Question

A journal bearing having a shaft diameter of 50 mm and length of 60 mm rotates at a speed of 1500 rpm. The loss of torque due to viscosity of lubricant is 0.5 N-m and radial clearance is 0.25 mm, then what will be the viscosity of the lubricant?

• A. 0.135 Pa-s
• B. None of these
• C. 0.230 Pa-s
• D. 0.048 Pa-s

Answer 1

The correct option is A 0.135 Pa-s

$Given,D=50mm,L=60mm,N=1500rpm,n=\frac{1500}{60}=25rev/s,T_f=0.5N-m$

$C=2C_1=2\times 0.025=0.05mm$

$\text{we know that,}{T}_f=\mu WR$

$0.5=\mu W\times (25\times {10}^{-3})$

$\mu W=20N$

$\text{we know that,}$

$\mu =2{\pi }^2\left(\frac{Z}{P}\right)\left(\frac{D}{C}\right)$

$\mu \times \left(\frac{W}{DL}\right)=2{\pi }^2\left(Zn\right)\left(\frac{D}{C}\right)$

$\Rightarrow \left(\frac{\mu W}{DL}\right)\times \left(\frac{C}{D}\right)\times \left(\frac{1}{2{\pi }^{2}n}\right)=Z$

$\left(\frac{20\times 0.05\times {10}^{-}3}{50\times 60\times {10}^{-6}\times 50\times {10}^{-3}}\right)\times \left(\frac{1}{2{\pi }^2\times 5}\right)=z$

$Z=0.0135095Pa-s$