Area-Mach number relation A - A - = f(M) — isentropic flow: Worked examples

3.1.6 · D3 · Physics › Compressible Flow & Aerodynamics › Area-Mach number relation A - A - = f(M) — isentropic flow

Yeh page Area–Mach relation ka drill deck hai. Parent note ne formula build kiya; yahan hum use har tarah ki situation mein push karte hain jo ek nozzle throw kar sakta hai — subsonic branch, supersonic branch, sonic point, tiny aur huge Mach limits, choking, real word problems, aur ek exam twist.

Poore note mein hum master formula use karte hain (air, $γ = 1.4$ , to exponent $\frac{γ + 1}{2 ( γ - 1 )} = 3$ ):

Examples se pehle, har case ka naam rakh lete hain taki kuch bhi surprise na kare.

Scenario matrix

$A / A^{*}$ curve ko ek U-shaped valley ki tarah socho (neeche dekho). Har problem in cells mein se kisi ek mein aati hai:

#	Case class	Uski khaas baat	Kaunsa example
C1	Sonic point $M = 1$	valley ka exact bottom, $A / A^{*} = 1$	Ex 1
C2	Subsonic branch $M < 1$	valley ki left wall; $A / A^{*} > 1$	Ex 2
C3	Supersonic branch $M > 1$	right wall; kisi subsonic $M$ jaisa hi $A / A^{*}$	Ex 2
C4	Ek area ke liye dono roots	geometry akele ambiguous hai	Ex 3
C5	Limit $M \to 0$	$A / A^{*} \to \infty$ (bada reservoir, still air)	Ex 4
C6	Limit $M \to \infty$	$A / A^{*} \to \infty$ (hypersonic, area blow up ho jaata hai)	Ex 4
C7	Choked mass flow	throat $M = 1$ par hai; $\overset{m}{˙}$ frozen	Ex 5
C8	Real-world word problem	wind-tunnel / rocket design phrasing	Ex 6
C9	Exam twist — area se property	chain $A / A^{*} \to M \to p, T, ρ, V$	Ex 7
C10	Exam twist — back-pressure branch choose karta hai	same duct, do flow states	Ex 8

Figure — Area-Mach number relation A - A - = f(M) — isentropic flow

Figure neeche ke saare examples ka map hai. Har example padhte waqt uspe ek dot point karo.

Worked examples

Ex 1 — Case C1: sonic point (ek sanity anchor)

Worked example Valley ka bottom

Ek stream exactly $M = 1$ par hai. $A / A^{*}$ kya hai? $M = 1.001$ aur $M = 0.999$ par $A / A^{*}$ kya hoga?

Forecast: Padhne se pehle guess karo. $M = 1$ par flow sonic reference hai hi, to zaroor $A / A^{*} = 1$ hoga. $M = 1$ se thoda hatne par, ratio upar jaayega ya neeche? (Sirf upar ja sakta hai — 1 ek minimum hai.)

Step 1 — $M = 1$ plug karo. $\frac{A}{A ^{*}} = \frac{1}{1} [\frac{1 + 0.2 ( 1 )}{1.2}]^{3} = [\frac{1.2}{1.2}]^{3} = 1.$ Yeh step kyun? Yeh $A^{*}$ ki definition hai jo arithmetic mein convert ho gayi: $M = 1$ par local area hi sonic area hai, isliye ratio exactly 1 hona hi chahiye. Agar tumhara formula yahan 1 nahi deta, to woh galat hai.

Step 2 — $M = 0.999$ aur $M = 1.001$ par nudge karo. $\frac{A}{A ^{*}} (0.999) \approx 1.0000003, \frac{A}{A ^{*}} (1.001) \approx 1.0000003.$ Yeh step kyun? Dono taraf 1 se upar jaate hain — yeh visual proof hai ki valley ka $M = 1$ par flat minimum hai (wahan derivative zero hai). Flat bottom = throat area mein bahut kam change hota hai bade Mach swing ke baad bhi near sonic, isliye nozzle throats itne sensitive hote hain.

