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

3.1.12 · D4 · Physics › Compressible Flow & Aerodynamics › Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂

Poore page mein, jab tak problem alag na kahe, gas air hai jisme $γ = 1.4$ hai (ratio of specific heats — gas kitni "springy" hai). State 1 = upstream (shock se pehle, supersonic); State 2 = downstream (baad mein, subsonic). Ek symbol jo tumhe abhi milega: $R$ hai specific gas constant — air ke liye $R = 287 J kg^{- 1} K^{- 1}$ — woh number jo pressure, density, aur temperature ko ideal-gas law $P = ρR T$ ke through link karta hai. Parent se paanch workhorse formulas, yahan restate ki gayi hain taaki yeh page apne aap mein complete ho:

Ek aur relation L2 trap mein naam se quote ki gayi hai, isliye hum ise yahan bhi restate karte hain — isentropic temperature ratio, jo directly energy conservation se follow karti hai ( $T_{0}$ conserved) aur shock tak aur shock ke baad hold karti hai lekin shock ke through kabhi nahi: $\frac{T_2}{T_1}=\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}\tag{5}$

Yahan $u$ flow ki speed hai (metres per second), $ρ$ density (mass per cubic metre), $P$ static pressure, $T$ absolute temperature (kelvin mein, K — kabhi celsius nahi, kyunki yeh formulas ratios use karti hain jo sirf absolute scale par kaam karti hain). $P_{0}$ aur $T_{0}$ stagnation (total) pressure aur temperature hain — woh values jo gas tab reach karti agar use smoothly rest par laya jata.

Climb karne se pehle map padho. Neeche ka figure is page ke har exercise ke peeche ki mental picture hai: ek single control volume jisme shock ek patli amber wall hai. Left par gas fast aur supersonic hai (State 1); wall cross karne par yeh slower aur subsonic nikali hai (State 2), jisme $P$ , $ρ$ , $T$ sab jump kar gaye hain. Har problem jo aage hai woh is picture mein ek arrow ya ek label quantify karne ke liye kahin hai — isliye baar baar ise dekhte raho.

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

Level 1 — Recognition

Goal: sahi formula pick karo aur plug in karo. Koi chaining nahi.

Exercise 1.1. Ek normal shock air flow mein $M_{1} = 3.0$ par baitha hai. Downstream Mach number $M_{2}$ find karo.

Recall Solution 1.1

Kaunsa tool aur kyun: $M_{2}$ sirf $M_{1}$ par depend karta hai — yahi exactly equation (7) deliver karti hai. Koi aur relation zaroorat nahi. $γ = 1.4$ ke saath do pieces compute karo, toh $\frac{γ - 1}{2} = 0.2$ :

Numerator: $1 + 0.2 (3^{2}) = 1 + 0.2 (9) = 1 + 1.8 = 2.8$ .
Denominator: $γ M_{1}^{2} - \frac{γ - 1}{2} = 1.4 (9) - 0.2 = 12.6 - 0.2 = 12.4$ . $M_{2}^{2} = \frac{2.8}{12.4} = 0.2258 \Rightarrow M_{2} = 0.2258 = 0.4752.$ Sanity check: $M_{2} < 1$ , toh flow ab subsonic hai — exactly wahi jo ek shock ko karna chahiye.

Exercise 1.2. Usi shock ( $M_{1} = 3.0$ ) ke liye, static pressure ratio $P_{2} / P_{1}$ find karo.

Recall Solution 1.2

Tool: equation (8) — yeh strength ( $M_{1}^{2} - 1$ ) ko directly pressure jump mein convert karti hai. $\frac{P _{2}}{P _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) = 1 + \frac{2.8}{2.4} (9 - 1) = 1 + \frac{2.8}{2.4} (8) .$ $\frac{2.8}{2.4} = 1.1667$ , toh $\frac{P _{2}}{P _{1}} = 1 + 1.1667 \times 8 = 1 + 9.333 = 10.33.$ Pressure apni upstream value se das guna se zyada jump kar jaati hai.

