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

Step 1 (WHAT): Compute the back-pressure ratio $p_b/p_0=600/800=0.750$. Step 2 (WHY): Compare to the critical ratio $0.5283$. The throat can only reach $M=1$ if the exit is asked to drop to or below $p^{\ast }/p_0$. Step 3: $0.750>0.5283$, so the required exit ratio is above the threshold. The flow stays subsonic everywhere; the exit pressure simply matches $p_b$. Answer: Not choked. Lowering $p_b$ further would still increase $\dot{m}$ (until we hit $0.528$).

Step 1: $p_b/p_0=300/800=0.375$. Step 2: $0.375<0.5283$: the back pressure is below critical. A converging nozzle cannot deliver an exit pressure below $p^{\ast }$ while remaining valid subsonic flow — the throat locks at $M=1$. Step 3: So the exit pressure is not $p_b$; it is locked at $p^{\ast }=0.5283\times 800\text{kPa}=422.6\text{kPa}.$ Answer: Choked. Exit pressure $=422.6\text{kPa}$ (higher than $p_b$); the remaining expansion to $300\text{kPa}$ happens outside the nozzle as an under-expanded jet.

Step 1 — the bracket (WHAT/WHY): exponent $\frac{\gamma +1}{2(\gamma -1)}=\frac{2.4}{0.8}=3$, and $\frac{2}{\gamma +1}=0.8333$. So the bracket is $0.8333^3=0.5787$. Step 2 — the square-root factor: $\sqrt{\frac{\gamma }{R{T}_0}}=\sqrt{\frac{1.4}{287\times 290}}=\sqrt{1.6816\times 10^{-5}}=4.101\times 10^{-3}.$ Step 3 — assemble: ${\dot{m}}_{\max }=A^{\ast }{p}_0\times (4.101\times 10^{-3})\times \mathrm{0.5787.}$ With $A^{\ast }{p}_0=1.5\times 10^{-4}\times 8\times 10^5=120$: ${\dot{m}}_{\max }=120\times 4.101\times 10^{-3}\times 0.5787=0.285\text{kg/s}.$ Answer: ${\dot{m}}_{\max }\approx 0.285\text{kg/s}$.

Step 1: It is already choked ($0.375<0.528$), so the throat is at $M=1$ and stays there. Step 2 (WHY): ${\dot{m}}_{\max }$ depends only on $p_0,T_0,A^{\ast },\gamma ,R$ — the formula contains no $p_b$. Downstream signals cannot travel upstream past a sonic throat, so the throat never "hears" the new $p_b$. Answer: $\dot{m}$ is unchanged: still $0.285\text{kg/s}$. $\Delta \dot{m}=0$.

Step 1 — Predict (WHY): In the formula the only $T_0$ is ${\dot{m}}_{\max }\propto 1/\sqrt{T_0}$. Hotter reservoir ⇒ less dense gas at the throat; the faster sound speed ($\propto \sqrt{T}$) doesn't compensate. So mass flow drops. Step 2 — Compute: $\frac{{\dot{m}}_{\max }(580)}{{\dot{m}}_{\max }(290)}=\sqrt{\frac{290}{580}}=\sqrt{0.5}=\mathrm{0.7071.}$ Answer: Mass flow falls to $70.7\%$ of its value, i.e. $0.285\times 0.7071=0.202\text{kg/s}$. Doubling $T_0$ cuts ${\dot{m}}_{\max }$ by $\approx 29\%$.

Step 1 — choked? $p_b/p_0=200/500=0.400<0.5283$ ⇒ choked ⇒ exit is at $M=1$? Careful: for a purely converging nozzle, yes — the maximum Mach it can reach at its own exit (the throat/exit are the same section) is $M=1$. Step 2 (WHY $M=1$ and not more): A converging duct can accelerate subsonic flow only up to $M=1$; to exceed it you need a diverging section (see Converging–Diverging (de Laval) Nozzle). The exit pressure locks at $p^{\ast }=0.5283\times 500=264\text{kPa}$. Answer: $M_{\text{exit}}=1$. (The gas keeps expanding from $264\text{kPa}$ down to $200\text{kPa}$ outside the nozzle.)

Step 1 (a) — mass flow is set by the throat: convert $A^{\ast }=4{\text{cm}}^2=4\times 10^{-4}{\text{m}}^2$, $p_0=10^6\text{Pa}$. Bracket $=0.8333^3=0.5787$; $\sqrt{\gamma /(R{T}_0)}=\sqrt{1.4/(287\times 350)}=\sqrt{1.3937\times 10^{-5}}=3.733\times 10^{-3}$. ${\dot{m}}_{\max }=4\times 10^{-4}\times 10^6\times 3.733\times 10^{-3}\times 0.5787=0.864\text{kg/s}.$ Step 2 (b) — continuity (WHY): Steady flow means $\dot{m}=\rho AV$ is identical at every cross-section — the same $0.864\text{kg/s}$ passes the throat and the exit. The larger exit area is balanced by faster, less dense supersonic gas there (the Area–Mach Relation (A/A*) fixes what that exit Mach is). The mass flow number is decided entirely at the throat, because that is where $M=1$. Answer: ${\dot{m}}_{\max }\approx 0.864\text{kg/s}$ through both sections.

