai — helium ko ek shock mein at most $4 \times$ compress kiya ja sakta hai, air ke $6 \times$ se kam. Zyada $γ$ matlab zyada stiff gas.

Step 2 — $M_{1} = 2$ par $M_{2}$ . $\frac{γ - 1}{2} = 0.3333$ aur $M_{1}^{2} = 4$ ke saath: $M_{2}^{2} = \frac{1 + 0.3333 ( 4 )}{1.6667 ( 4 ) - 0.3333} = \frac{2.3333}{6.3333} = 0.3684 \Rightarrow M_{2} = 0.607$ Yeh step kyun? Dikhata hai ki same upstream Mach number $M_{1}$ ek alag $M_{2}$ deta hai jab $γ$ change hota hai — sirf air-only numbers kabhi mat memorize karo.

Step 3 — $M_{1} = 2$ par Density ratio. $\frac{ρ _{2}}{ρ _{1}} = \frac{2.6667 ( 4 )}{0.6667 ( 4 ) + 2} = \frac{10.667}{4.6667} = 2.286$ Yeh step kyun? Same $M_{1}$ par air ki value $2.667$ se compare karte hue, helium kam compress hota hai — uski lower ceiling ke saath consistent.

Verify: $ρ_{2} / ρ_{1} = 2.286 < 4.0$ ceiling ✓; $M_{2} < 1$ ✓; helium ceiling $4 <$ air ceiling $6$ ✓ (stiffer gas, kam squeezable).

Example 8 — Cell H · Real-world word problem

Worked example Supersonic wind-tunnel model par ek pitot probe upstream conditions

$P_{1} = 40 kPa$ , $T_{1} = 250 K$ , air mein $M_{1} = 3.0$ par ek normal shock ke aage measure karta hai. Actual downstream pressure $P_{2}$ , temperature $T_{2}$ , aur downstream flow speed $u_{2}$ nikalo.

Forecast: $M_{1} = 3$ ek strong shock hai. Guess karo: kya $P_{2}$ lagbhag $200$ kPa par aata hai ya $400$ kPa par? Kya $T_{2}$ $600$ K se zyada hogi? Kya $u_{2}$ speed of sound se neeche aayegi (jo aana zaroori hai, kyunki $M_{2} < 1$ hoga)?

Step 1 — $M_{1} = 3$ se ratios. $M_{1}^{2} = 9$ ke saath: $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (8) = 1 + 9.333 = 10.333, \frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 9 )}{0.4 ( 9 ) + 2} = \frac{21.6}{5.6} = 3.857$ $\frac{T _{2}}{T _{1}} = \frac{10.333}{3.857} = 2.679$ Yeh step kyun? Pehle dimensionless ratios lo — woh saari physics carry karte hain; dimensions last mein aate hain.

Step 2 — Actual static values. Har ratio ko measured upstream value se multiply karo: $P_{2} = 10.333 \times 40 = 413.3 kPa, T_{2} = 2.679 \times 250 = 669.8 K$ Yeh step kyun? Yahan "sirf ek formula" ek real number ban jaata hai jo ek sensor read karta; Mach $3$ par ek shock static pressure ko das guna badhati hai aur temperature roughly teen guni kar deti hai.

Step 3 — Downstream flow speed. Pehle downstream Mach number: $M_{2}^{2} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{2.8}{12.4} = 0.2258$ , toh $M_{2} = 0.4752$ . Downstream sound speed hotter downstream temperature use karta hai, $a_{2} = γ R T_{2}$ jahan $R = 287 J/kg\cdotK$ : $a_{2} = 1.4 (287) (669.8) = 518.6 m/s, u_{2} = M_{2} a_{2} = 0.4752 (518.6) = 246.4 m/s$ Yeh step kyun? Flow speed = Mach number × local sound speed, aur local sound speed downstream (hotter) temperature par depend karti hai — ek common slip yahan $T_{1}$ use karna hai.