Verify: $A / A^{*} = 1$ exactly $M = 1$ par, aur dono neighbours $> 1$ hain aur lagbhag equal hain — smooth minimum se consistent. ✔

Ex 2 — Cases C2 & C3: ek subsonic, ek supersonic (direct evaluation)

Worked example Dono walls par chalo

(a) $M = 0.5$ (subsonic) aur (b) $M = 2.0$ (supersonic) par $A / A^{*}$ compute karo.

Forecast: Dono $M = 1$ se door hain, isliye dono $A / A^{*} > 1$ denge. Steep supersonic wall zyada fast climb karti hai, isliye $M = 2$ ki value $M = 0.5$ se kaafi upar hogi.

Step 1 — Subsonic, $M = 0.5$ . $\frac{A}{A ^{*}} = \frac{1}{0.5} [\frac{1 + 0.2 ( 0.25 )}{1.2}]^{3} = 2 [\frac{1.05}{1.2}]^{3} = 2 (0.875)^{3} \approx 1.340.$ Yeh step kyun? Direct substitution — koi ambiguity nahi jab $M$ diya hua ho. $1/ M = 2$ factor chhote bracket par dominate karta hai, isliye hum left wall par 1 se thoda upar hain.

Step 2 — Supersonic, $M = 2.0$ . $\frac{A}{A ^{*}} = \frac{1}{2} [\frac{1 + 0.2 ( 4 )}{1.2}]^{3} = \frac{1}{2} [\frac{1.8}{1.2}]^{3} = \frac{1}{2} (1.5)^{3} = \frac{3.375}{2} = 1.6875.$ Yeh step kyun? Ab bracket $(1.5)^{3}$ fast grow karta hai aur $1/2$ ko beat kar deta hai — hum right wall par hain, subsonic case se upar, jaise forecast kiya tha.

Verify: $1.340$ (sub) $< 1.6875$ (super), dono $> 1$ . s01 map par dono dots locate karo: same height rule yahan fail hoti hai kyunki yeh alag $M$ hain — lekin dono sonic minimum se upar hain. ✔

Ex 3 — Case C4: ek area, DO Mach numbers (inverse problem)

Worked example Geometry ambiguous hai

Ek duct station par $A / A^{*} = 2.0$ , $γ = 1.4$ hai. Saare Mach numbers nikalo.

Forecast: $2 > 1$ , isliye valley ko height 2 par ek horizontal line cut karti hai — woh do baar cross karti hai. Ek root 1 se kaafi neeche (near $0.3$ ), ek 1 se kaafi upar (near $2.2$ ).

Step 1 — Equation set up karo. $2 = \frac{1}{M} [\frac{1 + 0.2 M ^{2}}{1.2}]^{3} .$ Yeh step kyun? Hum $f (M)$ ko invert kar rahe hain. Koi closed-form inverse nahi hai (yeh $1/ M$ ko cubed polynomial ke saath mix karta hai), isliye iterate karo ya table dekho — exactly isliye compressible-flow tables exist karte hain.

Step 2 — Subsonic root by iteration. $M = 0.3$ try karne par $A / A^{*} \approx 2.035$ milta hai; $M = 0.31$ par $\approx 1.978$ milta hai. Interpolate karke: $M_{sub} \approx 0.306.$ Yeh step kyun? Do guesses ke beech bracket karo jinke outputs 2 ke dono taraf hain, phir squeeze karo — Newton's method ka hand-version.

Step 3 — Supersonic root by iteration. $M = 2.2$ par $A / A^{*} \approx 2.005$ milta hai; $M = 2.19$ par $\approx 1.985$ milta hai. To $M_{super} \approx 2.197.$ Yeh step kyun? Same height line right wall ko bhi hit karti hai. Kaunsa physical hai yeh back-pressure par depend karta hai (Ex 8 dekho) — formula akele choose nahi karega.

Verify: Dono plug back karo: $f (0.306) \approx 2.00$ aur $f (2.197) \approx 2.00$ . Ek area ke liye do valid roots — exactly wahi U-shape prediction jo parent note ne ki thi. ✔

Ex 4 — Cases C5 & C6: do limits (valley walls infinity tak kyun jaati hain)

Worked example Extremes par kya hota hai?

