Mass flow rate ṁ and its relation to throat area: Worked examples

3.3.7 · D3 · Physics › Rocket Propulsion › Mass flow rate ṁ and its relation to throat area

Yeh page parent topic ka exhaustive drill hai. Humne assumptions lena band karke listing shuru ki: mass-flow formula ko diye ja sakte har tarah ke number, sign, aur limit. Pehle hum saare cases ki ek table banate hain. Phir har case ka ek poora example kaam karte hain taaki exam mein koi bhi scenario tumhe surprise na kar sake.

Is page ko jo vocabulary chahiye (use karne se pehle define ki gayi)

Definition Density

$ρ$ , static pressure $p$ , velocity $v$ , Mach number $M$ , aur choking

Density $ρ$ (Greek letter "rho") = har cubic metre mein kitne kilograms gas packed hain, $kg/m^{3}$ mein. $\overset{m}{˙} = ρ A v$ mein yeh batata hai ki swept gas ka har chhota cylinder kitna bhaari hai. Garam patli gas mein $ρ$ chhota hota hai; thandi ghani gas mein $ρ$ bada hota hai. Ideal gas ke liye $ρ = p / (R T)$ .
Static pressure $p$ = gas ka actual pressure jis station par hum dekh rahe hain (jaise throat par), $Pa$ mein. Yeh chamber/stagnation pressure $p_{0}$ se alag hai, jo woh pressure hai jo gas ko agar rest par laya jaaye to hota. Jab gas speed up hoti hai, uska static $p$ $p_{0}$ se neeche gir jaata hai.
Velocity $v$ = gas duct mein kitni tezi se move kar rahi hai, $m/s$ mein. $\overset{m}{˙} = ρ A v$ mein yeh woh speed hai jis par "gas ka cylinder" har second cross-section se guzarta hai. Positive $v$ ka matlab gas baahir flow karti hai (chamber → exit).
Speed of sound $a$ = sabse tez speed jis par pressure signal gas ke andar travel kar sakta hai, $a = γ R T$ $m/s$ mein.
Mach number $M = \frac{v}{a}$ = flow speed local sound speed ke units mein measured. Toh $M = 1$ matlab "gas exactly speed of sound par move kar rahi hai"; $M < 1$ subsonic hai, $M > 1$ supersonic. Yeh pure number hai (koi units nahi).
Choking = throat par $M = 1$ ki state. Ek baar reach ho jaane par, downstream se pressure signals upstream nahi ja sakte, toh baahri pressure kam karne se $\overset{m}{˙}$ nahi badhta.

Baaki sab teen equations par tika hai. Hum inhe repeat karte hain taaki yeh page akele kaam kar sake:

Scenario matrix

Mass-flow problems arbitrary signs aur quadrants ke baare mein nahi hain (mass, area, pressure, temperature sab $\geq 0$ hain; sirf velocity sign carry karta hai, backflow case deta hai). Jo "cases" actually pareshaan karte hain woh kis regime mein gas hai aur kaunsi limit push kar rahe ho ke baare mein hain. Poori list yahan hai.

#	Case class	Kya special hai	Covered by
C1	Plain kinematic flow	Not choked; sirf $ρ A v$	Ex 1
C2	Full choked formula, real numbers	Paancho inputs diye	Ex 2
C3	Scaling / proportionality	Inputs badlo, $\overset{m}{˙}$ predict karo	Ex 3
C4	Degenerate: zero input	$A^{*} = 0$ ya $p_{0} = 0$	Ex 4
C5	Limiting behaviour of $γ$	$γ \to 1$ vs large $γ$	Ex 5
C6	Back-pressure DROP (choked rehta hai)	$p_{b}$ kam karo — $\overset{m}{˙}$ hilta hai?	Ex 6
C6b	Back-pressure RISE (un-chokes)	$p_{b}$ critical se upar uthao	Ex 6b
C7	Inverse problem	$\overset{m}{˙}$ diya, $A^{*}$ nikalo	Ex 7
C8	Real-world word problem	Target thrust ke liye throat size karo	Ex 8
C9	Exam twist: hidden units	Molar mass ki jagah $R$ ; kPa	Ex 9

Choking coefficient itni baar aata hai ki hum ise ek baar naam de dete hain: $Γ (γ) = γ (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}}$ toh choked formula compact $\overset{m}{˙} = \frac{A ^{*} p _{0}}{R T _{0}} Γ (γ)$ ban jaata hai.

