Converging-diverging (de Laval) nozzle — subsonic, supersonic flow: Exercises

Throughout, unless a problem says otherwise, take air with $γ = 1.4$ .

Definition Notation used on this page (defined once, here)

$V$ = the speed of the gas at a point (m/s).
$a$ = the local speed of sound, the speed a tiny pressure ripple travels, $a = γ R T$ . For air $R = 287 \frac{J}{kg K}$ (the specific gas constant) and $γ = c_{p} / c_{v} = 1.4$ .
$M = \frac{V}{a}$ = the Mach number: how many times faster than sound the gas moves. $M < 1$ subsonic, $M = 1$ sonic, $M > 1$ supersonic. See Mach Number.
$A$ = the tube's cross-sectional area at a point.
$A^{*}$ = the throat / sonic area — the area where $M = 1$ . A star superscript always means "the value at the sonic point."
Subscript $0$ = stagnation value: what you'd measure in the still reservoir where $V = 0$ (so $T_{0}$ , $p_{0}$ are reservoir temperature and pressure).
$p^{*}, T^{*}, ρ^{*}, a^{*}$ = pressure, temperature, density, sound speed at the throat ( $M = 1$ ).
$p_{b}$ = back pressure, the pressure of the surroundings the gas exhausts into.

Level 1 — Recognition

(Can you name the rule and read its sign?)

L1.1 In the master law $\frac{d A}{A} = (M^{2} - 1) \frac{d V}{V}$ , a gas is subsonic ( $M = 0.5$ ) and we want it to speed up ( $d V > 0$ ). Must the area grow or shrink? What everyday device does this match?

Recall Solution L1.1

Compute the sign of the bracket: $M^{2} - 1 = 0.25 - 1 = - 0.75 < 0$ . For $d V > 0$ (speed up), the right side is negative, so $d A < 0$ → the area must shrink (converging). This matches a garden hose / venturi: pinch it and the water speeds up.

L1.2 A de Laval nozzle runs at design condition. Where is $M = 1$ located, and why can it live only there?

Recall Solution L1.2

$M = 1$ lives only at the throat (the minimum area). Reason: set $M = 1$ in the master law → $(M^{2} - 1) = 0$ , so $\frac{d A}{A} = 0$ . A zero fractional area change means the area is momentarily flat — a minimum — which is exactly the throat.

L1.3 True or false: "For any area ratio $A / A^{*} > 1$ there is exactly one possible Mach number." Explain.

Recall Solution L1.3

False. The $A / A^{*}$ curve dips to a minimum of $1$ at $M = 1$ and rises on both sides. So each $A / A^{*} > 1$ is hit by two Mach numbers: one subsonic ( $M < 1$ ) and one supersonic ( $M > 1$ ). Which one nature picks depends on the back pressure. See Isentropic Flow Relations.

Level 2 — Application

(Plug into the isentropic formulas correctly.)

L2.1 Air flows at $M = 2.0$ . Find the temperature ratio $T_{0} / T$ and the pressure ratio $p_{0} / p$ .

Recall Solution L2.1

WHAT: use the stagnation ratios. WHY: they connect the reservoir (still) state to the moving state via energy + isentropic conservation. $\frac{T _{0}}{T} = 1 + \frac{γ - 1}{2} M^{2} = 1 + 0.2 \times 4 = 1.8$ $\frac{p _{0}}{p} = (1.8)^{\frac{1.4}{0.4}} = 1. 8^{3.5} \approx 7.82$ So the static pressure is about $1/7.82 \approx 0.128$ of reservoir pressure.

L2.2 A reservoir holds air at $p_{0} = 8$ bar. The nozzle is choked. What is the throat (critical) pressure $p^{*}$ ?

Recall Solution L2.2

WHY this formula: "choked" means the throat has reached exactly $M = 1$ . So we just take the general pressure ratio $\frac{p _{0}}{p} = (1 + \frac{γ - 1}{2} M^{2})^{γ / (γ - 1)}$ and set $M = 1$ — that special case is the critical (starred) pressure. It tells us the particular throat pressure at which choking begins. At $M = 1$ : $\frac{p ^{*}}{p _{0}} = (1 + \frac{γ - 1}{2})^{- \frac{γ}{γ - 1}} = (\frac{2}{γ + 1})^{\frac{γ}{γ - 1}} = (\frac{2}{2.4})^{3.5} \approx 0.5283.$ Thus $p^{*} = 0.5283 \times 8 \approx 4.23$ bar. See Choked Flow & Mass Flow Limit.

L2.3 Reservoir temperature $T_{0} = 600$ K, air. Find the throat temperature $T^{*}$ and the throat speed of sound $a^{*}$ .

Recall Solution L2.3

WHY invert the ratio: the formula gives $T_{0} / T$ (still-reservoir over local), but we know $T_{0}$ and want the local throat $T^{*}$ . So we rearrange to $T^{*} = T_{0} / (T_{0} / T^{*})$ — dividing the known reservoir temperature by the ratio evaluated at $M = 1$ . At $M = 1$ : $\frac{T _{0}}{T ^{*}} = 1 + 0.2 \times 1 = 1.2$ , so $T^{*} = 600/1.2 = 500$ K. Speed of sound there: $a^{*} = γ R T^{*} = 1.4 \times 287 \times 500 = 200900 \approx 448 m/s$ . Since $M = 1$ at the throat, the flow speed equals $a^{*} \approx 448$ m/s. See Speed of Sound in a Gas.

