Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations: Worked examples

3.1.11 · D3 · Physics › Compressible Flow & Aerodynamics › Normal shock waves — Rankine-Hugoniot relations (all 5) — de

Yeh page normal-shock relations ka drill hall hai. Parent note ne paanch formulas build kiye. Yahan hum har tarah ka input unpar throw karte hain — weak shocks, strong shocks, exact boundary, degenerate "no shock" case, ek real-world word problem, aur ek exam-style trap — taaki jab tum inhe wild mein milo, tum pehle hi unka judwa dekh chuke ho.

Shuru karne se pehle, paanch workhorses ko samne rakho. Ye sab ek input lete hain, upstream Mach number $M_{1}$ , ek calorically perfect gas ke liye jiska heat-capacity ratio $γ$ hai (air ke liye $γ = 1.4$ ).

Pore examples mein main inhe unke tags se cite karta hoon: R1 = downstream Mach, R2 = pressure ratio, R3 = density ratio, R4 = temperature ratio, R5 = stagnation-pressure loss.

Reminders jo tum skip nahi kar sakte:

$M$ ("Mach number") flow speed divided by the local speed of sound hai: $M = u / a$ , $a = γ R T$ . Dekho Speed of sound and Mach number.
Ek normal shock tabhi exist karta hai jab incoming flow supersonic ho, $M_{1} > 1$ . Yeh Second Law se enforce hota hai.
Subscript $0$ ka matlab stagnation hai (wo value jo flow tab reach karti jab use reversibly rest par laaya jaaye). Dekho Stagnation properties.

Scenario matrix

Har problem jo is topic mein aa sakti hai, yahan diye gaye cells mein se kisi ek mein fall karti hai. Neeche har worked example us cell se tagged hai jisme wo land karta hai.

Cell	Input class	Kya special hai	Example
C1	Degenerate: $M_{1} = 1$	Infinitely weak shock — saare ratios $= 1$	Ex 1
C2	Forbidden: $M_{1} < 1$	Shock allowed nahi — 2nd Law veto	Ex 2
C3	Weak shock: $M_{1}$ just above 1	Chhote jumps, near-isentropic	Ex 3
C4	Moderate shock: $M_{1} = 2$ – $3$	"Standard" case	Ex 4
C5	Strong-shock limit: $M_{1} \to \infty$	Density saturate hoti hai, $T$ blow up karti hai	Ex 5
C6	Total-pressure / entropy loss	$p_{0}$ drop karta hai par $T_{0}$ constant rehta hai	Ex 6
C7	Real-world word problem	Dimensional data, actual $p_{2}, T_{2}$ nikalo	Ex 7
C8	Exam twist: ek downstream quantity di hai, $M_{1}$ back-solve karo	Inverse problem	Ex 8
C9	Different gas ( $γ \neq = 1.4$ )	$γ$ -dependence check karo	Ex 9

Neeche ki picture chaar static ratios ko $M_{1}$ ke against plot karti hai taaki tum dekh sako ki har example kis region mein hai.

Figure — Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations

Figure s01 — bina image ke padhna. Horizontal axis upstream Mach number $M_{1}$ hai $1$ se $6$ tak; vertical axis shock ke across ek ratio ki value hai. Chaar curves sab point $(1, 1)$ par start karti hain — Ex 1 ka degenerate shock. Daayein jaate hue: blue $p_{2} / p_{1}$ curve bina limit ke upar jaati hai; red $T_{2} / T_{1}$ curve aur bhi tezi se upar jaati hai (like $M_{1}^{2}$ ); green $ρ_{2} / ρ_{1}$ curve upar jaati hai lekin $6$ par ek horizontal dashed cap ki taraf flatten hoti hai; orange $M_{2}$ curve $1$ se girti hai ek floor ke paas $0.38$ ki taraf, dotted sonic line $y = 1$ ke neeche rehti hai. $M_{1} = 1.2, 2, 3$ par marked dots dikhate hain kahan Ex 3, Ex 4 aur Ex 7 baithte hain. Ek hi message: pressure aur temperature bhag jaate hain, density saturate hoti hai, flow hamesha subsonic khatam hoti hai.

