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Unformatted text preview: Chapter 6 Fundamentals of Convection 6-39 The oil in a journal bearing is considered. The velocity and temperature distributions, the maximum temperature, the rate of heat transfer, and the mechanical power wasted in oil are to be determined. Assumptions 1 Steady operating conditions exist. 2 Oil is an incompressible substance with constant properties. 3 Body forces such as gravity are negligible. Properties The properties of oil at 50 C are given to be k = 0.17 W/m-K and = 0.05 N-s/m 2 Analysis ( a ) Oil flow in journal bearing can be approximated as parallel flow between two large plates with one plate moving and the other stationary. We take the x-axis to be the flow direction, and y to be the normal direction. This is parallel flow between two plates, and thus v = 0. Then the continuity equation reduces to Continuity : = + y v x u = x u u = u ( y ) Therefore, the x- component of velocity does not change in the flow direction (i.e., the velocity profile remains unchanged). Noting that u = u ( y ), v = 0, and / = x P (flow is maintained by the motion of the upper plate rather than the pressure gradient), the x- momentum equation reduces to x-momentum : x P y u y u v x u u - = + 2 2 2 2 = dy u d This is a second-order ordinary differential equation, and integrating it twice gives 2 1 ) ( C y C y u + = The fluid velocities at the plate surfaces must be equal to the velocities of the plates because of the no- slip condition. Taking x = 0 at the surface of the bearing, the boundary conditions are u (0) = 0 and u ( L ) = V , and applying them gives the velocity distribution to be V L y y u = ) ( The plates are isothermal and there is no change in the flow direction, and thus the temperature depends on y only, T = T ( y ). Also, u = u ( y ) and v = 0. Then the energy equation with viscous dissipation reduce to Energy : 2 2 2 + = y u y T k 2 2 2 - = L dy T d k V since L y u / / V = . Dividing both sides by k and integrating twice give 4 3 2 2 ) ( C y C L y k y T + + - = V Applying the boundary conditions T (0) = T and T ( L ) = T gives the temperature distribution to be - + = 2 2 2 2 ) ( L y L y k T y T V The temperature gradient is determined by differentiating T ( y ) with respect to y , - = L y kL dy dT 2 1 2 2 V The location of maximum temperature is determined by setting dT / dy = 0 and solving for y , 2 1 2 2 = - = L y kL dy dT V 2 L y = 6-11 12 m/s 6 cm 3000 rpm 20 cm Chapter 6 Fundamentals of Convection Therefore, maximum temperature will occur at mid plane in the oil. The velocity and the surface area are m/s 425 . 9 s 60 min 1 ) rev/min 3000 )( m 0.06 ( = = = n D V 2 m 0377...
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This homework help was uploaded on 03/19/2008 for the course ME 410 taught by Professor Benard during the Spring '08 term at Michigan State University.
- Spring '08
- Heat Transfer