Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions of M: Visual walkthrough

3.1.7 · D2 · Physics › Compressible Flow & Aerodynamics › Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions

Step 0 — Do words jo pehle kamane zaroori hain: "stagnation" aur "Mach"

KYA. Kisi bhi algebra se pehle, do ideas ko ek picture chahiye, kyunki baad ke har step unhi par tikhe hain.

Ek gas ki local properties hoti hain jahan woh actually flow kar rahi hai: temperature $T$ (kitni hot hai), pressure $P$ (walls par kitna hard push karta hai), density $ρ$ (molecules kitne tight pack hain), aur speed $V$ (poora parcel kitni tez move kar raha hai).
Agar hum us parcel ko dheere se rok sakein — bina kisi friction ya heat leak ke $V = 0$ tak slow kar dein — toh uski properties new values mein badal jaati hain jinhe hum stagnation (ya "total") values $T_{0}, P_{0}, ρ_{0}$ kehte hain. Chhota "0" matlab "rest par laya gaya."

KYUN. Flow ko rokne se uski motion, squeeze-aur-heat mein convert hoti hai. Toh $T_{0}, P_{0}, ρ_{0}$ hamesha flowing $T, P, ρ$ se bade hote hain. Poora chapter yeh poochta hai: kitna zyada?

PICTURE. Left mein parcel speed $V$ par streak karta hai; right mein woh ek wall ke against pile up ho gaya hai aur rest mein aa gaya hai, zyada hot aur dense.

Figure — Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions of M

Step 1 — Energy conserved hai: enthalpy + kinetic-energy budget

KYA. Steady flow ke liye jisme koi heat add nahi aur koi work nahi hota, ek quantity per kilogram gas ek streamline ke along constant rehti hai:

$enthalpy: stored heat energy h + kinetic energy of motion \frac{1}{2} V^{2} = constant total h_{0}$

KYUN yeh equation aur koi nahi? Yeh energy bookkeeping hai (from Stagnation Properties and Energy Equation). "Koi heat nahi, koi work nahi" matlab per kilogram total energy change nahi ho sakti — woh sirf "stored heat" pot $h$ aur "motion" pot $\frac{1}{2} V^{2}$ ke beech trade kar sakti hai. Speed up ⇒ $\frac{1}{2} V^{2}$ barta hai ⇒ $h$ shrink hona chahiye. Flow rok do ( $V = 0$ ) ⇒ sab kuch $h$ mein baith jaata hai, aur us value ka naam $h_{0}$ hai.

Har symbol, jahan woh rehta hai:

$h$ — enthalpy, gas ki internal heat content per kilogram. Bada $h$ ⇒ zyada hot gas.
$\frac{1}{2} V^{2}$ — kinetic energy per kilogram. $\frac{1}{2}$ aur $V^{2}$ wahi hain jo $\frac{1}{2} m v^{2}$ mein hain, bas per unit mass.
$h_{0}$ — frozen total. Kyunki yeh $h$ ke barabar hai jab $V = 0$ , toh yeh stagnation enthalpy hai.

PICTURE. Ek see-saw: ek taraf heat pot, doosri taraf motion pot. Plank total kabhi nahi badalti; sirf balance tilt hota hai.

Step 2 — Enthalpy ko temperature mein badlo

KYA. Ek perfect gas ke liye, enthalpy sirf temperature times ek constant hoti hai: $h = c_{p} T$ . Substitute karo:

$h c_{p} T + \frac{1}{2} V^{2} = h_{0} c_{p} T_{0}$

KYUN. Hum $T$ measure kar sakte hain but $h$ directly nahi, toh hum $h$ ko thermometer par kuch se swap karte hain. $c_{p}$ specific heat at constant pressure hai — ek kilogram ko ek degree warm karne mein kitne joules lagte hain. Ek perfect gas ke liye yeh ek fixed number hai, toh $h$ aur $T$ perfectly proportional hain.

Rearrange karo us ratio ko isolate karne ke liye jo hum ultimately chahte hain. Har term ko $c_{p} T$ se divide karo:

$\frac{T _{0}}{T} = 1 + \frac{V ^{2}}{2 c _{p} T}$

Padhne mein: left side yeh hai ki ruka hua gas kitni baar zyada hot hai. Right par, $1$ "pehle se wahan" wala part hai, aur fraction woh extra heat hai jo motion rokne se produce hoti hai. Agar $V = 0$ toh fraction gayab ho jaata hai aur $T_{0} / T = 1$ — kuch rokna kuch nahi badalta.

