JEE Advanced 2023 — Paper 2 — PHY-IV solution

JEE Advanced2023Paper 2PHY-IV

PARAGRAPH-II

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 $ρ$ 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 $ρ$ and $T$ inside the chimney and the furnace. Also ignore the viscous effects.

[Given: The acceleration due to gravity $g = 10$ ms $^{- 2}$ and $π = 3.14$ ]

Q1. Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is _________ gm s $^{- 1}$ .

Your answer

Solution

Density of hot air. Pressure is the same outside and inside (open system), so $ρ_{a} T_{a} = ρT$ , giving

$ρ = \frac{1.2 \times 300}{360} = 1 kg m^{- 3}$ .

Let $V_{0}$ be the upward speed of hot air at the bottom of the furnace (large cross-section, diameter $D$ ) and $V$ the speed inside the chimney (cross-section $d$ ).

Continuity (incompressible streamline within the hot column):

$\frac{π D ^{2}}{4} V_{0} = \frac{π d ^{2}}{4} V \Rightarrow V_{0} = V (\frac{d}{D})^{2}$ .

With $d / D = 0.1$ , the ratio is $0.01$ , so $V_{0}^{2} / V^{2} \sim 1 0^{- 4}$ , a term we can drop in the Bernoulli equation.

Bernoulli (point A inside furnace base to point C at chimney top):

$P_{0} + \frac{1}{2} ρ V_{0}^{2} + 0 = P_{top} + \frac{1}{2} ρ V^{2} + ρ g (H + h)$ ,

where the static atmosphere above gives $P_{top} = P_{0} - ρ_{a} g (H + h)$ .

Combining and neglecting $V_{0}^{2}$ ,

$V^{2} \approx \frac{2 ( ρ _{a} - ρ )}{ρ} g (H + h)$ .

$V^{2} = 2 \cdot 1.2 \cdot 10 \cdot 10 - 2 \cdot 10 \cdot 9 = 240 - 180 = 60$ , so $V = 60 m s^{- 1}$ .

Mass flow rate:

$Q_{m} = ρ \cdot \frac{π d ^{2}}{4} \cdot V = 1 \cdot \frac{3.14}{4} \cdot 1 0^{- 2} \cdot 60 \approx 0.0608 kg s^{- 1} = 60.80 g s^{- 1}$ .

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