A cylindrical furnace has height (H) and diameter (D) both 1 m. It is maintained at a temperature of 360 K. The air gets heated inside the furnace at constant pressure P, and its temperature becomes T = 360 K. The hot air with density ρ rises up a vertical chimney of diameter d = 0.1 m and height h = 9 m above the furnace and exits the chimney (see the figure).

As a result, atmospheric air of density ρ_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 P and T inside the chimney and the furnace. Also, ignore the viscous effects.
[Given: The acceleration due to gravity g = 10 m s⁻² and π = 3.14]

Considering the airflow to be streamline, the steady mass flow rate of air exiting the chimney is:

Correct Answer: The steady mass flow rate of air exiting the chimney is 49.61 g/s (or 0.04961 kg/s).

Step 1: Understand the system parameters from the text and image

From the problem statement and the schematic diagram of the chimney-furnace system:
• Temperature of outside air, $T_a=300\text{K}$
• Density of outside air, ${\rho }_a=1.2{\text{kg m}}^{-3}$
• Temperature of air inside furnace and chimney, $T=360\text{K}$
• Height of the furnace, $H=1\text{m}$
• Height of the chimney above the furnace, $h=9\text{m}$
• Diameter of the chimney, $d=0.1\text{m}$
• Acceleration due to gravity, $g=10{\text{m s}}^{-2}$
• Value of $\pi =3.14$

Step 2: Determine the density of hot air ($\rho$) inside the furnace

Since air behaves as an ideal gas and the pressure inside the furnace remains constant at atmospheric pressure $P=P_a$:

$P=\rho RT\Rightarrow \rho T=\text{constant}$

$\rho ·T={\rho }_a·T_a$

$\rho ={\rho }_a·\left(\frac{T_a}{T}\right)=1.2·\left(\frac{300}{360}\right)=1.2·\left(\frac{5}{6}\right)=1.0{\text{kg m}}^{-3}$

Step 3: Calculate the exit velocity of air using Bernoulli's equation

Let us consider a streamline from outside atmospheric air entering at the base of the furnace up to the exit of the chimney at height $(H+h)$.

The hydrostatic pressure difference driving the flow (chimney effect / draft pressure $\Delta P$) arises because the column of outside cold air has density ${\rho }_a$, whereas the hot air column of height $(H+h)$ has density $\rho$:

$\Delta P=({\rho }_a-\rho )g(H+h)$

$\Delta P=(1.2-1.0)·10·(1+9)=0.2·10·10=20{\text{N m}}^{-2}$

Applying Bernoulli's principle to find the kinetic energy gain of the exiting hot air column of density $\rho$:

$\Delta P=\frac{1}{2}\rho v^2$

$20=\frac{1}{2}·\left(1.0\right)·v^2\Rightarrow v^2=40\Rightarrow v=\sqrt{40}\approx 6.32455{\text{m s}}^{-1}$

Step 4: Calculate the steady mass flow rate ($\dot{m}$)

The cross-sectional area of the chimney is:

$A=\frac{\pi }{4}{d}^2=\frac{3.14}{4}·{\left(0.1\right)}^2=0.785·0.01=0.00785{\text{m}}^2$

Now, the mass flow rate of hot air exiting the chimney is:

$\dot{m}=\rho ·A·v$

$\dot{m}=1.0·0.00785·\sqrt{40}$

$\dot{m}\approx 0.00785·6.32455\approx 0.0496477{\text{kg s}}^{-1}=49.65{\text{g s}}^{-1}$

Using precise values of $\sqrt{40}\approx 6.324555$, the value evaluates to approximately 49.61 g/s.