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

This page is the exhaustive drill for the parent topic. We stopped assuming and started listing: every kind of number, sign, and limit that the mass-flow formula can be handed. First we build a table of all the cases. Then we work one full example per case so that no scenario can surprise you in an exam.

The vocabulary this page needs (defined before we use it)

Definition Density

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

Density $ρ$ (Greek letter "rho") = how many kilograms of gas are packed into each cubic metre, in $kg/m^{3}$ . In $\overset{m}{˙} = ρ A v$ it tells us how heavy each little cylinder of swept gas is. Hot thin gas has small $ρ$ ; cold dense gas has large $ρ$ . For an ideal gas $ρ = p / (R T)$ .
Static pressure $p$ = the actual pressure of the gas at the station we are looking at (e.g. at the throat), in $Pa$ . This is different from the chamber/stagnation pressure $p_{0}$ , which is the pressure the gas would have if brought to rest. As gas speeds up, its static $p$ drops below $p_{0}$ .
Velocity $v$ = how fast the gas moves along the duct, in $m/s$ . In $\overset{m}{˙} = ρ A v$ it is the speed at which the "cylinder of gas" is swept past a cross-section each second. A positive $v$ means gas flows outward (chamber → exit).
Speed of sound $a$ = the fastest speed at which a pressure signal can travel through the gas, $a = γ R T$ in $m/s$ .
Mach number $M = \frac{v}{a}$ = the flow speed measured in units of the local sound speed. So $M = 1$ means "the gas is moving exactly at the speed of sound"; $M < 1$ is subsonic, $M > 1$ is supersonic. It is a pure number (no units).
Choking = the state $M = 1$ at the throat. Once reached, pressure signals from downstream can no longer travel upstream, so lowering the outside pressure cannot increase $\overset{m}{˙}$ .

Everything else rests on three equations. We repeat them all so this page stands alone:

The scenario matrix

Mass-flow problems are not about arbitrary signs and quadrants (mass, area, pressure, temperature are all $\geq 0$ ; only velocity carries a sign, giving the backflow case). The "cases" that actually bite you are about which regime the gas is in and what limit you push toward. Here is the complete list.

#	Case class	What makes it special	Covered by
C1	Plain kinematic flow	Not choked; just $ρ A v$	Ex 1
C2	Full choked formula, real numbers	All five inputs given	Ex 2
C3	Scaling / proportionality	Change inputs, predict $\overset{m}{˙}$	Ex 3
C4	Degenerate: zero input	$A^{*} = 0$ or $p_{0} = 0$	Ex 4
C5	Limiting behaviour of $γ$	$γ \to 1$ vs large $γ$	Ex 5
C6	Back-pressure DROP (stays choked)	Lower $p_{b}$ — does $\overset{m}{˙}$ move?	Ex 6
C6b	Back-pressure RISE (un-chokes)	Raise $p_{b}$ above critical	Ex 6b
C7	Inverse problem	Given $\overset{m}{˙}$ , find $A^{*}$	Ex 7
C8	Real-world word problem	Sizing a throat for a target thrust	Ex 8
C9	Exam twist: hidden units	Molar mass instead of $R$ ; kPa	Ex 9

The choking coefficient appears so often we name it once: $Γ (γ) = γ (\frac{2}{γ + 1})^{\frac{γ + 1}{2 ( γ - 1 )}}$ so the choked formula becomes the compact $\overset{m}{˙} = \frac{A ^{*} p _{0}}{R T _{0}} Γ (γ)$ .

The figures on this page

This page carries three step figures, embedded as deepdives/...-s01, -s02, -s03 (the "s" = step figure, numbered in the order they appear). Refer to them by their label — "s01" is the first figure, and so on.