Ex 1 — Cell C1: degenerate shock $M_{1} = 1$

Worked example Air,

$γ = 1.4$ , $M_{1} = 1$ exactly. Saare ratios nikalo. Forecast: padhne se pehle guess karo. Exactly sonic speed par ek shock pressure ke saath kya karega? Kuch nahi, bahut kuch, ya beech mein kuch?

1. Pressure ratio (R2): $\frac{p _{2}}{p _{1}} = 1 + \frac{2 ( 1.4 )}{2.4} (1^{2} - 1) = 1 + 1.1667 \times 0 = 1.$ Yeh step kyun? Factor $(M_{1}^{2} - 1)$ , $M_{1} = 1$ par vanish ho jaata hai, poora increment khatam kar deta hai.

2. Density ratio (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 1 )}{0.4 ( 1 ) + 2} = \frac{2.4}{2.4} = 1.$ Yeh step kyun? Numerator $(γ + 1) M_{1}^{2} = 2.4$ aur denominator $(γ - 1) M_{1}^{2} + 2 = 2.4$ tab coincide karte hain jab $M_{1}^{2} = 1$ .

3. Downstream Mach (R1): $M_{2}^{2} = \frac{1 + 0.2 ( 1 )}{1.4 ( 1 ) - 0.2} = \frac{1.2}{1.2} = 1 \Rightarrow M_{2} = 1.$ Yeh step kyun? Boundary par do algebraic roots ( $M_{2} = M_{1}$ aur shock root) merge ho jaate hain — yahan wo algebra hai jo isko prove karta hai: shock relation R1 ko $M_{2}^{2} = f (M_{1}^{2})$ likha ja sakta hai, aur original quadratic ka doosra physical root hamesha $M_{2} = M_{1}$ hota hai. R1 mein $M_{1} = 1$ set karo: yeh $M_{2}^{2} = 1$ return karta hai, yaani $M_{2} = M_{1}$ . Dono roots ek hi number ban gaye hain. Yahi coincidence exactly "roots merge" ka matlab hai.

4. Temperature (R4): $T_{2} / T_{1} = 1/1 = 1$ .

Verify: saare chaar ratios 1 ke barabar hain — shock ki zero strength hai. Yeh figure ka left edge hai jahan saari curves $(1, 1)$ se guzarti hain. $M_{1} = 1$ par ek "shock" ek Mach wave hai jo koi jump nahi carry karta. ✓

Ex 2 — Cell C2: forbidden expansion shock $M_{1} < 1$

Worked example Maano koi claim karta hai ki ek shock air ko

$M_{1} = 0.5$ se kuch $M_{2} > 1$ tak le jaata hai. Isko test karo. Forecast: algebra khushi se ek number return karega. Kya Second Law ise allow karega?

1. $M_{1} = 0.5$ ko R1 mein daalo: $M_{2}^{2} = \frac{1 + 0.2 ( 0.25 )}{1.4 ( 0.25 ) - 0.2} = \frac{1.05}{0.15} = 7.0 \Rightarrow M_{2} = 2.65.$ Yeh step kyun? Algebra mein kuch bhi nahi roka hame subsonic value plug karne se — equation bas ek rational function hai.

2. To pure math kehta hai $0.5 \to 2.65$ : subsonic to supersonic. Ab pressure ratio (R2) compute karo: $\frac{p _{2}}{p _{1}} = 1 + 1.1667 (0.25 - 1) = 1 - 0.875 = 0.125.$ Yeh step kyun? Hume entropy test ke liye $p_{2} / p_{1}$ chahiye; yahan yeh $1$ se kam hai — pressure giregi.

