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

3.1.7 · D3 · Physics › Compressible Flow & Aerodynamics › Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions

Neeche sab kuch air ke liye hai, $γ = 1.4$ , isliye baar baar aane wale numbers hain:

Yahan $\frac{γ - 1}{2} = \frac{0.4}{2} = 0.2$ , isliye base factor hamesha $1 + 0.2 M^{2}$ hota hai.

Definition Is page pe use hone wale symbols aur conventions (pehle padho)

Subscript $0$ (jaise $T_{0}, P_{0}, ρ_{0}$ ) hamesha stagnation (total) value ko batata hai — woh property jo flow tab reach karti agar use isentropically rest pe laya jaata. Plain symbols $T, P, ρ$ woh local (static) values hain jo chalti hui gas actually mein hoti hain. To $T_{0} / T$ , $P_{0} / P$ , $ρ_{0} / ρ$ hamesha "stopped-over-flowing" hain aur $\geq 1$ hain.
$M = \frac{V}{a}$ Mach number hai: flow speed $V$ divided by local speed of sound $a$ . Perfect gas ke liye $a = γ R T$ (dekho Speed of Sound and Mach Number); jab problem velocity deta hai $M$ ki jagah tab $a$ local temperature se banate hain.
$R$ fluid ka specific gas constant hai (energy per unit mass per kelvin). Air ke liye $R = 287 J\cdotkg^{- 1} K^{- 1}$ ; ye $a = γ R T$ ke andar aata hai aur ideal-gas law $P = ρR T$ mein bhi.
Star superscript ( $T^{*}, P^{*}, ρ^{*}$ ) ==critical values at $M = 1$ == ko denote karta hai — woh "sonic" ya "choked" state jo ek nozzle apne throat par reach karta hai (dekho Converging-Diverging Nozzle & Choking).
$s$ entropy hai, irreversibility ka thermodynamic bookkeeper. Isentropic matlab " $s$ = constant"; hamare teen formulas exactly yahi assume karke derive hue the. Jab $s$ badhta hai (jaise ek shock ke paas) to ye formulas un do states ko connect nahi kar sakti.
Rounding convention: hum arithmetic mein full precision carry karte hain aur final answers 4 significant figures tak report karte hain (intermediate values bhi 4 s.f. tak dikhaye gaye). Kisi answer aur uske verification ke beech last-digit mein chhota sa difference sirf rounding ka kamal hai.

Scenario matrix

Is topic ka har sawaal inhi cells mein se kisi ek mein aata hai. Neeche ke examples cell ke label ke saath aate hain.

Cell	Kya cheez ise alag banati hai	Example
A. $M = 0$ (degenerate)	flow at rest: sabhi ratios $= 1$ , sanity anchor	Ex 1
B. Subsonic $0 < M < 1$	ratios slightly below 1, "gentle" regime	Ex 2
C. Sonic $M = 1$ (critical)	choke point, starred quantities $T^{}, P^{}, ρ^{*}$	Ex 3
D. Supersonic $M > 1$	badi drops, teeno ratios chhoti	Ex 4
E. Inverse: ratio diya → $M$ nikalo	formula invert karo (root, power nahi)	Ex 5
F. Limiting $M \to \infty$	asymptotic behaviour, ratios → 0	Ex 6
G. Real-world word problem	Pitot / air-intake, pehle $M$ banana padega	Ex 7
H. Exam twist — trap	ratios ek shock ke paas apply karna (illegal)	Ex 8
I. Ratios ke beech cross-check	$\frac{P _{0} / P}{ρ _{0} / ρ} = \frac{T _{0}}{T}$ use karo errors pakadne ke liye	Ex 9

Cell A — degenerate case $M = 0$

Worked example Example 1 — Reservoir mein still air

Ek bade tank mein air rest par hai: $T = 300$ K, $P = 101.3$ kPa. Local values ke relative $T_{0}, P_{0}, ρ_{0}$ kya hain?