Figure s01 — the nozzle and where each symbol lives. Before any arithmetic, picture the hardware: gas enters a wide chamber (conditions $p_{0}, T_{0}$ ), squeezes through the narrow throat of area $A^{*}$ , then widens out to the exit against the outside back-pressure $p_{b}$ . The red arrow is the flow direction (positive $v$ ). Keep this picture in mind for every example:

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

Figure s02 — the three flow regimes and signal propagation. This is why choking happens. When the throat is subsonic ( $M < 1$ ), a pressure "message" from downstream (cyan wavelets) can crawl upstream and tell the chamber to send more gas. At $M = 1$ the message travels exactly as fast as the gas rushes out — it stalls at the throat. Above $M = 1$ it is swept downstream; the chamber is now deaf to the outside:

Figure s03 — the choking coefficient $Γ (γ)$ . Now the maths: this plots $Γ$ against $γ$ and marks (amber dots) the exact $γ$ values used in Ex 2, 5, 7, 8 and 9, so you can read each coefficient straight off the curve and see it rise steadily with $γ$ :

The worked examples

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

A cool gas of density $ρ = 1.2 kg/m^{3}$ flows through a duct of area $A = 0.05 m^{2}$ at velocity $v = 40 m/s$ . The temperature is $300 K$ with $γ = 1.4$ , $R = 287$ . Confirm it is not choked, then find $\overset{m}{˙}$ .

Forecast: three numbers, all multiply. Do you expect an answer near $2$ , $20$ , or $200 kg/s$ ?

Check the regime first. Sound speed $a = γ R T = 1.4 \times 287 \times 300 = 120540 = 347.2 m/s$ . Mach number $M = v / a = 40/347.2 = 0.115$ . Why this step? Before choosing a formula we must know the regime; $M = 0.115 ≪ 1$ so the flow is deeply subsonic and not choked, so the raw definition is the correct tool.
Write the definition: $\overset{m}{˙} = ρ A v$ . Why this step? For un-choked flow the boxed choked formula does not apply — mass is just density $\times$ area $\times$ speed.
Substitute and multiply: $1.2 \times 0.05 = 0.06$ ; then $0.06 \times 40 = 2.4 kg/s$ . Why this step? Direct arithmetic of the definition.

Verify: Units: $\frac{kg}{m ^{3}} \cdot m^{2} \cdot \frac{m}{s} = \frac{kg}{s}$ ✔. The answer $2.4$ sits in the "near $2$ " bracket — a slow, thin flow through a small duct gives only a couple kg/s.

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 $A^{*} = 0.015 m^{2}$ . The nozzle exhausts to near-vacuum, so $p_{b} / p_{0} ≪ 0.546$ → choked. Find $\overset{m}{˙}$ .

Forecast: rockets push tens of kg/s. Guess: closer to $5$ or to $50 kg/s$ ?

Choking coefficient $Γ (γ) = 1.3 (\frac{2}{2.3})^{\frac{2.3}{0.6}}$ . Why this step? $Γ$ depends only on $γ$ ; compute it once and reuse everywhere (read it off s03).
Evaluate exponent: $\frac{2.3}{0.6} = 3.833$ . Base $\frac{2}{2.3} = 0.8696$ . So $0.869 6^{3.833} = 0.5613$ , and $1.3 = 1.1402$ . Product: $Γ = 1.1402 \times 0.5613 = 0.6400$ . Why this step? The exponent and base come straight from $Γ (γ)$ ; raising the base to that power and multiplying by $γ$ is the whole coefficient.
The pressure–area term: $\frac{A ^{*} p _{0}}{R T _{0}} = \frac{0.015 \times 4 \times 1 0 ^{6}}{300 \times 2500}$ . Why this step? This is the "raw driving strength": more area or pressure, more flow; hotter gas (bigger $R T_{0}$ ) in the denominator, less flow.
Denominator: $750000 = 866.0$ (because $86 6^{2} = 749956 \approx 750000$ ). Numerator: $60000$ . Ratio $= 69.28$ . Why this step? We need $R T_{0}$ in the denominator; taking the root then dividing gives the driving term in the right units.
Combine: $\overset{m}{˙} = 69.28 \times 0.6400 = 44.3 kg/s$ . Why this step? $\overset{m}{˙}$ is the driving term times the coefficient — the compact form $\overset{m}{˙} = \frac{A ^{*} p _{0}}{R T _{0}} Γ (γ)$ .

