Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations: Exercises

3.1.2 · D4 · Physics › Compressible Flow & Aerodynamics › Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

Teen tools jo tum baar baar use karoge, sab ke sab calorically perfect gas ke liye (constant $γ$ ):

Har symbol ka plain words mein reminder:

Definition Symbol glossary (agar koi letter unclear ho toh open karo)

$T$ = static temperature — woh temperature jo ek thermometer gas ke saath drift karte hue read karta hai (kelvin, K).
$P$ = static pressure — gas ke apne frame mein pressure (pascal, Pa; $1 kPa = 1000 Pa$ ).
$ρ$ = static density — mass per volume ( $kg/ m^{3}$ ).
Subscript $0$ = stagnation / total — woh value jo gas tab reach karti agar use smoothly rest par laya jaata.
$V$ = flow speed (m/s). $M = V / a$ = Mach number (dimensionless).
$γ$ = ratio of specific heats ( $= 1.4$ air ke liye). $\frac{γ}{γ - 1} = 3.5$ aur $\frac{1}{γ - 1} = 2.5$ air ke liye — yeh dono yaad kar lo.

Intuition Woh map jo tum poore page use karoge — pehle figure padho

Neeche di gayi figure master picture hai har exercise ke liye. Horizontal axis Mach number $M$ hai; vertical axis teen "stagnation over static" ratios mein se koi bhi hai. Teen curves ek hi point se upar jaati hain:

white = $T_{0} / T$ (exponent $1$ , sabse gentle climb),
cyan = $ρ_{0} / ρ$ (exponent $2.5$ ),
amber = $P_{0} / P$ (exponent $3.5$ , sabse steep).

Teeen cheezein jo dikhni chahiye: (1) teeno curves $M = 0$ par value $1$ se guzarti hain (rest par gas ke paas convert karne ke liye kuch nahi, isliye total = static); (2) yeh $1$ se neeche kabhi nahi jaatein aur sirf upar hi badhti hain — kyunki bracket $1 + \frac{γ - 1}{2} M^{2}$ hamesha $\geq 1$ hoti hai; (3) amber pressure curve baaki sabse upar khich jaati hai aur steep hoti rehti hai — jab $M \to \infty$ toh bracket $\to \infty$ , isliye teeno ratios infinity tak diverge karte hain, pressure sabse tez. Yeh picture dimag mein rakho: ek problem solve karna sirf inhi curves mein se ek ko read karna (ya invert karna) hai.

Figure — Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

Level 1 — Recognition

Q1.1 Kaun si stagnation quantity normal shock ke across conserved hoti hai, aur kaun si nahi? Har ek ke liye ek-word reason batao.

Q1.2 Air ke liye ( $γ = 1.4$ ), $P_{0} / P$ aur $ρ_{0} / ρ$ relations mein aane wale do numerical exponents likho.

Q1.3 Ek gas $M = 0$ par flow kar rahi hai (at rest). Koi calculation kiye bina, $T_{0} / T$ , $P_{0} / P$ , $ρ_{0} / ρ$ batao.

Recall Solutions L1

Q1.1 $== T_{0} ==$ conserved hoti hai (shock ke across flow adiabatic hai — koi heat bahar nahi jaati, aur $T_{0}$ ko sirf adiabaticity chahiye). $== P_{0} ==$ conserved nahi hoti: shock irreversible hai, entropy badhti hai, isliye total pressure drop hoti hai. Memory hook: temperature = energy = adiabatic; pressure = order = isentropic. Dekho Isentropic flow relations.

Q1.2 $\frac{γ}{γ - 1} = \frac{1.4}{0.4} = 3.5$ pressure ke liye; $\frac{1}{γ - 1} = \frac{1}{0.4} = 2.5$ density ke liye.

Q1.3 $M = 0$ par factor $(1 + \frac{γ - 1}{2} M^{2}) = 1$ hota hai, isliye teeno ratios exactly 1 ke barabar hain. Gas already rest par hai, isliye "static" aur "stagnation" ek hi hain — convert karne ke liye koi motion nahi hai.