Exercise 1.3. Ek ek sentence mein batao, kisi bhi normal shock ke across $T_{0}$ (stagnation temperature) aur $P_{0}$ (stagnation pressure) ka kya hota hai, aur kyun.

Recall Solution 1.3

$T_{0}$ unchanged rehta hai. Shock adiabatic hai (koi heat cross nahi karti) aur koi shaft work nahi karti, isliye total energy carried per kilogram conserved hai: $T_{02} = T_{01}$ .
$P_{0}$ drop karta hai ( $P_{02} < P_{01}$ ). Shock irreversible hai — entropy $s$ badhti hai — aur stagnation pressure entropy se $P_{02} / P_{01} = e^{- Δ s / R}$ ke through tied hai. Ek positive $Δ s$ ratio ko 1 se neeche force karta hai. Parent se mnemonic: "PRT up, M & P₀ down, T₀ stays."

Level 2 — Application

Goal: ek $M_{1}$ se kai properties nikalo, aur real numbers mein convert karo.

Exercise 2.1. Air ek normal shock mein $M_{1} = 2.5$ par enter karti hai, jisme upstream static conditions $P_{1} = 40 kPa$ aur $T_{1} = 250 K$ hain. $P_{2}$ , $T_{2}$ , aur $ρ_{2} / ρ_{1}$ find karo.

Recall Solution 2.1

Plan: (8), (9) se ratios nikalo, phir (10) se $T_{2} / T_{1}$ , phir diye gaye upstream values se multiply karo. $M_{1}^{2} = 6.25$ , $γ = 1.4$ ke saath:

Density: $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 ) M _{1}^{2}}{( γ - 1 ) M _{1}^{2} + 2} = \frac{2.4 ( 6.25 )}{0.4 ( 6.25 ) + 2} = \frac{15}{4.5} = 3.333.$
Pressure: $\frac{P _{2}}{P _{1}} = 1 + \frac{2.8}{2.4} (6.25 - 1) = 1 + 1.1667 (5.25) = 1 + 6.125 = 7.125.$ $\Rightarrow P_{2} = 7.125 \times 40 = 285 kPa .$
Temperature: $\frac{T _{2}}{T _{1}} = \frac{P _{2} / P _{1}}{ρ _{2} / ρ _{1}} = \frac{7.125}{3.333} = 2.1375.$ $\Rightarrow T_{2} = 2.1375 \times 250 = 534.4 K .$ Gas shock chodti hai pressure ke 7× se zyada par aur absolute temperature se double se zyada par.

Exercise 2.2. Usi shock ( $M_{1} = 2.5$ ) ke liye, stagnation-pressure ratio $P_{02} / P_{01}$ find karo aur stagnation pressure ka kitna percentage kho gaya yeh batao.

Recall Solution 2.2

Tool: equation (11), entropy fingerprint. Do bracketed factors, exponents $\frac{γ}{γ - 1} = 3.5$ aur $\frac{1}{γ - 1} = 2.5$ .

Pehla bracket: $\frac{\frac{( γ + 1 ) M _{1}^{2}}{2}}{1 + \frac{γ - 1}{2} M _{1}^{2}} = \frac{\frac{2.4 ( 6.25 )}{2}}{1 + 0.2 ( 6.25 )} = \frac{7.5}{2.25} = 3.3333.$ $3.5$ tak raise karo: $3.333 3^{3.5} = 67.72.$
Doosra bracket: $\frac{γ + 1}{2 γ M _{1}^{2} - ( γ - 1 )} = \frac{2.4}{2.8 ( 6.25 ) - 0.4} = \frac{2.4}{17.5 - 0.4} = \frac{2.4}{17.1} = 0.14035.$ $2.5$ tak raise karo: $0.1403 5^{2.5} = 0.007345.$
Product: $P_{02} / P_{01} = 67.72 \times 0.007345 = 0.4973.$ Toh is shock se stagnation pressure ka lagbhag $50%$ destroy ho jaata hai.