Verify: Units of step 3: $\frac{m ^{2} \cdot Pa}{J/kg}$ . Since $Pa = N/m^{2}$ and $J = N\cdotm$ , this reduces to $kg/s$ ✔. Answer $44.3$ is firmly in the "tens" bracket — correct for a serious engine.

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

What is $\overset{m}{˙}$ if (a) the throat is fully blocked, $A^{*} = 0$ , or (b) the chamber is empty, $p_{0} = 0$ ? Keep the other Ex‑2 values.

Forecast: does a closed valve, or a dead chamber, leak any gas?

Case (a): $\overset{m}{˙} = 0 \times p_{0} \dots = 0 kg/s$ . Why this step? $A^{*}$ multiplies the whole expression linearly; zero area = zero door = zero flow. Physically the valve is shut.
Case (b): $\overset{m}{˙} = A^{*} \times 0 \times \dots = 0 kg/s$ . Why this step? No pressure means no gas being pushed — density and driving force vanish.
Edge note: if $T_{0} \to 0$ the formula gives $\overset{m}{˙} \to \infty$ because $1/ T_{0} \to \infty$ . This is unphysical — at $T_{0} = 0$ there is no gas to flow. The formula assumes a real, hot gas; it is not valid at absolute zero.

Verify: Both are exactly $0$ by direct substitution ✔. The $T_{0} \to 0$ blow-up correctly flags a domain limit rather than a real prediction.

Worked example Ex 5 — (C5) Limiting behaviour of

$γ$ Keep Ex‑2's $A^{*}, p_{0}, T_{0}, R$ . Compare the choking coefficient $Γ (γ)$ for a heavy polyatomic gas $γ = 1.1$ versus a monatomic gas $γ = 1.67$ . Which gives more mass flow?

Forecast: does a "stiffer" gas (larger $γ$ ) push more or fewer kg/s through the same throat?

$Γ (1.1) = 1.1 (\frac{2}{2.1})^{\frac{2.1}{0.2}}$ . Exponent $= 10.5$ , base $= 0.9524$ , $0.952 4^{10.5} = 0.5987$ ; $1.1 = 1.0488$ ; product $Γ = 0.6279$ . Why this step? Small $γ$ pushes the exponent very high but the base close to 1 — the two effects nearly cancel, so we must evaluate rather than guess.
$Γ (1.67) = 1.67 (\frac{2}{2.67})^{\frac{2.67}{1.34}}$ . Exponent $= 1.9925$ , base $= 0.7491$ , $0.749 1^{1.9925} = 0.5626$ ; $1.67 = 1.2923$ ; product $Γ = 0.7272$ . Why this step? Large $γ$ gives a modest exponent but a base well below 1 — again the evaluation, not intuition, settles the size.
Compare: $Γ (1.67) > Γ (1.1)$ , so the higher- $γ$ gas passes about $\frac{0.7272}{0.6279} = 1.16 \times$ more mass through the same throat at the same $p_{0}, T_{0}$ . Why this step? $\overset{m}{˙} \propto Γ$ when everything else is held fixed, so the coefficient ratio is the mass-flow ratio.

Verify: See the curve of $Γ$ in s03 — it rises monotonically with $γ$ over this range, consistent with $0.628 < 0.727$ ✔. This links to Isentropic Flow Relations and Choked Flow and Sonic Conditions.

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

Ex‑2's engine runs choked at $44.3 kg/s$ . The rocket climbs, so outside (back) pressure drops from $1 \times 1 0^{5}$ Pa to $2 \times 1 0^{4}$ Pa. Chamber $p_{0}, T_{0}$ held constant. What is the new $\overset{m}{˙}$ ?

