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

3.1.7 · D4 · Physics › Compressible Flow & Aerodynamics › Isentropic flow tables — P - P₀, T - T₀, ρ - ρ₀ as functions

Vocabulary — har symbol use hone se pehle define kiya gaya

Exponents 1, 2.5, 3.5 yahan derive karna (koi external note zaroorat nahi)

Formula Step-by-step: kyun

$T$ ko exponent 1 milta hai, $ρ$ ko 2.5, $P$ ko 3.5 (a) Temperature — exponent 1. Energy equation hi temperature ka statement hai, toh directly: $\frac{T _{0}}{T} = B = B^{1} .$ (b) Density aur pressure — do laws combine karke. Ek isentropic (reversible, adiabatic) perfect gas $P v^{γ} = const$ follow karta hai, jahan $v = 1/ ρ$ ek kilogram ka volume hai. Yeh ideal-gas law $P v = R T$ bhi follow karta hai. Hum $v$ eliminate karte hain:

Ideal gas se: $v = \frac{R T}{P}$ . Isko $P v^{γ} = const$ mein daalte hain: $P (\frac{R T}{P})^{γ} = const \Rightarrow P^{1 - γ} T^{γ} = const \Rightarrow P \propto T^{\frac{γ}{γ - 1}} .$ Moving state ko stopped state se compare karte hain (same constant), $\frac{P _{0}}{P} = (\frac{T _{0}}{T})^{γ / (γ - 1)} = B^{γ / (γ - 1)}$ . Air ke liye $\frac{γ}{γ - 1} = \frac{1.4}{0.4} = 3.5$ .
Kyunki $ρ = \frac{P}{R T} \propto \frac{T ^{γ / (γ - 1)}}{T} = T^{\frac{1}{γ - 1}}$ , hume milta hai $\frac{ρ _{0}}{ρ} = B^{1/ (γ - 1)}$ . Air ke liye $\frac{1}{γ - 1} = \frac{1}{0.4} = 2.5$ .

Pressure ke numerator mein woh ek extra $γ$ exactly wahi reason hai ki pressure ka exponent sabse bada hota hai aur isliye sabse fast girti hai. (Yeh bhi Isentropic Process Relations for Perfect Gas mein hai.)

Figure — teen curves padhna

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

Intuition Is figure ko seedha problems solve karne ke liye kaise padhen

Horizontal axis Mach number $M$ hai; vertical axis teen local/stagnation ratios mein se koi ek hai. Teeno curves $M = 0$ pe 1 se start karti hain (koi motion nahi ⇒ kuch kharcha nahi ⇒ local $=$ stagnation) aur monotonically girती hain — koi wapas upar nahi jaati, kyunki $B = 1 + 0.2 M^{2}$ sirf $M$ ke saath badhta hai, aur har ratio $B$ ki ek decreasing power hai. Table ki tarah use karne ke liye: apna $M$ bottom axis pe pick karo, seedha upar jao, aur dekho kahaan har curve milti hai. Red curve ( $P / P_{0}$ ) hamesha sabse neeche rehti hai (exponent 3.5, sabse fast girti hai); dashed density curve beech mein hai (2.5); solid black temperature curve sabse upar hai (exponent 1, sabse dhire girti hai) — yeh ordering tum L1.1 mein prove karoge. $M = 1$ pe vertical dotted line sonic / critical point mark karti hai jahan red curve $P / P_{0} = 0.528$ padhti hai (woh value L3.2 mein choking govern karti hai). $M$ bade hone par teeno curves 0 ki taraf flatten hoti jaati hain. Neeche ka har exercise literally in curves pe ek point read karna, ya invert karna, hai.

Level 1 — Recognition

L1.1

Batao ki $M > 0$ ke liye har ratio $\geq 1$ hai ya $\leq 1$ , aur teen local-over-stagnation ratios ko smallest se largest order mein arrange karo. Yeh bhi batao ki $M \to 0$ aur $M \to \infty$ mein inki limits kya hain.

