Mass flow rate ṁ and its relation to throat area: Question bank

3.3.7 · D5 · Physics › Rocket Propulsion › Mass flow rate ṁ and its relation to throat area

Pehle, woh symbols jo yeh traps use karte hain

Koi bhi trap fair nahi ho sakta jab tak aap us har letter ke malik nahi bante jo woh aap par fire karta hai. Picture dekho: ek nozzle jo ek pinch tak narrow ho raha hai, phir bahar flare ho raha hai.

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

Definition Har question mein paanch characters

== $A^{*}$ == — throat area, nozzle ke sabse narrow point par cross-sectional area (in $m^{2}$ ) (star $^{*}$ ka matlab hamesha "sonic throat par" hota hai). Yeh figure mein pinch ki width hai.
== $a$ (speed of sound)== — ek tiny pressure ripple gas mein kitni tezi se travel karta hai: $a = γ R T$ in $m/s$ . Yeh woh top speed hai jis par "khabar" (ek pressure change) gas mein move kar sakti hai.
== $M$ (Mach number)== — ek pure ratio, $M = \frac{v}{a}$ = gas speed divided by local sound speed. $M = 1$ matlab gas exactly apni khud ki sound jitni tezi se move kar rahi hai. Yeh ek dial hai, speed nahi.
== $R$ (specific gas constant)== — $R = R / M_{molar}$ , universal gas constant divided by propellant ki molar mass, in $J/(kg\cdotK)$ . Yeh temperature ko "push per kilogram" mein convert karta hai.
$p_{0}, T_{0}$ — stagnation (chamber) pressure aur temperature: woh values jahan gas combustion chamber ke andar (lagbhag) rest par hoti hai, nozzle usse speed up karne se pehle.

Intuition

$γ / (R T_{0})$ aur choking coefficient kyun aata hai — formula ke peeche ki kahaani $\overset{m}{˙} = ρ A v$ se shuru karo aur teen facts feed karo:

Speed: throat par gas $M = 1$ hit karta hai, toh $v = a = γ R T$ — $γ$ aur $R T$ sirf sound-speed formula hain.
Density: ideal gas law se $ρ = p / (R T)$ — wahan se doosra $1/ R$ aur $1/ T$ aata hai. $ρ \cdot v$ multiply karo aur $R$ aur $T$ combine hokar $γ / (R T)$ shape mein aa jaate hain.
Chamber-to-throat drop: $M = 1$ reach karne ke liye gas ko chamber values se ek fixed amount tak thanda aur depressurise hona padta hai (sirf $γ$ se set hota hai). Throat par $p, T$ ko $p_{0}, T_{0}$ ke terms mein likhne se exactly factor $(\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}}$ produce hota hai — choking coefficient. Yeh kuch nahi sirf "gas ne sound speed tak pahunchne mein kitna kharch kar diya" hai.

Toh boxed formula magic nahi hai: $\overset{m}{˙} = A^{*} p_{0} \frac{γ}{R T _{0}} (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}}$ sirf $ρ A v$ hai jisme $v$ = sound speed aur $ρ$ = throat density hain, dono chamber terms mein re-express kiye gaye hain.

Neeche ki picture dikhati hai kyun $M = 1$ traffic jam hai: sonic se neeche, pressure ripples phir bhi upstream race karte hain; sonic par woh throat par atke rehte hain.

Har trap ke do engines:

Definition: $\overset{m}{˙} = ρ A v$ (density × area × speed).
Choked result: $\overset{m}{˙} = A^{*} p_{0} \frac{γ}{R T _{0}} (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}}$ , sirf tab valid jab throat sonic ho ( $M = 1$ ).

True ya False — justify karo

Throat area double karne se choked mass flow double ho jaata hai.

True. Boxed formula mein

\overset{m}{˙} \propto A^{*}

linearly hai, aur formula mein aur kuch

A^{*}

par depend nahi karta, toh double door double kg/s deta hai.

Chamber pressure

p_{0}

double karne se choked mass flow double ho jaata hai.

True.

\overset{m}{˙} \propto p_{0}

linearly — zyada

p_{0}

matlab throat par denser gas, aur density directly

ρ A v

mein enter karti hai jabki

T_{0}

se set hone wali sonic speed unchanged rehti hai.

Chamber temperature

T_{0}

double karne se choked mass flow double ho jaata hai.

False.

\overset{m}{˙} \propto 1/ T_{0}

, toh

T_{0}

double karne se

\overset{m}{˙}

ek factor

1/ 2 \approx 0.71

se kam ho jaata hai — hotter gas, density jitni tezi se girti hai, speed utni nahi badhti.

Jab throat choked ho jaata hai, exhaust ko hard vacuum mein kholne se

\overset{m}{˙}

badhta hai.

False.

M = 1

par pressure signals throat se aage upstream travel nahi kar sakti, toh chamber kabhi nahi "sunti" ki back-pressure kam hui;

\overset{m}{˙}

sirf

p_{0}, T_{0}, A^{*}, γ

se freeze hoti hai.

