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

Question

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 m s - 2 and π = 3 . 14 ] When the chimney is closed using a cap at the top, a pressure difference ∆ P 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 ∆ P is _____ N m − 2 .

TestHub · Accepted Answer

Applying Bernoulli’s theorem between top and bottom points of furnace :
$P_{a} + \frac{1}{2} ρ_{a} V^{2} = P_{a} + ρ g H + \frac{1}{2} ρ V^{2} . . . (1)$
Also, since $P M = ρ R T$
$\Rightarrow ρ_{a} \times 300 = ρ \times 360$
$\Rightarrow ρ = 1 kg m^{- 3} . . . (2)$
From equation(1), we can write
$\Rightarrow \frac{V^{2}}{2} (0.2) = 1 \times 10 \times 1$
$\Rightarrow V = 10 m s^{- 1}$
Now, let $V'$ be the speed of air in chimney.
$\Rightarrow$ By continuity equation,
$\frac{π D^{2}}{4} \cdot V = \frac{π d^{2}}{4} \cdot V'$
$\Rightarrow V' = 100 V = 1000 m s^{- 1}$
Therefore, pressure difference $∆ p$
$= ρ V'^{2} \cdot$ area
$= 1 \times {(1000)}^{2} \times \frac{π d^{2}}{4} = 1 \times {(1000)}^{2} \times \frac{3.14 \times {(0.1)}^{2}}{4} = 7850 N m^{2}$

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