PICTURE. See-saw ko thermometer scale ke saath redraw kiya: motion pot, jab heat pot mein khali hoti hai, mercury ko $T$ se $T_{0}$ tak utha deta hai.

Step 3 — Velocity ko Mach number se replace karo

KYA. Messy term $\frac{V ^{2}}{2 c _{p} T}$ mein abhi bhi $V$ , $c_{p}$ , $R$ chhupe hain. Hum ab ise do jaane-maane facts use karke single clean group $\frac{γ - 1}{2} M^{2}$ mein convert karte hain.

KYUN yeh do tools? Hum sab kuch $M$ ke terms mein chahte hain kyunki $M$ hi woh ek dial hai jo engineer set karta hai. Toh humein $V$ se $M$ tak ka bridge chahiye, aur $c_{p}$ ko erase karne ka tarika:

Speed of sound $a^{2} = γ R T$ (from Speed of Sound and Mach Number). Tab $V^{2} = (M a)^{2} = M^{2} a^{2} = M^{2} γ R T .$ Yahan $M = V / a$ toh $V = M a$ ; $γ$ (gamma) heat-capacity ratio $c_{p} / c_{v}$ hai, aur $R$ gas constant hai.
$c_{p}$ in terms of $γ$ aur $R$ : $c_{p} - c_{v} = R$ aur $γ = c_{p} / c_{v}$ se milta hai $c_{p} = \frac{γ R}{γ - 1} .$

Dono ko fraction mein substitute karo aur dekho $γ$ , $R$ , $T$ sab cancel ho jaate hain:

$\frac{V ^{2}}{2 c _{p} T} = \frac{M ^{2} γ R T}{2 \cdot \frac{γ R}{γ - 1} T} = \frac{γ - 1}{2} M^{2}$

Numerator $M^{2} γ R T$ $V^{2}$ se aaya.
Denominator ka $γ R T$ numerator ke $γ R T$ ko cancel karta hai; $2$ aur $\frac{1}{γ - 1}$ bach jaate hain, upar flip ho kar $\frac{γ - 1}{2}$ ban jaate hain.

PICTURE. Ek cancellation ladder: $γ$ , $R$ , $T$ mein se har ek top-and-bottom strike out ho jaata hai, tidy survivor $\frac{γ - 1}{2} M^{2}$ bachta hai.

Step 4 — Temperature se pressure aur density tak (isentropic bridge)

KYA — pehle missing character, $v$ , define karo. Bridge se pehle humein ek aur symbol chahiye. Specific volume $v$ simply "ek kilogram gas kitni jagah leta hai" hai. Yeh density ka exact opposite hai:

$v = \frac{1}{ρ} (cubic metres per kilogram) .$

Toh "bada $v$ " matlab gas spread thin hai (low $ρ$ ); "chhota $v$ " matlab tightly packed hai (high $ρ$ ). Ab hamare paas do facts hain jo dono $v$ involve karte hain:

Ideal-gas law $P v = R T$ — pressure times room-per-kilogram equals $R$ times temperature.
Isentropic (constant-entropy) law $P v^{γ} = const$ — ek perfect gas ko bina friction aur bina heat ke rokne ke liye, yeh combination kabhi nahi badalta (from Isentropic Process Relations for Perfect Gas).

KYUN yeh do, aur $v$ kaise eliminate hota hai. Hum $P$ , $ρ$ aur $T$ ke beech relations chahte hain sirf — $v$ scaffolding hai jise hatana hai. Yahan elimination hai, term by term:

Ideal-gas law se room-per-kilogram solve karo: $v = \frac{R T}{P}$ .
Use isentropic law mein dalo: $P (\frac{R T}{P})^{γ} = const$ .
$P$ ki powers collect karo: $P^{1 - γ} (R T)^{γ} = const$ , yani $P^{1 - γ} T^{γ} = const$ (constant $R^{γ}$ absorb ho jaata hai).
$P$ isolate karo power $\frac{1}{1 - γ}$ raise karke: yeh deta hai $P \propto T^{\frac{γ}{γ - 1}}$ . Exponent paida hota hai.
Density ke liye, $ρ = 1/ v$ aur $v = R T / P$ use karo, toh $ρ = P / (R T)$ . $P \propto T^{γ / (γ - 1)}$ substitute karne par milta hai $ρ \propto T^{γ / (γ - 1) - 1} = T^{\frac{1}{γ - 1}}$ .

