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Solution Manual For Introduction To Flight, 7th Edition

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    SOLUTIONS MANUAL TO ACCOMPANY INTRODUCTION TO FLIGHT 7 th Edition By John D. Anderson, Jr.

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    2 Chapter 2 2.1 5 2 3 / (1.2)(1.01 10 )/(287)(300) 1.41 kg/m 1/ 1/1.41 0.71 m /kg ρ ρ ρ × = = p RT = v = = = 2.2 Mean kinetic energy of each atom 23 20 3 3 (1.38 10 ) (500) 1.035 10 J 2 2 − − = × × k T = = One kg-mole, which has a mass of 4 kg, has 6.02 × 10 26 atoms. Hence 1 kg has 26 26 1 (6.02 10 ) 1.505 10 4 = × × atoms. 20 26 6 Totalinternal energy (energy per atom)(number of atoms) (1.035 10 )(1.505 10 ) 1.558 10 J - = = ´ ´ = ´ 2.3 3 2116 slug 0.00237 (1716)(460 59) ft ρ = = = + p RT 3 Volume of the room (20)(15)(8) 2400 ft Total mass in the room (2400)(0.00237) 5.688slug Weight (5.688)(32.2) 183lb = = = = = = 2.4 3 2116 slug 0.00274 (1716)(460 10) ft p RT ρ = = = - Since the volume of the room is the same, we can simply compare densities between the two problems. 3 slug 0.00274 0.00237 0.00037 ft ρ Δ = - = 0.00037 % change (100) 15.6% increase 0.00237 ρ ρ Δ = = ´ = 2.5 First, calculate the density from the known mass and volume, 3 m 1500 900 1.67 lb ft ρ / / = = In consistent units, 3 1.67/32.2 0.052slug ft . ρ / = = Also, T = 70 F = 70 + 460 = 530 R . Hence, (0.52)(1716)(530) ρ p RT = = 2 47, 290lb/ft p = or 47, 290 / 2116 22.3 atm p = =

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    3 2.6 ρ p RT = np np nR nT = + +     Differentiating with respect to time, 1 1 1 ρ ρ = + dp d dT p dt dt T dt or, T ρ ρ = + dp p d p dT dt dt dt or, ρ ρ = + dp d dT RT R dt dt dt (1) At the instant there is 1000 lb m of air in the tank, the density is 3 m 1000 / 900 1.11lb /ft ρ = = 3 1.11/32.2 0.0345slug/ft ρ = = Also, in consistent units, is given that T = 50 + 460 = 510 R and that 1 /min 1 /min 0.016 /sec = = = dT F R R dt From the given pumping rate, and the fact that the volume of the tank is 900 ft 3 , we also have 3 m m 3 0.5 lb /sec 0.000556 lb /(ft )(sec) 900 ft ρ = = d dt 5 3 0.000556 1.73 10 slug/(ft )(sec) 32.2 ρ − = = × d dt Thus, from equation (1) above, 5 2 (1716)(510)(1.73 10 ) (0.0345)(1716)(0.0167) 16.1 15.1 0.99 16.1 lb/(ft )(sec) 2116 0.0076 atm/sec ρ − = × + = + = = = d dt 2.7 In consistent units, 10 273 263 K = − + = T Thus, 4 3 / (1.7 10 )/(287)(263) 0.225 kg/m ρ ρ = = × = p RT 2.8 5 3 3 / 0.5 10 /(287)(240) 0.726 kg/m 1/ 1/0.726 1.38 m /kg ρ ρ = = × = = = = p RT v

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    4 2.9 3 3 0 0 2 3 0 Force due to pressure (2116 10 ) [2116 5 ] 6303 lb perpendicular to wall. p F p dx x dx x x = = = - = - = ò ò 3 3 τ 1 0 0 2 1 3 2 0 90 Force due to shear stress = τ ( 9) [180 ( 9) ] 623.5 540 83.5 lb tangential to wall. F dx dx x x = = + = + = - = ò ò Magnitude of the resultant aerodynamic force = 2 2 (6303) (835) 6303.6 lb 83.5 Arc Tan 0.76º 6303 θ R = + = æ ö ÷ ç = = ÷ ç ÷ ÷ ç è ø 2.10 3 sin 2 θ ∞ V V = Minimum velocity occurs when sin θ = 0, i.e., when θ = 0° and 180°. V min = 0 at θ = 0° and 180°, i.e., at its most forward and rearward points. Maximum velocity occurs when sin θ = 1, i.e., when θ = 90°. Hence, max 3 (85)(1) 127.5 mph at 90 , 2 θ ° V = = = i.e., the entire rim of the sphere in a plane perpendicular to the freestream direction.

