🚀 Compressible Flow Calculator
Isentropic flow relations, area ratios, and normal shock analysis for compressible gas dynamics.
Compressible Gas Dynamics
High-velocity gas flow exhibits expansion and compression effects. Under isentropic conditions, pressure and temperature drop as Mach number increases.
• Isentropic Expansion: Sonic threshold ($M=1$) occurs exactly at the minimum area throat ($A^*$).
• Normal Shock Wave: Discontinuous pressure rises that decelarate supersonic streams to subsonic speeds.
• Stagnation Conditions: Stagnation properties ($P_0, T_0$) represent fluid state brought to rest isentropically.
• Normal Shock Wave: Discontinuous pressure rises that decelarate supersonic streams to subsonic speeds.
• Stagnation Conditions: Stagnation properties ($P_0, T_0$) represent fluid state brought to rest isentropically.
📝 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 inputs and click Calculate Flow Properties to view results.
📘 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 ($M$):
$$\frac{T_0}{T} = 1 + \frac{\gamma - 1}{2} M^2$$
$$\frac{P_0}{P} = \left( 1 + \frac{\gamma - 1}{2} M^2 \right)^{\frac{\gamma}{\gamma - 1}}$$
$$\frac{A}{A^*} = \frac{1}{M} \left[ \frac{2}{\gamma+1} \left( 1 + \frac{\gamma-1}{2} M^2 \right) \right]^{\frac{\gamma+1}{2(\gamma-1)}}$$
Assumptions
- Ideal gas behavior with constant specific heat ratio ($\gamma$).
- Steady, one-dimensional, isentropic flow (except across shock waves).
- Standard air gas constant ($R = 287$ J/kg·K) is assumed for density and speed of sound calculations.
Academic References
- Anderson, J. D.: Modern Compressible Flow: With Historical Perspective, McGraw-Hill.
- Oosthuizen, P. H., and Carscallen, W. E.: Introduction to Compressible Fluid Flow, CRC Press.
Worked Engineering Example
Problem Statement:
Air ($\gamma = 1.4$) flows at Mach 2.0. 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. 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.