Is page ke figures

Is page mein teen step figures hain, deepdives/...-s01, -s02, -s03 ke roop mein embedded hain ("s" = step figure, jis order mein appear karte hain usi mein numbered). Inhe unke label se refer karo — "s01" pehla figure hai, aur aage isi tarah.

Figure s01 — nozzle aur jahan har symbol rehta hai. Koi bhi arithmetic se pehle, hardware ka picture karo: gas ek wide chamber mein enter hoti hai (conditions $p_{0}, T_{0}$ ), area $A^{*}$ ke narrow throat se squeeze hoti hai, phir baahri back-pressure $p_{b}$ ke against exit tak wide ho jaati hai. Red arrow flow direction hai (positive $v$ ). Har example ke liye yeh picture mind mein rakho:

Figure — Mass flow rate ṁ and its relation to throat area

Figure s02 — teen flow regimes aur signal propagation. Yahi wajah hai ki choking kyun hoti hai. Jab throat subsonic hai ( $M < 1$ ), downstream se pressure "message" (cyan wavelets) upstream crawl kar sakta hai aur chamber ko bata sakta hai ki aur gas bheji jaaye. $M = 1$ par message exactly utni hi tezi se travel karta hai jitni tezi se gas baahir bhagti hai — yeh throat par stall ho jaata hai. $M = 1$ se upar yeh downstream sweep ho jaata hai; chamber ab baahri duniya ke liye deaf hai:

Figure s03 — choking coefficient $Γ (γ)$ . Ab maths: yeh $Γ$ ko $γ$ ke against plot karta hai aur (amber dots) exact $γ$ values mark karta hai jo Ex 2, 5, 7, 8 aur 9 mein use hue hain, taaki tum har coefficient seedha curve se padh sako aur dekho ki yeh $γ$ ke saath steadily badhta hai:

Worked examples

Worked example Ex 1 — (C1) Plain kinematic flow, no choking

Ek cool gas jiska density $ρ = 1.2 kg/m^{3}$ hai, area $A = 0.05 m^{2}$ ke duct mein velocity $v = 40 m/s$ se flow kar rahi hai. Temperature $300 K$ hai, $γ = 1.4$ , $R = 287$ ke saath. Confirm karo ki yeh not choked hai, phir $\overset{m}{˙}$ nikalo.

Forecast: teen numbers hain, sab multiply karni hain. Kya answer $2$ , $20$ , ya $200 kg/s$ ke paas expect karte ho?

Pehle regime check karo. Sound speed $a = γ R T = 1.4 \times 287 \times 300 = 120540 = 347.2 m/s$ . Mach number $M = v / a = 40/347.2 = 0.115$ . Yeh step kyun? Formula choose karne se pehle hume regime jaanni chahiye; $M = 0.115 ≪ 1$ toh flow deeply subsonic hai aur not choked hai, toh raw definition sahi tool hai.
Definition likho: $\overset{m}{˙} = ρ A v$ . Yeh step kyun? Un-choked flow ke liye boxed choked formula apply nahi hota — mass sirf density $\times$ area $\times$ speed hai.
Substitute karo aur multiply karo: $1.2 \times 0.05 = 0.06$ ; phir $0.06 \times 40 = 2.4 kg/s$ . Yeh step kyun? Definition ka direct arithmetic.

Verify: Units: $\frac{kg}{m ^{3}} \cdot m^{2} \cdot \frac{m}{s} = \frac{kg}{s}$ ✔. Answer $2.4$ "near $2$ " bracket mein hai — ek chhote duct mein slow, thin flow sirf couple kg/s deta hai.

Worked example Ex 2 — (C2) Full choked formula, real numbers

Chamber: $p_{0} = 4 \times 1 0^{6} Pa$ , $T_{0} = 2500 K$ , $γ = 1.3$ , $R = 300 J/(kg\cdotK)$ , throat $A^{*} = 0.015 m^{2}$ . Nozzle near-vacuum mein exhaust karta hai, toh $p_{b} / p_{0} ≪ 0.546$ → choked. $\overset{m}{˙}$ nikalo.

Forecast: rockets tens of kg/s push karte hain. Guess: $5$ ke paas ya $50 kg/s$ ke paas?

