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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 '06
 COOL
 Standard Deviation, sampling rate

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