Choked flow — condition M = 1 at throat, maximum mass flow: Worked examples

3.1.8 · D3 · Physics › Compressible Flow & Aerodynamics › Choked flow — condition M = 1 at throat, maximum mass flow

Neeche sab kuch parent aur uske prerequisites mein banaye gaye tools use karta hai: Isentropic Flow Relations, Stagnation (Total) Properties, Speed of Sound & Mach Number, aur — aakhri twist ke liye — Converging–Diverging (de Laval) Nozzle.

Scenario matrix

Koi bhi arithmetic se pehle, har tarah ke questions list karo jo yeh topic pooch sakta hai. Pehle ek symbol, taaki table saaf padhe:

Cell	Case class	Kya decide karta hai	Example
A	Not choked (subsonic exit)	$p_{b} / p_{0} > r^{*}$	Ex. 1
B	Bilkul threshold par	$p_{b} / p_{0} = r^{*}$	Ex. 2
C	Choked (deep)	$p_{b} / p_{0} < r^{*}$	Ex. 3
D	Max mass flow number	plug $M = 1$ , $A = A^{*}$	Ex. 4
E	Limiting input: $T_{0}$ up	$\overset{m}{˙}_{m a x} \propto 1/ T_{0}$	Ex. 5
F	Limiting input: $p_{0}$ up	$\overset{m}{˙}_{m a x} \propto p_{0}$	Ex. 5
G	Degenerate: $p_{b} = p_{0}$ (koi drop nahi)	$\overset{m}{˙} = 0$	Ex. 6
H	Different gas ( $γ$ change)	$r^{*}$ recompute karo	Ex. 7
I	Real-world word problem	sahi cell pick karo	Ex. 8
J	Exam twist: C-D nozzle back pressure	throat phir bhi chokes	Ex. 9

Neeche ke figures $p_{b} / p_{0}$ se behaviour tak ka map anchor karte hain (Cells A–C) aur $\overset{m}{˙}$ -vs- $p_{b}$ saturation curve (Cells D–G).

Figure — Choked flow — condition M = 1 at throat, maximum mass flow

Jab tak kaha na jaye, throughout constants: air, $γ = 1.4$ , $R = 287 J kg^{- 1} K^{- 1}$ , toh $r^{*} = (2/2.4)^{3.5} \approx 0.528$ aur bracket $(\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}} = (0.8333)^{3} \approx 0.5787$ .

Worked example Example 1 — Cell A:

Not choked Air reservoir $p_{0} = 120$ kPa atmosphere $p_{b} = 101.3$ kPa mein exhaust karta hai ek converging nozzle ke through. Choked hai? Forecast: Pressure drop halka hai (sirf ~16%). Guess: not choked, subsonic exit. Step 1 — Back-pressure ratio banao. $p_{b} / p_{0} = 101.3/120 = 0.844$ . Yeh step kyun? Choking ka faisla is ek number ko $r^{*}$ se compare karke hota hai — aur kuch nahi. Step 2 — $r^{*} = 0.528$ se compare karo. $0.844 > 0.528$ . Yeh step kyun? Threshold se upar matlab exit pressure jo nozzle ko chahiye ( $= p_{b}$ ) abhi bhi $p^{*}$ se zyada hai, toh throat kabhi $M = 1$ tak pohonchne ki zaroorat nahi padti. Answer: Not choked. Exit pressure $= p_{b} = 101.3$ kPa, exit subsonic hai ( $M_{exit} < 1$ ), aur $p_{b}$ ko kam karna $\overset{m}{˙}$ badhata. Verify: Sanity — sirf 16% drop sonic flow produce nahi kar sakta; $r^{*} \approx 0.528$ ka poora point yahi hai ki pehle ~47% drop chahiye. Consistent. ✔

Worked example Example 2 — Cell B:

Bilkul threshold par $p_{0} = 192$ kPa, $p_{b} = 101.3$ kPa, converging nozzle, air. Choked hai, borderline hai, ya nahi? Forecast: $101.3/192 \approx 0.53$ — suspiciously 0.528 ke kareebi. Guess: bilkul edge par. Step 1 — Ratio. $p_{b} / p_{0} = 101.3/192 = 0.5276$ . Yeh step kyun? Cell A jaisa hi single-number test. Step 2 — Compare karo. $0.5276 \approx 0.528 = r^{*}$ . Yeh step kyun? Bilkul $r^{*}$ par exit $M = 1$ tak pohonchta hai lekin sirf abhi abhi; exit pressure ambient ke barabar BHI hai aur $p^{*}$ ke barabar BHI — last subsonic state aur first choked state ek saath coincide karte hain. Answer: Just choked. $M_{exit} = 1$ , $p_{exit} = p^{*} = p_{b}$ . $p_{b}$ mein koi bhi aur drop ise full choke mein lock kar dega. Verify: $0.528 \times 192 = 101.4 kPa = p_{b}$ (rounding tak). Exit pressure ambient se exactly match karta hai, threshold ki defining feature. ✔

