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

3.1.11 · D4 · Physics › Compressible Flow & Aerodynamics › Normal shock waves — Rankine-Hugoniot relations (all 5) — de

Throughout hum air ke saath $γ = 1.4$ use karte hain jab tak koi problem aur na kahe. Yahan complete toolkit hai jise tum reach karoge — har symbol parent note mein build kiya gaya tha, lekin chaliye unhe plain words mein restate karte hain taaki kuch bhi assumed na ho:

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

Figure dekho: har ratio ek curve hai jo $M_{1} = 1$ par 1 se upar jaati hai. Pressure aur temperature infinity ki taraf bhaagti hai; density apni ceiling $\frac{γ + 1}{γ - 1} = 6$ par flatten ho jaati hai; downstream Mach number apne floor ki taraf sagg karta hai. Yeh picture mind mein rakho — yeh tumhe turant bata deta hai ki koi answer plausible hai ya nahi.

Level 1 — Recognition

L1.1

Batao ki har quantity normal shock ke across badhti hai, girti hai, ya same rehti hai, aur physical reason bhi do: (i) static pressure $p$ , (ii) total temperature $T_{0}$ , (iii) total pressure $p_{0}$ , (iv) Mach number $M$ , (v) entropy $s$ .

Recall Solution

(i) $p$ badhta hai — shock ek compression ki wall hai.
(ii) $T_{0}$ same rehta hai — adiabatic, koi work nahi ⇒ total enthalpy conserved ⇒ $T_{0}$ constant.
(iii) $p_{0}$ girta hai — shock irreversible hai, entropy badhti hai, isliye $p_{02} / p_{01} = e^{- Δ s / R} < 1$ .
(iv) $M$ girta hai (supersonic → subsonic).
(v) $s$ badhta hai — Second Law; yahi reason hai ki sirf compression shocks exist karte hain.

L1.2

Air ( $γ = 1.4$ ) ke liye $M_{1} = 1$ (ek infinitely weak shock), compute karo $p_{2} / p_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ , aur $M_{2}$ . Kya notice karte ho?

Recall Solution

$M_{1}^{2} = 1$ plug karo:

$p_{2} / p_{1} = 1 + \frac{2.8}{2.4} (1 - 1) = 1$ .
$ρ_{2} / ρ_{1} = \frac{2.4 ( 1 )}{0.4 ( 1 ) + 2} = \frac{2.4}{2.4} = 1$ .
$T_{2} / T_{1} = \frac{[ 2.8 - 0.4 ] [ 0.4 + 2 ]}{2. 4 ^{2} ( 1 )} = \frac{2.4 \times 2.4}{5.76} = 1$ .
$M_{2}^{2} = \frac{1 + 0.2}{1.4 - 0.2} = \frac{1.2}{1.2} = 1 \Rightarrow M_{2} = 1$ . Notice karo: $M_{1} = 1$ par har ratio exactly 1 hai — koi shock hai hi nahi. Shock tabhi kuch karta hai jab $M_{1} > 1$ ho. Yeh figure ke saare curves ka anchor point hai.

Level 2 — Application

L2.1

Air ek normal shock mein $M_{1} = 3$ pe enter karta hai. $M_{2}$ , $p_{2} / p_{1}$ , $ρ_{2} / ρ_{1}$ , aur $T_{2} / T_{1}$ find karo.

Recall Solution

$M_{1}^{2} = 9$ , $γ = 1.4$ ke saath:

$M_{2}^{2} = \frac{1 + 0.2 ( 9 )}{1.4 ( 9 ) - 0.2} = \frac{2.8}{12.4} = 0.22581 \Rightarrow M_{2} = 0.4752$ .
$p_{2} / p_{1} = 1 + \frac{2.8}{2.4} (9 - 1) = 1 + 1.16667 \times 8 = 10.333$ .
$ρ_{2} / ρ_{1} = \frac{2.4 ( 9 )}{0.4 ( 9 ) + 2} = \frac{21.6}{5.6} = 3.857$ .
$T_{2} / T_{1} = \frac{p _{2} / p _{1}}{ρ _{2} / ρ _{1}} = \frac{10.333}{3.857} = 2.679$ . Sanity: $M_{2} < 1$ ✓, saare ratios $M_{1} = 2$ se zyaada hain ✓ (curves pe upar ki taraf steeper).

