Question 106: Thermodynamics & KTG

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 $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ and $\pi=3.14$ ]

Accepted Answer

We know, $$
\begin{aligned}
& \mathrm{P_a}=\frac{ho R \mathrm{T}}{M}
\end{aligned}
$$
For a gas R and M are constant. So, $$ho T$$ = Constant (for constant pressure).
The density of hot air inside the furnace is = $$ho $$
The air gets heated inside the furnace at constant pressure P
a
.
$$ 	herefore $$ $$
ho_{\mathrm{a}} \mathrm{T}_{\mathrm{a}}=ho \mathrm{T}
$$
$$ \Rightarrow $$ 1.2 $$ 	imes $$ 300 = $$ho $$ $$ 	imes $$ 360
$$\Rightarrow ho=1 \mathrm{~kg} / \mathrm{m}^3$$
Buoyant force applied on the hot air = $ho_{\mathrm{a}}$Vg (Upward direction)
Weight of the hot air = $$ho $$Vg (Downward direction)
$$ 	herefore $$ Net force on the hot air = $ho_{\mathrm{a}}$Vg - $$ho $$Vg
Let acceleration of the hot air in the upward direction = $a$
and mass of the hot air = $$ho $$V
$$ 	herefore $$ $$ho $$V$a$ = $ho_{\mathrm{a}}$Vg - $$ho $$Vg
$$ \Rightarrow $$ $a$ = $${{{ho _a}Vg - ho Vg} \over {ho V}}$$
= $${{{ho _a}g - ho g} \over ho }$$
= $${{1.2 	imes 10 - 1 	imes 10} \over 1}$$
= 2 m/s
2
$$ 	herefore $$ Velocity(v) of the hot air when exiting the chimney using formula $${v^2} = {u^2} + 2ah$$,
$${v^2} = 0 + 2 	imes 2 	imes 9$$
$$ \Rightarrow $$ v = 6 m/s
Mass flow rate = $${{dm} \over {dt}}$$ = $$ho $$Av
= $$
ho 	imes \frac{\pi \mathrm{d}^2}{4} 	imes\mathrm{v}=1 	imes \frac{\pi}{4} 	imes 10^{-2} 	imes 6
$$
= $${{471} \over {10000}}$$ kg/s = $${{471} \over {10000}} 	imes 1000$$ gm/s = 47.1

Thermodynamics & KTG

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