Exercises — Converging nozzle — subsonic flow, Mach 1 at exit

WHAT to compute: the ratio $p_b/p_0$ and compare it to $0.528$. $\frac{p_b}{p_0}=\frac{400}{600}=0.667$ WHY compare to 0.528: choking happens only when the back pressure is low enough to pull the exit up to $M=1$. That threshold is the critical ratio $0.528$. Since $0.667>0.528$, the room is not low-pressure enough. Not choked — the exit is subsonic and $p_e=p_b=400\text{kPa}$.

$\frac{p_b}{p_0}=\frac{200}{600}=0.333<0.528\Rightarrow \text{choked.}$ WHAT the exit pressure is: when choked, the exit locks at the critical pressure, not at $p_b$: $p_e=p^{\ast }=0.528p_0=0.528\times 600=316.9\text{kPa}.$ The leftover drop from $316.9\to 200$ kPa happens outside the nozzle as expansion.

WHAT drives flow at all: gas moves from high pressure to low pressure. The push is the difference $p_0-p_b$.

• $p_b=p_0$: the difference is zero, so there is no forward flow — the mass flow rate $\dot{m}=0$ and $M=0$ everywhere. The gas just sits balanced.
• $p_b>p_0$: the room is now at higher pressure than the tank, so gas would tend to flow backward into the reservoir (a converging nozzle is not designed for this — in practice it is simply "no useful forward flow"). Our forward-flow formulas only apply for $p_b<p_0$. WHY this matters: the whole ratio logic ($0.528$, choked/unchoked) lives in the window $0\le p_b/p_0<1$. At the top edge $p_b/p_0=1$ the flow switches off; that is the boundary of the picture.

First, are we in the flow window and unchoked? $p_b/p_0=0.70$, which is below $1$ (flow exists) and above $0.528$ (subsonic, unchoked). So $p_e=p_b=350$ kPa and we invert the master relation. $\frac{p_0}{p_e}=\frac{500}{350}=1.4286={\left(1+0.2M^2\right)}^{3.5}.$ Take the $3.5$-th root (i.e. raise to $1/3.5$) to peel off the exponent: $1.4286^{1/3.5}=1.1073\Rightarrow 0.2M^2=0.1073\Rightarrow M^2=0.5366\Rightarrow M=\pm \mathrm{0.732.}$ WHY only the positive, subsonic root: solving for $M^2$ gives two square roots, $M=+0.732$ and $M=-0.732$. Negative $M$ means flow in the reverse direction, which is not our forward-blowing nozzle, so we discard it. (Algebraically there is also a supersonic branch to the full nozzle problem, but a converging-only duct can only accelerate subsonic flow up to $M=1$ — it can never reach the supersonic root. So the only physical answer here is the subsonic one.) $M=0.732$.

WHERE $T^{\ast }$ comes from (unpacked): at $M=1$ the stagnation relation gives $\frac{T_0}{T^{\ast }}=1+\frac{\gamma -1}{2}(1{)}^2=\frac{2+\gamma -1}{2}=\frac{\gamma +1}{2}\Rightarrow T^{\ast }=\frac{2}{\gamma +1}{T}_0.$ Plug numbers: $T^{\ast }=\frac{2}{2.4}\times 320=266.7\text{K}.$ WHY the exit speed = speed of sound: $M=1$ means flow speed equals local sound speed $a=\sqrt{\gamma RT}$. $V_e=a^{\ast }=\sqrt{1.4\times 287\times 266.7}=327.3\text{m/s}.$

WHAT we build: at the exit $M=1$, so we use the starred values. The temperature factor is the unpacked one: $T^{\ast }=\frac{2}{\gamma +1}{T}_0=\frac{2}{2.4}(300)=250\text{K},p^{\ast }=0.528(500)=264.1\text{kPa}.$ Density at exit from the ideal gas law $\rho =p/(RT)$: ${\rho }^{\ast }=\frac{264100}{287\times 250}=3.681{\text{kg/m}}^3.$ Exit speed $V_e=a^{\ast }=\sqrt{1.4\times 287\times 250}=316.9\text{m/s}$. WHY $\dot{m}=\rho AV$: mass through the exit each second = (density)(area)(speed). $\dot{m}={\rho }^{\ast }{A}_e{V}_e=3.681\times 1.0\times 10^{-3}\times 316.9=1.167\text{kg/s}.$

At $p_b=264\text{kPa}$ the ratio is exactly $0.528$ — the exit just reaches $M=1$. From here on the flow is choked.

