Steady Mass Flow Rate of Air Exiting Vertical Chimney

To find the steady mass flow rate of air exiting the chimney, we proceed step-by-step:

Step 1: Calculate the density of hot air inside the furnace and chimney

Air is assumed to be an ideal gas. Since it is heated inside the furnace at a constant pressure $P_a$, we can write: $P_a = \frac{\rho_a R T_a}{M} = \frac{\rho R T}{M}$

Thus, the density of the hot air $\rho$ inside the furnace and chimney is given by: $\rho = \rho_a \left( \frac{T_a}{T} \right)$

Given values:

• $\rho_a = 1.2\text{ kg m}^{-3}$
• $T_a = 300\text{ K}$
• $T = 360\text{ K}$

$\rho = 1.2 \times \frac{300}{360} = 1.0\text{ kg m}^{-3}$

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Step 2: Apply Bernoulli's equation along a streamline

Consider a streamline starting from the entrance at the bottom of the furnace ($z = 0$) to the exit at the top of the chimney ($z = H + h$).

1. At the entrance ($z = 0$):

   • Pressure: $P_1 = P_a$
   • Height: $z_1 = 0$
   • Air velocity: $v_1 \approx 0$ (since the atmospheric air far from the furnace is stationary)

2. At the exit of the chimney ($z = H + h$):

   • Height: $z_2 = H + h = 1\text{ m} + 9\text{ m} = 10\text{ m}$
   • Air velocity: $v_2 = v$
   • External atmospheric pressure at height $z_2$: $P_{out}(H+h) = P_a - \rho_a g (H+h)$ Since the air exits into the surrounding atmosphere, the pressure at the chimney exit is equal to the local atmospheric pressure: $P_2 = P_a - \rho_a g (H+h)$

Applying Bernoulli's equation for the hot air flowing inside: $P_1 + \rho g z_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g z_2 + \frac{1}{2}\rho v_2^2$

Substituting the known values: $P_a + 0 + 0 = \left[ P_a - \rho_a g (H+h) \right] + \rho g (H+h) + \frac{1}{2}\rho v^2$

Simplifying the expression: $0 = - \rho_a g (H+h) + \rho g (H+h) + \frac{1}{2}\rho v^2$

$\frac{1}{2}\rho v^2 = (\rho_a - \rho) g (H+h)$

$v = \sqrt{\frac{2(\rho_a - \rho) g (H+h)}{\rho}}$

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Step 3: Calculate the exit velocity ($v$)

Substitute the given values into the velocity equation:

• $g = 10\text{ m s}^{-2}$
• $H + h = 10\text{ m}$
• $\rho_a = 1.2\text{ kg m}^{-3}$
• $\rho = 1.0\text{ kg m}^{-3}$

$v = \sqrt{\frac{2 \times (1.2 - 1.0) \times 10 \times 10}{1.0}} = \sqrt{2 \times 0.2 \times 100} = \sqrt{40} = 2\sqrt{10}\text{ m s}^{-1} \approx 6.32455\text{ m s}^{-1}$

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Step 4: Calculate the steady mass flow rate

The cross-sectional area of the chimney of diameter $d = 0.1\text{ m}$ is: $A = \frac{\pi d^2}{4} = \frac{3.14 \times (0.1)^2}{4} = 7.85 \times 10^{-3}\text{ m}^2$

The steady mass flow rate $\frac{dm}{dt}$ of air exiting the chimney is given by: $\frac{dm}{dt} = \rho A v$

$\frac{dm}{dt} = 1.0\text{ kg m}^{-3} \times (7.85 \times 10^{-3}\text{ m}^2) \times \sqrt{40}\text{ m s}^{-1}$

$\frac{dm}{dt} = 7.85 \times 10^{-3} \times 6.32455\text{ kg s}^{-1} \approx 0.049648\text{ kg s}^{-1}$

Converting to grams per second ($\text{gm s}^{-1}$): $\frac{dm}{dt} = 0.049648 \times 1000\text{ gm s}^{-1} = 49.65\text{ gm s}^{-1}$

(Rounding to two decimal places gives $49.65\text{ gm s}^{-1}$ or within the range of $49.60 - 49.70\text{ gm s}^{-1}$)