Level 2 — Application

Q2.1 Air at $M = 1.5$ , $T = 250 K$ . $T_{0}$ find karo.

Q2.2 Same flow, static pressure $P = 40 kPa$ . $P_{0}$ find karo.

Q2.3 Air at $M = 0.8$ . Density ratio $ρ_{0} / ρ$ find karo aur batao ki density apni stagnation value se kitne percentage kam hui hai.

Recall Solutions L2

Q2.1 KYA: adiabatic $T_{0}$ relation use karo. KYUN: sirf Mach aur temperature chahiye. $\frac{T _{0}}{T} = 1 + \frac{0.4}{2} (1.5)^{2} = 1 + 0.2 \times 2.25 = 1.45$ $T_{0} = 1.45 \times 250 = 362.5 K .$

Q2.2 KYA: same bracket ko power $3.5$ tak raise karo. KYUN: stagnation isentropic stop se define hoti hai. $\frac{P _{0}}{P} = (1.45)^{3.5} = 3.671 \dots$ $P_{0} = 3.671 \times 40 = 146.8 kPa .$

Q2.3 KYA: same bracket ko power $2.5$ tak raise karo. KYUN: density ratio isentropic $ρ_{0} / ρ$ relation hai, jiska exponent $\frac{1}{γ - 1} = 2.5$ hai — hum ise choose karte hain (pressure ya temperature relation nahi) kyunki question density ke baare mein pooch raha hai. Bracket $= 1 + 0.2 (0.8)^{2} = 1 + 0.128 = 1.128$ . $\frac{ρ _{0}}{ρ} = (1.128)^{2.5} = 1.352.$ Toh $ρ / ρ_{0} = 1/1.352 = 0.7396$ : flowing gas mein rest ki tulna mein $== 26% ==$ kam density hai ( $1 - 0.7396 = 0.2604$ ). Gas ko accelerate karna use expand (thin) karta hai — isliye hum compressibility ignore nahi kar sakte jab $M ≳ 0.3$ .

Level 3 — Analysis

Q3.1 (Inversion) Ek subsonic Pitot tube total $P_{0} = 125 kPa$ read karta hai static $P = 100 kPa$ ke against, $T = 300 K$ ke saath. $M$ aur speed $V$ find karo.

Q3.2 (Limit reasoning) Same numbers use karke, woh speed compute karo jo incompressible Bernoulli formula $P_{0} - P = \frac{1}{2} ρ V^{2}$ deta, aur Q3.1 ke versus percentage error batao. ( $ρ = P / R T$ .)

Q3.3 (Sign/behaviour) Calculator ke bina, argue karo ki $T_{0} / T$ , $P_{0} / P$ , $ρ_{0} / ρ$ increase ya decrease karte hain jab $M$ badhta hai, kaun sabse tez badhta hai, aur kya hota hai jab $M \to \infty$ . Yeh seedha master figure se padho.

Recall Solutions L3

Q3.1 KYA: pressure relation ko invert karo. KYUN: hume ratio diya gaya hai, $M$ chahiye. Yeh yahan valid hai kyunki Pitot subsonic stated hai — neeche warning box dekho ki yeh kyun matter karta hai. $\frac{P _{0}}{P} = 1.25 = (1 + 0.2 M^{2})^{3.5}$ Dono sides ka $1/3.5$ power lo: $1 + 0.2 M^{2} = 1.2 5^{1/3.5} = 1.2 5^{0.2857} = 1.0654$ $0.2 M^{2} = 0.0654 \Rightarrow M^{2} = 0.3272 \Rightarrow M = 0.5720.$ Speed of sound $a = 1.4 \times 287 \times 300 = 347.2 m/s$ , toh $V = M a = 0.5720 \times 347.2 = 198.6 m/s .$

Q3.2 KYA: incompressible estimate. $ρ = \frac{P}{R T} = \frac{100000}{287 \times 300} = 1.1614 kg/ m^{3}$ . $V_{Bern} = \frac{2 ( P _{0} - P )}{ρ} = \frac{2 \times 25000}{1.1614} = 43055 = 207.5 m/s .$ Error $= \frac{207.5 - 198.6}{198.6} = 4.5%$ zyada — Bernoulli ignore karta hai ki air rukne par compress hoti hai, isliye woh pressure rise ka credit speed ko zyada de deta hai. $M \approx 0.57$ par error already kaafi percent hai — full relation pe switch karne ki warning.

