Normal shock waves — Rankine-Hugoniot relations (all 5) — derivations: Question bank

3.1.11 · D5 · Physics › Compressible Flow & Aerodynamics › Normal shock waves — Rankine-Hugoniot relations (all 5) — de

Yeh bank un paanch Rankine–Hugoniot relations par aur unke supporting ideas par depend karta hai jo Speed of sound and Mach number, Isentropic flow relations, Second Law of Thermodynamics — entropy, aur Stagnation properties se aate hain. Yahan koi arithmetic grinding nahi hai — woh D3/D4 mein hai. Yeh understanding ke baare mein hai.

Zaroori vocabulary ka reminder. State 1 upstream hai (shock se pehle), state 2 downstream hai (baad mein). $M$ Mach number hai (flow speed ÷ local speed of sound). $p, ρ, T$ ordinary "static" pressure, density, temperature hain — jo ek thermometer ya gauge gas ke saath ride karte hue read karta. Subscript 0, jaise $p_{0}$ ya $T_{0}$ , ek stagnation (total) quantity ko mark karta hai — woh value jo gas tab reach karti agar use smoothly aur adiabatically rest par laya jaata.

True or false — justify

Recall

Ek normal shock hamesha supersonic flow ko subsonic tak slow karta hai. ::: True. Kisi bhi $M_{1} > 1$ ke liye shock equations ka surviving root $M_{2} < 1$ deta hai; dusra root ( $M_{2} = M_{1}$ ) sirf no-shock case hai. Ek shock subsonic flow ( $M_{1} < 1$ ) ko accelerate karke supersonic tak le ja sakta hai. ::: False. Woh "expansion shock" root entropy ko girayega, jo Second Law forbid karta hai. Real shocks hamesha compressive hote hain. Ek adiabatic shock mein total pressure $p_{0}$ conserved hoti hai. ::: False. Adiabatic total enthalpy conserve karta hai (isliye $T_{0}$ constant rehti hai), total pressure nahi. Shock irreversible hai, entropy badhti hai, aur isliye $p_{0}$ girti hai. Static temperature $T$ aur stagnation temperature $T_{0}$ dono shock ke across jump karte hain. ::: Aadha sach: static $T$ upar jump karta hai, lekin $T_{0}$ bilkul same rehti hai kyunki total enthalpy conserved hai ( $h_{1} + \frac{1}{2} u_{1}^{2} = h_{2} + \frac{1}{2} u_{2}^{2}$ ). Entropy har real normal shock ke across increase hoti hai. ::: True. Shock ke andar irreversible viscous/conductive dissipation $Δ s > 0$ guarantee karta hai; yeh exactly woh law hai jo compression root pick karta hai. Jab $M_{1} \to \infty$ hota hai toh density ratio $ρ_{2} / ρ_{1}$ bhi without bound grow karta hai. ::: False. Yeh $\frac{γ + 1}{γ - 1}$ par saturate ho jaata hai (air ke liye = 6). Ek perfect gas ko us ratio se zyada compress nahi kiya ja sakta, chahe shock kitni bhi strong ho. Jab $M_{1} \to \infty$ hota hai toh pressure ratio $p_{2} / p_{1}$ without bound grow karta hai. ::: True. $p_{2} / p_{1} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1)$ ki koi ceiling nahi hai — pressure aur temperature badhte rehte hain jab bhi density cap ho chuki ho. Exactly $M_{1} = 1$ Mach number ka shock finite pressure jump produce karta hai. ::: False. $M_{1} = 1$ par har ratio collapse hokar 1 ho jaata hai — yeh infinitely weak shock hai, yaani koi jump nahi. Ek "sonic shock" ki strength zero hoti hai. Flow shock ke across constant total temperature se cross karta hai lekin constant total pressure se nahi. ::: True. Yeh shock ki defining signature hai: $T_{0}$ preserved (adiabatic, koi heat/work nahi), $p_{0}$ lost (irreversible). Adiabatic $\neq =$ isentropic. Kyunki shock mass, momentum aur energy conserve karta hai, isliye woh reversible hona chahiye. ::: False. Woh teen conservation laws kisi bhi control volume mein hold karte hain, chahe reversible ho ya nahi. Reversibility ek alag condition hai (constant $s$ ), jo shock violate karta hai.

Spot the error

Recall

"Gas compress ho raha hai, isliye $ρ_{2} / ρ_{1} = p_{2} / p_{1}$ ." ::: Galat — yeh constant temperature assume karta hai. Kyunki $T$ bhi jump karta hai, $ρ_{2} / ρ_{1} = (p_{2} / p_{1}) (T_{1} / T_{2})$ ; density saturate hoti hai jabki pressure nahi. "Flow adiabatic hai, isliye shock ke across isentropic hona chahiye." ::: Galat — adiabatic ka matlab hai koi heat add nahi, lekin internal viscous dissipation phir bhi entropy generate karta hai. Adiabatic + reversible isentropic deta; shock adiabatic + irreversible hai. "Downstream Mach number: $M_{2} = 1/ M_{1}$ ." ::: Galat — koi simple reciprocal rule nahi hai. Sahi relation hai $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{1}^{2}}{γ M _{1}^{2} - \frac{γ - 1}{2}}$ , jo sirf $M_{1} = 1$ ke paas roughly 1 ko mirror karta hai. "Kyunki $T_{0}$ unchanged hai, flow ko bina kisi loss ke rest par laya ja sakta hai." ::: Galat — $T_{0}$ unchanged hai lekin $p_{0}$ gir gayi hai. Stagnation state degrade ho gayi hai; aap same temperature recover karoge lekin lower pressure par, yaani usable work ka loss hua. " $ρ u^{2}$ ke units sirf coincidence se pressure ke units hain." ::: Galat — $ρ u^{2}$ ek genuine momentum flux hai aur physics ki wajah se pressure ke units share karta hai, isliye $p + ρ u^{2}$ (momentum equation) meaningful hai. $ρ u^{2} = γ p M^{2}$ use karna derivation ka engine hai. "Ek weak shock ( $M_{1}$ sirf 1 se thoda upar) phir bhi bahut saara total pressure lose karta hai." ::: Galat — weak shocks ke liye $p_{0}$ loss $(M_{1}^{2} - 1)^{3}$ ki tarah scale karta hai, isliye woh third-order small hai. Weak shocks nearly isentropic hote hain; bade losses ke liye strong shocks chahiye. "Kyunki velocity shock ke across drop hoti hai, kinetic energy destroy ho jaati hai." ::: Galat — kinetic energy destroy nahi hoti balki enthalpy (thermal energy) mein convert hoti hai; total enthalpy $h + \frac{1}{2} u^{2}$ conserved hai. Us conversion ki disorder hi entropy raise karti hai.