Recall Solution

Har local property wahi bachi hai jo "bank account" mein motion pe kharchne ke baad rahi, toh local $<$ stagnation: $\frac{T}{T _{0}}, \frac{P}{P _{0}}, \frac{ρ}{ρ _{0}}$ sab $\leq 1$ hain. Inke reciprocals $B^{1}, B^{2.5}, B^{3.5}$ ke barabar hain jahan $B > 1$ ; bada exponent matlab bada reciprocal, isliye chota ratio. Toh: $\frac{P}{P _{0}} < \frac{ρ}{ρ _{0}} < \frac{T}{T _{0}} .$ Pressure ka exponent sabse bada hai (3.5) toh yeh sabse fast girti hai. Limits: $M \to 0$ par, $B \to 1$ aur teeno ratios $\to 1$ (kuch kharcha nahi). $M \to \infty$ par, $B \to \infty$ aur teeno $\to 0$ — figure mein curves axis par flatten hoti dikhti hain.

L1.2

Air ke liye $M = 0.5$ par, base factor $B$ compute karo aur phir $T / T_{0}$ .

Recall Solution

$B = 1 + 0.2 (0.5)^{2} = 1 + 0.2 (0.25) = 1 + 0.05 = 1.05$ . $\frac{T}{T _{0}} = \frac{1}{B} = \frac{1}{1.05} = 0.9524$ . Parent table ke row $M = 0.5$ se match karta hai. ✓

Level 2 — Application

L2.1

Air ek duct mein $T_{0} = 320$ K, $P_{0} = 600$ kPa ke saath enter karti hai. Jahan $M = 0.8$ hai wahan $T$ , $P$ , aur $ρ / ρ_{0}$ nikalo.

Recall Solution

$B = 1 + 0.2 (0.8)^{2} = 1 + 0.2 (0.64) = 1.128$ .

$T = T_{0} / B = 320/1.128 = 283.7$ K.
$P = P_{0} / B^{3.5} = 600/ (1.128)^{3.5} = 600/1.5243 = 393.6$ kPa.
$ρ / ρ_{0} = B^{- 2.5} = (1.128)^{- 2.5} = 0.7400$ . Note karo $B^{- 2.5} = 1/ B^{2.5}$ — negative exponent sirf inversion hai, kuch nahi.

L2.2

Ek wind tunnel $M = 1.5$ par air supply karta hai. Agar local static temperature $T = 200$ K hai, toh stagnation temperature $T_{0}$ kya hai?

Recall Solution

$B = 1 + 0.2 (1.5)^{2} = 1 + 0.2 (2.25) = 1.45$ . $T_{0} = B \cdot T = 1.45 \times 200 = 290$ K. (Yahan hum multiply karte hain, kyunki $T_{0} / T = B$ toh $T_{0} = B T$ .)

L2.3

$M = 2.0$ air ke liye, Feynman consistency check numerically verify karo: dikhao ki $\frac{P _{0} / P}{ρ _{0} / ρ} = \frac{T _{0}}{T}$ .

Recall Solution

$B = 1 + 0.2 (4) = 1.8$ . $P_{0} / P = 1. 8^{3.5} = 7.824$ , $ρ_{0} / ρ = 1. 8^{2.5} = 4.347$ , $T_{0} / T = 1.8$ . $\frac{7.824}{4.347} = 1.800 = B$ . ✓ Yeh sirf ideal-gas law $P / ρ \propto T$ ka wapas aana hai — aur yeh exactly isliye kaam karta hai kyunki pressure exponent $\frac{γ}{γ - 1}$ minus density exponent $\frac{1}{γ - 1}$ equals $\frac{γ - 1}{γ - 1} = 1$ , yeh temperature exponent hai.

Level 3 — Analysis

L3.1 (inverse problem)

Ek nozzle station par measured static pressure reservoir pressure ka $40%$ hai: $P / P_{0} = 0.40$ . Mach number nikalo (air).

Recall Solution

$\frac{P _{0}}{P} = \frac{1}{0.40} = 2.5 = B^{3.5}$ . Exponent undo karo $1/3.5$ se raise karke: $B = 2. 5^{1/3.5} = 2. 5^{0.2857} = 1.30931$ . Speed recover karo kinetic-energy relation $0.2 M^{2} = B - 1$ use karke (yaad karo: $B - 1$ Mach units mein spent kinetic energy hai): $0.2 M^{2} = 0.30931 \Rightarrow M^{2} = 1.54657 \Rightarrow M = 1.2436$ .

