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

3.1.7 · D5 · Physics › Compressible Flow & Aerodynamics › Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions

Shuru karne se pehle, teen reminders taaki har "kyunki" kisi concrete jagah point kar sake.

Neeche ki figure poori cheez ek hi nazar mein dikhati hai: dekho kaise teen curves $M$ badhne ke saath alag-alag ho jaati hain, aur $M = 1$ par vertical line note karo jahan starred critical values rehte hain.

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

Pressure ka "sabse tezi se girna" sirf paper par bade exponent ki wajah se nahi hai — yeh energetic hai. Figure usi base factor $b = 1 + \frac{γ - 1}{2} M^{2}$ ko powers $1$ , $2.5$ , $3.5$ par raise karte dikhata hai: exponent tower jitna uncha, ratio utna hi tezi se zero ki taraf khinchta hai.

True or false — justify karo

Zyada Mach number ka matlab hamesha bade stagnation-to-local ratios hota hai.

True,

M > 0

ke liye: har ratio monotonically increasing function hai factor

1 + \frac{γ - 1}{2} M^{2}

ki, jo

M

ke saath badhta hai. Tez flow "zyada pile up" hoti hai jab roke jaate hain.

T / T_{0}

supersonic flow mein 1 se zyada ho sakta hai.

False. Kyunki

T_{0} / T = 1 + \frac{γ - 1}{2} M^{2} \geq 1

, local

T

hamesha

\leq T_{0}

hota hai; chalti gas apni stagnation value se thandi hoti hai kyunki kinetic energy "kharach" ho gayi hai.

M = 0

par teenon ratios exactly 1 ke barabar hote hain.

True.

M = 0

ke saath base factor

1 + 0 = 1

hai, aur

1

kisi bhi power par

1

hi hota hai — ek ruka hua fluid apna khud ka stagnation state hai.

Same

M

ke liye, pressure density se zyada girta hai, aur density temperature se zyada girti hai.

True, aur iski wajah algebraic nahi balki energetic hai: same base factor ko bade exponent (

P

ke liye

3.5

ρ

ke liye

2.5

T

ke liye

1

) par raise karna same "piling up" ko zyada baar compound karta hai, isliye

P / P_{0} < ρ / ρ_{0} < T / T_{0}

— bilkul figure s01 mein curves ka fanning-out.

Tables ko normal shock ke across apply kiya ja sakta hai.

False. Shock entropy badhata hai, isliye

P_{0}

discontinuously girta hai; isentropic ratios constant entropy assume karte hain aur sirf shock-free points ke beech hold karte hain. Dekho Normal Shock Relations.

Stagnation temperature

T_{0}

shock ke through bhi constant rehta hai.

True.

T_{0}

sirf stagnation enthalpy par depend karta hai, jise (adiabatic) energy equation shock ke across conserve karti hai — sirf

P_{0}

aur

ρ_{0}

girte hain. Dekho Stagnation Properties and Energy Equation.

Fixed

M

ke liye,

γ

change karne se ratios unchanged rehte hain.

False. Factor

1 + \frac{γ - 1}{2} M^{2}

aur exponents dono

γ

par depend karte hain; ek monatomic gas (

γ = 5/3

) air (

γ = 1.4

) se alag table values dega.

Consistency identity

\frac{P _{0} / P}{ρ _{0} / ρ} = \frac{T _{0}}{T}

sirf

γ = 1.4

ka coincidence hai.

False. Yeh exactly ideal-gas law

P / ρ \propto T

hai aur kisi bhi

γ

ke liye hold karta hai, kyunki

\frac{γ}{γ - 1} - \frac{1}{γ - 1} = 1

M = 1

par pressure

P_{0}

ke lagbhag aadhe tak gir chuka hota hai.

Kafi sahi —

P^{*} / P_{0} = 1. 2^{- 3.5} \approx 0.528

, toh sonic conditions tak pahunchne ke liye pressure ko stagnation ka ~53% tak girna padta hai. Yeh choking threshold hai, dekho Converging-Diverging Nozzle & Choking.

Ratios sirf accelerating (nozzle) flow ke liye valid hain, decelerating (diffuser) flow ke liye nahi.

