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MECH 660 Advanced Fluid Mechanics March 17, 2004 American University of Beirut, Spring 2005 Handout # PS4 Solution Problem Set 4 Solution: Viscous Flow “Advanced Fluid Mechanics Problems” by Shapiro and Sonin Problems 6.1, 6.3, 6.7, 6.10, 6.16, 6.20, 6.21. Problem 6.1 Refer to Figure 1 for the schematic. The flow is steady. M OIL L h D p p h Q R z r Figure 1: Schematic of Problem 6.1 (a) We assume that the flow is steady and fully developed in the channel between the piston and cylinder. The z - component of the momentum equation is 0 = - ∂p ∂z + μ 2 u ∂r 2 + 1 r ∂u ∂r where gravity is neglected. We can neglect the term 1 r ∂u ∂r because 1 r ∂u ∂r 2 u ∂r 2 h R << 1 where R = D/ 2. From the momentum conservation in the r - direction, we can show that ∂p ∂r 0 p = p ( z ) and ∂p ∂z = dp dz Therefore equation (1) reduces to 2 u ∂r 2 = 1 μ dp dz subject to the boundary conditions u = 0 at r = R - h u = 0 at r = R 1

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The velocity is then given by u ( r ) = 1 2 μ dp dz r 2 - (2 R - h ) r + R ( R - h ) The volume flow rate is then given by Q = 2 π R - h 2 h R - h u ( r ) dr = - 2 π R - h 2 h 3 12 μ dp dz We conclude from the last equation that dp/dz is constant, dp dz = - p 1 - p a L = - Mg πR 2 L where the shear force on the piston acting vertically upwards has been neglected. The leakage volume flow rate is finally given as Q = Mgh 3 6 μR 2 L R - h 2 Mgh 3 6 μRL 2
Problem 6.3 Refer to Figure 2 for the schematic. The flow is steady. P o Q D F h d r z Figure 2: Schematic of Problem 6.3 (a) We assume that the flow is steady and fully developed in the channel between the upper and lowe plates away from the center, i.e. for r > d/ 2. The momentum equation

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