Crocco's theorem relates vorticity $\vec{\omega} = \nabla \times \vec{V}$, entropy gradients $\nabla s$, and total enthalpy gradients $\nabla h_0$ in steady, inviscid fluid flow as $\vec{V} \times \vec{\omega} = \nabla h_0 - T \nabla s$. Which of the following conclusions directly follow from this theorem?

(A) In a steady, homenthalpic (constant $h_0$) and homentropic (constant $s$) flow, the flow field is necessarily irrotational everywhere

(B) Across a curved shock wave, entropy gradients are generated normal to streamlines, creating vorticity and making the downstream flow field rotational

(C) Across a planar oblique shock wave in a uniform upstream supersonic flow, the downstream flow field is rotational because entropy increases

(D) In a steady, inviscid flow where both stagnation temperature and stagnation pressure are uniform everywhere, the vorticity is zero

Answer Key:

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### 1. Strategy & Theoretical Principles Crocco's Theorem states: $$\vec{V} \times \vec{\omega} = \nabla h_0 - T \nabla s$$ where $\vec{\omega} = \nabla \times \vec{V}$ is the vorticity vector. ### 2. Analytical Evaluation of Statements - **Statement A:** **CORRECT.** If $h_0$ is constant ($\nabla h_0 = 0$) and $s$ is constant ($\nabla s = 0$), then $\vec{V} \times \vec{\omega} = 0$. Since $\vec{V} \neq 0$, the flow is irrotational ($\|\vec{\omega}\| = 0$) in the direction perpendicular to streamlines. - **Statement B:** **CORRECT.** A curved shock wave has varying shock angles $\beta$ along its span. Since entropy jump $\Delta s$ depends on $M_{1n} = M_1 \sin \beta$, the entropy downstream varies across streamlines, creating a transverse entropy gradient $\nabla s \neq 0$. By Crocco's theorem, this transverse entropy gradient produces vorticity ($\vec{\omega} \neq 0$), making the downstream flow field intrinsically rotational. - **Statement C:** **INCORRECT.** Across a *planar* (straight) oblique shock in a uniform upstream flow, the shock wave angle $\beta$ is constant everywhere along the wave. Therefore, the entropy increase $\Delta s$ is identical on every streamline. Since $s$ is uniform downstream, $\nabla s = 0$, and the downstream flow remains irrotational! (Vorticity requires an entropy *gradient*, not just an entropy increase). - **Statement D:** **CORRECT.** Uniform stagnation temperature means $\nabla h_0 = 0$, and uniform stagnation pressure means $\nabla s = 0$ (since $s - s_0 = -R \ln(p_0/p_{0,\text{ref}})$). Thus $\vec{\omega} = 0$. ### 3. Exam Trap & Standard Reference - **Trap Alert:** Straight shock waves create entropy, but NOT an entropy gradient; therefore downstream flow behind straight shocks is irrotational. Only curved shocks generate vorticity. - **Standard Reference:** Anderson, J.D., *Modern Compressible Flow*, 4th Edition, Chapter 6 (Three-Dimensional and Rotational Flow).

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