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 𝑃𝑎 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 𝜌𝑎 = 1.2 kg m−3 , pressure 𝑃𝑎 and temperature 𝑇𝑎 = 300 K enters the furnace. Assume air as an ideal gas, neglect 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] When the chimney is closed using a cap at the top, a pressure difference Δ𝑃 develops between the top and the bottom surfaces of the cap. If the changes in the temperature and density of the hot air, due to the stoppage of air flow, are negligible then the value of Δ𝑃 is ______ N m−2

 To determine the pressure difference 

$\mathrm{\Delta }P$ between the top and bottom surfaces of the cap when the chimney is closed, we need to consider the hydrostatic pressure difference due to the column of hot air in the chimney.

Step 1: Understanding the System

• Furnace Dimensions: Height $H=1\text{m}$, Diameter $D=1\text{m}$.

• Furnace Temperature: $T=360\text{K}$.

• Chimney Dimensions: Diameter $d=0.1\text{m}$, Height $h=9\text{m}$.

• Atmospheric Conditions: Temperature $T_a=300\text{K}$, Density ${\rho }_a=1.2{\text{kg/m}}^3$, Pressure $P_a$.

• Assumptions: Air is an ideal gas, density $\rho$ and temperature $T$ are constant inside the chimney and furnace, viscous effects are negligible.

Step 2: Calculating the Density of Air Inside the Furnace

Using the ideal gas law:

$$PV=nRT\Rightarrow \rho =\frac{P}{RT}$$

Since the pressure inside the furnace is constant and equal to atmospheric pressure $P_a$, and the temperature is $T=360\text{K}$, the density $\rho$ of the air inside the furnace is:

$$\rho =\frac{P_a}{R\cdot 360}$$

However, we can also express the density in terms of the atmospheric density ${\rho }_a$ at $T_a=300\text{K}$:

$${\rho }_a=\frac{P_a}{R\cdot 300}\Rightarrow \rho ={\rho }_a\cdot \frac{300}{360}={\rho }_a\cdot \frac{5}{6}=1.2\cdot \frac{5}{6}=1.0{\text{kg/m}}^3$$

Step 3: Calculating the Pressure Difference

When the chimney is closed, the pressure difference $\mathrm{\Delta }P$ between the top and bottom surfaces of the cap is due to the hydrostatic pressure of the column of hot air in the chimney. The hydrostatic pressure difference is given by:

$$\mathrm{\Delta }P=\rho gh$$

Where:

• $\rho =1.0{\text{kg/m}}^3$ (density of hot air)

• $g=10{\text{m/s}}^2$ (acceleration due to gravity)

• $h=9\text{m}$ (height of the chimney)

Plugging in the values:

$$\mathrm{\Delta }P=1.0\cdot 10\cdot 9=90{\text{N/m}}^2$$

Final Answer

The pressure difference $\mathrm{\Delta }P$ between the top and bottom surfaces of the cap is 90 N/m².