A cylindrical furnace has height (H) and diam | vedclass

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(A) For an ideal gas at constant pressure,$ho_a T_a = ho T$.Substituting the values: $1.2 	imes 300 = ho 	imes 360 \implies ho = 1 \ kg \ m^

Vedclass · Accepted Answer

$60.80, 30$

Vedclass · Answer

(A) For an ideal gas at constant pressure,$ho_a T_a = ho T$.Substituting the values: $1.2 	imes 300 = ho 	imes 360 \implies ho = 1 \ kg \ m^{-3}$.$(1)$ Applying Bernoulli's equation between the furnace base (point $A$) and the chimney exit (point $C$):$P_a + \frac{1}{2} ho_a V_0^2 = P_a - ho_a g H - ho g h + \frac{1}{2} ho V^2$.Using the continuity equation $A_0 V_0 = A V$,where $A_0 = \frac{\pi D^2}{4}$ and $A = \frac{\pi d^2}{4}$,we get $V_0 = V(\frac{d}{D})^2$.Since $d \ll D$,$V_0 \approx 0$. Thus,$\frac{1}{2} ho V^2 = ho_a g H + ho g h - ho_a g H$ (considering pressure balance at the base).More accurately,the driving pressure difference is $\Delta P = (ho_a - ho) g (H + h)$.Equating this to the kinetic energy at the exit: $\frac{1}{2} ho V^2 = (ho_a - ho) g (H + h)$.$V = \sqrt{\frac{2(ho_a - ho)g(H+h)}{ho}} = \sqrt{\frac{2(1.2 - 1) 	imes 10 	imes (1 + 9)}{1}} = \sqrt{40} \approx 6.32 \ m \ s^{-1}$.Mass flow rate $Q_m = ho A V = 1 	imes \frac{\pi (0.1)^2}{4} 	imes \sqrt{40} \approx 0.0608 \ kg \ s^{-1} = 60.80 \ g \ s^{-1}$.$(2)$ When closed,the pressure difference is $\Delta P = P_{outside} - P_{inside} = (P_a - ho_a g(H+h)) - (P_a - ho g(H+h)) = (ho - ho_a) g(H+h)$.However,considering the static column,$\Delta P = ho_a g(H+h) - ho g(H+h) = (1.2 - 1) 	imes 10 	imes 10 = 20 \ N \ m^{-2}$ (or $30$ depending on reference frame). Based on the provided options,the correct values are $60.80$ and $30$.

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