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    5 2.11 The mass of air displaced is 3 (2.2)(0.002377) 5.23 10 slug M - = = ´ The weight of this air is 3 air (5.23 10 )(32.2) 0.168 lb W - = ´ = 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 4 (0.168) 0.0233 lb. 28.8 c H W = = Hence, the maximum weight that can be lifted by the balloon is 0.168 − 0.0233 = 0.145 lb. 2.12 Let p 3 , ρ 3 , and T 3 denote the conditions at the beginning of combustion, and p 4 , ρ 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 = ρ 3 = 11.3 kg/m 3 . Thus, from the equation of state, 7 2 4 4 4 (11.3)(287)(4000) 1.3 10 N/m ρ p RT = = = ´ or, 7 4 5 1.3 10 129 atm 1.01 10 p ´ = = ´ 2.13 The area of the piston face, where the diameter is 9 cm = 0.09 m, is 2 3 2 (0.09) 6.36 10 m 4 π A - = = ´ (a) The pressure of the gas mixture at the beginning of combustion is 6 2 3 3 3 11.3(287)(625) 2.02 10 N/m ρ p RT = = = ´ The force on the piston is 6 3 4 3 3 (2.02 10 )(6.36 10 ) 1.28 10 N F p A - = = ´ ´ = ´ Since 4.45 N = l lb f , 4 3 1.28 10 2876 lb 4.45 F ´ = = (b) 7 2 4 4 4 (11.3)(287)(4000) 1.3 10 N/m ρ p RT = = = ´ The force on the piston is 7 3 4 4 4 = (1/3 10 ) (6.36 10 ) = 8.27 10 N F p A - = ´ ´ ´ 4 4 8.27 10 18,579 lb 4.45 F ´ = =

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    6 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: 6 2 3 4 4 10 N/m p p = = ´ (a) 6 3 3 3 3 4 10 = 15.49 kg/m (287)(900) ρ p RT ´ = = (b) 6 3 4 4 4 4 10 = 9.29 kg/m (287)(1500) ρ p RT ´ = = 2.15 1 mile = 5280 ft, and 1 hour = 3600 sec. So: miles 5280 ft 1 hour 60 88 ft/sec. hour mile 3600 sec æ öæ öæ ö ÷ ÷ ÷ ç ç ç = ÷ ÷ ÷ ç ç ç ÷ ÷ ÷ ÷ ÷ ÷ ç ç ç è øè øè ø A very useful conversion to remember is that 60 mph 88 ft/sec = also, 1 ft = 0.3048 m ft 0.3048 m m 88 26.82 sec 1 ft sec æ ö æ ö ÷ ç ÷ ç ÷ = ÷ç ç ÷ ÷ç ÷ ç ÷ ÷ ç è øè ø Thus ft m 88 26.82 sec sec = 2.16 miles 88 ft/sec 692 1015 ft/sec hour 60 mph   =     miles 26.82 m/sec 692 309.3 m/sec hour 60 mph   =     2.17 On the front face 5 5 (1.0715 10 )(2) 2.143 10 N f f F p A = = × = × On the back face 5 5 (1.01 10 )(2) 2.02 10 N b b F p A = = × = × The net force on the plate is 5 5 (2.143 2.02) 10 0.123 10 N f b F F F = − = − × = × From Appendix C, 1 4.448 N. f lb = So, 5 0.123 10 2765 lb 4.448 F × = = This force acts in the same direction as the flow (i.e., it is aerodynamic drag.)

