Question

A full hydroodynamic journal bearing, with following specification for machine tool application: Journal diameter = 75 mm, coefficient of friction variable CFV = 3.22, radial load = 15 kN, journal speed = 1440 rpm, radial clearance = 0.0375 mm. Oil used for lubrication has density equals to $860kg/m^3$ and specific heat equals to $1.76kJ/k{g}^{\circ }C$. The flow rate of lubricant is $10960m{m}^3/s$. What is the average temperature of oil, if inlet temperature was ${41}^{\circ }C?$

• A. ${50}^{\circ }C$
• B. ${42}^{\circ }C$
• C. ${49.23}^{\circ }C$
• D. ${59.23}^{\circ }C$

Answer 1

The correct option is C ${49.23}^{\circ }C$

Given; D = 75 mm, CFV = 3.22, W = 15 kN, N = 1440 rpm, c = 0.0375 mm, $\rho$ = $860kg/m^3$,

$c_p=1.76kJ/k{g}^{\circ }C$, Q = 10960 $m{m}^2/s$, $T_0={41}^{\circ }C$

Heat generated in bearing is = $f\times w\times v$

Coefficient of flow variable, $CFV=\frac{r}{c}f$

$3.22=\frac{75\times f}{2\times 0.0375}$

$f=3.22\times {10}^{-3}$

$H_g=f\times w\times \frac{\pi D\times N}{60}=3.22\times {10}^{-3}\times 15\times {10}^3\times \pi \times \frac{0.075\times 1440}{60}$

$H_g=273.13W$

Heat gained by lubricating oil = $m\times c_p\times \Delta T$

$=\rho \times Q\times c_p\times \Delta T=860\times 10960\times {10}^{-9}\times 1.76\times {10}^3\times \Delta T=16.589\Delta T$

So, $H_3=16.589\Delta T$

$273.13=16.589\Delta T$

$\Delta T={16.464}^{\circ }C$

$T_{avg}=T_0+\frac{\Delta T}{2}=41+\frac{16.464}{2}={49.23}^{\circ }C$