Question

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_𝑎 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 enter the furnace. Assume air as an ideal gas, neglecting the variations in 𝜌 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 streamlined, the steady mass flow rate of air exiting the chimney is _______ gm s⁻¹.

Answer 1

Applying Bernoulli's theorem,
\(P_a+\frac{1}{2}\rho_aV^2=P_a+\rho gH+\frac{1}{2}\rho V^2\,\,\,\,...........(1)\)
Also, since PM = \(\rho\)RT
\(\rho_{a}\times300=\rho\times360\)
\(\Rightarrow \rho=1\,kg/m^3\,\,\,..........(2)\)
\(\frac{v^2}{2}(0.2)=1\times10\times1\)
\(\Rightarrow v=10\,\,m/s\)
Mass flow rate = \(V\times \frac{\pi D^2}{4}\times \rho=10\times\frac{3.14}{4}\times\,\,1 kg/s=7.85\,\,kg/s=7850\,gm/s\)
 

Answer 2

We're given:

1. Gas constants R and M remain constant, so MR​ is constant (for constant pressure).
2. The density of hot air inside the furnace is 1.2 kg/m³ at a temperature of 300K and pressure P_a​.
3. The air heats up at a constant pressure P_a​.

Given the buoyant force applied on the hot air is V_g in the upward direction and the weight of the hot air is also Vg in the downward direction, the net force on the hot air is 𝑉_𝑔−𝑉_𝑔

Let the acceleration of the hot air in the upward direction be 𝑎a, and the mass of the hot air be 𝑉V.

We have:

\(a = \frac{V_g - V_g}{V} = \frac{g - g}{1.2} = \frac{0.8 \, \text{m/s}^2}{1.2} = 0.67 \, \text{m/s}^2\)

Using the formula for velocity 𝑣v of the hot air when exiting the chimney:

\(v = 2gh\)

Given \(v=6m/s\), we can solve for h:
\(6 = \sqrt{2 \times 9.8 \times h}\)
\(36=19.6h\) 
\(h = \frac{36}{19.6} \quad h \approx 1.83 \, \text{m}\)

The mass flow rate is given by:

Mass flow rate=\(\frac{\rho Av}{1000}\)

Given 𝐴=1 m², we can calculate:

Mass flow rate=\(1.2 \times 1 \times \frac{6 }{ 1000} = 0.0072 \, \text{kg/s} = 7.2 \, \text{g/s} = 72 \, \text{gm/s}\)

So, the mass flow rate is \(72g/s.\)