🔺 Converging-Diverging Nozzle Flow

Analyze C-D nozzle performance: operating regimes, choking, shock location, and pressure distribution along the nozzle.

📝 Configuration

Gas Properties

Stagnation Conditions

Nozzle Geometry

Back Pressure

$P_b/P_0$ = 0.203
Area-Mach Relation:
$A/A^* = \frac{1}{M} \left[\frac{2}{\gamma+1}\left(1 + \frac{\gamma-1}{2} M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$

Choked Mass Flow:
$\dot{m} = \frac{P_0 A^* \sqrt{\gamma}}{\sqrt{R T_0}} \left(\frac{2}{\gamma+1}\right)^{\frac{\gamma+1}{2(\gamma-1)}}$

Normal Shock:
$M_2^2 = \frac{(\gamma-1) M_1^2 + 2}{2\gamma M_1^2 - (\gamma-1)}$
$P_2/P_1 = 1 + \frac{2\gamma}{\gamma+1}(M_1^2-1)$

📊 Results & Visualization

Configure the inputs and click Calculate to see results.

ℹ️ About Converging-Diverging Nozzles

A C-D nozzle accelerates gas from subsonic to supersonic speeds. The flow behavior depends on the back pressure ratio P_b/P₀.

Operating regimes:
Subsonic: P_b close to P₀; flow never reaches sonic speed
Choked: M = 1 at throat; maximum mass flow rate reached
Shock in nozzle: Normal shock forms in diverging section
Overexpanded: Oblique shocks at exit plane
Design: Perfectly expanded isentropic flow
Underexpanded: Expansion fans at exit plane

📘 Calculation Methodology

Mathematical Model & Theory

Compressible flow in a converging-diverging (de Laval) nozzle accelerates gas to supersonic velocities. The relationship between local cross-sectional area and Mach number is governed by the Area-Mach relation:

$$\frac{A}{A^*} = \frac{1}{M} \left[ \frac{2}{\gamma + 1} \left( 1 + \frac{\gamma - 1}{2} M^2 \right) \right]^{\frac{\gamma + 1}{2(\gamma - 1)}}$$

Under choked flow conditions, Mach number equals 1.0 at the throat ($A = A^*$), and mass flow rate is maximized.

Worked Engineering Example

Problem Statement:
Air ($\gamma = 1.4$) flows through a CD nozzle. The exit area is 2 times the throat area ($A_e/A^* = 2.0$). Find the isentropic supersonic exit Mach number.

Step-by-step Solution:
1. Solve the Area-Mach relation for $A/A^* = 2.0$ (supersonic branch, $M > 1$):
$$2.0 = \frac{1}{M} \left[ \frac{2}{2.4} \left( 1 + 0.2 M^2 \right) \right]^{3.0}$$ 2. By numerical solver / bisection:
$$M \approx 2.197$$
Final Result:
Supersonic exit Mach number is 2.20.