Instructions

A cylindrical furnace has height (H) and diameter (D) both 1 m. It is maintained at temperature 360 K. The air gets heated inside the furnace at constant pressure $$P_a$$ and its temperature becomes $$T = 360$$ K. The hot air with density $$\rho$$ rises up a vertical chimney of diameter $$d = 0.1$$ m and height $$h = 9$$ m above the furnace and exits the chimney. As a result, atmospheric air of density $$\rho_a = 1.2$$ kg m$$^{-3}$$, pressure $$P_a$$ and temperature $$T_a = 300$$ K enters the furnace. Assume air as an ideal gas, neglect the variations in $$\rho$$ and $$T$$ inside the chimney and the furnace. Also ignore the viscous effects.
[Given: The acceleration due to gravity $$g = 10$$ ms$$^{-2}$$ and $$\pi = 3.14$$]

Question 33

Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is ____ gm s$$^{-1}$$.

Correct Answer: 47.1

Using ideal gas relation at constant pressure:

$$\rho T = \text{constant} \implies \rho_a T_a = \rho T$$

$$1.2 \times 300 = \rho \times 360 \implies \rho = 1\text{ kg/m}^3$$

From Newton's second law:

$$a = \frac{\rho_a V g - \rho V g}{\rho V} = \frac{\rho_a g - \rho g}{\rho}$$

$$a = \frac{1.2 \times 10 - 1 \times 10}{1} = 2\text{ m/s}^2$$

Using equations of motion:

$$v^2 = u^2 + 2ah$$

$$v^2 = 0 + 2 \times 2 \times 9 = 36 \implies v = 6\text{ m/s}$$

Mass flow rate:

$$\frac{dm}{dt} = \rho A v = \rho \left(\frac{\pi d^2}{4}\right)v$$

$$\frac{dm}{dt} = 1 \times \frac{\pi}{4} \times 10^{-2} \times 6 = \frac{471}{10000}\text{ kg/s} = 47.1\text{ g/s}$$

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