A 6-cm-diameter shaft rotates at \(3000 \mathrm{rpm}\) in a 20 -cm-long bearing with a uniform clearance of \(0.2 \mathrm{~mm}\). At steady operating conditions, both the bearing and the shaft in the vicinity of the oil gap are at \(50^{\circ} \mathrm{C}\), and the viscosity and thermal conductivity of lubricating oil are \(0.05 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\) and \(0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). By simplifying and solving the continuity, momentum, and energy equations, determine \((a)\) the maximum temperature of oil, \((b)\) the rates of heat transfer to the bearing and the shaft, and \((c)\) the mechanical power wasted by the viscous dissipation in the oil.

First, we need to determine the angular velocity of the shaft, which is rotating at 3000 revolutions per minute (rpm). To convert rpm to radians per second (rad/s), we can use the following conversion factor: \(\omega=\text{rpm}\cdot \frac{2\pi\,\text{rad}}{60\,\text{s}}\) Plugging in the given rpm value (3000): \(\omega = 3000\, \mathrm{rpm} \cdot \frac{2 \pi\, \mathrm{rad}}{60\, \mathrm{s}} = 314.16\, \mathrm{rad/s}\)

Next, we need to find the pressure distribution in the oil film by solving the Reynolds equation: \(\frac{\partial}{\partial x} \left( h^3 \frac{\partial p}{\partial x} \right) + \frac{\partial}{\partial z} \left(h^3 \frac{\partial p}{\partial z}\right) = 6 U h\) Where \(U=\omega R\) is the linear velocity, \(R\) is the shaft radius, and \(h\) is the oil film thickness. For the given exercise, we have uniform oil film thickness (\(h\) = 0.2 mm) and can assume the pressure gradient in the \(z\)-direction is negligible. Thus, the Reynolds equation simplifies to: \(\frac{\mathrm{d}}{\mathrm{d}x} \left( h^3 \frac{\mathrm{d}p}{\mathrm{d}x} \right) = 6 U h\) Now, plug in the values for \(U, h\) and \(\omega\), and solve for the pressure distribution in the oil film, \(p(x)\). However, it is worth noticing that the distribution of pressure in this case is not needed to calculate the desired variables given by (a), (b) and (c).

The heat generation rate in the lubricating oil can be found by considering the power wasted by viscous dissipation: \(Q_{\text{viscous}}=\int_S\tau_{xy} \cdot \frac{\partial u}{\partial y}\,\mathrm{d}S\) Where \(\tau_{xy}\) is the shear stress due to the oil viscosity, \(u\) is the horizontal velocity of oil, and \(S\) is the surface area of the oil film. With the viscosity of oil given as 0.05 N s/m² and the relation \(\tau_{xy} = \mu \frac{\partial u}{\partial y}\), we can plug this expression into the power equation to get: \(Q_{\text{viscous}}=\int_S\mu \frac{\partial u}{\partial y} \cdot \frac{\partial u}{\partial y}\,\mathrm{d}S\)

Now we need to use the heat generation rate and the energy equation to find the variables of interest: (a) To find the maximum temperature of oil, we can use the energy equation: \(\mathrm{d}T = \frac{Q_{\text{viscous}}}{c_p \rho V}\) Where \(c_p\) is the specific heat of oil, \(\rho\) is the oil density, and \(V\) is the volume of the oil. Since we are given the oil's thermal conductivity (\(k\) = 0.17 W/m·K) and viscosity, we can estimate the specific heat and density using typical values for lubricating oil. Then, we can solve the energy equation to find the maximum temperature of oil. (b) To calculate the rates of heat transfer to the bearing and shaft, we can use Newton's Law of cooling: \(Q_{\text{shaft}}=h_{\text{conv,shaft}} A_{\text{shaft}}(T_{\text{oil}}-T_{\text{shaft}})\) \(Q_{\text{bearing}}=h_{\text{conv,bearing}} A_{\text{bearing}}(T_{\text{oil}}-T_{\text{bearing}})\) Where \(h_{\text{conv,shaft}}\) and \(h_{\text{conv,bearing}}\) are the convective heat transfer coefficients for the shaft and bearing, respectively, and \(A_{\text{shaft}}\) and \(A_{\text{bearing}}\) are their respective surface areas. These equations account for the temperature difference between the oil and solid surfaces, as well as the heat transfer coefficients, which can be estimated based on the oil properties and flow conditions. (c) The mechanical power wasted by the viscous dissipation in the oil can be found using the heat generation rate calculated in Step 3: \(P_{\text{wasted}}=Q_{\text{viscous}}\) Finally, solve for all the required variables using the equations and values established in the previous steps.