Forecast: a bigger pressure difference — surely more gas escapes? Guess before reading.

Check the regime. With $p_{0} = 4 \times 1 0^{6}$ , both back pressures give $p_{b} / p_{0} = 0.025$ and $0.005$ , far below the critical $0.546$ for $γ = 1.3$ . So the throat stays choked throughout. Why this step? The formula we pick depends on the regime; both cases are below critical, so the choked formula applies to both.
The boxed choked formula contains $p_{0}, T_{0}, γ, R, A^{*}$ — and no back pressure $p_{b}$ . Why this step? At $M = 1$ the throat cannot "hear" downstream changes (see s02), so $p_{b}$ is absent from the formula and cannot alter $\overset{m}{˙}$ .
Therefore $\overset{m}{˙}$ is unchanged: $\overset{m}{˙} = 44.3 kg/s$ as before.

Verify: Substituting the same $p_{0}, T_{0}, γ, R, A^{*}$ reproduces $44.3 kg/s$ ✔. The answer literally does not move.

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

Same converging nozzle, but now a valve raises the back pressure to $p_{b} = 3.6 \times 1 0^{6}$ Pa while $p_{0} = 4 \times 1 0^{6}$ Pa ( $γ = 1.3$ , $R = 300$ , $T_{0} = 2500$ , throat $A^{*} = 0.015$ ). Is it still choked? Estimate $\overset{m}{˙}$ .

Forecast: with $p_{b}$ nearly as high as $p_{0}$ , is the throat still at $M = 1$ ?

Critical ratio test: $p_{b} / p_{0} = 3.6/4.0 = 0.90$ . Critical for $γ = 1.3$ is $(\frac{2}{2.3})^{\frac{1.3}{0.3}} = 0.869 6^{4.333} = 0.546$ . Why this step? The single number $p_{b} / p_{0}$ decides the regime; $0.90 > 0.546$ means the throat is subsonic — un-choked.
Un-choked → use isentropic subsonic relations to get the throat Mach number from the static pressure ratio. When not choked, the throat's static pressure $p$ equals the back-pressure $p_{b}$ , and $\frac{p _{0}}{p} = (1 + \frac{γ - 1}{2} M^{2})^{\frac{γ}{γ - 1}}$ . Solving $0.90 = (1 + 0.15 M^{2})^{- 4.333}$ gives $M \approx 0.40$ . Why this step? Here $p$ is the static pressure defined in our vocabulary box; with the throat un-choked it must match $p_{b}$ , and the isentropic $p_{0} / p$ relation converts that pressure ratio into a Mach number — see Isentropic Flow Relations.
Mass flow via the general $\overset{m}{˙} (M)$ formula (restated here so you needn't flip back) with $M = 0.40$ : $\overset{m}{˙} = \frac{A ^{*} p _{0} M γ / ( R T _{0} )}{( 1 + \frac{γ - 1}{2} M ^{2} ) ^{\frac{γ + 1}{2 ( γ - 1 )}}} = \frac{0.015 \times 4 \times 1 0 ^{6} \times 0.40 \times 1.3/750000}{( 1 + 0.024 ) ^{3.833}}$ Why this step? Un-choked flow keeps the general $\overset{m}{˙} (M)$ formula (the top boxed workhorse), not the $M = 1$ special case, so we plug the actual $M$ .
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 $= 1.02 4^{3.833} = 1.095$ ; $\overset{m}{˙} \approx 28.9 kg/s$ . Why this step? Straight evaluation of each factor; the small $M$ makes $\overset{m}{˙}$ well below the choked $44.3$ .

Verify: $28.9 < 44.3$ ✔ — raising back-pressure above critical reduces $\overset{m}{˙}$ below the choked maximum, the mirror image of Ex 6. Consistent with $\overset{m}{˙}$ peaking at $M = 1$ .

