Air Pressure Difference Around Moving Train inside Tunnel

To solve the problem, we consider the frame of reference attached to the moving train. In this reference frame, the flow of air is steady and laminar.

Step 1: Continuity Equation

In the train's frame of reference:

• Far ahead of the train, the air approaches the train with a relative speed equal to the speed of the train, $v_1 = v_t$.
• The cross-sectional area far ahead of the train is the tunnel's area, $S_0 = 4S_t$.
• In the region between the sides of the train and the tunnel walls, the cross-sectional area available for the air flow is: $S = S_0 - S_t = 4S_t - S_t = 3S_t$

Let $v_2$ be the speed of the air relative to the train in this region. Applying the equation of continuity: $S_0 v_1 = S v_2$ $(4S_t) v_t = (3S_t) v_2$ $v_2 = \frac{4}{3}v_t$

Step 2: Bernoulli's Equation

Since the flow is steady and laminar in the frame of the train, we can apply Bernoulli's principle along a streamline extending from far ahead of the train to the region between the train and the tunnel walls:

$p_0 + \frac{1}{2}\rho v_1^2 = p + \frac{1}{2}\rho v_2^2$

Rearranging the terms to find the pressure difference $(p_0 - p)$:

$p_0 - p = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Substituting $v_1 = v_t$ and $v_2 = \frac{4}{3}v_t$:

$p_0 - p = \frac{1}{2}\rho \left[ \left(\frac{4}{3}v_t\right)^2 - v_t^2 \right]$ $p_0 - p = \frac{1}{2}\rho v_t^2 \left( \frac{16}{9} - 1 \right)$ $p_0 - p = \frac{1}{2}\rho v_t^2 \left( \frac{7}{9} \right) = \frac{7}{18}\rho v_t^2$

Step 3: Finding the value of $N$

Comparing our derived result with the given expression:

$p_0 - p = \frac{7}{2N}\rho v_t^2$

Equating the two coefficients:

$\frac{7}{2N} = \frac{7}{18}$ $2N = 18 \implies N = 9$