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

3.1.2 · D2 · Physics › Compressible Flow & Aerodynamics › Stagnation (total) quantities — T₀, P₀, ρ₀ — derivations

Step 1 — Ek chalti gas mein energy ke do buckets

KYA HAI. Socho ek chhota sa air ka parcel right taraf ud raha hai. Uske paas energy do alag buckets mein hai:

ek thermal bucket — uske molecules ki invisible jiggling, jise hum temperature $T$ ki tarah feel karte hain;
ek motion bucket — poora parcel speed $V$ (metres per second) par sail kar raha hai.

YE YAHAN SE KYU SHURU. Temperature literally stored microscopic energy hai. Agar hum parcel ko rok dein, toh motion bucket ko kahin na kahin khaali hona hoga — aur woh sirf thermal bucket mein ja sakta hai (heat escape nahi hone di). Toh temperature zaroor badhegi. Is story ko numbers mein badalne ke liye humein energy accounting rule chahiye.

PICTURE. Blue arrow bulk motion hai (bucket 2); orange fuzz molecular jiggling hai (bucket 1). Dekho total fixed rehta hai jabki split badalta hai.

Step 2 — Energy ledger: enthalpy + motion = constant

KYA HAI. Parcel ko ek pipe mein follow karo. Hum ek rule likhte hain jo kehta hai ki uski total energy kabhi nahi badlegi. Woh rule hai steady-flow energy equation:

$thermal energy per kg (enthalpy) h + motion energy per kg \frac{V ^{2}}{2} = the total, a constant h_{0}$

Enthalpy $h$ kyun, plain heat kyun nahi? Ek flowing pipe mein gas aage waali gas par push-work bhi karta hai. Woh quantity jo already us push-work ko include karti hai woh hai enthalpy $h$ (isliye parent ka Mistake #3 warn karta hai: $c_{p}$ use karo, $c_{v}$ nahi). Toh ek flow ke liye honest "thermal bucket" $h$ hai, internal energy nahi.

Right side constant kyun hai? Pipe wall se koi heat cross nahi karti (adiabatic) aur koi shaft gas par work nahi karta. In do switches ke off hone par, First Law collapse ho jaata hai "do buckets ka sum nahi badlega." Us unchanging sum ko $h_{0}$ kaho — stagnation enthalpy, woh value jo tumhare paas hogi agar saari motion thermal mein convert ho jaaye ( $V = 0$ ).

PICTURE. Jab pipe narrow hoti hai toh parcel speed up hota hai (motion bucket bharta hai) aur thanda hota hai (thermal bucket drain hota hai) — lekin dono heights hamesha same dashed line $h_{0}$ mein add hoti hain.

Step 3 — Enthalpy ko temperature mein trade karo

KYA HAI. Hum temperature chahte hain, lekin ledger enthalpy mein bolta hai. Ek ideal gas ke liye constant specific heat ke saath ek simple exchange rate hai:

$h = c_{p} T \Rightarrow thermal, K units mein c_{p} T + \frac{V ^{2}}{2} = c_{p} T_{0}$

Yeh swap kyun. $c_{p}$ (specific heat at constant pressure) exactly woh "ek kelvin raise karne ke liye kitne joules per kg" number hai. Yeh energy (joules/kg) aur temperature (kelvin) ke beech currency converter hai. Yeh $c_{p}$ hona chahiye (naa ki $c_{v}$ ) kyunki hum enthalpy convert kar rahe hain.

$c_{p} T$ se divide karo aur rearrange karo:

$T_{0} = T + \frac{V ^{2}}{2 c _{p}}$

Padho: stagnation temperature $T_{0}$ , static temperature $T$ hai plus ek bonus $\frac{V ^{2}}{2 c _{p}}$ jo poora motion bucket se aata hai.

PICTURE. Roki hui gas mein ek thermometer zyada read karta hai ( $T_{0}$ ) bante flow ke saath drift karne wale thermometer se ( $T$ ); gap woh khaali hua motion bucket hai.

