p0650 - 6.50. CHAPTER 6, PROBLEM 50 749 6.50 Chapter 6,...

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Unformatted text preview: 6.50. CHAPTER 6, PROBLEM 50 749 6.50 Chapter 6, Problem 50 Problem: A water jet of cross-sectional area A with velocity Uj and density ρ causes a cart to move at a constant velocity U = 1 Uj . Use a Galilean transformation and a 5 stationary control volume to solve. (a) What is the rolling resistance of the cart? ˆ (b) The jet velocity is changed to Uj and we observe that the cart velocity doubles, 2 i.e., U = 5 Uj , where Uj is the original jet velocity. What must the incident jet velocity be? NOTE: The rolling resistance is independent of cart velocity. 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U NOTE: The f ow s s eady as seen by a car -f xed observer Solution: This problem is most conveniently solved by using a Galilean transformation so that the cart is at rest as shown in the figure below. The transformed velocity that accomplishes this is ˜ u = u − Ui Since we are given that a cart-fixed observer sees a steady flow, the flow is steady in this coordinate frame. Uj − U d (a) The mass principle is of no interest here as it involves the unknown areas and outflow velocities at the sides of the cart. Rather, it is sufficient to address the x component of the momentum principle. Taking the dot product of the net pressure integral with a unit 750 CHAPTER 6. CONTROL-VOLUME METHOD vector in the x direction, i, to extract the x component, the x-momentum principle is S ρu (u · n) dS = −i · ˜ S (p − pa ) n dS + Rx where the rolling resistance, Rx , appears as a reaction force. Because the pressure is atmospheric everywhere on the chosen control-volume surface, we can say S (p − pa ) n dS = 0 The only part of the control-volume surface across which x momentum passes is through the incident jet. Thus, π ρ (Uj − U ) − (Uj − U ) d2 = Rx 4 Therefore, for U = 1 Uj , the rolling resistance of the cart is 5 R=− 4 πρ Uj2 d2 i 25 (b) As in Part (a), we know that the rolling resistance balances the x-momentum flux, wherefore 2 2 π ˆ R = − ρ Uj − Uj d2 i 4 5 Since the rolling resistance is independent of cart velocity, we can equate this result with the final equation in Part (a). Thus, 2 π ˆ ρ Uj − Uj 4 5 2 d2 = 4 πρ Uj2 d2 25 =⇒ 2 ˆ Uj − Uj 5 2 = 16 2 U 25 j The velocity has two possible values, viz., 6 2 ˆ Uj = Uj , − Uj 5 5 The negative velocity is not physically meaningful. Hence, the jet velocity is 6 ˆ Uj = Uj 5 ...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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