A journal and bearing are to be designed for a shaft that tu

A journal and bearing are to be designed for a shaft that turns at 250 rpm. Suppose ISO VG100 (SAE Engine 30) is to be used as lubricant and the bearing length is to be equal to 1.2 times the diameter. If the no-load power loss is not to exceed 2.5 Times 10^-4 horsepower and the diametral clearance is 0.0045 times the diameter, estimate the maximum diameter that can be used for the journal, and the allowable temperature limit.

Solution

Sol:

Torque = r x A x Fn(t)

Given Power = 2.5*10^-4 HP = 0.186425 W

Torque = 60P/(2*pi*N)

Torque = 7.12 *10^-3 N.m

(r=radius, A = brg surface area, t = clearance, Fn = \"some function of\")
Torque = D/2 x (Pi x D x 1.2D) x Fn(0.0045D)

7.12 *10^-3 = 8.48*10^-3 D⁴

D⁴ = 0.83939

D= 0.957 m