L3.2 (choking check)

Ek converging nozzle ko feed karne wala reservoir $P_{0} = 250$ kPa par hai. Back pressure $150$ kPa hai. Kya nozzle choked hai?

Recall Solution

Air ke liye critical ratio: $P^{*} / P_{0} = B^{- 3.5}$ at $M = 1$ , jahan $B = 1.2$ , deta hai $P^{*} / P_{0} = 1. 2^{- 3.5} = 0.5283$ . Critical exit static pressure $= 0.5283 \times 250 = 132.1$ kPa. Back pressure $150$ kPa $> 132.1$ kPa, toh exit pressure ko $P^{*}$ tak girne ki zaroorat nahi; flow sirf $M < 1$ tak pahunchti hai. Choked nahi hai. (Dekho Converging-Diverging Nozzle & Choking.)

L3.3 (density se do measurements)

Ek point par $ρ / ρ_{0} = 0.50$ hai. $M$ nikalo aur phir corresponding $T / T_{0}$ (air).

Recall Solution

$\frac{ρ _{0}}{ρ} = \frac{1}{0.50} = 2 = B^{2.5}$ . Exponent undo karo $1/2.5$ se raise karke: $B = 2^{1/2.5} = 2^{0.4} = 1.31951$ . Speed recover karo same kinetic-energy relation $0.2 M^{2} = B - 1$ se: $0.2 M^{2} = 0.31951 \Rightarrow M^{2} = 1.59754 \Rightarrow M = 1.2640$ . $T / T_{0} = 1/ B = 1/1.31951 = 0.7579$ .

Level 4 — Synthesis

L4.1 (mass-flow-ish, ratios combine karna)

Air $M = 1.2$ , static $T = 250$ K, static $P = 80$ kPa ke saath flow karti hai. Local speed $V$ , phir $T_{0}$ aur $P_{0}$ compute karo. Specific gas constant $R = 287 J\cdotkg^{- 1} K^{- 1}$ (air ke liye) aur speed of sound $a = γ R T$ use karo.

Recall Solution

$R$ kya hai? Specific gas constant, universal constant $8.314 J\cdotmol^{- 1} K^{- 1}$ ko air ke molar mass ( $0.02897 kg/mol$ ) se divide karke aata hai, deta hai $287 J\cdotkg^{- 1} K^{- 1}$ . Iske units $γ R T$ ko $(m/s)^{2}$ banate hain, toh $a = γ R T$ ek speed hai (details Speed of Sound and Mach Number mein). $a = 1.4 \times 287 \times 250 = 100450 = 316.9$ m/s. $V = M a = 1.2 \times 316.9 = 380.3$ m/s. $B = 1 + 0.2 (1.2)^{2} = 1 + 0.288 = 1.288$ . $T_{0} = B T = 1.288 \times 250 = 322.0$ K. $P_{0} = P B^{3.5} = 80 \times (1.288)^{3.5} = 80 \times 2.4303 = 194.4$ kPa.

L4.2 (do stations, isentropic)

Ek isentropic nozzle mein, station 1 par $M_{1} = 0.3$ hai, station 2 par $M_{2} = 1.8$ hai. Reservoir dono ko feed karta hai. Do stations ke beech pressure ratio $P_{2} / P_{1}$ nikalo.

Recall Solution

Kyunki $P_{0}$ same hai (isentropic, shock-free) aur $P / P_{0} = B^{- 3.5}$ , $\frac{P _{2}}{P _{1}} = \frac{P _{0} B _{2}^{- 3.5}}{P _{0} B _{1}^{- 3.5}} = \frac{B _{2}^{- 3.5}}{B _{1}^{- 3.5}} = (\frac{B _{1}}{B _{2}})^{3.5} .$ $B_{1} = 1 + 0.2 (0.09) = 1.018$ , $B_{2} = 1 + 0.2 (3.24) = 1.648$ . $\frac{B _{1}}{B _{2}} = \frac{1.018}{1.648} = 0.6177$ . $P_{2} / P_{1} = 0.617 7^{3.5} = 0.1855$ .