L2.2

Upstream conditions hain $p_{1} = 50 kPa$ , $T_{1} = 250 K$ , $M_{1} = 2.5$ . Actual downstream static pressure $p_{2}$ aur temperature $T_{2}$ find karo (sirf ratios nahi).

Recall Solution

$M_{1}^{2} = 6.25$ ke saath:

$p_{2} / p_{1} = 1 + \frac{2.8}{2.4} (6.25 - 1) = 1 + 1.16667 \times 5.25 = 7.125 \Rightarrow p_{2} = 7.125 \times 50 = 356.25 kPa$ .
$ρ_{2} / ρ_{1} = \frac{2.4 ( 6.25 )}{0.4 ( 6.25 ) + 2} = \frac{15}{4.5} = 3.3333$ .
$T_{2} / T_{1} = 7.125/3.3333 = 2.1375 \Rightarrow T_{2} = 2.1375 \times 250 = 534.4 K$ .

Level 3 — Analysis

L3.1

Air ke liye, upstream Mach number $M_{1}$ find karo jo static pressure ratio $p_{2} / p_{1} = 4.5$ produce karta hai. (Relation 3 ko invert karo.)

Recall Solution

Set karo $4.5 = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1)$ . $M_{1}^{2}$ ke liye solve karne ke liye rearrange karo — kyun algebra, guessing nahi? Relation 3 $M_{1}^{2}$ mein linear hai, isliye cleanly invert hoti hai: $M_{1}^{2} - 1 = (4.5 - 1) \cdot \frac{γ + 1}{2 γ} = 3.5 \cdot \frac{2.4}{2.8} = 3.5 \times 0.857143 = 3.0.$ Toh $M_{1}^{2} = 4 \Rightarrow M_{1} = 2.0$ . (Parent note ke worked example (a) se consistent hai.)

L3.2

Air mein $M_{1} = 2$ par ek normal shock hai. Entropy rise $Δ s$ (J kg⁻¹K⁻¹) aur total-pressure ratio $p_{02} / p_{01}$ compute karo.

Recall Solution

Example (a) se: $p_{2} / p_{1} = 4.5$ , $T_{2} / T_{1} = 1.6875$ . Yeh $Δ s$ mein kyun enter karte hain? Ek perfect gas ka entropy change sirf $T$ aur $p$ ke do end-state ratios par depend karta hai: $Δ s = c_{p} ln \frac{T _{2}}{T _{1}} - R ln \frac{p _{2}}{p _{1}} .$

$c_{p} = 1005$ : $1005 ln (1.6875) = 1005 \times 0.523248 = 525.86 J kg^{- 1} K^{- 1}$ .
$R ln (p_{2} / p_{1}) = 287 ln (4.5) = 287 \times 1.504077 = 431.67$ .
$Δ s = 525.86 - 431.67 = 94.19 J kg^{- 1} K^{- 1} > 0$ ✓ (shock allowed hai).
$p_{02} / p_{01} = exp (- Δ s / R) = exp (- 94.19/287) = exp (- 0.32819) = 0.7203$ . Toh flow apna lagbhag $28%$ total pressure lose kar deta hai. Compare karo Stagnation properties se.

Level 4 — Synthesis

L4.1

Ek aircraft $M_{1} = 3$ par fly karta hai jahan ambient air $T_{1} = 220 K$ hai. Ek pitot probe ke saamne ek normal shock khada hai. Find karo (i) shock ke bilkul peeche temperature $T_{2}$ , aur (ii) stagnation temperature $T_{0}$ jo probe read karta hai. Kya woh same hain? Explain karo.

