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] Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is ______

To determine the steady mass flow rate of air exiting the chimney, we need to analyze the flow of air through the system using principles of fluid dynamics and thermodynamics. Here's a step-by-step breakdown of the solution:

1. Understanding the System

Furnace Dimensions: Height $H = 1 m$ , Diameter $D = 1 m$ .
Furnace Temperature: $T = 360 K$ .
Chimney Dimensions: Diameter $d = 0.1 m$ , Height $h = 9 m$ .
Atmospheric Conditions: Temperature $T_{a} = 300 K$ , Density $ρ_{a} = 1.2 {kg/m}^{3}$ , Pressure $P_{a}$ .
Assumptions: Air is an ideal gas, density $ρ$ and temperature $T$ are constant inside the chimney and furnace, viscous effects are negligible, flow is streamline and steady.

2. Calculating the Density of Air Inside the Furnace

Using the ideal gas law:

P V = n R T \Rightarrow ρ = \frac{P}{R T}

Since the pressure inside the furnace is constant and equal to atmospheric pressure $P_{a}$ , and the temperature is $T = 360 K$ , the density $ρ$ of the air inside the furnace is:

ρ = \frac{P_{a}}{R \cdot 360}

However, we can also express the density in terms of the atmospheric density $ρ_{a}$ at $T_{a} = 300 K$ :

ρ_{a} = \frac{P_{a}}{R \cdot 300} \Rightarrow ρ = ρ_{a} \cdot \frac{300}{360} = ρ_{a} \cdot \frac{5}{6} = 1.2 \cdot \frac{5}{6} = 1.0 {kg/m}^{3}

3. Applying Bernoulli's Equation

Bernoulli's equation for streamline flow between the furnace (point 1) and the top of the chimney (point 2) is:

P_{1} + \frac{1}{2} ρ v_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ v_{2}^{2} + ρ g h_{2}

Assuming the pressure at the top of the chimney is atmospheric pressure $P_{a}$ , and the velocity at the furnace (point 1) is negligible ( $v_{1} \approx 0$ ), the equation simplifies to:

P_{a} + ρ g H = P_{a} + \frac{1}{2} ρ v_{2}^{2} + ρ g (H + h)

Simplifying further:

ρ g H = \frac{1}{2} ρ v_{2}^{2} + ρ g (H + h)

0 = \frac{1}{2} ρ v_{2}^{2} + ρ g h

\frac{1}{2} ρ v_{2}^{2} = - ρ g h

This leads to an inconsistency, indicating a need to reconsider the application of Bernoulli's equation. Instead, we should consider the pressure difference driving the flow.

4. Calculating the Pressure Difference

The pressure difference driving the flow is due to the difference in air density inside the furnace and the atmospheric air. The buoyant force causes the hot air to rise.

The pressure difference $Δ P$ over the height $h$ of the chimney is:

Δ P = (ρ_{a} - ρ) g h

Substituting the known values:

Δ P = (1.2 - 1.0) \cdot 10 \cdot 9 = 0.2 \cdot 90 = 18 Pa

5. Determining the Velocity of Air Exiting the Chimney

Using the pressure difference to find the exit velocity $v$ of the air:

Δ P = \frac{1}{2} ρ v^{2}

Solving for $v$ :

v = \sqrt{\frac{2 Δ P}{ρ}} = \sqrt{\frac{2 \cdot 18}{1.0}} = \sqrt{36} = 6 m/s

6. Calculating the Mass Flow Rate

The mass flow rate $\dot{m}$ is given by:

\dot{m} = ρ \cdot A \cdot v

Where $A$ is the cross-sectional area of the chimney:

A = π {(\frac{d}{2})}^{2} = 3.14 \cdot {(\frac{0.1}{2})}^{2} = 3.14 \cdot 0.0025 = 0.00785 m^{2}

Now, calculate $\dot{m}$ :

\dot{m} = 1.0 \cdot 0.00785 \cdot 6 = 0.0471 kg/s = 47.1 g/s

Final Answer

The steady mass flow rate of air exiting the chimney is 47.1 grams per second.

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