Level 5 — Mastery

L5.1 (shock ke across — forbidden move sahi se karna)

$M_{1} = 2.0$ , $P_{0, 1} = 500$ kPa wali air ek normal shock se guzarti hai. Downstream $M_{2} = 0.5774$ hai aur shock ke across stagnation-pressure ratio $P_{0, 2} / P_{0, 1} = 0.7209$ hai (Normal Shock Relations se). Downstream ek probe static pressure read karta hai. Downstream static pressure $P_{2}$ nikalo.

Recall Solution

Key idea: isentropic table har side ke andar alag kaam karta hai, lekin downstream correct (post-shock) $P_{0}$ use karna hoga. $P_{0, 2} = 0.7209 \times 500 = 360.45$ kPa. Downstream base factor $B_{2} = 1 + 0.2 (0.5774)^{2} = 1 + 0.2 (0.33339) = 1.06668$ . $P_{2} = P_{0, 2} \cdot B_{2}^{- 3.5} = 360.45 \times (1.06668)^{- 3.5} = 360.45 \times 0.79371 = 286.06$ kPa.

L5.2 (variable $γ$ )

Sonic (critical) static pressure ratio $P^{*} / P_{0}$ — phir se $*$ matlab $M = 1$ par value — diatomic-vs-monatomic comparison ke liye recompute karo: (a) $γ = 1.4$ (air), (b) $γ = 1.667$ (argon). Kaun zyada $P^{*} / P_{0}$ par choke karta hai?

Recall Solution

$M = 1$ par: $B = 1 + \frac{γ - 1}{2}$ aur $P^{*} / P_{0} = B^{- γ / (γ - 1)}$ . Yahan exponent abhi bhi isentropic law wala pressure exponent $\frac{γ}{γ - 1}$ hai — yeh change hota hai kyunki $γ$ change hota hai. (a) $γ = 1.4$ : $B = 1.2$ , exponent $= 3.5$ , $P^{*} / P_{0} = 1. 2^{- 3.5} = 0.5283$ . (b) $γ = 1.667$ : $B = 1 + 0.3335 = 1.3335$ , exponent $= γ / (γ - 1) = 1.667/0.667 = 2.4993$ , $P^{*} / P_{0} = 1.333 5^{- 2.4993} = 0.4871$ . Air ( $γ = 1.4$ ) zyada ratio $0.528$ par choke karta hai; monatomic gas ko $\approx 0.487$ tak girna padta hai. (Isentropic Process Relations for Perfect Gas use karta hai.)

L5.3 (full inverse synthesis)

Subsonic air mein ek Pitot-static setup $P_{0} = 105$ kPa, static $P = 101$ kPa, static $T = 288$ K read karta hai. $M$ aur flow speed $V$ nikalo. Upar ki tarah $R = 287 J\cdotkg^{- 1} K^{- 1}$ use karo.

Recall Solution

$\frac{P _{0}}{P} = \frac{105}{101} = 1.03960 = B^{3.5}$ . Exponent undo karo: $B = 1.0396 0^{1/3.5} = 1.0396 0^{0.2857} = 1.01116$ . Speed recover karo kinetic-energy relation $0.2 M^{2} = B - 1 = 0.01116$ se: $M^{2} = 0.05581 \Rightarrow M = 0.2362$ . $a = 1.4 \times 287 \times 288 = 115718 = 340.2$ m/s. $V = M a = 0.2362 \times 340.2 = 80.35$ m/s. (Subsonic hai, toh plain isentropic Pitot formula valid hai — probe ke saamne koi shock nahi.)

Connections

Parent: Isentropic flow tables
Speed of Sound and Mach Number — $a = γ R T$ aur $R$ L4.1, L5.3 mein use hue.
Stagnation Properties and Energy Equation — kyun $T_{0} = B T$ aur $\frac{γ - 1}{2} M^{2}$ kahan janam leta hai.
Normal Shock Relations — L5.1 mein $M_{2}$ aur $P_{0, 2} / P_{0, 1}$ deta hai.
Converging-Diverging Nozzle & Choking — L3.2, L5.2 mein critical ratio.
Area-Mach Relation A/A* — same stagnation framework.
Isentropic Process Relations for Perfect Gas — exponents $\frac{γ}{γ - 1}$ , $\frac{1}{γ - 1}$ deta hai.