Step 4 — Bonus ko Mach number se rewrite karo

KYA HAI. Clumsy $\frac{V ^{2}}{2 c _{p}}$ clean ho jaata hai agar hum speed ko Mach number mein measure karein. Humein do known facts chahiye:

$a^{2} = γ R T (sound speed), c_{p} = \frac{γ R}{γ - 1} (ideal-gas identity) .$

Yeh do kyun. Pehla $T$ ko ek speed mein convert karta hai jisse hum $V$ compare kar sakein. Doosra $c_{p}$ ko $γ$ aur $R$ ke terms mein rewrite karta hai, jo magically $a^{2} = γ R T$ ke against cancel ho jaayega.

$T_{0} / T = 1 + \frac{V ^{2}}{2 c _{p} T}$ mein substitute karo:

$\frac{T _{0}}{T} = 1 + \frac{V ^{2}}{2 \cdot \frac{γ R}{γ - 1} \cdot T} = 1 + \frac{γ - 1}{2} = V^{2} / a^{2} = M^{2} \frac{V ^{2}}{γ R T}$

Under-brace poora trick hai: $γ R T$ hi $a^{2}$ hai, toh $V^{2} / (γ R T) = M^{2}$ .

PICTURE. Ek rising curve: $T_{0} / T$ vs $M$ , $M^{2}$ ki wajah se parabola ki tarah dheere dheere upar curve karta hua.

Step 5 — Temperature se pressure tak smoothness chahiye

KYA HAI. Temperature ne sirf energy ki parwah ki. Pressure parwah karta hai kaise gas ko rokaa jaaye. Smoothly roko (koi rubbing nahi) aur tumhe zyada se zyada pressure milega; roughly roko aur kuch hamesha ke liye kho jaata hai. Smooth ideal hai isentropic — reversible aur adiabatic.

Naya ingredient kyun. Do parcels same $T_{0}$ share kar sakte hain phir bhi alag $P_{0}$ reach kar sakte hain agar ek ko friction ke saath slow kiya gaya ho. $P_{0}$ pin down karne ke liye humein clean, reversible stop demand karni chahiye. Isentropic relations temperature ko pressure aur density se link karte hain:

$\frac{P _{0}}{P} = (\frac{T _{0}}{T})^{\frac{γ}{γ - 1}}, \frac{ρ _{0}}{ρ} = (\frac{T _{0}}{T})^{\frac{1}{γ - 1}}$

Yeh kahan se aate hain. Isentropic ka matlab hai $P v^{γ} = const$ (ek stiff spring law); ideal-gas law $P v = R T$ ke saath combine karo volume $v$ eliminate karne ke liye, aur woh exponents nikal aate hain.

PICTURE. Static se stagnation tak do ramps: smooth green ramp poore $P_{0}$ tak chadhti hai; jagged red ramp (friction) neeche top out hoti hai — same $T_{0}$ , chhota $P_{0}$ .

Step 6 — Do boxed pressure/density laws assemble karo

KYA HAI. Result 1 ko isentropic exponents mein daalo:

Exponents is tarah order kyun karte hain. Gas ko squeeze karo aur pressure sabse tezi se jump karta hai, density baad mein, temperature sabse kam — isliye "Temp is Tame, Press is Power." Ek consistency check jo hold hona chahiye (ideal gas law): $\frac{P _{0}}{P} = \frac{ρ _{0}}{ρ} \cdot \frac{T _{0}}{T}$ , kyunki $3.5 = 2.5 + 1$ . ✓

PICTURE. Teen curves $M = 0$ se upar uthte hue: temperature (gentle), density (steeper), pressure (steepest) — exponent ranking visible ho gayi.