Toh dono power laws derive hue hain, assert nahi kiye:

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

Kyunki stagnation aur local states usi constant-entropy line par hain, "const" dono jagah identical hai, toh divide karne par exactly yahi ratios milte hain.

PICTURE. Elimination visualized: $v$ dono laws se enter karta hai, middle mein cancel ho jaata hai, aur do clean power laws nikalakar aate hain.

$T_{0} / T = F$ substitute karo:

Term-by-term. Teeno ratios ek identical base $F$ share karte hain; woh sirf exponent mein alag hain:

ratio	exponent	air ke liye $γ = 1.4$
$T_{0} / T$	$1$	$1$
$ρ_{0} / ρ$	$\frac{1}{γ - 1}$	$2.5$
$P_{0} / P$	$\frac{γ}{γ - 1}$	$3.5$

PICTURE. Ek base factor $F$ ek "power fork" mein enter karta hai aur teen exponents mein split ho jaata hai; jitna bada exponent, utna uuncha resulting stack.

Step 5 — Pressure sabse fast kyun girta hai (ordering, visualized)

KYA. Kyunki $F > 1$ har $M > 0$ ke liye, ek bada exponent matlab bada ratio $X_{0} / X$ hai, isliye chhoti local value $X / X_{0}$ . Kyunki $3.5 > 2.5 > 1$ :

$\frac{P}{P _{0}} < \frac{ρ}{ρ _{0}} < \frac{T}{T _{0}} (for all M > 0)$

KYUN aisa hona zaroori hai, sirf "numbers aisa kehte hain" nahi. 1 se bade number ko zyada power par raise karna use aur inflate karta hai — $1. 8^{3.5} \approx 7.82$ lekin $1. 8^{1} = 1.8$ . Pressure sabse bada exponent carry karta hai, toh uska reciprocal sabse fast collapse karta hai.

PICTURE. $X / X_{0}$ versus $M$ ke teen curves: pressure (plum) sabse neeche girata hai, density (teal) beech mein, temperature (orange) sabse uupar rehta hai — vertical ordering kabhi cross nahi hoti.

Step 6 — Degenerate & edge cases (koi gap mat chhodna)

KYA & KYUN. Ek formula tabhi trustworthy hai jab extremes par sahi behave kare. Corners ko $F = 1 + \frac{γ - 1}{2} M^{2}$ mein plug karo.

$M = 0$ (koi flow nahi). $F = 1$ , toh $T / T_{0} = P / P_{0} = ρ / ρ_{0} = 1^{- anything} = 1$ . Matlab: rest par gas pehle se apna stagnation state hi hai — kuch rokna nahi, kuch nahi badalata. Sanity ✓.
$M = 1$ (sonic / choking). Air ke liye $F = 1 + 0.2 = 1.2$ , jisse critical ratios milte hain $T^{*} / T_{0} = 0.8333$ , $ρ^{*} / ρ_{0} = 0.6339$ , $P^{*} / P_{0} = 1. 2^{- 3.5} = 0.5283$ . Yeh woh numbers hain jo Converging-Diverging Nozzle & Choking mein choking decide karte hain: back-pressure ko $\approx 0.528 P_{0}$ tak girna padta hai pehle throat $M = 1$ tak pahunche.
$M \to \infty$ (hypersonic limit). $F \to \infty$ , toh har ratio $X / X_{0} \to 0$ . Matlab: "bank account" ka lagbhag sab kuch kinetic hai; local $T, P, ρ$ kuch nahi ke paas jaate hain. Physically gas pehle ionize ho jaati, lekin trend sahi hai.
Shock ke across — formula ka forbidden zone. Yeh pillars isentropic (constant-entropy) stopping assume karte hain. Ek normal shock (Normal Shock Relations) entropy jump karta hai, toh $P_{0}$ actually iske across drop karta hai. Tum inhe abhi bhi har side par alag se use kar sakte ho, lekin kabhi ek single $P_{0}$ se jo shock straddle kare.