False. Yeh sirf local

M

aur

γ

par depend karte hain, is baat par nahi ki flow speed up ho rahi hai ya slow down, jab tak flow isentropic rehti hai.

Error dhundho

"Kyunki

T_{0} / T = 1 + \frac{γ - 1}{2} M^{2}

, toh

T / T_{0} = 1 + \frac{γ - 1}{2} M^{2}

bhi hai."

Error:

T / T_{0}

reciprocal hai,

(1 + \frac{γ - 1}{2} M^{2})^{- 1} \leq 1

. Yaadgaar factor

T_{0} / T

ke barabar hai, jo

\geq 1

hai; local

T

chota wala hona chahiye.

"Density ratio mein bhi wahi exponent 3.5 use hota hai jaise pressure mein, kyunki dono same base factor se aate hain."

Error: density ka exponent

\frac{1}{γ - 1} = 2.5

hai,

3.5

nahi. Sirf pressure mein

\frac{γ}{γ - 1}

exponent hota hai; exponents ko mix karna sabse common numerical galti hai.

"Flow shock mein decelerate karti hai, toh uska stagnation pressure badhta hai."

Error: shock ke through deceleration irreversible hai. Entropy badhti hai, isliye

P_{0}

girta hai; ek reversible stop

P_{0}

preserve karta, lekin shock reversible nahi hai.

M = 2

par,

T / T_{0} = 1/1. 8^{3.5}

Error: temperature ka exponent

1

hai, isliye

T / T_{0} = 1/1.8 = 0.556

3.5

exponent pressure ka hai, temperature ka nahi.

"Kyunki

P_{0}

'total pressure' hai, yeh har case mein sum

P + \frac{1}{2} ρ V^{2}

hai."

Error: woh Bernoulli-style sum low-speed (incompressible) approximation hai. Compressible flow mein

P_{0} = P (1 + \frac{γ - 1}{2} M^{2})^{γ / (γ - 1)}

hai, jo sirf

M \to 0

par sum mein reduce hota hai.

P_{0}

nikalne ke liye main local

P

use karta hoon lekin mujhe stagnation state ke liye alag sound speed measure karni padegi."

Error:

P_{0} / P

ke liye sirf local

M

aur

γ

chahiye. Stagnation state ek hypothetical reference hai; tum yahan uski sound speed independently kabhi measure nahi karte.

"Tables absolute temperature deti hain, toh main

T

Celsius mein plug in kar sakta hoon."

Error: ratios

a^{2} = γ R T

aur

h = c_{p} T

se aate hain, jo absolute (Kelvin) temperature require karte hain. Celsius proportionality tod deta hai. Dekho Speed of Sound and Mach Number.

Why questions

Har property ratio sirf

M

(aur

γ

) ka function kyun ban jaata hai?

Kyunki energy equation velocity ko

M

mein convert karti hai

a^{2} = γ R T

ke zariye, aur isentropic relations

P, ρ

T

se tie karte hain; sound ke relative flow ki speed hi ek matra free dial bachi rehti hai.

Pressure ratio

M

badhne ke saath sabse tezi se kyun girta hai?

Base factor

b > 1

ko teen exponents par raise karne ki imagine karo (figure s02):

b^{3.5}

b^{2.5}

aur

b

se aage nikal jaata hai, toh uska reciprocal

P / P_{0}

sabse tezi se zero ki taraf khinchta hai. Physically, pressure ko dono temperature drop aur density drop compound feel hoti hai, isliye woh sabse zyada shed karta hai.

Nozzle ka back-pressure lagbhag

0.528 P_{0}

tak kyun girna chahiye tabhi throat

M = 1

tak pahunchti hai?

Kyunki

P^{*} / P_{0} = 1. 2^{- 3.5} \approx 0.528

exactly woh local-to-stagnation pressure hai jo sonic flow ke liye chahiye; uske neeche throat choke ho jaata hai aur mass flow max ho jaata hai.

T_{0}

adiabatic flow mein friction present hone par bhi conserve hota hai, lekin

P_{0}

nahi — kyun?