Forecast: gas chal nahi rahi, to "use rokne" se kuch nahi badlega — har ratio exactly 1 hona chahiye.

Base factor $= 1 + 0.2 \cdot 0^{2} = 1$ . Ye step kyun? Kyunki $V = 0$ hai, Mach number $M = V / a = 0$ hai (jahan $a = γ R T$ speed of sound hai, upar define ki gayi); pehle isse plug in karo.
$\frac{T _{0}}{T} = 1 \Rightarrow T_{0} = T = 300$ K. Kyun? 1 ka reciprocal 1 hota hai — convert karne ke liye koi kinetic energy nahi.
$\frac{P _{0}}{P} = 1^{3.5} = 1$ aur $\frac{ρ _{0}}{ρ} = 1^{2.5} = 1$ . Kyun? 1 ki koi bhi power 1 hi hoti hai.

Verify: rest par fluid ke liye stagnation = local, jo ek reservoir ki definition hai — uski properties hi stagnation properties HAIN. Ye woh anchor hai jis se har table row shuru hoti hai. ✓

Cell B — subsonic flow

Worked example Example 2 — Wing ke upar air

$M = 0.5$ par Free-stream: $T = 288$ K, $P = 100$ kPa, $M = 0.5$ . Stagnation (total) values nikalo.

Forecast: subsonic aur moderate hai, isliye ratios 1 ke karib honge — shayad 5–15% shifts.

Base factor $= 1 + 0.2 (0.5)^{2} = 1 + 0.2 (0.25) = 1.05$ . Ye step kyun? Sab kuch is ek number par depend karta hai.
$T_{0} = T \cdot 1.05 = 288 \times 1.05 = 302.4$ K. Kyun? $T_{0} / T$ base factor ke barabar hota hai (exponent 1), isliye multiply karo.
$P_{0} = P \cdot 1.0 5^{3.5} = 100 \times 1.1867 = 118.7$ kPa. Kyun? Pressure exponent 3.5 carry karta hai.
$ρ_{0} / ρ = 1.0 5^{2.5} = 1.1290$ , isliye $ρ_{0} = 1.129 ρ$ . Kyun? Density exponent 2.5 hai.

Verify: teeno stagnation values local values se zyada hain (flow rokne se sab kuch raise hota hai), aur pressure sabse zyada badha (18.7%) jabki temperature sabse kam badha (5%) — parent ke promise ki hui ordering. ✓

Cell C — sonic / critical case $M = 1$

Worked example Example 3 — Nozzle throat (choked)

Ek converging nozzle ko ek reservoir se feed kiya ja raha hai $T_{0} = 500$ K, $P_{0} = 800$ kPa par. Throat par flow $M = 1$ reach karta hai. $T^{*}, P^{*}, ρ^{*}$ nikalo (star superscript critical, $M = 1$ , values denote karta hai).

Forecast: $M = 1$ par pressure ratio famous $\approx 0.528$ hoti hai, temperature $\approx 0.833$ .

Base factor $= 1 + 0.2 (1)^{2} = 1.2$ . Ye step kyun? $M = 1$ critical/star condition deta hai jo Converging-Diverging Nozzle & Choking mein use hoti hai.
$T^{*} = T_{0} /1.2 = 500/1.2 = 416.7$ K. Kyun? $T^{*} / T_{0} = 1/1.2 = 0.8333$ ; stagnation value ko neeche divide karo.
$P^{*} = P_{0} \cdot 1. 2^{- 3.5} = 800 \times 0.5283 = 422.6$ kPa. Kyun? $P^{*} / P_{0} = 0.5283$ choke threshold hai — back pressure itna girna chahiye $M = 1$ reach karne ke liye.
$ρ^{*} = ρ_{0} \cdot 1. 2^{- 2.5} = 0.6339 ρ_{0}$ . Kyun? Density exponent 2.5.

