p0462 - (the room) is 4 students/sec. Therefore, in terms...

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436 CHAPTER 4. KINEMATICS 4.62 Chapter 4, Problem 62 We can apply the Reynolds Transport Theorem to this problem if we consider the students to be a system and the room a control volume. Since the room is stationary, u rel = u . Thus, the Reynolds Transport Theorem tells us dN dt = d dt 888 V ρ dV + 8 s 8 S ρ ( u · n ) dS where N is the total number of students, and ρ is the number density of students, i.e., number of students per unit volume. We must have dN/dt =0 as we are neither creating nor destroying students (it was not a “killer exam”). Also, by definition, the number of students in the room, N R ,is N R = 888 V ρ dV So, the Reynolds Transport Theorem simplifies to dN R dt = 8 s 8 S ρ ( u · n ) dS In words, this equation says the rate of change of the number of students remaining in the room is minus the net flux of students out of the room. Now, we know that the rate of decrease of the number of students in the control volume
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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.

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