A nozzle is a tube that converts pressure/thermal energy into kinetic energy (speed). For a slow (subsonic) gas, you speed it up by squeezing it (converging) . But once the gas hits the speed of sound, the rules flip: to go faster than sound you must let it expand (diverging) . So to take a gas from rest all the way to supersonic, you need both : a converging section, then a throat (where it hits Mach 1), then a diverging section.
Definition de Laval nozzleA duct whose cross-sectional area A A A first decreases (converging), reaches a minimum at the throat , then increases (diverging). It is the standard device to accelerate a compressible gas from subsonic to supersonic speed, used in rockets and steam turbines.
The state of the flow is tracked by the Mach number :
M = V a , a = γ R T M = \frac{V}{a}, \qquad a = \sqrt{\gamma R T} M = a V , a = γ R T V V V a a a γ = c p / c v \gamma=c_p/c_v γ = c p / c v R R R T T T
M < 1 M<1 M < 1 subsonic M = 1 M=1 M = 1 sonic (this happens only at the throat — called choked flow)M > 1 M>1 M > 1 supersonic
This is the heart of the topic. We DERIVE it from first principles for steady, 1-D, isentropic (no friction, no heat transfer) flow.
Intuition WHY does supersonic flow speed up when it expands?When M > 1 M>1 M > 1 faster than its area grows. To keep ρ A V \rho A V ρ A V V V V
From energy conservation h 0 = h + 1 2 V 2 h_0 = h + \tfrac12 V^2 h 0 = h + 2 1 V 2 h = c p T h=c_pT h = c p T a 2 = γ R T a^2=\gamma RT a 2 = γ R T stagnation-to-static ratios (subscript 0 0 0 V = 0 V=0 V = 0
Intuition The two-solution factFor any diverging area there are two possible flows: a subsonic decelerating one and a supersonic accelerating one. Which one Nature picks depends on the back pressure p b p_b p b
Intuition Operating regimes (lowerp b p_b p b
High p b p_b p b flow stays subsonic everywhere; converging speeds it up, diverging slows it down (acts like a venturi). Throat is not choked.p b p_b p b M = 1 M=1 M = 1 choked — m ˙ \dot m m ˙ p b p_b p b Slightly lower: flow goes supersonic just past throat, then a normal shock stands in the diverging part, jumping it back to subsonic.Design pressure: smooth supersonic flow all the way to the exit, perfectly matched — no shocks.p b p_b p b
Once M = 1 M=1 M = 1 maximum and fixed : m ˙ m a x = ρ ∗ A ∗ a ∗ \dot m_{max}=\rho^* A^* a^* m ˙ ma x = ρ ∗ A ∗ a ∗ p b p_b p b
Worked example Example 1 — Exit Mach from area ratioA nozzle has exit area A e = 4 A ∗ A_e = 4A^* A e = 4 A ∗ γ = 1.4 \gamma=1.4 γ = 1.4 M e M_e M e Solve A / A ∗ = 4 A/A^* = 4 A / A ∗ = 4 M e ≈ 2.94 M_e \approx 2.94 M e ≈ 2.94 Why this step? A / A ∗ > 1 A/A^*>1 A / A ∗ > 1 supersonic root because the diverging section past a choked throat accelerates the gas.
Then exit pressure: p 0 / p e = ( 1 + 0.2 ⋅ 2.94 2 ) 3.5 ≈ 33 p_0/p_e=(1+0.2\cdot2.94^2)^{3.5}\approx 33 p 0 / p e = ( 1 + 0.2 ⋅ 2.9 4 2 ) 3.5 ≈ 33 p e ≈ p 0 / 33 p_e\approx p_0/33 p e ≈ p 0 /33
Worked example Example 2 — Is it choked?Reservoir p 0 = 10 p_0=10 p 0 = 10 γ = 1.4 \gamma=1.4 γ = 1.4 p ∗ p 0 = ( 2 γ + 1 ) γ γ − 1 = ( 2 2.4 ) 3.5 ≈ 0.528 \frac{p^*}{p_0}=\left(\frac{2}{\gamma+1}\right)^{\frac{\gamma}{\gamma-1}}=\left(\frac{2}{2.4}\right)^{3.5}\approx 0.528 p 0 p ∗ = ( γ + 1 2 ) γ − 1 γ = ( 2.4 2 ) 3.5 ≈ 0.528 p ∗ ≈ 5.28 p^*\approx 5.28 p ∗ ≈ 5.28 Why? At M = 1 M=1 M = 1 ≤ 5.28 \le 5.28 ≤ 5.28 m ˙ \dot m m ˙ Throat temperature: T ∗ / T 0 = 1 / ( 1 + 0.2 ) = 0.833 T^*/T_0 = 1/(1+0.2)=0.833 T ∗ / T 0 = 1/ ( 1 + 0.2 ) = 0.833
Worked example Example 3 — Why a converging-only nozzle can't go supersonicA converging nozzle alone: the smallest area is the exit , where at best M = 1 M=1 M = 1 d A > 0 dA>0 d A > 0 M > 1 M>1 M > 1 p b p_b p b Why? The Area–Velocity law forbids passing through M = 1 M=1 M = 1 d A = 0 dA=0 d A = 0
Common mistake "Squeezing always speeds gas up."Why it feels right: for water/subsonic air (garden hose, your everyday experience) narrowing does speed flow. The fix: that's only true for M < 1 M<1 M < 1 ( M 2 − 1 ) (M^2-1) ( M 2 − 1 ) expanding speeds it up. Memorise the master equation, not the hose.
