An ideal gas of density ρ = 0 . 2   kg   m - 3 enters a chimney of height h at the rate of α = 0 . 8   kg   s - 1 from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is A 1 = 0 . 1   m 2 and the…

Question

An ideal gas of density &#961; = 0 . 2 &#160; kg &#160; m - 3 enters a chimney of height h at the rate of &#945; = 0 . 8 &#160; kg &#160; s - 1 from its lower end, and escapes through the upper end as shown in the figure. The cross-sectional area of the lower end is A 1 = 0 . 1 &#160; m 2 and the upper end is A 2 = 0 . 4 &#160; m 2 . The pressure and the temperature of the gas at the lower end are 600 &#160; Pa and 300 &#160; K , respectively, while its temperature at the upper end is 150 &#160; K . The chimney is heat insulated so that the gas undergoes adiabatic expansion. Take g = 10 &#160; m &#160; s - 2 and the ratio of specific heats of the gas &#947; = 2 . Ignore atmospheric pressure. Which of the following statement(s) is(are) correct?

Aicharya · Accepted Answer

Here the gas undergoes adiabatic expansion,
$P^{1 - γ} T^{γ} =$ constant
$\Rightarrow \frac{P_{2}}{P_{1}} = {(\frac{T_{1}}{T_{2}})}^{\frac{γ}{1 - γ}}$
$\Rightarrow P_{2} = {(\frac{300}{150})}^{\frac{2}{1 - 2}} \times 600$
$\Rightarrow P_{2} = \frac{600}{4} = 150 Pa$
Thus, pressure of the gas at the upper end of the chimney is $P_{2} = 150 Pa$
From ideal gas equation, $ρ = \frac{PM}{RT} \Rightarrow ρ \propto \frac{P}{T}$
Then, $\frac{ρ_{1}}{ρ_{2}} = (\frac{P_{1}}{P_{2}}) (\frac{T_{1}}{T_{2}}) = (\frac{150}{600}) (\frac{300}{150}) = \frac{1}{2}$
$\Rightarrow ρ_{2} = (0.2) \frac{1}{2} = 0.1 kg m^{- 3}$
Rate of mass flow $α = 0.8 kg s^{- 1}$
$\Rightarrow$ Volume flow rate at top $= \frac{0.8}{0.1} = 8 m^{3} s^{- 1}$
Velocity of gas at lower end $v_{1} = \frac{V_{1}}{A_{1}} = \frac{0.8}{0.2 \times 0.1} = 40 m s^{- 1}$
Velocity of gas at upper end $v_{2} = \frac{0.8 \times 2}{0.2 \times 0.4} = 20 m s^{- 1}$
Work done $W_{o n g a s} = ∆ K + ∆ U_{g} + internal energy$
$P_{1} A_{1} ∆ x_{1} - P_{2} A_{2} ∆ x_{2} = \frac{1}{2} ∆ m {v_{1}}^{2} - \frac{1}{2} ∆ m {v_{2}}^{2} + m g ∆ h + \frac{f}{2} (P_{2} ∆ V_{2} - P_{1} ∆ V_{1})$
Solving we get $h = 360 m$

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An ideal gas of density ρ = 0 . 2   kg   m - 3 enters a chimney of height h at the rate of…

Solution

An ideal gas of density &#961; = 0 . 2 &#160; kg &#160; m - 3 enters a chimney of height h at the rate of…

Solution

An ideal gas of density ρ = 0 . 2 kg m - 3 enters a chimney of height h at the rate of…