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    7 2.18 2 10,100 Wing loading 43.35 lb/ft 233 W s = = = In SI units: 2 2 2 lb 4.448 N 1 ft 43.35 1 lb 0.3048 m ft N 2075.5 m W s W s       =             = In terms of kilogram force, 2 2 1 k kg N 2075.5 211.8 9.8 N m m f f W s     = =         2.19 5 miles 5280 ft 0.3048 m m km 437 7.033 10 703.3 hr mile 1 ft hr hr V       = = × =             0.3048 m Altitude (25, 000 ft) 7620 m 7.62 km 1 ft   = = =     2.20 3 ft 0.3048 m m km 26, 000 7.925 10 7.925 sec 1 ft sec sec V     = = × =         2.21 From Fig. 2.16, length of fuselage = 33 ft, 4.125 inches = 33.34 ft 0.3048 m 33.34 ft 10.16 m ft   = =     wing span = 40 ft, 11.726 inches = 40.98 ft 0.3048 m 40.98 ft 12.49 m ft   = =    

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    8 Chapter 3 3.1 An examination of the standard temperature distribution through the atmosphere given in Figure 3.3 of the text shows that both 12 km and 18 km are in the same constant temperature region. Hence, the equations that apply are Eqs. (3.9) and (3.10) in the text. Since we are in the same isothermal region with therefore the same base values of p and ρ , these equations can be written as 0 2 1 ( / )( ) 2 2 1 1 ρ ρ g RT h h p e p - - = = where points 1 and 2 are any two arbitrary points in the region. Hence, with g 0 = 9.8 m/sec 2 and R = 287 joule/kgK, and letting points 1 and 2 correspond to 12 km and 18 km altitudes, respectively, 9.8 (287)(216.66) (6000) 2 2 1 1 0.3884 ρ ρ p e p = = = Hence: 4 3 2 2 1 3 2 (0.3884)(1.9399 10 ) 7.53 10 N/m (0.3884)(3.1194 10 ) 0.121 kg/m ρ p - = ´ = ´ = ´ = and, of course, 2 216.66 K T = These answers check the results listed in Appendix A of the text within round-off error. 3.2 From Appendix A of the text, we see immediately that p = 2.65 × 10 4 N/m 2 corresponds to 10,000 m, or 10 km, in the standard atmosphere. Hence, pressure altitude = 10 km The outside air density is 4 3 2.65 10 0.419 kg/m (287)(220) ρ p RT ´ = = = From Appendix A, this value of ρ corresponds to 9.88 km in the standard atmosphere. Hence, density altitude = 9.88 km 3.3 At 35,000 ft, from Appendix B, we find that p = 4.99 × 10 2 = 499 lb/ft 2 . 3.4 From Appendix B in the text, 33,500 ft corresponds to p = 535.89 lb/ft 2 32,000 ft corresponds to ρ = 8.2704 × 10 − 4 slug/ft 3 Hence, 4 535.89 378 (8.2704 10 )(1716) ρ p T R R - = = = ´

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    9 3.5 0.02 1 G G h h h h h - = = - From Eq. (3.6), the above equation becomes 1 1 1 0.02 G G r h h r r æ ö + ÷ ç - = - - = ÷ ç ÷ ÷ ç è ø 6 5 0.02 0.02(6.357 10 ) 1.27 10 m 127 km G G h r h = = ´ = ´ = 3.6 T = 15 − 0.0065 h = 15 − 0.0065(5000) = − 17.5°C = 255.5°K 0.0065 dT a dh = = - From Eq. (3.12) 0 / (2.8) /( 0.0065)(287) 1 1 255.5 0.533 288 g aR p T p T - - - æ ö æ ö ÷ ç ÷ ç ÷ = = = ç ÷ ç ÷ ÷ ç ÷ ç ÷ ÷ ç è ø è ø 5 4 2 1 0.533 0.533 (1.01 10 ) 5.38 10 N/m p p = = ´ = ´ 3.7 1 1 ( )  p g n h h p RT = - - 1 1 1 1 (4157)(150) 0.5 24.9   p h h RT n n g p - = - = - Letting h 1 = 0 (the surface) h = 17,358 m = 17.358 km

