Bravo! That is the right answer. The Equation of Continuity is

Let the subscripts "1" represent the duct and "2" represent the room. Even then, it looks like we might be in trouble for we don't know anything about v_{2}the velocity in the room and it's not at all clear what we would take as A_{2}, some sort of cross-sectional area for the room. Nonetheless, let's start with this and see what happens.A _{1}v_{1}= A_{2}( l_{2}/ t)A

_{1}v_{1}= (A_{2}l_{2}) / tA

_{1}v_{1}= V_{2}/ tV

_{2}is the total volume of the room and t is thetimeof 20 minutes,V _{2}= (2.4 m) (3.0 m) (4.0 m)V

_{2}= 28.8 m^{3}A

_{1}v_{1}= (28.8 m^{3}) / (20 min)A

_{1}v_{1}= 1.44 (m^{3}/min)A

_{1}= (10 cm) (30 cm) = 300 cm^{2}As always,

be careful with the units!It is probably easier tokeepeverything inmetersfrom the very beginning,A _{1}= (0.10 m) (0.30 m) = 0.03 m^{2}A

_{1}v_{1}= (0.03 m^{2}) v_{1}= 1.44 (m^{3}/min)v

_{1}= [1.44 (m^{3}/min)] / [ 0.03 m^{2}]

v_{1}= 48 m/minThat answer is fine but we often or usually express speeds like this in m/s or cm/s so let's go ahead and make that conversion,

v _{1}=^{48 m}/_{min}[^{min}/_{60 s}]

v_{1}= 0.8^{m}/_{s}That is the speed of the air from the duct or vent.

(c) 2000, Doug Davis; all rights reserved.