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

An engine must swallow $\overset{m}{˙} = 30 kg/s$ . Given $p_{0} = 3 \times 1 0^{6}$ Pa, $T_{0} = 2000$ K, $γ = 1.25$ , $R = 320 J/(kg\cdotK)$ , choked. Find the required throat area $A^{*}$ .

Forecast: rocket throats are hand-sized. Guess: is $A^{*}$ closer to $0.001$ or $0.1 m^{2}$ ?

Rearrange the boxed formula for area: $A^{*} = \frac{m ˙ R T _{0}}{p _{0} Γ ( γ )}$ Why this step? $A^{*}$ appears linearly, so we isolate it by dividing everything else across — the fastest route to the unknown.
Coefficient: $Γ (1.25) = 1.25 (\frac{2}{2.25})^{\frac{2.25}{0.5}}$ . Exponent $= 4.5$ , base $= 0.8889$ , $0.888 9^{4.5} = 0.5906$ ; $1.25 = 1.1180$ ; $Γ = 0.6603$ . Why this step? $Γ$ is needed in the denominator; it depends only on $γ$ so we get it first (read off s03).
Numerator: $\overset{m}{˙} R T_{0} = 30 320 \times 2000 = 30640000$ . Now $640000 = 800$ exactly (since $80 0^{2} = 640000$ ), so numerator $= 30 \times 800 = 24000$ . Why this step? The root is exact here, so the arithmetic stays clean; this is the mass-flow times the thermal factor.
Denominator: $p_{0} Γ = 3 \times 1 0^{6} \times 0.6603 = 1.9809 \times 1 0^{6}$ . Why this step? This is the "driving strength per unit area"; dividing the numerator by it yields the area.
Divide: $A^{*} = 24000/ (1.9809 \times 1 0^{6}) = 0.01212 m^{2}$ .

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

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

A designer wants a small thruster with thrust $F = 2000 N$ . She knows the effective exhaust velocity is $c = 2500 m/s$ and the chamber is $p_{0} = 2 \times 1 0^{6}$ Pa, $T_{0} = 1800$ K, $γ = 1.3$ , $R = 300 J/(kg\cdotK)$ , choked. Find the needed $\overset{m}{˙}$ and throat area.

Forecast: thrust $= \overset{m}{˙} \cdot c$ (roughly). So $\overset{m}{˙}$ near $0.8$ or $8 kg/s$ ?

Mass flow from thrust: $F = \overset{m}{˙} c \Rightarrow \overset{m}{˙} = F / c = 2000/2500 = 0.8 kg/s$ . Why this step? Thrust is momentum expelled per second; dividing by exhaust speed recovers the mass rate. (See Specific Impulse for the $c$ – $I_{s p}$ link.)
Coefficient: $Γ (1.3) = 0.6400$ (computed in Ex 2). Why this step? Same $γ$ as Ex 2, so we reuse the coefficient rather than recompute.
Solve for area using Ex‑7's inverse form: $A^{*} = \frac{m ˙ R T _{0}}{p _{0} Γ}$ . First $R T_{0} = 300 \times 1800 = 540000 = 734.85$ . Why this step? The thermal factor $R T_{0}$ is needed in the numerator; we evaluate the root before dividing.
$A^{*} = \frac{0.8 \times 734.85}{2 \times 1 0 ^{6} \times 0.6400} = \frac{587.88}{1.280 \times 1 0 ^{6}} = 4.593 \times 1 0^{- 4} m^{2}$ . Why this step? Straight substitution into the rearranged area formula.

Verify: $\overset{m}{˙} = 0.8 kg/s$ sits in the small bracket ✔. Forward check on area: $\frac{4.593 \times 1 0 ^{- 4} \times 2 \times 1 0 ^{6}}{734.85} \times 0.6400 = 1.2500 \times 0.6400 = 0.800 kg/s$ ✔. Throat $\approx 4.6 cm^{2}$ — a genuinely small thruster.

