{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW1CASE4 - EML 3005:Homework#1 SOLUTION Nagaraj Arakere...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EML 3005:Homework #1, SOLUTION, Nagaraj Arakere CASE 4: Hot Day (heat soak) (Oil inlet temp = 240F, 800 rpm) Select Journal Radius (Inch), Length (Inch), Load (lbf), and Oil Inlet Temp (F) r := 0.75 2 L := .75 W := 51.0 T1 := 240.0 rpm := 800 Define Radial Clearance Range (0.0002 - 0.003 inch) c := 0.0002 , 0.0004 .. 0.0022 Define Journal Speed (rev/sec) N := rpm 60 Define Average Oil Temperature (F), i.e., Tavg = T1 + DT/2 (Guess on oil DT, and iterate on calculated value) DTGUESS( c) := 25⋅ 0.0002 c 1.2 T( c) := T1 + DTGUESS( c) 2 Define Viscosity (Reyns) vs. Temp for 10W30 oil µ ( c) := 0.7323⋅ T( c) − 2.4735 Define Unit Load Capacity (P) P := W 2⋅ L⋅ r Define Sommerfeld Number (S) as a function of clearance (c), since c is a design variable r 2 N S( c) := ⋅ µ ( c) ⋅ P c Define bearing performance parameters in terms of curve fits provided (a) Min Film Thickness, ho ho( c) := c⋅ 0.0247 + 4.2606⋅ S( c) − 10.2144⋅ S( c) + 11.4556⋅ S( c) − 4.664⋅ S( c) ( 2 3 4 ) (b) Friction Factor, f f( c) := c r ⋅ 0.7316 + 18.9931⋅ S( c) + 0.1877⋅ S( c) ( 2 ) 2 3 (c) Flow Variable Q Q( c) := ( r⋅ c⋅ N⋅ L) ⋅ 4.8281 − 4.6055⋅ S( c) + 5.9194⋅ S( c) − 2.7516⋅ S( c) ( ) (d) Side Flow Qs Qs( c) := Q( c) ⋅ 0.9614 − 2.6056⋅ S( c) + 3.4272⋅ S( c) − 1.6012⋅ S( c) ( 2 3 ) CASE 1 Calculate Oil Temp Rise DT( c) := 0.103⋅ P 1 − 0.5⋅ Qs( c) c Q( c) ⋅ r ⋅ f( c) Q ( c) ( r⋅ c⋅ N⋅ L) Print the variable values c= 2·10-4 4·10-4 6·10-4 8·10-4 0.001 0.0012 0.0014 0.0016 0.0018 0.002 0.0022 S( c) = 0.4327 0.11603 0.05268 0.02993 0.01926 0.01343 0.00989 0.00758 0.006 0.00487 0.00403 ho( c) = 1.44079·10-4 1.59441·10-4 1.33458·10-4 1.14689·10-4 1.03061·10-4 9.61047·10-5 9.21806·10-5 9.02933·10-5 8.98269·10-5 9.03862·10-5 9.17082·10-5 f( c) = 0.00479 0.00313 0.00277 0.00277 0.00293 0.00316 0.00343 0.00374 0.00406 0.00439 0.00474 Q( c) = 0.00279 0.00655 0.01035 0.01409 0.01778 0.02145 0.02511 0.02876 0.0324 0.03604 0.03968 Qs( c) = 9.65256·10-4 0.00461 0.00863 0.01249 0.01622 0.01989 0.0235 0.02709 0.03065 0.0342 0.03773 Print Oil Temperature Rise (guess and calculated), Average Oil Temp, and oil Viscosity (Calculated DT) DTGUESS( c) = 25 10.88188 6.68951 4.73661 3.6239 2.91178 2.42004 2.06173 1.78998 1.57739 1.40692 DT( c) = 27.26827 9.68129 6.0284 4.64434 3.97526 3.60251 3.37417 3.22439 3.12094 3.04653 2.99125 Avg oil temp T( c ) = 252.5 245.44094 243.34476 242.36831 241.81195 241.45589 241.21002 241.03087 240.89499 240.7887 240.70346 µ ( c) ⋅ 10 = 0.83694 0.89775 0.917 0.92616 0.93144 0.93485 0.9372 0.93893 0.94024 0.94127 0.94209 6 Plot Variables CASE 1 Min Film Thickness Vs. Clearance 160 150 140 130 6 ho( c) ⋅ 10 120 110 100 90 80 0.2 0.37 0.53 0.7 0.87 1.03 1.2 c⋅ 10 3 1.37 1.53 1.7 1.87 2.03 2.2 Oil Temp Rise (F) Vs. Clearance 30 24 18 DT( c) 12 6 0 0.2 0.37 0.53 0.7 0.87 1.03 1.2 c⋅ 10 3 1.37 1.53 1.7 1.87 2.03 2.2 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online