Question 3 of 13: Stagnation and Side Pressure on a Cylindrical Chimney

Question 3: Stagnation and Side Pressure on a Cylindrical Chimney (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. Velocity and gauge pressure at θ = 0° (point 1, stagnation) and θ = 90° (point 2, side).

Approach. Apply Bernoulli's equation between the undisturbed free stream and each surface point (inviscid, no elevation change), using the given surface-velocity law to fix V at each point.

1. Velocities from the surface-velocity law. At the stagnation point (θ=0°) and the side point (θ=90°): $$V_1 = 2V_0\sin(0^\circ) = 0, \qquad V_2 = 2V_0\sin(90^\circ) = 2V_0 = 2(27.78) = \boxed{55.56\ \text{m/s}}$$
2. Bernoulli between free stream and the surface (gauge, no elevation change). $$p(\theta) - p_\infty = \tfrac{1}{2}\rho\left(V_0^2 - V(\theta)^2\right)$$
3. Gauge pressure at the stagnation point (V=0): $$p_1 = \tfrac{1}{2}(1.19)(27.78^2 - 0) = \boxed{459\ \text{Pa} = 0.459\ \text{kPa}}$$
4. Gauge pressure at the side point (V=55.56 m/s): $$p_2 = \tfrac{1}{2}(1.19)(27.78^2 - 55.56^2) = \tfrac{1}{2}(1.19)(-2314.8) = \boxed{-1377\ \text{Pa} = -1.377\ \text{kPa}}$$