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Solutions to Midterm Exam, Spring 2009

# Solutions to Midterm Exam, Spring 2009 - Solutions to...

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Solutions to Midterm Exam, Spring 2009 1. Fluid Mechanics (A) The power is given by the product of the drag force and the velocity, P = F D V F D = C D A ρV 2 /2 V = 85 mph, or V = (85 mi/hr)(5280 ft/mi)(0.305 m/ft)(1 hr/3600 s)=38 m/s ρ = (0.074 lb/ft 3 )(0.454 kg/lb)/(0.305) 3 m 3 = 1.184 kg/m 3 ρV 2 /2 = (1.184)(38) 2 /2 = 855 N/m 2 F D = (0.38)(20 ft 2 )(0.305 m/ft) 2 (855 N/m 2 ) = 604 N P = (604 N)(38 m/s) = 22970 watts = 30.8 hp . (B) d 2 = 4(16.5)(.305) 2 /π = 1.95 m 2 ; d = 1.4 m Re d =Vd/ ע ) = 38 m/s)(1.4 m)/1.56x10 -5 m 2 /s) = 3.4x10 6 (C) The boundary layer should be turbulent because this Reynolds number is an order of magnitude larger than the value for transition over a sphere (or cylinder); see Figs. 4.6 and 4.7 in your course packet. 2. Error analysis The mean length measurement is L m =19.980 mm. The standard deviation for each length measurements is ΔL=0.039 mm. The standard deviation in the mean length is ΔL/√N = 0.039/3.16 = 0.012 mm The mean diameter for each measurement is D m =4.905 mm. The standard deviation for the diameter measurements is ΔD=0.044 mm The standard deviation in the mean diameter is ∆D/√N = 0.044/3.16 = 0.0139 mm The mean volume is V(L m ,D m ) = πL m D m 2 /4 = 377.5 (mm) 3 ln(V) = ln(π/4) + ln(L) + 2ln(D)

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