Q3.3 Teeno factors $(1 + positive)^{positive}$ hain, isliye teeno $M$ ke saath increase karte hain (figure mein curves sirf upar hi jaati hain); $M = 0$ par har ek $1$ ke barabar hai (common starting point). Exponents steepness ka order dete hain: pressure $3.5$ carry karta hai, density $2.5$ , temperature $1$ . Isliye $== P_{0} / P ==$ sabse tez badhta hai (amber), phir $ρ_{0} / ρ$ (cyan), phir $T_{0} / T$ (white) — exactly teeno curves ka vertical ordering. Limit $M \to \infty$ : bracket $1 + \frac{γ - 1}{2} M^{2} \to \infty$ , toh teeno ratios $\infty$ par diverge karte hain; kyunki exponents alag hain amber curve baaki ko bina bound ke outrun karta hai. Physically: ek arbitrarily fast stream ko rokna arbitrarily large kinetic energy dump karta hai, toh total temperature, pressure aur density sab bina limit ke badhte hain.

Level 4 — Synthesis

Q4.1 Ek supersonic wind tunnel air ko ek reservoir (stagnation) se $T_{0} = 500 K$ , $P_{0} = 800 kPa$ par expand karta hai test section mein $M = 2.0$ tak. Wahan static $T$ , $P$ , aur flow speed $V$ find karo. (Assume isentropic nozzle.)

Q4.2 (Chain through a shock) Ek normal shock phir test section mein baith jaati hai. Downstream static values re-measure hoti hain aur total pressure $P_{0, 2} = 575 kPa$ tak gir jaata hai, jabki total temperature unchanged rehta hai. Entropy rise per unit mass $Δ s = - R ln (P_{0, 2} / P_{0, 1})$ compute karo aur uska sign confirm karo. Downstream $T_{0}$ bhi batao.

Recall Solutions L4

Q4.1 KYA: stagnation se static ki taraf ulta kaam karo same relations use karke, kyunki reservoir values hi isentropic nozzle mein har jagah stagnation values hain. $M = 2$ par bracket: $1 + 0.2 (2)^{2} = 1 + 0.8 = 1.8$ . $T = \frac{T _{0}}{1.8} = \frac{500}{1.8} = 277.8 K .$ $P = \frac{P _{0}}{( 1.8 ) ^{3.5}} = \frac{800}{7.824} = 102.3 kPa .$ Test section par speed of sound: $a = 1.4 \times 287 \times 277.8 = 334.1 m/s$ , toh $V = M a = 2.0 \times 334.1 = 668.2 m/s .$

Q4.2 KYA: generate hui entropy akele total pressure ki drop se padhi jaati hai (total temperature adiabatic shock ke across fixed rehta hai). $Δ s = - R ln \frac{P _{0, 2}}{P _{0, 1}} = - 287 ln \frac{575}{800} = - 287 ln (0.71875) .$ $ln (0.71875) = - 0.33024, Δ s = - 287 \times (- 0.33024) = 94.8 J k g^{- 1} K^{- 1} .$ $Δ s > 0$ ✓ — entropy badhta hai, jaisa ki Second Law irreversible shock ke liye demand karta hai. Downstream $T_{0, 2} = T_{0, 1} = 500 K$ (shock adiabatic hai; total temperature untouched rehti hai chahe total pressure giri ho).

Level 5 — Mastery

Q5.1 (Compressibility correction, quantitative) Air mein $M = 0.5$ par Pitot ke liye, exact $\frac{P _{0} - P}{\frac{1}{2} ρ V ^{2}}$ compute karo aur ise low-speed prediction $1$ se compare karo. Dikhao ki yeh series $1 + \frac{M ^{2}}{4} + \dots$ se first order tak match karta hai. (Yaad karo $\frac{1}{2} ρ V^{2} = \frac{γ}{2} P M^{2}$ .)