Choking coefficient $Γ (γ) = 1.3 (\frac{2}{2.3})^{\frac{2.3}{0.6}}$ . Yeh step kyun? $Γ$ sirf $γ$ par depend karta hai; ise ek baar compute karo aur har jagah reuse karo (s03 se padh lo).
Exponent evaluate karo: $\frac{2.3}{0.6} = 3.833$ . Base $\frac{2}{2.3} = 0.8696$ . Toh $0.869 6^{3.833} = 0.5613$ , aur $1.3 = 1.1402$ . Product: $Γ = 1.1402 \times 0.5613 = 0.6400$ . Yeh step kyun? Exponent aur base seedhe $Γ (γ)$ se aate hain; base ko us power tak raise karna aur $γ$ se multiply karna poora coefficient hai.
Pressure–area term: $\frac{A ^{*} p _{0}}{R T _{0}} = \frac{0.015 \times 4 \times 1 0 ^{6}}{300 \times 2500}$ . Yeh step kyun? Yeh "raw driving strength" hai: zyaada area ya pressure, zyaada flow; denominator mein garam gas (bada $R T_{0}$ ), kam flow.
Denominator: $750000 = 866.0$ (kyunki $86 6^{2} = 749956 \approx 750000$ ). Numerator: $60000$ . Ratio $= 69.28$ . Yeh step kyun? Hume denominator mein $R T_{0}$ chahiye; root lekar phir divide karne se driving term sahi units mein aata hai.
Combine: $\overset{m}{˙} = 69.28 \times 0.6400 = 44.3 kg/s$ . Yeh step kyun? $\overset{m}{˙}$ driving term times coefficient hai — compact form $\overset{m}{˙} = \frac{A ^{*} p _{0}}{R T _{0}} Γ (γ)$ .

Verify: Step 3 ke units: $\frac{m ^{2} \cdot Pa}{J/kg}$ . Kyunki $Pa = N/m^{2}$ aur $J = N\cdotm$ , yeh $kg/s$ mein reduce hota hai ✔. Answer $44.3$ firmly "tens" bracket mein hai — serious engine ke liye sahi.

Worked example Ex 3 — (C3) Scaling / proportionality (Forecast-then-Verify)

Ex 2 se start karte hue, engineer chamber pressure triple karta hai aur throat area quadruple karta hai. Chamber temperature aur $γ$ unchanged. Naaya $\overset{m}{˙}$ ?

Forecast: $\overset{m}{˙} \propto A^{*} p_{0}$ . Compute karne se pehle factors multiply karo — kya number aata hai?

Formula se, $\overset{m}{˙} \propto A^{*} p_{0}$ (temperature aur $γ$ fixed). Yeh step kyun? Sirf linear factors change hote hain, toh answer unke product se scale hota hai; $Γ$ dubara compute karne ki zaroorat nahi.
Scale factor $= 3 \times 4 = 12$ . Yeh step kyun? Do independent linear factors add nahi, multiply hote hain.
Naaya $\overset{m}{˙} = 12 \times 44.3 = 531.6 kg/s$ .

Verify: Scratch se recompute karo $p_{0} = 1.2 \times 1 0^{7}$ , $A^{*} = 0.06$ ke saath: driving term $= \frac{0.06 \times 1.2 \times 1 0 ^{7}}{866.0} = 831.4$ , times $Γ = 0.6400$ gives $532.1 kg/s$ ✔ (thodi si rounding). Prediction held.

Worked example Ex 4 — (C4) Degenerate inputs: zero throat, zero pressure

$\overset{m}{˙}$ kya hai agar (a) throat poori tarah blocked hai, $A^{*} = 0$ , ya (b) chamber empty hai, $p_{0} = 0$ ? Baaki Ex‑2 ki values rakho.

Forecast: kya ek band valve, ya dead chamber, koi bhi gas leak karta hai?

Case (a): $\overset{m}{˙} = 0 \times p_{0} \dots = 0 kg/s$ . Yeh step kyun? $A^{*}$ poore expression ko linearly multiply karta hai; zero area = zero door = zero flow. Physically valve band hai.
Case (b): $\overset{m}{˙} = A^{*} \times 0 \times \dots = 0 kg/s$ . Yeh step kyun? Koi pressure nahi matlab gas ko push karne wala koi driving force nahi — density aur driving force vanish ho jaate hain.
Edge note: agar $T_{0} \to 0$ toh formula $\overset{m}{˙} \to \infty$ deta hai kyunki $1/ T_{0} \to \infty$ . Yeh unphysical hai — $T_{0} = 0$ par flow karne ke liye koi gas hi nahi hai. Formula ek real, hot gas assume karta hai; yeh absolute zero par valid nahi hai.