Same numerical

\overset{m}{˙}

chamber, throat, aur exit plane se cross karta hai.

True. Steady flow + conservation of mass (continuity) force karti hai ki

ρ A v

har station par identical ho; sirf

ρ

A

v

individually change hote hain.

Throat par gas pure nozzle mein apne fastest point par move kar rahi hai.

False. Throat sonic hai (

M = 1

); downstream diverging section mein flow supersonic aur tez ho jaata hai. Maximum mass flux per area throat par hai, maximum speed nahi.

Agar flow choked NAHI hai, toh bhi choked-flow boxed formula sahi

\overset{m}{˙}

deta hai.

False. Boxed formula

M = 1

set karke derive kiya gaya tha; subsonic (un-choked) throats ke liye aapko actual local Mach number ke saath general

\overset{m}{˙} (M)

expression use karna hoga, aur

\overset{m}{˙}

phir back-pressure par depend karta hai.

Zyada wide nozzle exit (bada exit area) ek choked engine ke liye

\overset{m}{˙}

badhata hai.

False.

\overset{m}{˙}

throat par

A^{*}

se fix hoti hai; diverging exit sirf exit velocity aur pressure change karta hai, kitne kg/s pass hote hain nahi. Exit area thrust affect karta hai, throughput nahi.

Error dhundho

\overset{m}{˙} = ρ A v

, aur speed of sound

T_{0}

ke saath badhti hai, toh hotter chambers zyada mass flow karte honge."

Error density ko ignore karna hai.

ρ

(fixed

p

par

1/ T

ki tarah girta hai) aur

v = a \propto T

dono change hote hain; product

1/ T_{0}

jaisa hota hai, toh

\overset{m}{˙}

kam hoti hai.

"Kyunki zyada pressure difference zyada flow drive karta hai, exit pressure ko

p_{0} /2

se neeche drop karte rehne se

\overset{m}{˙}

badhti rehti hai."

Sirf tab tak sach hai jab tak throat choke na ho. Critical pressure ratio se aage throat sonic hai aur

\overset{m}{˙}

plateau ho jaati hai — exit pressure mein aur drops

\overset{m}{˙}

ke liye kuch nahi karte.

"Main throat ka static pressure

p

aur temperature

T

boxed formula mein plug karunga."

Boxed formula stagnation (chamber) values

p_{0}, T_{0}

mein likha gaya hai; Mach-number correction already baked in hai. Static values feed karne se woh correction double-count ho jaati hai aur galat number milta hai.

\overset{m}{˙} = ρ A v

ka matlab hai agar main

A

half karu aur

v

double karun, toh

\overset{m}{˙}

unchanged rehega — hamesha."

Sirf tab agar

ρ

fixed rahe, jo compressible nozzle mein nahi hota. Area squeeze karna isentropic relations ke through density aur speed dono ko saath change karta hai, toh aap

ρ

ko constant treat nahi kar sakte.

"Choking hoti hai kyunki friction narrow point par flow ko block karta hai."

Galat mechanism. Choking friction nahi hai; baat yeh hai ki pressure information sound speed

a

par travel karti hai, toh

M = 1

par downstream signals aage flow badhane ke liye upstream propagate nahi kar sakti.

"Mach number

M

ek speed hai, toh

M = 1

matlab

1 m/s

M

ek ratio hai:

M = v / a

, speed divided by local speed of sound

a

M = 1

matlab gas exactly local sound speed par move kar rahi hai, jo hundreds of m/s ho sakti hai.

Why questions

\overset{m}{˙}

exactly

M = 1

par kyun freeze hoti hai aur kisi aur Mach number par kyun nahi?

Kyunki pressure disturbances sound speed

a

par travel karti hain;

M = 1

par flow apne khud ke signals jitni tezi se move karta hai, toh koi "aur gas bhejo" message upstream iske khilaaf travel nahi kar sakta — throat downstream changes par respond karna band kar deta hai.

Throat choke point kyun hai instead of chamber ya exit?

Throat ki minimum area

A^{*}

hoti hai, toh ek fixed

\overset{m}{˙}

ke liye mass flux

ρ v = \overset{m}{˙} / A

wahan sabse bada hota hai; woh flux

M = 1

par peak karta hai, toh minimum-area station wahan hota hai jahan sonic conditions pehle reach hoti hain.

\overset{m}{˙}

kyu

A^{*}

ke saath linearly scale karta hai lekin temperature ke saath sirf

1/ T_{0}

ke roop mein?

Area directly

ρ A v

ko kisi side effect ke bina multiply karta hai, toh yeh linear hai. Temperature do baar enter karti hai — density ke through (

\propto 1/ T

) aur sound speed

a

(

\propto T

) — jinka combination

1/ T

hai.

Final formula mein throat values ki jagah stagnation (chamber) conditions

p_{0}, T_{0}

kyun use karte hain?

