Solution Manual For Introduction to Flight, 8th Edition by Anderson

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Chapter 2 2.1 ρ= p RT/ = (1.2)(1.01 10 )/(287)(300)× 5 ρ=1.41kg/m2 v=1/ρ=1/1.41= 0.71m /kg3 3 kT= 3 (1.38× 10−23)(500)=1.035× 10−20J 2.2 Mean kinetic energy of each atom = 2 2 On... e kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has (6.02 ×1026)=1.505 ×1026 atoms. Totalinternalenergy = (energyperatom)(numberof atoms) = (1.035´10-20)(1.505´1026) = 1.558 ´106J RTp (1716)(4602116 + 0.00237 slug3 2.3 ρ= = = 59) ft Volumeof theroom = (20)(15)(8) = 2400ft3 Total mass in the room = (2400)(0.00237) = 5.688slug Weight = (5.688)(32.2) = 183lb p 2116 0.00274 slug3 2.4 ρ= RT = (1716)(460 - 10) = ft Since the volume of the room is the same, we can simply compare densities between the two problems. ∆ρ= 0.00274 - 0.00237 = 0.00037 slugft3 %change = = 0.000370.00237 ´ (100) = 15.6% increase 2.5 First, calculate the density from the known mass and volume, ρ= 1500 900/ = 1.67lbm/ft3 In consistent units, ρ= 1.67/32.2 = 0.052slug ft ./ 3 Also, T = 70 F = 70 + 460 = 530 R. Hence, p = ρRT = (0.52)(1716)(530) p = 47,290lb/ft2 or p = 47,290/ 2116 = 22.3 atm 2.6 p =ρRT np = np+ nR+ nT Differentiating with respect to time, 1 dp 1 dρ 1 dT = + p dt ρ dt T dt dp p dρ p dT or, = + dt ρ dt T dt or, dp = RT dρ ρR dT+ (1) dt dt dt At the instant there is 1000 lbm of air in the tank, the density is ρ=1000/900 1.11lb /ftm =3 ρ=1.11/32.2 0.0345slug/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 ft3, we also have ddtρ 0.5900lb /secmft 0.000556 lb /(ft )(sec=m 3 ) = 3 dρ 0.000556 1.73 10= −5slug/(ft )(sec× 3 ) = dt 32.2 Thus, from equation (1) above, dρ= (1716)(510)(1.73 10 )−5 (0.0345)(1716)(0.0167)× + dt =15.1+ 0.99= 16.1lb/(ft )(sec2 )= = 0.0076 atm/sec 2.7 In consistent units, T −= 10+ 273= 263 K Thus, ρ= p RT/ (1.7 10 )/(287)(263)= 4 × 0.225 kg/m3 ρ= 2.8 ρ= pRT/ = 0.5 10 /(287)(240)5 0.726 kg/m3 v =1/ρ 1/0.726= 1.38 m /kg3= 2.9 3 3 Fp = Forceduetopressure =ò pdx =ò (2116 - 10 )x dx 0 0 = [2116x - 5x2 3]0= 6303 lb perpendicular to wall. 3 3 Fτ = Force due to shear stress=ò0 τ dx =ò0 90 1 dx (x ) 9)2 = [180 (x ) 9) ] 30= 623.5 - 540 = 83.5 lbtangentialtowall. Magnitude of the resultant aerodynamic force = R = (6303)2 ) (835)2 = 6303.6 lb θ = Arc Tan 溺ºè 630383.5 ö÷÷÷ø = 0.76º 2.10 Minimum velocity occurs when sin θ = 0, i.e., when θ = 0° and 180°. Vmin = 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, Vmax = 32 (85)(1) = 127.5 mph atθ = 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 Wair = (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 WHc = (0.168) = 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 p3, ρ3, and T3 denote the conditions at the beginning of combustion, and p4, ρ4, and T4 denote conditions at the end of combustion. Since the volume is constant, and the mass of the gas is constant, then p4 = ρ3 = 11.3 kg/m3. Thus, from the equation of state, 129 atm p4 = ρ4 RT4 = (11.3)(287)(4000) = 1.3 ´107 N/m2 or, p4 = 1.011.3 ´´101075 = [Show More]

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