Verify: Dono direct substitution se exactly $0$ hain ✔. $T_{0} \to 0$ blow-up sahi tarike se ek domain limit flag karta hai na ki real prediction.

Worked example Ex 5 — (C5)

$γ$ ka Limiting behaviour Ex‑2 ka $A^{*}, p_{0}, T_{0}, R$ rakho. Ek heavy polyatomic gas $γ = 1.1$ versus monatomic gas $γ = 1.67$ ke liye choking coefficient $Γ (γ)$ compare karo. Kaunsa zyaada mass flow deta hai?

Forecast: kya "stiffer" gas (bada $γ$ ) same throat se zyaada ya kam kg/s push karta hai?

$Γ (1.1) = 1.1 (\frac{2}{2.1})^{\frac{2.1}{0.2}}$ . Exponent $= 10.5$ , base $= 0.9524$ , $0.952 4^{10.5} = 0.5987$ ; $1.1 = 1.0488$ ; product $Γ = 0.6279$ . Yeh step kyun? Chhota $γ$ exponent bahut high push karta hai lekin base ko 1 ke paas — dono effects almost cancel karte hain, toh guess ki jagah evaluate karna padega.
$Γ (1.67) = 1.67 (\frac{2}{2.67})^{\frac{2.67}{1.34}}$ . Exponent $= 1.9925$ , base $= 0.7491$ , $0.749 1^{1.9925} = 0.5626$ ; $1.67 = 1.2923$ ; product $Γ = 0.7272$ . Yeh step kyun? Bada $γ$ moderate exponent deta hai lekin base 1 se kaafi neeche — yahan bhi evaluation, intuition nahi, size settle karta hai.
Compare: $Γ (1.67) > Γ (1.1)$ , toh higher- $γ$ gas same throat se same $p_{0}, T_{0}$ par roughly $\frac{0.7272}{0.6279} = 1.16 \times$ zyaada mass pass karta hai. Yeh step kyun? Jab sab kuch fixed ho toh $\overset{m}{˙} \propto Γ$ , toh coefficient ratio hi mass-flow ratio hai.

Verify: s03 mein $Γ$ ka curve dekho — yeh is range mein $γ$ ke saath monotonically rise karta hai, $0.628 < 0.727$ ke consistent ✔. Yeh Isentropic Flow Relations aur Choked Flow and Sonic Conditions se link karta hai.

Worked example Ex 6 — (C6) Back-pressure DROP: still choked

Ex‑2 ka engine $44.3 kg/s$ par choked chal raha hai. Rocket climb karta hai, toh baahri (back) pressure $1 \times 1 0^{5}$ Pa se $2 \times 1 0^{4}$ Pa tak drop hoti hai. Chamber $p_{0}, T_{0}$ constant rakho. Naaya $\overset{m}{˙}$ kya hai?

Forecast: bada pressure difference — zaroor zyaada gas escape hogi? Padhne se pehle guess karo.

Regime check karo. $p_{0} = 4 \times 1 0^{6}$ ke saath, dono back pressures $p_{b} / p_{0} = 0.025$ aur $0.005$ dete hain, $γ = 1.3$ ke critical $0.546$ se kaafi neeche. Toh throat poore time choked rehta hai. Yeh step kyun? Hum jo formula pick karte hain woh regime par depend karta hai; dono cases critical se neeche hain, toh dono par choked formula apply hota hai.
Boxed choked formula mein $p_{0}, T_{0}, γ, R, A^{*}$ hain — aur koi back pressure $p_{b}$ nahi. Yeh step kyun? $M = 1$ par throat downstream changes "sun" nahi sakta (dekho s02), toh $p_{b}$ formula mein absent hai aur $\overset{m}{˙}$ change nahi kar sakta.
Isliye $\overset{m}{˙}$ unchanged hai: $\overset{m}{˙} = 44.3 kg/s$ pehle jaisa.

Verify: Same $p_{0}, T_{0}, γ, R, A^{*}$ substitute karne par $44.3 kg/s$ reproduce hota hai ✔. Answer literally nahi hilta.

