Question

A \(4-\mathrm{cm}\)-diameter shaft rotates at \(5200 \mathrm{rpm}\) in a \(15-\mathrm{cm}-\) long, 8 -cm-outer-diameter cast iron bearing $(k=70 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( with a uniform clearance of \)0.6 \mathrm{~mm}\( filled with lubricating oil \)\left(\mu=0.03 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\right.\( and \)\left.k=0.14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)$. The bearing is cooled externally by a liquid, and its outer surface is maintained at \(40^{\circ} \mathrm{C}\). Disregarding heat conduction through the shaft and assuming one-dimensional heat transfer, determine \((a)\) the rate of heat transfer to the coolant, \((b)\) the surface temperature of the shaft, and (c) the mechanical power wasted by the viscous dissipation in oil. Treat this problem as parallel flow between two large plates with one plate moving at constant velocity and the other stationary.

Short answer

Answer: The rate of heat transfer to the coolant is \(0.010756 \; \mathrm{W}\), the surface temperature of the shaft is \(38.76^\circ \mathrm{C}\), and the mechanical power wasted by viscous dissipation in oil is \(0.010756 \; \mathrm{W}\).

Step-by-step solution

Calculate the relative velocity between the shaft and bearing

To find the relative velocity between the shaft and bearing, we will convert the shaft rotation rate (5200 rpm) to linear velocity. To do this, we first convert the rotation rate to revolutions per second by dividing by 60: $$ \omega = \frac{5200 \textrm{ revolutions per minute}}{60 \textrm{ seconds per minute}} = 86.67 \textrm{ revolutions per second} $$ Now, we can find the linear velocity of the shaft surface (in meters per second) as: $$ v = \omega \pi D = 86.67 \times \pi \times 0.04 \textrm{ m/s} = 10.96 \textrm{ m/s} $$

Calculate the heat transfer coefficient for the bearing

We will now find the heat transfer coefficient using the equation for heat transfer between two parallel plates: $$ h = \frac{k}{\delta} $$ where: - \(h\) is the heat transfer coefficient, - \(k\) is the thermal conductivity of the oil, and - \(\delta\) is the clearance of the bearing. In this case, \(k = 0.14 \; \mathrm{W/m.K}\) and \(\delta = 0.6 \mathrm{~mm} = 0.0006 \mathrm{~m}\), so: $$ h = \frac{0.14 \; \mathrm{W/m.K}}{0.0006 \mathrm{~m}} = 233.33 \; \mathrm{W/m} \cdot \mathrm{K} $$

Calculate the rate of heat transfer to the coolant

To find the rate of heat transfer to the coolant, we will use the equation for heat transfer in one-dimensional flow: $$ q = A h \Delta T $$ where: - \(q\) is the rate of heat transfer, - \(A\) is the contact area between the shaft and the bearing, and - \(\Delta T\) is the temperature difference between the shaft surface and the coolant. Using the given data, we find the contact area as \(A = \pi D L\), where \(D = 0.04 \mathrm{~m}\) and \(L = 0.15 \mathrm{~m}\): $$ A = \pi \times 0.04 \mathrm{~m} \times 0.15 \mathrm{~m} = 0.01885 \mathrm{~m}^{2} $$ The temperature difference is given as \(\Delta T = T_f - T_c = 40^\circ \mathrm{C} - T_s\), where \(T_f = 40^\circ \mathrm{C}\) is the temperature of the coolant and \(T_s\) is the surface temperature of the shaft (unknown). For now, we will only determine the expression for \(q\): $$ q = 0.01885 \mathrm{~m}^{2} \times 233.33 \; \mathrm{W/m} \cdot \mathrm{K} \times (40^\circ \mathrm{C} - T_s) = 4.40188 (40 - T_s) \; \mathrm{W} $$

Calculate the mechanical power wasted by viscous dissipation in oil

The mechanical power wasted by viscous dissipation in oil can be found using the equation: $$ W = \mu A v \frac{\Delta h}{L} $$ where \(W\) is the mechanical power wasted by viscous dissipation, \(\mu = 0.03 \; \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) is the dynamic viscosity of the oil, and \(\Delta h = 0.0006 \; \mathrm{m}\) is the height difference between the stationary plate and the moving plate. Substituting the known values, we get: $$ W = 0.03 \left( \frac{\mathrm{N} \cdot \mathrm{s}}{\mathrm{m}^{2}} \right) \times 0.01885 \mathrm{~m}^{2} \times 10.96 \textrm{ m/s} \times \frac{0.0006 \; \mathrm{m}}{0.15 \mathrm{~m}} = 0.010756 \; \mathrm{W} $$

Determine the surface temperature of the shaft

At steady-state condition, the rate of heat transfer to the coolant is equal to the mechanical power wasted by viscous dissipation in oil: $$ q = W $$ Using the expression for \(q\) from Step 3 and the value of \(W\) from Step 4: $$ 4.40188 (40 - T_s) \; \mathrm{W} = 0.010756 \; \mathrm{W} $$ Solving for \(T_s\), we get: $$ T_s = 40 - \frac{0.010756 \; \mathrm{W}}{4.40188 \; \mathrm{W}} = 38.76^\circ \mathrm{C} $$ Now we have determined all three required quantities: - (a) The rate of heat transfer to the coolant is \(0.010756 \; \mathrm{W}\). - (b) The surface temperature of the shaft is \(38.76^\circ \mathrm{C}\). - (c) The mechanical power wasted by viscous dissipation in oil is \(0.010756 \; \mathrm{W}\).