$M \to 0$ aur $M \to \infty$ par $A / A^{*}$ estimate karo. $M = 0.05$ aur $M = 5$ ko stand-in ki tarah use karo.

Forecast: Dono limits blow up karte hain. Near-still air ( $M \to 0$ ) ko same mass dhheere dhheere carry karne ke liye huge pipe chahiye; hypersonic flow ( $M \to \infty$ ) ko huge pipe chahiye kyunki density collapse ho jaati hai. Isliye dono walls $\infty$ tak climb karti hain.

Step 1 — $M \to 0$ tail. Tiny $M$ ke liye, bracket $\to (1/1.2)^{3} \approx 0.5787$ constant rehta hai, jabki $1/ M \to \infty$ : $\frac{A}{A ^{*}} \approx \frac{0.5787}{M} M \to 0 \infty.$ $M = 0.05$ check karo: $\frac{1}{0.05} (0.5787) \approx 11.57.$ Yeh step kyun? $1/ M$ factor culprit hai — yeh kehta hai "almost-stationary gas mein fixed mass flow squeeze karne ke liye enormous area chahiye." Physically: $\overset{m}{˙} = ρ A V$ mein $V \to 0$ hone par $A \to \infty$ ho jaata hai.

Step 2 — $M \to \infty$ tail. Bracket $\sim (\frac{0.2}{1.2} M^{2})^{3} = (0.1 \overline{6} M^{2})^{3}$ grow karta hai like $M^{6}$ , $M$ se divide karne par $\sim M^{5} \to \infty$ milta hai. $\frac{A}{A ^{*}} (5) = \frac{1}{5} [\frac{1 + 0.2 ( 25 )}{1.2}]^{3} = \frac{1}{5} (5)^{3} = \frac{125}{5} = 25.0.$ Yeh step kyun? High Mach par, density $ρ$ zero ki taraf crash karti hai, isliye continuity phir se large area demand karti hai. Right wall infinity ki race bahut faster jeetती hai (power $M^{5}$ vs. $1/ M$ ).

Verify: $A / A^{*} (0.05) \approx 11.57$ aur $A / A^{*} (5) = 25.0$ , dono large hain aur rising hain — "dono limits $\to \infty$ " se match karta hai. ✔

Ex 5 — Case C7: choked mass flow (throat frozen hai)

Worked example Is nozzle se kitni air pass ho sakti hai?

Reservoir $p_{0} = 500 kPa$ , $T_{0} = 300 K$ , throat $A^{*} = 10 cm^{2} = 1 0^{- 3} m^{2}$ , air ( $R = 287 J/kg\cdotK$ , $γ = 1.4$ ). Choked mass flow nikalo.

Forecast: Half-megapascal reservoir aur $10 cm^{2}$ throat ke saath, order $\sim 1 kg/s$ air expect karo — itne chhote hole ke liye bahut zyada, kyunki throat Mach 1 par run kar raha hai.

Step 1 — Choked-flow formula use karo. Choked Flow & Maximum Mass Flow dekho: $\overset{m}{˙} = \frac{p _{0} A ^{*}}{T _{0}} \frac{γ}{R} (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}} .$ Yeh step kyun? Jab throat $M = 1$ hit karta hai ( $A = A^{*}$ ), mass flow purely reservoir state se set hoti hai — exit pressure drop karne se aur zyada air nahi aayegi. Yahi $A^{*}$ ke constant yardstick hone ka physical meaning hai.

Step 2 — Bracket factor. $\frac{2}{γ + 1} = \frac{2}{2.4} = 0.8333$ , exponent $= 3$ , isliye $(0.8333)^{3} = 0.5787$ . Yeh step kyun? Yeh wahi sonic constant hai jo $M \to 0$ tail mein mila tha — koi coincidence nahi: dono sonic reference state describe karte hain.

Step 3 — Numbers. $\overset{m}{˙} = \frac{500000 \cdot 1 0 ^{- 3}}{300} \frac{1.4}{287} (0.5787) \approx 1.157 kg/s .$ Yeh step kyun? Pieces assemble karo. $300 \approx 17.32$ , $1.4/287 \approx 0.06983$ .