Q5.2 (Design synthesis) Ek aircraft $M = 0.85$ par cruise karta hai jahan ambient (static) air $T = 223 K$ , $P = 26.5 kPa$ hai. Ek leading-edge sensor flow ko isentropically rok leta hai. Temperature $T_{0}$ find karo jo sensor tip reach karta hai aur total pressure $P_{0}$ jo use feel hoti hai. Comment karo ki de-icing kyun zaroori hai chahe $T_{0} > T$ ho.

Recall Solutions L5

Q5.1 KYA: exact dynamic-pressure-normalised aur incompressible ka ratio banao. Exact: $P_{0} - P = P [(1 + 0.2 M^{2})^{3.5} - 1]$ . $M = 0.5$ ke saath: bracket $= 1 + 0.2 (0.25) = 1.05$ , aur $1.0 5^{3.5} = 1.1783$ , toh $P_{0} - P = 0.17827 P$ . Reference $\frac{1}{2} ρ V^{2} = \frac{γ}{2} P M^{2} = 0.7 \times 0.25 P = 0.175 P$ . Ratio $= \frac{0.17827}{0.175} = 1.0187$ . Series prediction: $1 + \frac{M ^{2}}{4} + \frac{2 - γ}{24} M^{4} + \dots = 1 + \frac{0.25}{4} = 1 + 0.0625 = 1.0625$ first order tak — trend mein $1.0187$ se agree karta hai jab tum notice karo ki $M^{4}$ term $\frac{2 - γ}{24} M^{4} = \frac{0.6}{24} (0.0625) = 0.00156$ ise wapas khichta hai, aur higher terms ise aur trim karti hain. Toh $M = 0.5$ par bhi Bernoulli pressure rise ko lagbhag $1.9%$ kam read karta hai; $M^{2} /4$ term leading compressibility correction hai.

Q5.2 Bracket $= 1 + 0.2 (0.85)^{2} = 1 + 0.2 \times 0.7225 = 1.1445$ . $T_{0} = 1.1445 \times 223 = 255.2 K .$ $P_{0} = 26.5 \times (1.1445)^{3.5} = 26.5 \times 1.6026 = 42.5 kPa .$ Stagnation tip $223 K (- 5 0^{\circ} C)$ se $255 K (- 1 8^{\circ} C)$ tak warm hoti hai — ek $+ 32 K$ ram-heating jump. Lekin $255 K$ abhi bhi paani ke $273 K$ freezing point se neeche hai, toh super-cooled droplets impact par abhi bhi freeze hote hain: de-icing zaroori hai heating ke bawajood. Pressure tip par almost double ho jaata hai ( $26.5$ se $42.5 kPa$ tak), jo ek compressible Pitot ko account karna hi padta hai.

Self-test summary

Aage badhne se pehle yeh memory se answer karo:

Kaun si stagnation quantity shock ke baad bachti hai, aur kyun?

T_{0}

, kyunki shock adiabatic hai (energy conserved);

P_{0}

girta hai kyunki shock irreversible hai (entropy badhti hai).

Air ke liye,

P_{0} / P

aur

ρ_{0} / ρ

par exponent kya hai?

3.5

aur

2.5

respectively.

Subsonic Pitot ratio se

M

nikalne ke liye,

P_{0} / P

ko kaunsa operation invert karta hai?

Ratio ko power

\frac{γ - 1}{γ} = \frac{1}{3.5}

tak raise karo, 1 ghataao,

\frac{γ - 1}{2}

se divide karo, square-root lo — valid sirf tab jab free stream subsonic ho.

Bernoulli

M = 0.5

par pressure rise ko roughly kitne percent kam read karta hai?

Lagbhag

2%

(leading

M^{2} /4

compressibility correction, next term se trim kiya gaya).

Jab

M \to \infty

toh teeno stagnation ratios ka kya hota hai?

Sab infinity tak diverge karte hain,

P_{0} / P

sabse tez badhta hai.