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Unformatted text preview: (the room) is 4 students/sec. Therefore, in terms of the notation above, we have dN R dt = 4 students sec Also, since the space surrounding each student is 40 ft 3 , the number density is = 1 40 students ft 3 Finally, the area of the only part of the control volume through which students pass, the door, is A = (7 . 5 ft) (3 . 5 ft) = 26 . 25 ft 2 Substituting into the Reynolds Transport equation as specialized above for this problem, and denoting the speed of the students by U = u n , we have 4 students sec = UA = X 1 40 students ft 3 ~ U p 26 . 25 ft 2 Q Solving for U , we conclude that U = 6 . 1 ft sec Moving this rapidly, we can reasonably conclude that they are racing home to boast about how well they did on the exam!...
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This note was uploaded on 10/12/2009 for the course AME 309 at USC.