Level 3 — Analysis

Goal: backwards reason karo, aur probe karo kyun numbers aise behave karte hain.

Exercise 3.1. Ek pressure gauge air mein normal shock ke across $P_{2} / P_{1} = 4.5$ read karta hai. Upstream Mach number $M_{1}$ find karo jo ise produce karta hai, phir $M_{2}$ find karo.

Recall Solution 3.1

Idea: equation (8) ko invert karo. Kyunki $P_{2} / P_{1}$ monotonically $M_{1}$ ke saath badhta hai, ek pressure ratio exactly ek $M_{1}$ pin karta hai. $\frac{P _{2}}{P _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) \Rightarrow M_{1}^{2} - 1 = \frac{γ + 1}{2 γ} (\frac{P _{2}}{P _{1}} - 1) .$ $M_{1}^{2} - 1 = \frac{2.4}{2.8} (4.5 - 1) = 0.85714 \times 3.5 = 3.0 \Rightarrow M_{1}^{2} = 4 \Rightarrow M_{1} = 2.0.$ Phir equation (7): numerator $1 + 0.2 (4) = 1.8$ ; denominator $1.4 (4) - 0.2 = 5.4$ ; $M_{2}^{2} = 1.8/5.4 = 0.3333$ , toh $M_{2} = 0.5774.$ Yeh parent ka Example 1 hai, doosri taraf se reach kiya gaya.

Exercise 3.2. Numerically dikhao ki $M_{1} = 5$ par density ratio (5.00) apni infinite- $M_{1}$ ceiling ke kitna karib hai, aur pressure ratio kisi bhi finite cheez ke kitne karib hai. Density ceiling kya hai, aur $M_{1} = 5$ par uska kitna fraction reach hota hai?

Recall Solution 3.2

Ceiling: (9) mein $M_{1} \to \infty$ lo. Top aur bottom ko $M_{1}^{2}$ se divide karo: $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 )}{( γ - 1 ) + 2/ M _{1}^{2}} M_{1} \to \infty \frac{γ + 1}{γ - 1} = \frac{2.4}{0.4} = 6.0.$ $M_{1} = 5$ par: $\frac{ρ _{2}}{ρ _{1}} = \frac{2.4 ( 25 )}{0.4 ( 25 ) + 2} = \frac{60}{12} = 5.0.$ Yeh ceiling ka $5.0/6.0 = 83.3%$ hai. Contrast: $P_{2} / P_{1} = 1 + \frac{2.8}{2.4} (24) = 29$ aur yeh $M_{1} \to \infty$ ke saath bina bound ke badhta rehta hai. Toh density saturate hoti hai jabki pressure run away karta hai. Kyun: mass conservation ( $ρ_{2} u_{2} = ρ_{1} u_{1}$ ) plus ek velocity jo sirf ek finite floor tak drop kar sakti hai woh $ρ_{2}$ cap kar deti hai; bacha hua kinetic energy sirf $P$ aur $T$ mein ja sakta hai.

Level 4 — Synthesis

Goal: shock relations ko ek bahari idea (speed of sound, inlets, entropy) ke saath combine karo.

Exercise 4.1. Ek supersonic aircraft $T_{1} = 216.7 K$ (tropopause) wali jagah fly karta hai. Uske inlet ke aage $M_{1} = 2.0$ par ek normal shock khada hai. Actual flow speeds $u_{1}$ aur $u_{2}$ (m/s mein) dono taraf find karo. $R = 287 J kg^{- 1} K^{- 1}$ use karo.

Recall Solution 4.1

Bridge tool — speed of sound. $u = M a$ jahan $a = γ R T$ (dekhein Speed of Sound and Mach Number). Humein $a$ dono taraf chahiye, isliye $T_{2}$ chahiye.