3. Density (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 0.25 )}{0.4 ( 0.25 ) + 2} = \frac{0.6}{2.1} = 0.2857.$ Temperature (R4): $T_{2} / T_{1} = 0.125/0.2857 = 0.4375.$

4. Entropy change (R5): $Δ s / R = \frac{γ}{γ - 1} ln \frac{T _{2}}{T _{1}} - ln \frac{p _{2}}{p _{1}} = 3.5 ln (0.4375) - ln (0.125) .$ $= 3.5 (- 0.8267) - (- 2.0794) = - 2.8935 + 2.0794 = - 0.8141.$ Yeh step kyun? Yeh entropy test hai. Ek allowed adiabatic process mein $Δ s \geq 0$ hona chahiye.

Verify: $Δ s / R = - 0.814 < 0$ . Negative entropy → forbidden. Expansion shock exist nahi kar sakta. Ek hi physical direction hai: supersonic→subsonic. Is cell ka koi valid answer nahi, aur yahi answer hai. ✓

Ex 3 — Cell C3: weak shock $M_{1} = 1.2$

Worked example Air,

$M_{1} = 1.2$ . $M_{2}, p_{2} / p_{1}, ρ_{2} / ρ_{1}, T_{2} / T_{1}$ nikalo, aur comment karo ki yeh kitna "gentle" hai. Forecast: Mach 1 se thoda upar. Chhote jumps expect karo — lekin $M_{1} = 2$ se compare mein kitne chhote?

1. $M_{1}^{2} = 1.44$ . Pressure (R2): $\frac{p _{2}}{p _{1}} = 1 + 1.1667 (1.44 - 1) = 1 + 1.1667 (0.44) = 1.513.$ Yeh step kyun? Increment $M_{1}^{2} - 1 = 0.44$ ke saath scale karta hai, jo chhota hai.

2. Density (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 1.44 )}{0.4 ( 1.44 ) + 2} = \frac{3.456}{2.576} = 1.342.$ Yeh step kyun? R3 mein direct substitution.

3. Temperature (R4): $\frac{T _{2}}{T _{1}} = \frac{1.513}{1.342} = 1.128.$ Yeh step kyun? Ideal-gas identity $T = p / ρ$ ratio.

4. Downstream Mach (R1): $M_{2}^{2} = \frac{1 + 0.2 ( 1.44 )}{1.4 ( 1.44 ) - 0.2} = \frac{1.288}{1.816} = 0.7093 \Rightarrow M_{2} = 0.842.$

Verify: saare ratios 1 ke karib hain aur $M_{2} = 0.842$ barely subsonic hai — ek weak shock, almost isentropic. Ex 4 se compare karo jahan $M_{1} = 2$ deta hai $p_{2} / p_{1} = 4.5$ . Figure par yeh point left mein baitha hai, curves $(1, 1)$ corner se muskil se uthe hain. ✓

Ex 4 — Cell C4: standard shock $M_{1} = 2$

Ex 5 — Cell C5: strong-shock limit $M_{1} = 10$ aur $M_{1} \to \infty$

Worked example Air,

$M_{1} = 10$ , phir mathematical limit $M_{1} \to \infty$ . Density aur temperature compare karo. Forecast: Mach 10 par, kya density pressure ki tarah bina bound ke climb karti rehti hai?

1. $M_{1} = 10$ par Density (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 100 )}{0.4 ( 100 ) + 2} = \frac{240}{42} = 5.714.$ Yeh step kyun? R3 mein $M_{1}^{2} = 100$ ke saath.

2. Infinite-Mach cap: $M_{1} \to \infty lim \frac{ρ _{2}}{ρ _{1}} = \frac{γ + 1}{γ - 1} = \frac{2.4}{0.4} = 6.$ Yeh step kyun? R3 ke top aur bottom ko $M_{1}^{2}$ se divide karo; $M_{1}^{2}$ -independent terms vanish ho jaate hain, $\frac{γ + 1}{γ - 1}$ bachta hai. Density saturate hoti hai.

