The technicians with correct approach.

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Chapter 2, Problem 6MC

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The technicians with correct approach.

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You are working as an engineer in a bearing systems design company. The flow of lubricant inside a hydrodynamic bearing (p = 0.001 kg m-1 s-1) can be approximated as a parallel, steady, two-dimensional, incompressible flow between two parallel plates. The top plate, representing the moving part of the bearing, travels at a constant speed, U, while the bottom plate remains stationary (Figure Q1). The plates are separated by a distance of 2h = 1 cm and are W = 20 cm wide. Their length is L = 10 cm. By applying the above approximations to the Navier-Stokes equations and assuming that end effects can be neglected, the horizontal velocity profile can be shown to be y = +h I 2h = 1 cm x1 y = -h u(y) 1 dP 2μ dx -y² + Ay + B moving plate stationary plate U 2 I2 L = 10 cm Figure Q1: Flow in a hydrodynamic bearing. The plates extend a width, W = 20 cm, into the page.

Question 1 You are working as an engineer in a bearing systems design company. The flow of lubricant inside a hydrodynamic bearing (µ = 0.001 kg m¯¹ s¯¹) can be approximated as a parallel, steady, two-dimensional, incompressible flow between two parallel plates. The top plate, representing the moving part of the bearing, travels at a constant speed, U, while the bottom plate remains stationary (Figure Q1). The plates are separated by a distance of 2h = 1 cm and are W = 20 cm wide. Their length is L = 10 cm. By applying the above approximations to the Navier-Stokes equations and assuming that end effects can be neglected, the horizontal velocity profile can be shown to be 1 dP u(y) = 2μ dx -y² + Ay + B y= +h Ꮖ 2h=1 cm 1 x1 y = −h moving plate stationary plate 2 X2 L = 10 cm Figure Q1: Flow in a hydrodynamic bearing. The plates extend a width, W = 20 cm, into the page. (a) By considering the appropriate boundary conditions, show that the constants take the following forms: U U 1 dP A =…