Question

A journal bearing 4 inches in diameter and 4 inches long has a radial clearance of 0.002 inches. It rotates at 2000 rpm and is lubricated is SAE 10 oil at 200F. In the following, use an applied load of 225 lbs and neglect temperature rise.

a. Estimate the power loss and the friction torque using the Petrov equation. Calculate the coefficient of friction.
b. Calculate the minimum clearance, the power loss, friction torque and the coefficient of friction using the charts in section 12-8 of the text.

Figures in section 12-8:

I/d=00 (dimensionless) Minimum film thickness variable Eccentricity ratio e (dimensionless) 1.0 10 0.01 0.02 0.04 0.06 0.08 0.1 0.2 0.4 0.6 0.8 1.0 6 8 Bearing characteristic number, S=0 " Figure 12-15 Chart for minimum film thickness variable and eccentricity ratio. (Source: A A. Raimondi and John Boyd, “A Solution for the Finite Journal Bearing and Its Application to Analysis and Design, Parts I, II, and III," Trans. ASLE. vol. 1, no. 1, in Lubrication Science and Technology, Pergamon, New York, 1958, pp. 159-209.)

Answer 1

The detailed solution is given below:

ama Giveno dmex = 4in , N= 2000rpm Cmin = 0.002 in b- Ain | W= 225665 ) T= 200°F # Cm = bundmax = 0.002 .: bmin = 4.004. #r= - = 4 = din # 3 - 6.002 = 1000 # N = 2000 rev = 33.33 rev/s #P-W, = 225— - 14.0625 psiº 2 nev # P = M ad = LAKAT # 1 = 0.74 seyn (for SAE loat:200F) ^ - The Sommerfeld numbers (even SA Elo 0.74 7 DOO = (1000)?6.74X106) x (33.33) (14.0625) • S= 1.75 T (°F)

# The coefficient of friction : -> f = 27°UN X Y =(22)X(0.74x10^^) x (33.33)x (looo) (14.0625) -0.0346 # The fractional torque: → T= fxwx8 = (0.0346 )x(225)x(2)= 15.57 lbfi in * The power loss: → Poss = TN - (15.57]X[ 33.3) - 10.494 KP] b. [lld=1 # Cmon = 0.002in Sgiven) # The coefficient of friction Variable : YE = 3.33 => f = (3-33X C) / = (3.33 X (0.002)/(x2) = 0.00333) loso 1050 7 3.33 eld=/ 11.75

# The frictional torque: → T=fxwx8 = (0.00333) x (225)x(2) - 1.4985 8bf. # The power loss; → Pas = TN = (14985 1X(33.33) - 10.0475 hp] loso loso