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Heat Chap06-039

# Heat Chap06-039 - Chapter 6 Fundamentals of Convection 6-39...

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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 μ = 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 : 0 = + y v x u → 0 = 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 0 / = 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 - μ = + ρ 2 2 → 0 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 0 μ + = y u y T k → 2 2 2 μ - = 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 + + μ - = V Applying the boundary conditions T (0) = T 0 and T ( L ) = T 0 gives the temperature distribution to be - μ + = 2 2 2 0 2 ) ( L y L y k T y T V The temperature gradient is determined by differentiating T ( y ) with respect to y , - μ = 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 , 0 2 1 2 2 = - μ = L y kL dy dT V → 2 L y = 6-11 12 m/s 6 cm 3000 rpm 20 cm

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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 ( = = = π π n D V 2 m 0377 . 0 ) m 20 . 0 )( m 0.06 ( = π = π = bearing DL A The maximum temperature is C 53.3 ° = ° + ° = μ + = - μ + = = m/s N 1 W 1 ) C W/m 8(0.17 m/s) 425 . 9 )( s/m N 05 . 0 ( C 50 8 ) 2 / ( 2 / 2 ) 2 / ( 2 2 2 0 2 2 2 0 max k T L L L L k T L T T V V
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