HW1soln - EAS 4101 solutions by Damian Ricci and Mark...

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EAS 4101 solutions by Damian Ricci and Mark Sheplak 1 HW 1 Problem 1 Given: A jet of water issuing into a moving cart, declined at an angle θ as shown. Find: 1. Draw the appropriate control volume, explicity stating unit normal vetors and flux areas. 2. List your assumptions required to solve this problem. 3. What are the pressure boundary conditions on your control volume? 4. What is the time rate of change of the liquid level? 5. What is the thrust force acting on the car by the jet of water? 6. What is the frictional force acting on the car? Schematic: Control Volume (a) Control Volume (b) Basic Eqns: Integral form of the continuity equation: 0 dV t ρ = () CV CS Vd A +⋅ ∫∫ J KJ J K Integral form of the momentum equation: CV CS F VdV V V dA t ρρ Σ= + JK JK JK J JK Assumptions: a) Incompressible flow b) Inviscid flow c) Flow in cart has no velocity relative to CV
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EGM 6812 F03 solutions by Damian Ricci, Miguel Palaviccini, and Mark Sheplak 2 HW 3 d) Uniform jet area and velocity e) Neglect body forces in x-direction f) Cart has constant velocity (no acceleration) Solution: 1. The control volume and normal area vectors are shown in the schematic. For this problem we will take a look at two different control volumes and solve the problem each way with the same results. The CV on the left will be labeled as a) and the CV on the right will be labeled as b). 2. The proper assumptions are listed above. 3. Pressure Boundary conditions: o The pressure is atmospheric around the entire control surface and therefore cancels out its own effects. 4. Here we will look at the solution for both CV after a quick explanation of the terms in the continuity equation. o We know that the volume inside of the cart is given by: ) ( 4 ) ( 2 t h D t Ah π = = o We therefore have that: ) ( 4 2 t dh D d = o Plugging this into the first term of the continuity equation: = = ∫∫∫ ∫∫∫ ) ( 4 ) ( 4 2 2 t h D t t dh D t d t CV CV ρ o We know that the the density is not a function of time (we assumed an incompressible flow) and can therefore pull the density term out of the partial derivative: t t h D t h D t = )) ( ( 4 ) ( 4 2 2
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EGM 6812 F03 solutions by Damian Ricci, Miguel Palaviccini, and Mark Sheplak 3 HW 3 a) The components of the velocity jet U in the control volume are given by: ˆˆ cos sin jet jet jet UU U i U j θ ⎡⎤ =− + ⎣⎦ ± Using the continuity equation we have: ( ) () 22 0s i n 44 s i n jet dh t Dd U dt ππ ρ ρθ ± Which can be rearranged to ( ) 2 2 jet dh t d U dt D = b) Since the jet velocity and area are perpendicular to one another, we do not need to decompose the velocity into i and j components. ± We can take a look at the continuity equation to get (note the negative sign in the second term is due to the fact that the normal area vector and velocity vector are in opposite directions): 4 )) ( ( 4 0 2 2 d U t t h D jet π = ± Which can be rearranged to ( ) 2 2 jet dh t d U dt D = 5.
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This note was uploaded on 03/18/2008 for the course EAS 4101 taught by Professor Sheplak during the Spring '08 term at University of Florida.

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HW1soln - EAS 4101 solutions by Damian Ricci and Mark...

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