3. $M_{1} = 10$ par Temperature (R4): $\frac{T _{2}}{T _{1}} = \frac{[ 2.8 ( 100 ) - 0.4 ] [ 0.4 ( 100 ) + 2 ]}{2. 4 ^{2} ( 100 )} = \frac{279.6 \times 42}{576} = 20.39.$ Yeh step kyun? R4 ko pressure aur density relations ke product ke roop mein likha, $M_{1}^{2} = 100$ par evaluate kiya.

4. Jab $M_{1} \to \infty$ , $T_{2} / T_{1} \sim \frac{2 γ ( γ - 1 )}{( γ + 1 ) ^{2}} M_{1}^{2} \to \infty.$ Yeh step kyun? Dono bracket terms $M_{1}^{2}$ ki tarah grow karte hain lekin denominator sirf ek baar $M_{1}^{2}$ ki tarah — net growth $\propto M_{1}^{2}$ . Temperature unbounded hai.

Verify: density $5.71$ pehle hi apne hard cap $6$ ke 5% ke andar hai, jabki $T$ balloon karke $20 \times$ ho gayi hai. Extra shock energy heat mein jaati hai, compression mein nahi — re-entry heating ki physics yahi hai. ✓

Ex 6 — Cell C6: total-pressure loss aur entropy rise ( $M_{1} = 2$ )

Worked example

$M_{1} = 2$ air shock (Ex 4) ke liye, $Δ s / R$ aur $p_{02} / p_{01}$ nikalo. Confirm karo ki $T_{0}$ unchanged hai. Forecast: flow adiabatic hai. Kya stagnation pressure bachta hai? Kya stagnation temperature?

1. Entropy (R5): $\frac{Δ s}{R} = \frac{γ}{γ - 1} ln \frac{T _{2}}{T _{1}} - ln \frac{p _{2}}{p _{1}} = 3.5 ln (1.687) - ln (4.50) .$ $= 3.5 (0.5232) - 1.5041 = 1.8312 - 1.5041 = 0.3272.$ Yeh step kyun? Yeh irreversibility measure karta hai. Positive ⇒ physically allowed shock (Ex 2 se contrast karo).

2. Stagnation-pressure ratio (R5): $\frac{p _{02}}{p _{01}} = exp (- Δ s / R) = exp (- 0.3272) = 0.7209.$ Yeh step kyun? Kyunki $T_{0}$ conserved hai (agla step), stagnation pressure loss poori tarah entropy se governed hai: $p_{02} / p_{01} = e^{- Δ s / R}$ .

3. Stagnation temperature: total enthalpy shock ke across conserved hai (energy equation with $q = 0$ ), isliye $T_{02} = T_{01}$ , yaani $T_{02} / T_{01} = 1$ exactly. Yeh step kyun? Adiabatic + no work ⇒ constant total enthalpy ⇒ constant $T_{0}$ . Isliye "adiabatic ≠ isentropic."

Verify: $p_{0}$ apni upstream value ka $72%$ tak girta hai (28% loss = drag penalty), phir bhi $T_{0}$ untouched hai. Dekho Isentropic flow relations ki "no loss" kaisa dikhta: wahan $p_{0}$ constant rehta. Gap wo entropy hai jo shock ke andar generate hoti hai. ✓

Ex 7 — Cell C7: real-world word problem

Worked example Ek supersonic wind tunnel air ko

$M_{1} = 3$ par run karta hai, static pressure $p_{1} = 20 kPa$ , static temperature $T_{1} = 220 K$ . Test section mein ek normal shock khadi hai. Physical $p_{2}$ (kPa), $T_{2}$ (K), aur downstream speed $u_{2}$ (m/s) nikalo. $R = 287 J/kg\cdotK$ lo. Forecast: upstream air thandi aur patli hai. Shock ke baad — kitni garam, kitni dense, aur kya yeh local sound speed ke neeche slow ho jaati hai?