Why questions

Recall

Supersonic flow ko kisi obstacle se milne ke liye smooth pressure rise ki jagah shock kyun use karni padti hai? ::: Pressure signals speed of sound par travel karte hain; sound se tez move karne wale upstream molecules obstacle ko "sun" nahi sakte, isliye information aage propagate nahi ho sakti aur adjustment abruptly hona padta hai. Shock compression root select karta hai aur expansion root kyun nahi? ::: Dono roots mass, momentum aur energy satisfy karte hain, lekin sirf compression $Δ s > 0$ deta hai. Second Law tie-breaker hai jo teen conservation laws akele nahi provide kar sakte. Density saturate kyun hoti hai lekin temperature $M_{1} \to \infty$ par badhti rehti hai? ::: Ek perfect gas mein molecules ko in relations ke under pack karne ki fixed limit hoti hai, isliye extra incoming kinetic energy sirf random thermal motion mein jaati hai — yaani temperature mein, aur compression mein nahi. Shock ke across $T_{0}$ constant kyun rehti hai jab baaki sab kuch change ho raha hai? ::: Kyunki shock koi heat add nahi karta aur koi work nahi karta, isliye total enthalpy $h_{0} = c_{p} T_{0}$ conserved hai; $T_{0}$ us conserved total enthalpy ko directly measure karta hai. Re-entry heating shock se aati hai, sirf surface friction se kyun nahi? ::: Bow shock vehicle ki enormous relative kinetic energy ko $T_{2} / T_{1} \sim M_{1}^{2}$ heating mein convert karta hai; hypersonic $M_{1}$ par yeh thermal jump boundary-layer friction se kahin zyada hoti hai. Substitution $ρ u^{2} = γ p M^{2}$ algebra ko sirf $M$ ke functions mein kyun collapse kar deta hai? ::: Yeh momentum equation mein sirf velocity-aur-density term ko purely $p$ aur $M$ ke terms mein rewrite karta hai, isliye jab $T$ ratios $a = γ R T$ se handle ho jaate hain, sab kuch Mach number mein reduce ho jaata hai. Total-pressure loss ko shock ki "price" kyun kehte hain? ::: Lost $p_{0}$ useful work (thrust, recovery) karne ki lost ability hai; ek strong shock wale engine inlet mein stagnation pressure waste hoti hai, jo drag aur reduced efficiency ke roop mein dikhta hai.

Edge cases

Recall

$M_{1} \to 1^{+}$ limit mein paanch ratios ke saath kya hota hai? ::: Har ratio ( $p_{2} / p_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ , $p_{02} / p_{01}$ ) $\to 1$ ho jaata hai aur $M_{2} \to 1$ . Shock ek infinitely weak Mach wave mein disappear ho jaati hai. $M_{1} \to \infty$ par limiting downstream Mach number $M_{2}$ kya hai? ::: $M_{2}^{2} \to \frac{γ - 1}{2 γ}$ ( $\approx 0.378$ , isliye air ke liye $M_{2} \approx 0.615$ ). Ek infinitely strong shock bhi finite subsonic exit Mach number chhod jaati hai. Kya $M_{1} = 1$ exactly ek shock produce kar sakta hai? ::: Nahi — $M_{1} = 1$ degenerate boundary hai jahan shock strength zero hoti hai. Ek genuine shock ke liye strictly $M_{1} > 1$ chahiye. Kya normal shock purely subsonic flow mein kahi bhi possible hai? ::: Nahi. Ek stationary normal shock ke liye upstream flow supersonic hona zaroori hai; $M_{1} < 1$ ke saath compression root Second Law violate karta. Strong-shock limit mein, kya $p_{02} / p_{01}$ zero approach karta hai ya koi finite floor hai? ::: Yeh zero approach karta hai — entropy generation without bound badhta hai, isliye $p_{02} / p_{01} = exp (- Δ s / R) \to 0$ . Bahut strong shocks ke liye total-pressure recovery collapse ho jaati hai. Same $M_{1}$ par normal shock ke downstream state aur oblique shock ke downstream state mein kya fark hai? ::: Ek oblique shock sirf flow ke normal component ko process karta hai, isliye effectively yeh $M_{1} sin β$ par ek weaker "normal shock" hai; tangential velocity unchanged pass hoti hai, milder jump deta hai aur downstream flow possibly still-supersonic reh sakti hai. Agar aap $γ \to 1$ insist karo, toh density-saturation limit ka kya hota hai? ::: $\frac{γ + 1}{γ - 1} \to \infty$ , matlab ek gas jisme bahut saare internal energy modes hain (low $γ$ ) arbitrarily compress ho sakti hai — "6× cap" specific to $γ = 1.4$ hai.