Question

A rotational velocity field in an air flow is given as \( \vec{V} = a y \hat{i} + b x \hat{j} \), with \( a = 10 \, \text{s}^{-1} \), \( b = 20 \, \text{s}^{-1} \). The air density is 1.0 kg/m\(^3\) and the pressure at \( (x, y) = (0, 0) \) is 100 kPa. Neglecting gravity, the pressure at \( (x, y) = (6 \, \text{m}, 8 \, \text{m}) \) is _________ kPa (rounded off to the nearest integer).

Hint:

In rotating flows, Bernoulli’s equation can be modified to account for the rotational velocity field. The pressure drop is proportional to the square of the velocity magnitude.

Answer 1

Correct Answer: 88

The pressure difference in a rotating flow can be found using Bernoulli's equation for rotating flows: \[ \Delta P = -\frac{1}{2} \rho \left( |\vec{V}|^2 \right). \] At the point \( (6, 8) \), the velocity magnitude \( |\vec{V}| \) is: \[ |\vec{V}| = \sqrt{(a y)^2 + (b x)^2} = \sqrt{(10 \times 8)^2 + (20 \times 6)^2} = \sqrt{6400 + 14400} = \sqrt{20800} \approx 144.22 \, \text{m/s}. \] Now, compute the pressure difference: \[ \Delta P = -\frac{1}{2} \times 1.0 \, \text{kg/m}^3 \times (144.22)^2 = -\frac{1}{2} \times 1.0 \times 20880.3 = -10440.15 \, \text{Pa} = -10.44 \, \text{kPa}. \] The pressure at \( (x, y) = (6, 8) \) is: \[ P = 100 \, \text{kPa} - 10.44 \, \text{kPa} = 89.56 \, \text{kPa}. \] Thus, the pressure is approximately: \[ \boxed{90\ \text{kPa}}. \]