1. $M_{1}^{2} = 9$ . Pressure ratio (R2): $\frac{p _{2}}{p _{1}} = 1 + 1.1667 (9 - 1) = 1 + 9.333 = 10.33 \Rightarrow p_{2} = 10.33 \times 20 = 206.7 kPa .$ Yeh step kyun? Ratios dimensionless hain; real units ke liye diye gaye $p_{1}$ se multiply karo.

2. Density ratio (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 9 )}{0.4 ( 9 ) + 2} = \frac{21.6}{5.6} = 3.857.$ Temperature (R4): $\frac{T _{2}}{T _{1}} = \frac{10.33}{3.857} = 2.679 \Rightarrow T_{2} = 2.679 \times 220 = 589.4 K .$ Yeh step kyun? $T_{2} = T_{1} \times (T_{2} / T_{1})$ ; air $220$ K se garam hokar nearly $590$ K tak pahunch jaati hai.

3. Downstream Mach (R1): $M_{2}^{2} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{2.8}{12.4} = 0.2258 \Rightarrow M_{2} = 0.4752.$ Yeh step kyun? Real speed mein convert karne ke liye hume $M_{2}$ aur $a_{2}$ chahiye.

4. Downstream sound speed: $a_{2} = γ R T_{2} = 1.4 \times 287 \times 589.4 = 236, 800 = 486.6 m/s .$ Phir $u_{2} = M_{2} a_{2} = 0.4752 \times 486.6 = 231.3 m/s .$ Yeh step kyun? $u = M a$ dimensionless Mach number ko m/s mein turn karta hai. Units: $(J/kg\cdotK) (K) = m^{2} / s^{2} = m/s$ . ✓

Verify: $p_{2} \approx 207$ kPa (10× jump), $T_{2} \approx 589$ K (air absolute temp mein double se zyada), aur flow $231$ m/s tak slow ho jaati hai, jo subsonic hai kyunki $u_{2} < a_{2} = 487$ m/s ( $M_{2} = 0.475 < 1$ ). Sab physically sane hai. ✓

Ex 8 — Cell C8: exam twist (inverse problem)

Worked example Air mein ek normal shock ke downstream

measured static pressure ratio $p_{2} / p_{1} = 7.0$ hai. Ulta solve karke $M_{1}$ aur hence $M_{2}$ nikalo. Forecast: normally $M_{1}$ input hota hai. Yahan yeh unknown hai — kya hum pressure formula ko cleanly invert kar sakte hain?

1. R2 invert karo: $\frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) \Rightarrow M_{1}^{2} = 1 + \frac{γ + 1}{2 γ} (\frac{p _{2}}{p _{1}} - 1) .$ Yeh step kyun? R2 $M_{1}^{2}$ mein linear hai, isliye yeh algebraically invert ho jaata hai — koi quadratic ki zaroorat nahi.

2. Plug in karo: $M_{1}^{2} = 1 + \frac{2.4}{2.8} (7.0 - 1) = 1 + 0.8571 (6) = 1 + 5.143 = 6.143 \Rightarrow M_{1} = 2.478.$ Yeh step kyun? $γ = 1.4$ ke saath numeric substitution.

3. Ab $M_{2}$ ke liye aage jaao (R1): $M_{2}^{2} = \frac{1 + 0.2 ( 6.143 )}{1.4 ( 6.143 ) - 0.2} = \frac{2.229}{8.400} = 0.2653 \Rightarrow M_{2} = 0.5151.$ Yeh step kyun? $M_{1}$ recover hone ke baad, R1 downstream Mach deta hai.

Verify: $M_{1} = 2.478$ ko R2 mein wapas daalo: $1 + 1.1667 (6.143 - 1) = 1 + 1.1667 (5.143) = 7.00$ ✓ — diya gaya data match karta hai. Aur $M_{2} = 0.515 < 1$ , ek real compression shock ke consistent. ✓

Ex 9 — Cell C9: ek alag gas ( $γ = 1.667$ , monatomic)

Worked example Argon (monatomic,

$γ = 5/3 \approx 1.667$ ) $M_{1} = 2$ par. $ρ_{2} / ρ_{1}$ aur uska saturation cap nikalo, aur air se compare karo. Forecast: air se bhaari $γ$ . Kya density zyada ya kam compress hoti hai?