Worked example Ex 6b — (C6b) Back-pressure RISE: flow UN-chokes

Same converging nozzle, lekin ab ek valve back pressure badhata hai $p_{b} = 3.6 \times 1 0^{6}$ Pa tak jabki $p_{0} = 4 \times 1 0^{6}$ Pa ( $γ = 1.3$ , $R = 300$ , $T_{0} = 2500$ , throat $A^{*} = 0.015$ ). Kya yeh abhi bhi choked hai? $\overset{m}{˙}$ estimate karo.

Forecast: jab $p_{b}$ almost $p_{0}$ jaisa high ho, kya throat abhi bhi $M = 1$ par hai?

Critical ratio test: $p_{b} / p_{0} = 3.6/4.0 = 0.90$ . $γ = 1.3$ ke liye critical hai $(\frac{2}{2.3})^{\frac{1.3}{0.3}} = 0.869 6^{4.333} = 0.546$ . Yeh step kyun? Single number $p_{b} / p_{0}$ regime decide karta hai; $0.90 > 0.546$ matlab throat subsonic hai — un-choked.
Un-choked → isentropic subsonic relations use karo static pressure ratio se throat Mach number nikalne ke liye. Jab choked nahi hota, throat ki static pressure $p$ back-pressure $p_{b}$ ke equal hoti hai, aur $\frac{p _{0}}{p} = (1 + \frac{γ - 1}{2} M^{2})^{\frac{γ}{γ - 1}}$ . $0.90 = (1 + 0.15 M^{2})^{- 4.333}$ solve karne par $M \approx 0.40$ milta hai. Yeh step kyun? Yahan $p$ hamare vocabulary box mein define ki gayi static pressure hai; throat un-choked hone par ise $p_{b}$ match karna hota hai, aur isentropic $p_{0} / p$ relation us pressure ratio ko Mach number mein convert karta hai — dekho Isentropic Flow Relations.
General $\overset{m}{˙} (M)$ formula se mass flow (yahan restate kiya taaki flip na karna pade) $M = 0.40$ ke saath: $\overset{m}{˙} = \frac{A ^{*} p _{0} M γ / ( R T _{0} )}{( 1 + \frac{γ - 1}{2} M ^{2} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}} = \frac{0.015 \times 4 \times 1 0 ^{6} \times 0.40 \times 1.3/750000}{( 1 + 0.024 ) ^{3.833}}$ Yeh step kyun? Un-choked flow general $\overset{m}{˙} (M)$ formula (top boxed workhorse) use karta hai, $M = 1$ special case nahi, toh hum actual $M$ plug karte hain.
Numerically: $1.3/750000 = 1.3166 \times 1 0^{- 3}$ ; numerator $= 0.015 \times 4 \times 1 0^{6} \times 0.40 \times 1.3166 \times 1 0^{- 3} = 31.6$ ; denominator $= 1.02 4^{3.833} = 1.095$ ; $\overset{m}{˙} \approx 28.9 kg/s$ . Yeh step kyun? Har factor ki straight evaluation; chhota $M$ $\overset{m}{˙}$ ko choked $44.3$ se kaafi neeche rakhta hai.

Verify: $28.9 < 44.3$ ✔ — back-pressure ko critical se upar raise karna $\overset{m}{˙}$ ko choked maximum se neeche reduce karta hai, Ex 6 ka mirror image. $\overset{m}{˙}$ ka $M = 1$ par peak hona consistent hai.

Worked example Ex 7 — (C7) Inverse problem: throat size karo

Ek engine ko $\overset{m}{˙} = 30 kg/s$ swallow karna hai. Diya hai $p_{0} = 3 \times 1 0^{6}$ Pa, $T_{0} = 2000$ K, $γ = 1.25$ , $R = 320 J/(kg\cdotK)$ , choked. Required throat area $A^{*}$ nikalo.

Forecast: rocket throats hand-sized hote hain. Guess: kya $A^{*}$ $0.001$ ya $0.1 m^{2}$ ke paas hai?

