EAS4101_S11_HW3S

# EAS4101_S11_HW3S - EAS 4101 solutions by Damian Ricci and...

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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 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: F U D d jet U () ht Control Volume (a) Control Volume (b) Basic Eqns: Integral form of the continuity equation: 0 dV t  CV CS Vd A    Integral form of the momentum equation: CV CS F VdV V V dA t        Assumptions: a) Incompressible flow b) Inviscid flow c) Flow in cart has no velocity relative to CV

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EAS 4101 S11 solutions by Damian Ricci, Miguel Palaviccini, Matt Williams, 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). The CV on the left is the easier approach. 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 CVs after a quick explanation of the terms in the continuity equation. o It is also important to note that when dealing with problems in which a relative velocity exists between the fluid and control volume, the velocity of interest in the conservation equations is the relative velocity: "" f CV f CV VV   a) If we attach an x-y coordinate system to the lower left hand side of the CV, we can see that the volume of the fluid can be found from the below, with   0 yt ht  . (We are effectively defining a deforming CV.)  2 1 4 D dA d y t d y t  The first term in the continuity equation is then
EAS 4101 S11 solutions by Damian Ricci, Miguel Palaviccini, Matt Williams, and Mark Sheplak 3 HW 3  22 2 0 () 44 4 a ht CV DD D h dd y h t tt t t         The components of the fluid velocity with respect to the control volume at the upper control surface are given by: ˆˆ cos sin fC V jet jet VU U i U j     Using the continuity equation we have:   0s i n s i n jet dh t Dd U dt  Which can be rearranged to   2 2 jet dh t d U dt D b)

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## This note was uploaded on 09/05/2011 for the course EAS 4101 taught by Professor Sheplak during the Spring '08 term at University of Florida.

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EAS4101_S11_HW3S - EAS 4101 solutions by Damian Ricci and...

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