Consider a two-dimensional potential flow over a cylinder. If the freestream speed
is Uஶ, the maximum speed on the cylinder surface is
Options:
(A) Uஶ
2
(B) 3Uஶ
2
(C) 2Uஶ
(D) 4Uஶ
3
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2022 Q.15)**
**1. Intuitive Physical Setup & Given Parameters**:
We consider 2D incompressible, irrotational potential flow over a circular cylinder of radius $R$ in a uniform stream of velocity $U_\infty$.
**2. Step-by-Step Mathematical Derivation**:
1. **Potential Flow Superposition**:
Flow past a non-lifting cylinder is modeled as a uniform flow plus a doublet of strength $\kappa = 2\pi U_\infty R^2$:
$\psi(r, \theta) = U_\infty r \sin\theta \left(1 - \frac{R^2}{r^2}\right)$
2. **Surface Velocity Field ($r = R$)**:
On the cylinder boundary $r = R$, the radial velocity is zero ($v_r = 0$, flow tangency).
The tangential velocity on the surface is:
$v_\theta(R, \theta) = -\left. \frac{\partial \psi}{\partial r} \right|_{r=R} = -2 U_\infty \sin\theta$
3. **Maximizing Surface Speed**:
The speed magnitude is $|v_\theta| = 2 U_\infty |\sin\theta|$.
The maximum speed occurs where $|\sin\theta| = 1$ (at the top $\theta = 90^\circ$ and bottom $\theta = 270^\circ$ of the cylinder):
$V_{\max} = 2 U_\infty$
4. **Minimum Pressure Coefficient**:
By Bernoulli's equation, $C_p = 1 - (V/U_\infty)^2 = 1 - 4\sin^2\theta$, which reaches a minimum of $C_{p,\min} = 1 - 4 = -3$ at these crest points.
**3. Comprehensive Options & Distractor Analysis**:
- **(A) U_∞**: ❌ **INCORRECT**. Free-stream velocity.
- **(B) 1.5 U_∞**: ❌ **INCORRECT**. Maximum surface velocity for a 3D sphere is $1.5 U_\infty$.
- **(C) 2 U_∞**: ✅ **CORRECT**. For a 2D circular cylinder, the maximum surface speed is exactly twice the free-stream speed ($2 U_\infty$).
- **(D) 4 U_∞**: ❌ **INCORRECT**. Confuses with pressure drop factor ($4\sin^2\theta$).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **C**
- **Exam Takeaway**: Memorize the peak potential velocities: 2D circular cylinder maximum surface velocity $= 2 U_\infty$ ($C_{p,\min} = -3$). 3D sphere maximum surface velocity $= 1.5 U_\infty$ ($C_{p,\min} = -1.25$).
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