Worked example Example 3 — Cell C:

Choked (deep) $p_{0} = 500$ kPa, $p_{b} = 101.3$ kPa, air, converging nozzle. Choked hai? Exit pressure kya hai? Forecast: Bada drop (80%). Guess: choked, exit ambient se upar locked. Step 1 — Ratio. $p_{b} / p_{0} = 101.3/500 = 0.2026$ . Yeh step kyun? Choke test. Step 2 — Compare karo. $0.2026 < 0.528$ → choked. Yeh step kyun? Ek converging exit $p^{*}$ se neeche nahi ja sakta smooth subsonic-to-sonic flow rehte hue; yeh $M = 1$ par clamp ho jaata hai. Step 3 — Locked exit pressure. $p_{exit} = p^{*} = 0.528 \times 500 = 264 kPa$ . Yeh step kyun? Ek baar choked ho jaaye, exit pressure $p^{*}$ par freeze ho jaata hai, chahe $p_{b}$ kitna bhi low ho. Answer: Choked. Exit pressure $= 264$ kPa (ambient 101.3 kPa se kaafi upar → under-expanded jet bahar expand karta rehta hai). Verify: $264 > 101.3$ , toh jet under-expanded hai — exactly wahi jo "choked with low $p_{b}$ " predict karta hai. ✔

Worked example Example 4 — Cell D:

Maximum mass-flow number Ex. 3 jaisa hi nozzle: throat area $A^{*} = 2 \times 1 0^{- 4} m^{2}$ , $T_{0} = 300$ K, $p_{0} = 500$ kPa, air. $\overset{m}{˙}_{m a x}$ nikalo. Forecast: Order of magnitude? Chhota sa hole, half-MPa reservoir — guess ek fraction of a kg/s. Step 1 — Bracket factor. $(\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}} = (0.8333)^{3} = 0.5787$ . Yeh step kyun? Yeh dimensionless factor sonic throat ke liye "efficiency penalty" hai; air ke liye exponent $\frac{γ + 1}{2 ( γ - 1 )} = 3$ hai. Step 2 — Root factor. $γ / (R T_{0}) = 1.4/ (287 \times 300) = 1.626 \times 1 0^{- 5} = 4.032 \times 1 0^{- 3}$ . Yeh step kyun? Yeh units $kg / J = s/m$ carry karta hai jo final answer kg/s mein laata hai. Step 3 — Assemble karo. $\overset{m}{˙}_{m a x} = A^{*} p_{0} γ / (R T_{0}) (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}} = (2 \times 1 0^{- 4}) (5 \times 1 0^{5}) (4.032 \times 1 0^{- 3}) (0.5787)$ . Yeh step kyun? Yeh parent se boxed choked-flow formula hai jisme $M = 1$ , $A = A^{*}$ substitute kiya gaya hai. Answer: $\overset{m}{˙}_{m a x} \approx 0.233 kg/s$ . Verify: Units: $m^{2} \cdot Pa \cdot (s/m) = m^{2} \cdot \frac{kg}{m s ^{2}} \cdot \frac{s}{m} = kg/s$ . ✔ Magnitude ~0.2 kg/s "fraction of a kg/s" forecast se match karta hai. ✔

Worked example Example 5 — Cells E & F:

Input knobs ghoomao Ex. 4 se start karte hue ( $\overset{m}{˙}_{m a x} = 0.233$ kg/s): (a) $T_{0}$ double karo 600 K; (b) iske bajaay $p_{0}$ double karo 1000 kPa. Naya $\overset{m}{˙}_{m a x}$ dono mein? Forecast: Zyada garm gas → guess kam flow (density loss). Zyada $p_{0}$ → guess zyada flow, linearly. Step 1 — (a) Temperature scaling. $\overset{m}{˙}_{m a x} \propto 1/ T_{0}$ , toh $T_{0}$ double karne par $1/ 2$ se multiply hota hai: $0.233/ 2 = 0.165 kg/s$ . Yeh step kyun? Sirf denominator mein $T_{0}$ badalta hai; baaki sab factors hold kiye jaate hain. Zyada garm gas kam dense hoti hai, aur sound speed sirf $T$ se badhti hai — density loss jeetta hai. Step 2 — (b) Pressure scaling. $\overset{m}{˙}_{m a x} \propto p_{0}$ , toh $p_{0}$ double karne par double ho jaata hai: $2 \times 0.233 = 0.466 kg/s$ . Yeh step kyun? $p_{0}$ pehli power mein aata hai; zyada upstream pressure matlab throat par proportionally dense gas. Answer: (a) $0.165$ kg/s (kam); (b) $0.466$ kg/s (zyada). Verify: $0.233/1.4142 = 0.1648$ ✔ aur $0.233 \times 2 = 0.466$ ✔. Direction forecast se match karta hai: heat hurt karta hai, pressure help karta hai. ✔

Worked example Example 6 — Cell G:

Degenerate — koi pressure drop nahi $p_{0} = 500$ kPa, $p_{b} = 500$ kPa (same reservoir par valve band), converging nozzle. $\overset{m}{˙}$ nikalo. Forecast: Koi drop nahi → guess zero flow. Step 1 — Ratio. $p_{b} / p_{0} = 500/500 = 1$ . Yeh step kyun? Choke test phir bhi apply hota hai lekin apni ceiling par. Step 2 — Interpret karo. Ratio $= 1 > 0.528$ , toh not choked; exit Mach isentropic relation se solve hota hai $p_{exit} / p_{0} = 1$ ke saath, jo $M_{exit} = 0$ deta hai. Yeh step kyun? $\frac{p _{0}}{p} = 1 \Rightarrow 1 + \frac{γ - 1}{2} M^{2} = 1 \Rightarrow M = 0$ : pressure difference ke bina gas ko accelerate karne ke liye kuch nahi hai. Step 3 — Mass flow. $\overset{m}{˙} = ρ A V$ with $V = M a = 0 \Rightarrow \overset{m}{˙} = 0$ . Yeh step kyun? Master $\overset{m}{˙}$ formula mein $M$ ka factor hai; $M = 0$ par yeh zero ho jaata hai. Answer: $\overset{m}{˙} = 0$ . Yeh figure s02 mein saturation curve ka far-left end hai. Verify: Parent ke master formula mein $M$ -factor $\frac{M}{( \dots )}$ hai, jo $M = 0$ par exactly 0 hai. ✔

Worked example Example 7 — Cell H:

Different gas ( $γ$ change) Helium ( $γ = 5/3 \approx 1.667$ ) reservoir $p_{0} = 500$ kPa, $p_{b} = 101.3$ kPa, converging nozzle. Choked hai? Helium ka critical ratio do. Forecast: Ex. 3 jaisa hi bada drop; guess phir bhi choked hai, lekin threshold number 0.528 se alag hogi. Step 1 — Helium ka critical ratio. $r^{*} = (\frac{2}{γ + 1})^{\frac{γ}{γ - 1}}$ with $γ = 1.667$ : $\frac{2}{2.667} = 0.75$ , exponent $\frac{1.667}{0.667} = 2.5$ , toh $r^{*} = 0.7 5^{2.5} = 0.487$ . Yeh step kyun? $r^{*}$ gas-specific hai $γ$ ke through; air ka 0.528 reuse nahi kar sakte. Step 2 — Compare karo. $p_{b} / p_{0} = 101.3/500 = 0.2026 < 0.487$ → choked. Yeh step kyun? Same threshold logic; sirf number badla. Answer: Choked. Helium ka critical ratio $\approx 0.487$ hai (air ke 0.528 se kam, kyunki zyada $γ$ sonic point ko deeper pressure drop par push karta hai). Verify: $0.7 5^{2.5} = 0.4871$ aur $0.2026 < 0.4871$ . ✔ Saath hi $r^{*}$ decrease hota hai jab $γ$ increase hota hai (air 0.528 → helium 0.487). ✔

Worked example Example 8 — Cell I:

Real-world word problem Ek compressed-air tank $p_{0} = 800$ kPa, $T_{0} = 290$ K par hai jo ek chhote pneumatic tool ko throat area $A^{*} = 1.5 \times 1 0^{- 5} m^{2}$ ke converging orifice ke through feed karta hai. Room 101 kPa par hai. Tank kitna mass flow deliver karta hai? Forecast: Bada drop → choked → $\overset{m}{˙}_{m a x}$ formula use karo. Guess ~0.03 kg/s (tiny hole). Step 1 — Choke check. $p_{b} / p_{0} = 101/800 = 0.126 < 0.528$ → choked, $\overset{m}{˙}_{m a x}$ use karo. Yeh step kyun? Formula choose karne se pehle case classify karna zaroori hai; yahan yeh Cell C/D mein land karta hai. Step 2 — Root factor. $γ / (R T_{0}) = 1.4/ (287 \times 290) = 1.682 \times 1 0^{- 5} = 4.101 \times 1 0^{- 3}$ . Yeh step kyun? Ex. 4 jaisa hi units-carrying factor, ab $T_{0} = 290$ ke saath. Step 3 — Assemble karo. $\overset{m}{˙}_{m a x} = (1.5 \times 1 0^{- 5}) (8 \times 1 0^{5}) (4.101 \times 1 0^{- 3}) (0.5787)$ . Yeh step kyun? Is tank ke numbers ke saath choked formula. Answer: $\overset{m}{˙}_{m a x} \approx 0.0285 kg/s$ (≈ 28.5 g/s). Verify: Compute $(1.5 \times 1 0^{- 5}) (8 \times 1 0^{5}) = 12$ ; $12 \times 4.101 \times 1 0^{- 3} = 0.04921$ ; $\times 0.5787 = 0.02849$ kg/s. ✔ ~0.03 kg/s forecast se match karta hai. ✔

Worked example Example 9 — Cell J:

Exam twist — C-D nozzle, throat phir bhi chokes Ek Converging–Diverging (de Laval) Nozzle ka throat $A^{*} = 2 \times 1 0^{- 4} m^{2}$ , $p_{0} = 500$ kPa, $T_{0} = 300$ K, air hai. Uska back pressure set kiya gaya hai taaki flow fully choked ho (throat sonic). Kya $\overset{m}{˙}$ Ex. 4 ke plain converging nozzle se alag hai? Forecast: Diverging part flow ko supersonic tak accelerate karta hai — zaroor zyada mass nikalta hoga? Guess: nahi, mass flow throat par set hoti hai. Step 1 — $\overset{m}{˙}$ kahan fix hoti hai? Continuity se $\overset{m}{˙} = ρ A V$ har section par same hoti hai; choking condition ( $M = 1$ ) minimum area = throat par impose hoti hai. Yeh step kyun? Mass flow poore steady nozzle ke liye ek single number hai; yeh sonic throat se cap hoti hai, exit se nahi. Step 2 — Compute karo. Ex. 4 jaisi identical inputs → $\overset{m}{˙}_{m a x} = 0.233 kg/s$ . Yeh step kyun? $\overset{m}{˙}_{m a x}$ formula sirf throat area aur stagnation state use karta hai — diverging section appear hi nahi karta. Answer: Same $\overset{m}{˙}_{m a x} = 0.233$ kg/s. Diverging section exit speed aur pressure (supersonic exit) change karta hai, mass flow nahi. Agar back pressure galat hai, toh diverging part mein ek normal shock appear hogi — lekin $\overset{m}{˙}$ phir bhi $0.233$ kg/s rahega kyunki throat unchanged hai. Verify: Ex. 4 jaisi same numbers → $0.233$ kg/s, diverging geometry ya downstream shock se independent. ✔

Recall Matrix ke across quick self-test

Air ke liye threshold ratio ::: $r^{*} \approx 0.528$ ; choked jab $p_{b} / p_{0} < r^{*}$ . Kya $\overset{m}{˙}_{m a x}$ choked hone ke baad $p_{b}$ par depend karta hai? ::: Nahi — yeh sirf $p_{0}, T_{0}, A^{*}, γ, R$ par depend karta hai. $T_{0}$ double karne ka $\overset{m}{˙}_{m a x}$ par effect? ::: $1/ 2 \approx 0.707$ se multiply karo (kam flow). $p_{0}$ double karne ka effect? ::: $\overset{m}{˙}_{m a x}$ double ho jaata hai. Helium ka critical ratio? ::: $\approx 0.487$ (air se kam kyunki $γ$ zyada hai). Kya diverging section $\overset{m}{˙}$ badhata hai? ::: Nahi — sonic throat ise cap karta hai.