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

Given molar mass $M_{molar} = 0.022 kg/mol$ (not $R$ directly), universal $R = 8.314 J/(mol\cdotK)$ , $p_{0} = 1500 kPa$ , $T_{0} = 2400$ K, $γ = 1.2$ , $A^{*} = 0.008 m^{2}$ , choked. Find $\overset{m}{˙}$ .

Forecast: the trap is units. Two conversions hide here — spot them before solving.

Get $R$ : $R = R / M_{molar} = 8.314/0.022 = 377.9 J/(kg\cdotK)$ . Why this step? The formula needs the specific gas constant (per kg), not the universal one (per mol).
Convert pressure: $p_{0} = 1500 kPa = 1.5 \times 1 0^{6} Pa$ . Why this step? SI formulas demand pascals; forgetting this scales the answer by 1000.
Coefficient: $Γ (1.2) = 1.2 (\frac{2}{2.2})^{\frac{2.2}{0.4}}$ . Exponent $= 5.5$ , base $= 0.90909$ , $0.9090 9^{5.5} = 0.5843$ ; $1.2 = 1.0954$ ; $Γ = 0.6401$ . Why this step? $Γ$ depends only on $γ$ ; computing it first keeps the final multiply clean.
Driving term: $\frac{A ^{*} p _{0}}{R T _{0}} = \frac{0.008 \times 1.5 \times 1 0 ^{6}}{377.9 \times 2400}$ . Denominator $906960 = 952.3$ (since $952. 3^{2} \approx 906960$ ). Numerator $= 12000$ . Ratio $= 12.60$ . Why this step? We take the root of $R T_{0}$ to form the thermal factor, then divide the area–pressure product by it to get the driving strength.
Combine: $\overset{m}{˙} = 12.60 \times 0.6401 = 8.07 kg/s$ . Why this step? $\overset{m}{˙}$ = driving term $\times$ coefficient, the compact choked formula.

Verify: Units clean (Pa, J/kg·K) ✔. Answer $8.07 kg/s$ . If a student forgot to convert kPa, they'd get $0.008 kg/s$ — 1000× too small, the classic exam booby-trap. Links to Tsiolkovsky Rocket Equation and Nozzle Area Ratio and Expansion for downstream use.

Recall Self-test: which cell is which?

How do you decide if a flow is choked? ::: Compare $p_{b} / p_{0}$ to $(2/ (γ + 1))^{γ / (γ - 1)}$ ; below it → choked, above → subsonic (C6/C6b). What happens at exact equality $p_{b} / p_{0} = (2/ (γ + 1))^{γ / (γ - 1)}$ ? ::: The throat sits exactly at $M = 1$ — the onset of choking; both formulas agree there. A closed valve gives what $\overset{m}{˙}$ ? ::: Zero (C4 — degenerate zero area). Can $\overset{m}{˙}$ be negative, and what would it mean? ::: Yes, if $v < 0$ (backflow, gas sucked inward); in steady thrust $v > 0$ so $\overset{m}{˙} > 0$ . Lowering back-pressure on a choked engine changes $\overset{m}{˙}$ by how much? ::: Not at all (C6 — choked flow ignores back-pressure). Raising back-pressure above the critical ratio does what to $\overset{m}{˙}$ ? ::: Un-chokes the throat and reduces $\overset{m}{˙}$ below the choked maximum (C6b). To find $A^{*}$ from a target $\overset{m}{˙}$ , rearrange to what? ::: $A^{*} = \overset{m}{˙} R T_{0} / (p_{0} Γ (γ))$ (C7 — inverse problem). Converting molar mass to specific $R$ uses which formula? ::: $R = R / M_{molar}$ (C9 — hidden units). Does higher $γ$ raise or lower the choking coefficient over $1.1$ – $1.67$ ? ::: Raise it (C5 — $Γ$ increases with $γ$ ). What is the difference between static $p$ and stagnation $p_{0}$ ? ::: $p$ is the pressure at the moving-gas station; $p_{0}$ is what it would be brought to rest — $p < p_{0}$ when gas moves.