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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
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