Yeh woh quantities hain jo ek engineer chamber mein actually control aur measure karta hai, aur yeh isentropic flow ke along constant hain, ek single clean formula dete hain jo local station se independent ho.

Chamber ko heat karna rocket ki madad phir bhi kyun karta hai, chahe

\overset{m}{˙}

kam ho?

Heat exit velocity

v_{e}

badhata hai (

a \propto T

aur expansion ke through), aur thrust

\approx \overset{m}{˙} v_{e}

velocity gain se zyada benefit karta hai jo small mass-flow drop se milta hai. Dekho Thrust Equation and Effective Exhaust Velocity.

"Choking coefficient"

(\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}}

ek given propellant ke liye fixed number kyun hai?

Yeh sirf

γ

par depend karta hai, specific heats ka ratio, jo gas mixture ki property hai; ek baar propellant choose ho gaya, woh factor

p_{0}

T_{0}

, ya

A^{*}

se regardless lock ho jaata hai.

Same throat area wale do nozzles bahut alag thrusts push kar sakte hain lekin same

\overset{m}{˙}

kyun?

\overset{m}{˙}

throat par fix hai, lekin thrust exit velocity aur exit pressure par bhi depend karta hai, jo diverging section (area ratio) control karta hai — dekho Nozzle Area Ratio and Expansion.

Edge cases

Agar throat area

A^{*} \to 0

ho toh

\overset{m}{˙}

kya hoga?

\overset{m}{˙} \to 0

: koi area nahi, koi gas pass nahi hoti. Ek pinched-shut throat, chahe pressure kitna bhi ho, zero kilogram per second transmit karta hai.

Jab chamber pressure

p_{0} \to 0

ho toh

\overset{m}{˙}

ka kya hota hai?

\overset{m}{˙} \to 0

linearly; koi pressure nahi matlab gas ko kuch bhi push nahi kar raha, aur throat par density zero collapse ho jaati hai.

Formula

T_{0} \to 0

par kya kehta hai, aur kya yeh physical hai?

Kyunki

\overset{m}{˙} \propto 1/ T_{0}

, formula blow up ho jaata hai (

\overset{m}{˙} \to \infty

) jab

T_{0} \to 0

; lekin woh limit unphysical hai —

0 K

ke paas real gas liquefy ho jaata hai aur ideal-gas, sonic-throat assumptions collapse ho jaati hain, toh formula wahan simply apply nahi hota.

Agar back-pressure chamber pressure ke equal ho (

p_{exit} = p_{0}

), toh

\overset{m}{˙}

kya hai?

Zero — koi pressure difference nahi toh flow bilkul nahi, throat subsonic-flowing bhi nahi hai, aur choked formula apply nahi hota.

Agar back-pressure chamber pressure se zyada ho (

p_{exit} > p_{0}

Pressure gradient reverse ho jaata hai, toh gas chamber mein ulti taraf flow karne ki tendency hogi; forward-flow model aur uska

\overset{m}{˙}

formula invalid hai, aur ek real engine exhaust expel karne ki jagah ingest karta.

Critical pressure ratio se neeche (throat abhi bhi subsonic), kya

\overset{m}{˙}

back-pressure par depend karta hai?

Haan — un-choked regime mein back-pressure kam karne se genuinely

\overset{m}{˙}

badhta hai, kyunki signals upstream travel kar sakte hain aur flow response mein speed up ho jaata hai.

Jab propellant ka

γ \to 1

(bahut complex, many-atom molecule), toh choking coefficient ka kya hota hai?

Exponent

\frac{γ + 1}{2 ( γ - 1 )}

blow up ho jaata hai, lekin base

\frac{2}{γ + 1} \to 1

; limit finite hai (

\to e^{- 1/2} \approx 0.607

), toh

\overset{m}{˙}

well-behaved rehta hai.

Agar flow throat par supersonic ho, kya woh ek valid steady choked state hai?

Nahi — converging-diverging nozzle ke liye sonic point exactly minimum area par baithta hai; supersonic-at-throat ek stable steady solution nahi hai, throat

M = 1

pin karta hai.

Kya choked

\overset{m}{˙}

tab change hoti hai jab rocket sea level se vacuum mein fly kare?

Nahi — external (ambient) pressure choked throat ke downstream hai, toh

\overset{m}{˙}

flight mein unchanged rehti hai chahe altitude ke saath thrust aur exit conditions change kyun na ho.

Recall Aage badhne se pehle ek-line self-test

Reveals cover karo aur teen items phir se answer karo jinhe aapne sabse slow kiya. Agar aapki reasoning abhi bhi bhatakti hai, parent ka Step 4 (kyun $M = 1$ flux maximize karta hai) aur "hotter chamber" ki do galtiyan dobara padho.

Upar ke traps ke liye prerequisite links: Choked Flow and Sonic Conditions, Isentropic Flow Relations, Specific Impulse, Tsiolkovsky Rocket Equation.