Verify: $P^{*} / P_{0} = 0.5283 < ρ^{*} / ρ_{0} = 0.6339 < T^{*} / T_{0} = 0.8333$ , strict ordering hold karta hai, aur $0.5283 \approx 0.528$ universal air choke ratio se match karta hai. ✓

Cell D — supersonic flow

Figure 1 neeche poore $M$ axis par teeno local-over-stagnation ratios trace karta hai aur chaar worked cells (A, B, C, D) unke curves par pin karta hai. Har curve directly label ki gayi hai aur legend mein bhi, taaki grayscale mein bhi theek se padhe: pressure curve ( $P / P_{0}$ , exponent 3.5) sabse neeche hai kyunki ye sabse tezi se girta hai, density curve ( $ρ / ρ_{0}$ , exponent 2.5) beech mein hai, aur temperature curve ( $T / T_{0}$ , exponent 1) sabse upar rehti hai — $M > 0$ ke liye har jagah.

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

Figure 1 — Mach number ke against local/stagnation ratios, worked cells A–D mark kiye hue. Curves label aur colour-coded hain (lavender = $T / T_{0}$ , mint = $ρ / ρ_{0}$ , coral = $P / P_{0}$ ).

Cell E — inverse problem (ratio diya, $M$ nikalo)

Worked example Example 5 — Pitot

$P / P_{0} = 0.3$ read karta hai. $M$ kya hai? Forecast: $0.3$ sonic value $0.528$ se neeche hai, isliye flow supersonic hoga ( $M > 1$ ).

Pressure formula likho aur flip karo: $(1 + 0.2 M^{2})^{3.5} = \frac{P _{0}}{P} = \frac{1}{0.3} = 3.333$ . Ye step kyun? Hum jaante hain left side base factor hai 3.5 power tak; right side diya hua ratio inverted hai.
Power ko 3.5-waa root lekar undo karo (raise to $1/3.5$ ): $1 + 0.2 M^{2} = 3.33 3^{1/3.5} = 1.4053$ . Ye step kyun? Ye isentropic version hai "arctan undoes tan" ka — root power ko undo karta hai, base factor recover hota hai.
$M^{2}$ isolate karo: $0.2 M^{2} = 0.4053 \Rightarrow M^{2} = 2.026$ . Kyun? 1 subtract karo, $0.2$ se divide karo.
$M = 2.026 = 1.424$ . Kyun? Mach number positive hota hai, isliye positive root lo.

Verify: $M = 1.424 > 1$ forecast se match karta hai (supersonic, kyunki ratio $0.528$ se neeche gaya), aur $M = 1.424$ forward feed karne par $P / P_{0} = 1.405 3^{- 3.5} = 0.300$ milta hai. ✓

Cell F — limiting behaviour $M \to \infty$

Worked example Example 6 — Ratios

$M$ badhne par kya karte hain? Forecast: bahut badi speed par, flow rokne se enormous kinetic energy release hoti hai, isliye local $T, P, ρ$ sab stagnation ke relative zero ki taraf shrink honge.

Large $M$ ke liye base factor $\approx 0.2 M^{2}$ ("+1" negligible ho jaata hai). Ye step kyun? Jab $0.2 M^{2} ≫ 1$ tab constant term dab jaata hai.
$\frac{T}{T _{0}} = \frac{1}{1 + 0.2 M ^{2}} \to 0$ like $\frac{5}{M ^{2}}$ . Kyun? Denominator $M^{2}$ ki tarah badhta hai, isliye ratio $1/ M^{2}$ ki tarah decay karta hai.
$\frac{P}{P _{0}} = (1 + 0.2 M^{2})^{- 3.5} \to 0$ like $M^{- 7}$ ; $\frac{ρ}{ρ _{0}} \to 0$ like $M^{- 5}$ . Kyun? Badhte base ko negative power par raise karna use zero ki taraf le jaata hai — exponent jitna bada, utna tezi se.

