JEE Advanced 2023 Paper 2 · Q16 · Friction & NLM
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 $\rho$ rises up a vertical chimney of diameter $d = 0.1$ m and height $h = 9$ m above the furnace and exits the chimney. As a result, atmospheric air of density $\rho_a = 1.2$ kg·m$^{-3}$, pressure $P_a$ and temperature $T_a = 300$ K enters the furnace. Considering air flow streamline and ignoring viscous effects, the steady mass flow rate of air exiting the chimney is ___ gm·s$^{-1}$. [$g = 10$ m·s$^{-2}$, $\pi = 3.14$]
Reveal answer + step-by-step solution
Correct answer:60.8
Solution
Density of hot air: $\rho_{\text{hot}} = \rho_{a}\dfrac{T_{a}}{T} = 1.2\cdot\dfrac{300}{360} = 1.0$ kg/m³.
Take section (1) at the furnace inlet at ground level (atmospheric, $v_{1}\approx 0$, $P_{1}=P_{a}$) and section (2) at the chimney exit, height $H+h = 10$ m above ground. The pressure of the atmosphere just outside the exit is $$P_{2} = P_{a} - \rho_{a}g(H+h) = P_{a} - \rho_{a}g(10).$$ Inside, the hot-air column above the furnace has length $h=9$ m. Apply Bernoulli for the flowing hot air from (1) to (2): $$P_{a} + 0 + 0 = \big[P_{a} - \rho_{a}g(10)\big] + \tfrac{1}{2}\rho_{\text{hot}}v_{2}^{2} + \rho_{\text{hot}}g(9).$$ $$\tfrac{1}{2}(1)v_{2}^{2} = \rho_{a}g(10) - \rho_{\text{hot}}g(9) = 120 - 90 = 30 \Rightarrow v_{2}^{2}=60,\ v_{2}=\sqrt{60}\ \text{m/s}.$$
Cross-section of chimney: $A = \pi(d/2)^{2} = \pi(0.05)^{2} = 7.85\times 10^{-3}$ m². Mass flow rate: $\dot m = \rho_{\text{hot}}\,A\,v_{2} = (1)(7.85\times 10^{-3})(\sqrt{60}) \approx 0.0608$ kg/s $= 60.8$ g/s.
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