Area ke liye boxed formula rearrange karo: $A^{*} = \frac{m ˙ R T _{0}}{p _{0} Γ ( γ )}$ Yeh step kyun? $A^{*}$ linearly appear karta hai, toh hum baaki sab divide karke isolate karte hain — unknown tak pahunchne ka sabse tez raasta.
Coefficient: $Γ (1.25) = 1.25 (\frac{2}{2.25})^{\frac{2.25}{0.5}}$ . Exponent $= 4.5$ , base $= 0.8889$ , $0.888 9^{4.5} = 0.5906$ ; $1.25 = 1.1180$ ; $Γ = 0.6603$ . Yeh step kyun? $Γ$ denominator mein chahiye; yeh sirf $γ$ par depend karta hai toh pehle nikalo (s03 se padh lo).
Numerator: $\overset{m}{˙} R T_{0} = 30 320 \times 2000 = 30640000$ . Ab $640000 = 800$ exactly (kyunki $80 0^{2} = 640000$ ), toh numerator $= 30 \times 800 = 24000$ . Yeh step kyun? Root yahan exact hai, toh arithmetic clean rehti hai; yeh mass-flow times thermal factor hai.
Denominator: $p_{0} Γ = 3 \times 1 0^{6} \times 0.6603 = 1.9809 \times 1 0^{6}$ . Yeh step kyun? Yeh "driving strength per unit area" hai; numerator ko isse divide karne par area milta hai.
Divide karo: $A^{*} = 24000/ (1.9809 \times 1 0^{6}) = 0.01212 m^{2}$ .

Verify: $A^{*} = 0.01212$ forward plug karo: $\frac{0.01212 \times 3 \times 1 0 ^{6}}{800} \times 0.6603 = 45.45 \times 0.6603 = 30.0 kg/s$ ✔. Answer $\approx 0.012 m^{2}$ (ek throat lagbhag $12 cm$ across) — " $0.1$ -ish, small" bracket.

Worked example Ex 8 — (C8) Real-world word problem: thrust se throat tak

Ek designer $F = 2000 N$ thrust wala small thruster banana chahti hai. Usse pata hai ki effective exhaust velocity $c = 2500 m/s$ hai aur chamber $p_{0} = 2 \times 1 0^{6}$ Pa, $T_{0} = 1800$ K, $γ = 1.3$ , $R = 300 J/(kg\cdotK)$ , choked hai. Zaroori $\overset{m}{˙}$ aur throat area nikalo.

Forecast: thrust $= \overset{m}{˙} \cdot c$ (roughly). Toh $\overset{m}{˙}$ $0.8$ ya $8 kg/s$ ke paas?

Thrust se mass flow: $F = \overset{m}{˙} c \Rightarrow \overset{m}{˙} = F / c = 2000/2500 = 0.8 kg/s$ . Yeh step kyun? Thrust har second expelled momentum hai; exhaust speed se divide karne par mass rate milta hai. ( $c$ – $I_{s p}$ link ke liye dekho Specific Impulse.)
Coefficient: $Γ (1.3) = 0.6400$ (Ex 2 mein compute kiya). Yeh step kyun? Same $γ$ Ex 2 jaisa hai, toh recompute ki jagah coefficient reuse karte hain.
Area ke liye solve karo Ex‑7 ke inverse form se: $A^{*} = \frac{m ˙ R T _{0}}{p _{0} Γ}$ . Pehle $R T_{0} = 300 \times 1800 = 540000 = 734.85$ . Yeh step kyun? Thermal factor $R T_{0}$ numerator mein chahiye; divide karne se pehle root evaluate karte hain.
$A^{*} = \frac{0.8 \times 734.85}{2 \times 1 0 ^{6} \times 0.6400} = \frac{587.88}{1.280 \times 1 0 ^{6}} = 4.593 \times 1 0^{- 4} m^{2}$ . Yeh step kyun? Rearranged area formula mein straight substitution.

Verify: $\overset{m}{˙} = 0.8 kg/s$ small bracket mein hai ✔. Area ka forward check: $\frac{4.593 \times 1 0 ^{- 4} \times 2 \times 1 0 ^{6}}{734.85} \times 0.6400 = 1.2500 \times 0.6400 = 0.800 kg/s$ ✔. Throat $\approx 4.6 cm^{2}$ — genuinely small thruster.

Worked example Ex 9 — (C9) Exam twist: hidden units (molar mass, kPa)

Diya hai molar mass $M_{molar} = 0.022 kg/mol$ (seedha $R$ nahi), universal $R = 8.314 J/(mol\cdotK)$ , $p_{0} = 1500 kPa$ , $T_{0} = 2400$ K, $γ = 1.2$ , $A^{*} = 0.008 m^{2}$ , choked. $\overset{m}{˙}$ nikalo.