Recall Solution

(i) L2.1 se, $T_{2} / T_{1} = 2.679$ , toh $T_{2} = 2.679 \times 220 = 589.4 K$ . (ii) Stagnation temperature isentropic $T_{0}$ formula use karta hai (Isentropic flow relations), aur kyunki $T_{0}$ shock ke across conserved hai hum ise upstream se evaluate kar sakte hain: $T_{0} = T_{1} (1 + \frac{γ - 1}{2} M_{1}^{2}) = 220 (1 + 0.2 \times 9) = 220 \times 2.8 = 616 K .$ Same nahi hain. $T_{2} = 589$ K woh static temperature hai jo shock ke bilkul peeche hai, jahan flow abhi bhi $M_{2} = 0.475$ par move kar rahi hai. Probe us residual motion ko rest pe laata hai, kinetic energy ka aakhiri hissa add karke $T_{0} = 616$ K tak pahunchta hai. Kyunki $T_{0}$ dono sides par same hai, hum ise state 2 se bhi compute kar sakte the — check karo: $T_{0} = T_{2} (1 + 0.2 M_{2}^{2}) = 589.4 (1 + 0.2 \times 0.22581) = 616$ K ✓.

L4.2

Usi $M_{1} = 3$ shock ke liye air mein, upstream total pressure $p_{01} = 400 kPa$ hai. Shock ke peeche read hone wala total pressure $p_{02}$ find karo. (Yahi exactly hai jo ek supersonic pitot correction handle karta hai.)

Recall Solution

Pehle L2.1 ratios se $Δ s$ ( $p_{2} / p_{1} = 10.333$ , $T_{2} / T_{1} = 2.679$ ):

$c_{p} ln (T_{2} / T_{1}) = 1005 ln (2.679) = 1005 \times 0.985637 = 990.57$ .
$R ln (p_{2} / p_{1}) = 287 ln (10.333) = 287 \times 2.335375 = 670.25$ .
$Δ s = 990.57 - 670.25 = 320.32 J kg^{- 1} K^{- 1}$ .
$p_{02} / p_{01} = exp (- 320.32/287) = exp (- 1.11609) = 0.32765$ .
$p_{02} = 0.32765 \times 400 = 131.1 kPa$ . Shock jitna strong, utna zyaada total pressure destroy hota hai — $M_{1} = 3$ par hum sirf lagbhag ek third bachate hain.

Level 5 — Mastery

L5.1 (Strong-shock limit)

Prove karo ki $M_{1} \to \infty$ (air) par $ρ_{2} / ρ_{1} \to 6$ aur $M_{2}^{2} \to \frac{γ - 1}{2 γ}$ . Phir dono ko numerically $M_{1} = 20$ par evaluate karo aur comment karo ki limits kitne close hain.

Recall Solution

Density limit. $M_{1}^{2}$ se kyun divide karein? Dominant terms expose karne ke liye: $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 ) M _{1}^{2}}{( γ - 1 ) M _{1}^{2} + 2} = \frac{γ + 1}{( γ - 1 ) + 2/ M _{1}^{2}} M_{1} \to \infty \frac{γ + 1}{γ - 1} = \frac{2.4}{0.4} = 6.$ Mach limit. $M_{2}^{2}$ ke top aur bottom ko $M_{1}^{2}$ se divide karo: $M_{2}^{2} = \frac{1/ M _{1}^{2} + \frac{γ - 1}{2}}{γ - \frac{γ - 1}{2} / M _{1}^{2}} M_{1} \to \infty \frac{( γ - 1 ) /2}{γ} = \frac{0.4/2}{1.4} = 0.142857 \Rightarrow M_{2} = 0.3780.$ $M_{1} = 20$ par ( $M_{1}^{2} = 400$ ):

$ρ_{2} / ρ_{1} = \frac{2.4 ( 400 )}{0.4 ( 400 ) + 2} = \frac{960}{162} = 5.926$ — ceiling 6 se $1.2%$ ke andar.
$M_{2}^{2} = \frac{1 + 0.2 ( 400 )}{1.4 ( 400 ) - 0.2} = \frac{81}{559.8} = 0.14469 \Rightarrow M_{2} = 0.3804$ — limit se $0.6%$ ke andar. Comment: density essentially saturate ho chuki hai jabki temperature (aur pressure) badhte rehte hain — ek ever-faster flow ki extra energy heat mein jaati hai, compression mein nahi. Yahi reason hai ki hypersonic re-entry ek thermal problem hai, compression problem nahi.

