A train with cross-sectional area $S_t$ is moving with speed $v_t$ inside a long tunnel of cross-sectional area $S_0$ ($S_0 = 4S_t$). Assume that almost all the air (density $\rho$) in front of the train flows back between its sides and the walls of the tunnel. Also, the air flow with respect to the train is steady and laminar. Take the ambient pressure and that inside the train to be $p_0$. If the pressure in the region between the sides of the train and the tunnel walls is $p$, then $p_0 - p = \frac{7}{2N}\rho v_t^{2}$. The value of $N$ is ______.

Step 1: Define the reference frame.

In the reference frame of the train, the train is stationary and the air far ahead flows backward with speed $v_t$ at pressure $p_0$.

Step 2: Calculate the cross-sectional area of the gap.

The air flows through the annular gap between the train and the tunnel walls.

$$S_{\text{gap}} = S_0 - S_t$$

Substitute $S_0 = 4S_t$.

$$S_{\text{gap}} = 4S_t - S_t$$ $$S_{\text{gap}} = 3S_t$$

Step 3: Apply the continuity equation.

The mass flow rate of the air is conserved.

$$S_0 v_t = (S_0 - S_t) v$$

Solve for the air speed $v$ in the gap.

$$v = \frac{S_0}{S_0 - S_t} v_t$$ $$v = \frac{4S_t}{3S_t} v_t$$ $$v = \frac{4}{3}v_t$$

Step 4: Apply Bernoulli's equation.

Apply Bernoulli's equation along a streamline from the region far ahead of the train to the gap.

$$p_0 + \frac{1}{2}\rho v_t^2 = p + \frac{1}{2}\rho v^2$$

Rearrange the terms to find the pressure difference.

$$p_0 - p = \frac{1}{2}\rho \left(v^2 - v_t^2\right)$$

Step 5: Substitute the velocity into the pressure difference equation.

Substitute $v = \frac{4}{3}v_t$.

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

Step 6: Solve for $N$.

Compare the derived pressure difference with the given expression.

$$\frac{7}{2N}\rho v_t^2 = \frac{7}{18}\rho v_t^2$$ $$2N = 18$$ $$N = 9$$

Answer: 9