Rayleigh Flow Calculator | Compressible Flow with Heat Addition

📝 Configuration

Inlet Mach number ($M_1$):

Specific Heat Ratio ($\gamma$):

Inlet Stagnation Temp ($T_{01}$) [K]:

Inlet Stagnation Pressure ($P_{01}$) [kPa]:

Heat Added per unit mass ($q$) [kJ/kg]:

Quick Presets:

📊 Results & Diagrams

Configure inputs and click Calculate Rayleigh Flow to view results.

📘 Calculation Methodology

Mathematical Model & Theory

Rayleigh flow refers to steady 1D compressible flow of an ideal gas in a constant-area duct with heat addition or extraction, assuming wall friction is negligible. Heat addition increases the stagnation temperature ($T_0$) and drives the Mach number towards $1.0$ in both subsonic and supersonic flow regimes.

The Rayleigh flow relations are governed by the following ratio equations:

$$\frac{T_0}{T_0^*} = \frac{2(\gamma+1)M^2(1 + \frac{\gamma-1}{2}M^2)}{(1 + \gamma M^2)^2}$$

$$\frac{T}{T^*} = \left[\frac{(\gamma+1)M}{1 + \gamma M^2}\right]^2, \quad \frac{P}{P^*} = \frac{\gamma+1}{1 + \gamma M^2}$$

Static temperature achieves its maximum value at $M = 1/\sqrt{\gamma}$ ($M \approx 0.845$ for $\gamma = 1.4$), creating a region between $1/\sqrt{\gamma} < M < 1.0$ where heat addition actually causes a decrease in static temperature.

Academic References

Shapiro, Ascher H.: The Dynamics and Thermodynamics of Compressible Fluid Flow, Ronald Press.
Anderson, John D.: Modern Compressible Flow, McGraw-Hill.

Worked Engineering Example

Problem Statement:
Air ($\gamma = 1.4, C_p = 1.005\text{ kJ/kg-K}$) enters a duct at $M_1 = 0.5$ and $T_{01} = 300\text{ K}$. Calculate the maximum heat addition before thermal choking ($q_{max}$) and the exit Mach $M_2$ for a heat addition $q = 50.0\text{ kJ/kg}$.

Step-by-step Solution:
1. Evaluate the Rayleigh stagnation temperature ratio at $M_1 = 0.5$:
$$\frac{T_{01}}{T_0^*} = \frac{2 \times 2.4 \times 0.25 \times (1 + 0.2 \times 0.25)}{(1 + 1.4 \times 0.25)^2} = \frac{1.26}{1.8225} = 0.6914$$ 2. Calculate the choking stagnation temperature $T_0^*$:
$$T_0^* = \frac{300\text{ K}}{0.6914} = 433.9\text{ K}$$ 3. Calculate the maximum heat addition before choking $q_{max}$:
$$q_{max} = C_p (T_0^* - T_{01}) = 1.005 \times (433.9 - 300) = 134.6\text{ kJ/kg}$$ 4. Find the exit stagnation temperature $T_{02}$ for $q = 50.0\text{ kJ/kg}$:
$$T_{02} = 300 + \frac{50.0}{1.005} = 349.75\text{ K}$$ 5. Evaluate the exit stagnation temperature ratio:
$$\frac{T_{02}}{T_0^*} = \frac{349.75}{433.9} = 0.8061$$ 6. Solve the Rayleigh temperature relation for $T_0/T_0^* = 0.8061$ on the subsonic branch:
$$M_2 = 0.6033$$ Final Results:
• Maximum Heat Allowed: 134.6 kJ/kg
• Exit Mach: 0.6033