Verify: $P_{2} = 413$ kPa, $T_{2} = 670$ K, $u_{2} = 246$ m/s. Sanity: upstream speed hai $u_{1} = M_{1} γ R T_{1} = 3 1.4 (287) (250) = 3 (316.9) = 950.6$ m/s. Mass conservation kehta hai $u_{1} / u_{2}$ ko $ρ_{2} / ρ_{1} = 3.857$ equal karna chahiye: check karo $950.6/246.4 = 3.858$ ✓ (rounding). Khoobsoorat closure.

Example 9 — Cell I · Exam twist (inlet design)

Worked example Ek supersonic engine inlet air ko

$M_{1} = 3$ se subsonic tak decelerate karta hai. Design A ek normal shock use karta hai $M_{1} = 3$ par. Design B do shocks series mein use karta hai: ek weak normal shock $M = 1.5$ par aur phir ek normal shock $M = 2$ par. Total stagnation-pressure recovery $P_{02} / P_{01}$ compare karo aur explain karo kaun better hai. (Har stage ki recovery multiplying maano.)

Forecast: Parent note ne hint diya tha "multiple weak shocks preferred hain." Guess karo: kya Design B ki total recovery Design A se thodi zyada hogi ya bahut zyada?

Step 1 — Design A recovery. $M_{1} = 3$ par ek single normal shock ka (two-bracket formula se): $\frac{P _{02}}{P _{01}}_{A} = 0.3283$ Yeh step kyun? Yeh baseline hai — ek violent shock stagnation pressure ka lagbhag do-tihai throw away kar deti hai.

Step 2 — Design B, stage recoveries. Pehla (weak) shock $M = 1.5$ par aur doosra $M = 2$ par — dono ki apni single-shock recovery hai: $\frac{P _{02}}{P _{01}}_{M = 1.5} = 0.9298, \frac{P _{02}}{P _{01}}_{M = 2} = 0.7209$ Yeh step kyun? Har stage ki entropy generation choti hai kyunki $M = 1$ ke paas entropy cube $(M^{2} - 1)^{3}$ ki tarah badhti hai — deceleration spread karna har jump gentle rakhta hai.

Step 3 — Stage recoveries multiply karo. $\frac{P _{02}}{P _{01}}_{B} = 0.9298 \times 0.7209 = 0.6703$ Yeh step kyun? Stagnation-pressure recovery successive stages ke through multiply hoti hai, toh do gentle shocks ( $0.67$ ) ek harsh shock ( $0.33$ ) ko crush karte hain.

Step 4 — Verdict. Design B stagnation pressure ka Design A se lagbhag do guna recover karta hai. Yahi reason hai ki real supersonic inlets ek final weak normal shock se pehle Oblique Shock Waves ka ek ramp use karte hain (dekho Supersonic Inlets and Diffusers). Yeh step kyun? Numeric result ko engineering reason se jodhta hai jis liye yeh matter karta hai — thrust recovered stagnation pressure ke proportional hota hai.

Verify: $0.6703 > 0.3283$ (staged better hai ✓); har single-stage factor $< 1$ hai (2nd law ✓); sabse harsh single shock sabse zyada khota hai ✓. Staged recovery $1$ se zyada nahi ho sakti ( $0.67 < 1$ ✓).

Recall Quick self-test (answers cover karo)

Kaun sa cell shock forbid karta hai, aur kyun? ::: Cell E — subsonic inflow ( $M_{1} < 1$ ) ko $Δ s < 0$ chahiye hoga, jo 2nd Law ne ban kiya hai. Air mein $M_{1} \to \infty$ ke saath density ratio kis number ke paas jaata hai? ::: $\frac{γ + 1}{γ - 1} = 6$ . Helium ( $γ = 5/3$ ) ke liye density ceiling kya hai? ::: $4.0$ . Air mein $M_{1} = 2$ par $P_{2} / P_{1}$ aur $M_{2}$ ? ::: $4.5$ aur $0.577$ . Inlet mein do weak shocks ek strong shock ko kyun beat karte hain? ::: Entropy $(M^{2} - 1)^{3}$ ki tarah badhti hai, toh gentle jumps stagnation pressure bahut kam waste karte hain; recoveries multiply hoti hain.