A Pitot tube placed in a uniform supersonic air stream ($\gamma = 1.4$) with upstream Mach number $M_1 = 3.0$ and static pressure $p_1 = 20\text{ kPa}$ records the pitot stagnation pressure $p_{02}$ behind the detached bow shock. According to the Rayleigh Pitot tube formula $\frac{p_{02}}{p_1} = \left(\frac{(\gamma+1)^2 M_1^2}{4\gamma M_1^2 - 2(\gamma-1)}\right)^{\frac{\gamma}{\gamma-1}} \left(\frac{1 - \gamma + 2\gamma M_1^2}{\gamma + 1}\right)$, the ratio $p_{02} / p_1$ is closest to:

(A) 12.06

(B) 36.73

(C) 6.82

(D) 22.15

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles In supersonic flow, a detached normal bow shock stands immediately ahead of a Pitot tube nose. - The flow decelerates across the normal shock from $(M_1, p_1)$ to $(M_2, p_2)$. - The downstream flow then undergoes isentropic deceleration from state 2 to the stagnation point at the probe tip ($p_{02}$). Combining normal shock pressure jump and isentropic stagnation relations yields the **Rayleigh Supersonic Pitot Formula**: $$\frac{p_{02}}{p_1} = \left[ \frac{(\gamma + 1)^2 M_1^2}{4\gamma M_1^2 - 2(\gamma - 1)} \right]^{\frac{\gamma}{\gamma - 1}} \left[ \frac{1 - \gamma + 2\gamma M_1^2}{\gamma + 1} \right]$$ ### 2. Analytical Formulation & Calculation Given $M_1 = 3.0$ and $\gamma = 1.4$: 1. First bracketed term: $$\frac{(\gamma + 1)^2 M_1^2}{4\gamma M_1^2 - 2(\gamma - 1)} = \frac{(2.4)^2 \times 9}{4(1.4)(9) - 2(0.4)} = \frac{5.76 \times 9}{50.4 - 0.8} = \frac{51.84}{49.6} \approx 1.04516$$ Raise to power $\frac{\gamma}{\gamma - 1} = 3.5$: $$(1.04516)^{3.5} \approx 1.1668$$ 2. Second bracketed term: $$\frac{1 - \gamma + 2\gamma M_1^2}{\gamma + 1} = \frac{1 - 1.4 + 2(1.4)(9)}{2.4} = \frac{-0.4 + 25.2}{2.4} = \frac{24.8}{2.4} = \frac{31}{3} \approx 10.3333$$ 3. Multiply terms: $$\frac{p_{02}}{p_1} = 1.1668 \times 10.3333 \approx 12.057 \approx 12.06$$ ### 3. Distractor Forensics - **Option A (12.06):** Correct Rayleigh pitot ratio. - **Option B (36.73):** Isentropic stagnation pressure ratio $p_{01}/p_1 = (1 + 0.2 \times 9)^{3.5} = 2.8^{3.5} = 36.73$, completely ignoring the normal shock wave loss. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Never use isentropic relations ($p_{01}/p_1$) to measure supersonic Pitot pressures because the bow shock creates massive entropy loss. - **Standard Reference:** Anderson, J.D., *Modern Compressible Flow*, Chapter 3.

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