Forecast: trap units mein hai. Yahan do conversions chupi hain — solve karne se pehle unhe spot karo.

$R$ nikalo: $R = R / M_{molar} = 8.314/0.022 = 377.9 J/(kg\cdotK)$ . Yeh step kyun? Formula ko specific gas constant (per kg) chahiye, universal wala (per mol) nahi.
Pressure convert karo: $p_{0} = 1500 kPa = 1.5 \times 1 0^{6} Pa$ . Yeh step kyun? SI formulas ko pascals chahiye; yeh bhool jaane par answer 1000 se scale ho jaata hai.
Coefficient: $Γ (1.2) = 1.2 (\frac{2}{2.2})^{\frac{2.2}{0.4}}$ . Exponent $= 5.5$ , base $= 0.90909$ , $0.9090 9^{5.5} = 0.5843$ ; $1.2 = 1.0954$ ; $Γ = 0.6401$ . Yeh step kyun? $Γ$ sirf $γ$ par depend karta hai; pehle compute karne par final multiply clean rehti hai.
Driving term: $\frac{A ^{*} p _{0}}{R T _{0}} = \frac{0.008 \times 1.5 \times 1 0 ^{6}}{377.9 \times 2400}$ . Denominator $906960 = 952.3$ (kyunki $952. 3^{2} \approx 906960$ ). Numerator $= 12000$ . Ratio $= 12.60$ . Yeh step kyun? Thermal factor banane ke liye $R T_{0}$ ka root lete hain, phir driving strength nikalne ke liye area–pressure product ko isse divide karte hain.
Combine: $\overset{m}{˙} = 12.60 \times 0.6401 = 8.07 kg/s$ . Yeh step kyun? $\overset{m}{˙}$ = driving term $\times$ coefficient, compact choked formula.

Verify: Units clean hain (Pa, J/kg·K) ✔. Answer $8.07 kg/s$ . Agar koi student kPa convert karna bhool jaata, toh $0.008 kg/s$ milta — $1000 \times$ bahut chhota, classic exam booby-trap. Tsiolkovsky Rocket Equation aur Nozzle Area Ratio and Expansion se downstream use ke liye link karta hai.

Recall Self-test: kaunsa cell kaunsa hai?

Aap kaise decide karte ho ki flow choked hai ya nahi? ::: $p_{b} / p_{0}$ ko $(2/ (γ + 1))^{γ / (γ - 1)}$ se compare karo; usse neeche → choked, upar → subsonic (C6/C6b). Exact equality $p_{b} / p_{0} = (2/ (γ + 1))^{γ / (γ - 1)}$ par kya hota hai? ::: Throat exactly $M = 1$ par hota hai — choking ka onset; dono formulas yahan agree karte hain. Band valve kya $\overset{m}{˙}$ deta hai? ::: Zero (C4 — degenerate zero area). Kya $\overset{m}{˙}$ negative ho sakta hai, aur iska kya matlab hoga? ::: Haan, agar $v < 0$ ho (backflow, gas andar khichi jaaye); steady thrust mein $v > 0$ toh $\overset{m}{˙} > 0$ . Choked engine par back-pressure kam karne se $\overset{m}{˙}$ kitna change hota hai? ::: Bilkul nahi (C6 — choked flow back-pressure ignore karta hai). Back-pressure ko critical ratio se upar raise karne par $\overset{m}{˙}$ ka kya hota hai? ::: Throat un-choke ho jaata hai aur $\overset{m}{˙}$ choked maximum se neeche aa jaata hai (C6b). Target $\overset{m}{˙}$ se $A^{*}$ nikalne ke liye, kis form mein rearrange karo? ::: $A^{*} = \overset{m}{˙} R T_{0} / (p_{0} Γ (γ))$ (C7 — inverse problem). Molar mass ko specific $R$ mein convert karne ke liye kaunsa formula use hota hai? ::: $R = R / M_{molar}$ (C9 — hidden units). $1.1$ – $1.67$ range mein higher $γ$ choking coefficient badhata hai ya ghatata hai? ::: Badhata hai (C5 — $Γ$ $γ$ ke saath increase hota hai). Static $p$ aur stagnation $p_{0}$ mein kya fark hai? ::: $p$ moving-gas station par pressure hai; $p_{0}$ woh hai jo rest par laya jaaye to hota — jab gas move kare $p < p_{0}$ .