L5.2 (Reverse-shock check)

Suppose koi claim karta hai ki ek shock air ko $M_{1} = 0.6$ se ek downstream $M_{2} > 1$ tak le jaata hai ("expansion shock"). Ise entropy relation se test karo aur explain karo ki Second Law kya kehta hai.

Recall Solution

Formally $M_{1} = 0.6$ ( $M_{1}^{2} = 0.36$ ) ratios mein plug karo:

$p_{2} / p_{1} = 1 + \frac{2.8}{2.4} (0.36 - 1) = 1 + 1.16667 \times (- 0.64) = 0.25333$ (pressure drop ho jaata).
$T_{2} / T_{1} = \frac{[ 2.8 ( 0.36 ) - 0.4 ] [ 0.4 ( 0.36 ) + 2 ]}{2. 4 ^{2} ( 0.36 )} = \frac{0.608 \times 2.144}{2.0736} = 0.62864$ .
$Δ s = 1005 ln (0.62864) - 287 ln (0.25333) = 1005 (- 0.46428) - 287 (- 1.37333)$ $= - 466.60 + 394.15 = - 72.45 J kg^{- 1} K^{- 1} < 0$ . Entropy decrease karta hai — yeh Second Law violate karta hai. Aisa "expansion shock" impossible hai. Ek subsonic flow ko shock karke supersonic nahi banaya ja sakta; physically real branch sirf compression hai, $M_{1} > 1 \to M_{2} < 1$ . Supersonic tak accelerate karne ke liye tumhe smooth isentropic expansion chahiye (ek nozzle ya ek Prandtl–Meyer fan), kabhi shock nahi.

L5.3 (Cross-check via Fanno/Rayleigh intuition)

Confirm karo ki $M_{1} = 2$ air shock correct branch par hai yeh check karke ki total enthalpy (isliye $T_{0}$ ) dono sides par identical hai, state 1 se aur state 2 se $T_{0} / T_{1}$ compute karo aur dikhao ki woh match karte hain.

Recall Solution

Example (a) se: $M_{2}^{2} = 1/3$ , $T_{2} / T_{1} = 1.6875$ .

Upstream: $T_{0} / T_{1} = 1 + 0.2 (4) = 1.8$ .
Downstream: $T_{0} / T_{1} = (T_{2} / T_{1}) (1 + 0.2 M_{2}^{2}) = 1.6875 (1 + 0.2/3) = 1.6875 \times 1.066667 = 1.8$ . Woh exactly agree karte hain — total temperature conserved hai, energy equation confirm karta hai. Yeh Rayleigh & Fanno flow ke saath shared backbone hai: ek shock woh point hai jahan Fanno aur Rayleigh lines cross karti hain, mass, momentum, aur total enthalpy conserve karte hue supersonic se subsonic intersection par jump karte hue.

Recall

Recall Kaun sa relation reach karna hai? (answers cover karo)

$M_{1}$ diya ho, downstream Mach kaise nikaalun? ::: $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{1}^{2}}{γ M _{1}^{2} - \frac{γ - 1}{2}}$ Ek target $p_{2} / p_{1}$ diya ho, $M_{1}$ kaise find karein? ::: Linear-in- $M_{1}^{2}$ pressure relation ko invert karo. Har shock ke across kya conserved hai? ::: mass, momentum, total enthalpy (isliye $T_{0}$ ). $p_{02} / p_{01}$ kaise nikaalun? ::: $Δ s = c_{p} ln \frac{T _{2}}{T _{1}} - R ln \frac{p _{2}}{p _{1}}$ compute karo, phir $e^{- Δ s / R}$ . Ek impossible shock ko kaise reject karein? ::: $Δ s \geq 0$ check karo; agar negative hai, toh Second Law violate hota hai.