de Laval nozzle

Example

A de Laval nozzle is a duct whose cross-section area $A(x)$ first decreases to a minimum $A^*$ at the throat and then increases. Model the flow as steady, quasi-one-dimensional and isentropic, with velocity $u(x)$, density $\rho(x)$, pressure $p(x)$, local speed of sound $c(x)$ and Mach number $M = u/c$. Conservation of mass gives $\rho u A = \text{const}$, so $\frac{d\rho}{\rho} + \frac{du}{u} + \frac{dA}{A} = 0$. The momentum equation gives $u\, du = -dp/\rho = -c^2\, d\rho/\rho$, so $\frac{d\rho}{\rho} = -M^2 \frac{du}{u}$. Together these give the area–velocity relation $$\frac{dA}{A} = (M^2 - 1)\frac{du}{u}.$$ So for $M < 1$ the flow speeds up where the duct narrows, and for $M > 1$ it speeds up where the duct widens. A smooth acceleration from rest to supersonic speed is therefore possible only if $M = 1$ exactly at the throat, where $dA = 0$. For an ideal gas with ratio of specific heats $\gamma$ the area at Mach number $M$ is then given by $$\frac{A}{A^*} = \frac{1}{M}\left[\frac{2}{\gamma + 1}\left(1 + \frac{\gamma - 1}{2} M^2\right)\right]^{(\gamma + 1)/(2(\gamma - 1))}.$$ Check for air, $\gamma = 1.4$, at $M = 2$: the bracket is $\frac{2}{2.4} \cdot 1.8 = 1.5$, the exponent is $3$, and $A / A^* = 1.5^3 / 2 = 1.6875$. So a nozzle whose exit area is about $1.69$ times its throat area produces a Mach $2$ jet.