This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 6 Fundamentals of Convection 639 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/mK and = 0.05 Ns/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 xaxis 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 xmomentum : x P y u y u v x u u  = + 2 2 2 2 = dy u d This is a secondorder 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 = 611 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...
View
Full
Document
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
 BENARD
 Heat Transfer

Click to edit the document details