1. $γ - 1 = 0.6667, γ + 1 = 2.6667$ . Density (R3): $\frac{ρ _{2}}{ρ _{1}} = \frac{2.6667 ( 4 )}{0.6667 ( 4 ) + 2} = \frac{10.667}{4.6667} = 2.286.$ Yeh step kyun? R3 argon ke $γ$ ke saath; har $γ$ explicitly enter karta hai, isliye hum bas re-substitute karte hain.

2. Saturation cap: $\frac{γ + 1}{γ - 1} = \frac{2.6667}{0.6667} = 4.0.$ Yeh step kyun? Ex 5 jaisi $M_{1} \to \infty$ limit; monatomic gas $4 \times$ par cap hoti hai, air ki tarah $6 \times$ nahi.

3. Context ke liye Pressure (R2): $\frac{p _{2}}{p _{1}} = 1 + \frac{2 ( 1.667 )}{2.6667} (4 - 1) = 1 + 1.25 (3) = 4.75.$ Yeh step kyun? R2 argon ke $γ$ ke saath; air ke $4.50$ se thoda zyada.

4. Poornata ke liye Temperature (R4): $\frac{T _{2}}{T _{1}} = \frac{p _{2} / p _{1}}{ρ _{2} / ρ _{1}} = \frac{4.75}{2.286} = 2.078.$ Yeh step kyun? Wahi ideal-gas identity, isliye hum argon ki heating ko air ke $T_{2} / T_{1} = 1.687$ se same $M_{1} = 2$ par compare kar sakte hain.

Verify: argon kam compress hota hai ( $2.286$ vs air ka $2.667$ ) aur uska density cap lower hai ( $4$ vs $6$ ), phir bhi yeh zyada heat karta hai ( $T_{2} / T_{1} = 2.078$ vs air ka $1.687$ ). Ek gas jisme fewer molecular degrees of freedom hain (larger $γ$ ), shock energy ka bada hissa temperature mein daal deti hai. Formulas $γ$ ko correctly track karte hain. ✓

Recall

Recall Active recall — answers cover karo

$M_{1} = 1$ par, saare chaar static ratios kya hain? ::: Sab 1 ke barabar — zero strength ka shock (dono roots merge ho jaate hain). Ek shock subsonic flow ko supersonic kyun nahi le ja sakta? ::: Isse $Δ s < 0$ milega, jo Second Law ne forbid kiya hai. Air mein $M_{1} = 2$ par $ρ_{2} / ρ_{1}$ kya hai? ::: $2.667$ . Air aur argon ke liye $M_{1} \to \infty$ par density cap kya hai? ::: $6$ (air, $γ = 1.4$ ) aur $4$ (argon, $γ = 5/3$ ). Shock ke across kaun sa stagnation quantity conserved hai aur kaun sa girta hai? ::: $T_{0}$ conserved hai, $p_{0}$ girta hai (yahan $M_{1} = 2$ par $0.72$ tak). $M_{1}$ ko measured $p_{2} / p_{1}$ se nikalne ke liye tum kaise invert karte ho? ::: $M_{1}^{2} = 1 + \frac{γ + 1}{2 γ} (\frac{p _{2}}{p _{1}} - 1)$ (linear, koi quadratic nahi).

Yeh bhi dekho: Oblique shock waves (wahi jumps velocity component par apply hote hain jo ek tilted shock ke normal ho), Rayleigh & Fanno flow (heat aur friction flow ko $M = 1$ ki taraf drive karte hain), aur Isentropic flow relations (loss-free limit jisse yeh shocks depart karte hain).