Upstream: $a_{1} = 1.4 \times 287 \times 216.7 = 87069 = 295.1 m/s$ . $u_{1} = M_{1} a_{1} = 2.0 \times 295.1 = 590.2 m/s .$
Downstream numbers: Ex. 3.1 se $M_{1} = 2$ par, $M_{2} = 0.5774$ , $T_{2} / T_{1} = 1.6875$ (parent Example 1). $T_{2} = 1.6875 \times 216.7 = 365.7 K$ ; $a_{2} = 1.4 \times 287 \times 365.7 = 146930 = 383.3 m/s$ . $u_{2} = M_{2} a_{2} = 0.5774 \times 383.3 = 221.3 m/s .$ Mass se check: $ρ_{2} / ρ_{1} = u_{1} / u_{2} = 590.2/221.3 = 2.667$ — equation (9) ki value $2.667$ se exactly match karta hai. Note karo ki flow $590$ se $221$ m/s tak slow hui, phir bhi $M_{2} < 1$ sirf isliye nahi ki $u$ gira balki isliye bhi ki $a$ badha (hotter gas ⇒ faster sound). Dono effects $M = u / a$ ko neeche push karti hain.

Exercise 4.2. Ek engineer ko $M_{1} = 2.0$ intake flow decelerate karni hai. Option A: ek normal shock. Option B: do successive weaker shocks, pehla effective $M_{1} = 1.5$ par jo $M \approx 1.34$ chhodta hai, phir $M = 1.34$ par ek normal shock. Dono options ka stagnation-pressure survival $P_{02} / P_{01}$ compare karo aur batao kaun behtar hai. (Dekhein Supersonic Inlets and Diffusers.)

Recall Solution 4.2

Tool: equation (11), A ke liye ek baar aur B ke liye do baar apply ki (multiply ki), kyunki stagnation ratios successive shocks ke across multiplicatively chain hote hain.

Option A, single shock at $M_{1} = 2.0$ : (11) use karke (parent Example 1e) $P_{02} / P_{01} = 0.7209.$
Option B, shock 1 at $M_{1} = 1.5$ :
- Pehla bracket: $\frac{2.4 ( 2.25 ) /2}{1 + 0.2 ( 2.25 )} = \frac{2.7}{1.45} = 1.8621$ ; $3.5$ tak raise karo toh $8.884$ milta hai.
- Doosra bracket: $\frac{2.4}{2.8 ( 2.25 ) - 0.4} = \frac{2.4}{5.9} = 0.40678$ ; $2.5$ tak raise karo toh $0.10559$ milta hai.
- Product: $0.9382.$
Option B, shock 2 at $M = 1.34$ : (same formula, $M^{2} = 1.7956$ )
- Pehla bracket: $\frac{2.4 ( 1.7956 ) /2}{1 + 0.2 ( 1.7956 )} = \frac{2.1547}{1.3591} = 1.5854$ ; raise $3.5$ → $5.022$ .
- Doosra bracket: $\frac{2.4}{2.8 ( 1.7956 ) - 0.4} = \frac{2.4}{4.6277} = 0.51861$ ; raise $2.5$ → $0.19365$ .
- Product: $0.9725.$
Combined B: $0.9382 \times 0.9725 = 0.9125.$ Verdict: Option B stagnation pressure ka lagbhag $91%$ rakhta hai jabki single strong shock ke liye sirf $72.1%$ hai. Do weak shocks ek strong shock se better hain — multi-shock supersonic inlets ke peeche yahi engineering rationale hai.

Level 5 — Mastery

Goal: ek result derive karo, limits aur second law ke baare mein reason karo.

Exercise 5.1. Equations (7) aur (8) use karke prove karo ki $M_{1} \to 1^{+}$ par har static ratio $\to 1$ ho jaata hai (shock ek sound wave ban jaata hai). Phir $P_{2} / P_{1}$ , $ρ_{2} / ρ_{1}$ , $T_{2} / T_{1}$ ko $M_{1} = 1.001$ par evaluate karo confirm karne ke liye.

