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STD 11 - 9.1. fluid mechanics — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 9.1. fluid mechanics4 Marks
Question
A cylindrical furnace has height $(H)$ and diameter $(D)$ both $1 \mathrm{~m}$. It is maintained at temperature $360 \mathrm{~K}$. The air gets heated inside the furnace at constant pressure $P_a$ and its temperature becomes $T=360 \mathrm{~K}$. The hot air with density $\rho$ rises up a vertical chimney of diameter $d=0.1 \mathrm{~m}$ and height $h=9 \mathrm{~m}$ above the furnace and exits the chimney (see the figure). As a result, atmospheric air of density $\rho_a=1.2 \mathrm{~kg} \mathrm{~m}^{-3}$, pressure $P_a$ and temperature $T_a=300 \mathrm{~K}$ enters the furnace. Assume air as an ideal gas, neglect the variations in $\rho$ and $T$ inside the chimney and the furnace. Also ignore the viscous effects.

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

(image)

($1$) Considering the air flow to be streamline, the steady mass flow rate of air exiting the chimney is

. . . . .$\mathrm{gm} \mathrm{s}^{-1}$.

($2$) When the chimney is closed using a cap at the top, a pressure difference $\Delta P$ 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 $\Delta P$ is. . . . .$\mathrm{Nm}^{-2}$.

✓

Answer

(image)

$\rho_0 T_0=\rho T$

$\Rightarrow 1.2 \times 300=\rho(360) \therefore \rho=1$

Between $A$ & $B$

$\mathrm{P}_0+\frac{1}{2} \rho \mathrm{V}_0^2=\mathrm{P}+\frac{1}{2} \rho^2+\rho \mathrm{gh}$

$\frac{\pi \mathrm{D}^2}{4} \mathrm{~V}_0=\frac{\pi \mathrm{d}^2}{4} \mathrm{~V} \ldots \ldots \ldots \text { (2) }$. . . . .

Between $B$ & $C$

$\mathrm{P}+\frac{1}{2} \rho^2=\mathrm{P}_0-\rho_0 \mathrm{~g}(\mathrm{H}+\mathrm{h})+\frac{1}{2} \rho^2 \text {. }$

from $(1)$ & $(2)$ :

$\Rightarrow P_0+\frac{1}{2} \rho\left(V \frac{d^2}{D^2}\right)^2=P+\frac{1}{2} \rho^2+\rho g h$

$\Rightarrow \rho_0 g(H+h)=\frac{1}{2} \rho V^2\left[1-\frac{d^4}{D^4}\right]+\rho g h$

$\Rightarrow V^2 \simeq \frac{2 \rho_0}{\rho} g(H+h)-2 g h$

$=2 \times 1.2 \times 10 \times 10-2 \times 10 \times 9$

$=240-180=60 \quad \therefore V=\sqrt{60 m} \mathrm{~m} / \mathrm{s}$

$Q_m=\rho \frac{\pi d^2}{4} V=1 \times \frac{\pi}{4} \times 10^{-2} \times \sqrt{60}=60.80$

(image)

$\mathrm{P}=\text { constant }$

$\Rightarrow \rho_2 \mathrm{~T}_2=\rho \mathrm{T}$

$1.2 \times 300=\rho \times 360$

$\rho=1 \mathrm{~kg} / \mathrm{m}^3$

$\Delta \mathrm{P}=\rho_2 \mathrm{~g}(\mathrm{~h}+\mathrm{H})-\rho \mathrm{gh}$

$=1.2 \times 10 \times 10-1 \times 10 \times 9$

$=120-90=30 \mathrm{~N} / \mathrm{m}^2$

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