Question

A train with a cross-sectional area $s_t$ is moving with speed $v_t$ inside a long tunnel of cross-sectional area $s_0\left(s_0=4s_t\right)$. 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 airflow 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 $P_0-P=\left(\frac{7}{2N}\right)\rho v_t^2$. The value of $N$ is ________.

Answer 1

Step 1: Given data and assumptions.

The cross-section area of the train$=s_t$

The cross-section area of the tunnel, $s_0=4s_t$

The density of air in front of the train$=\rho$

Ambient pressure$=p_0$

Pressure between the sides of the train and the tunnel walls$=p$

Given equation, $P_0-P=\left(\frac{7}{2N}\right)\rho v_t^2$

Step 2: Formula used:

$P_1+\frac{1}{2}\rho {v_1}^2+\rho g{h}_1=P_2+\frac{1}{2}\rho {v_2}^2+\rho g{h}_2$

Where,

$\rho =$ fluid density

$g=$ acceleration due to gravity

$P_1=$ pressure at elevation $1$

$v_1=$ velocity at elevation $1$

$h_1=$ height at elevation $1$

$P_2=$ pressure at elevation $2$

$v_2=$ velocity at elevation $2$

$h_2=$ height at elevation $2$

Step 3: Find the value of $N$ in the given equation.

According to Bernoulli's equation.

$P_0+\frac{1}{2}\rho v_t^2=P+\frac{1}{2}\rho v^2$

$\Rightarrow$ $P_0-P=\frac{1}{2}\rho \left(v^2-v_t^2\right)$ ……$\left(i\right)$

Where,

$P_0$ is the ambient pressure

$P$ is the pressure between the sides of the train and the tunnel walls

$\rho$ is the density of air in front of the train

$v_t$ is the speed inside the tunnel

$v$ is the speed outside the tunnel

According to the equation of continuity,

$A_1\times V_1=A_2\times V_2$

$\Rightarrow$$4s_t{v}_t=v\times 3v_t$

$\Rightarrow$ $v=\frac{4}{3}{v}_t$ …..$\left(ii\right)$

Substitute equation $\left(ii\right)$ in $\left(i\right)$ we get.

$P_0-P=\frac{1}{2}\rho \left[{\left(\frac{4}{3}{v}_t\right)}^2-v_t^2\right]$

$\Rightarrow$$P_0-P=\frac{1}{2}\rho \left[\frac{16}{9}{v}_t^2-v_t^2\right]$

$\Rightarrow$$P_0-P=\frac{7}{2\times 9}\rho v_t^2$ ……$\left(iv\right)$

compare the equation $\left(iv\right)$ from the given equation, we get

$N=9$

Hence, the value of $N$ is $\mathbf{9}$.