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JEE Advanced 2023Paper 2
Exam
JEE Advanced
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Advanced 2023 · Paper 2 · Physics · Mechanical Properties of Fluids

A cylindrical furnace has height (H) and diameter (D) both 1m . It...

A cylindrical furnace has height (H)\left(H\right) and diameter (D)\left(D\right) both 1m1m. It is maintained at temperature 360K360K. The air gets heated inside the furnace at constant pressure Pa{P}_{a} and its temperature becomes T=360KT = 360K. The hot air with density ρρ rises up a vertical chimney of diameter d=0.1md = 0 . 1m and height h=9mh = 9m above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density ρa=1.2kgm−3{ρ}_{a} = 1 . 2kg{m}^{- 3}, pressure Pa{P}_{a} and temperature Ta=300K{T}_{a} = 300K enters the furnace. Assume air as an ideal gas, neglect the variations in ρandTρ and T inside the chimney and the furnace. Also ignore the viscous effects.

[Given: The acceleration due to gravity g=10ms−2andπ=3.14g = 10m{s}^{- 2} and \pi = 3 . 14]

A cylindrical furnace has height (H) and diameter (D) both 1m . It is maintained at temperature 360K . The air gets


Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is _____ gms−1gm{s}^{- 1}.

Correct answer

0

Step-by-step solution

Applying Bernoulli’s theorem between top and bottom points of furnace :

Pa+12ρaV2=Pa+ρgH+12ρV2...(1){P}_{a} + \dfrac{1}{2}{ρ}_{a}{V}^{2} = {P}_{a} + ρgH + \dfrac{1}{2}ρ{V}^{2} . . . \left(1\right)

Also, since PM=ρRTPM = ρRT

⇒ρa×300=ρ×360⇒ {ρ}_{a} \times 300 = ρ \times 360

⇒ρ=1kgm−3...(2)⇒ ρ = 1kg{m}^{- 3} . . . \left(2\right)

From equation(1), we can write

⇒V22(0.2)=1×10×1⇒ \dfrac{{V}^{2}}{2}\left(0 . 2\right) = 1 \times 10 \times 1

⇒V=10ms−1⇒ V = 10m{s}^{- 1}

Required mass flow rate =V×πD24×ρ= V \times \dfrac{\pi{D}^{2}}{4} \times ρ

=10×3.144×12×1kgs−1= 10 \times \dfrac{3 . 14}{4} \times {1}^{2} \times 1kg{s}^{- 1}

=7.85kgs−1= 7 . 85kg{s}^{- 1}

=7850gms−1= 7850gm{s}^{- 1}

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