2.25 Advanced Fluid Mechanics, Solution to Problem 4.09

MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Problem 4.09 This problem is from “Advanced Fluid Mechanics Problems” by A.H. Shapiro and A.A. Sonin T a , ρa d h g d«H«h 1 2 H Ta Ta + ΔT Consider a furnace of height H with a tall cylindrical smoke stack of diameter d (d « H) and height h (h » H). Air, an ideal gas (P = ρRT ), enters the furnace at atmospheric density and temperature and at local atmospheric pressure. Between stations 1 and 2, heat is added at constant pressure and the air temperature is raised by an amount ΔT . Thereafter, heat addition is negligible and the air rises through the stack at a sensibly constant density. (a) On the assumption that viscous effects are negligible, derive an expression for the steady mass flow rate of air drawn by a stack of given height, h, in terms of the temperature rise in the furnace. (b) If the chimney were capped off at the top, what would be the pressure differntial across the cap, assuming that ΔT would not be altered by the flow stoppage? Note: The height h of the stack is small compared with the length RTa /g over which the atmosphere density falls by 1/e (see Problem 1.8). Hence, gravitational density changes can be neglected. 1 Solution: (a) First consider the effect of heat addition on the air density. Since the pressure at stations 1 and 2 is Pa , the heat is added at constant pressure such that Pa = ρ1 RT1 = ρ2 RT2 ρa Ta = ρ2 (Ta + ΔT ) ρa ⇒ ρ2 = 1 + ΔT /Ta 3 d h g Consider a streamline from station 2 to station 3: 1 2 H Ta Ta + ΔT As stated, the density is reasonably constant (i.e. ρ2 = ρ3 ) and viscous effects are negligible so we can apply Bernoulli’s equation along the streamline shown above. 1 1 P2 + ρ2 v2 2 + ρ2 gh2 = P3 + ρ2 v3 2 + ρ2 gh3 2 2 If we assume that the streamlines at the exit of the smoke stack at station 3 are parallel, then we also set P3 equal to the local ambient pressure at the top of the stack. Provided the air outside of the furnace and stack is isothermal at Ta and has roughly constant density ρa , then we can relate the pressure at station 3, P3 , to the ambient pressure, Pa , at ground level by stations 1 and 2 using our knowledge of hydrostatic pressure, such that Pa = P2 = P3 + ρa gh. Hence the pressure at the top of the smoke stack, P3 , is below the pressure at ground level, Pa = P1 = P2 . Accordingly, 1 ρ2 (v3 2 − v2 2 ) = (P2 − P3 ) + ρ2 g(h2 − h3 ) 2 2 � v3 2 − v2 2 = (P P� 3 + ρa gh − � 3 ) − 2gh ρ2 � ρa v3 2 − v2 2 = 2 − 1 gh ρ2 Conservation of mass from stations 2 to 3 tells us that v2 ∼ d H 2 v3 Given that (d « H), we can neglect v2 such that v3 = 2 ρa − 1 gh ρ2 ⇒ v3 = 2gh ΔT Ta (b) If the chimney were capped, there would be no flow and we can apply our knowledge of fluid statics. 2 cap P3 Consider a control volume around the air in the chimney. Note that this control volume is just below the cap such that the pressure at the top of the CV is not the local atmospheric pressure, but an unkown pressure P3 . z h W 1 2 Static equilibrium gives: Fz = −WCV + P2 (πd2 /4) − P3 (πd2 /4) = 0 −ρ2 gh(πd2 /4) + Pa (πd2 /4) − P3 (πd2 /4) = 0 ρa gh + Pa − P3 = 0 − 1 + ΔT /Ta ρa P3 = Pa − gh 1 + ΔT /Ta The ambient pressure above the cap is calculated from our knowledge of hydrostatic pressure as before, Pa,cap = Pa − ρa gh. Hence the pressure differential ΔPcap = P3 − Pa,cap across the cap is ΔPcap = ρa ghΔT Ta + ΔT D 3