Recall Solution 5.1

Proof. (8) mein, driver $(M_{1}^{2} - 1)$ hai. Jaise $M_{1} \to 1$ , $M_{1}^{2} - 1 \to 0$ , toh $\frac{P _{2}}{P _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1) \to 1 + 0 = 1.$ (9) mein, $M_{1}^{2} = 1$ set karo: $\frac{ρ _{2}}{ρ _{1}} = \frac{( γ + 1 )}{( γ - 1 ) + 2} = \frac{γ + 1}{γ + 1} = 1.$ Phir (10) se, $\frac{T _{2}}{T _{1}} = \frac{P _{2} / P _{1}}{ρ _{2} / ρ _{1}} \to \frac{1}{1} = 1.$ Teeno ratios 1 par collapse ho jaate hain — koi jump nahi, koi entropy rise nahi, "shock" sirf ek infinitesimal sound wave hai (isentropic, dekhein Isentropic Flow Relations). $M_{1} = 1.001$ par numerical confirmation ( $M_{1}^{2} = 1.002001$ ):

$P_{2} / P_{1} = 1 + 1.1667 (0.002001) = 1.002335.$
$ρ_{2} / ρ_{1} = \frac{2.4 ( 1.002001 )}{0.4 ( 1.002001 ) + 2} = \frac{2.404802}{2.400800} = 1.001667.$
$T_{2} / T_{1} = 1.002335/1.001667 = 1.000667.$ Sab 1 ke fraction of a percent ke andar — precisely "vanishingly weak shock."

Exercise 5.2. $M_{1} = 2.0$ ke liye specific entropy rise $Δ s = c_{p} ln \frac{T _{2}}{T _{1}} - R ln \frac{P _{2}}{P _{1}}$ compute karo, $c_{p} = 1004.5 J kg^{- 1} K^{- 1}$ , $R = 287$ use karke. Phir verify karo ki yeh $P_{02} / P_{01} = e^{- Δ s / R}$ ke through stagnation-pressure ratio reproduce karta hai, aur confirm karo ki sign second law obey karta hai.

Recall Solution 5.2

Tool + kyun: entropy static ratios (jo hamare paas hain) aur stagnation-pressure loss ke beech bridge hai. Yeh Entropy and Second Law aur Rankine–Hugoniot Relations ke saath loop close karta hai. Parent Example 1 se: $T_{2} / T_{1} = 1.6875$ , $P_{2} / P_{1} = 4.5$ . $Δ s = 1004.5 ln (1.6875) - 287 ln (4.5) .$

$ln (1.6875) = 0.52343 \Rightarrow 1004.5 \times 0.52343 = 525.79.$
$ln (4.5) = 1.50408 \Rightarrow 287 \times 1.50408 = 431.67.$ $Δ s = 525.79 - 431.67 = 94.12 J kg^{- 1} K^{- 1} .$ Second law: $Δ s = + 94.1 > 0$ — shock irreversible hai, entropy badhti hai, jaisi ki honi chahiye. ✓ $P_{0}$ ratio reproduce karo: $\frac{P _{02}}{P _{01}} = e^{- Δ s / R} = e^{- 94.12/287} = e^{- 0.32796} = 0.7204.$ Yeh direct formula (11) ki value $0.7209$ se match karta hai (thoda rounding difference) — dono roads milti hain, jaisi ki milni chahiye. Lesson yeh hai: tumne jo positive entropy rise compute ki wahi hai stagnation-pressure loss, doosre andaaz mein likhi hui. Jab bhi koi question "irreversibility," "loss," ya "wasted work" mention kare, yahi quantity uske peeche hai.

Recall One-line self-test (answers cover karo)

$M_{1} = 3$ gives $M_{2}$ ::: $\approx 0.475$ $M_{1} = 2.5$ gives $P_{2} / P_{1}$ ::: $\approx 7.125$ Density ceiling as $M_{1} \to \infty$ (air) ::: $\frac{γ + 1}{γ - 1} = 6$ Sign of $Δ s$ across any real shock ::: strictly positive Two weak shocks vs one strong (at overall $M_{1} = 2$ ): better for $P_{0}$ ? ::: two weak (91% vs 72%)