Step 7 — Edge case A: $M \to 0$ Bernoulli mein collapse ho jaata hai

KYA HAI. Speed tiny set karo. Small $M^{2}$ ke liye pressure bracket expand karo:

$\frac{P _{0}}{P} \approx 1 + \frac{γ}{2} M^{2} + \frac{γ}{8} M^{4} + \dots \Rightarrow P_{0} - P \approx \frac{1}{2} ρ V^{2}$

Yeh hona hi chahiye. Crawl speed par gas barely compress hoti hai, toh compressible law ko familiar Bernoulli result $P_{0} = P + \frac{1}{2} ρ V^{2}$ mein fold ho jaana chahiye. Hota hai. Bache hue $M^{4}$ piece ko compressibility correction kehte hain — $M \approx 0.3$ se neeche negligible, usse upar essential.

PICTURE. Exact pressure curve aur Bernoulli line $M = 0$ ke paas ek doosre se chipke hain, phir jaise jaise $M$ badhta hai alag ho jaate hain — exactly dikhate hain ki "incompressible" kahan toot ta hai.

Step 8 — Edge case B: irreversibility $P_{0}$ ko girata hai (shock trap)

KYA HAI. Gas ko ek normal shock (ya kisi bhi friction) se bhejo. Ab stop rough hai, entropy badhti hai, aur $P_{0}$ girta hai — jabki $T_{0}$ nahi hiltaa.

Asymmetry kyun. $T_{0}$ energy conservation par ride karta hai, jise ek shock respect karta hai (still adiabatic). $P_{0}$ reversibility par ride karta hai, jise ek shock violate karta hai. Toh:

$T_{0}$ : ek shock ke across conserved. ✓
$P_{0}$ : ek shock ke across decrease hota hai. ✗

Yahi parent ka Mistake #1 hai visual banaya gaya — kabhi mat maano ki $P_{0}$ har adiabatic process mein survive karta hai.

PICTURE. Red shock line ke across: $T_{0}$ bar level rehti hai, $P_{0}$ bar loss $Δ P_{0}$ se girta hai.

Ek-picture summary

Upar saari cheez ek single map par: shared bracket teen exponents ko feed karta hai; smooth road poora $P_{0}$ rakhta hai, rough road (shock/friction) kuch khota hai.

Recall Feynman retelling — poora walkthrough plain words mein

Ek chalti gas energy do tarike se carry karti hai: warmth aur rush. Dheere se roko aur rush warmth mein badal jaata hai, toh ruki hui gas zyada hot hoti hai — woh zyada hot reading hai total temperature $T_{0}$ , aur yeh sirf maangta hai ki koi heat leak na ho. Kyunki temperature actually sirf stored energy hai, ek clean energy ledger ( $h + \frac{1}{2} V^{2} =$ constant) deta hai $T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}$ , jahan $M$ speed hai sound-speeds mein measured. Pressure zyada choosy hai: yeh depend karta hai kaise tum gas ko rokate ho. Smoothly roko aur tumhe poora total pressure $P_{0}$ milta hai; roughly roko (ek shock, friction) aur kuch squeeze hamesha ke liye kho jaata hai chahe warmth untouched rahe. Woh smoothness rule (isentropic) same bracket ko pressure aur density laws mein turn karta hai, bas bade exponents ke saath — pressure sabse bada, density baad mein, temperature sabse chhota. Dheere chalo aur sab kuch plain Bernoulli mein melt ho jaata hai. Yahi poori story hai: ek bracket, teen exponents, rough stops ke baare mein ek warning.

Recall Term-by-term self-test

Kaun si quantity ek shock mein survive karti hai, $T_{0}$ ya $P_{0}$ ? ::: $T_{0}$ (adiabatic); $P_{0}$ girta hai (irreversible). Teeno formulas mein shared bracket kya hai? ::: $1 + \frac{γ - 1}{2} M^{2}$ . $T_{0} = T + V^{2} /2 c_{p}$ mein $c_{p}$ kyun, $c_{v}$ kyun nahi? ::: Kyunki humne enthalpy $h = c_{p} T$ convert ki, internal energy nahi. Bernoulli expansion mein $M^{4}$ term kya represent karta hai? ::: Compressibility correction. Air ke liye $P_{0} / P$ , $ρ_{0} / ρ$ , $T_{0} / T$ par exponents? ::: $3.5, 2.5, 1$ .