Common mistake "Lowering back pressure more increases mass flow more."Why it feels right: more suction sounds like more flow. The fix: once choked (M = 1 M=1 M = 1 m ˙ \dot m m ˙ locked at m ˙ m a x \dot m_{max} m ˙ ma x
Common mistake "Maximum speed is reached at the throat."Why it feels right: throat is narrowest, and narrow = fast for subsonic intuition. The fix: at the throat M = 1 M=1 M = 1 highest speed is at the exit , in the diverging section.
M = 1 M=1 M = 1 Why it feels right: pressure controls speed. The fix: d A A = ( M 2 − 1 ) d V V \frac{dA}{A}=(M^2-1)\frac{dV}{V} A d A = ( M 2 − 1 ) V d V d A = 0 dA=0 d A = 0 M = 1 M=1 M = 1 only at a local area minimum — the throat.
Recall Feynman: explain to a 12-year-oldImagine blowing through a straw to push a toy car. If you pinch the straw, the air shoots out faster — that's the "squeeze to speed up" trick, and it works while the air is slow. But there's a magic speed: the speed of sound. To make air go faster than sound , you have to do the opposite — let the tube get wider , not narrower! So a rocket nozzle is shaped like an hourglass: it pinches in the middle (the throat, where air hits exactly the speed of sound), then flares out so the air can blast out super fast. The pinched throat also acts like a one-way gate: once air there hits the sound barrier, no matter how hard you suck from the outside, the same amount of air flows through — it's "choked."
Mnemonic Remember the flip"Sub Squeezes, Super Spreads."
M < 1 M<1 M < 1 Squeeze to speed up. M > 1 M>1 M > 1 Spread (expand) to speed up. The handover happens at the throat (M = 1 M=1 M = 1 d A = 0 dA=0 d A = 0
What is the Area–Velocity relation for compressible flow? d A A = ( M 2 − 1 ) d V V \frac{dA}{A}=(M^2-1)\frac{dV}{V} A d A = ( M 2 − 1 ) V d V For subsonic flow, how do you accelerate the gas? Decrease area (converging section), since
M 2 − 1 < 0 M^2-1<0 M 2 − 1 < 0 .
For supersonic flow, how do you accelerate the gas? Increase area (diverging section), since
M 2 − 1 > 0 M^2-1>0 M 2 − 1 > 0 .
Where can the flow reach exactly Mach 1 in a nozzle? Only at the throat, where
d A = 0 dA=0 d A = 0 (area minimum).
What does "choked flow" mean? M = 1 M=1 M = 1 at the throat; mass flow rate is maximum and fixed regardless of further lowering back pressure.
Critical pressure ratio p ∗ / p 0 p^*/p_0 p ∗ / p 0 γ = 1.4 \gamma=1.4 γ = 1.4 ( 2 / ( γ + 1 ) ) γ / ( γ − 1 ) ≈ 0.528 (2/(\gamma+1))^{\gamma/(\gamma-1)}\approx 0.528 ( 2/ ( γ + 1 ) ) γ / ( γ − 1 ) ≈ 0.528 .
Stagnation temperature ratio T 0 / T T_0/T T 0 / T 1 + γ − 1 2 M 2 1+\frac{\gamma-1}{2}M^2 1 + 2 γ − 1 M 2 .
Stagnation pressure ratio p 0 / p p_0/p p 0 / p ( 1 + γ − 1 2 M 2 ) γ / ( γ − 1 ) (1+\frac{\gamma-1}{2}M^2)^{\gamma/(\gamma-1)} ( 1 + 2 γ − 1 M 2 ) γ / ( γ − 1 ) .
Why does a converging-only nozzle max out at M = 1 M=1 M = 1 Its minimum area is the exit; with no diverging part,
d A > 0 dA>0 d A > 0 needed for
M > 1 M>1 M > 1 is impossible, so exit chokes at sonic.
Where is the highest speed in a properly running de Laval nozzle? At the exit (end of the diverging section), not the throat.
What stands in the diverging section if back pressure is between choke and design? A normal shock that jumps flow from supersonic back to subsonic.
Definition of Mach number and speed of sound? M = V / a M=V/a M = V / a ,
a = γ R T a=\sqrt{\gamma R T} a = γ R T .
Accelerate gas to supersonic
Subsonic sonic supersonic
Speed of sound a^2=dp/drho
Sign of M^2-1 sets area change
Intuition Hinglish mein samjhoDekho, de Laval nozzle ek hourglass jaisa pipe hota hai — pehle patla hota jaata hai (converging), beech mein sabse patli jagah aati hai jise throat kehte hain, phir wapas chaura ho jaata hai (diverging). Iska kaam hai gas ki pressure/heat energy ko speed mein badalna. Rockets aur steam turbines mein yahi use hota hai.
Sabse important baat yeh hai: subsonic gas (M < 1 M<1 M < 1 dabao (area kam karo), bilkul garden hose jaise. Lekin jab gas sound ki speed pakad leti hai (M = 1 M=1 M = 1 badhana padta hai (diverging part). Yeh master equation se samajh aata hai: d A A = ( M 2 − 1 ) d V V \frac{dA}{A}=(M^2-1)\frac{dV}{V} A d A = ( M 2 − 1 ) V d V M < 1 M<1 M < 1 ( M 2 − 1 ) (M^2-1) ( M 2 − 1 ) M > 1 M>1 M > 1 M = 1 M=1 M = 1 d A = 0 dA=0 d A = 0
Ek aur cheez — choking . Jab throat par M = 1 M=1 M = 1
Yaad rakho mnemonic: "Sub Squeezes, Super Spreads." Yeh ek line pure topic ka dil hai. Exam mein agar aap area-velocity relation derive kar paate ho aur sign analysis samajh lete ho, toh 80% marks pakke. Baaki isentropic formulas (T0/T, p0/p, A/A*) sirf usi se nikalte hain.
