AME-309 Exam1-soln

# 4 16h 12h 21h problem 3 to begin we must define the

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Unformatted text preview: substituting into the moment-balance equation with Fs = 0 yields 4ρgH (H + h)w H + 1 H2 3H +h = 6ρgH 2 w 7 H 6 or, 3H + 3h + H = 7ρgH 3 w =⇒ 3 Solving for h, there follows 5 h= H 12 4ρgH 2 w Hence, since 1 H < 3 5 12 H , the LEFT gate will open first. 4 16H + 12h = 21H Problem 3. To begin, we must define the system and the control volume. Let the system be all living harmless tropical fish and define the control volume to be the aquarium. Also, let N denote the number of living harmless tropical fish in the system and Ncv the number of living harmless tropical fish in the control volume. Then, since our control volume is stationary so that ucv = u, letting ρ denote living harmless tropical fish number density, the Reynolds Transport Theorem tells us that d dN = dt dt ρ dV + V S ρ u · n dS By definition of ρ, necessarily Ncv = ρ dV . Also, the net flux of harmless tropical fish out of the control volume is (they’re leaving the tank regardless of partition chosen): S ρ u · n dS = nU A + nU A = 2nU A So, our basic conservation law is dN dNcv = + 2nU A dt dt Now, the rate of change of the number of harmless tropical fish in the system is half the rate at which the number of harmless tropical fish, Ncv , is changing. Thus, dN dNcv =2 dt dt and our equation becomes dN dN =2 + 2nU A dt dt =⇒ dN = −2nU A dt Therefore, we conclude that the shark is feasting at a rate of 2nU A harmless tropical fish per second. 5...
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