Verify (units): $\frac{Pa \cdot m ^{2}}{K} \frac{1}{J/kg\cdotK} = \frac{N}{K} \frac{kg\cdotK}{J} = \frac{N}{K} \cdot \frac{K kg}{N\cdotm} = N \frac{kg}{N\cdotm} = N\cdotkg/m = kg^{2} / s^{2} = kg/s .$ ✔ Value $\approx 1.16 kg/s$ parent Example 3 se match karti hai. ✔

Ex 6 — Case C8: real-world word problem (wind-tunnel design)

Worked example Tunnel size karo

Tumhe air ke liye $M = 3.0$ par ek supersonic wind tunnel test section chahiye. Tumhara throat $A^{*} = 25 cm^{2}$ hai. Tumhe kaun si test-section area $A_{e}$ machine karni padegi?

Forecast: $M = 3$ supersonic wall par deep hai — $A / A^{*}$ around 4 expect karo, isliye $A_{e}$ throat se kaafi zyada, roughly $100 cm^{2}$ .

Step 1 — $f (3)$ evaluate karo. $\frac{A _{e}}{A ^{*}} = \frac{1}{3} [\frac{1 + 0.2 ( 9 )}{1.2}]^{3} = \frac{1}{3} [\frac{2.8}{1.2}]^{3} = \frac{1}{3} (2.3333)^{3} = \frac{12.7037}{3} \approx 4.235.$ Yeh step kyun? Task $M$ deta hai, isliye yeh ek forward evaluation hai — koi ambiguity nahi. Hum design intent se supersonic branch par hain (flow throat par choke hoti hai phir diverging section mein expand hoti hai — Converging-Diverging (de Laval) Nozzle dekho).

Step 2 — Throat se scale karo. $A_{e} = 4.235 \times 25 cm^{2} \approx 105.9 cm^{2} .$ Yeh step kyun? $A / A^{*}$ dimensionless hai; real hardware ke millimetres paane ke liye real throat area se multiply karo.

Verify: $105.9/25 = 4.235 = f (3)$ . Throat se bada hai, "supersonic diverging se accelerate hota hai" ke saath consistent. ✔

Ex 7 — Case C9 exam twist: area se seedha $p, T, ρ, V$ tak

Worked example Poori toolkit chain karo

Ex 6 jaisa hi tunnel: $M_{e} = 3.0$ , $T_{0} = 300 K$ , $p_{0} = 500 kPa$ , air. Test-section $T_{e}$ , $p_{e}$ , aur speed $V_{e}$ nikalo.

Forecast: $M = 3$ par temperature kaafi drop hoti hai (kinetic energy thermal se steal ho jaati hai), isliye $T_{e}$ 300 K se bahut neeche hogi — shayad ~100 K; pressure $p_{0}$ ke kuch percent tak crash kar jaayegi.

Step 1 — Stagnation relation se temperature (Isentropic Stagnation Relations): $\frac{T _{e}}{T _{0}} = \frac{1}{1 + 0.2 ( 9 )} = \frac{1}{2.8} = 0.3571 \Rightarrow T_{e} = 107.1 K .$ Yeh step kyun? $M$ known hai ⇒ har thermodynamic ratio follow karta hai. Temperature pehle aati hai kyunki pressure aur density ussi se build hoti hain.

Step 2 — Pressure. $\frac{p _{e}}{p _{0}} = (\frac{T _{e}}{T _{0}})^{γ / (γ - 1)} = (0.3571)^{3.5} = 0.02722 \Rightarrow p_{e} = 13.6 kPa .$ Yeh step kyun? Isentropic flow $p$ aur $T$ ko exponent $γ / (γ - 1) = 3.5$ se link karta hai — koi heat add nahi, isliye entropy relation fix karta hai.

