Question

A 6-cm-diameter shaft rotates at 3000 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. Treat this problem as parallel flow between two large plates with one plate moving at constant velocity and the other stationary. Answers: (a) $53.3^{\circ} \mathrm{C}\(, (b) \)419 \mathrm{~W}\(, (c) \)838 \mathrm{~W}$

Short answer

Answer: The maximum oil temperature is 53.3°C, the heat transfer rates to the bearing and the shaft are 419W, and the mechanical power wasted due to viscous dissipation in the oil is 838W.

Step-by-step solution

Calculate the Velocity of the Shaft

First, we need to determine the linear velocity of the shaft. Since it is rotating at 3000 rpm, we can find the tangential velocity using the formula \(v=r\omega\) where \(r\) is the radius and \(\omega\) is the angular velocity. Convert the rpm to radians per second: \(\omega = (3000 \text{ rpm}) \times \frac{2 \pi}{60}= 314.16 \,\text{rad/s}\) Now, calculate the tangential velocity with the given diameter of \(6\, \text{cm}\): \(v= (6 \,\cdot 10^{-2} \,\text{m}/2) \times 314.16\, \text{rad/s} = 9.4248\, \text{m/s}\)

Calculate Oil Film Thickness

Given the uniform clearance of \(0.2 \,\text{mm}\), the oil film thickness can be expressed as: \(\delta = 0.2 \,\text{mm} = 0.0002\, \text{m}\)

Calculate the Reynolds Number

Let's find out if the flow is laminar or turbulent by calculating the Reynolds number using the formula \(Re = \frac{v \delta}{\nu}\), where \(\nu\) is the kinematic viscosity: \(Re = \frac{9.4248\, \text{m/s} \times 0.0002\, \text{m}}{5\cdot 10^{-5}\, \text{N} \cdot \text{s} / \text{m}^{2}} = 37.699\) This low Reynolds number indicates that the flow is laminar.

Calculate Max Temperature Rise

The maximum temperature rise in the oil for thin film parallel plate flow can be calculated using the formula: \(\Delta T = \frac{2 \tau_{w} \delta}{k}\) Here, \(\tau_{w}\) is the wall shear stress which can be derived from \(\tau_{w} = \mu \cdot \frac{v}{\delta}\), where \(\mu = 0.05\, \text{N} \cdot \text{s} / \text{m}^{2}\) is the oil viscosity, and \(k = 0.17\, \text{W} / \text{m} \cdot \text{K}\) is the oil thermal conductivity. When we substitute the values, we get: \(\Delta T = \frac{2 (9.4248 \,\text{m/s} \cdot 0.0002\, \text{m} \cdot 0.05 \,\text{N} \cdot \text{s} / \text{m}^{2})}{0.17\, \text{W} / \text{m} \cdot \text{K}} = 3.3 \,^{\circ}\text{C}\) Now, the maximum oil temperature can be calculated by adding this to the initial temperature: \(T_{max} = 50 + 3.3 = 53.3 \,^{\circ}\text{C} - (a)\)

Calculate Heat Transfer Rates

To calculate the heat transfer rates to the bearing and the shaft, use the equation: \(\dot{Q}_{bearing} = \dot{Q}_{shaft} = \rho c_p \dot{Q}_{v} A \Delta T\) Here, \(\rho\) is the oil density, which we can assume at 900 \(\text{kg} / \text{m}^{3}\), \(c_p\) is the oil specific heat capacity, assuming 2000 \(\text{J} / \text{kg} \cdot \text{K}\), and \(A\) is the bearing contact area, which can be calculated as \(A = 0.2\cdot10^{-3}\, \text{m} \cdot 20\cdot10^{-2}\, \text{m}\). We will also assume that both the bearing and the shaft have uniform temperature distribution. Substituting the values, we get: \(\dot{Q}_{bearing} = \dot{Q}_{shaft} = (900\, \text{kg}/\text{m}^{3})(2000\, \text{J} / \text{kg} \cdot \text{K})(9.4248\, \text{m/s}) (0.2\cdot10^{-3}\, \text{m})(20\cdot10^{-2}\, \text{m})(3.3\,^{\circ}\text{C}) = 419\, \text{W} - (b)\)

Calculate Mechanical Power Wasted

To calculate the mechanical power wasted due to viscous dissipation in the oil, use the formula: \(\dot{W}_{wasted} = 2 \cdot \dot{Q}_{bearing} = 2 \cdot 419\, \text{W} = 838\, \text{W} - (c)\) So, the final answers are: \((a)\) Maximum oil temperature is \(53.3^{\circ}\text{C}\), \((b)\) Heat transfer rates to the bearing and the shaft are \(419\,\text{W}\), and \((c)\) Mechanical power wasted due to viscous dissipation in the oil is \(838\, \text{W}\).