PICTURE. Teen curves apni limits tak extended, sonic point marked aur ek hatched "shock — do not cross" band ke saath.

Ek picture summary

Derivation ka har arrow ek single map par: energy conservation $F$ ko janam deta hai; speed-of-sound aur $c_{p}$ substitutions ise clean karte hain; isentropic bridge ( $v$ eliminate karke) ise exponents $1, 2.5, 3.5$ ke saath teen pillars mein fork karta hai.

Recall Feynman retelling — plain words mein poora walkthrough

Socho ek parcel of air zoom kar rahi hai aur tum use dheere se dead stop par laate ho. Uski motion energy gayab nahi hoti — woh heat aur squeeze mein badal jaati hai, toh ruki hui air zyada hot, denser hoti hai aur zyada hard push karti hai. Yahi stagnation state $T_{0}, P_{0}, ρ_{0}$ hai. Step 1 ne energy see-saw likha: heat plus motion ek fixed total hai. Step 2 ne heat ko temperature se swap kiya. Step 3 ne awkward speed $V$ ko Mach number $M$ (sound ki comparison mein speed) se trade kiya, aur sab kuch ugly cancel ho gaya, ek neat lump $F = 1 + \frac{γ - 1}{2} M^{2}$ bachha — "stopping ne kitna badla" factor. Step 4 ne specific volume $v$ (room per kilogram) introduce kiya, ideal-gas aur constant-entropy laws use karke $v$ cancel kiya, aur power laws nikle: pressure aur density temperature ko fixed powers se follow karte hain — temperature ke liye $1$ , density ke liye $2.5$ , pressure ke liye $3.5$ . Step 5 ne pecking order explain kiya: sabse bada power (pressure) sabse fast girta hai. Step 6 ne corners check kiye — resting gas apna stagnation hi hai, sonic flow choking ke liye famous $0.528$ pressure ratio deta hai, aur tumhe yeh kabhi ek shock ke across nahi kheenchna chahiye, jahan smoothness assumption break ho jaati hai. Ek factor, teen exponents, aur poori isentropic table nikal aati hai.

Active-recall

M, γ

ke terms mein base factor

F

kya hai?

F = 1 + \frac{γ - 1}{2} M^{2}

, aur

F = T_{0} / T

Kaun se do substitutions

V^{2} / (2 c_{p} T)

\frac{γ - 1}{2} M^{2}

mein convert karte hain?

V^{2} = M^{2} γ R T

(

a^{2} = γ R T

se) aur

c_{p} = \frac{γ R}{γ - 1}

Specific volume

v

kya hai aur yeh density se kaise relate karta hai?

v

= ek kilogram ke liye occupied room;

v = 1/ ρ

v

eliminate karke

P \propto T^{γ / (γ - 1)}

kaise milta hai?

v = R T / P

P v^{γ} =

const mein dalo,

P

ki powers collect karo.

Hum

T_{0} / T

ko ek power par raise karke

P_{0} / P

kyun paa sakte hain?

Stopping isentropic hai, toh

P \propto T^{γ / (γ - 1)}

usi constant-entropy line par hai.

M = 0

par teeno ratios kya hain?

Sab

1

ke barabar — rest par gas apna stagnation state hai.

M

ke saath

P / P_{0}

sabse fast kyun girta hai?

Uska exponent

γ / (γ - 1) = 3.5

sabse bada hai, toh uska reciprocal sabse jaldi shrink karta hai.

Yeh formulas kahan break hote hain, aur kyun?

Ek shock ke across — entropy badhti hai toh

P_{0}

conserved nahi; har side alag use karo.

Connections

Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions of M (index 3.1.7) — woh parent jise yeh walkthrough build karta hai.
Stagnation Properties and Energy Equation — Step 1 ka energy budget.
Speed of Sound and Mach Number — Step 3 ka $a = γ R T$ aur $M$ ki definition.
Isentropic Process Relations for Perfect Gas — Step 4 ka power-law bridge.
Converging-Diverging Nozzle & Choking — Step 6 ke critical ratios use karta hai.
Normal Shock Relations — Step 6 ka forbidden zone.
Area-Mach Relation A/A* — wahi stagnation framework, extended.