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Solutions to Midterm Exam

Solutions to Midterm Exam - 4x10-6 s(C neglecting the...

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Solutions to Midterm Exam AEP 2640, Fall 2008 (1) (a) L=mg=(320,000 kg)x(9.8 m/s 2 )= 3.136x10 6 N (b) Dp+Di = Cd S (½ρV 2 ) + kL 2 /(½ρV 2 SπAR) (1) Here ρ=ρ 0 exp(-h/a); ρ 0 =Air density at sea level = 1.225 kg/m 3 , a=8.42 km, h=10 km. This gives: ρ=0.352 kg/m 3 AR=b 2 /S=24.8 Dp+Di = 0.031x(525 m 2 )(0.176 kg/m 3 )(250 m/s) 2 + 1.1x(9.83x10 12 N 2 )/ [0.176 kg/m 3 )(250 m/s) 2 (525 m 2 )(24.8 π)] = 179,025 N + 10.81x10 12 /4.50x10 8 N = 203,000 N (c) This is below the maximum thrust and so at this altitude a larger payload is possible. (B) Differentiate Eq. (1) w/r V to find the air speed for minimum drag: Cd S (ρV)= 2kL 2 /(½ρV 3 SπAR) or, V 4 = 4kL 2 /(Cd S 2 ρ 2 πAR) (C) V 4 =4.4x(9.83x10 12 )/(0.031x(525) 2 x0.124xπx24.8) = 5.24x10 8 (m/s) 4 V= 151 m/s optimal air speed 2. (A) 250,000 samples/s (B) each 16 bit conversion requires
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Unformatted text preview: 4x10-6 s . (C) neglecting the start-conversion cycle, the SAR requires 16 clock cycles for the conversion; each the period of each cycle must therefore be 0.25x10-6 s, or a clock frequency of 4 MHz. (D) 10.00000 volts represents 65536 SAR units, therefore 9.52925 V corresponds to 62450.89 SAR units. The indicated digital voltage will therefore be 62450x10.00000/65536= 9.52911 V 3. P=I 2 R ~ T 4 If R is constant, then F = T 2 /T 1 = (I 2 /I 1 ) 1/2 = 1.062 (∆lnF) 2 = (∂lnF/∂I 1 )(∆I 1 ) 2 + (∂lnF/∂I 2 )(∆I 2 ) 2 (∆F/F) 2 = (½∆ 1 /I 1 ) 2 + (½∆ 2 /I 2 ) 2 = ¼[ (0.05) 2 + (0.05) 2 ] = 0.00125 ∆F/F = 0.0354 . 4. The completed arrays of the front panel are shown below:...
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