Question:

A scramjet engine features an intake, isolator, combustor, and a nozzle, as shown. Station 3 indicates the combustor entry point. Assume stagnation enthalpy is constant between Stations 1 and 3, and air is a calorically perfect gas with specific heat ratio $\gamma$. Select the correct expression for Mach number $M_3$ at the inlet to the combustor from the options given.

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Always start with the isentropic relation $T_0/T = 1 + \tfrac{\gamma-1}{2}M^2$ and use conservation of stagnation enthalpy. This is the most direct path to Mach number relations in compressible flows.
Updated On: Aug 31, 2025
  • $M_3 = M_\infty \sqrt{\dfrac{2}{\gamma -1}\left(\dfrac{T_0}{T_3}-1\right)}$
  • $M_3 = \sqrt{\dfrac{2}{\gamma-1}\left[\dfrac{T_0}{T_3}\left(1+\dfrac{\gamma-1}{2}M_\infty^2\right)-1\right]}$
  • $M_3 = M_\infty \sqrt{\dfrac{T_\infty}{T_3}}$
  • $M_3 = \sqrt{\dfrac{\gamma+1}{2}\left(\dfrac{T_0}{T_3}-1\right)M_\infty^2}$
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The Correct Option is B

Solution and Explanation

Step 1: Energy relation.
Between station 1 (free stream) and station 3 (combustor entry), stagnation enthalpy is constant: \[ h_0 = c_p T_0 = \text{constant} \] Thus, stagnation temperature is preserved: \[ T_{0,1} = T_{0,3} = T_0 \]

Step 2: Temperature–Mach relation.
For a calorically perfect gas: \[ \frac{T_0}{T} = 1 + \frac{\gamma -1}{2} M^2 \] At station 3: \[ \frac{T_0}{T_3} = 1 + \frac{\gamma-1}{2}M_3^2 \] At free stream: \[ \frac{T_0}{T_\infty} = 1 + \frac{\gamma-1}{2}M_\infty^2 \]

Step 3: Express $M_3$.
From station 3 relation: \[ M_3^2 = \frac{2}{\gamma -1}\left(\frac{T_0}{T_3}-1\right) \] But $T_0$ is related to free stream values: \[ \frac{T_0}{T_3} = \frac{T_0}{T_\infty}\cdot \frac{T_\infty}{T_3} \] \[ = \left(1+\frac{\gamma-1}{2}M_\infty^2\right)\frac{T_\infty}{T_3} \] Substitute into $M_3^2$: \[ M_3^2 = \frac{2}{\gamma -1}\left[\frac{T_0}{T_3}\left(1+\frac{\gamma-1}{2}M_\infty^2\right)-1\right] \]

Step 4: Take square root.
\[ M_3 = \sqrt{\frac{2}{\gamma -1}\left[\frac{T_0}{T_3}\left(1+\frac{\gamma-1}{2}M_\infty^2\right)-1\right]} \] This matches option (B). \[ \boxed{M_3 = \sqrt{\dfrac{2}{\gamma-1}\left[\dfrac{T_0}{T_3}\left(1+\dfrac{\gamma-1}{2}M_\infty^2\right)-1\right]}} \]

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