Step 3 — Speed. Local sound speed (Speed of Sound a = sqrt(gamma R T)): $a_{e} = γ R T_{e} = 1.4 \cdot 287 \cdot 107.1 = 207.5 m/s, V_{e} = M_{e} a_{e} = 3 (207.5) = 622.5 m/s .$ Yeh step kyun? $V = M a$ — Mach ek ratio hai, isliye real velocity m/s mein recover karne ke liye tumhe hamesha local sound speed chahiye (jo local, thandi, $T_{e}$ par depend karti hai).

Verify: $T_{e} \approx 107$ K (thanda, jaise forecast kiya), $p_{e} \approx 13.6$ kPa ( $p_{0}$ ka $2.7%$ ), $V_{e} \approx 623$ m/s. Sab internally consistent: $V_{e} / a_{e} = 622.5/207.5 = 3.0 = M_{e}$ . ✔

Ex 8 — Case C10 exam twist: back-pressure branch choose karta hai

Worked example Same duct, do alag flows

Ek CD nozzle mein $A_{e} / A^{*} = 2.0$ hai. (a) Agar yeh fully subsonic run kare (high back-pressure, throat NOT sonic), to exit subsonic root par hogi. (b) Agar yeh choked & fully expanded supersonic run kare, to exit supersonic root par hogi. Dono cases mein exit Mach batao aur explain karo ki number alag kyun hai jabki shape identical hai.

Forecast: Ex 3 se hi $A_{e} / A^{*} = 2$ ke liye dono roots already known hain: $M \approx 0.306$ aur $M \approx 2.197$ . Sirf pressure boundary condition alag hai, metal nahi.

Step 1 — Case (a) subsonic exit. Back-pressure high hai; flow diverging part mein decelerate hoti hai (subsonic branch), throat par $M = 1$ kabhi nahi aata. Exit root: $M_{e} = 0.306.$ Yeh step kyun? Subsonic wall par, diverging area flow ko slow karta hai ( $d A / A = (M^{2} - 1) d V / V$ with $M < 1$ ⇒ $d A$ aur $d V$ opposite signs hote hain) — Area-Velocity Relation dA-A = (M^2-1) dV-V dekho.

Step 2 — Case (b) supersonic exit. Back-pressure itna low hai ki throat choke ho jaaye ( $M = 1$ wahan); diverging section ab flow ko accelerate karta hai. Exit root: $M_{e} = 2.197.$ Yeh step kyun? Throat ke baad jab $M > 1$ hota hai, same diverging area flow speed up karta hai — area–velocity relation mein sign flip. Isliye geometry akele (Ex 3) decide nahi kar sakti thi.

Step 3 — Tie-break. Physical selector downstream reservoir ki demand ki exit pressure ratio $p_{e} / p_{0}$ hai. High $p_{b}$ → subsonic; sufficiently low $p_{b}$ → supersonic (in-between pressures ke liye andar normal shock bhi ho sakti hai). Yeh step kyun? Yeh "kaunsa root?" wala sawaal close karta hai jo U-shape ne open chhoda tha.

Verify: Dono $M = 0.306$ aur $M = 2.197$ satisfy karte hain $f (M) = 2.0$ (Ex 3 mein check kiya). Same area, do legal flows, back-pressure se choose hote hain — geometry se nahi. ✔

Recall Quick self-test

$M = 1$ par $A / A^{*}$ ? ::: Exactly 1 (Ex 1, valley floor). $M = 0.5$ aur $M = 2.0$ par $A / A^{*}$ ? ::: $1.340$ aur $1.6875$ (Ex 2). Kitne Mach numbers $A / A^{*} = 2$ dete hain? ::: Do — $M \approx 0.306$ aur $M \approx 2.197$ (Ex 3). Kaunsa root physical hai yeh kya decide karta hai? ::: Downstream back-pressure, shape nahi (Ex 8). $p_{0} = 500$ kPa, $T_{0} = 300$ K, $A^{*} = 10 cm^{2}$ ke liye choked $\overset{m}{˙}$ ? ::: $\approx 1.16$ kg/s (Ex 5). $A^{*} = 25 cm^{2}$ ho to $M = 3$ ke liye test-section area? ::: $\approx 106 cm^{2}$ (Ex 6).