Level 3 — Analysis

(Compare competing effects; pick the right branch.)

L3.1 A nozzle has exit area $A_{e} = 4 A^{*}$ , air, supersonic flow. Find the exit Mach number $M_{e}$ and the ratio $p_{0} / p_{e}$ .

Recall Solution L3.1

WHAT: solve $A / A^{*} = 4$ for $M$ . WHY the supersonic root: past a choked throat the diverging section satisfies $d A > 0$ with $M > 1$ , so we take the supersonic root. Solving $\frac{1}{M} [\frac{2}{2.4} (1 + 0.2 M^{2})]^{3} = 4$ numerically gives $M_{e} \approx 2.94$ . Then $\frac{p _{0}}{p _{e}} = (1 + 0.2 \times 2.9 4^{2})^{3.5} = (1 + 1.729)^{3.5} \approx 33.0$ , so $p_{e} \approx p_{0} /33$ .

L3.2 Look at Figure 1. For the same area ratio $A / A^{*} = 2$ , list the two Mach numbers and state which physical situation selects each.

Figure — Converging-diverging (de Laval) nozzle — subsonic, supersonic flow

Recall Solution L3.2

Solving $A / A^{*} = 2$ gives a subsonic root $M \approx 0.31$ and a supersonic root $M \approx 2.20$ .

The subsonic root ( $0.31$ ) occurs when the back pressure is high: the diverging part acts as a diffuser and slows the flow (venturi mode).
The supersonic root ( $2.20$ ) occurs at (or below) design back pressure: the diverging part accelerates an already-sonic throat flow. Which one nature picks is set by the exit back pressure $p_{b}$ .

L3.3 A converging-only nozzle is fed from a reservoir and the back pressure is dropped lower and lower. What is the maximum exit Mach number achievable, and why can it never exceed it?

Recall Solution L3.3

Maximum is $M_{e} = 1$ . In a converging-only duct the smallest area is the exit itself. The master law forbids crossing $M = 1$ except where $d A = 0$ . Since there is no diverging ( $d A > 0$ ) section, the flow can reach sonic at best at the exit; dropping $p_{b}$ further merely chokes it — the exit stays at $M = 1$ and $\overset{m}{˙}$ locks at its maximum. See Choked Flow & Mass Flow Limit.

Level 4 — Synthesis

(Combine mass flow, shocks and geometry.)

L4.1 Reservoir: $p_{0} = 10$ bar, $T_{0} = 500$ K, air. Throat area $A^{*} = 5 cm^{2} = 5 \times 1 0^{- 4} m^{2}$ . Nozzle is choked. Find the maximum mass flow rate $\overset{m}{˙}_{m a x} = ρ^{*} A^{*} a^{*}$ .

Recall Solution L4.1

Step 1 — throat temperature: $T^{*} = T_{0} /1.2 = 500/1.2 = 416.7$ K. Step 2 — throat pressure: $p^{*} = 0.5283 \times 10 \times 1 0^{5} = 5.283 \times 1 0^{5}$ Pa. Step 3 — throat density (ideal gas $ρ = p / R T$ ): $ρ^{*} = \frac{5.283 \times 1 0 ^{5}}{287 \times 416.7} = 4.418 kg/m^{3}$ . Step 4 — throat sound speed: $a^{*} = 1.4 \times 287 \times 416.7 = 409.2$ m/s (this equals $V$ since $M = 1$ ). Step 5 — combine: $\overset{m}{˙}_{m a x} = ρ^{*} A^{*} a^{*} = 4.418 \times 5 \times 1 0^{- 4} \times 409.2 \approx 0.904 kg/s$ . Lowering $p_{b}$ below choke does not raise this. See Conservation of Mass (Continuity).

L4.2 Air enters a normal shock at $M_{1} = 2.0$ (sitting in the diverging section). Using the normal-shock relations $M_{2}^{2} = \frac{1 + \frac{γ - 1}{2} M _{1}^{2}}{γ M _{1}^{2} - \frac{γ - 1}{2}}, \frac{p _{2}}{p _{1}} = 1 + \frac{2 γ}{γ + 1} (M_{1}^{2} - 1),$ find the downstream Mach number $M_{2}$ and static pressure jump $p_{2} / p_{1}$ .

Recall Solution L4.2

WHY a shock appears: when back pressure is too high for full supersonic exit but too low for pure subsonic flow, the gas must jump abruptly back to subsonic — a normal shock. $M_{2}^{2} = \frac{1 + 0.2 \times 4}{1.4 \times 4 - 0.2} = \frac{1.8}{5.4} = 0.3333 \Rightarrow M_{2} \approx 0.577.$ $\frac{p _{2}}{p _{1}} = 1 + \frac{2 \times 1.4}{2.4} (4 - 1) = 1 + 1.1667 \times 3 = 4.50.$ So the flow drops to $M_{2} \approx 0.58$ and static pressure rises $4.5 \times$ across the shock.

