Introduction to Flight 8th Edition Anderson Solutions Manual

GENERAL ORGANIC AND BIOLOGICAL CHEMISTRY STRUCTURES OF LIFE 4TH EDITION TIMBERLAKE TEST BANK SOLUTIONS MANUAL TO ACCOMPANY INTRODUCTION TO FLIGHT 8 th Edition By John D. Anderson, Jr. Chapter 2 2.1  = p/RT = (1.2)(1.0110 5 )/(287)(300)   1.41 kg/m 2 v = 1/ = 1/1.41= 0.71 m 3 /kg 2.2 Mean kinetic energy of each atom  3 k T = 3 (1.38  10 23 ) (500) = 1.035  10 20 J 2 2 One kg-mole, which has a mass of 4 kg, has 6.02 × 10 26 atoms. Hence 1 kg has 1 (6.02  10 26 ) = 1.505  10 26 atoms. 4 Total internal energy = (energy per atom)(number of atoms) = (1.035 ´ 10 - 20 )(1.505 ´ 10 26 ) = 1.558 ´ 10 6 J p 2116 slug 2.3     0.00237 RT (1716)(460  59) ft 3 Volume of the room = (20)(15)(8) = 2400 ft 3 Total mass in the room = (2400)(0.00237) = 5.688 slug Weight = (5.688)(32.2) = 183 lb p 2116 slug 2.4  = = RT (1716)(460 - = 10) 0.00274 ft 3 Since the volume of the room is the same, we can simply compare densities between the two problems.  = 0.00274 - 0.00237 = 0.00037 slug ft 3 % change =  =  0.00037 ´ 0.00237 (100) = 15.6% increase 2.5 First, calculate the density from the known mass and volume,  = 1500/ 900 = 1.67 lb m /ft 3 In consistent units,  = Hence, 1.67/32.2 = 0.052 slug/ft 3 . Also, T = 70 F = 70 + 460 = 530 R. p =  RT = (0.52)(1716)(530) p = or p = 47, 290 lb/ft 2 47, 290 / 2116 = 22.3 atm p =  RT 2.6 np  np  nR  nT 3 Differentiating with respect to time, 1 dp  1 d   1 dT p dt  dt T dt or, dp  p d    p dT dt  dt T dt or, dp  RT d     R dT (1) dt dt dt At the instant there is 1000 lb m of air in the tank, the density is   1000 / 900  1.11lb m /ft   1.11/32.2  0.0345 slug/ft 3 Also, in consistent units, is given that T = 50 + 460 = 510 R and that dT  1F/min  1R/min  0.016R/sec dt From the given pumping rate, and the fact that the volume of the tank is 900 ft 3 , we also have d   0.5 lb m /sec  0.000556 lb /(ft 3 )(sec) dt 900 ft 3 m d   0.000556  1.7310 5 slug/(ft 3 )(sec) dt 32.2 Thus, from equation (1) above, d   (1716)(510)(1.73  10 5 )  (0.0345)(1716)(0.0167) dt  15.1  0.99  16.1 lb/(ft 2 )(sec)  16.1 2116  0.0076 atm/sec 2.7 In consistent units, T  10  273  263 K Thus,   p/RT  (1.7 10 4 )/(287)(263)   0.225 kg/m 3 2.8   p/RT  0.5 10 5 /(287)(240)  0.726 kg/m 3 v  1/  1/0.726  1.38 m 3 /kg 2.9 0 0 ÷ F p = Force due to pressure = 3 3 ò p dx = ò (2116 - 10x) dx = [2116x - 5x 2 ] 3 = 0 0 6303 lb perpendicular to wall. F τ = Force due to shear stress = ò 3 3 τ dx = ò 90 1 dx 0 0 (x + 1 9) 2 = [180 (x + 9) 2 ] 3 = 623.5 - 540 = 83.5 lb tangential to wall. Magnitude of the resultant aerodynamic force = R = (6303) 2 + (835) 2 = 6303.6 lb æ 83.5 ö  = Arc Tan ç ÷ = 0.76º 6303 ø 2.10 V = 3 2 V  sin  Minimum velocity occurs when sin  = 0, i.e., when  = 0° and 180°. V min = 0 at  = 0° and 180°, i.e., at its most forward and rearward points. Maximum velocity occurs when sin  = 1, i.e., when  = 90°. Hence, V max = 3 (85)(1) = 2 127.5 mph at  = 90, i.e., the entire rim of the sphere in a plane perpendicular to the freestream direction. 