Ideal Gas Flow and Adiabatic Expansion in Chimney

To determine the correct statement(s), we analyze the adiabatic flow of the ideal gas through the chimney step-by-step.

1. Velocity at the Lower End ($v_1$)

The mass flow rate $\alpha$ at the lower end is given by: $\alpha = \rho_1 A_1 v_1$

Given:

• $\alpha = 0.8 \text{ kg s}^{-1}$
• $\rho_1 = 0.2 \text{ kg m}^{-3}$
• $A_1 = 0.1 \text{ m}^2$

Substituting the values: $0.8 = 0.2 \times 0.1 \times v_1 \implies v_1 = \frac{0.8}{0.02} = 40 \text{ m s}^{-1}$

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2. Pressure at the Upper End ($P_2$)

For a reversible adiabatic process of an ideal gas, the relation between pressure $P$ and temperature $T$ is given by: $P^{1-\gamma} T^\gamma = \text{constant} \implies P \propto T^{\frac{\gamma}{\gamma - 1}}$

Given $\gamma = 2$: $P \propto T^{\frac{2}{2-1}} = T^2$

Thus, $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^2$

Given $P_1 = 600 \text{ Pa}$, $T_1 = 300 \text{ K}$, and $T_2 = 150 \text{ K}$: $P_2 = 600 \times \left(\frac{150}{300}\right)^2 = 600 \times \frac{1}{4} = 150 \text{ Pa}$

Hence, Option A is incorrect.

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3. Density at the Upper End ($\rho_2$)

For an adiabatic process, pressure and density are related by: $P \propto \rho^\gamma \implies \frac{P_2}{P_1} = \left(\frac{\rho_2}{\rho_1}\right)^\gamma$

Given $\gamma = 2$: $\frac{\rho_2}{\rho_1} = \sqrt{\frac{P_2}{P_1}} = \sqrt{\frac{150}{600}} = \frac{1}{2}$

$\rho_2 = \frac{\rho_1}{2} = \frac{0.2}{2} = 0.1 \text{ kg m}^{-3}$

Hence, Option D is incorrect.

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4. Velocity at the Upper End ($v_2$)

Using the continuity equation for mass flow rate at the upper end: $\alpha = \rho_2 A_2 v_2$

Given $A_2 = 0.4 \text{ m}^2$ and $\rho_2 = 0.1 \text{ kg m}^{-3}$: $0.8 = 0.1 \times 0.4 \times v_2 \implies v_2 = \frac{0.8}{0.04} = 20 \text{ m s}^{-1}$

Thus, $v_1 = 40 \text{ m s}^{-1}$ and $v_2 = 20 \text{ m s}^{-1}$. Hence, Option B is correct.

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5. Height of the Chimney ($h$)

Applying the Steady Flow Energy Equation (SFEE) for an insulated chimney with no external work done: $h_1 + \frac{1}{2}v_1^2 + g z_1 = h_2 + \frac{1}{2}v_2^2 + g z_2$

where $h_1$ and $h_2$ represent the specific enthalpy of the gas ($h = C_p T$). Taking $z_1 = 0$ at the lower end and $z_2 = h$ at the upper end: $C_p T_1 + \frac{1}{2}v_1^2 = C_p T_2 + \frac{1}{2}v_2^2 + gh$

$gh = C_p (T_1 - T_2) + \frac{1}{2}\left(v_1^2 - v_2^2\right)$

First, we determine the specific gas constant $R_s$: $P_1 = \rho_1 R_s T_1 \implies R_s = \frac{P_1}{\rho_1 T_1} = \frac{600}{0.2 \times 300} = 10 \text{ J kg}^{-1} \text{K}^{-1}$

The specific heat capacity at constant pressure $C_p$ is: $C_p = \frac{\gamma}{\gamma - 1} R_s = \frac{2}{2 - 1} \times 10 = 20 \text{ J kg}^{-1} \text{K}^{-1}$

Now, substituting $C_p$, velocities, and $g = 10 \text{ m s}^{-2}$: $10 \times h = 20 \times (300 - 150) + \frac{1}{2}\left(40^2 - 20^2\right)$

$10 h = 20 \times 150 + \frac{1}{2}(1600 - 400)$

$10 h = 3000 + 600 = 3600 \implies h = 360 \text{ m}$

Hence, Option C is incorrect.

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Conclusion

The correct option is B.