Edge-case warning — stagnation pressure is NOT conserved through a shock. A shock is irreversible (entropy rises), so while the static pressure jumps up, the stagnation pressure drops: $p_{0, 2} < p_{0, 1}$ . For $M_{1} = 2$ the ratio is $p_{0, 2} / p_{0, 1} \approx 0.721$ — about a $28%$ loss. This matters when you chain relations along one flow path: use the upstream $p_{0}$ only up to the shock; after it, switch to the reduced $p_{0, 2}$ . (The static-only ratios $T, p, ρ$ across the shock are fine; it is the reservoir/stagnation reference that resets.)

Level 5 — Mastery

(Design-level reasoning across the whole regime.)

L5.1 A rocket nozzle is designed so the exit exactly matches ambient pressure $p_{b} = p_{e} = 0.5$ bar with reservoir $p_{0} = 50$ bar, air. (a) Find the design exit Mach number. (b) Find the required area ratio $A_{e} / A^{*}$ .

Recall Solution L5.1

(a) WHY: "perfectly matched" means the isentropic exit static pressure equals ambient — no shocks, no expansion fans. $\frac{p _{0}}{p _{e}} = \frac{50}{0.5} = 100 = (1 + 0.2 M_{e}^{2})^{3.5} .$ Take both sides to the $1/3.5$ power: $10 0^{1/3.5} = 1 + 0.2 M_{e}^{2}$ . Now $10 0^{0.2857} = 3.727$ , so $0.2 M_{e}^{2} = 2.727$ , giving $M_{e}^{2} = 13.63$ and $M_{e} \approx 3.69$ . (b) WHY this branch and this form: the area–Mach relation $\frac{A}{A ^{*}} = \frac{1}{M} [\frac{2}{γ + 1} (1 + \frac{γ - 1}{2} M^{2})]^{\frac{γ + 1}{2 ( γ - 1 )}}$ compares any station's area to the sonic throat area; a rocket runs supersonic in the cone, so we feed the supersonic $M_{e} = 3.69$ (the larger root) into it. $\frac{A _{e}}{A ^{*}} = \frac{1}{3.69} [\frac{2}{2.4} (1 + 0.2 \times 13.63)]^{3} = \frac{1}{3.69} [0.8333 \times 3.727]^{3} = \frac{( 3.106 ) ^{3}}{3.69} \approx 8.12.$ So the exit is about $8.1 \times$ the throat area.

L5.2 Using Figure 2, explain the five back-pressure regimes of a de Laval nozzle in order of decreasing $p_{b}$ , and state what makes each transition happen.

Recall Solution L5.2

Reading the pressure-vs-position curves from top (high $p_{b}$ ) to bottom (low $p_{b}$ ):

Venturi (high $p_{b}$ ): fully subsonic; converging speeds up, diverging slows down. Throat not choked.
Choke onset: $p_{b}$ drops until $M = 1$ at throat. $\overset{m}{˙}$ now maxed — the transition is the throat reaching sonic.
Shock in diverging section: slightly lower $p_{b}$ ; flow goes supersonic then a normal shock jumps it back to subsonic to meet the still-high $p_{b}$ . Lowering $p_{b}$ pushes the shock toward the exit.
Design (perfectly expanded): $p_{e} = p_{b}$ ; smooth supersonic flow to the exit, no shocks.
Under-expanded ( $p_{b} <$ design): flow keeps expanding outside via expansion waves. The controlling knob is always $p_{b}$ ; the throat sonic condition (choke) is the pivot that separates purely-subsonic operation from all the supersonic regimes. See Steam Turbine Nozzles for the same picture in turbomachinery.

L5.3 A nozzle is choked with $\overset{m}{˙}_{m a x}$ fixed. The reservoir temperature $T_{0}$ is doubled (pressure $p_{0}$ held fixed, same throat area). Does $\overset{m}{˙}_{m a x}$ rise, fall, or stay the same? Justify with the scaling.

Recall Solution L5.3

$\overset{m}{˙}_{m a x} = ρ^{*} A^{*} a^{*}$ . With $p_{0}$ fixed, $p^{*}$ is fixed (it's just $0.528 p_{0}$ ), so $ρ^{*} = p^{*} / (R T^{*}) \propto 1/ T^{*} \propto 1/ T_{0}$ . Meanwhile $a^{*} = γ R T^{*} \propto T_{0}$ . Product: $\overset{m}{˙}_{m a x} \propto \frac{1}{T _{0}} \times T_{0} = \frac{1}{T _{0}}$ . Doubling $T_{0}$ multiplies $\overset{m}{˙}_{m a x}$ by $1/ 2 \approx 0.707$ — the mass flow falls to about $70.7%$ . Hotter gas is thinner faster than it is faster, so less mass moves.

Recall One-line self-test

Which single quantity, once fixed by the throat reaching $M = 1$ , "freezes" the mass flow no matter how low the back pressure goes? ::: The mass flow rate $\overset{m}{˙}_{m a x} = ρ^{*} A^{*} a^{*}$ — the flow is choked.