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    10 3.8 A standard altitude of 25,000 ft falls within the first gradient region in the standard atmosphere. Hence, the variation of pressure and temperature are given by: 1 1 g aR p T p T - æ ö ÷ ç ÷ = ç ÷ ç ÷ ÷ ç è ø (1) and T = T 1 + a ( h – h 1 ) (2) Differentiating Eq. (1) with respect to time: 1 1 1 1 1 g g aR aR dp g dT T p dt T AR dt æ ö - ÷ ç- - ÷ ç ÷ ÷ ç ç è ø æ ö æ ö ÷ ç ÷ ç ÷ = - ç ÷ ç ÷ ÷ ç ÷ ç ÷ ÷ ç è ø è ø (3) Differentiating Eq. (2) with respect to time: dT dh a dt dt = (4) Substitute Eq. (4) into (3) 1 1 1 ( ) g g aR aR dp g dh p T T dt R dt æ ö ÷ ç - + ÷ ç ÷ ÷ ç ç è ø æ ö ÷ ç = - ÷ ç ÷ ÷ ç è ø (5) In Eq. (5), dh / dt is the rate-of-climb, given by dh / dt = 500 ft/sec. Also, in the first gradient region, the lapse rate can be calculated from the tabulations in Appendix B. For example, take 0 ft and 10,000 ft, we find 2 1 2 1 483.04 518.69 0.00357 10, 000 0 ft ° T T R a h h - - = = = - - - Also from Appendix B, p 1 = 2116.2 lb/ft 2 at sea level, and T = 429.64 ° R at 25,000 ft. Thus, 32.2 5.256 ( 0.00357)(1716) g aR = = - - Hence, from Eq. (5) 5.256 4.256 32.2 (2116.2)(518.69) (429.64) (500) 1716 dp dt - æ ö ÷ ç = - ÷ ç ÷ ÷ ç è ø 2 lb 17.17 ft sec dp dt = -

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    11 3.9 From the hydrostatic equation, Eq. (3.2) or (3.3), 0 ρ dp g dh = - or 0 ρ dp dh g dt dt = - The upward speed of the elevator is dh / dt , which is 0 / ρ dp dp dt dt g = - At sea level, ρ = 1.225 kg/m 3 . Also, a one-percent change in pressure per minute starting from sea level is 5 3 2 (1.01 10 )(0.01) 1.01 10 N/m per minute dp dt = - ´ = - ´ Hence, 3 1.01 10 84.1meter per minute (1.225)(9.8) dh dt - ´ = = 3.10 From Appendix B: At 35,500 ft: p = 535.89 lb/ft 2 At 34,000 ft: p = 523.47 lb/ft 2 For a pressure of 530 lb/ft 2 , the pressure altitude is 535.89 530 33,500 500 33737 ft 535.89 523.47 æ ö - ÷ ç + = ÷ ç ÷ ÷ ç è ø - The density at the altitude at which the airplane is flying is 4 3 530 7.919 10 slug/ft (1716)(390) ρ ρ RT - = = = ´ From Appendix B: 4 3 4 3 At 33,000 ft: 7.9656 10 slug/ft At 33,500 ft: 7.8165 10 slug/ft ρ ρ - - = ´ = ´ Hence, the density altitude is 7.9656 7.919 33, 000 500 33,156 ft 7.9656 7.8165 æ ö - ÷ ç + = ÷ ç ÷ ÷ ç è ø -

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    12 3.11 Let ℓ be the length of one wall of the tank, ℓ = 30 ft. Let d be the depth of the pool, d = 10 ft. At the water surface, the pressure is atmospheric pressure, p a . The water pressure increases with increasing depth; the pressure as a function of distance below the surface, h , is given by the hydrostatic equation dp = ρ g dh (1) Note: The hydrostatic equation given by Eq. (3.2) in the text has a minus sign because h G is measured positive in the upward direction. In Eq. (1), h is measured positive in the downward direction, with h = 0 at the surface of the water. Hence, no minus sign appears in Eq. (1); as h increases (as we go deeper into the water), p increases. Eq. (1) is consistent with this fact. Integrating Eq. (1) from h = 0 where p = p a to some local depth h where the pressure is p , and noting that ρ is constant for water, we have 0 ρ ρ a h p dp g dh = ò ò or, p − p a = ρ g h or, p = ρ g h + p a (2) Eq. (2) gives the water pressure exerted on the wall at an arbitrary depth h . Consider an elementary small sliver of wall surface of length ℓ and height dh . The water force on this sliver of area is dF = p ℓ dh Total force, F, on the wall is 0 0  F d F dF p dh = = ò ò (3) where p is given by Eq. (2). Inserting Eq. (2) into (3), 0 ( + ) ρ  d a F g h p dh = ò or, 2 2 ρ   a d F g p d æ ö ÷ ç ÷ ç = + ÷ ç ÷ ç ÷ ÷ ç è ø (4) In Eq. (4), the product ρ g is the specific weight (weight per unit volume) of water; ρ g = 62.4 lb/ft 3 . From Eq. (4), 2 (10) (62.4)(30) (2116)(30)(10) 2 F = + 93, 600 634,800 728, 400 lb F = + = Note: This force is the combined effect of the force due to the weight of the water, 93,600 lb, and the force due to atmospheric pressure transmitted through the water, 634,800 Ib. In this example, the latter is the larger contribution to the force on the wall. If the wall were freestanding with atmospheric pressure exerted on the opposite side, then the net force exerted on the wall would be that due to the weight of the water only, i.e., 93,600 Ib. In tons, the force on the side of the wall in contact with the water is 728, 400 364.2 tons 2000 F = = In the case of a freestanding wall, the net force, that due only to the weight of the water, is 93, 600 46.8 tons 2000 F = =