• $p_b=150\text{kPa}$: ratio $0.30<0.528$ → choked, $\dot{m}$ stays at $1.167\text{kg/s}$.
• $p_b=50\text{kPa}$: ratio $0.10<0.528$ → still choked, $\dot{m}$ unchanged at $1.167\text{kg/s}$. WHAT the plot shows (coral curve, right of the dashed line): $\dot{m}$ is flat — a horizontal ceiling. The reservoir cannot "hear" the falling room pressure because signals cannot swim upstream through sonic flow. See Mass Flow Rate & Choking.

WHY a shortcut exists: in the choked mass-flow formula from the parent Step 4, $\dot{m}=A{p}_0\sqrt{\frac{\gamma }{R{T}_0}}\frac{M}{{\left(1+\frac{\gamma -1}{2}{M}^2\right)}^{\frac{\gamma +1}{2(\gamma -1)}}}$ at $M=1$ everything except $p_0$ and $T_0$ is fixed. So $\dot{m}\propto p_0$ at constant $T_0$. Doubling $p_0$ doubles $\dot{m}$: ${\dot{m}}_{\text{new}}=2\times 1.167=2.334\text{kg/s}.$ Verify from scratch: ${\rho }^{\ast }$ doubles (since $p^{\ast }=0.528p_0$ doubles and $T^{\ast }$ is unchanged), $V_e=a^{\ast }$ is unchanged (depends only on $T^{\ast }$), area unchanged $\Rightarrow \dot{m}$ doubles. ✓

From the same formula at $M=1$: $\dot{m}\propto p_0/\sqrt{T_0}$. Doubling $T_0$ divides $\dot{m}$ by $\sqrt{2}$: ${\dot{m}}_{\text{new}}=\frac{1.167}{\sqrt{2}}=0.825\text{kg/s}.$ WHY it drops even though the gas is faster: hotter gas means higher $a^{\ast }$ (faster exit), but also lower density ${\rho }^{\ast }=p^{\ast }/(R{T}^{\ast })$, and density loses. Net: $\dot{m}\propto 1/\sqrt{T_0}$.

(a) $p_b/p_0=300/800=0.375$, which is below $1$ (flow exists) and below $0.528$ → choked. (b) Choked, so exit is at $M=1$: $p_e=p^{\ast }=0.528(800)=422.5\text{kPa}.$ Unpack the temperature factor as before: at $M=1$, $\frac{T_0}{T^{\ast }}=1+\frac{\gamma -1}{2}=\frac{\gamma +1}{2}$, so $T_e=T^{\ast }=\frac{2}{\gamma +1}{T}_0=\frac{2}{2.4}(500)=416.7\text{K}.$ $V_e=a^{\ast }=\sqrt{1.4\times 287\times 416.7}=409.2\text{m/s}.$ (c) ${\rho }^{\ast }=p^{\ast }/(R{T}^{\ast })=422500/(287\times 416.7)=3.533{\text{kg/m}}^3$. $\dot{m}={\rho }^{\ast }{A}_e{V}_e=3.533\times 8\times 10^{-4}\times 409.2=1.157\text{kg/s}.$ (d) In a converging-only duct, the mass-flow factor $M/(1+\frac{\gamma -1}{2}{M}^2{)}^{(\gamma +1)/[2(\gamma -1)]}$ peaks at $M=1$ — a shrinking area can only push subsonic flow up to Mach 1, never past. To get $M=1.5$ you must let the area increase again after the throat: a Converging-Diverging (de Laval) Nozzle. Only there does the diverging section keep accelerating the already-sonic flow into the supersonic regime. (If $p_b$ is not low enough, that supersonic flow can jump back down via a normal shock — a separate story.)

WHY start from $p$ and $T$: we already have $p^{\ast }/p_0$ and $T^{\ast }/T_0$; use the ideal gas law $\rho =p/(RT)$ to combine them: $\frac{{\rho }^{\ast }}{{\rho }_0}=\frac{p^{\ast }/p_0}{T^{\ast }/T_0}=\frac{{\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma }{\gamma -1}}}{\frac{2}{\gamma +1}}={\left(\frac{2}{\gamma +1}\right)}^{\frac{\gamma }{\gamma -1}-1}={\left(\frac{2}{\gamma +1}\right)}^{\frac{1}{\gamma -1}}.$ For air, $\gamma =1.4$: ${\left(\frac{2}{2.4}\right)}^{1/0.4}=0.8333^{2.5}=0.634$. Matches the parent's "sanity anchor" ${\rho }^{\ast }=0.634{\rho }_0$. ✓

Reservoir $\to$ exit at choke: $p^{\ast }=0.528p_0$, $T^{\ast }=0.833T_0$, ${\rho }^{\ast }=0.634{\rho }_0$. Everything in this page is built from these plus $\dot{m}=\rho AV$.