3000 N = 0.8- 75(50) mm 33 0.014(1600) +0.002 = +0.002 0.01124 108 0.8 0.001 m = 4.189 60 S v = dN (0.05) (1600) Q=0.01124(3000) (4.189)=141.25 W Heat dissipated by a bearing, Q = CA (t t) = Cld (t t) For unventilated bearings (Still air), C = 140 to 420 W (t 1) ==- (t1) ==-(60-15. say, C=300 (60-15.5)=22.25C 2 Q = 300 (75)(50) (22.25)=25 W (1000) OW/m -C Sample Problem 1 [prob.2/p.1003] A journal bearing is proposed for a steam engine. The load on the journal is 3 kN, diameter 50 mm, length 75 mm, speed 1600 rpm, diametral clearance ratio 0.001 mm, ambient temperature 15.5C. Oil SAE 10 is used and the film temperature is 60C. Determine the heat generated and heat dissipated. Take absolute viscosity of SAE10 at 60C = 0.014 kg/m-s. [Ans. 141.3 J/s ; 25 J/s] Given: W = 3kN = 3000N; d = 50mm; 1 = 75mm; a N =1600rpm; % = 0.001mm; t =15.5C; t = 60C; = 0.014 kg/m-s Q = fWv p = f g W W = A ld 33 | N = 108 P 60 53 63