🚀 Compressible Flow Calculator
Isentropic flow relations, area ratios, and normal shock analysis for compressible gas dynamics.
📝 Configuration
Isentropic Relations:
$T_0/T = 1 + \frac{\gamma-1}{2} M^2$
$P_0/P = (T_0/T)^{\frac{\gamma}{\gamma-1}}$
$A/A^* = \frac{1}{M} \left[\frac{2}{\gamma+1}\left(1 + \frac{\gamma-1}{2} M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$
Normal Shock:
$M_2^2 = \frac{(\gamma-1) M_1^2 + 2}{2\gamma M_1^2 - (\gamma-1)}$
$P_2/P_1 = 1 + \frac{2\gamma}{\gamma+1}(M_1^2-1)$
$T_0/T = 1 + \frac{\gamma-1}{2} M^2$
$P_0/P = (T_0/T)^{\frac{\gamma}{\gamma-1}}$
$A/A^* = \frac{1}{M} \left[\frac{2}{\gamma+1}\left(1 + \frac{\gamma-1}{2} M^2\right)\right]^{\frac{\gamma+1}{2(\gamma-1)}}$
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 Compressible Flow
Compressible flow occurs when fluid density changes significantly, typically at Mach numbers above 0.3.
Key concepts:
• Subsonic (M < 1): Pressure disturbances propagate upstream
• Sonic (M = 1): Flow at the speed of sound; A* = minimum area
• Supersonic (M > 1): Shock waves form; area increases accelerate flow
• Normal shock: Abrupt deceleration from supersonic to subsonic; entropy increases
Compressible flow occurs when fluid density changes significantly, typically at Mach numbers above 0.3.
Key concepts:
• Subsonic (M < 1): Pressure disturbances propagate upstream
• Sonic (M = 1): Flow at the speed of sound; A* = minimum area
• Supersonic (M > 1): Shock waves form; area increases accelerate flow
• Normal shock: Abrupt deceleration from supersonic to subsonic; entropy increases
📘 Calculation Methodology
Mathematical Model & Theory
For steady, 1D isentropic flow of an ideal gas with constant specific heats, local properties are related to stagnation properties via Mach number ($M$):
$$\frac{T_0}{T} = 1 + \frac{\gamma - 1}{2} M^2$$
$$\frac{P_0}{P} = \left( 1 + \frac{\gamma - 1}{2} M^2 \right)^{\frac{\gamma}{\gamma - 1}}$$
$$\frac{\rho_0}{\rho} = \left( 1 + \frac{\gamma - 1}{2} M^2 \right)^{\frac{1}{\gamma - 1}}$$
Worked Engineering Example
Problem Statement:
Air ($\gamma = 1.4$) flows at Mach 2.0. The stagnation pressure is 300 kPa and stagnation temperature is 400 K. Find static pressure and static temperature.
Step-by-step Solution:
1. Calculate temperature ratio:
$$\frac{T_0}{T} = 1 + \frac{1.4 - 1}{2} \times 2.0^2 = 1.8$$ $$T = \frac{T_0}{1.8} = \frac{400}{1.8} = 222.22 \text{ K}$$ 2. Calculate pressure ratio:
$$\frac{P_0}{P} = (1.8)^{\frac{1.4}{1.4-1}} = 1.8^{3.5} = 7.824$$ $$P = \frac{P_0}{7.824} = \frac{300}{7.824} = 38.34 \text{ kPa}$$
Final Result:
Static temperature is 222.2 K and pressure is 38.3 kPa.
Air ($\gamma = 1.4$) flows at Mach 2.0. The stagnation pressure is 300 kPa and stagnation temperature is 400 K. Find static pressure and static temperature.
Step-by-step Solution:
1. Calculate temperature ratio:
$$\frac{T_0}{T} = 1 + \frac{1.4 - 1}{2} \times 2.0^2 = 1.8$$ $$T = \frac{T_0}{1.8} = \frac{400}{1.8} = 222.22 \text{ K}$$ 2. Calculate pressure ratio:
$$\frac{P_0}{P} = (1.8)^{\frac{1.4}{1.4-1}} = 1.8^{3.5} = 7.824$$ $$P = \frac{P_0}{7.824} = \frac{300}{7.824} = 38.34 \text{ kPa}$$
Final Result:
Static temperature is 222.2 K and pressure is 38.3 kPa.