Numeric check $M = 10$ par: base $= 1 + 0.2 (100) = 21$ ; $T / T_{0} = 1/21 = 0.04762$ , $P / P_{0} = 2 1^{- 3.5} = 2.472 \times 1 0^{- 5}$ , $ρ / ρ_{0} = 2 1^{- 2.5} = 5.191 \times 1 0^{- 4}$ .

Verify: teeno 0 ki taraf collapse karte hain, pressure sabse tezi se — ek hypersonic vehicle ki surface pressure stagnation pressure ka ek tiny fraction hoti hai, isliye hypersonic stagnation-point heating (from $T_{0}$ ) design dominate karta hai. ✓ Dekho Speed of Sound and Mach Number ki $M$ khud kaise scale karta hai.

Cell G — real-world word problem

Worked example Example 7 — Aircraft par Pitot tube

Ek jet wahan cruise karta hai jahan ambient (static) air $T = 223$ K, $P = 26.5$ kPa hai, aur wo $V = 250$ m/s par fly karta hai. Nose Pitot port air ko isentropically rok leta hai. Wo kya total pressure aur total temperature read karta hai? ( $R = 287$ J/kg·K.)

Forecast: subsonic-ish airliner speed, isliye $M < 1$ ; expect karo total pressure shayad static se 40–50% upar.

Speed of sound: $a = γ R T = 1.4 \times 287 \times 223 = 299.3$ m/s. Ye step kyun? Ratios ko $M$ chahiye, aur $M = V / a$ — pehle local temperature se $a$ banana padega (dekho Speed of Sound and Mach Number).
Mach number: $M = V / a = 250/299.3 = 0.8353$ . Kyun? Ye ek number har ratio feed karta hai.
Base factor $= 1 + 0.2 (0.8353)^{2} = 1 + 0.2 (0.6977) = 1.1395$ . Kyun? Standard step.
$T_{0} = T \cdot 1.1395 = 223 \times 1.1395 = 254.1$ K. Kyun? Temperature exponent 1.
$P_{0} = P \cdot 1.139 5^{3.5} = 26.5 \times 1.5793 = 41.85$ kPa. Kyun? Pressure exponent 3.5; ye wahi hai jo Pitot gauge read karta hai.

Verify: $M = 0.8353 < 1$ subsonic confirm karta hai (forecast ✓), aur $P_{0}$ static se 58% upar hai — high-subsonic flight ke liye plausible. Units: $(J/kg\cdotK)(K) = J/kg = m^{2} / s^{2} =$ m/s ✓. Yahan stagnation framework exactly Stagnation Properties and Energy Equation hai.

Cell H — exam trap (ratios across a shock)

Worked example Example 8 — Tables wahan apply karna jahan illegal hai

$M_{1} = 2$ par flow ka stagnation pressure $P_{0, 1} = 500$ kPa hai. Ye ek normal shock se guzarta hai. Ek student downstream static pressure compute karta hai same $P_{0}$ ke against $M_{2}$ isentropic ratio lookup karke. Ye galat kyun hai, aur fix kya hai?

Forecast: shock ke paas entropy $s$ jump karta hai, isliye $P_{0}$ conserve nahi ho sakta — naive lookup $P_{0, 2}$ overestimate karta hai.

Pehchano ki shock isentropic nahi hai: entropy $s$ (upar define hua irreversibility bookkeeper) badhta hai. Ye step kyun? Hamare teen formulas ek isentropic path ( $s$ constant) assume karke local se stagnation tak derive hue the — derivation tab hi toot jaati hai jab entropy badhti hai.
Consequence: $P_{0, 2} < P_{0, 1}$ . $M_{1} = 2$ ke liye (Normal Shock Relations se) stagnation-pressure ratio $P_{0, 2} / P_{0, 1} = 0.7209$ hai. Kyun? Shock available pressure ko entropy ke roop mein "waste" karta hai; sirf $T_{0}$ bachta hai (energy abhi bhi conserved hai).
Sahi procedure: downstream stagnation pressure use karo. $P_{0, 2} = 0.7209 \times 500 = 360.4$ kPa, aur tabhi downstream Mach number ke saath isentropic ratio apply karo ( $M_{2} = 0.5774$ for $M_{1} = 2$ ). Kyun? Shock ke har taraf ka apna alag isentropic bank account hota hai.
Check ke roop mein, $T_{0}$ unchanged rehta hai: $T_{0, 2} = T_{0, 1}$ . Kyun? Stagnation enthalpy $h_{0}$ ek shock ke through conserved hoti hai (adiabatic), isliye $T_{0}$ bina change ke cross karta hai.