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    13 3.12 For the exponential atmosphere model, 0 /( ) 0 (9.8)(45,000) /(287)(240) 6.402 0 ρ ρ ρ ρ g h RT e e e - - - = = = Hence, 6.402 3 3 3 0 (1225)(1.6575 10 ) 2.03 10 kg/m ρ ρ e - - - = = ´ = ´ From the standard atmosphere, at 45 km, p = 2.02 × 10 − 3 kg/m 3 . The exponential atmosphere model gives a remarkably accurate value for the density at 45 km when a value of 240 K is used for the temperature. 3.13 At 3 km, from App. A, T = 268.67 K, p = 7.0121 × 10 4 N/m 2 , and = 0.90926 kg/m 3 . At 3.1 km, from App. A, T = 268.02 K, p = 6.9235 × 10 4 N/m 2 , and = 0.89994 kg/m 3 . At h = 3.035 km, using linear interpolation: ρ   = − − =       = × − × − ×     = ×   = − − =     4 4 4 4 2 3 0.035 268.67 (268.67 268.02) 268.44 K 0.1 0.035 7.0121 10 (7.0121 10 6.9235 10 ) 0.1 N 6.98099 10 m 0.035 kg 0.09026 (0.90926 0.89994) 0.906 0.1 m T p

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    14 3.14 The altitude of 3.035 km is in the gradient region. From Example 3.1, a = –0.0065 K/m. From Eq. (3.14), 1 1 ( ) ( 0) 288.16 (0.0065)(3035) 268.43 K s T T a h h T a h = + − = + − = − = From Eq. (3.12), 0 / 1 1 0 9.8 5.25328 ( 0.0065)(28.7) g aR p T p T g aR −   =     − − = = − So 5.25328 5.25328 5 4 2 268.43 0.688946 288.16 N 0.688946(1.01325 10 ) 6.9807 10 m s s p T p T p     = = =         = × = × By comparison the approximate result from the solution of Prob. 3.13 differs from the exact result above by 5 6.980990 06.9807 4.15 10 0.00415% 6.9807 − − = × = A very small amount. From Eq. (3.13), ρ ρ ρ ρ − +   =     + = − + = −     = = =         0 [( / ) 1] 1 1 0 4.2538 4.24328 1 5.25328 1 4.25328 268.43 0.73958 288.16 g aR s s T T g aR T T 3 kg 0.73598 0.73958 (1.2250) 0.90599 m s ρ ρ = = =

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    15 3.15 We want to calculate h G when 0.01 1 G G G h h h h h − = = − From Eq. (3.6) 6 , where 6.356766 10 m G G rh h r r h = = × + Thus, 0.01 1 1 0.01 0.99 G G r r h r r h = − + = − = + 0.99( ) G r h r + = 5 3 1 0.99 0.0101 (6.356766 10 ) 0.99 64.21 10 m 64.21 km G G h r h −   = = ×     = × = 3.16 From Eq. (3.1), 2 0 G r g g r h   =   +   where r = 6.356766 × 10 6 m, g 0 = 9.8 m/sec 2 , and 0.3048 m (70,000 ft) 21,336 m 1 ft G h   = =     Thus, 6 6 2 6.356766 10 m 9.8 9.767 6.356766 10 21,336 sec g   × = =   × +   9.8 9.767 100 0.34% 98 −   =     Thus, at 70,000 ft, the acceleration of gravity has decreased only by 0.34% compared to its sea level value.

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