T_{0}

stagnation enthalpy track karta hai, jise adiabatic energy balance conserve karta hai;

P_{0}

entropy bhi track karta hai, aur friction entropy badhati hai, isliye

P_{0}

kho jaata hai.

Flow ko stagnation properties define karne ke liye isentropically rest par kyun laate hain, kisi bhi tarike se nahi?

Ek isentropic (adiabatic + reversible) stop ek unique, path-independent reference state deta hai; ek real messy stop losses par depend karta aur cleanly tabulate nahi hota. Dekho Isentropic Process Relations for Perfect Gas.

Consistency check

\frac{P _{0} / P}{ρ _{0} / ρ} = \frac{T _{0}}{T}

Feynman sanity test ki tarah kyun kaam karta hai?

Yeh do independent ratios se ideal-gas law

P / ρ \propto T

reproduce karta hai, toh agar yeh fail ho toh pata chalta hai koi exponent galat enter hua hai.

Same ratios ek converging nozzle aur diffuser dono describe kar sakte hain — kyun?

Yeh sirf local aur stopped states ke beech isentropic relationship encode karte hain; acceleration ki direction enter nahi hoti, sirf local

M

enter hota hai.

Edge cases

Ratios

M \to 0

par kya dete hain?

Teenon 1 ki taraf jaate hain: vanishingly slow flow apne stagnation state se alag nahi ki ja sakti, toh kuch bhi "pile up" nahi hota.

M \to \infty

(hypersonic limit) par

T / T_{0}

ρ / ρ_{0}

, aur

P / P_{0}

ka kya hota hai?

Teenon 0 ki taraf jaate hain — base factor blow up ho jaata hai, isliye local static values stagnation ke negligible fractions ban jaati hain. Lekin woh alag-alag rates par vanish hoti hain:

P / P_{0}

sabse tezi se (

b^{- 3.5}

), phir

ρ / ρ_{0}

(

b^{- 2.5}

), phir

T / T_{0}

(

b^{- 1}

), isliye figure s01 mein curves puri tarah baahar tak ordered rehti hain.

Kya

M = 1

isentropic formulas mein khud mathematically special hai?

Mathematically nahi — formulas

M = 1

ke through smooth hain. Yeh physically special sirf isliye hai kyunki

M = 1

choking/sonic condition hai, jo starred critical ratios deta hai.

Air ke liye

M = 1

par exactly critical ratios kya hain?

T^{*} / T_{0} = 1/1.2 = 0.833

ρ^{*} / ρ_{0} = 1. 2^{- 2.5} = 0.634

P^{*} / P_{0} = 1. 2^{- 3.5} = 0.528

— reference values (star = sonic) jo choking mark karte hain.

Agar ek flow ke do points ka same

M

ho lekin woh ek shock ke opposite sides par hoon, toh kya unka local

P

equal hai?

Nahi. Same

M

same ratio

P / P_{0}

deta hai, lekin

P_{0}

shock ke across alag hota hai, isliye absolute

P

alag hoga. Ek shock ke across kabhi ek

P_{0}

share mat karo.

Kya

M

in formulas mein negative ho sakta hai, aur kya fark padega?

M

sirf

M^{2}

ke form mein aata hai, isliye sign reversal (flow direction) har ratio ko unchanged chhod deta hai; tables sirf sound ke relative speed magnitude ki parwah karti hain, direction ki nahi.

Stagnation point par hi (

V = 0

M = 0

), tables kaunsa state report karti hain?

Trivial wala — local equals stagnation, saare ratios 1 — kyunki convert hone ke liye koi kinetic energy bachi hi nahi.

Connections

Speed of Sound and Mach Number — kyun absolute $T$ aur $a = γ R T$ har ratio ke neeche hain.
Stagnation Properties and Energy Equation — kyun $T_{0}$ friction survive karta hai lekin $P_{0}$ nahi.
Normal Shock Relations — woh boundary jahan yeh tables toot jaati hain.
Converging-Diverging Nozzle & Choking — jahan $0.528$ critical ratio bite karta hai.
Area-Mach Relation A/A* — same stagnation framework, geometry add karke.
Isentropic Process Relations for Perfect Gas — $γ / (γ - 1)$ aur $1/ (γ - 1)$ exponents ka source.