Verify: $P_{0, 2} = 360.4$ kPa $< 500$ kPa (loss ✓), jabki $T_{0}$ preserve hota hai — exactly woh split jo parent ka teesra "mistake" callout warn karta hai. Kisi bhi $P_{0}$ ko shock ke aage-peeche reuse mat karo.

Cell I — ratios ke beech cross-check

Worked example Example 9 — Apni arithmetic error khud pakdo

$M = 2$ ke liye tumne compute kiya $P / P_{0} = 0.1278$ aur $ρ / ρ_{0} = 0.2301$ . Ideal-gas cross-check use karke confirm karo ki ye $T / T_{0} = 0.5556$ ke saath mutually consistent hain.

Forecast: pressure ratio ko density ratio se divide karne par temperature ratio milna chahiye (kyunki $P = ρR T$ ).

Compute karo $\frac{P / P _{0}}{ρ / ρ _{0}} = \frac{0.1278}{0.2301} = 0.5554$ . Ye step kyun? Ideal-gas law $P = ρR T$ se, stagnation-normalised pressure ko stagnation-normalised density se divide karne par $\frac{T}{T _{0}}$ milta hai — ek internal consistency law jo kisi bhi correct triple ko satisfy karna hi hoga (specific gas constant $R$ cancel ho jaata hai).
Direct temperature ratio $T / T_{0} = 1/1.8 = 0.5556$ se compare karo. Ye step kyun? Agar ye do numbers rounding se zyada differ karein, to tumhare teen ratios mein se kisi mein mis-typed exponent hai — ye sabse fast self-audit hai, parent ka "Feynman test".

Verify: $0.5554 \approx 0.5556$ (3 decimals tak agree; tiny gap sirf tabulated $0.1278$ aur $0.2301$ ki rounding hai). Base factor $1.8$ se exactly recompute karne par $1. 8^{- 3.5} /1. 8^{- 2.5} = 1. 8^{- 1} = 0.5556$ milta hai, $T / T_{0}$ se perfectly match. ✓

Recall Which cell is which? (quick self-test)

$M = 1$ throat conditions kaun si cell mein hain ::: C (sonic / critical, starred $T^{*}, P^{*}, ρ^{*}$ ) "Diya $P / P_{0}$ , nikalo $M$ " kaun si cell hai ::: E (inverse — root lo, power nahi) Normal shock se pehle aur baad ek hi $P_{0}$ reuse karna kaun si cell hai ::: H (illegal, entropy $s$ badhta hai) $M \to \infty$ par teeno local/stagnation ratios kahan jaate hain ::: 0 (cell F)

Connections

Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions of M (index 3.1.7) — parent formulas jo har example use karta hai.
Speed of Sound and Mach Number — Ex 6–7 mein $V$ se $a = γ R T$ ke zariye $M$ build karne ke liye zaroori.
Stagnation Properties and Energy Equation — kyun $T_{0}$ survive karta hai (Ex 7, 8).
Normal Shock Relations — Ex 8 mein $P_{0, 2} / P_{0, 1}$ aur $M_{2}$ supply karta hai.
Converging-Diverging Nozzle & Choking — Ex 3 ke critical/star values.
Area-Mach Relation A/A* — same stagnation framework, agla step.
Isentropic Process Relations for Perfect Gas — $P \propto T^{γ / (γ - 1)}$ jo har jagah use hota hai.