2.11 The mass of air displaced is M = (2.2)(0.002377) = 5.23 ´ 10 - 3 slug The weight of this air is W air = (5.23´ 10 - 3 )(32.2) = 0.168 lb This is the lifting force on the balloon due to the outside air. However, the helium inside the balloon has weight, acting in the downward direction. The weight of the helium is less than that of air by the ratio of the molecular weights W H c = (0.168) 4 = 28.8 0.0233 lb. Hence, the maximum weight that can be lifted by the balloon is 0.168  0.0233 = 0.145 lb. 2.12 Let p 3 ,  3 , and T 3 denote the conditions at the beginning of combustion, and p 4 ,  4 , and T 4 denote conditions at the end of combustion. Since the volume is constant, and the mass of the gas is constant, then p 4 =  3 = 11.3 kg/m 3 . Thus, from the equation of state, or, p 4 =  4 RT 4 = (11.3)(287)(4000) = 1.3 ´ 10 7 N/m 2 p 4 = 1.3 ´ 1.01 ´ 10 7 = 10 5 129 atm 2.13 The area of the piston face, where the diameter is 9 cm = 0.09 m, is 2 A = (0.09) = 4 6.36 ´ 10 - 3 m 2 (a) The pressure of the gas mixture at the beginning of combustion is p 3 =  3 RT 3 = 11.3(287)(625) = 2.02 ´ 10 6 N/m 2 The force on the piston is 6 - 3 4 F 3 = p 3 A = (2.02 ´ 10 )(6.36 ´ 10 ) = 1.28 ´ 10 N Since 4.45 N = l lbf, 1.28 ´ 10 4 F 3 = 4.45 = 2876 lb (b) p 4 =  4 RT 4 = (11.3)(287)(4000) = 1.3 ´ 10 7 N/m 2 The force on the piston is 7 - 3 4 F 4 = p 4 A = (1/3 ´ 10 ) (6.36 ´ 10 ) = 8.27 ´ 10 N 8.27 ´ 10 4 F 4 = 4.45 = 18, 579 lb   f f 2.14 Let p 3 and T 3 denote conditions at the inlet to the combustor, and T 4 denote the temperature at the exit. Note: p 3 = p 4 = 4 ´ 10 6 N/m 2 (a)  3 = p 3 = 4 ´ 10 6 = 15.49 kg/m 3 RT 3 (287)(900) (b)  4 = p 4 = 4 ´ 10 6 = 9.29 kg/m 3 RT 4 (287)(1500) 2.15 1 mile = 5280 ft, and 1 hour = 3600 sec. So: æ ÷ öæ= ÷ öæ= ÷ ö miles 5280 ft ç 60 ÷ ç ÷ ç 1 hour ÷ = 88 ft/sec. hour ÷ ø mile ÷ øè ç 3600 sec ÷ ø A very useful conversion to remember is that 60 mph = 88 ft/sec also, 1 ft = 0.3048 m ç 88 ft æ 0.3048 m ö 26.82 m æ ÷ ö ç è ç sec ÷ ø ç è ÷ = 1 ft ø ÷ sec ft m Thus 88 = 26.82 sec sec 2.16 692 miles  88 ft/sec   1015 ft/sec hour  60 mph  692 miles  26.82 m/sec   309.3 m/sec hour   60 mph    2.17 On the front face F  p A  (1.0715 10 5 )(2)  2.14310 5 N On the back face F  p A  (1.01  10 5 )(2)  2.02  10 5 N b b The net force on the plate is F  F f  F b  (2.143  2.02) 10 5  0.123 10 5 N From Appendix C, 1 lb f  4.448 N. So, 0 . 123 10 5 F   2765 lb 4.448 This force acts in the same direction as the flow (i.e., it is aerodynamic drag.)   hr 60 2.18 Wing loading  W  10,100  43.35 lb/ft 2 In SI units: s 233 W  lb   4.448 N   1 ft  2  43.35 s ft 2 1 lb 0.3048 m W  2075.5 N s m 2 In terms of kilogram force, W  N   1 k f   kg f  2075.5 s m 2 9.8 N  211.8 m 2 2.19 V   437 miles   5280 ft   0.3048 m   7.033  10 5 m  703.3 km hr mile 1 ft hr hr Altitude  (25, 000 ft)  0.3048 m   7620 m  7.62 km 1 ft 2.20 V   26, 000 ft   0.3048 m   7.925  10 3 m  7.925 km sec 1 ft sec sec 2.21 From Fig. 2.16, length of fuselage = 33 ft, 4.125 inches = 33.34 ft  33.34 ft  0.3048 m   10.16 m ft wing span = 40 ft, 11.726 inches = 40.98 ft  40.98 ft  0.3048 m   12.49 m ft 2.22 (a) From App. C 1 ft.  0.3048 m. Thus, 354,200 ft  (354,000)(0.3048)  107,960 m  107.96 km (b) From Example 2.6: 60 mph  26.82 m/sec Thus,  26.82 m   4520 miles  4520 miles   sec  2020.4 m/sec hr hr  mi       2.23 m  34, 000 lb  1055.9 slug 32.2 lb/slug From Newton’s 2 nd Law F  ma a  F  57, 000  53.98 ft/sec 2 M 1055.9 2.24 # of g’s  53.98  1.68 32.2 2.25 From Appendix C, one pound of force equals 4.448 N. Thus, the thrust of the Rolls-Royce Trent engine in pounds is 373.7  10 3 N T   84, 015 lb 4.448 N/ lb 2.26 (a) (b)