Step 1 (WHY combine both scalings): ${\dot{m}}_{\max }\propto p_0/\sqrt{T_0}$ (all other factors identical). Form the ratio: $\frac{{\dot{m}}_B}{{\dot{m}}_A}=\frac{p_{0,B}}{p_{0,A}}\sqrt{\frac{T_{0,A}}{T_{0,B}}}=\frac{900}{600}\sqrt{\frac{300}{675}}.$ Step 2: $\frac{900}{600}=1.5$; $\sqrt{300/675}=\sqrt{0.4444}=0.6667$. Product $=1.5\times 0.6667=1.000$. Answer: They pass the same mass flow (${\dot{m}}_B/{\dot{m}}_A=1.000$). B's higher pressure is exactly cancelled by its higher temperature.

Step 1 — onset of choking (WHAT): choking begins the instant $p_b/p_0=0.5283$, i.e. $p_{b,\text{choke}}=0.5283\times 400=211.3\text{kPa}.$ Step 2 — the two regimes (WHY):

• For $p_b$ from $400$ down to $211.3\text{kPa}$: subsonic, unchoked. $\dot{m}$ rises as $p_b$ falls (bigger pressure drop ⇒ more flow), climbing from $0$ toward ${\dot{m}}_{\max }$.
• For $p_b$ from $211.3$ down to $50\text{kPa}$: choked. $\dot{m}$ is flat at ${\dot{m}}_{\max }$ — the throat is deaf to $p_b$. Step 3 — ${\dot{m}}_{\max }$: bracket $=0.5787$; $\sqrt{1.4/(287\times 320)}=\sqrt{1.5244\times 10^{-5}}=3.905\times 10^{-3}$. ${\dot{m}}_{\max }=3\times 10^{-4}\times 4\times 10^5\times 3.905\times 10^{-3}\times 0.5787=0.2712\text{kg/s}.$ Answer: Choking starts at $p_b=211.3\text{kPa}$; ${\dot{m}}_{\max }\approx 0.271\text{kg/s}$. The $\dot{m}$–$p_b$ curve rises then goes flat at the knee — see the figure below.

Step 1 — invert the formula (WHY): solve ${\dot{m}}_{\max }=A^{\ast }{p}_0\sqrt{\gamma /(R{T}_0)}(0.5787)$ for $A^{\ast }$: $A^{\ast }=\frac{{\dot{m}}_{\max }}{p_0\sqrt{\gamma /(R{T}_0)}(0.5787)}.$ Step 2 — numbers: $\sqrt{1.4/(287\times 400)}=\sqrt{1.2195\times 10^{-5}}=3.492\times 10^{-3}$. Denominator $=6\times 10^5\times 3.492\times 10^{-3}\times 0.5787=1212.5$. $A^{\ast }=\frac{0.500}{1212.5}=4.124\times 10^{-4}{\text{m}}^2.$ Answer: $A^{\ast }\approx 4.12\times 10^{-4}{\text{m}}^2$ ($\approx 4.12{\text{cm}}^2$, i.e. a throat of diameter $\approx 2.29\text{cm}$).

Step 1 (WHY the claim is wrong): the throat is upstream of the shock and remains at $M=1$; a downstream shock cannot send information back through a sonic throat. Mass flow is set at the throat. Step 2: A normal shock is adiabatic — it conserves $\dot{m}$ and $T_0$ across itself; it only raises entropy (drops $p_0$ downstream of the shock). It does not touch the throat. Answer: False. $\dot{m}$ is unchanged. The shock changes downstream pressure/velocity and reduces stagnation pressure past it, but the choked throat fixes $\dot{m}$ regardless.

Onset-of-choking back pressure for $p_0=400$ kPa air? ::: $0.5283\times 400=211$ kPa. Does a downstream normal shock change $\dot{m}$ in a choked nozzle? ::: No — throat stays sonic. Which area sets ${\dot{m}}_{\max }$ in a C-D nozzle? ::: The throat area $A^{\ast }$. Double $T_0$: what happens to ${\dot{m}}_{\max }$? ::: Falls by factor $1/\sqrt{2}=0.707$. The choked-flow bracket exponent for air? ::: $\frac{\gamma +1}{2(\gamma -1)}=3$.